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Special Right Triangles: 45-45-90
07 Apr
I gave my students our learning progression for SMP 8 a few weeks ago as we started a unit on Right Triangles and had a lesson specifically on 45-45-90 Special Right Triangles.
The Geometry Nspired Activity Special Right Triangles contains an Action-Consequence document that focuses students attention on what changes and what stays the same. The big idea is this: students take some kind of action on an object (like grabbing and dragging a point or a graph). Then they pay attention to what happens. What changes? What stays the same? Through reflection and conversation, students make connections between multiple representations of the mathematics to make sense of the mathematics.
Looking at the side lengths in a chart helps students notice and note what changes and what stays the same:
The legs of the triangle are always the same length.
As the legs increase, the hypotenuse increases.
The hypotenuse is always the longest side.
Students begin to identify and describe patterns and regularities:
All of the hypotenuses have √2.
The ratio of the hypotenuse to the leg is √2.
Students practice look for and express regularity in repeated reasoning as they generalize what is true:
To get from the leg to the hypotenuse, multiply by √2.
To get from the hypotenuse to the leg, divide by √2.
hypotenuse = leg * √2
Teachers and students have to be careful with look for and express regularity in repeated reasoning. Are we providing students an opportunity to work with diagrams and measurements that make us attend to precision as we express the regularity in repeated reasoning that we notice?
In a Math Practice journal, Kaci writes about “look for regularity in repeated reasoning”. We figured out that half of a square is a 45-45-90 triangle, and students were trying to determine the other two sides of the triangle given one side length of the triangle. She says “To find the length of the hypotenuse, you take the length of a side and multiply by √2. The √2 will always be in the hypotenuse even though it may not be seen like √2. In her examples, the triangle to the left has √2 shown in the hypotenuse, but the triangle to the right has √2 in the answer even though it isn’t shown, since 3√2√2 is not in lowest form. She says, “I looked for regularity in repeated reasoning and found an interesting answer.”
What opportunities can we provide our students this week to look for and express regularity in repeated reasoning and find out something interesting?
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
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# Differences of Integers Using Absolute Value
## Subtract positive and negative numbers by using the distance a value is from zero.
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Practice Differences of Integers Using Absolute Value
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Differences of Integers Using Absolute Value
Remember Cameron and the diving from the Differences of Integers Using a Number Line Concept?
While on the plane, Cameron looked at photos from his Dad's deep sea dive. On a shark dive, Cameron's Dad had gone down to a depth of 80 feet with hopes of seeing a shark. After ten minutes or so, he had spotted a beautiful shark swimming above him. Cameron’s Dad went up about 20 feet to try to catch a picture of the shark.
He did get a few good shots before the shark swam away.
“What depth did you see the shark at?” Cameron asked his Dad showing him the picture.
Do you know? To figure this out, you will need to subtract integers. Subtracting integers is the focus of this Concept. By the end of it, you will know the depth of the shark.
### Guidance
Previously we worked on subtracting integers using a number line. Another strategy for subtracting integers involves using opposites. Remember, you can find the opposite of an integer by changing the sign of an integer. The opposite of any integer, b\begin{align*}b\end{align*}, would be b\begin{align*}-b\end{align*}.
For any two integers, a\begin{align*}a\end{align*} and b\begin{align*}b\end{align*}, the difference of ab\begin{align*}a-b\end{align*} can be found by adding a+(b)\begin{align*}a+(-b)\end{align*} So, to subtract two integers, take the opposite of the integer being subtracted and then add that opposite to the first integer.
Sure.
For any two integers, a\begin{align*}a\end{align*} and b\begin{align*}b\end{align*}, the difference of ab\begin{align*}a-b\end{align*} can be found by adding a+(b)\begin{align*}a+(-b)\end{align*}. So, to subtract two integers, take the opposite of the integer being subtracted and then add that opposite to the first integer.
Write this down in your notebook and then continue with the Concept.
Find the difference of 5(8)\begin{align*}5-(-8)\end{align*}.
The integer being subtracted is -8. The opposite of that integer is 8, so add 8 to 5.
5(8)=5+8=13\begin{align*}5-(-8)=5+8=13\end{align*}.
So, the difference of 5(8)\begin{align*}5-(-8)\end{align*} is 13.
Find the difference of 12(2)\begin{align*}-12-(-2)\end{align*}.
The integer being subtracted is -2. The opposite of that integer is 2, so add 2 to -12.
12(2)=12+2\begin{align*}-12-(-2)=-12+2\end{align*}.
|12|=12\begin{align*}|-12|=12\end{align*} and |2|=2\begin{align*}|2|=2\end{align*}, so subtract the lesser absolute value from the greater absolute value.
122=10\begin{align*}12-2=10\end{align*}
Give that answer the same sign as the integer with the greater absolute value. 12>2\begin{align*}12>2\end{align*}, so -12 has a greater absolute value than 2. Give the answer a negative sign.
So, the difference of 12(2)\begin{align*}-12-(-2)\end{align*} is -10.
Find the difference of 203\begin{align*}-20-3\end{align*}.
The integer being subtracted is 3. The opposite of that integer is -3, so add -3 to -20.
203=20+(3)\begin{align*}-20-3=-20+(-3)\end{align*}.
Add as you would add any integers with the same sign––a negative sign.
|20|=20\begin{align*}|-20|=20\end{align*} and |3|=3\begin{align*}|-3|=3\end{align*}, so add their absolute values:
20+3=23\begin{align*}20+3=23\end{align*}
Give that answer the same sign as the two original integers, a negative sign.
So, the difference of 203\begin{align*}-20-3\end{align*} is -23.
Now take a few minutes to practice what you have learned. Find the differences using opposites.
#### Example A
57\begin{align*}-5 - 7\end{align*}
Solution: 12\begin{align*}-12\end{align*}
#### Example B
8(4)\begin{align*}8 - (-4)\end{align*}
Solution: 12\begin{align*}12\end{align*}
#### Example C
12(8)\begin{align*}-12 - (-8)\end{align*}
Solution: 4\begin{align*}-4\end{align*}
Here is the original problem once again.
While on the plane, Cameron looked at photos from his Dad's deep sea dive. On a shark dive, Cameron's Dad had gone down to a depth of 80 feet with hopes of seeing a shark. After ten minutes or so, he had spotted a beautiful shark swimming above him. Cameron’s Dad went up about 20 feet to try to catch a picture of the shark.
He did get a few good shots before the shark swam away.
“What depth did you see the shark at?” Cameron asked his Dad showing him the picture.
To find the depth that Cameron’s Dad saw the shark, we need to write a subtraction problem and solve it. Remember that depth has to do with below the surface, so we use negative integers to represent different depths.
-80 was his starting depth, then he went up -20 so we take away 20 feet.
80(20)=60 feet\begin{align*}-80 - (-20) = -60 \ feet\end{align*}
Cameron’s Dad saw the shark at 60 feet below the surface.
### Vocabulary
Difference
the answer in a subtraction problem.
Integer
the set of whole numbers and their opposites.
### Guided Practice
Here is one for you to try on your own.
The temperature inside a laboratory freezer was 10\begin{align*}-10^\circ\end{align*} Celsius. A scientist at the lab then lowered the temperature inside the freezer so it was 5\begin{align*}5^\circ\end{align*} Celsius less. What was the new temperature inside the freezer?
The problem says that the temperature was lowered. This means that the temperature decreased, so you should subtract. To find the new temperature, you can subtract the amount by which the temperature was lowered from the original temperature, using one of these equations.
10C5Cor105=?=?\begin{align*}-10^\circ C-5^\circ C &= ?\\ \text{or} \qquad -10-5&=?\end{align*}
The integer being subtracted is 5. The opposite of that integer is -5, so add -5 to -10.
105=10+(5)\begin{align*}-10-5=-10+(-5)\end{align*}.
Add as you would add any integers with the same sign––a negative sign.
|10|=10\begin{align*}|-10|=10\end{align*} and |5|=5\begin{align*}|-5|=5\end{align*}, so add their absolute values.
Give that answer the same sign as the two original integers, a negative sign.
10+5=15\begin{align*}10+5=15\end{align*}
So, the difference of 105\begin{align*}-10-5\end{align*} is -15.
This means that the new temperature inside the freezer must be 15\begin{align*}-15^\circ\end{align*} Celsius.
### Practice
Directions: Find each difference using opposites.
1. 157\begin{align*}15-7\end{align*}
2. 712\begin{align*}-7-12\end{align*}
3. 04\begin{align*}0-4\end{align*}
4. 13(9)\begin{align*}13-(-9)\end{align*}
5. 214\begin{align*}-21-4\end{align*}
6. 33(4)\begin{align*}33-(-4)\end{align*}
7. 11(8)\begin{align*}-11-(-8)\end{align*}
8. 1828\begin{align*} 18-28\end{align*}
9. \begin{align*} 28-(-8)\end{align*}
10. \begin{align*}13-18\end{align*}
11. \begin{align*}-21-(-8)\end{align*}
12. \begin{align*}-9-(-38)\end{align*}
13. \begin{align*}22-8\end{align*}
14. \begin{align*}25-38\end{align*}
15. \begin{align*}19-(-19)\end{align*}
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
Difference
The result of a subtraction operation is called a difference.
Integer
The integers consist of all natural numbers, their opposites, and zero. Integers are numbers in the list ..., -3, -2, -1, 0, 1, 2, 3...
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# How Much Is 2/3 of a Cup?
Approximately 2.68 ounces or one-third of a cup equals half of two-thirds of a cup. This presupposes that you are dividing two-thirds of a typical 8-ounce cup by half to determine the quantity.
## Reasons to Split a Cup in Half
It’s useful to know how to modify a recipe to account for different yields or to adapt it to your measuring utensils if you’re making homemade cleaning products or meals. Imagine you are creating a batch of homemade punch. Two-thirds of a cup of fruit juice is required for the dish.
You want to divide the recipe in half because you don’t need the entire quantity. You’ll need to change the amounts for each of the specified denominations, including the two-thirds cup of fruit juice, to accomplish this.
The concepts listed below can be applied to both wet and dry measures. The computations that follow, however, are all based on wet measures.
## Cup’s Ounces in Number
You must first know how many ounces are in the cup in order to calculate how much is half or two-thirds of a cup. The typical measuring cup for recipes has a capacity of 8 ounces.
Calculate what is two-thirds of 8 ounces to start. Specifically, multiply 8 by 0.67. (the approximate decimal form of two-thirds). Once the math is done, you’ll discover that two-thirds of a cup is 5.36 ounces. Divide this sum by two as your next step. The result of two divided by 5.36 is 2.68 ounces.
## Reviewing the Solution Using Fractions
You can frame the issue in terms of fractions if you don’t need to know the exact volume of the cup in ounces. You divide a cup into thirds, for instance, when you take two-thirds of it. Then you divide each of these thirds into two by taking two. This leaves you with one-third as your response because you’re dividing these two thirds by two.
## Ounces to Tablespoons Conversion
Two-thirds of a cup and a half can also be thought of in terms of tablespoons. It takes two tablespoons to make one ounce. This indicates that a typical cup contains 16 tablespoons. Multiply two by eight to get this information. 10.72 tablespoons is the same as two-thirds of 16 tablespoons. 10.72 divided by 2 equals 5.35 tablespoons.
## Ounces to Teaspoons Conversion
A cup can also be thought of in terms of teaspoons. A cup has 48 teaspoons. 32 of these 48 teaspoons make up two-thirds of them. Divide 32 by two to get half of it. You can see that half of two-thirds of a cup is equivalent to 16 teaspoons by multiplying by two.
## Cups and Ounces to Milliliters Conversion
The volume of a liquid is measured in millilitres, which are metric units of measurement. The metric system is frequently employed in lab settings even though it isn’t widely used in the United States. Metric measurements are also available on some measuring cups.
Each ounce is equal to 29.57 millilitres, making an 8-ounce cup have a volume of 236.59 millilitres. Two-thirds of a cup has 158.52 millimetres, which can be calculated by multiplying 236.59 by 0.67. Two-thirds of a cup has this many millilitres in it. After that, divide this number by two. The volume of two-thirds of a cup is 79.26 millilitres.
Misha Khatri
Misha Khatri is an emeritus professor in the University of Notre Dame's Department of Chemistry and Biochemistry. He graduated from Northern Illinois University with a BSc in Chemistry and Mathematics and a PhD in Physical Analytical Chemistry from the University of Utah.
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# How many faces, edges, and vertices does a cube have?
Last Updated : 18 Feb, 2024
Cube is a 3-Dimensional Figure in which all dimensions are equal. A cube has 6 Square faces as all the sides of a cube are equal. The boundary where the faces of the cube meet are called the cube edges. The point at which the cube edges meet is called the cube vertices. A cube has 12 Edges and 8 vertices. In this article, we will learn about cube edges faces vertices in detail with a brief introduction to cubes.
## What is a Cube?
A Cube is a 3-Dimensional Solid figure whose all faces are square shaped. We can also say that a cube can be visualized in the form of a square prism. This is because the faces of a cube are in the form of a square and are also platonic solid in nature. The faces of a cube are also known as planes.
### Properties of a Cube
The properties of a cube are mentioned below:
• All the faces are square-shaped, which implies that the length, breadth, and height are the same.
• The angles between any two faces or surfaces are equivalent to 90°.
• The opposite planes are parallel to each other.
• The opposite edges are parallel to each other.
• Each of the faces forms an intersection with four faces.
• Each of the vertices intersects with three faces and three edges.
### Cube Examples
Examples of Cube include, Rubik’s Cube, Ice Cube, Die used in Ludo, Cubical Box Etc. A picture of examples of a Cube is attached below:
## How many Faces, Edges, and Vertices does a Cube have?
There are 6 faces, 12 edges, and 8 vertices in a cube. Let’s look into them in detail:
### Faces in Cube
There are six faces in a cube. The faces in a cube are in the shape of a square. Faces are flat surfaces bounded by line segments on four sides called edges. We can realize there are six faces in a cube by seeing the number written 1 to 6 on the faces of the die of Ludo.
### Edges in cube
There are 12 Edges in a Cube. Edges are the boundary of a flat surface. Edges are the line segment where are two faces of a geometrical figure. Edges meet each other at a point called Vertices.
### Vertices in Cube
There are 8 vertices in a Cube. Vertices are the points where edges meet. In a cube, a minimum of three edges meet at a vertex. Vertices are the corners of the cube. Vertices are dimensionless.
## Formula of Cube
A cube is a 3D figure. Hence, it will occupy space which is called the volume of the cube. Each face has an area that combines to give up the surface area of the cube. Let’s learn the Formula of the Cube. Let us assume each side of the cube measures ‘a’ units. Hence formulas for this cube are given as:
• Volume of Cube = (Side)3 = a3 cubic units
• Total Surface Area of the Cube = 6 ⨯ (side)2 = 6a2 square units
• Lateral Surface Area of the Cube = 4 ⨯ (side)2 = 4a2 square units
• Diagonal of Cube = √3 ⨯ side = √3 a units
## Sample Problems on Cube Faces Edges and Vertices
Problem 1: Find the surface area of the cube if its side is 6 cm
Solution:
Given:
Side of the cube = 6 cm
As we know that
Surface area of the cube = 6 × side × side
⇒ Surface area of the cube = 6 × side2
⇒ Surface area of the cube = 6 × 62
⇒ Surface area of the cube = 216 cm2
Therefore,
Surface area of the cube is 216 cm2.
Problem 2: Find the volume of the cube if its side is 4 m2.
Solution:
Here we need to find the volume of the cube
Given:
Side of the cube = 4 m2
As we know that
Volume of the cube = Side × Side × Side
⇒ Volume of the cube = Side3
⇒ Volume of the cube = 43
⇒ Volume of the cube = 4 × 4 × 4
⇒ Volume of the cube = 64 m3
Therefore,
Volume of the cube is 64 m3.
Problem 3: Find how many small cubes can be made from a big cube of side 16 m in small cubes of side 4 m
Solution:
Here we need to find out how many small cubes can be made out of one big cube.
As we know that
Volume of cube = Side3
⇒ Volume of big cube = Side × Side × Side
⇒ Volume of big cube = 16 × 16 × 16
⇒ Volume of big cube = 163
⇒ Volume of big cube = 4096 m3
Further,
Volume of small cube = Side × Side × Side
⇒ Volume of small cube = 4 × 4 × 4
⇒ Volume of small cube = 43
⇒ Volume of small cube = 64 m3
Now,
Number of small cubes that can be made from the big cubes = Volume of big cube/Volume of small cube
⇒ Number of small cubes = 4096/64
⇒ Number of small cubes = 64
Therefore,
64 small cubes will be made out of the big cube.
Problem 4. If the surface area of a cube is 486 m2. Then find the volume of the cube.
Solution:
Here we need to find the volume of the cube from a given surface area
Given that Surface area of the cube = 486 m2
As we know that
Surface area of the cube = 6 × Side2
⇒ 486 = 6 × Side2
⇒ Side2 = 486/6
⇒ Side2 = 81
⇒ Side = √81
⇒ Side = 9 m
Now,
Volume of cube = Side3
⇒ Volume of cube = 93
⇒ Volume of cube = 9 × 9 × 9
⇒ Volume of cube = 729 m3
Therefore,
Volume of the cube is 729 m3.
## FAQs on Cube Faces Edges and Vertices
### Q1: Define Cube.
A Cube is a three-dimensional figure whose each face is a square.
### Q2: How many Faces are there in a Cube?
In a cube there are six faces.
### Q3: How many Edges are there in a Cube?
There are 12 Edges in a Cube.
### Q4: How many Vertices are there in a Cube?
A cube has 8 vertices.
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# Can the sides 30, 40, 50 be a right triangle?
Jun 10, 2015
If a right angled triangle has legs of length $30$ and $40$ then its hypotenuse will be of length $\sqrt{{30}^{2} + {40}^{2}} = 50$.
#### Explanation:
Pythagoras's Theorem states that the square of the length of the hypotenuse of a right-angled triangle is equal to the sum of the squares of the lengths of the other two sides.
${30}^{2} + {40}^{2} = 900 + 1600 = 2500 = {50}^{2}$
Actually a $30$, $40$, $50$ triangle is just a scaled up $3$, $4$, $5$ triangle, which is a well known right angled triangle.
Jun 10, 2015
Yes it can.
#### Explanation:
To find out whether the triangle with sides 30, 40, 50, you would need to use the Pythagoras theorem ${a}^{2} + {b}^{2} = {c}^{2}$ (equation for calculating unknown side of a triangle).
Substituting the variables we get the equation ${30}^{2} + {40}^{2} = {c}^{2}$ we won't substitute 50. because we are trying to find whether this equals 50
${30}^{2} + {40}^{2} = {c}^{2}$
$2500 = {c}^{2}$
$\sqrt{2500} = c$
$50 = c$
Therefore because 'c' equals 50 we know that this triangle is a right triangle.
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# Inference on Proportions. Assumptions: SRS Normal distribution np > 10 & n(1-p) > 10 Population is at least 10n.
## Presentation on theme: "Inference on Proportions. Assumptions: SRS Normal distribution np > 10 & n(1-p) > 10 Population is at least 10n."— Presentation transcript:
Inference on Proportions
Assumptions: SRS Normal distribution np > 10 & n(1-p) > 10 Population is at least 10n
Formula for Confidence interval: Normal curve Note: For confidence intervals, we DO NOT know p – so we MUST substitute p-hat for p in both the SD & when checking assumptions.
A May 2000 Gallup Poll found that 38% of a random sample of 1012 adults said that they believe in ghosts. Find a 95% confidence interval for the true proportion of adults who believe in ghost.
Assumptions: Have an SRS of adults np =1012(.38) = 384.56 & n(1-p) = 1012(.62) = 627.44 Since both are greater than 10, the distribution can be approximated by a normal curve Population of adults is at least 10,120. We are 95% confident that the true proportion of adults who believe in ghost is between 35% and 41%. Step 1: check assumptions! Step 2: make calculations Step 3: conclusion in context
Another Gallop Poll istaken in order to measure the proportion of adults who approve of attempts to clone humans. What sample size is necessary to be within + 0.04 of the true proportion of adults who approve of attempts to clone humans with a 95% Confidence Interval? To find sample size: However, since we have not yet taken a sample, we do not know a p-hat (or p) to use!
What p-hat (p) do you use when trying to find the sample size for a given margin of error?.1(.9) =.09.2(.8) =.16.3(.7) =.21.4(.6) =.24.5(.5) =.25 By using.5 for p-hat, we are using the worst- case scenario and using the largest SD in our calculations.
Another Gallop Poll is taken in order to measure the proportion of adults who approve of attempts to clone humans. What sample size is necessary to be within + 0.04 of the true proportion of adults who approve of attempts to clone humans with a 95% Confidence Interval? Use p-hat =.5 Divide by 1.96 Square both sides Round up on sample size
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### Course: Grade 6 (VA SOL)>Unit 3
Lesson 2: Multiplication as scaling
# Multiplication as scaling with fractions
Learn about the concept of multiplication as scaling. Watch and understand how multiplying fractions can be seen as scaling, or resizing, the value of a number. Visualize this concept with various examples, reinforcing their understanding of multiplication as a scaling process. Created by Sal Khan.
## Want to join the conversation?
• hi is this easy for anyone because sadly i DO NOT GET IT!
• Hi! I don't know if your question has been answered yet, but if not, maybe this will help. What Sal is trying to explain in the video is that anytime you multiply a number, let's say x, by another number, let's say m, that is less than 1, then the product will be less than the original number x. That is, given a positive integer x, and another positive integer m, and m<1, then mx < x.
Now if you multiply a number, x, by a number, m, that is greater than 1, then their product will be greater than x. That is, if x is a positive integer, and m is a positive integer and m>1, then mx > x.
Finally, if you multiply a number x by a number m that is equivalent to 1, then the product of the numbers will be equivalent to x. That is, if x is a positive integer, and m=1, then mx = x.
SO..... to relate it to the video, in each scenario Sal gives, x = 2/3. So that is the positive integer that you will multiply by other positive integers to compare their products. But the point he is trying to make is that you don't actually have to multiply in order to know if the product will be less than, greater than, or equal to 2/3.
In the first example, 2/3 (remember, x) is being multiplied by 7/8. So the m I was talking about before is 7/8. Now 7/8 is less than 1, so 2/3*7/8 < 2/3 (if you want to prove it to yourself, 2/3*7/8 = 14/24 = 7/12, and so you can compare 7/12 with 2/3, 2/3=8/12, so we can now see that 7/12 is in fact less than 8/12).
In the second example, 2/3 is being multiplied by 8/7. So m=8/7 now. Since 8/7 > 1, then 2/3*8/7 > 2/3. Again, you can prove that to yourself by actually doing the math like I did in the paragraph above.
Finally, in the last example, even though it doesn't initially look like it, 2/3 is being multiplied by 5/5. Now 5/5 = 1, so 2/3*5/5 = 2/3.
Sorry for the lengthy explanation, but hopefully that's helpful to someone out there!
• I don't want to watch this but i have to i have no choice.
• Yup, The videos are kind of confusing, wish they changed that.
• I am stuck in this section and really worried, what is the exact way for knowing that which number is greater than 1 or less than
1/3 x 575, 4/3 x 575, 5/6 x 575. how can we know that which one of them greater or lesser (1/3, 4/3, 5/6)I am stuck in this section and really worried, what is the exact way for knowing that which number is greater than 1 or less than
1/3 x 575, 4/3 x 575, 5/6 x 575. how can we know that which one of them greater or lesser (1/3, 4/3, 5/6)I am stuck in this section and really worried, what is the exact way for knowing that which number is greater than 1 or less than
1/3 x 575, 4/3 x 575, 5/6 x 575. how can we know that which one of them greater or lesser (1/3, 4/3, 5/6)
• Here's how to find if a fraction is greater than 1 or less than one. Let's look at your fractions. You have 1/3, 4/3, and 5/6. There are 2 numbers in each fraction. Let's look at the first one. 1/3 has 1 as the numerator, and 3 as the denominator. If the numerator is smaller than the denominator, it's less than 1. If the numerator is larger than the denominator, it's larger than 1. So 1/3 and 5/6 are both less than 1, and 4/3 is greater than 1. Now time for the comparison. 5/6 is the odd one out when we're looking at denominators. Let's look at the LCM (Least Common Multiple). 6 is the LCM, so let's convert the denominators with 3 to 6, by multiplying both the numerator and denominator by 2, so the new fractions are 2/6, 8/6, and 5/6. Now we just order all of the fractions by their numerator, so the least to greatest is 2/6, 5/6, and 8/6.
Sorry for the lengthy explanation, but it's the easiest way to give the answer, and to the point where something is learned!
(1 vote)
• Wait is this basically just multiplying fractions?
• Yes you are right.
• I don't understand this, What is scaling.
• A scale factor is a number which scales, or multiplies, some quantity. In the equation y = Cx, C is the scale factor for x. C is also the coefficient of x, and may be called the constant of proportionality of y to x.
• but 3x5=15 so it is equal for 15/15 that's a whole
• It is a whole number because 7/7 well you divide 7 and 7 so you get 1 but if somewone asked what 1 x 1/1 that one will be 1/1 x 1/2 so it equals 1/2
• Whats 9 plus 10?
• its 21
• I dont get it
• If you break it down and start by crossing out 2/3 from each expression, you can see the first expression had the lowest value because its factor 7/8 was less than 1. The second expression had the highest value because its factor 8/7 was greater than 1, and the third expression was in the middle because it was equal to 1, being greater than 7/8 and less than 8/7. So from least to greatest, it would be 7/8, 1, 8/7. Then don't forget to multiply 2/3 back to each number to get the full expression back.
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# MODULE 2.
CONGRUENCE
2.2
Lecture 2
Lecture 2
Preamble: In this lecture, we will derive a necessary and sucient condition for existence of solutions of an equation of the form ax b mod n. We will also discuss
uniqueness of solutions.
## Keywords: linear congruence
2.2.1
Linear Congruence
## Let n be a positive integer. Consider the following linear congruence
ax b mod n,
where a is an integer which is not divisible by n. We want to nd all integers x which
satisfy the above congruence. It is clear that if r is a solution, so is any s r modulo n.
So by a solution we mean a congruence class mod n whose members satisfy the equation.
We would also like to know when we have a unique solution. If we look at a few examples,
we nd that it is possible to have linear congruence which has no solutions, only one
solution or more than one solutions. For example, the linear congruence
2x 5 mod 6
has no solution: if r is a solution, then 6 must divide 2r 5, which implies in particular
that 2r 5 must be even. But that is not possible as 2r is even but 5 is odd. Now
consider
2x 1 mod 3.
If we look at three congruence classes modulo 3, we nd that [0] and [1] are not solutions, but [2] is a solution. Therefore, this congruence has a unique equivalence class of
solutions. Now consider the congruence
4x 2 mod 6.
We can check the 6 elements of a complete residue system of 6, and observe that both
[2] and [5] are solutions.
We will rst nd a necessary and sucient condition for existence of solutions of a linear
congruence. Then we will investigate the number of inequivalent solutions.
31
MODULE 2. CONGRUENCE
Lecture 2
## THEOREM 2.10. The congruence
ax b mod n
has a solution if and only if gcd(a, n) divides b.
Proof: Let gcd(a, n) = d. First assume that the above congruence has a solution r.
Then,
ar b mod n
= n
(b ar)
= d
(b ar),
= d
(b ar + ar)
= d
b.
d|a
Conversely, suppose d divides b. We will now exhibit a solution for the above congruence.
We can write b = db1 for some integer b1 . By Euclids algorithm, we can nd integers
r1 and s1 such that
ar1 + ns1 = d
= b1 (ar1 + ns1 ) = db1
= a(b1 r1 ) + n(b1 s1 ) = b
= a(b1 r1 ) b mod n.
The examples that we saw above are consistent with the theorem. The congruence
2x 5 mod 6 had no solution as the gcd(2, 6) = 2 does not divide 5. But 2x 1 mod 3
has a solution as the gcd of 2 and 3 divides 1. In the third example too, the gcd of 4
and 6 divides 2, and we could nd solutions.
THEOREM 2.11. Consider the congruence
ax b mod n,
where the gcd(a, n) = d divides b. Let x0 be a solution. Then all the other solutions are
precisely given by the following set:
x0 , x 0 +
n
2n
(d 1)n
, x0 +
, , x0 +
.
d
d
d
32
MODULE 2. CONGRUENCE
Lecture 2
## Proof: It is a trivial exercise to verify that for all i with 0 i (d 1), x0 +
solution ax b mod n.
in
d
is a
Next, we show that any two distinct elements in the above set are inequivalent modulo
n. As d divides n, we can write n = dk for some integer k. Consider i, j such that
0 i, j (d 1). Then ,
in
jn
x0 +
mod n
d
d
in
jn
=
mod n
d
d
= ik jk mod dk
(n = dk)
x0 +
= dk
k(i j)
= d
(i j).
## But (d 1) (i j) (d 1), hence d | (i j) implies i = j. Thus, two distinct
elements in the above set can not be congruent modulo n.
We still have to show that any solution x1 must be congruent to one of the d elements
in the set modulo n. We have n = dk and a = da1 , where gcd(k, a1 ) = 1.
ax1 b ax0 mod n
=
dk
| da1 (x1 x0 )
x1
x1
= x0 + ik
n
= x0 + i .
d
## It is enough to consider the above integer i in the range {0, 1, , (d 1)}, as
i i mod d = x0 +
in
i n
x0 +
mod n.
d
d
## COROLLARY 2.12. The congruence
ax b mod n
has a unique solution if and only if a and n are coprime.
In the examples that we have discussed in this lecture, we saw that 2x 1 mod 3 has a
unique solution, namely [2], as 2 and 3 are coprime. On the other hand, 4x 2 mod 6
has more than one solution, as 4 and 6 are not coprime.
33
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# 2 women and 5 men can together finish a piece of embroidery in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the embroidery and that taken by 1 man alone.
Given:
2 women and 5 men can together finish a piece of embroidery in 4 days, while 3 women and 6 men can finish it in 3 days.
To do:
We have to find the time taken by 1 woman alone to finish the embroidery and that taken by 1 man alone.
Solution:
Let the number of days taken by one man alone to finish a piece of embroidery be $x$.
This implies,
The amount of work done by one man in a day $=\frac{1}{x}$.
Let the number of days taken by one woman alone to finish the embroidery be $y$.
This implies,
The amount of work done by one woman in a day $=\frac{1}{y}$.
In the first case, 2 women and 5 men can together finish the piece of embroidery in 4 days.
The amount of work done by 5 men in 1 day $=5\times\frac{1}{x}=\frac{5}{x}$.
The amount of work done by 2 women in 1 day $=2\times\frac{1}{y}=\frac{2}{y}$.
According to the question,
$4(\frac{5}{x}+\frac{2}{y})=1$
$\frac{20}{x}+\frac{8}{y}=1$....(i)
In the second case, 6 men and 3 women finish the work in 3 days.
The amount of work done by 6 men in 1 day $=6\times\frac{1}{x}=\frac{6}{x}$.
The amount of work done by 3 women in 1 day $=3\times\frac{1}{y}=\frac{3}{y}$.
According to the question,
$3(\frac{6}{x}+\frac{3}{y})=1$
$\frac{18}{x}+\frac{9}{y}=1$....(ii)
Multiplying equation (i) by 9 and equation (ii) by 8, we get,
$9(\frac{20}{x}+\frac{8}{y})=9(1)$
$\frac{180}{x}+\frac{72}{y}=9$.....(iii)
$8(\frac{18}{x}+\frac{9}{y})=8(1)$
$\frac{144}{x}+\frac{72}{y}=8$.....(iv)
Subtracting (iv) from (iii), we get,
$\frac{180}{x}+\frac{72}{y}-(\frac{144}{x}+\frac{72}{y})=9-8$
$\frac{180-144}{x}=1$
$x=36$
Substituting $x=36$ in (i), we get,
$\frac{20}{36}+\frac{8}{y}=1$
$\frac{8}{y}=1-\frac{5}{9}$
$\frac{8}{y}=\frac{9-5}{9}$
$\frac{8}{y}=\frac{4}{9}$
$y=\frac{8\times9}{4}$
$y=18$
Therefore, one man alone takes 36 days to finish the work and one woman alone takes 18 days to finish the work.
Tutorialspoint
Simply Easy Learning
Updated on: 10-Oct-2022
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|
# How do you tell if a relation is a function on a map?
## How do you tell if a relation is a function on a map?
In order for a relation to be a function, each x must correspond with only one y value. Use mapping to determine if the relation is a function by listing all the x-values in a column and all the y-values in a column. Draw a line to match the domain value with the corresponding range value.
### How do you determine whether the relation is a function?
A relation is a function only if it relates each element in its domain to only one element in the range. When you graph a function, a vertical line will intersect it at only one point.
#### What is the rule for determining if the mapping is a function?
Mapping or Functions: If A and B are two non-empty sets, then a relation ‘f’ from set A to set B is said to be a function or mapping, If f is a function from A to B and x ∈ A, then f(x) ∈ B where f(x) is called the image of x under f and x is called the pre image of f(x) under ‘f’.
What is function or mapping?
A function is a special type of relation in which each element of the domain is paired with exactly one element in the range . A mapping shows how the elements are paired. Its like a flow chart for a function, showing the input and output values. A mapping diagram consists of two parallel columns.
Is a circle a function?
A circle can be described by a relation (which is what we just did: x2+y2=1 is an equation which describes a relation which in turn describes a circle), but this relation is not a function, because the y value is not completely determined by the x value.
## How do you determine what is a function?
Use the vertical line test to determine whether or not a graph represents a function. If a vertical line is moved across the graph and, at any time, touches the graph at only one point, then the graph is a function. If the vertical line touches the graph at more than one point, then the graph is not a function.
### What are the 3 types of relation?
The types of relations are nothing but their properties. There are different types of relations namely reflexive, symmetric, transitive and anti symmetric which are defined and explained as follows through real life examples.
#### What is a function or mapping?
A function is a special type of relation in which each element of the domain is paired with exactly one element in the range . A mapping shows how the elements are paired. Lines or arrows are drawn from domain to range, to represent the relation between any two elements.
What is a mapping formula?
Mapping Rule A mapping rule has the following form (x,y) → (x−7,y+5) and tells you that the x and y coordinates are translated to x−7 and y+5. Translation A translation is an example of a transformation that moves each point of a shape the same distance and in the same direction. Translations are also known as slides.
What are the example of mapping?
An example of mapping is creating a map to get to your house. An example of mapping is identifying which cell on one spreadsheet contains the same information as the cell on another speadsheet.
## What are the types of mapping?
Types of Maps
• Political Map. A political map shows the state and national boundaries of a place.
• Physical Map. A physical map is one which shows the physical features of a place or country, like rivers, mountains, forests and lakes.
• Topographic Map.
• Climatic Map.
• Economic or Resource Map.
• Scale of a Map.
• Symbols.
### What is the standard form of a circle?
The standard form of a circle is given below: (x – h)2 + (y – k)2 = r2, where the center is located at (h, k) and r is the length of the radius.
#### How to determine if a relationship is a function?
How To: Given a relationship between two quantities, determine whether the relationship is a function. Identify the input values. Identify the output values. If each input value leads to only one output value, classify the relationship as a function. If any input value leads to two or more outputs, do not classify the relationship as a function.
Is the relationship given in the mapping diagram a function?
Because the input value 3 is paired with more than one output value, the relationship given in the above mapping diagram is not a function. Determine whether the relationship given in the mapping diagram is a function.
How to determine if a table is a function?
How To: Given a table of input and output values, determine whether the table represents a function. Identify the input and output values. Check to see if each input value is paired with only one output value. If so, the table represents a function.
## Which is the output of a mapping diagram?
A function assigns only output to each input. The value that is put into a function is the input. The result is the output. A mapping diagram can be used to represent a relationship between input values and output values.
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# FINDING THE LOG VALUE OF A NUMBER
The procedure to find the log value of a number involves three major steps. They are: 1) finding characteristic, 2) finding mantissa, and 3) finding anti-logarithm. The integral part of a common logarithm is called characteristic and the fractional part is called mantissa. Note that the characteristics can be zero, positive or negative, but the mantissa is always positive. Now let us discuss these three steps in detail.
1. Finding Characteristic: In the first stage, we have to find out the characteristic. As discussed earlier, if the digits in the number are more than one, the will be one less than the number of digits to the left of the decimal place. For example, the characteristic of 415.42 is 2, as the number of digits to the left of the decimal place is 3. Similarly, characteristic of 17.23 is 1 and 7.23 is 0. In the case of the numbers which are less the characteristic is equal to one more than the number of zeros after the decimal point and before any significant digit. Thus, characteristic of 0.98 is −1,10.098 is −2, 0.00908 is −3 so on and so forth.
2. Finding Mantissa: To find out the mantissa of a number, you have to use logarithm table. Logarithm tables are presented at the end of this unit. For example, you want to find mantissa of the number 3451. First you have to look at the log tables at the row corresponding to 34 (the first two digits of the given number) and the column corresponding to 5 (the third digit of the given number). The mantissa is 5378. Now look at the mean difference 1 (the fourth digit in the given number) in the same row. The value is Add this 1 to 5378 to obtain 5379. So, for the number 3451, the mantissa part is 0.5379. You already know that the characteristic is 3 for this number. So the log 3451 is .5379.
Note : that mantissa is always positive. It is not affected by the position of the decimal point. That is to say, the mantissa of would be the same. Looking at the table, it can be seen that the mantissa value of 245 is 0.3892. The characteristic of a number can be decided upon by looking at the digits in that number itself and the mantissa can be obtained from the table using the first four significant digits. Look at the following table and observe how the characteristic is changing without a change in the mantissa value.
Number Log Value
2450.0 2.3892
245.0 3.3892
24.5 1.3892
2.45 0.3892
0.245 1.3892
0.0245 12.3892
0.00245 13.3892
Note: For some log values, you can find a bar over the characteristic. Putting bar over the characteristic indicates that the part where the bar appeared is negative and mantissa (the decimal part) is positive.
3. Anti Logarithms: As you know the logarithm tables give the value of mantissa in the logarithms of Whereas the antilog tables give the value of the number whose log value is known. Suppose in the above example, log value 3.3892 is known. We are now interested in finding out the corresponding actual number whose log value is 3.3892 the number 2450. Here, we can say that the antilog of 3.3892 is 2450. Now let us learn how this antilog value is found from antilog tables. In order to find the antilog of 3.3892, first consider only the mantissa part, Look at the antilog tables at the row corresponding to .38 and column corresponding to number is 2449. Look at the mean column at 2 in the same row, and the value is 1. By adding 1 to 2449, the digits in the antilog value will be 2450. he next task is to decide the decimal position. In the log value of 3.3892 the characteristic is 3. So according to rules earlier, there should be four digits in the antilog number. Therefore, place a decimal value after four digits. That means, 2450.0 is the original value. To find the number corresponding to log 2.3892, the digits in antilog value obtained from the table will have to be the same as in the earlier case. Only the position of decimal point will change, which will have to be decided by the characteristic. In this case, characteristic is So according to rules given earlier, the antilog must be less than ‘1’ and there must be one zero after the decimal and before the first significant digit in the result. Thus antilog 2.3892 would be 0.0245.
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# Finding Partial Derivatives: The Meaning of Holding a Variable Constant
I'm having some difficulty understanding the meaning of "hold y (or x) constant" when finding the partial derivative of an equation such as:
$z=x^2+xy+y^2$
Please show all intermediate steps using partial derivative notation and explain what it means in practical terms to hold a variable constant using actual numbers.
Thanks!
One way of understanding it is to introduce a one-variable function and then differentiate that function in the way you already know. For instance, if you are asked to compute $\frac{\partial z}{\partial x}$ at $(x,y)=(1,2)$, you can introduce $r(x)=z(x,2)=x^2+2x+4$. Then $\frac{\partial z}{\partial x}(1,2)=r'(1)$.
If you want to get $\frac{\partial z}{\partial x}$ for general $(x,y)$, you can imagine having a different function $r$ for each value of $y$, and differentiating that function. This is not a very convenient way to do calculations, but it is a simple way of thinking about what's going on.
As a practical matter, when you compute $\frac{\partial z}{\partial x}$, you treat $y$ as if it were a number. Thus for instance $\frac{\partial}{\partial x}(xy)=y$, for the same reason that $\frac{d}{dx}(2x)=2$. Similarly $\frac{\partial}{\partial x}(y^2)=0$, for the same reason that $\frac{d}{dx}(4)=0$.
• So, the example in my question would become: $\frac{∂z}{∂x}(x^2+xy+y^2)=\frac{d}{dx}(x^2)+\frac{d}{dx}(xy)+\frac{d}{dx}(y^2)$ And, if we work those out individually: $\frac{d}{dx}(x^2)=2x$ (x term, but no y term, therefore: differentiate with respect to x, $x^2$ = 2x) $\frac{d}{dx}(xy)=y$ (x term and a y term, therefore: differentiate with respect to x, x = 1 and y remains untouched, as it is held constant) $\frac{d}{dx}(y^2)=0$ (y term but no x term, therefore: differentiate with respect to x, $y^2$ = 0) Yielding: $\frac{∂z}{∂x}(x^2+xy+y^2)=2x+y$ Is this correct? – user913304 Oct 20 '15 at 23:04
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## LCM and HCF
#### LCM and HCF
1. The least number which when divided by 4, 6, 8 and 9 leave zero remainder in each case and when divided by 13 leaves a remainder of 7 is :
1. LCM of 4, 6, 8, 9
∴ LCM = 2 × 2 × 3 × 2 × 3 = 72
##### Correct Option: B
LCM of 4, 6, 8, 9
∴ LCM = 2 × 2 × 3 × 2 × 3 = 72
∴ Required number = 72, because it is exactly divisible by 4, 6, 8 and 9 and it leaves remainder 7 when divided by 13.
1. The smallest number, which when divided by 12 and 16 leaves remainder 5 and 9 respectively, is :
1. As we know that when a number is divided by a, b leaving remainders p, q respectively such that the difference between divisor and remainder in each case is same i.e., (a – p) = (b – q) = t (say) then that (least) number must be in the form of (k – t), where k is LCM of a and b .
Here, t = 12 – 5 = 7, 16 – 9 = 7
##### Correct Option: B
As we know that when a number is divided by a, b leaving remainders p, q respectively such that the difference between divisor and remainder in each case is same i.e., (a – p) = (b – q) = t (say) then that (least) number must be in the form of (k – t), where k is LCM of a and b .
Here, t = 12 – 5 = 7, 16 – 9 = 7
∴ Required number = k - t = (L.C.M. of 12 and 16) – 7
Required number = 48 – 7 = 41
1. A number which when divided by 10 leaves a remainder of 9, when divided by 9 leaves a remainder of 8, and when divided by 8 leaves a remainder of 7, is :
1. We know that when a number is divided by a, b or c leaving remainders p, q and r respectively such that the difference between divisor and remainder in each case is same i.e., (a – p) = (b – q) = (c – r) = t (say) then that (least) number must be in the form of (k – t), where k is LCM of a , b and c .
Here, t = Divisor – remainder = 1
t = 10 – 9 = 1, 9 – 8 = 1, 8 – 7 = 1
##### Correct Option: C
We know that when a number is divided by a, b or c leaving remainders p, q and r respectively such that the difference between divisor and remainder in each case is same i.e., (a – p) = (b – q) = (c – r) = t (say) then that (least) number must be in the form of (k – t), where k is LCM of a , b and c .
Here, t = Divisor – remainder = 1
t = 10 – 9 = 1, 9 – 8 = 1, 8 – 7 = 1
∴ Required number = k - t = (L.C.M. of 10, 9, 8) – 1
Hence , Required number = 360 – 1 = 359
1. What is the smallest number which leaves remainder 3 when divided by any of the numbers 5, 6 or 8 but leaves no remainder when it is divided by 9 ?
1. We find LCM of 5, 6 and 8
5 = 5
6 = 3 × 2
8 = 23
LCM of 5, 6 and 8 = 23 × 3 × 5 = 8 × 15 = 120
Required number = 120K + 3
when K = 2,
##### Correct Option: D
We find LCM of 5, 6 and 8
5 = 5
6 = 3 × 2
8 = 23
LCM of 5, 6 and 8 = 23 × 3 × 5 = 8 × 15 = 120
Required number = 120K + 3
when K = 2,
∴ Required number = 120 × 2 + 3 = 243
It is completely divisible by 9
1. What least number must be subtracted from 1936 so that the resulting number when divided by 9, 10 and 15 will leave in each case the same remainder 7 ?
1. LCM of 9, 10 and 15 = 90
⇒ The multiple of 90 are also divisible by 9, 10 or 15.
∴ 21 × 90 = 1890 will be divisible by them.
∴ Now, 1897 will be the number that will give remainder 7.
∴ Required number = 1936 – 1897
##### Correct Option: C
LCM of 9, 10 and 15 = 90
⇒ The multiple of 90 are also divisible by 9, 10 or 15.
∴ 21 × 90 = 1890 will be divisible by them.
∴ Now, 1897 will be the number that will give remainder 7.
∴ Required number = 1936 – 1897 = 39
|
Section 9.1 The Ellipse. Overview Conic sections are curves that result from the intersection of a right circular cone—think ice cream cone—and a plane—think.
Presentation on theme: "Section 9.1 The Ellipse. Overview Conic sections are curves that result from the intersection of a right circular cone—think ice cream cone—and a plane—think."— Presentation transcript:
Section 9.1 The Ellipse
Overview Conic sections are curves that result from the intersection of a right circular cone—think ice cream cone—and a plane—think sheet of paper. Two of the sections, the circle and the parabola, have been discussed previously. We will re-introduce the parabola later on in the chapter.
The Ellipse An ellipse is the set of all points in a plane the sum of whose distances from two fixed points is a constant. The two fixed points are called foci (plural of focus). The midpoint of the line segment containing the foci is the center of the ellipse.
A Picture
Parts of the ellipse The line that passes through the foci intercepts the ellipse at two points, called vertices (plural of vertex). The line segment with vertices for endpoints is called the major axis. The line segment with endpoints on the ellipse, through the center and perpendicular to the major axis is called the minor axis.
Another Picture
Nomenclature The distance from the center of the ellipse to either of the vertices is a (it follows that the length of the major axis is 2a). The distance from the center to either endpoint of the minor axis is b (it follows that the length of the minor axis is 2b). The distance from the center to a focus is c.
Two equations The standard form of the equation of an ellipse depends on whether the major axis is horizontal Or vertical
One More Picture
Important information b 2 = a 2 – c 2 (equivalently, c 2 = a 2 – b 2 ) When all else fails, draw a picture!
Examples Graph the ellipse and locate the foci:
More Examples—Draw The Picture! Find the standard form of the equation of the ellipse satisfying the following conditions: 1.Foci: (-5, 0) and (5, 0); vertices: (-8, 0) and (8, 0) 2.Major axis vertical with length 12; length of minor axis = 10; center: (-7, -1).
One More…. Convert the equation to standard form by completing the square on x and y. Then graph the ellipse and give the location of its foci.
Download ppt "Section 9.1 The Ellipse. Overview Conic sections are curves that result from the intersection of a right circular cone—think ice cream cone—and a plane—think."
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You are watching: 85 as a fraction in simplest form
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Sometimes it's helpful to convert one kind of number right into another. For example, it's much easier to include 1/4 and 0.5 if you revolve 0.5 into a fraction. Learning just how to transform fractions, decimals, and percents will certainly also assist you together you learn much more advanced math.
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Then we'll move the decimal suggest two spaces come the left.
78% is same to 0.78.
Let's watch at another example. This time we'll turn 8% right into a decimal.
First, we'll change the percent sign v a decimal point.
Then, we'll relocate the decimal suggest two spaces to the left.
Then, we'll relocate the decimal point two spaces come the left.
Notice there is an extra an are next to the 8. We can't just leave an open space with nothing in it. Due to the fact that zero amounts to nothing, we'll change the room with zero.
Notice there is one extra room next come the 8. We can't just leave one open an are with nothing in it. Due to the fact that zero equals nothing, we'll replace the room with zero.
8% is same to .08
Why go this work?
Converting percents into decimals is so simple that you might feel prefer you've let go something. Yet don't worry—it really is the simple! Here's why the an approach we proved you works.
When we rotate a percent right into a decimal, we're in reality doing 2 steps. First, we transform our percent right into a fraction. Because all percents room out the 100, we simply put the percent over 100, choose this:
78% = 78/100
In the 2nd step, we transform 78/100 into a decimal. You already know this way we'll division the numerator through the denominator, favor this:
78 ÷ 100 = 0.78
So why didn't we show you these steps in the slideshow? due to the fact that you can get the answer without them. You understand that all percents room out that 100, therefore you deserve to skip do the percent right into a fraction. You need to divide the percent by 100 to obtain a decimal, yet there's a quick way to do that. Simply move the decimal point two spaces come the left! This way, friend can get the same answer with just one basic step.
Try This!
Convert these percents into decimals.
Converting a decimal right into a percent
Now we'll reverse what you simply learned. Let's transform a decimal into a percent.
Click with the slideshow come see just how to transform a decimal right into a percent.
We're walk to convert 0.45 right into a percent.
We'll reverse what we did in the last section. This time, we'll relocate the decimal suggest two places to the right.
We'll reverse what us did in the last section. This time, we'll move the decimal suggest two areas to the right.
Now we'll change the decimal allude with a percent sign.
We've perfect converting our decimal into a percent. 0.45 is equal to 45%.
Let's shot another example. This time our decimal has actually three numbers to the best of the decimal point.
But we're quiet going to move the decimal point two spaces come the right.
But we're quiet going to move the decimal point two spaces to the right.
We still have actually a number come the right the the decimal point. The number isn't a 0, so we can't fall it.
Instead, we'll keep the decimal suggest and add a percent authorize at the end of the number.
So 0.635 is equal to 63.5%.
Try This!
Calculate this decimals as percents.
### Percents and also fractions
Knowing how to create percents as fractions and also vice versa can help you in your day-to-day life. For example, let's say you earn a grade of 80% top top a test. Friend can transform 80% right into a fraction to find out how many of your answers were correct. When your teacher qualities the test, she may do the opposite. If a student got 8 out of 10 inquiries right, the teacher can convert 8/10 come a percent to offer the student a grade.
Converting a percent right into a fraction
When you're convert a percent into a fraction, it help to remember the percents are always out that 100. You deserve to practice with percents in our introduction to Percentages lesson.
Click with the slideshow come learn exactly how to convert a percent into a fraction.
We're walking to convert 30% right into a fraction.
So any type of percent is same to itself end 100. In our example, 30% is equal to 30/100.
Now we've convert 30% into a fraction, yet we still have to ryellowcomic.comce it.
10 is the biggest number that goes evenly into 30 and 100. For this reason we deserve to divide both components of the fraction by 10.
We'll division the numerator of the fraction first. 30 divided by 10 amounts to 3.
Now we'll divide the denominator. 100 split by 10 equals 10.
So 30% is same to 3/10.
Try This!
Write these percentages as fractions. Make certain to ryellowcomic.comce each fraction to its easiest form.
Converting a portion into a percent
Converting a fraction uses 2 of the an abilities you just learned: creating a fraction as a decimal, and also writing a decimal together a percent. Let's see exactly how we deserve to use these an abilities to transform a portion into a percent.
Click v the slideshow come learn exactly how to convert a portion into a percent.
Let's convert 3/6 into a percent.
Just like as soon as we converted a fraction into a decimal, we'll divide the numerator by the denominator.
3 split by 6 amounts to 0.5.
We've turn our portion into a decimal.
Now we'll rotate the decimal right into a percent by relocating the decimal allude two spaces come the right.
Now we'll turn the decimal into a percent by moving the decimal point two spaces to the right.
See more: What Should You Do When Operating In Conditions Of Reduced Visibility?
We'll also adjust the decimal suggest into a percent sign. 0.50 is equal to 50%.
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Guided Lessons
# Place Value Tower
Students will learn how to find the actual value of three-digit numbers with this activity that has them use manipulatives to create the expanded form of a number.
No standards associated with this content.
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Which set of standards are you looking for?
Students will understand place values of two and three-digit numbers and be able to write the expanded form of given numbers.
(10 minutes)
• Remind students of what they have already learned about place value. This should include how 2-digit numbers have a ones and a tens place.
• On the board draw a place value chart, and in it write a two-digit number.
• Using base ten blocks, show your students this number.
• Use craft sticks to count the students in the class. For example, If you have 24 students you should also have 24 craft sticks.
• Count out groups of 10 sticks, wrapping each group with a rubber band.
• Have your students count the number of groups. Remind them that there are ten sticks in each of those groups, so even if they counted two groups, that means there are 20 sticks total.
• Use ten frames to represent the same number.
• Using linking cubes to represent this same number again.
(15 minutes)
• Model another two-digit number using any of the above methods.
• Once your students seem to have a grasp on these representations, use the base ten blocks to model a three-digit number.
• Talk about how you are modeling the number, including what the different kinds of blocks represent, and write the number out in expanded form.
• After that choose a few more numbers to model for the students.
• Explain to students that Expanded numbersShow the value on each digit individually.
• Write an example number in expanded form to show students this technique. Example: 300+20+1
(15 minutes)
• Have two bins containing the linking cubes and the base ten blocks set out our for each group.
• Ask the students to work together to make numbers you will give them.
• Write a few numbers on the board for your groups to work on.
• Ask your students to write down the expanded form of the given numbers on a separate piece of paper.
(10 minutes)
• Tell students that they are now going to do the same thing, but individually.
• Put up several numbers on the board and have students use the base ten blocks or linking cubes to represent the numbers and then have them write the number's expanded form on a piece of paper.
• Enrichment:Let advanced students use base ten blocks to try to find the expanded form of four-digit numbers.
• Support:Give struggling students one-on-one assistance. Pair struggling students with peer mentors.
(5 minutes)
• Throughout the lesson, make notes about your students to determine whether or not they understand the material.
(5 minutes)
• Review the fact that two and three-digit numbers contain different place values.
• Draw a place value chart to demonstrate your point with a two-digit number, then with a couple of three-digit numbers.
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Update all PDFs
# Folding a square into thirds
Alignments to Content Standards: G-SRT.B.5
Suppose we take a square piece of paper and fold it in half vertically and diagonally, leaving the creases shown below:
Next a fold is made joining the top of the vertical crease to the bottom right corner, leaving the crease shown below: the point $P$ is the intersection of this new crease with the diagonal.
In the diagram below, some additional points are labelled:
1. Show that $|AP| = 2|CP|$.
2. Using part (a), explain how to use the point $P$ in order to fold the square into equal thirds.
## IM Commentary
The goal of this task is to apply knowledge about similar triangles in order to understand an origami construction: trisection of the side of a square. The task stem provides the basic method of construction and then students will need to explain how it works. There are many other closely related approaches based upon similarity of different pairs of triangles: for example, in the picture below $\triangle DPE$ is similar to $\triangle CPB$
Or, if $H$ is the intersection of $\overleftrightarrow{PD}$ with $\overleftrightarrow{AF}$, then $\triangle AHP$ is similar to $\triangle CBP$. In order to give a flavor of the different similarity arguments available, we present an alternate proof of part (b) using a horizontal line through $P$ (shown in purple in the picture above) instead of a vertical line.
An algebraic version of this task, using coordinates, is given in www.illustrativemathematics.org/illustrations/1571. Teachers may wish to compare and contrast the solution given there for part (b) which is briefly mentioned here without details.
One practice standard closely related to this task is MP7, Look For and Make Use of Structure. Students will need to idenitfy, amongst the many triangles made by the paper folds, similar triangles and then use these to set up the appropriate proprotions which reveal the location of the point $P$.
Here is an image of a modified omega origami star that begins by folding the paper into thirds:
## Solutions
Solution: 1
1. To get our ratio between side lengths we will study $\triangle APB$ and $\triangle CPD$, showing that they are similar with a scale factor of $\frac{1}{2}$ to go from $\triangle APB$ to $\triangle CPD$. First, to see that they are similar, note that <\p>
• $m(\angle BAP) = (\angle PCD)$ as these are alternate interior angles formed by the transverse $\overleftrightarrow{AC}$ meeting opposite (parallel) sides of the square.
• $m(\angle PBA) = (\angle PDC)$ as these are alternate interior angles formed by the transverse $\overleftrightarrow{DB}$ meeting opposite (parallel) sides of the square.
By the AA triangle similarity criterion, it follows that $\triangle APB$ and $\triangle CPD$.
To find the scale factor between $\triangle APB$ and $\triangle CPD$, we know that $|AB| = 2|CD|$ because $D$ is on the fold bisecting opposite sides of the square. Since $\overline{AB}$ and $\overline{CD}$ are corresponding parts of similar triangles, this means that the scale factor to go from $\triangle APB$ to $\triangle CPD$ is $\frac{1}{2}$. Since $\overline{AP}$ and $\overline{CP}$ are corresponding parts of the similar triangles $\triangle APB$ and $\triangle CPD$ we can conclude that $|CP| = \frac{1}{2}|AP|$ or $$|AP| = 2|CP|$$ as desired.
2. Intuitively, we know from the first part of the problem that $P$ is two thirds of the way from $A$ to $B$. This means that it is two thirds of the way from the lower edge to the upper edge of the square and also two thirds of the way from the left edge to the right edge of the square. Hence folding the left edge of the square over to $P$ will produce a vertical crease $\frac{1}{3}$ of the width of the square.
To make this argument rigorous, we could use an explicit coordinate system. Instead, to reinforce the similarity argument of part (a), we add a vertical line segment through $P$:
We know that $\overline{PQ}$ is an altitude of $\triangle CPD$. We also know that $\overline{PR}$ is the corresponding altitude of $\triangle APB$. Since the two triangles are similar with scale factor $\frac{1}{2}$ going from $\triangle APB$ to $\triangle CPD$, this means that $|PQ| = \frac{1}{2}|PR|$ of $|PR| = 2|PQ|$. Since $|BC| = |PQ| + |PR|$, we see that $|BC| = 3|PQ|$. Thus to fold the square in equal thirds horizontally, we can fold side $\overline{AB}$ up to $P$ (making a crease one third of the way from $B$ to $C$) and then fold the opposite side down to the new crease (making a crease two thirds of the way from $B$ to $C$).
Solution: 2 Alternate solution for part (b)
We can add a horizontal line, through $P$, perpendicular to $\overleftrightarrow{BC}$, and make another use of similar triangles.
We claim that $\triangle FPA$ is similar to $\triangle GPC$ with scale factor $\frac{1}{2}$. We have
• $m(\angle PFA) = (\angle PGC)$ since these are both right angles: this is true of $\angle PGC$ by construction and $\overleftrightarrow{AF}$ is parallel to $\overleftrightarrow{BC}$ so $\overleftrightarrow{FG}$ is also perpendicular to $\overleftrightarrow{AF}$.
• $m(\angle FPA) = (\angle GPC)$ since these are vertical angles.
By the AA criterion for similarity, $\triangle FPA$ is similar to $\triangle GPC$. The scale factor is $$\frac{|CP|}{|AP|} = \frac{1}{2}$$ from part (a). Hence $|GC| = \frac{1}{2}|FA|$ or, equivalently $|FA| = 2|GC|$. Since $|GB| = |FA|$ we get
\begin{align} |BC| &= |GC| + |GB| \\ &= |GC| + 2|GC| \\ &= 3|GC|. \end{align}
This means that the crease $\overline{FG}$ trisects the vertical sides of the square. In order to get the other vertical division, $\frac{2}{3}$ of the way from $C$ to $B$, we can fold $\overline{AB}$ up to $\overline{FG}$ and the new crease will be $\frac{2}{3}$ of the way from $C$ to $B$.
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## Proposition 49
To find the second binomial line. Set out two numbers AC and CB such that the sum of them AB has to BC the ratio which a square number has to a square number, but does not have to AC the ratio which a square number has to a square number. Set out a rational straight line D, and let EF be commensurable in length with D, therefore EF is rational. Let it be contrived then that as the number CA is to AB, so is the square on EFto the square on FG, therefore the square on EF is commensurable with the square on FG. Therefore FG is also rational. X.6,Cor. X.6 Now, since the number CA does not have to AB the ratio which a square number has to a square number, neither does the square on EF have to the square on FG the ratio which a square number has to a square number. Therefore EF is incommensurable in length with FG. Therefore EF and FG are rational straight lines commensurable in square only. Therefore EG is binomial. X.9 X.36 It is next to be proved that it is also a second binomial straight line. Since, inversely, the number BA is to AC as the square on GF is to the square on FE, while BA is greater than AC, therefore the square on GF is greater than the square on FE. V.7.Cor Let the sum of the squares on EF and H equal the square on GF. Then, in conversion, AB is to BC as the square on FG is to the square on H. V.19,Cor. But AB has to BC the ratio which a square number has to a square number, therefore the square on FG also has to the square on H the ratio which a square number has to a square number. Therefore FG is commensurable in length with H, so that the square on FG is greater than the square on FE by the square on a straight line commensurable with FG. X.9 And FG and FE are rational straight lines commensurable in square only, and EF, the lesser term, is commensurable in length with the rational straight line D set out. Therefore EG is a second binomial straight line. Q.E.D.
(Forthcoming)
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## The Fathers' Day card
Achievement Objectives
NA3-1: Use a range of additive and simple multiplicative strategies with whole numbers, fractions, decimals, and percentages.
Student Activity
Niko and Kaia buy their father a card for Fathers’ Day.
The card costs \$5.90. Niko puts in 40c more than Kaia.
How much do they each contribute to the card?
Specific Learning Outcomes
Subtract using decimals
Find ½ of a decimal number
Devise and use problem solving strategies (act it out, draw a picture, guess and improve, make a table)
Description of Mathematics
This problem involves a two-digit subtraction and division by 2 and can be solved using a number of strategies. The greater challenge is in the extensions to the problem which require the students to think about the multiples of numbers.
Required Resource Materials
Activity
### The Problem
Niko and Kaia buy their father a card for Fathers’ Day. The card costs \$5.90. Niko puts in 40c more than Kaia. How much do they each contribute to the card?
### Teaching Sequence
1. Pose the problem and ask students to restate it in their own words.
2. Circulate as students work on the problem, asking them to explain the strategy that they are using.
3. Share solutions.
#### Extension to the problem
Have students:
• write a similar problem with higher values.
• write a problem in which three people share the cost.
• write a similar problem in which a fraction of the cost is given instead of a \$ value.
#### Other contexts for the problem
Purchasing a gift, sharing the cost of a meal
### Solution
If Niko had not put in the extra 40c, they would have only paid \$5.50.
Divide by 2 to find Kaia's share: \$5.50 ÷ 2 = \$2.75
Niko contributes \$2.75 + \$0.40 = \$3.15.
Attachments
Printed from https://nzmaths.co.nz/resource/fathers-day-card at 4:08pm on the 20th January 2021
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# CBSE Class 10 Maths Chapter 6 Triangles Objective Questions
CBSE Class 10 Maths Chapter 6 Triangles Objective Questions explain concepts such as Introduction to Triangles, Similarity of Triangles, Areas of Similar Triangles, Pythagoras Theorem, and so on. An important part of Unit 3 – Geometry, Chapter 6 – Triangles have nine theorems in total. Having a clear understanding of the concepts, theorems, and problem-solving methods in this chapter is mandatory to score well in the board examination of Class 10 Maths. Taking this into consideration, along with the fact that the changed exam pattern will include more MCQs, we have compiled the CBSE Class 10 Maths Chapter 6 – Triangles Objective Questions for the students to solve and practise. The more they practise, the more they master the concepts.
### List of Sub-Topics Covered in Chapter 6
6.1 Areas of Similar Triangles (4 MCQs from the Topic)
6.2 Basic Proportionality Theorem (4 MCQs from the Topic)
6.3 Criteria for Similarity of Triangles (4 MCQs from the Topic)
6.4 Pythagoras Theorem (4 MCQs from the Topic)
6.5 Similar Triangles (4 MCQs from the Topic)
The CBSE Class 10 Maths Objective Questions covering the above topics will help the students to prepare well for the board exams.
## Download CBSE Class 10 Maths Chapter 6 – Triangles Objective Questions Free PDF
### Areas of Similar Triangles
1. If â–³ ABC ~ â–³ DEF such that AB = 12 cm and DE = 14 cm, find the ratio of areas of â–³ ABC and â–³ DEF.
1. 49/9
2. 36/49
3. 49/16
4. 25/49
Solution: We know that the ratio of areas of two similar triangles is equal to the ratio of the squares.
For any two corresponding sides,
The area of â–³ ABC / area of â–³ DEF = (AB/DE) 2= (12/14) 2= 36/49
1. D and E are points on the sides AB and AC, respectively, of a â–³ABC such that DE || BC. Which of the following statement is true?
1. △ ADE ~ △ ABC
1. only (iii)
2. only (i)
3. only (i) and (ii)
4. all (i) , (ii) and (iii)
Solution:
In △ ADE and △ ABC, we have
∠ ADE = ∠ B
[Since DE || BC ∠ADE = ∠B (Corresponding angles)], and ∠A = △ A [Common]
In the figure, PB and QA are perpendicular to segment AB. If OA = 5 cm, PO = 7cm, and area (ΔQOA) = 150 cm2, find the area of ΔPOB.
1. 233 cm2
2. 294 cm2
3. 300 cm2
4. 420 cm2
Solution: Consider Δ~QOA and Δ POB
QA || PB,
Therefore, ∠ AQO = ∠ PBO [Alternate angles]
∠ QAO = ∠ BPO [Alternate angles]
and
∠ QOA = ∠ BOP [Vertically opposite angles]
Δs QOA ~ BOP [by AAA similarity]
Therefore, (OQ/ OB) = (OA/OP)
Now, area (POB)/ area (QOA) = (OP) 2/ (OA) 2= 72/ 52
Since area (QOA) = 150cm2
⇒area (POB) = 294cm2
1. Two isosceles triangles have equal angles, and their areas are in the ratio 16:25. The ratio of corresponding heights is:
1. 4:5
2. 5:4
3. 3:2
4. 5:7
Solution: For similar isosceles triangles,
Area (Δ1) / Area (Δ2) = (h1)2 / (h2)2
(h1 / h2) = 4/5
### Basic Proportionality Theorem
1. In △ABC, AB = 3 and AC = 4 cm, and AD is the bisector of ∠A. Then, BD:DC is:
1. 9:16
2. 4:3
3. 3:4
4. 16:9
Solution:
The angle bisector theorem states that if a ray bisects an angle of a triangle, then it divides the opposite side into segments that are proportional to the other two sides (It may be similar or may not depending on the type of triangle it divides).
In â–³ABC, as per the statement AB/ AC= BD/DC, i.e., a/b= c/d
So, BD/DC= AB/AC= ¾
So, BD:DC = 3:4
1. ABCD is a parallelogram with diagonals AC if a line XY is drawn such that XY ∥ AB.
BX/XC=?
1. (AY/AC)
2. DZ/AZ
3. AZ/ZD
4. AC/AY
Solution: In the Δ ABC,
AB ∥ XZ
AB ∥ XY
∴ BX/ XC= AY/YC…. (By BPT)….. (1)
In parallelogram ABCD,
AB ∥ CD
AB ∥ CD ∥ XZ
In the Δ ACD,
CD ∥ YZ
∴ AY/YC= AZ/ZD … (By BPT)…… (2)
From 1 & 2,
BX/XC= AY/YC= AZ/ZD
BX/XC= AZ/ZD
In ABC, given that DE//BC, D is the midpoint of AB, and E is the midpoint of AC.Â
The ratio AE:EC is ____.
         A. 1:3
         B. 1:1
         C. 2:1
         D. 1:2
Solution:
DE is parallel to BC
∠DAE = ∠ECF {Alternate angles}
∠ADE = ∠EFC {Alternate angles}
∠BAC = ∠DAE
⇒ AD/DB= AE/EC (Basic Proportionality Theorem)
Since D is the midpoint of AB,
⇒ AE/EC= 1 /1
∴ AE:EC = 1:1
1. In ΔABC, AC = 15 cm and DE || BC. If AB/AD=3, find EC.
1. 5 cm
2. 10 cm
3. 2.5 cm
4. 9 cm
Solution: Given: DE∥BC
From the basic proportionality theorem,
⇒2AE=EC⇒AC=AE+EC−−−−−−− (1)
On substituting the value of EC in (1), we get
15=3AE⇒5=AE⇒EC=10 cm
### Criteria for Similarity of Triangles
1. â–³ ABC is an acute-angled triangle. DE is drawn parallel to BC, as shown. Which of the following is always true?
i) △ ABC ∼ △ ADE
iii) DE= BC/2
1. Only (i)
2. (i) and (ii) only
3. (i), (ii) and (iii)
4. (ii) and (iii) only
Answer: (B) (i) and (ii) Only
DE = BC/2 only if D and E are the midpoints of AB and AC, respectively. So this may not always be true.
1. The triangles ABC and ADEÂ are similar
Which of the following is true?
2. BC/BD=CE/DE
4. All of the Above
Solution: Since the given triangles are similar, the ratios of the corresponding sides are equal.
1. If in △ CAB and △ FED, AB/ EF=BC/FD=AC/ED, then:
1. △ ABC∼△ DEF
2. △ CAB∼△ DEF
3. △ ABC∼△ EFD
4. △ CAB∼△ EFD
Solution: If two triangles are similar, the corresponding sides are proportional.
Therefore, △ABC∼△EFD.
1. A tower of height 24m casts a shadow of 50 m, and at the same time, a girl of height 1.8 m casts a shadow. Find the length of the shadow of the girl.
1. 3.75 m
2. 3.5 m
3. 3.25 m
4. 3 m
Solution: In △ABC and △DEC
∠ABC=∠DEC=90°
∠C=∠C (common)
Therefore, △ABC∼△DEC    [by AA similarity]
So, DE/AB=EC/BC
EC=DE × (BC/AB)
EC= 1.8× (50/24) ⇒EC=3.75 m
### Pythagoras Theorem
1. In the adjoining figure, if BC = a, AC = b, AB = c and ∠CAB = 120°, then the correct relation is:
1. a2 = b2 +  c2 – bc
2. a2Â =Â b2Â +Â Â c2Â + bc
3. a2Â =Â b2Â +Â Â c2Â –Â 2bc
4. a2Â =Â b2Â +Â Â c2Â + 2bc
Answer: (B) a2Â =Â b2Â +Â Â c2Â + bc
Solution: In △CDB,
BC2Â =Â Â CD2Â +Â BD2Â Â Â Â Â [By Pythagoras Theorem]
BC2Â =Â Â CD2Â +Â (DA+AB)2Â
BC2 =  CD2 + DA2 + AB2 + (2×DA×AB)     (i)
CD2Â +Â DA2Â =Â AC2Â Â Â (ii)Â [By Pythagoras Theorem]
AC = 2ADÂ Â Â Â (iii)
Putting the values from (ii) and (iii) in (i), we get
BC2 =  AC2 + AB2 + (AC×AB)
a2Â =Â b2Â +Â Â c2Â + bc
Alternatively,
Since ∠A is an obtuse angle in ΔABC,
BC2Â =Â Â AB2Â +Â Â AC2Â + 2AB . AD
= AB2 +  AC2 + 2×AB× ½ ×AC
[∵ AD = AC cos  60∘ = 1/2AC]
= AB2 +  AC2 + AB × AC
a2Â =Â b2Â +Â Â c2Â + bc.
1. If the distance between the top of two trees 20 m and 28 m tall is 17 m, then the horizontal distance between the trees is:
1. 11 m
2. 31 m
3. 15 m
4. 9 m
Solution: Let AB and CD be two trees such that AB = 20 m, CD = 28 m & BD = 17 m
Draw BEÂ parallel to CD. Then, ED = 8 m.
By applying the Pythagoras theorem:
BE2+DE2=BD2
∴
=15m
∴ AC = BE = 15 m
In the figure, â–³ABC is a right-angled triangle with a right angle at B, and BD is perpendicular to AC. Then which of the following options will hold true?
Solution:
∠A=∠A (common angle)
Therefore, △ABC∼△ADB   [by AA similarity]
1. In a right â–³ABC, a perpendicular BD is drawn to the largest side from the opposite vertex. Which of the following does not give the ratio of the areas of â–³ABD and â–³ACB?
1. (AB/AC)2
4. (BD/CB)2
Solution:
Consider ΔABD and ΔACB:
∠BAD = ∠BAC   [common angle]
∠BDA = ∠ABC   [ 90°]
By AA similarity criterion, △ABD ~ △ACB
Hence,
### Similar Triangles
1. â–³ ABC is such that AB = 3 cm, BC = 2 cm, and CA = 2.5 cm. â–³ DEF is similar to â–³ABC. If EF = 4 cm, then the perimeter of â–³DEF is:
1. 7.5 cm
2. 15 cm
3. 30 cm
4. 22.5 cm
Solution: AB/DE= AC/DF=BC/EF=2/4=1/2
DE = 2×AB = 6 cm, DF = 2×AC = 5 cm
∴ The perimeter of △DEF = (DE + EF + DF) = 15 cm.
1. In △ ABC and △ DEF, ∠A = ∠E = 40∘ and AB/ED=AC/EF. Find ∠B if ∠F is 65°.
1. 85°
2. 75°
3. 35°
4. 65°
Solution:
AB/ED= AC/EF (Given)
∠A = ∠E = 40°
Since the ratio of adjacent sides and the included angles are equal,
â–³ABC is similar to â–³EDF by the SAS similarity criterion.
Now, ∠C = ∠F = 65°    [Corresponding angles of similar triangles are equal]
∴ ∠B = 180°− (∠A+∠C)
=180° – (40° + 65°) = 75°
1. The ratio of the corresponding sides of two similar triangles is 1:3. The ratio of their corresponding heights is:
1. 1:3
2. 3:1
3. 1:9
4. 9:1
Solution: Ratio of heights = Ratio of sides = 1:3.
1. If Δ ABC and Δ DEF are similar such that 2AB = DE and BC = 8 cm, then find EF.
1. 16 cm
2. 12 cm
3. 8 cm
4. 4 cm
Solution:
2AB = DE
2BC = EF
⇒ 2×8 = EF
⇒ EF = 16 cm
## CBSE Class 10 Maths Chapter 6 Extra MCQs
1. A square and a rhombus are always _____.
(a) congruent
(b) similar but congruent
(c) similar
(d) neither similar nor congruent
2. What is the length of each side of a rhombus whose diagonals have lengths of 10 cm and 24 cm?
(a) 13 cm
(b) 26 cm
(c) 25Â cm
(d) 34 cm
Triangle is the most interesting and exciting chapter of Unit 3 – Geometry, as it takes the students through the different aspects and concepts related to the geometrical figure triangle. A triangle is a plane figure with three sides and three angles. This chapter deals with several topics and sub-topics related to triangles.
Get more such CBSE Class 10 Maths and Science notes, at BYJU’S. Also, access Class 10 question papers, sample papers, and other study materials to prepare for the board exam in a more effective way.
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# Exploring the Math:
## Magic Squares: 'Multiplication' in a new context.
### How can we call the result a 'product'?
Suzanne's Magic Squares || Multiplying Magic Squares: Contents
### Statement:
The first magic square, which will be called A, is 3x3, and the second one, which will be called B, is 4x4. The 'product' of A and B, denoted by A*B, will be a 12x12 magic square.
### Question 1:
How can this be called a product?
### Explanation 1:
The words product and multiplication are intended to convey the notion that two magic squares are being combined to get another magic square. Since we are not dealing with numbers, however, we cannot claim that the word 'multiplication' already means something here; the use of the word in this new context is an invention.
Having used the word, it is reasonable to ask whether it has anything else in common with the usual notion of multiplication for numbers besides the fact that things get combined.
### Question 2:
What must be true to have the operation qualify as multiplication?
### Explanation 2:
1. The operation * , which we defined for magic squares, satisfies the associative law.
In other words, if A,B,C are magic squares, then A*(B*C)=(A*B)*C.
2. There is an identity element for this multiplication, namely the 1x1 magic square whose only entry is 1:
3. One way this is different from multiplication of numbers is that the commutative law does NOT hold.
For example, with the 3x3 square A and the 4x4 square B, the two 12x12 squares A*B and B*A are not equal.
However, that doesn't mean that the word multiplication is not appropriate. For example, we also speak of multiplication of matrices, and that notion of multiplication is not commutative.
4. Another way this is different from multiplication is that there is no corresponding notion of addition, but that is not a serious problem.
For example, the set of all powers of 2 is closed under ordinary multiplication, even though it is not closed under addition. It still makes sense to call the operation multiplication even though the set has no addition. In the case of the powers of 2, we can enlarge the set in various ways (by putting in other numbers) to get something where the multiplication still makes sense and where we also have addition, so addition is still there in the background and we don't need to have it explicitly before us. Similarly, it is possible to enlarge the set of magic squares to a larger set on which the multiplication makes sense and in which one also has addition.
### Other Properties:
1. Once we accept that this is a somewhat reasonable notion of multiplication, we can study its other properties. In multiplication of positive numbers, we have the notion of prime versus composite, with the 'unit' 1
being excluded from these two categories. In the same way, we can define a magic square to be composite if it is the product of two strictly smaller magic squares.
2. Thus, the 12x12 magic square A*B is composite. On the other hand, the 4x4 magic square B is not composite, because the only way it could be the product of two strictly smaller magic squares would be if both were 2x2, and it is easy to see that there are no 2x2 magic squares.
3. A magic square will be called prime if it is not composite and is not the 'unit', i.e. the 1x1 magic square whose only entry is 1.
One of the remarkable facts that justifies this use of the words prime and composite is that every magic square can be written in one and only one way as a product of prime magic squares. (When we say only one way, this includes the ordering of the factors, since the multiplication is not commutative.)
So not only do we have some general properties such as associativity and identity element, we even have a property analogous to unique factorization of numbers, and this above all justifies the use of the word 'multiplication'. Furthermore, the proof that this property holds is strikingly similar to the proof of unique factorization of numbers into primes in ordinary arithmetic.
### Definition of Terms:
1. Monoid:
A set M with an associative operation * and an identity element 1 for that operation is called a monoid.
Numbers under multiplication form a monoid, and under our extended concept of multiplication, magic squares form a monoid.
2. Freely generated:
Let (M,*,1) be a monoid and let P be a subset of M. We say that M is freely generated by P if the unique factorization theorem holds with respect to P, i.e. every element of M can be written in one and only one way as a product (in the sense of *) of elements of P, including order.
3. Theorem:
The set of all magic squares is freely generated by the set of all prime magic squares.
We say a monoid (M,*,1) is free if there is a subset P of M such that M is freely generated by P, so this fundamental theorem of arithmetic for magic squares also says that
The set of all magic squares is a free monoid.
The analogous result holds for magic cubes, magic tessaracts, and higher dimensional analogues, and the same applies to the multiplication itself.
4. Unique factorization of numbers:
Since the unique factorization of magic squares has this interpretation, it is interesting and useful to see what the corresponding interpretation is for unique factorization of numbers.
It is basically the same, but instead of using the notion of monoid, we use the notion of commutative monoid. Thus, a commutative monoid is a monoid (M,*,1) in which the operation * is commutative. If P is a subset of M, we say that the commutative monoid M is freely generated by P if every element of M can be written in one and only one way as a product (in the sense of *) of elements of P. Here, when we say 'one and only one way', we don't care about the ordering of the factors since we are assuming commutativity. If (M,*,1) is a commutative monoid and if M has a subset P such that M is freely generated by P as a commutative monoid, we say that M is a free commutative monoid.
So the usual unique factorization theorem for positive integers can be expressed by saying that the set of all positive integers, under multiplication, is a free commutative monoid.
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# College Algebra II BGSU chapter one
The name we use for a set of ordered pairs is a ___. relation
We often use the notation (x,y) to name an ordered pair. When given a set of ordered pairs, the set containing all of the x's is called the ___. domain
We often use the notation (x,y) to name an ordered pair. When given a set of ordered pairs, the set containing all of the y's is called the ___. range
A relation between two sets so that each member of the first set goes with exactly one member of the second set is called a ___. function
A function is a ___ between ___ ____ so that each member of the ___ corresponds to ___ ___ member of the ___. relation, two sets, domain, exactly one, range
We can determine whether a graph is that of a function by applying the Vertical Line Test. The graph is that of a function if ___. every vertical line crosses the graph at most once
The easiest way to show that a relation is not a function is to find one member of the ___ that goes with two members of the ___. domain, range
For a function f, f(a)=b means the point __ is a point on the graph. (a,b)
For a function f, if the point (a,b) is a point on the graph, then f(__)=___. a,b
If f(a)=b, then the function sends the x-value ___ to the y-value ___. a, b
We read the symbols "f(x)" as___. f of x
If you are given the graph of f, and are asked to find f(a) for some number a, you find the point on the graph where the ___-value is a, then find the ___-value of that point. The answer to f(a) is that ___-value. x,y,y
If you are given the graph of f, and are asked to solve f(x)=b for some number b, you find the point(s) on the graph where the ___-value is b, then find the ___-value of that/those point(s). The answer to f(x)=b is that/those ___-value(s). y,x,x
The slope of the line containing the points (x1, y1) and (x2, y2) is ___. (y2 – y1)/(x2 – x1)
If f is a linear function, then the slope of the function between x1 and x2 is ___. (f(x2)-f(x1))/(x2-x1)
If a function can be written in the form f(x)=ax+b, where a and b are numbers, it is called a ___ function. linear
For a linear function f(x)=ax+b, a is the ___ and b is the __. slope, y-intercept
The point where a graph crosses the y-axis is called the ___ of the graph. y-intercept
We say a graph is increasing on the interval a < x < b if, as you move along the graph from left to right, between x=a and x=b, the y-values get ___. bigger
We say a graph is decreasing on the interval a < x < b if, as you move along the graph from left to right, between x=a and x=b, the y-values get ___. smaller
What is the interval notation for the set of x's that are greater than a and less than b? (a,b)
What is the interval notation for the set of x's that are greater than or equal to a and less than or equal to b? [a,b]
What is the interval notation for the set of x's that are greater than or equal to a and less than b? [a,b)
What is the interval notation for the set of x's that are greater than a and less than or equal to b? (a,b]
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# Taks Objective 2 Properties and attributes of function.
## Presentation on theme: "Taks Objective 2 Properties and attributes of function."— Presentation transcript:
Taks Objective 2 Properties and attributes of function
Simplify To reduce to basic essentials 2+5[9+9] 2+5[18] 2+90 =92
Numerical Expressions
An expression that consists of numbers, operations, and grouping symbols Example 6 – 4 = 2 (3 + 3) – 4 = 2 (3 x 2) – 4 = 2
Algebraic expression Example 2x + 3y – 20 5x-2y+38 4x+5y-72 x-4y+108
An expression that consists of numbers, variables, operations, and grouping symbols. Example 2x + 3y – 20 5x-2y+38 4x+5y-72 x-4y+108 3x-y+39
Commutative Property Commutative property states that numbers can be added or multiplied in any order Commutative property of Addition states that changing the order of addends does not change the sum. That is, a + b = b+ a Commutative property of Multiplication states that changing the order of factors does not change the product. That is, a x b = b x a
Associative Property The associative property of addition says that when we add more than two numbers the grouping of the addends does not change the product. The associative property of multiplication says that when we multiply more that two numbers the grouping of the factors does not change product.
Example
Distributive Property
Is an algebra property which is used to multiply a single term and two or more terms inside a set of parentheses. 4 = ( -6 + ( -10) ) 4=4
Terms that have the same variables parts.
Like Terms Terms that have the same variables parts.
Examples Combine like terms. Subtract 12 from each side. Simplify
Divide each side by 3 Use the commutative Property of Addition. Combine like terms Subtract 16 from each side Divide each side by 6 a.2c + c + 12 = 78 2c + c + 12 = 78 3c + 12 = 78 3c – 12 = 78 3c/3 = 66/3 c = 22 b.4b b = 46 4b b = 46 4b + 2b + 16 = 46 6b + 16 = 46 6b + 16 – 16 = 6b/6 = 30/6 b = 5
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# Question: Are Planes Infinite?
## Can 3 planes intersect in one line?
Each plane cuts the other two in a line and they form a prismatic surface.
The second and third planes are coincident and the first is cuting them, therefore the three planes intersect in a line..
## How do you know if 4 points lie on the same plane?
Once you have the equation of the plane, put the coordinates of the fourth point into the equation to see if it is satisfied. If the three points you chose do happen to lie on a single line then you are done- any fourth point will determine a plane that all four points lie on.
## How do you know if a line lies on a plane?
We observe that a straight line will lie in a plane if every point on the line, lie in the plane and the normal to the plane is perpendicular to the line. We observe that a straight line will lie in a plane if every point on the line, lie in the plane and the normal to the plane is perpendicular to the line.
## What is a normal to a plane?
In three dimensions, a surface normal, or simply normal, to a surface at point P is a vector perpendicular to the tangent plane of the surface at P. The word “normal” is also used as an adjective: a line normal to a plane, the normal component of a force, the normal vector, etc.
## How many numbers are needed to uniquely determine a plane?
Determining a plane is the fancy, mathematical way of saying “showing you where a plane is.” There are four ways to determine a plane: Three non-collinear points determine a plane. This statement means that if you have three points not on one line, then only one specific plane can go through those points.
## Can two planes intersect at one point?
They cannot intersect at only one point because planes are infinite. Furthermore, they cannot intersect over more than one line because planes are flat. One way to think about planes is to try to use sheets of paper, and observe that the intersection of two sheets would only happen at one line.
## How many vectors are normal to a plane?
Any nonzero vector can be divided by its length to form a unit vector. Thus for a plane (or a line), a normal vector can be divided by its length to get a unit normal vector. Example: For the equation, x + 2y + 2z = 9, the vector A = (1, 2, 2) is a normal vector. |A| = square root of (1+4+4) = 3.
## Can a plane have 3 points?
In a three-dimensional space, a plane can be defined by three points it contains, as long as those points are not on the same line.
## Does a plane consist of an infinite set of points?
A plane may be considered as an infinite set of points forming a connected flat surface extending infinitely far in all directions. A plane has infinite length, infinite width, and zero height (or thickness). It is usually represented in drawings by a four‐sided figure.
## Do two parallel lines determine a plane?
to determine a plane: two parallel lines determine a plane. … DEFINITION OF PARALLEL AND SKEW LINES Two lines are parallel if they lie in the same plane and do not intersect. Lines that do not intersect and are not contained in any single plane are called skew lines.
## Does a plane have infinite length?
Two lines that meet in a point are called intersecting lines. A plane extends infinitely in two dimensions. It has no thickness. An example of a plane is a coordinate plane.
## Do planes go on forever?
A plane is a flat surface with no thickness. A plane has no thickness, and goes on forever.
## Does a point lie on a plane?
In a Euclidean space of any number of dimensions, a plane is uniquely determined by any of the following: Three non-collinear points (points not on a single line). A line and a point not on that line. Two distinct but intersecting lines.
## What do two rays make?
When two rays share an endpoint, an angle is formed. Angles can be described as acute, right, obtuse, or straight, and are measured in degrees.
## What is a ray in math 4th grade?
In geometry, a ray can be defined as a part of a line that has a fixed starting point but no end point. It can extend infinitely in one direction. On its way to infinity, a ray may pass through more than one point. … The vertex of the angles is the starting point of the rays.
## What is it called when points lie on the same line?
Three or more points that lie on the same line are collinear points .
## How do you determine if a point is above or below a plane?
Take the dot product of the perpendicular and the coordinates of the point. If it is less than zero, the point is below the plane. If v is the vector that points ‘up’ and p0 is some point on your plane, and finally p is the point that might be below the plane, compute the dot product v⋅(p−p0).
## How do you find the normal to a plane?
The normal to the plane is given by the cross product n=(r−b)×(s−b).
## What is a normal to a curve?
The normal to the curve is the line perpendicular (at right angles) to the tangent to the curve at that point. Remember, if two lines are perpendicular, the product of their gradients is -1.
## Do any two planes intersect?
There are three possible relationships between two planes in a three-dimensional space; they can be parallel, identical, or they can be intersecting. Comparing the normal vectors of the planes gives us much information on the relationship between the two planes.
## When two infinite planes intersect what do they form?
If two planes intersect, they intersect in a line. The line and the plane intersect in a single point. The line and the plane do not interesect. In this case they are called parallel.
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# Chapter 6 - Inverse Circular Functions and Trigonometric Equations - Section 6.2 Trigonometric Equations I - 6.2 Exercises - Page 273: 22
The solution set to this problem is $$\{\frac{\pi}{4},\frac{3\pi}{4},\frac{7\pi}{6},\frac{11\pi}{6}\}$$
#### Work Step by Step
$$(\csc x+2)(\csc x-\sqrt2)=0$$ over interval $[0,2\pi)$ 1) Solve the equation: $$(\csc x+2)(\csc x-\sqrt2)=0$$ $$\csc x+2=0\hspace{1cm}\text{or}\hspace{1cm}\csc x-\sqrt2=0$$ $$\csc x=-2\hspace{1cm}\text{or}\hspace{1cm}\csc x=\sqrt2$$ 2) Apply the inverse function: - $\csc x=-2$: $\csc\lt0$ means that the angle of $x$ lies either in quadrant III or IV. In quadrant I, $\csc x=2$ refers to angle $\frac{\pi}{6}$; therefore, in quadrant III and IV, $\csc x=-2$ would refer to angle $\frac{7\pi}{6}$ and $\frac{11\pi}{6}$ - $\csc x=\sqrt2$: $\csc\gt0$ means that the angle of $x$ lies either in quadrant I or II. In quadrant I, $\csc x=\sqrt2$ refers to angle $\frac{\pi}{4}$; in quadrant II, it refers to angle $\frac{3\pi}{4}$ Therefore, $$x\in\{\frac{\pi}{4},\frac{3\pi}{4},\frac{7\pi}{6},\frac{11\pi}{6}\}$$ In other words, the solution set to this problem is $$\{\frac{\pi}{4},\frac{3\pi}{4},\frac{7\pi}{6},\frac{11\pi}{6}\}$$
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
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Breaking News
# What is 9 factorial ?
Steps to calculate factorial of 9
To find 9 factorial, or 9!, simply use the formula that multiplies the number 9 by all positive whole numbers less than it.
Let’s look at how to calculate the Factorial of 9:
9! is exactly :
362880
Factorial of 9 can be calculated as:
9! = 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1
## What is Factorial?
Factorial, symbolized by an exclamation mark (!), is a mathematical concept that represents the product of all positive integers from 1 up to a given number. The factorial of 9, denoted as 9!, is particularly significant in mathematics because it reflects the number of ways you can arrange or permute 9 distinct items. Understanding factorials is fundamental in various areas of mathematics, including combinatorics and probability theory.
## Formula to Calculate the Factorial of 9
The basic formula to calculate the factorial of a number n is n! = n × (n-1) × … × 1. Applying this formula to the factorial of 9, we follow these steps:
• Multiply 9 by every natural number less than itself down to 1.
• 9! = 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1.
Following the calculation, the factorial of 9 equals 362,880.
## What is the Factorial of 9 Used For?
Factorials serve as a key tool in several mathematical fields. The factorial of 9, or 9!, has its applications in areas such as:
• Combinatorics: Determining the number of possible combinations in a lottery or a card game.
• Probability Theory: Calculating the odds of certain events occurring within a fixed set of outcomes.
• Algebra: In the expansion of binomial expressions and solving series related problems.
These applications make the factorial of 9 a useful concept beyond theoretical math, extending into real-world scenarios and problem-solving.
## Exercises
• 1. Calculate the number of different ways to arrange 9 books on a shelf.
• 2. If 9 people are running in a race, how many different ways can the first three winners be chosen?
## Solutions to Exercises
1. There are 9! or 362,880 different arrangements for the 9 books on a shelf.
2. For the race with 9 runners, the first three winners can be chosen in 9!/(9-3)! or 9! / 6! ways, which is 9 × 8 × 7, or 504 different ways.
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### Area of a Parallelogram Practice
#### Question 1
Given a parallelogram with the base of 4cm and the height of 3cm. Find the area of this parallelogram.
A. 12 cm2
B. 12 cm
C. 7 cm2
D. 7 cm
### Step by Step Solution
• #### Step 1
The picture below shows the parallelogram with the base of 4cm and the height of 3cm.
• #### Step 2
To calculate the area, we can start with the formula for the area of a parallelogram:
• #### Step 3
Since the base is given as 4cm, we can substitute b with 4. Similarly, since the height is given as 3cm, we can substitute h with 3. After doing so, we can calculate for A, as shown below:
• #### Step 4
Now, the calculated number 12 has no meaning unless we include the unit for it. Since the base is in cm, the unit for the area will be cm2. Hence:
A = 12 cm2
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What is the derivative of y=sqrt(x^2+3y^2)?
Oct 29, 2016
$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{x}{2 y}$
Explanation:
Although you could differentiate it in it's present form using $y = {\left({x}^{2} + 3 {y}^{2}\right)}^{\frac{1}{2}}$, it is much easier to manipulate it as follows:
$y = \sqrt{{x}^{2} + 3 {y}^{2}} \implies {y}^{2} = {x}^{2} + 3 {y}^{2}$
$\therefore {x}^{2} + 2 {y}^{2} = 0$
We can now very easily differentiate this implicitly and we have the advantage of not having messy $\frac{1}{2}$ powers to deal with:
Differentiating wrt $x$ gives us:
$\frac{d}{\mathrm{dx}} \left({x}^{2}\right) + \frac{d}{\mathrm{dx}} \left(2 {y}^{2}\right) = 0$
$\therefore 2 x + 2 \frac{d}{\mathrm{dx}} \left({y}^{2}\right) = 0$
$\therefore x + \frac{\mathrm{dy}}{\mathrm{dx}} \frac{d}{\mathrm{dy}} \left({y}^{2}\right) = 0$ (this is the implicit differentiation)
$\therefore x + \frac{\mathrm{dy}}{\mathrm{dx}} \left(2 y\right) = 0$
$\therefore x + 2 y \frac{\mathrm{dy}}{\mathrm{dx}} = 0$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{x}{2 y}$
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# Binary Numbers
### Peter Oye Sagay
A Binary Number System is a number system created with only two different symbols . The binary number system created with the digits 0 and 1 is very dominant in computer science and is considered the binary number system. In essence, any number system created with only two symbols is a binary number system.
A Hexadecimal Number System is a number system created with only sixteen symbols. The hexadecimal number system created with the decimal digits ( 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 ) and the first six letters of the alphabet ( a, b, c, d, e, f ) are also very dominant in computer science.
The binary and hexadecimal number systems are easily understood when the decimal number system is used as a frame of reference. This is because the key concepts of the decimal number system are also applicable in the binary and hexadecimal number systems. The first objective is to establish the counting numbers of the number system of interest relative to the decimal number system. In other words how do we express 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, in the binary number system? This objective is easy to realize once we understand that a unit in the binary number system is the same as a unit in the decimal number system and that only the digits 0 and 1 are used to compose the binary number system. The first ten numbers of the binary number system relative to the first ten numbers of the decimal number system are as follows:
However, this is usually not necessary because there is a simple conversion method that can be used to convert a decimal number to a binary number and a binary number to a decimal number. Each decimal number ≥ 1 can be obtained by multiplying the digits of the corresponding binary number by a power of 2 and summing the result . The power of 2 to use for a particular digit, equals the position number of the digit. Like decimal numbers, each digit in a binary number has a position number . The rightmost digit (assuming the number is a whole number) has a position number equal to 0 and the leftmost digit has a position number equal to n-1 (n is the number of digits in the number). For example, suppose we want to convert the binary number 1010 to a decimal number, we will carry out the conversion as follows:
(1) determine the position numbers of the digits of the number 1010
(2) multiply each digit with 2 raised to a power equal to their respective position number
(3) sum the result of the multiplication
So for our example we will have: (1x23) + (0x22 + (1x21) + (0x20) = 8 +2 = 10.
The steps to convert an arbitrary decimal number c , to a binary number are as follows:
(1) Determine the maximum 2 n that is ≤ to c
If 2 n = c , the leftmost digit of the binary number is 1 and its position value equals n. Assign n zeroes to the remaining positions. For example, suppose c = 8, then 2 n= 2 3. Consequently, the binary equivalent of 8 is 1000.
(2) Calculate c - 2 n = y. Form the set A = { n-1, n-2...1, 0}.
(3) From A, select exponents to use to determine powers of 2 whose sum equals y
(4) For each such exponent, place the digit 1 in the position with the position value equal to the exponent.
(5) If there are open positions, place zeroes in them. The resulting string of 0 and 1 digits is the binary equivalent of the decimal number c.
Conversion of a decimal to binary can also be done by repeated division of the decimal by 2 until the result of the division is no longer divisible by 2. The remainders of the divisions are selected with the last remainder being the leftmost binary digit.
Binary strings of 0 and 1 digits can be very long. The computer has no serious problem with lengthy binary numbers. But humans do. So in the computing space, hexadecimal numbers are used as compact representations of binary numbers. The decimal numbers 10, 11, 12, 13, 14, 15 are equivalent to the hexadecimal numbers a, b, c, d, e, f and the binary numbers 1010, 1011, 1100, 1101, 1110, 1111 respectively.
Now, it turns out that we can easily convert a lengthy binary number to a compact hexadecimal number when we group the digits of the binary number into 4s while still keeping their respective position values. We just replace each group of 4 with its hexadecimal equivalent. For example, 3d6e33be3ac5f is the hexadecimal number of the binary number 00111101011011110001100111011111000111010110001011111. The binary digits are sequentially grouped into 4s and then each group is replaced with their equivalent hexdecimal numbers. Note that when working with four digit hexadecimal numbers, 0 is 0000, 1 is 0001, 2 is 0010 and so on up until 7. In other words, if the hexadecimal digits are less than four, prepend zeroes.
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## Solve Lesson 55 Page 63 SBT Math 10 – Kite>
Topic
Uncle Nam plans to make a rectangular picture frame so that the inner part of the frame is a rectangle with dimensions of 6 cm x 11 cm, the width of the border is $$x$$ cm (Figure 27). The area of the frame border should not exceed $$38c{m^2}$$. What is the width of the largest frame border in cm?
Solution method – See details
Set the width of the frame border to $$x$$(cm) ($$x > 0$$). Representing the area of the frame border and solving the inequalities
Detailed explanation
Set the width of the frame border to $$x$$(cm) ($$x > 0$$).
We have the frame border area $$\left( {11 + 2x} \right)\left( {6 + 2x} \right) – 66 = 4{x^2} + 34x$$ ($$c{) m^2}$$)
According to the problem we have: $$4{x^2} + 34x \le 38 \Leftrightarrow 4{x^2} + 34x – 38 \le 0$$
The quadratic triangle $$4{x^2} + 34x – 38$$ has two solutions $${x_1} = \frac{{ – 19}}{2};{x_2} = 1$$ and has coefficients $$a = 4 > 0$$
Using the sign theorem of quadratic triangles, we see that the set of values of $$x$$ such that the triangle $$4{x^2} + 34x – 38$$ bears the sign “-” is \ (\left[ {\frac{{ – 19}}{2};1} \right]\)
So $$0 < x \le 1$$
So the maximum width of the frame border is 1 cm.
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# Area of Square
## Theorem
A square has an area of $L^2$ where $L$ is the length of a side of the square.
Thus we have that the area is a function of the length of the side:
$\forall L \in \R_{\ge 0}: \map \Area L = L^2$
where it is noted that the domain of $L$ is the set of non-negative real numbers.
## Proof 1
### Integer Side Length
In the case where $L = 1$, the statement follows from the definition of area.
If $L \in \N, L > 1$, then we can divide the square into smaller squares, each of side length one.
Since there will be $L$ squares of side length one on each side, it follows that there will be $L \cdot L = L^2$ squares of side length one.
Thus, the area of the square of side length $L$ is $L^2 \cdot 1 = L^2$.
$\Box$
### Rational Side Length
Let $A$ be the area of the square $S$ with side length $L$.
If $L$ is a rational number, then:
$\exists p, q \in \N: L = \dfrac p q$
Create a square of side length $p$.
From the integer side length case, its area equals $p^2$.
Divide the sides into $q$ equal parts.
Thus the square of side length $p$ is divided into $q^2$ small squares.
As they all have side length $\dfrac p q$, each of them equals $S$.
It follows arithmetically:
$A = \dfrac {p^2} {q^2} = \paren {\dfrac p q}^2 = L^2$
$\Box$
### Irrational Side Length
Let $L$ be an irrational number.
Then from Rationals are Everywhere Dense in Topological Space of Reals we know that within an arbitrarily small distance $\epsilon$ from $L$, we can find a rational number less than $L$ and a rational number greater than $L$.
In formal terms, we have:
$\forall \epsilon > 0: \exists A, B \in \Q_+: A < L < B: \left|{A - L}\right| < \epsilon, \left|{B - L}\right| < \epsilon$
Thus:
$\displaystyle \lim_{\epsilon \to 0^+} A = L$
$\displaystyle \lim_{\epsilon \to 0^+} B = L$
Since a square of side length $B$ can contain a square of side length $L$, which can in turn contain a square of side length $A$, then:
$\operatorname {area} \Box B \ge \operatorname {area} \Box L \ge \operatorname {area}\Box A$
By the result for rational numbers:
$\operatorname {area}\Box B = B^2$
$\operatorname {area}\Box A = A^2$
We also note that:
$\displaystyle \lim_{B \to L} B^2 = L^2 = \lim_{A \to L} A^2$
Thus:
$\displaystyle \lim_{B \to L} \operatorname {area} \Box B = \lim_{B \to L} B^2 = L^2$
$\displaystyle \lim_{A \to L} \operatorname {area} \Box A = \lim_{A \to L} A^2 = L^2$
Finally:
$L^2 \ge \operatorname {area}\Box L \ge L^2$
so:
$\operatorname {area}\Box L = L^2$
$\blacksquare$
## Proof 2
Let $\Box ABCD$ be a square whose side $AB$ is of length $L$.
Let $\Box EFGH$ be a square whose side $EF$ is of length $1$.
From the Axioms of Area, the area of $\Box EFGH$ is $1$.
By definition, $AB : EF = L : 1$.
From Similar Polygons are composed of Similar Triangles, the ratio of the areas of $\Box ABCD$ to $\Box EFGH$ is the duplicate ratio of the ratio of $AB$ to $EF$.
Thus by definition of duplicate ratio:
$\Box ABCD : \Box EFGH = \left({AB : EF}\right)^2$
That is:
$\dfrac {\Box ABCD} {\Box EFGH} = \left({\dfrac L 1}\right)^2 = L^2$
That is, the area of $\Box ABCD$ has $L^2$ as many units as $\Box EFGH$.
Hence the result.
$\blacksquare$
## Proof 3
Let a square have a side length $a \in \R$.
This square is equivalent to the area under the graph of $\map f x = a$ from $0$ to $a$.
Thus from the geometric interpretation of the definite integral, the area of the square will be the integral:
$\ds A = \int_0^a a \rd l$
Thus:
$\ds A$ $=$ $\ds \int_0^a a \rd l$ $\ds$ $=$ $\ds \bigintlimits {l \cdot a} 0 a$ Integral of Constant $\ds$ $=$ $\ds a \cdot a - 0 \cdot a$ $\ds$ $=$ $\ds a^2$
$\blacksquare$
### Warning
This proof is circular.
The use of the definite integral to represent area is based on the fact that the area of a rectangle is the product of the rectangle's width and height.
That fact is in turn derived from this one.
However, this demonstration neatly parallels the integration based proofs of the areas of other figures, for example Area of Circle.
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# How To Write An Inequality Using A Graph References
How To Write An Inequality Using A Graph. 2000 2200 2400 2600 2800 3000 graph the. 62/87,21 write a compound inequality from the graph.
A region that satisfies the inequality, and a region that does not. Add 5 to 5 and add 5 to 2.
### Compound Inequalities Foldable Solving Graphing And
But we want to make sure we’re adding five to every part of the inequality. By using the above two information we can easily get.
### How To Write An Inequality Using A Graph
Graph the solution set and write it in interval notation.Here we have x minus five.How to write an inequality :If one point on one side of the line satisfies our inequality, the.
If the graph contains the dotted line, then we have to use one of the signs < or >.If the graph contains the solid line, then we have to use one of the signs ≤ or ≥.If the inequality is ≤ or ≥, graph the equation as a solid line.If the symbol is (> or <) then you do.
If the symbol is (≥ or ≤) then you fill in the dot, like the top two examples in the graph below.If the values are included we draw a solid line as before.In order to graph an inequality we work in 3 steps:Inequalities in two variables from graph.
Inequality t ≤ 2600 an inequality is t ≤ 2600.Is less than x is less than 10.Make sure that you can draw a graph from an inequality and write an inequality when given a graph.Next, pick a point not on the line.
Now an inequality uses a greater than, less than symbol, and all that we have to do to graph an inequality is find the the number, ‘3’ in this case and color in everything above or below it.One may also ask, what is an inequality for a graph?Pay attention to open and closed circles.Second we test a point in each region.
Since it is starting from −20 with a closed circle, the inequality is x ≤ −20.So to undo the subtraction, we’re going to add five.Solve the inequality and write the solution set using interval notation;Some specific examples of solutions to the inequality include 1, 2, 3, 7, and 10.5.
Step 1 replace the inequality symbol with an equal sign and graph the resulting line.Step 3 starting at (0,b), use the slope m to locate a second point.Step 4 connect the two points with a straight line.Terms in this set (30) x > 5.
Testing solutions to inequalities (basic) plotting inequalities.The inequality for the above graph is x ≥ 1.The mountain is at least 985 feet tall.The movie will be no more than 90 minutes in length.
Then determine what can be added or subtracted from each term soThen graph the solution set on a number line.Then, s ≤ 140 example #2 the height of a human being is not greater than 10 feet.This implies that | x + c | < a.
This indicates that all real numbers greater than or equal to 0 are solutions to the inequality.This is a value we can easily.This is the currently selected item.This will leave us with seven.
To find linear inequalities in two variables from graph, first we have to find two information from the graph.To graph a linear inequality:To graph an inequality, treat the <, ≤, >, or ≥ sign as an = sign, and graph the equation.To sketch the graph of a line using its slope:
We dash the line if the values on the line are not included in the boundary.What is the inequality for the graph?What is the inequality for the graph?What is the inequality for the graph?
Words temperatures up to and including 2600°f variable let t be the temperatures the probe can withstand.Write a compound inequality from the graph.Write an inequality describing all the real numbers on the number line that are less than 2, then draw the corresponding graph.Write an inequality representing the amount of sheet metal that can be used to make a.
Write an inequality that models this situation.Write an inequality that models this situation.Write and graph an inequality that represents the temperatures the probe can withstand.Write each inequality using interval notation, and graph each inequality on the real mumber line.
Write the inequality for the graph given below.Write the inequality for the graph.🚨 claim your spot here.
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# Class 11 RD Sharma Solutions – Chapter 5 Trigonometric Functions – Exercise 5.1 | Set 1
• Last Updated : 21 Feb, 2021
### Question 1. sec4θ – sec2θ = tan4θ + tan2θ
Solution:
We have
sec4θ – sec2θ = tan4θ + tan2θ
Taking LHS
= sec4θ – sec2θ
= sec2θ(sec2θ – 1)
Using sec2 θ = tan2θ + 1, we get
= (1 + tan2θ)tan2θ
= tan2θ + tan4θ
Hence, LHS = RHS (Proved)
### Question 2. sin6θ + cos6θ = 1 – 3sin2θcos2θ
Solution:
We have
sin6θ + cos6θ = 1 – 3sin2θcos2θ
Taking LHS
= sin6θ + cos6θ
= (sin2θ)3 + (cos2θ)3
Using a3 + b3 = (a + b)(a2 + b2 – ab), we get
= (sin2θ + cos2θ)(sin4θ + cos4θ – sin2θcos2θ)
Using a2 + b2 = (a + b)2 – 2ab and sin2θ + cos2θ = 1, we get
= (1)[(sin2θ + cos2θ)2 – 2sin2θcos2θ – sin2θcos2θ]
= (1)[(1)2 – 3sin2θcos2θ]
= 1 – 3sin2θcos2θ
Hence, LHS = RHS (Proved)
### Question 3. (cosecθ – sinθ)(secθ – cosθ)(tanθ + cotθ) = 1
Solution:
We have
(cosecθ – sinθ)(secθ – cosθ)(tanθ + cotθ) = 1
Taking LHS
= (cosecθ – sinθ)(secθ – cosθ)(tanθ + cotθ)
Using cosecθ = 1/sinθ and secθ = 1/cosθ
= 1
Hence, LHS = RHS (Proved)
### Question 4. cosecθ(secθ – 1) – cotθ(1 – cosθ) = tanθ – sinθ
Solution:
We have
cosecθ(secθ – 1) – cotθ(1 – cosθ) = tanθ – sinθ
Taking LHS
Hence, LHS = RHS(Proved)
### Question 5.
Solution:
We have
Taking LHS
Using a2 – b2 = (a + b)(a – b) and a3 + b3 = (a + b)(a2 + b2ab), we get
= sinA
Hence, LHS = RHS(Proved)
### Question 6.
Solution:
We have
Taking LHS
Using tanA = sinA/cosA and cotA = cosA/sinA, we get
Using a3 – b3 = (a – b)(a2 + b2 + ab), we get
Using cosecA = 1/sinA and secA = 1/cosA, we get
= secAcosecA + 1
Hence, LHS = RHS(Proved)
### Question 7.
Solution:
We have
Taking LHS
Using a3 ± b3 = (a ± b)(a2 + b2 ± ab), we get
Using sin2θ + cos2θ = 1, we get
= 1 – sinAcosA + 1 + sinAcosA
= 2
Hence, LHS = RHS(Proved)
### Question 8. (secAsecB + tanAtanB)2 – (secAtanB + tanAsecB)2 = 1
Solution:
We have
(secAsecB + tanAtanB)2 – (secAtanB + tanAsecB)2 = 1
Taking LHS
= (secAsecB + tanAtanB)2 – (secAtanB + tanAsecB)2
Expanding the above equation using the formula
(a + b)2 = a2 + b2 + 2ab
= (secAsecB)2 + (tanAtanB)2 + 2(secAsecB)(tanAtanB) –
(secAtanB)2 – (tanAsecB)2 – 2(secAtanB)(tanAsecB)
= sec2Asec2B + tan2Atan2B – sec2Atan2B – tan2Asec2B
= sec2A(sec2B – tan2B) – tan2A(sec2B – tan2B)
= sec2A – tan2A -(Using sec2θ – tan2θ = 1)
= 1
Hence, LHS = RHS(Proved)
### Question 9.
Solution:
We have
Taking RHS
×
×
Hence, RHS = LHS(Proved)
### Question 10.
Solution:
We have
Taking LHS
Using 1 + tan2x = sec2x and 1 + cot2x = cosec2x, we get
Using a2 + b2 = (a + b)2 – 2ab, we get
Hence, LHS = RHS (Proved)
### Question 11.
Solution:
We have
Taking LHS
By using the formulas cotθ = cosθ/sinθ and tanθ = sinθ/cosθ, we get
Using a3+b3 = (a + b)(a2 + b2 – ab), we get
= 1 – (sin2θ + cos2θ) + sinθcosθ
= 1 – 1 + sinθcosθ
= sinθcosθ
Hence, LHS = RHS (Proved)
### Question 12.
Solution:
We have
Taking LHS
Hence, LHS = RHS(Proved)
### Question 13. (1 + tanαtanβ)2 + (tanα – tanβ)2 = sec2αsec2β
Solution:
We have
(1 + tanαtanβ)2 + (tanα – tanβ)2 = sec^2αsec2β
Taking LHS
= (1 + tanαtanβ)2 + (tanα – tanβ)2
= (1 + tan2αtan2β + 2tanαtanβ) + (tan2α + tan2β – 2tanαtanβ)
= 1 + tan2αtan2β + tan2α + tan2β
= (1 + tan2β) + tan2α(1 + tan2β)
= (1 + tan2β)(1 + tan2α)
= sec2αsec2β
Hence, LHS = RHS (Proved)
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## Lesson 9.1 Understanding the concept of relations
### 9.101 Introduction
• Relationships between elements of sets occur in many contexts.
• We deal with relationships in a daily basis:
• A relationship between a person and a relative.
• A relationship between an employee and their salary.
• A relationship between a business and its telephone number.
• A relationship between a computer language and a valid statement in this language will often arise in computer science.
• In maths, we study relationships like:
• the relation between a positive integer and one it divides.
• relation between a real number and one that is larger than it.
• relation between a business and its telephone number.
• relation between a computer language and a valid statement
• relation between a real number x and the value f(x) where f is a function, and so on.
• Relations in maths
• We can define relationship between elements of 2 sets
• We can also define the relationship between 2 elements of a set.
### 9.103 Definition of a relation: relation versus function
• Relation
• A relation can be defined between elements of set A and elements of another set B.
• Can also be defined between elements of the same set.
• We use the letter $R$ to refer to relation.
• Let $A$ and $B$ be two sets.
• Let $R$ be a relation linking elements of set $A$ to elements of set $B$.
• Let $x \in A$ and $y \in B$
• We say that $x$ is related to $y$ with respect to relation $R$ and we write $x \ R \ y$
• A relation can also be between elements of the same set.
• A relation is a link between two elements of a set
• For example
• A person x is a SON OF' a person y.
• SON OF is a relation that links x to y
• Usually use the letter $R$ to refer to a relation:
• In this case $R$ = 'SON OF'
• If $x$ is SON OF $y$ we write $x \ R \ y$
• If $y$ is NOT a SON OF $x$ we write $y \ \not R \ x$
• A relation can be defined as a link or connection between elements of set A to set B.
• Example
• A is the set of students in a Comp Science class
• $A = \{Sofia, Samir, Sarah\}$
• B is the courses the department offers
• $B = \{Maths, Java, Databases, Art\}$
• Let $R$ be a relation linking students in set $A$ to classes they are enrolled in: A student is related to the course if the student is enrolled in the course.
• Examples:
• Sofia is enrolled in Math and Java
• Samir is enrolled in Java and Databases
• Sarah is enrolled in Math and Art
• Sofia is not enrolled in Art
• Notations:
• Sofia $R$ Maths
• Sofia $R$ Java
• Samir $R$ Java
• Samir $R$ Databases
• Sarah $R$ Maths
• Sarah $R$ Art
• Sofia $\not R$ Art
• Cartesian product
• Let $A$ and $B$ be 2 sets.
• The Cartesian product A x B is defined by a set of pairs (x, y) such that $x \in A$ and $y \in B$
• $A \ x \ B = \{(x, y): x \in A \text{ and } y \in B \}$
• For example:
• $A = \{ a_1, a_2 \}$ and $B = \{b_1, b_2, b_3\}$
• $A \ x \ B = \{(a_1, b_1), (a_1, b_2), (a_1, b_3), (a_2, b_1), (a_2, b_2), (a_2, b_3)\}$
• Definition of relation
• Let $A$ and $B$ be two sets.
• A binary relation from A to B is a subset of a Cartesian product $A \ x \ B$
• $R \subseteq A \ x \ B$ means $R$ is a set of ordered pairs of the form (x, y) where $x \in A$ and $y \in B$.
• $(x, y) \in R$ means $x \ R \ y$ (x is related to y)
• For example:
• $A = \{a, b, c\}$ and $B = \{1, 2, 3\}$
• The following is a relation defined from A to B:
• $R = \{(a, 1), (b, 2), (c, 3)\}$
• This means that: $a\ R \ 1$, $b \ R \ 2$ and $c \ R \ 3$
• Relations on a set
• When A = B
• A relation R on the set A is a relation from A to A
• $R \subseteq A \ x \ A$
• We will be studying relations of this type.
• Example
• A = {1, 2, 3, 4}
• Let R be a relation on the set A:
• $x, y \in A$, $x \ R \ y$ if and only if $x < y$
• We have 1 R 2, 1 R 3, 1 R 4, 2 R 3, 2 R4, 3 R 4
• $R = \{(1,2), (1,3),(1,4), (2, 3), (2, 4), (3, 4)\}$
### 9.105 Matrix and graph representatins of a relation
• Relations using matrices
• Given a relation R from a set A to set B.
• List the elements of sets A and B in a particular order
• Let $n_a = |A|$ and $n_b = |B|$
• The matrix of R is an $n_a \ \text{x} \ n_b$ matrix.
• $M_r = [m_{ij}] n_a \ x \ n_b$
• In matrix store a 1 if $(a_i, b_j) \in R$ otherwise, 0
• Example 1
• Let A = {Sofia, Samir, Sarah}
• Let B = {CS100, CS101, CS102, CS103}
• Consider the relation of who is enrolled in which class
• R = { (a,b) | person a is enrolled in course b }
CS100 CS101 CS102
Sofia x x
Samir x x
Sarah x x
$M_r = \begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \end{bmatrix}$
• Example 2
• Let A = { 1, 2, 3, 4, 5 }
• Consider a relation: $< (x, y) \in R \text{ if and only if } x < y$
• Every element is not related to itself (hence the diagonal 0s).
$M_r = \begin{bmatrix} 0 & 1 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 & 1 \\ 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 \\ \end{bmatrix}$
• Example 3
• Let A = { 1, 2, 3, 4, 5 }
• Consider a relation : $\leq (x, y) \in R$ if and only if $x \leq y$
• Note the diagonal is all 1s.
$M_r = \begin{bmatrix} 1 & 1 & 1 & 1 & 1 \\ 0 & 1 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 & 1 \\ 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 & 1 \\ \end{bmatrix}$
• Combining relations
• Union
• The Union of 2 relations is a new set that contains all of the pairs of elements that are in at least one of the two relations.
• The union is written as $R \ U \ S$ or "R or S".
• $R \ U \ S$ = { $(a, b): (a, b) \in R$ or $(a, b) \in S$ }
• Intersection
• The intersection of 2 relations is a new set that contains all of the pairs that are in both sets.
• The intersection is written as $R \cap S$ or "R and S"
• $R \cap S$ = { $(a, b): (a, b) \in R$ and $(a, b) \in S$ }
• Combining relations: via Boolean operators
• Let $M_{R} = \begin{bmatrix}1 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix} \\ M_{s} = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 0 & 0 \end{bmatrix}$
• Join $M_{R \cup S} = M_R \vee M_S = \begin{bmatrix}1 & 0 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 0\end{bmatrix}$
• Meet $M_{R \cap S} = M_R \land M_S = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$
• Representing relations using directed graphs
• When a relation is defined on a set, we can represent with a digraph.
• Building the digraph
• First, the elements of A are written down,
• When $(a, b) \in R$ arrows are drawn from a to b.
• Example 1
• A = { 1, 2,3 ,4}
• Let R be relation on A defined as follows:
• R = { $(x, y)$ | x divides y}
• R can be represented by this digraph
• Each value divides itself, hence the loop at each vertex.
• One divides all other elements, so it has a link to each elements.
• 2 only divides into 4
• So on...
• Example 2
• Let A = { 1, 2, 3, 4, 5 }
• Consider relation: $\leq (x, y) \in R$ if and only if $x \leq y$
• Each element is equal to itself.
• One is less than or equal to all elements.
• 5 is greater than all elements except itself.
• So on.
### 9.107 The properties of a relation: reflexive, symmetric and anti-symmetric
• Reflexivity
• A relation R in a set S is said to be reflexive if and only if $x \ R \ x$, $\forall x \in S$
• Or, for all x in the set, if the pairs (x, x) is in the relation, then it's reflexive.
• Example 1 (reflexive example)
• Let R be a relation of elements in Z:
• $R = \{ (a, b) \in Z^2 | a \leq b \}$
• For all x elements of Z, we have $x \leq x$, hence $x \ R \ x$
• This implies that R is reflexive.
• Example 2 (non-reflexive)
• $R = \{ (a, b) \in Z^2 | a \lt b \}$
• Digraph of reflexive relation
• Every element will have a loop.
• In this example, S = {1, 2, 3, 4} and R is a relation of elements S $R = \{ (a, b) \in S^2 | a \leq b \}$
• Matrix of reflexive relation
• Same example as above.
• Note that all the values in the diagonal are 1.
• Definition of Symmetry
• A relation is said to be symmetric if and only if:
• $\forall a, b \in S$, if $a \ R \ b$ then $b \ R \ a$.
• Proof: let $a, b \in Z$ with $a \ R \ b$:
• a mod 2 = b mod 2
• b mod 2 = a mod 2
• b R a
• R is symmetric
• Diagram of a symmetric relation
• Example
• Let S = {1, 2, 3, 4} and R be relation of elements in S
• R = { (a,b) \in S^2 | a mod 2 = b mod 2 }
• Matrix of symmetric relation
• Example
• Let S = {1, 2, 3, 4} and let R be relation of elements in S
• R = { (a, b) \in S^2 | a mod 2 = b mod 2 }
• $M_R = \begin{bmatrix}1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 \\ 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1\end{bmatrix}$
• Definition of anti-symmetric
• A relation R on a set S is said to be anti-symmetric if and only if $\forall a, b \in S$, if $a \ R \ b$ and $b \ R \ a$ then $a = b$
• In other words, no 2 diff elements are related both ways.
• Examples
• Let R be a relation on elements in Z:
• $R = \{ (a, b) \in Z^2 | a \leq b \}$
• Let $a, b \in Z$, $a \ R \ b$ and $b \ R \ a$
• $a \leq b$ and $b \leq a$
• $a = b$
• R is anti-symmetric
• Digraph of anti-symmetric relation
• Digraph contains no parallel edges between any 2 different vertices.
• Example
• Let S = {1, 2, 3, 4} and R be relation on elements in S
• $R = \{ (a, b) \in S^2 | \text{ a divides b } \}$
• Matrix of an anti-symmetric relation
• Exercise
• Let R be the relation defined by the Matrix M_r
• $M_r = \begin{bmatrix} 0 & 0 & 1 \\ 1 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix}$
• Is R reflexive? Symmetric? Anti-symmetric?
• Clearly R is not reflexive: $m_{1,1} = 0$
• It is not symmetric because $m_{2,1}=1, m_{1, 2}=0$
• However, it is anti-symmetric.
### 9.109 Relation properties: transitivity
• Definition of transitivty
• A relation $R$ on a set S is called transitive if and only if:
• $\forall a, b, c \in S$, if ($a \ R \ b$ and $b \ R \ c$) then $a \ R \ c$
• Examples of transitive relations
• $R = \{ (x , y) \in N^{2} | x \leq y \}$
• It is transitive as $\forall x, y, z \in \mathbf{N}$ if $x \leq y$ and $y \leq z$ then $x \leq z$
• R = { (a, b) | a is an ancestor of b }
• It is transitive because if a is an acestory of b and b is an ancestor of c, then a is an ancestor of c.
• Example of non-transitive relations
• R = { (2, 3), (3, 2), (2, 2) }
• It is not transitive because 3 R 2 and 2 R 3 but 3 \not R 3
• Not transitive as:
• $a \ R \ c$ and $b \ R \ c$, however $a \ \not R \ c$
• Also: $b \ R \ c$ and $c \ R \ d$ however $b \ \not R \ d$
• What edges need to be added to make it transitive?
• The result enhanced relation is called the "transitive closure of the original relation"
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# MATHEMATICS V
Date: ___________
I. Objectives:
Cognitive: Visualized addition of dissimilar fractions without and with regrouping
## II. Learning Content
Skills: Visualized addition of dissimilar fractions without and with regrouping
References: BEC PELC II B 1.2
Enfolding Mathematics V
Materials: flashcards, game boards for square deal, fraction chart, strips
## III. Learning Experiences:
A. Preparatory Activities:
1. Drill on adding similar fractions will be flashed to the class. The pupils give the correct
2. Motivation:
Can we mix oil with water? Why?
Similarly, we cannot just put together dissimilar fractions, can we?
B. Developmental Activities:
1. Presentation:
Strategy: Modeling
## Using a problem opener
Mother has one whole cake. First she sliced 1/3 and then 1/6 if the cake. What part of the
cake did she slice?
i. 1
3 6
Ask: What parts of the cake had been sliced off? What was the total part of the cake that was
sliced off?
1 1
3 6
1
2
2. Practice Exercises
Use diagrams or fractions regions to add the following.
1. 2 + 1 = 3. 2 + 5 = 5. 5 + 1 =
3 4 3 9 8 2
2. 2 + 1 = 4. 3 + 1 =
6 3 8 4
3. Generalization:
How can we add fractions if they are dissimilar? (We make them similar)
IV. Evaluation:
Complete the diagrams by shading them correctly showing the given addition statements.
V. Assignment
Find the sum
1. 11 + 5 = 3. 2 + 7 = 5. 5 + 1 =
12 6 3 8 6 5
2. 1 + 3 = 4. 7 + 3 =
4 5 10 4
MATHEMATICS V
Date: ___________
I. Objectives:
Values: Obedience
## II. Learning Content
References: BEC PELC II B 1.3
Enfolding Mathematics V
Materials: flashcards, concrete objects
## II. Learning Experiences:
A. Preparatory Activities:
1. Mental Computation
Drill on finding the LCM of some given numbers.
Strategy: Relay game
Mechanics:
a. Divide the class into 2 (boys and girls). One representative from each group stands at he
back of the room.
b. Teacher flashes card with 2 to 3 written numbers. (Be sure numbers are manageable by
the pupils)
c. Pupils give the LCM orally6 and pupil who gives the 1st correct answer gets the point.
d. The game continues until all the 10 participants from each group have participated.
e. The group having the most points wins the game.
B. Developmental Activities:
a. Presentation:
Problem opener through cut outs (drawing picture of a circular pizza)
Faith ate 3/6 of a pizza. Mark ate 2/ 12 of the same pizza. How many parts of the pizza did
they eat in all?
2. What are given?
3. What kind of fractions are 3/6 and 2/12?
4. What operation is, needed to solve the problem?
5. Can we easily add 3/6 and 1/12? Why?
6. How can we add them? (Rename 3/6 into a fraction similar to 1/12)
7. Lets solve the problem.
2. Practice Exercises
Find the sum
## + 4/8 + 2/10 + 6/16 + 4/6 + 2/20
3. Generalization:
How do we add dissimilar fractions?
In adding dissimilar fractions, find the LCD first. Then rename them to similar fractions. Add
IV. Evaluation:
Rename these fractions as similar fractions. Add then express the sum in lowest term if possible
1. 2 + 3 = 3. 1 + 3 = 5. 5 + 1 =
8 4 4 6 8 4
2. 2 + 1 = 4. 6 + 1 =
8 2 10 2
V. Assignment
Find the sum and if necessary reduce the answer in its simplest form.
1. 3 + 4 = 3. 6 + 7 = 5. 5 + 10 =
6 10 15 10 9 15
2. 8 + 5= 4. 2 + 3 =
12 9 10 4
MATHEMATICS V
Date: ___________
I. Objectives:
Cognitive: Add dissimilar fraction and whole number
Values: Industry
## II. Learning Content
Skills: Adding the dissimilar fractions and whole numbers
References: BEC PELC II B 1.4
Enfolding Mathematics V
Materials: fraction cards, fraction strips, cut-outs
## II. Learning Experiences:
A. Preparatory Activities:
1. Mental Computation
Drill on giving the LCD of given fractions
Example:
4 , 2 4 , 2 4 , 2
5 3 5 3 5 3
2. Motivation:
Who among you have tasted sweet tamarind candies? Do you have an idea what
ingredients they have?
B. Developmental Activities:
1. Presentation:
Strategy: Modeling Paper Folding
Use a problem Opener
Last week, Mr. Sanchez worked three days in his vegetable garden. He worked 1/3 hour
en the first day, 3/6 hour on the second day and 2 hours on the third day. How long did he
work in all?
1. Do as in strategy l-numbers 1 and 2 you may further as: What good trait do you think has
Mr. Sanchez for having a vegetable garden at home? How can such garden help in
sustaining a family's day to day expenses? What other benefits can you get for
maintaining such garden at home?
2. Divide the class in-groups. Give each group circular cutouts of uniform sizes.
3. Focus their attention on the number sentence they have written on the board.
5. Lead the pupils to notice that the fractions have ' different denominators and are therefore
unlike fractions.
2. Practice Exercises
Find the sum. Express answer in simplest form if possible
1) 4 + 6 + 2 + 3 3) 2 + 1 + 2 + 9 5) 8 + 6 + 3 + 4
3 4 8 2 8 6
2) 5 + 3 + 15 4) 10 + 6 + 1
10 6 12 3
3. Generalization:
How do we add dissimilar fractions and whole numbers?
- Change the dissimilar fractions to similar fractions then add following the rules in adding
similar fractions
- Express the answer in lowest tem of possible
IV. Evaluation:
Find the sum. Express the answer in lowest term of possible
1) 7 + 12 + 3 + 2 = 3) 9 + 3 + 7 + 11 = 5) 15 + 9 + 3 =
10 6 15 6 14 8
2) 9+5 +4 = 4) 6+7+ 4 +3 =
12 8 20 8
V. Assignment
Find the sum: Write the answer in the lowest term if possible
1) 8 + 10 + 2 + 4 = 3) 8 +3+6+4 = 5) 18 + 6 + 4 =
12 9 10 8 15 10
2) 6 +2 +7+2+3 = 4) 12 + 2 + 7 + 3 =
4 9 10 6
MATHEMATICS V
Date: ___________
I. Objectives:
Cognitive: Add whole numbers and mixed forms
## II. Learning Content
Skills: Adding whole numbers and mixed forms
References: BEC PELC II B 1.5
Enfolding Mathematics V
Materials: cut-outs, cardboard/cartolina, pair of scissors
## II. Learning Experiences:
A. Preparatory Activities:
1. Mental Computation
Drill on changing fractions to simplest form
2. Review on adding mixed forms and similar fractions.
B. Developmental Activities:
1. Presentation:
Strategy: Cut-it-Out (Modeling)
Mechanics:
1. The class will be divided into groups of five members.
2. Pupils will cut figures (whole and fractions) from the cartolina.
3. After cutting figures, pupils will construct problem exercise using the cutouts
4. The groups will exchange each others work and do the excises.
2. Practice Exercises
1) 4 + 2 7 = 3) 5 + 5 3 = 5) 9 + 3 4 =
8 4 5
2) 5 + 10 1 = 4) 7 5 + 3 =
2 6
3. Generalization:
What kind of numbers did we add today?
How do we add mixed forms and whole numbers?
IV. Evaluation:
1) 6 + 3 1 = 3) 9+1 2 = 5) 6 + 4 =
10 3 7
2) 4 +5 = 4) 18 + 5 3 =
5 8
V. Assignment
Think of an addition statement that would give the following as the answer.
(Guess and check)
1. ______ + ______ = 11 3
4
2. ______ + ______ = 16 5
8
3. ______ + ______ = 9 4
9
4. ______ + ______ = 16 7
10
5. ______ + ______ = 13 5
11
MATHEMATICS V
Date: ___________
I. Objectives:
Cognitive: Add a mixed form and a dissimilar fraction
Values: Thoughtfulness
## II. Learning Content
Skills: Adding of Mixed Form and Dissimilar Fraction
References: BEC PELC II B 1.6
Enfolding Mathematics V
Materials: fraction cards, cut-outs, number line model
## II. Learning Experiences:
A. Preparatory Activities:
1. Mental Computation.
2. Review on giving LCD of 2 or more fractions
B. Developmental Activities:
1. Presentation:
Strategy: Developing a concept Using Models
Using a Problem Opener
1. Do as in Strategy I - numbers 1 and 2
2. You may ask further: What kind of children do you think are Tina and Ayen? Why is it
important to remember our old folks? What are other ways of showing our love and
concern for them?
3. Post this activity:
a. Represent 1 112 and 3f4 = N by actually putting together 1 1/2 and 3f4 (Guide the
pupils in cutting and pasting the parts together as shown)
b. What value did get for N as shown by the models? (2 1/4 )
4. Using the cut-outs, let the pupils discover the rule in" adding a mixed form and a fraction
by asking some leading questions such as: What did we do with the pair of dissimilar
fractions before we did addition? (change / rename them into similar fractions)
5. Provide more exercises.
2. Practice Exercises
Find the sum
1) 9 1 + 4 2) 4 3 + 1 3) 2 1 + 2
3 4 6 3 4 6
4) 5 2 + 1 5) 1 3 + 3
10 4 4 8
3. Generalization:
How do we add a mixed form and dissimilar fractions?
First rename the fractions into similar fractions. Add as we do with similar fractions.
Express the answer in simplest form if possible.
IV. Evaluation:
1) 6 2 + 1 = 2) 8 5 + 1 = 3) 2 1 + 2 =
3 6 10 4 4 6
4) 10 5 + 3 = 5) 7 7 + 2 =
8 6 10 5
V. Assignment
Find the sum.
1) 3 2 + 1 = 2) 9 4 + 3 = 3) 17 3 + 3 =
7 3 16 4 6 8
4) 4 8 + 3 = 5) 7 + 1 + 3 =
10 4 12 8
|
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# Solving quadratic equations part 2
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• ### Solving quadratic equations part 2
1. 1. Solving Quadratic Equations PART 2
2. 2. Solving Quadratic Equations when they do NOT equal ZERO• So far all the quadratic equations we’ve solve have been equal to zero but not all are so nice. If they don’t equal zero, make them equal to zero!• View this Cool Math website to how to solve Quadratic Equations when the original equation is not equal to 0.• There are 3 pages to view.• Complete the Your Turn problems in your notebook and check your answers on the next slide.
3. 3. Page 1 - Finish it• Solve. 2 x − 3x − 18 = 0
4. 4. Page 1 - Finish it• Solve. 2 x − 3x − 18 = 0 ( x + 3) ( x − 6 ) = 0
5. 5. Page 1 - Finish it• Solve. 2 x − 3x − 18 = 0 ( x + 3) ( x − 6 ) = 0 x+3= 0 or x−6=0
6. 6. Page 1 - Finish it• Solve. 2 x − 3x − 18 = 0 ( x + 3) ( x − 6 ) = 0 x+3= 0 or x−6=0 −3 −3 +6 +6
7. 7. Page 1 - Finish it• Solve. 2 x − 3x − 18 = 0 ( x + 3) ( x − 6 ) = 0 x+3= 0 or x−6=0 −3 −3 +6 +6 x = −3 or x=6
8. 8. Page 1 - Finish it• Solve. 2 x − 3x − 18 = 0 ( x + 3) ( x − 6 ) = 0 x+3= 0 or x−6=0 −3 −3 +6 +6 x = −3 or x=6 Answer: {−3, 6}
9. 9. Page 1 - Your Turn 2• Solve. x + 7x = 170
10. 10. Page 1 - Your Turn 2• Solve. x + 7x = 170 −170 −170
11. 11. Page 1 - Your Turn 2• Solve. x + 7x = 170 −170 −170 2 x + 7x − 170 = 0
12. 12. Page 1 - Your Turn 2• Solve. x + 7x = 170 −170 −170 2 x + 7x − 170 = 0 ( x + 17 ) ( x − 10 ) = 0
13. 13. Page 1 - Your Turn 2• Solve. x + 7x = 170 −170 −170 2 x + 7x − 170 = 0 ( x + 17 ) ( x − 10 ) = 0 x + 17 = 0 or x − 10 = 0
14. 14. Page 1 - Your Turn 2• Solve. x + 7x = 170 −170 −170 2 x + 7x − 170 = 0 ( x + 17 ) ( x − 10 ) = 0 x + 17 = 0 or x − 10 = 0 −17 −17 +10 +10
15. 15. Page 1 - Your Turn 2• Solve. x + 7x = 170 −170 −170 2 x + 7x − 170 = 0 ( x + 17 ) ( x − 10 ) = 0 x + 17 = 0 or x − 10 = 0 −17 −17 +10 +10 x = −17 or x = 10
16. 16. Page 1 - Your Turn 2• Solve. x + 7x = 170 −170 −170 2 x + 7x − 170 = 0 ( x + 17 ) ( x − 10 ) = 0 x + 17 = 0 or x − 10 = 0 −17 −17 +10 +10 x = −17 or x = 10 Answer: {−17,10}
17. 17. Page 2 - Finish it• Solve. 2 2x − 5x − 3 = 0
18. 18. Page 2 - Finish it• Solve. 2 2x − 5x − 3 = 0 ( 2x + 1) ( x − 3) = 0
19. 19. Page 2 - Finish it• Solve. 2 2x − 5x − 3 = 0 ( 2x + 1) ( x − 3) = 0 2x + 1 = 0 or x−3= 0
20. 20. Page 2 - Finish it• Solve. 2 2x − 5x − 3 = 0 ( 2x + 1) ( x − 3) = 0 2x + 1 = 0 or x−3= 0 −1 −1 +3 +3
21. 21. Page 2 - Finish it• Solve. 2 2x − 5x − 3 = 0 ( 2x + 1) ( x − 3) = 0 2x + 1 = 0 or x−3= 0 −1 −1 +3 +3 2x = −1 or x=3
22. 22. Page 2 - Finish it• Solve. 2 2x − 5x − 3 = 0 ( 2x + 1) ( x − 3) = 0 2x + 1 = 0 or x−3= 0 −1 −1 +3 +3 2x = −1 or x=3 2 2
23. 23. Page 2 - Finish it• Solve. 2 2x − 5x − 3 = 0 ( 2x + 1) ( x − 3) = 0 2x + 1 = 0 or x−3= 0 −1 −1 +3 +3 2x = −1 or x=3 2 2 x = − 12
24. 24. Page 2 - Finish it• Solve. 2 2x − 5x − 3 = 0 ( 2x + 1) ( x − 3) = 0 2x + 1 = 0 or x−3= 0 −1 −1 +3 +3 2x = −1 or x=3 2 2 x = − 12 ⎧ 1 ⎫ Answer: ⎨− , 3⎬ ⎩ 2 ⎭
25. 25. Page 2 - Your Turn• Solve. 2 11x + 6 − 10x = 0
26. 26. Page 2 - Your Turn• Solve. 11x + 6 − 10x = 02 ( −1 11x + 6 − 10x 2 ) = 0 ⋅ −1
27. 27. Page 2 - Your Turn• Solve. 11x + 6 − 10x = 02 ( −1 11x + 6 − 10x 2 ) = 0 ⋅ −1 2 −11x − 6 + 10x = 0
28. 28. Page 2 - Your Turn• Solve. 11x + 6 − 10x = 02 ( −1 11x + 6 − 10x 2 ) = 0 ⋅ −1 2 −11x − 6 + 10x = 0 2 10x − 11x − 6 = 0
29. 29. Page 2 - Your Turn• Solve. 11x + 6 − 10x = 02 ( −1 11x + 6 − 10x 2 ) = 0 ⋅ −1 2 −11x − 6 + 10x = 0 2 10x − 11x − 6 = 0 ( 5x + 2 ) ( 2x − 3) = 0
30. 30. Page 2 - Your Turn• Solve. 11x + 6 − 10x = 02 ( −1 11x + 6 − 10x 2 ) = 0 ⋅ −1 2 −11x − 6 + 10x = 0 2 10x − 11x − 6 = 0 ( 5x + 2 ) ( 2x − 3) = 0 5x + 2 = 0 or 2x − 3 = 0
31. 31. Page 2 - Your Turn• Solve. 11x + 6 − 10x = 02 ( −1 11x + 6 − 10x 2 ) = 0 ⋅ −1 2 −11x − 6 + 10x = 0 2 10x − 11x − 6 = 0 ( 5x + 2 ) ( 2x − 3) = 0 5x + 2 = 0 or 2x − 3 = 0 −2 −2 +3 +3
32. 32. Page 2 - Your Turn• Solve. 11x + 6 − 10x = 02 ( −1 11x + 6 − 10x 2 ) = 0 ⋅ −1 2 −11x − 6 + 10x = 0 2 10x − 11x − 6 = 0 ( 5x + 2 ) ( 2x − 3) = 0 5x + 2 = 0 or 2x − 3 = 0 −2 −2 +3 +3 5x = −2 or 2x = 3
33. 33. Page 2 - Your Turn• Solve. 11x + 6 − 10x = 02 ( −1 11x + 6 − 10x 2 ) = 0 ⋅ −1 2 −11x − 6 + 10x = 0 2 10x − 11x − 6 = 0 ( 5x + 2 ) ( 2x − 3) = 0 5x + 2 = 0 or 2x − 3 = 0 −2 −2 +3 +3 5x = −2 or 2x = 3 5 5 2 2
34. 34. Page 2 - Your Turn• Solve. 11x + 6 − 10x = 02 ( −1 11x + 6 − 10x 2 ) = 0 ⋅ −1 2 −11x − 6 + 10x = 0 2 10x − 11x − 6 = 0 ( 5x + 2 ) ( 2x − 3) = 0 5x + 2 = 0 or 2x − 3 = 0 −2 −2 +3 +3 5x = −2 or 2x = 3 5 5 2 2 x = −25 or x= 3 2
35. 35. Page 2 - Your Turn• Solve. 11x + 6 − 10x = 02 ( −1 11x + 6 − 10x 2 ) = 0 ⋅ −1 2 −11x − 6 + 10x = 0 2 10x − 11x − 6 = 0 ( 5x + 2 ) ( 2x − 3) = 0 5x + 2 = 0 or 2x − 3 = 0 −2 −2 +3 +3 5x = −2 or 2x = 3 5 5 2 2 ⎧ 2 3 ⎫ Answer: ⎨− , ⎬ x = −25 or x= 3 2 ⎩ 5 2 ⎭
36. 36. Page 3 - Your Turn 3• Solve. −16x = −4x
37. 37. Page 3 - Your Turn 3• Solve. −16x = −4x +4x 3 +4x 3
38. 38. Page 3 - Your Turn 3• Solve. −16x = −4x +4x 3 +4x 3 3 4x − 16x = 0
39. 39. Page 3 - Your Turn 3• Solve. −16x = −4x +4x 3 +4x 3 3 4x − 16x = 0 ( 2 4x x − 4 = 0)
40. 40. Page 3 - Your Turn 3• Solve. −16x = −4x +4x 3 +4x 3 3 4x − 16x = 0 ( 2 4x x − 4 = 0 ) 4x ( x + 2 ) ( x − 2 ) = 0
41. 41. Page 3 - Your Turn 3• Solve. −16x = −4x +4x 3 +4x 3 3 4x − 16x = 0 ( 2 4x x − 4 = 0 ) 4x ( x + 2 ) ( x − 2 ) = 0 4x = 0 or x+2=0 or x−2=0
42. 42. Page 3 - Your Turn 3• Solve. −16x = −4x +4x 3 +4x 3 3 4x − 16x = 0 ( 2 4x x − 4 = 0) 4x ( x + 2 ) ( x − 2 ) = 0 4x = 0 or x+2=0 or x−2=0 4 4 −2 −2 +2 +2
43. 43. Page 3 - Your Turn 3• Solve. −16x = −4x +4x 3 +4x 3 3 4x − 16x = 0 ( 2 4x x − 4 = 0) 4x ( x + 2 ) ( x − 2 ) = 0 4x = 0 or x+2=0 or x−2=0 4 4 −2 −2 +2 +2 x=0 or x = −2 or x=2
44. 44. Page 3 - Your Turn 3• Solve. −16x = −4x +4x 3 +4x 3 3 4x − 16x = 0 ( 2 4x x − 4 = 0) 4x ( x + 2 ) ( x − 2 ) = 0 4x = 0 or x+2=0 or x−2=0 4 4 −2 −2 +2 +2 x=0 or x = −2 or x=2 Answer: {0, −2, 2}
45. 45. Do you want to Play a GAME?• Check your knowledge on solving Quadratic Equations by playing Jeopardy. Ok, technically it’s called Challenge Board but it’s the same idea!• You have the option to play alone or against a friend or maybe a family member!• You could even arrange a time with a classmate to meet on Pronto to play. Try the App Share feature to see the same game board!• Finding the x-intercepts is the same as solving!
46. 46. Fantastic Job!• You’ve finished reviewing Solving Quadratic Equations Part 2.• Exit and proceed to the Homework Assignment.
|
# How do you find all zeros of the function f(x) = x^2 - 12x + 20?
Apr 2, 2016
zeros of $f \left(x\right)$ are $x = 2$ and $x = 10$
#### Explanation:
$f \left(x\right) = {x}^{2} - 12 x + 20$
We look for two numbers which when multiplied together equal $20$
and which when added together equal $\left(- 12\right)$
With a bit of though we come up with $\left(- 2\right)$ and $\left(- 10\right)$
which allows us to factor:
$f \left(x\right) = {x}^{2} - 12 x + 20 = \left(x - 2\right) \left(x - 10\right)$
For $f \left(x\right)$ to be zero
either
$\textcolor{w h i t e}{\text{XXX}} \left(x - 2\right) = 0 \rightarrow x = 2$
or
$\textcolor{w h i t e}{\text{XXX}} \left(x - 10\right) = 0 \rightarrow x = 10$
|
# Orbital Eccentricity
### Problem
What determines the shapes of the planets’ orbits?
### Materials
• Pencil
• Ruler
• 2 pushpins
• 12-inch (31-cm), square piece of poster board
• 12-inch (31-cm), square piece of thick cardboard
• 10-inch (26-cm) length of string
### Procedure
1. Draw a straight line through the centre of the poster board.
2. Mark two dots on the line 4 inches apart at the centre of the board.
3. Place the poster board on top of the cardboard and stick the pushpins into the dots.
4. Tie the loose ends of the string together to make a loop.
5. Drape the string loop around the pushpins. Place the pencil inside the loop as well and use it stretch the loop into a triangle.
6. Keeping the string taut, guide the pencil around the pushpins while tracing out a path on the poster board below. Continue around the pins until you’ve drawn a closed loop. Describe the shape of the curve you’ve drawn. Is the diameter across the loop always the same, or does it change? Where is it widest? Where is it narrowest?
1. Repeat the above steps on the same poster board with the pins at different distances from each other: 0 inches (just use one pin), 3 inches, and 5 inches. How does the shape of the loop change? How does the distance across the widest and narrowest parts change as the pins get closer together?
### Results
You have four ellipses on the poster board. Each pushpin marks a point in the ellipse called a Focus (plural: Foci). The widest diameter across the ellipse is called the Major axis; the narrowest diameter is known as the Minor axis. The foci sit along the major axis, equidistant from the centre of the ellipse. As the foci get closer together, the ellipse looks more like a circle. An ellipse with only one focus is a circle (the major and minor axes are the same length).
### Why?
In 1609, Johannes Kepler figured out that the planets travel along elliptical paths with the sun sitting at one focus of the ellipse. He called this his First Law of Planetary Motion. As a planet moves along its orbit, the distance between it and the sun changes. The point on the ellipse where the planet is closest to the sun is called the Perihelion; the point where it is farthest is the Aphelion. The Earth passes its perihelion in early January and goes through aphelion in early July. On your ellipses, can you mark where the perihelion and aphelion might be?
The eccentricity of an orbit is a single number, between 0 and 1, which describes how stretched out the orbit is. Zero means the orbit is perfectly circular. An eccentricity close to 1 means the orbit is extremely elongated; only comets coming from the outer reaches of the solar system get close to this value.
You can calculate the eccentricity of your ellipses using the following equation:
• Where EIs the eccentricity,
• AIs the aphelion distance, and
• PIs the perihelion distance.
For each ellipse, pick a focus where the sun should sit (either one will do), then measure the aphelion and perihelion distances. Calculate the eccentricity, and record your results in Table 1.
Distance between foci Aphelion Perihelion Eccentricity 0 inches 3 inches 4 inches 5 inches
Table 1. Calculating the eccentricity of your ellipses
How does the eccentricity change as the foci get farther apart?
Now you know enough to figure out the eccentricities of the planets in the solar system. Table 2 lists the aphelion and perihelion distance of all the major planets (plus one famous comet). The distances are in Astronomical units(AU), where 1 AU is the average distance between the Earth and Sun (93 million miles).
Name Aphelion (AU) Perihelion (AU) Eccentricity Mercury 0.47 0.31 Venus 0.73 0.72 Earth 1.02 0.98 Mars 1.67 1.38 Jupiter 5.46 4.95 Saturn 10.12 9.05 Uranus 20.08 18.38 Neptune 30.44 29.77 Comet Halley 35.1 0.59
Table 2. Orbits in the solar system
### Going Further
Draw scale models of the orbits of all the planets! The first thing you’ll need to do is decide what sort of scale to use. For the inner planets (Mercury, Venus, Earth, and Mars), using 1 AU = 10 cm should work pretty well. For the outer planets, you will need to use either a much larger sheet of poster board or switch to a different scale. Try and figure out what scale would be useful for drawing Jupiter, Saturn, Uranus, and Neptune on a single poster board. What challenges do you run in to if you try to draw all the orbits using one scale?
You also need to calculate how far apart the foci need to be based on the orbital data in Table 2. The distance between the foci is simply the difference between the aphelion and perihelion distances: F= A – p.
1. Convert the aphelion and perihelion distances of the planets to centimeters and record your results in Table 3. Example: Earth aphelion = 1.02 AU x 10 cm/AU = 10.2 cm.
2. Calculate the distance between the foci for each orbit. Record these distances in Table 3.
3. Follow the steps from the original drawings to place the pushpins the right distance apart and set up the string and pencil. To get the string to be the right length, you will need to cut it to be equal to twice the sum of the aphelion and perihelion (plus a few centimeters extra to make room for the knot). Record the length of the string you’ll need in Table 3.
4. Draw the orbits for each of the inner planets on a single poster board. Which is the most eccentric? Which is the least?
Name Aphelion (cm) Perihelion (cm) Foci distance (cm) String length (cm) Mercury Venus Earth Mars Jupiter Saturn Uranus Neptune
Table 3.
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RD Sharma Class 8 Solutions Chapter 9 Linear Equations in One Variable Ex 9.2
These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Linear Equations in One Variable Ex 9.2
Other Exercises
Solve each of the following equations and also check your result in each case :
Question 1.
$$\frac { 2x + 5 }{ 3 }$$ = 3x – 10
Solution:
$$\frac { 2x + 5 }{ 3 }$$ = $$\frac { 3x – 10 }{ 1 }$$
By cross multiplication
⇒ 2x + 5 = 3 (3x – 10)
⇒ 2x + 5 = 9x – 30
⇒ 5 + 30 = 9x – 2x (By transposition)
⇒ 35 = 7x
⇒ x = 5
Question 2.
$$\frac { a – 8 }{ 3 }$$ = $$\frac { a – 3 }{ 2 }$$
Solution:
Question 3.
$$\frac { 7y + 2 }{ 5 }$$ = $$\frac { 6y – 5 }{ 11 }$$
Solution:
Question 4.
x – 2x + 2 – $$\frac { 16 }{ 3 }$$ x + 5 = 3 – $$\frac { 7 }{ 2 }$$ x.
Solution:
Question 5.
$$\frac { 1 }{ 2 }$$ x + 7x – 6 = 7x + $$\frac { 1 }{ 4 }$$
Solution:
Question 6.
$$\frac { 3 }{ 4 }$$ x + 4x = $$\frac { 7 }{ 8 }$$ + 6x – 6
Solution:
Question 7.
$$\frac { 7 }{ 2 }$$ x – $$\frac { 5 }{ 2 }$$ x = $$\frac { 20 }{ 3 }$$ x + 10
Solution:
$$\frac { 7 }{ 2 }$$ x – $$\frac { 5 }{ 2 }$$ x = $$\frac { 20 }{ 3 }$$ x + 10
Question 8.
$$\frac { 6x + 1 }{ 2 }$$ + 1 = $$\frac { 7x – 3 }{ 3 }$$
Solution:
Question 9.
$$\frac { 3a – 2 }{ 3 }$$ + $$\frac { 2a + 3 }{ 2 }$$ = a + $$\frac { 7 }{ 6 }$$
Solution:
Question 10.
x – $$\frac { x – 1 }{ 2 }$$ = 1 – $$\frac { x – 2 }{ 3 }$$
Solution:
Question 11.
$$\frac { 3x }{ 4 }$$ – $$\frac { x – 1 }{ 2 }$$ = $$\frac { x – 2 }{ 3 }$$
Solution:
Question 12.
$$\frac { 5x }{ 3 }$$ – $$\frac { x – 1 }{ 4 }$$ = $$\frac { x – 3 }{ 5 }$$
Solution:
$$\frac { 5x }{ 3 }$$ – $$\frac { x – 1 }{ 4 }$$ = $$\frac { x – 3 }{ 5 }$$
Question 13.
Solution:
Question 14.
Solution:
Question 15.
Solution:
Question 16.
0.18 (5x – 4) = 0.5x + 0.8
Solution:
Question 17.
Solution:
Question 18.
Solution:
Question 19.
Solution:
Question 20.
Solution:
Question 21.
Solution:
Question 22.
Solution:
Question 23.
Solution:
Question 24.
(3x – 8) (3x + 2) – (4x – 11) (2x + 1) = (x – 3) (x + 7)
Solution:
(3x – 8) (3x + 2) – (4x – 11) (2x + 1) = (x – 3) (x + 7)
⇒ (9x² + 6x – 24x – 16) – (8x² + 4x – 22x – 11) = x² + 7x – 3x – 21
⇒ 9x² + 6x – 24x – 16 – 8x² – 4x + 22x + 11 = x² + 4x – 21
⇒ 9x² – 8x² – x² + 6x – 24x + 22x – 4x – 4x = -21 + 16 – 11
⇒ 28x – 32x = -32 + 16
⇒ -4x = -16
⇒ x = 4
Verification:
L.H.S. = (3x – 8) (3x + 2) – (4x – 11) (2x + 1)
= (3 x 4 – 8) (3 x 4 + 2) – (4 x 4 – 11) (2 x 4 + 1)
= (12 – 8) (12 + 2) – (16 – 11) (8 + 1)
= 4 x 14 – 5 x 9 = 56 – 45 = 11
R.H.S. = (x – 3) (x + 7) = (4 – 3) (4 + 7) = 1 x 11 = 11
L.H.S. = R.H.S.
Question 25.
[(2x + 3) + (x + 5)]² + [(2x + 3) – (x + 5)]² = 10x² + 92
Solution:
[(2x + 3) + (x + 5)]² + [(2x + 3) – (x + 5)]² = 10x² + 92
⇒ (2x + 3 + x + 5)² + (2x + 3 – x – 5)² = 10x² + 92
⇒ (3x + 8)² + (x – 2)² = 10x² + 92
⇒ 9x² + 2 x 3x x 8 + 64 + x² – 2 x x x 2 + 4 = 10x² + 92
⇒ 9x² + 48x + 64 + x² – 4x + 4 = 10x² + 92
⇒ 9x² + x² – 10x² + 48x – 4x = 92 – 64 – 4
⇒ 44x = 24
Hope given RD Sharma Class 8 Solutions Linear Equations in One Variable Ex 9.2 are helpful to complete your math homework.
If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.
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# Math in Focus Grade 5 Chapter 7 Practice 6 Answer Key Real-World Problems: More Ratios
Practice the problems of Math in Focus Grade 5 Workbook Answer Key Chapter 7 Practice 6 Real-World Problems: More Ratios to score better marks in the exam.
## Math in Focus Grade 5 Chapter 7 Practice 6 Answer Key Real-World Problems: More Ratios
Question 1.
For a school fair, Lolita’s parents donated 4 bottles of orange juice, 10 bottles of fruit punch and 8 bottles of apple juice. Find the ratio of the number of bottles of orange juice to the number of bottles of fruit punch to the number of bottles of apple juice Lolita’s parents donated.
Answer: 2 : 5 : 4
Explanation:
Given,
Lolita’s parents donated 4 bottles of orange juice,
10 bottles of fruit punch,
8 bottles of apple juice,
By dividing 4 , 10 and 8 with 2 we get 2 : 5 : 4 to get into simplest form.
Question 2.
A company gave a total of $900 to three charities. Charity A received$200, Charity B received $400 and Charity C received the remaining amount. What is the ratio of the amount Charity A received to the amount Charity B received to the amount Charity C received? Answer: 2 : 4 : 3 Explanation: Given, A company gave a$900 to three charities,
Charity A received $200, Charity B received$400,
Charity C received the remaining amount,
By subtracting Charity A’s and Charity B’s amount from the total amount,
900 – 600 = 300
The ratio of amount Charity A received to the amount Charity B received to the amount Charity C is 200 : 400 : 300 to get into the simplest form we get 2 : 4 : 3
Question 3.
Ruth cuts a piece of string into three parts. Their lengths are in the ratio 2:3:5. The longest part is 35 centimeters long. How long is the shortest part?
Answer: The shortest part is 14 cm
Explanation:
Given,
The lengths are in the ratio of 2 : 3 : 5,
The longest part is 35 centimeters long,
By multiplying all the three parts with 7 we get the ratio as 14 : 21 : 35,
Therefore, it states that the longest part is 35 cm and shortest part is 14 cm.
Question 4.
The ages of three brothers, Dave, Randy, and Martin, are in the ratio 1 : 2 : 3. Dave is 7 years old. Find the total age of all three brothers.
Answer: The total age of all three brothers is 42 years
Explanation:
Given,
Dave, Randy and Martin are in the ratio of 1 : 2 : 3,
Dave is 7 years old,
By multiplying all the three ratios with 7 we get the ratio as 7 : 14 : 21,
By adding 7 with 14 we get 21,
By adding 21 with 21 we get 42 as total.
Question 5.
The number of movies that Lisa, Mia, and Nina have seen are in the ratio 6 : 4 : 7. Nina has seen 21 movies this year.
a. How many movies has Lisa seen?
Answer: Lisa has seen 18 movies
Explanation:
Given,
The numbers of movies that Lisa, Mia and Nina have seen are in the ratio of 6 : 4 : 7,
Nina has seen 21 movies,
By multiplying 6, 4 and 7 with 3 we get 18, 12 and 21.
b. What is the total number of movies that the three girls have seen?
Answer: The total number of movies that three girls have seen is 51
Explanation:
Given,
Lisa watched 18 movies,
Mia watched 12 movies,
Nina watched 21 movies,
By adding 18, 12 and 21 we get 51,
So the total number of movies three girls watched is 51.
Question 6.
Amin, Barb, and Curt collected seashells in the ratio of 10 : 12 : 7. Curt collected 98 seashells. How many seashells did they collect together?
Answer: They collected 406 shells together
Explanation:
Given,
Amin, Barb and Curt collected seashells in the ratio of 10 : 12 : 7,
Curt collected 98 seashells,
By multiplying 10, 12 and 7 with 14 we get 140 : 168 : 98,
By adding 140, 168 and 98 we get 406.
Question 7.
By the end of a year, Kieran’s savings is $$\frac{9}{2}$$ of Simon’s savings.
a. What is the ratio of Kieran’s savings to Simon’s savings to their total savings?
Answer: 9 : 2 : 11
Explanation:
Given,
The ratio of Kieran’s savings to Simon’s savings to their total savings is 9 : 2 : 11
b. How many times the total amount of money saved is Kieran’s savings?
Explanation:
Given,
Kieran’s savings is 9,
Total amount of money is 11,
The ratio of Kieran’s savings to the total amount of money is 9 : 11
c. How many times the total amount of money saved is Simon’s savings?
Explanation:
Given,
Simon’s savings is 2,
Total amount of money is 11,
The ratio of Simon’s savings to the total amount of money is 2 : 11
d. Simon saves $28 less than Kieran. How much do both of them save altogether? Answer: They saved 44 altogether Explanation: Given, Simon saves$28 less than Kieran,
Kieran and Simon’s savings are in the ratio of 9 : 2,
By multiplying 9 with 4 we get 36, 2 with 4 we get 8,
Kieran has $36 and Simon has$8,
By adding 36 with 8 we get 44.
Question 8.
Lita, Kala, and Rose entered a typing competition. Lita typed 2 times as fast as Kala. The ratio of the number of words Kala typed to the number of words Rose typed was 4 : 1. If Rose typed 48 words, how many words did Lita type?
Explanation:
Given,
Lita typed 2 times as fast as Kala which is 8,
Kala and Rose typed in the ratio of 4 : 1,
By multiplying 8, 4 and 1 with 48 we get 384 : 192 : 48,
Therefore, Lita typed 384 words in a typing competition
Question 9.
Young’s Dairy Bar produces yogurt in three flavors: vanilla, strawberry, and banana. The amount of vanilla yogurt they produce in a day is 2 times the amount of banana yogurt. The amount of banana yogurt they produce in a day is 3 times the amount of strawberry yogurt.
a. What is the ratio of the amount of vanilla yogurt to the amount of banana yogurt to the amount of strawbery yogurt it produces in a day?
Answer: The ratio of the amount of vanilla yogurt to the amount of banana yogurt to the amount of strawberry yogurt is 6 : 3 : 1
Explanation:
Given,
The amount of vanilla yogurt they produce is 2 times the amount of banana yogurt,
The amount of banana yogurt they produce is 3 times the amount of strawberry amount.
b. How many times the total amount of yogurt produced is the amount of vanilla yogurt produced?
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# How to solve 1/x+1+2/x+2=5/x+4?
hello friends,
Welcome to my article 1/x+1+2/x+2=5/x+4.
This article has been taken from the simplification lesson, in this article we have been told to simplify with the action of addition, subtraction, multiplication, division and fractions.
The solution of this question has been explained in a very simple way by a well-known teacher by doing addition, subtraction, and fractions.
For complete information on how to solve this question 1/x+1+2/x+2=5/x+4, read and understand it carefully till the end.
First of all we should write the article on the page of the notebook.
## 1/x+1+2/x+2=5/x+4
Let’s solve it by placing it up and down,
\displaystyle \frac{1}{x}+1+\frac{2}{x}+2=\frac{5}{x}+4
First of all, we need to understand such questions and solve them in a simple way by keeping the terms of variable amount with variable amount and the terms of constant amount with constant amount.
\displaystyle \frac{1}{x}+\frac{2}{x}+1+2=\frac{5}{x}+4
Now if the denominators are equal, then add their numerators together.
\displaystyle \frac{{1+2}}{x}+3=\frac{5}{x}+4
\displaystyle \frac{3}{x}+3=\frac{5}{x}+4
\displaystyle \frac{3}{x}-\frac{5}{x}=4-3
\displaystyle \frac{{3-5}}{x}=1
\displaystyle \frac{{-2}}{x}=1
\displaystyle -2=x
This article has been completely solved by tireless effort from our side, still if any error remains in it then definitely write us your opinion in the comment box. If you like or understand the methods of solving all the questions in this article, then send it to your friends who are in need.
Note: If you have any such question, then definitely send it by writing in our comment box to get the answer.
Similarly, as an example, see / understand the solution of the question –
how to solve 18/x-7/x+6-2/x-9=3/x+2?
First of all we should write the article on the page of the notebook.
18/x-7/x+6-2/x-9=3/x+2
Let’s solve it by placing it up and down,
\displaystyle \frac{{18}}{x}-\frac{7}{x}+6-\frac{2}{x}-9=\frac{3}{x}+2
First of all, we need to understand such questions and solve them in a simple way by keeping the terms of variable amount with variable amount and the terms of constant amount with constant amount.
\displaystyle \frac{{18}}{x}-\frac{7}{x}-\frac{2}{x}+6-9=\frac{3}{x}+2
\displaystyle \frac{{18-7-2}}{x}-3=\frac{3}{x}+2
\displaystyle \frac{9}{x}-3=\frac{3}{x}+2
Now let’s transpose –
\displaystyle \frac{9}{x}-\frac{3}{x}=3+2
Now if the denominators are equal, then add their numerators together.
\displaystyle \frac{{9-3}}{x}=5
\displaystyle \frac{6}{x}=5
\displaystyle 6=5x
\displaystyle \frac{6}{5}=x
or,
This article has been completely solved by tireless effort from our side, still if any error remains in it then definitely write us your opinion in the comment box. If you like or understand the methods of solving all the questions in this article, then send it to your friends who are in need.
Note: If you have any such question, then definitely send it by writing in our comment box to get the answer.
Similarly, as an example, see / understand the solution of the question
how to solve ,3/x-4+6/x+3=13+7/x+6?
First of all we should write the article on the page of the notebook.
3/x-4+6/x+3=13+7/x+6
Let’s solve it by placing it up and down,
\displaystyle \frac{3}{x}-4+\frac{6}{x}+3=13+\frac{7}{x}+6
First of all, we need to understand such questions and solve them in a simple way by keeping the terms of variable amount with variable amount and the terms of constant amount with constant amount.
\displaystyle \frac{3}{x}+\frac{6}{x}-4+3=\frac{7}{x}+6+13
\displaystyle \frac{{3+6}}{x}-1=\frac{7}{x}+19
\displaystyle \frac{9}{x}-1=\frac{7}{x}+19
Now let’s transpose –
\displaystyle \frac{9}{x}-\frac{7}{x}=1+19
Now if the denominators are equal, then add their numerators together.
\displaystyle \frac{{9+(-7)}}{x}=1+19
\displaystyle \frac{{9-7}}{x}=20
\displaystyle \frac{2}{x}=20
\displaystyle 2=20x
\displaystyle \frac{2}{{20}}=x
\displaystyle \frac{1}{{10}}\text{x }\frac{2}{2}=x
\displaystyle \frac{1}{{10}}=x
or,
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# What Is A Function Rule For Input Output Table?
## Whats a function and not a function?
A function is a relation in which each input has only one output.
: y is a function of x, x is not a function of y (y = 9 has multiple outputs).
: y is not a function of x (x = 1 has multiple outputs), x is not a function of y (y = 2 has multiple outputs)..
## What does a function look like on a table?
A function table displays the inputs and corresponding outputs of a function. Function tables can be vertical (up and down) or horizontal (side to side). In this lesson, we are using horizontal tables.
## Does the input output table represent a function?
A function is a relation where there is only one output for every input. In other words, for every value of x, there is only one value for y. An input-output table is a table that shows how a value changes according to a rule.
## What is the input and output of a function?
In mathematics, a function is any expression that produces exactly one answer for any given number that you give it. The input is the number you feed into the expression, and the output is what you get after the look-up work or calculations are finished.
## What is a function rule example?
A function rule describes how to convert an input value (x) into an output value (y) for a given function. An example of a function rule is f(x) = x^2 + 3.
## What is a function table?
A function table displays the relationship between the inputs and outputs of a specified function. A function table will also follow the rules of a function in that every input only produces one output.
## What is input and output in math definition?
In mathematics, input and output are terms that relate to functions. Both the input and output of a function are variables, which means that they change. You can choose the input variables yourself, but the output variables are always determined by the rule established by the function.
## What is the rule of a function table?
A function table has values of input and output and a function rule. In the function rule, if we plug in different values for the input, we get corresponding values of output. There is always a pattern in the way input values x and the output values y are related which is given by the function rule.
## When the input is what is the output?
An input is data that a computer receives. An output is data that a computer sends. Computers only work with digital information. Any input that a computer receives must be digitised.
## What is the rule for the pattern of numbers?
A numbers pattern is a sequence of numbers that grows or repeats according to a specific rule. For example, the following number pattern starts at 2 and follows the rule add 3: 2, 5, 8, 11, 14….and so forth.
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# Differentiation of Implicit Function
Here you will learn what is the differentiation of implicit function with examples.
Let’s begin –
## Differentiation of Implicit Function
If the variables x and y are connected by a relation of the form f(x,y) = 0 and it is not possible or convenient to express y as a function x in the form y = $$\phi (x)$$, then y is said to be an implicit function of x. To find $$dy\over dx$$ in such a case, we differentiate both sides of the given relation with respect to x, keeping in mind that the derivative of $$\phi (y)$$ with respect to x is $$d\phi\over dy$$.$$dy\over dx$$.
for example, $$d\over dx$$ (sin y) = cos y. $$dy\over dx$$ , $$d\over dx$$ ($$y^2$$) = 2y$$dy\over dx$$
It should be noted that $$d\over dy$$ (sin y) = cos y but $$d\over dx$$ (sin y) = cos y. $$dy\over dx$$.
Similarly , $$d\over dy$$ $$(y^3)$$ = $$3y^2$$ whereas $$d\over dx$$ $$(y^3)$$ = $$3y^2$$ $$dy\over dx$$.
Example : If $$ax^2 + 2hxy + by^2 + 2gx + 2fy + c$$ = 0, find $$dy\over dx$$.
Solution : We have,
$$ax^2 + 2hxy + by^2 + 2gx + 2fy + c$$ = 0
Differentiating both sides of this with respect to x, we get
$$d\over dx$$ $$(ax^2)$$ + $$d\over dx$$(2hxy) + $$d\over dx$$$$(by^2)$$ + $$d\over dx$$(2gx) + $$d\over dx$$(2fy) + $$d\over dx$$(c) = $$d\over dx$$(0)
$$\implies$$ a$$d\over dx$$$$(x^2)$$ + 2h$$d\over dx$$(xy) + b$$d\over dx$$$$(y^2)$$ + 2g$$d\over dx$$(x) + 2f$$d\over dx$$(y) + 0 = 0
$$\implies$$ 2ax + 2h$$(x{dy\over dx} + y)$$ + b 2y$$dy\over dx$$ + 2g.1 + 2f.$$dy\over dx$$ = 0
= $$dy\over dx$$(2hx + 2by + 2f) + 2ax + 2hy + 2g = 0
$$dy\over dx$$ = -$$2(ax + hy + g)\over 2(hx + by + f)$$
$$dy\over dx$$ = -$$(ax + hy + g)\over (hx + by + f)$$
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2 Different Ways to Find a Circle’s Arc Length
Determining the arc length of a circle is easy with these simple formulas
If you’re learning about arc lengths in geometry, your teacher probably just assigned you a bunch of problems for homework. You’ve got the circle’s radius and the central angle, so how do you find the length of the arc? Well, you’ve come to the right place! Arc length is the distance between one endpoint of an arc on a circle to the other. In this article, we’ll tell you what formulas you need and how to use them to find a circle's arc length. Read on to learn more!
Things You Should Know
• When the circle’s central angle is measured in degrees, use the formula ${\displaystyle {\text{arc length}}=2\pi (r)({\frac {\theta }{360}})}$.
• If the central angle is in radians, use the formula ${\displaystyle {\text{arc length}}=\theta (r)}$.
• Plug in the circle’s radius and the central angle’s measurement to solve either formula.
Method 1
Method 1 of 2:
Solving When the Central Angle is in Degrees
1. 1
Set up the formula for arc length. The formula is ${\displaystyle {\text{arc length}}=2\pi (r)({\frac {\theta }{360}})}$, where ${\displaystyle r}$ equals the radius of the circle and ${\displaystyle \theta }$ equals the measurement of the arc’s central angle, in degrees.[1]
2. 2
Plug the length of the circle’s radius into the formula. This information is typically given to you in a problem. Otherwise, measure the circle’s radius with a ruler or protractor. Simply substitute the radius’ value for the variable ${\displaystyle r}$.
• For example, if the circle’s radius is 10 cm, set up the formula like this: ${\displaystyle {\text{arc length}}=2\pi (10)({\frac {\theta }{360}})}$.
3. 3
Insert the value of the arc’s central angle into the formula. Typically, the problem you’re working on provides this information. Make sure to convert the angle to degrees if it’s currently in radians. Then, substitute the central angle’s measurement for ${\displaystyle \theta }$ in the formula.
• For example, if the arc’s central angle is 135 degrees, your formula now looks like: ${\displaystyle {\text{arc length}}=2\pi (10)({\frac {135}{360}})}$.
4. 4
Multiply the radius by . If you are not using a calculator, use the approximation ${\displaystyle \pi =3.14}$ for your calculations. Rewrite the formula using this new value, which represents the circle’s circumference.[2]
• For example, your formula now looks like:
${\displaystyle {\text{arc length}}=2\pi (10)({\frac {135}{360}})}$
${\displaystyle =2(3.14)(10)({\frac {135}{360}})}$
${\displaystyle =(62.8)({\frac {135}{360}})}$
5. 5
Divide the arc’s central angle by 360 degrees. Since a circle has 360 degrees total, dividing the central angle by 360 degrees gives you the portion of the circle that the sector represents. Using this information, find what portion of the circumference the arc length represents.
• For example, simplify the formula to get:
${\displaystyle {\text{arc length}}=(62.8)({\frac {135}{360}})}$
${\displaystyle =(62.8)(.375)}$
6. 6
Multiply the two numbers together. This gives you the length of the arc.
• Solve the formula:
${\displaystyle {\text{arc length}}=(62.8)(.375)=23.55}$
So, the arc length of a circle with a radius of 10 cm and a central angle of 135 degrees is about 23.55 cm.
Method 2
Method 2 of 2:
Solving When the Central Angle is in Radians
1. 1
Set up the formula for arc length. The formula is ${\displaystyle {\text{arc length}}=\theta (r)}$, where ${\displaystyle \theta }$ equals the arc’s central angle in radians, and ${\displaystyle r}$ equals the length of the circle’s radius.[3]
2. 2
Plug the length of the circle’s radius into the formula. The math problem you’re working on typically provides this information. Just substitute the length of the radius for the variable ${\displaystyle r}$.
• For example, if the circle’s radius is 10 cm, your formula looks like this: ${\displaystyle {\text{arc length}}=\theta (10)}$.
3. 3
Plug the measurement of the arc’s central angle into the formula. When using this formula, the arc’s central angle needs to be in radians. If the central angle is in degrees, just convert it into radians.
• For example, if the arc’s central angle is 2.36 radians, your formula now looks like this: ${\displaystyle {\text{arc length}}=2.36(10)}$.
4. 4
Multiply the radius by the arc’s central angle. The product gives you the length of the arc.
• For example:
${\displaystyle {\text{arc length}}=2.36(10)=23.6}$
So, the length of an arc of a circle with a radius of 10 cm and a central angle of 23.6 radians, is about 23.6 cm.
Search
• Question
How do you calculate the length of an arc?
Mario Banuelos, PhD
Assistant Professor of Mathematics
Mario Banuelos is an Assistant Professor of Mathematics at California State University, Fresno. With over eight years of teaching experience, Mario specializes in mathematical biology, optimization, statistical models for genome evolution, and data science. Mario holds a BA in Mathematics from California State University, Fresno, and a Ph.D. in Applied Mathematics from the University of California, Merced. Mario has taught at both the high school and collegiate levels.
Assistant Professor of Mathematics
Support wikiHow by unlocking this expert answer.
To find the arc length, set up the formula Arc length = 2 x pi x radius x (arc's central angle/360), where the arc's central angle is measured in degrees.
• Question
How do l find the arc length if I'm not given the central angle?
wikiHow Staff Editor
This answer was written by one of our trained team of researchers who validated it for accuracy and comprehensiveness.
wikiHow Staff Editor
Support wikiHow by unlocking this staff-researched answer.
If you're not given the central angle, you'll typically be given the sector area of the arc. To find the arc length with a sector area, multiply the sector area by 2. Then, divide the product by the radius squared ((SA*2)/r^2). Your answer gives you the central angle in radians. You now have the central angle in radians, so simply multiply the central angle by the radius to find the arc length.
• Question
Is the measurement of an inscribed angle half the measurement of the central angle that intercepts the same arc?
wikiHow Staff Editor
This answer was written by one of our trained team of researchers who validated it for accuracy and comprehensiveness.
wikiHow Staff Editor
Support wikiHow by unlocking this staff-researched answer.
That is correct. The vertex of an inscribed angle is on the outside of a circle. When an inscribed angle and a central angle share the same arc on a circle, the inscribed angle is equal to half of the measure of the central angle. The inscribed angle is also equal to half the measure of the arc.
200 characters left
Tips
• If you only know the diameter of the circle, just divide the diameter by 2 to get the radius. A circle’s radius is half of its diameter.[4] For example, if the diameter of a circle is 14 cm, divide 14 by 2. ${\displaystyle 14\div 2=7}$. So, the radius of the circle is 7 cm.
⧼thumbs_response⧽
References
1. Mario Banuelos, PhD. Assistant Professor of Mathematics. Expert Interview. 19 January 2021.
3. http://www.mathopenref.com/diameter.html
Co-authored by:
Assistant Professor of Mathematics
This article was co-authored by Mario Banuelos, PhD and by wikiHow staff writer, Devin McSween. Mario Banuelos is an Assistant Professor of Mathematics at California State University, Fresno. With over eight years of teaching experience, Mario specializes in mathematical biology, optimization, statistical models for genome evolution, and data science. Mario holds a BA in Mathematics from California State University, Fresno, and a Ph.D. in Applied Mathematics from the University of California, Merced. Mario has taught at both the high school and collegiate levels. This article has been viewed 919,669 times.
Co-authors: 19
Updated: April 25, 2023
Views: 919,669
Categories: Featured Articles | Geometry
Article SummaryX
To find arc length, start by dividing the arc's central angle in degrees by 360. Then, multiply that number by the radius of the circle. Finally, multiply that number by 2 × pi to find the arc length. If you want to learn how to calculate the arc length in radians, keep reading the article!
Thanks to all authors for creating a page that has been read 919,669 times.
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# How to Break Data Into Quartiles in Excel
Mathematical statistics have a significant role and are frequently used. In statistics, the idea of quarters is crucial. Educate yourself further about quarters and the Quartile Formula.
### Quartile Calculation
Quartile is a statistical term that divides the data into four quarters, just as it sounds phonetically. On the number line, it essentially separates the data points into a data set into four quarters. One thing to remember is that data points can be random, so we must first arrange those numbers in ascending order on the number line before dividing them into quartiles. It essentially serves as an extended middle. Data is divided into two equal parts by the median, and into four parts by the quartiles.
A quartile divides a group of observations into four equally sized segments. The median value for the first phrase makes up the first quartile. Second quartile is the median. The third quartile is the midpoint between the median and the final term.
A set of data’s quartiles can be determined manually, however using the quartile formula in Microsoft Excel makes the process easier. Statistics that divide a collection of data into four equal sections are known as quartiles. The first quartile is reached by 75% of the numbers in the set, whereas the third quartile is reached by 25% of the numbers.
More than half of the values are greater than or equal to the median, or second quartile. (See First Reference) The Excel quartile formula is formatted as follows: QUARTILE (range, quart). The words “range” and “quart” in the formula stand for the range of cells containing the data and the quartile number, respectively.
### How do you split data into Quartiles in Excel?
1. Step 1: In Excel, enter a collection of data that you wish to find the quartiles for in a range of cells in the order of smallest to largest. For instance, select cell A1. In cells A1 through A8, type “2,” “3,” “4”, “5”, “5,” “6,” “8,” and “9,” in that order. After typing in each cell, hit “Enter.”
2. Step 2: In cell C1, click.
3. Step 3: In cells C1 through C3, type “First Quartile,” “Second Quartile,” and “Third Quartile,” correspondingly. After typing in each cell, hit “Enter.”
4. Step 4: In cell D1, click.
5. Step 5: To find the first quartile of the set of data, type “=” and the quartile formula, along with the range cells that hold the data and “1” for the quartile number. After finishing typing, hit “Enter.” Type “=QUARTILE(A1:A8,1),” for instance, and hit “Enter.” The first quartile of the data set in cells A1 through A8 is equal to this, which is 3.75. (See Bibliography (2)
6. Step 6: In cell D2, click.
7. Step 7: To find the first quartile of the set of data, enter “=” and the quartile formula, along with the range cells that hold the data and “2” for the quartile number. After finishing typing, hit “Enter.” Type “=QUARTILE(A1:A8,2),” for instance, and hit “Enter.” The second quartile of the data set in cells A1 through A8 is 5, which is equivalent to this.
8. Step 8: In cell D3, click.
9. Step 9: To find the third quartile of the set of data, type “=” and the quartile formula, along with the range cells that contain the data and “3” for the quartile number. After finishing typing, hit “Enter.” Put in “=QUARTILE(A1:A8,3),” for instance, and hit “Enter.” The third quartile of the data set in cells A1 through A8 is equal to 6.5, which is what this equals.
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# System of equations with complex numbers-circles
The system of equations \begin{align*} |z - 2 - 2i| &= \sqrt{23}, \\ |z - 8 - 5i| &= \sqrt{38} \end{align*} has two solutions $z_1$ and $z_2$ in complex numbers. Find $(z_1 + z_2)/2$.
So far I have gotten the two original equations to equations of circles, $(a-2)^2 +(b-2)^2=23$ and $(a-8)^2+(b-5)^2=38$. From here how do I find the solutions? Thanks.
• Why do you need the original circle equations? – Koba Jun 12 '15 at 21:42
• The point of this exercise is that you realize that $(z_1+z_2)/2$ must lie on the line that connects the centres of the two circles. – M. Wind Jun 12 '15 at 22:01
• Gotcha. I was writing an answer solving those by letting $z=a+bi$. – Koba Jun 12 '15 at 22:07
The first thing to do is to make a drawing of the two circles. Label their centres $C_1$ and $C_2$, and the two intersections of the circles $I_1$ and $I_2$. Now connect these four points by straight lines, and label as $P$ the point half-way between $I_1$ and $I_2$, which is on the line connecting $C_1$ and $C_2$.
We recognize that we have four triangles, all with a $90$ degree angle. So we can apply Pythagoras. For convenience we label the line-piece $C_1$ to $P$ as $a$; $C_2$ to $P$ as $b$; $I_1$ (or $I_2$) to $P$ as $c$. Now:
$$(a+b)^2 = 45$$
$$a^2 + c^2 = 23$$
$$b^2 + c^2 = 38$$
We eliminate $c^2$ by subtracting the second equation from the third, yielding:
$$(b-a)(b+a) = 15$$
Dividing the first equation by this result gives
$$\frac {b+a}{b-a} = 3$$
From which it follows that $b = 2a$. So $a$ is one-third of the distance between $C_1$ and $C_2$. Therefore the position of point $P$ is given by $2+2i$ + $(6+3i)/3$ = $4+3i$.
• Sorry if this should be clear, but I'm not really sure how we know we have four right triangles. – jjhh Jan 15 '18 at 0:01
You're given the distances of the solution points to $A = 2+2i$ and $B = 8+5i.$
One of the solutions and these two complex numbers give you a triangle in the complex plane, and you know the lengths of all the sides.
We can use Law of Cosines to find the angle $\theta$ that has $2+2i$ as its vertex:
$$38 = 45 + 23 - 2 \sqrt{45} \sqrt{23} \cos \theta \to \cos \theta \approx. 0.466252.$$
The projection of $AZ_1$ onto $AB$ is then $\sqrt{23} \cdot 0.466252 \approx 2.23606$.
Why calculate this? Because the other solution is symmetric on the other side of $AB$, and what you're asked to find, in essence, is the average of the two solutions, which lies on $AB$.
The unit vector from $A$ to $B$ is $(6 + 3i)/\sqrt{45} = (2+i)/\sqrt{5}.$
So, the solution is
$$(Z_1 + Z_2)/2 = 2 + 2i + 2.23606 \cdot (6 + 3i)/\sqrt{45} \approx 4 + 3i.$$
(The fact that this worked out so nicely makes me think there was an easier way, but oh well ...)
• nice solution. Though one of the tags is algebra precalculus, so I would think there must be an easier way. I think it definitely has to do with finding the midway point between the centers of two circles. – Koba Jun 12 '15 at 22:25
We can also treat this in a way that has little to do with complex numbers. The line connecting the centers of the circles at $\ (2, \ 2) \$ and $\ (8, \ 5 ) \$ has a slope of $\ \frac{5 - 2}{8 - 2} \ = \ \frac{1}{2} \$ and has the equation
$$y \ - \ 2 \ = \ \frac{1}{2} \ (x - 2) \ \ \Rightarrow \ \ y \ = \ \frac{1}{2} \ x \ + \ 1 \ \ .$$
Using a theorem from classical geometry, the two intersection points of the two circles lie on a mutual chord, of which the line connecting the centers of the circles is its perpendicular bisector (a fact also used in some way by M. Wind and John). Thus, we expect the chord to have a slope of $\ -2 \$ .
A relation for the coordinates of the intersection points of the circles can be found by subtracting one circle equation from the other; thus,
$$\ x^2 \ - \ 16x \ + \ 64 \ + \ y^2 \ - \ 10y \ + \ 25 \ = \ 38$$ $$- \ ( \ x^2 \ - \ 4x \ + \ 4 \ + \ y^2 \ - \ 4y \ + \ 4 \ = \ 23 \ )$$ $$--------------------$$ $$-12x \ + \ 60 \ - \ 6y \ + \ 21 \ = \ 15$$
$$\Rightarrow \ \ y \ = \ -2x \ + \ 11 \ \ .$$
The chord and its perpendicular bisector meet at
$$\frac{1}{2} \ x \ + \ 1 \ = \ -2x \ + \ 11 \ \ \Rightarrow \ \ \frac{5}{2} \ x \ = \ 10 \ \ \Rightarrow \ \ x \ = \ 4 \ \ \Rightarrow \ \ y \ = \ 3 \ \ ,$$
found by inserting the value for $\ x \$ into either of the linear equations we've determined. Hence, the midpoint of the chord containing the intersections of the circles lies at $\ z \ = \ 4 \ + \ 3i \$ in the complex plane.
Though we aren't asked to find them, we can also locate $\ z_1 \ \ \text{and} \ \ z_2 \$ . Inserting $\ y \ = \ -2x \ + \ 11 \$ into the first circle equation, we produce
$$(x - 2)^2 \ + \ (-2x + 11 - 2)^2 \ = \ 23 \ \ \Rightarrow \ \ 5x^2 \ - \ 40x \ + \ 62 \ = \ 0 \ \ ,$$
for which the roots are $\ x \ = \ 4 \ \pm \ \frac{3}{5} \ \sqrt{10} \$ . [Using the other circle equation instead yields exactly the same quadratic equation.]
We may then calculate $\ y \ = \ -2 \ (4 \ \pm \ \frac{3}{5} \ \sqrt{10}) \ + \ 11 \ = \ 3 \ \mp \ \frac{6}{5} \ \sqrt{10} \$ . Thus, we find
$$z_1 \ = \ (4 \ - \ \frac{3}{5} \ \sqrt{10}, \ 3 \ + \ \frac{6}{5} \ \sqrt{10}) \ \ , \ \ z_2 \ = \ (4 \ + \ \frac{3}{5} \ \sqrt{10}, \ 3 \ - \ \frac{6}{5} \ \sqrt{10}) \ \ .$$
(Some brief work with right triangles then tells us that the length of the chord between $\ z_1 \ \ \text{and} \ \ z_2 \$ is $\ \sqrt{\frac{36 + 144}{25} \ \cdot \ 10} \ = \ \sqrt{72} \ = \ 6 \ \sqrt{2} \$ .)
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# Difference between revisions of "2017 AMC 10A Problems/Problem 4"
## Problem
Mia is "helping" her mom pick up $30$ toys that are strewn on the floor. Mia’s mom manages to put $3$ toys into the toy box every $30$ seconds, but each time immediately after those $30$ seconds have elapsed, Mia takes $2$ toys out of the box. How much time, in minutes, will it take Mia and her mom to put all $30$ toys into the box for the first time?
$\textbf{(A)}\ 13.5\qquad\textbf{(B)}\ 14\qquad\textbf{(C)}\ 14.5\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 15.5$
## Solution
Every $30$ seconds, $3$ toys are put in the box and $2$ toys are taken out, so the number of toys increases by $3-2=1$ every $30$ seconds. Then after $27 \times 30 = 810$ seconds (or $13 \frac{1}{2}$ minutes), there are $27$ toys in the box. Mia's mom will then put the remaining $3$ toys into the box after $30$ more seconds, so the total time taken is $27\times 30+30=840$ seconds, or $\boxed{(\textbf{B})\ 14}$ minutes.
~savannahsolver
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Negative exponents tell united state that the power of a number is an unfavorable and it applies to the mutual of the number. We recognize that one exponent refers to the number of times a number is multiplied by itself. For example, 32 = 3 × 3. In the instance of positive exponents, we quickly multiply the number (base) by itself, however what happens once we have an adverse numbers as exponents? A negative exponent is characterized as the multiplicative station of the base, increased to the strength which is opposite to the provided power. In an easy words, we write the mutual of the number and also then resolve it like confident exponents. Because that example, (2/3)-2 deserve to be written as (3/2)2.
You are watching: 2 to the negative 2nd power
1 What are negative Exponents? 2 Negative Exponent Rules 3 Why are an unfavorable Exponents Fractions? 4 Multiplying an adverse Exponents 5 How to Solve negative Exponents? 6 FAQs on an unfavorable Exponents
## What are negative Exponents?
We know that the exponent the a number tells united state how numerous times we should multiply the base. For example, think about 82, 8 is the base, and also 2 is the exponent. We understand that 82 = 8 × 8. A an adverse exponent tells us, how numerous times we need to multiply the reciprocal of the base. Take into consideration the 8-2, here, the basic is 8 and we have actually a negative exponent (-2). 8-2 is expressed as 1/82 = 1/8×1/8.
### Numbers and also Expressions with an adverse Exponents
Here are a few examples i m sorry express negative exponents with variables and also numbers. Watch the table to see just how the number is created in that is reciprocal type and just how the sign of the powers changes.
Negative ExponentResult
2-11/2
3-21/32=1/9
x-31/x3
(2 + 4x)-21/(2+4x)2
(x2+ y2)-31/(x2+y2)3
## Negative Exponent Rules
We have a collection of rules or laws for an unfavorable exponents which do the process of simplification easy. Given below are the an easy rules for solving an unfavorable exponents.
Rule 1: The an adverse exponent ascendancy states that for every number 'a' with the negative exponent -n, take it the mutual of the base and multiply it follow to the worth of the exponent: a(-n)=1/an=1/a×1/a×....n timesRule 2: The dominion for a negative exponent in the denominator argues that for every number 'a' in the denominator and also its negative exponent -n, the result can be composed as: 1/a(-n)=an=a×a×....n times
Let us use these rules and see just how they work-related with numbers.
Example 1: Solve: 2-2 + 3-2
Solution:
Use the negative exponent dominion a-n=1/an2-2 + 3-2 = 1/22 + 1/32 = 1/4 + 1/9
Therefore, 2-2 + 3-2 = 13/36
Example 2: Solve: 1/4-2 + 1/2-3
Solution:
Use the second rule with a negative exponent in the denominator: 1/a-n =an1/4-2 + 1/2-3 = 42 + 23 =16 + 8 = 24
Therefore, 1/4-2 + 1/2-3 = 24.
## Why are an adverse Exponents Fractions?
A an adverse exponent takes united state to the train station of the number. In various other words, a-n = 1/an and also 5-3 i do not care 1/53 = 1/125. This is how an adverse exponents adjust the number to fractions. Let united state take another example to view how negative exponents change to fractions.
Example: fix 2-1 + 4-2
Solution:
2-1 have the right to be written as 1/2 and 4-2 is written as 1/42. Therefore, an adverse exponents get changed to fractions when the sign of their exponent changes.
Multiplication of an unfavorable exponents is the exact same as the multiplication of any kind of other number. As we have currently discussed that negative exponents can be expressed together fractions, for this reason they can easily be addressed after they room converted come fractions. After ~ this conversion, us multiply an adverse exponents utilizing the very same rules the we use for multiplying hopeful exponents. Let's know the multiplication of negative exponents with the complying with example.
Example: Solve: (4/5)-3 × (10/3)-2
The an initial step is to write the expression in its mutual form, which changes the an adverse exponent come a positive one: (5/4)3×(3/10)2Now open the brackets: (frac5^3 imes 3^24^3 imes 10^2)(∵102=(5×2)2 =52×22)Check the usual base and simplify: (frac5^3 imes 3^2 imes 5^-24^3 imes 2^2)(frac5 imes 3^24^3 imes 4)45/44 = 45/256
## How come Solve an adverse Exponents?
Solving any kind of equation or expression is all around operating ~ above those equations or expressions. Similarly, solving an unfavorable exponents is around the leveling of terms with an adverse exponents and also then applying the given arithmetic operations.
Solution:
First, we transform all the an unfavorable exponents to optimistic exponents and then simplify
Given: (frac7^3 imes 3^-421^-2)Convert the an unfavorable exponents to hopeful by composing the reciprocal of the certain number:(frac7^3 imes 21^23^4)Use the rule: (ab)n = one × bn and also split the required number (21).(frac7^3 imes 7^2 imes 3^23^4)Use the rule: to be × one = a(m+n) to combine the usual base (7).75/32 =16807/9
Important Notes:
Note the complying with points which must be remembered while we occupational with an unfavorable exponents.
See more: How Bottles Of Water Is A Gallon S? How Many Bottles Of Water Are In A Gallon, How
Exponent or power method the variety of times the base needs to be multiply by itself.am = a × a × a ….. M timesa-m = 1/a × 1/a × 1/a ….. M timesa-n is likewise known together the multiplicative train station of an.If a-m = a-n climate m = n.The relation between the exponent (positive powers) and also the an unfavorable exponent (negative power) is expressed as ax=1/a-x
### Topics associated to an adverse Exponents
Check the provided articles similar or regarded the negative exponents.
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# Mathematical Vocabulary Page
• This is a list of mathematical vocabulary words to assist students with understanding computational and reasoning concepts. Students will use this vocabulary listing to add to their glossary of terms during the school year.
• Absolute Value - the distance a number is away from zero
Area - the covering of a two-dimensional shape
Coordinate Plane - the plane containing the x- and y- axes
Difference - the solution to a subtraction problem
Dividend - the number that is being split into groups
Divisor - a number that divides into another
Equivalent Ratio - two ratios that express the same relationship; all equivalent ratios can be simplified to the same ratio value
Inequality - states that two values are not equal
Integer - a positive or negative whole number
Lateral Surface Area - the area of the sides (faces) of a three-dimensional object
Net - a two-dimensional representation of a three-dimensional object
Opposite - two integers can be opposites if they are equal distance away from zero on the number line but on different sides of zero
Origin - the point (0, 0); where the x- and y- axes intersect
Parallelogram - a quadrilateral with opposite sides parallel and opposite sides congruent
Product - the solution to a multiplication problem
Proportion - two equal ratios
Quadrant - four regions of the coordinate plane
Quotient - the solution to a division problem
Ratio - a comparison between two different quantities
Ratio Table - a table which includes values that are equivalent ratios
Rational Number - a number that can be written as a fraction, as a terminating decimal or as a repeating number
Rectangular Prism - a three-dimensional object with two rectangular bases
Scale Factor - the ratio of two corresponding sides in a set of similar figures
Sum - the solution to an addition problem
Total Surface Area - the area of the faces and bases of a three-dimensional object
Trapezoid - a quadrilateral with one set of parallel sides
Triangle - a three-sided polygon with a sum of 180 degrees in the angle measures
Volume - the amount of space a three-dimensional object occupies
Whole Number - a postive counting number starting with zero
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Fraction calculator
The calculator performs basic and advanced operations with fractions, expressions with fractions combined with integers, decimals, and mixed numbers. It also shows detailed step-by-step information about the fraction calculation procedure. Solve problems with two, three, or more fractions and numbers in one expression.
Result:
53/4 - 35/6 = 23/12 = 1 11/12 ≅ 1.9166667
Spelled result in words is twenty-three twelfths (or one and eleven twelfths).
How do you solve fractions step by step?
1. Conversion a mixed number 5 3/4 to a improper fraction: 5 3/4 = 5 3/4 = 5 · 4 + 3/4 = 20 + 3/4 = 23/4
To find new numerator:
a) Multiply the whole number 5 by the denominator 4. Whole number 5 equally 5 * 4/4 = 20/4
b) Add the answer from previous step 20 to the numerator 3. New numerator is 20 + 3 = 23
c) Write a previous answer (new numerator 23) over the denominator 4.
Five and three quarters is twenty-three quarters
2. Conversion a mixed number 3 5/6 to a improper fraction: 3 5/6 = 3 5/6 = 3 · 6 + 5/6 = 18 + 5/6 = 23/6
To find new numerator:
a) Multiply the whole number 3 by the denominator 6. Whole number 3 equally 3 * 6/6 = 18/6
b) Add the answer from previous step 18 to the numerator 5. New numerator is 18 + 5 = 23
c) Write a previous answer (new numerator 23) over the denominator 6.
Three and five sixths is twenty-three sixths
3. Subtract: 23/4 - 23/6 = 23 · 3/4 · 3 - 23 · 2/6 · 2 = 69/12 - 46/12 = 69 - 46/12 = 23/12
For adding, subtracting, and comparing fractions, it is suitable to adjust both fractions to a common (equal, identical) denominator. The common denominator you can calculate as the least common multiple of both denominators - LCM(4, 6) = 12. In practice, it is enough to find the common denominator (not necessarily the lowest) by multiplying the denominators: 4 × 6 = 24. In the next intermediate step, the fraction result cannot be further simplified by canceling.
In words - twenty-three quarters minus twenty-three sixths = twenty-three twelfths.
Rules for expressions with fractions:
Fractions - use the slash “/” between the numerator and denominator, i.e., for five-hundredths, enter 5/100. If you are using mixed numbers, be sure to leave a single space between the whole and fraction part.
The slash separates the numerator (number above a fraction line) and denominator (number below).
Mixed numerals (mixed fractions or mixed numbers) write as non-zero integer separated by one space and fraction i.e., 1 2/3 (having the same sign). An example of a negative mixed fraction: -5 1/2.
Because slash is both signs for fraction line and division, we recommended use colon (:) as the operator of division fractions i.e., 1/2 : 3.
Decimals (decimal numbers) enter with a decimal point . and they are automatically converted to fractions - i.e. 1.45.
The colon : and slash / is the symbol of division. Can be used to divide mixed numbers 1 2/3 : 4 3/8 or can be used for write complex fractions i.e. 1/2 : 1/3.
An asterisk * or × is the symbol for multiplication.
Plus + is addition, minus sign - is subtraction and ()[] is mathematical parentheses.
The exponentiation/power symbol is ^ - for example: (7/8-4/5)^2 = (7/8-4/5)2
Examples:
adding fractions: 2/4 + 3/4
subtracting fractions: 2/3 - 1/2
multiplying fractions: 7/8 * 3/9
dividing Fractions: 1/2 : 3/4
exponentiation of fraction: 3/5^3
fractional exponents: 16 ^ 1/2
adding fractions and mixed numbers: 8/5 + 6 2/7
dividing integer and fraction: 5 ÷ 1/2
complex fractions: 5/8 : 2 2/3
decimal to fraction: 0.625
Fraction to Decimal: 1/4
Fraction to Percent: 1/8 %
comparing fractions: 1/4 2/3
multiplying a fraction by a whole number: 6 * 3/4
square root of a fraction: sqrt(1/16)
reducing or simplifying the fraction (simplification) - dividing the numerator and denominator of a fraction by the same non-zero number - equivalent fraction: 4/22
expression with brackets: 1/3 * (1/2 - 3 3/8)
compound fraction: 3/4 of 5/7
fractions multiple: 2/3 of 3/5
divide to find the quotient: 3/5 ÷ 2/3
The calculator follows well-known rules for order of operations. The most common mnemonics for remembering this order of operations are:
PEMDAS - Parentheses, Exponents, Multiplication, Division, Addition, Subtraction.
BEDMAS - Brackets, Exponents, Division, Multiplication, Addition, Subtraction
BODMAS - Brackets, Of or Order, Division, Multiplication, Addition, Subtraction.
GEMDAS - Grouping Symbols - brackets (){}, Exponents, Multiplication, Division, Addition, Subtraction.
Be careful, always do multiplication and division before addition and subtraction. Some operators (+ and -) and (* and /) has the same priority and then must evaluate from left to right.
Fractions in word problems:
• Cupcakes
In a bowl was some cupcakes. Janka ate one third and Danka ate one quarter of cupcakes. a) How many of cookies ate together? b) How many cookies remain in a bowl? Write the results as a decimal number and in notepad also as a fraction.
• Algebra problem
This is algebra. Let n represent an unknown number. 1. Eight more than the number n 2. Three times the number n 3. The product of the number n and eight 4. Three less than the number n 5. Three decreased by the number n
• Toilets
Federal law requires that all residential toilets sold in the United States use no more than 1 3/5 gallons of water per flush. Before this legislation, conventional toilets used 3 2/5 gallons of water per flush. Find the amount of water saved in one year
• Savings
Eva borrowed 1/3 of her savings to her brother, 1/2 of savings spent in the store and 7 euros left. How much did she save?
• Square metal sheet
We cut out four squares of 300 mm side from a square sheet metal plate with a side of 0,7 m. Express the fraction and the percentage of waste from the square metal sheet.
• Difference of two fractions
What is the difference between 1/2 and 1/6? (Write the answer as a fraction in lowest terms. )
• Package
The package was 23 meters of textile. The first day sold 12.3 meters. How many meters of textile remained in the package?
• Michael
Michael had a bar of chocolate. He ate 1/2 of it and gave away 1/3. What fraction had he left?
• You leave
You leave school and the end of the day and walk 3/8 of a mile away before realizing that you left your backpack and immediately turn around you then walk 1/6 of a mile back towards school at this point assuming you walked in a straight line how many mile
• Akpan
Akpan spent 3/8 of his time in school during the week. What fraction of his time does he spend at home during the week?
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# Decimals - NCERT Solutions for Class 6 Math
NCERT Solutions for Class 6 Math Chapter 8 Decimals are provided here with simple step-by-step explanations. These solutions for Decimals are extremely popular among class 6 students for Math Decimals Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the NCERT Book of class 6 Math Chapter 8 are provided here for you for free. You will also love the ad-free experience on Meritnation’s NCERT Solutions. All NCERT Solutions for class 6 Math are prepared by experts and are 100% accurate.
#### Question 1:
Write the following as numbers in the given table.
(a)
(b)
Hundreds(100) Tens (10) Ones (1) Tenths
It may be observed that
Row Hundreds Tens Ones Tenths a. 0 3 1 2 b. 1 1 0 4
#### Question 2:
Write the following decimals in the place value table.
(a) 19.4 (b) 0.3
(c) 10.6 (d) 205.9
Decimal Hundreds Tens Ones Tenths 19.4 0 1 9 4 0.3 0 0 0 3 10.6 0 1 0 6 205.9 2 0 5 9
#### Question 3:
Write each of the following as decimals:
(a) Seven-tenths (b) Two tens and nine-tenths
(c) Fourteen point six (d) One hundred and two ones
(e) Six hundred point eight
(a) Seven-tenths = = 0.7
(b) Two tens and nine-tenths = 20 + = 20.9
(c) Fourteen point six = 14.6
(d) One hundred and two ones = 100 + 2 = 102.0
(e) Six hundred point eight = 600.8
#### Question 4:
Write each of the following as decimals:
(a) (b)
(c) (d)
(e) (f)
(g) (h)
(i) (j)
(k)
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
(k)
#### Question 5:
Write the following decimals as fractions. Reduce the fractions to lowest form.
(a) 0.6 (b) 2.5 (c) 1.0 (d) 3.8
(e) 13.7 (f) 21.2 (g) 6.4
(a)
(b)
(c) 1.0 = 1
(d)
(e)
(f)
(g)
#### Question 6:
Express the following as cm using decimals.
(a) 2 mm (b) 30 mm
(c) 116 mm (d) 4 cm 2 mm
(e) 162 mm (f) 83 mm
It is known that 1cm = 10 mm
(a)
(b)
(c)
(d)
(e)
(f)
#### Question 7:
Between which two whole numbers on the number line are the given numbers lie? Which of these whole numbers is nearer the number?
(a) 0.8 (b) 5.1
(c) 2.6 (d) 6.4
(e) 9.1 (f) 4.9
(a) 0.8 lies between 0 and 1, and is nearer to 1.
(b) 5.1 lies between 5 and 6, and is nearer to 5.
(c) 2.6 lies between 2 and 3, and is nearer to 3.
(d) 6.4 lies between 6 and 7, and is nearer to 6.
(e) 9.1 lies between 9 and 10, and is nearer to 9.
(f) 4.9 lies between 4 and 5, and is nearer to 5.
#### Question 8:
Show the following numbers on the number line.
(a) 0.2 (b) 1.9
(c) 1.1 (d) 2.5
(a) 0.2 represents a point between 0 and 1 on number line, such that the
space between 0 and 1 is divided into 10 equal parts. Hence, each equal part will be equal to one-tenth. Now, 0.2 is the second point between 0 and 1.
(b) 1.9 represents a point between 1 and 2 on number line, such that the
space between 1 and 2 is divided into 10 equal parts. Hence, each equal part will be equal to one-tenth. Now, 1.9 is the ninth point between 1 and 2.
(c) 1.1 represents a point between 1 and 2 on number line, such that the
space between 1 and 2 is divided into 10 equal parts. Hence, each equal part will be equal to one-tenth. Now, 1.1 is the first point between 1 and 2.
(d) 2.5 represents a point between 2 and 3 on number line, such that the
space between 2 and 3 is divided into 10 equal parts. Hence, each equal part will be equal to one-tenth. Now, 2.5 is the fifth point between 2 and 3.
#### Question 9:
Write the decimal number represented by the points A, B, C, D on the given number line?
Point A represents 0.8.
Point B represents 1.3.
Point C represents 2.2.
Point D represents 2.9.
##### Video Solution for decimals (Page: 168 , Q.No.: 9)
NCERT Solution for Class 6 math - decimals 168 , Question 9
#### Question 10:
(a) The length of Ramesh’s notebook is 9 cm 5 mm. What will be its length in cm?
(b) The length of a young gram plant is 65 mm. Express its length in cm.
(a) The length of Ramesh’s notebook is 9 cm 5 mm.
Therefore, the length in cm is
(b) The length of a gram plant is 65 mm.
Therefore, the length in cm is
#### Question 1:
Complete the table with the help of these boxes and use decimals to write the number.
(a)
(b)
(c)
Ones Tenths Hundredths Number (a) - - - - (b) - - - - (c) - - - -
Row Ones Tenths Hundredths Numbers (a) 0 2 6 0.26 (b) 1 3 8 1.38 (c) 1 2 8 1.28
#### Question 2:
Write the numbers given in the following place value table in decimal form.
Hundreds 100 Tens 10 Ones 1 Tenths Hundredths Thousandths (a) 0 0 3 2 5 0 (b) 1 0 2 6 3 0 (c) 0 3 0 0 2 5 (d) 2 1 1 9 0 2 (e) 0 1 2 2 4 1
(a)
(b)
(c)
(d)
(e)
#### Question 3:
Write the following decimals in the place value table.
(a) 0.29 (b) 2.08 (c) 19.60 (d) 148.32
(e) 200.812
(a)
(b)
(c)
(d)
(e)
Row Hundreds Tens Ones Tenths Hundredths Thousandths (a) 0 0 0 2 9 0 (b) 0 0 2 0 8 0 (c) 0 1 9 6 0 0 (d) 1 4 8 3 2 0 (e) 2 0 0 8 1 2
#### Question 4:
Write each of the following decimals.
(a) (b)
(c) (d)
(e)
(a)
(b)
(c)
(d)
(e)
#### Question 5:
Write each of the following decimals in words.
(a) 0.03 (b) 1.20
(c) 108.56 (d) 10.07
(e) 0.032 (f) 5.008
(a) 0.03 = zero point zero three
(b) 1.20 = one point two zero
(c) 108.56 = one hundred eight point five six
(d) 10.07 = ten point zero seven
(e) 0.032 = zero point zero three two
(f) 5.008 = five point zero zero eight
#### Question 6:
Between which two numbers in tenths place on the number line does each of the given number lie?
(a) 0.06 (b) 0.45
(c) 0.19 (d) 0.66
(e) 0.92 (f) 0.57
(a) 0.06 0 and 0.1
(b) 0.45 0.4 and 0.5
(c) 0.19 0.1 and 0.2
(d) 0.66 0.6 and 0.7
(e) 0.92 0.9 and 1.0
(f) 0.57 0.5 and 0.6
#### Question 7:
Write as fractions in lowest terms.
(a) 0.60 (b) 0.05
(c) 0.75 (d) 0.18
(e) 0.25 (f) 0.125
(g) 0.066
(a)
(b)
(c)
(d)
(e)
(f)
(g)
#### Question 1:
Which is greater?
(a) 0.3 or 0.4 (b) 0.07 or 0.02
(c) 3 or 0.8 (d) 0.5 or 0.05
(e) 1.23 or 1.2 (f) 0.099 or 0.19
(g) 1.5 or 1.50 (h) 1.431 or 1.490
(i) 3.3 or 3.300 (j) 5.64 or 5.603
(a) 0.3 or 0.4
The whole parts of these numbers are same. It can be seen that the tenth part of 0.4 is greater than that of 0.3.
Hence, 0.4 > 0.3
(b) 0.07 and 0.02
Here, both numbers have same parts up to the tenth place. However, the hundredth part of 0.07 is greater than that of 0.02.
Hence, 0.07 > 0.02
(c) 3 or 0.8
It can be seen that the whole part of 3 is greater than that of 0.8.
Hence, 3 > 0.8
(d) 0.5 or 0.05
The whole parts of these numbers are same. It can be seen that the tenth part of 0.5 is greater than that of 0.05.
Hence, 0.5 > 0.05
(e) 1.23 or 1.20
Here, both numbers have same parts up to the tenth place. However, the hundredth part of 1.23 is greater than that of 1.20.
Hence, 1.23 > 1.20
(f) 0.099 or 0.19
The whole parts of these numbers are same. It can be seen that the tenth part of 0.19 is greater than that of 0.099.
Hence, 0.099 < 0.19
(g) 1.5 or 1.50
Here, both numbers have the same parts up to the tenth place. Also, there is no digit at hundredth place of 1.5. This implies that this digit will be 0, which is same as the digit at the hundredth place of 1.50. Therefore, both these numbers are equal.
(h) 1.431 or 1.490
Here, both numbers have the same parts up to the tenth place. However, the hundredth part of 1.490 is greater than that of 1.431.
Hence, 1.431 < 1.490
(i) 3.3 or 3.300
Here, both numbers have the same parts up to the tenth place. Also, there is no digit at hundredth and thousandth place of 3.3. This implies that these digits are 0, which are the same as the digits at the hundredth and thousandth place of 3.300. Therefore, both these numbers are equal.
(j) 5.64 or 5.603
Here, both numbers have the same parts up to the tenth place. However, the hundredth part of 5.64 is greater than that of 5.603.
Hence, 5.640 > 5.603
#### Question 1:
Express as rupees using decimals.
(a) 5 paise (b) 75 paise
(c) 20 paise (d) 50 rupees 90 paise
(e) 725 paise
It is known that there are 100 paise in 1 rupee.
(a)
(b)
(c)
(d)
(e)
#### Question 2:
Express as metres using decimals.
(a) 15 cm (b) 6 cm
(c) 2 m 45 cm (d) 9 m 7 cm
(e) 419 cm
It is known that there are 100 cm in 1 metre.
(a)
(b)
(c)
(d)
(e)
#### Question 3:
Express as cm using decimals.
(a) 5 mm (b) 60 mm
(c) 164 mm (d) 9 cm 8 mm
(e) 93 mm
It is known that there are 10 mm in 1 cm.
(a)
(b)
(c)
(d)
(e)
#### Question 4:
Express as km using decimals.
(a) 8 m (b) 88 m
(c) 8888 m (d) 70 km 5 m
It is known that there are 1000 metres in 1 km.
(a)
(b)
(c)
(d)
#### Question 5:
Express as kg using decimals.
(a) 2 g (b) 100 g
(c) 3750 g (d) 5 kg 8 g
(e) 26 kg 50 g
It is known that there are 1000 grams in 1 kg.
(a)
(b)
(c)
(d)
(e)
#### Question 1:
Find the sum in each of the following:
(a) 0.007 + 8.5 + 30.08
(b) 15 + 0.632 + 13.8
(c) 27.076 + 0.55 + 0.004
(d) 25.65 + 9.005 + 3.7
(e) 0.75 + 10.425 + 2
(f) 280.69 + 25.2 + 38
(a) 0.007 + 8.5 + 30.08
(b) 15 + 0.632 + 13.8
(c) 27.076 + 0.55 + 0.004
(d) 25.65 + 9.005 + 3.7
(e) 0.75 + 10.425 + 2
(f) 280.69 + 25.2 + 38
#### Question 2:
Rashid spent Rs 35.75 for Maths book and Rs 32.60 for Science book. Find the total amount spent by Rashid.
Price of Maths book = Rs. 35.75
Price of Science book = Rs. 32.60
Total amount spent by Rashid is
Therefore, the amount spent by Rashid is Rs 68.35.
#### Question 3:
Radhika’s mother gave her Rs 10.50 and her father gave her Rs 15.80, find the total amount given to Radhika by the parents.
Amount given by mother = Rs. 10.50
Amount given by mother = Rs. 15.80
Total amount given by parents is
Therefore, the amount given by her parents is Rs 26.30.
#### Question 4:
Nasreen bought 3 m 20 cm cloth for her shirt and 2 m 5 cm cloth for her trouser. Find the total length of cloth bought by her.
Cloth for shirt = 3 m 20 cm
Cloth for trouser = 2 m 5 cm
Total length of cloth is
Hence, the total length of cloth bought by her is 5.25 m.
#### Question 5:
Naresh walked 2 km 35 m in the morning and 1 km 7 m in the evening. How much distance did he walk in all?
Distance walked in the morning = 2 km 35 m
= 2.035 km
Distance walked in the evening = 1 km 7 m
= 1.007 km
Total distance walked by him is
##### Video Solution for decimals (Page: 179 , Q.No.: 5)
NCERT Solution for Class 6 math - decimals 179 , Question 5
#### Question 7:
Ravi purchased 5 kg 400 g rice, 2 kg 20 g sugar and 10 kg 850 g flour. Find the total weight of his purchases.
Weight of rice = 5 kg 400 g =
Weight of sugar = 2 kg 20 g =
Weight of flour = 10 kg 850 g =
Total weight of his purchases is
#### Question 6:
Sunita travelled 15 km 268 m by bus, 7 km 7 m by car and 500 m on foot in order to reach her school. How far is her school from her residence?
Distance travelled by bus = 15 km 268 m
= 15.268 km
Distance travelled by car = 7 km 7 m
= 7.007 km
Distance travelled on foot = 500 m
= 0.500 km
Total distance of school from her residence is
##### Video Solution for decimals (Page: 180 , Q.No.: 6)
NCERT Solution for Class 6 math - decimals 180 , Question 6
#### Question 1:
Subtract:
(a) Rs 18.25 from Rs 20.75
(b) 202.54 m from 250 m
(c) Rs 5.36 from Rs 8.40
(d) 2.051 km from 5.206 km
(e) 0.314 kg from 2.107 kg
(a) Rs 20.75 − Rs 18.25
(b) 250 m − 202.54 m
(c) Rs 8.40 − Rs 5.36
(d) 5.206 km − 2.051 km
(e) 2.107 kg − 0.314 kg
#### Question 2:
Find the value of:
(a) 9.756 − 6.28
(b) 21.05 − 15.27
(c) 18.5 − 6.79
(d) 11.6 − 9.847
(a)
(b)
(c)
(d)
#### Question 3:
Raju bought a book for Rs 35.65. He gave Rs 50 to the shopkeeper. How much money did he get back from the shopkeeper?
Money that Raju will get back will be the difference of these two.
Hence, money that Raju will get back is
Therefore, he will get back Rs 14.35.
#### Question 4:
Rani had Rs 18.50. She bought one ice-cream for Rs 11.75. How much money does she have now?
The money left with Rani will be the difference of these two.
Hence, the money left is
##### Video Solution for decimals (Page: 182 , Q.No.: 4)
NCERT Solution for Class 6 math - decimals 182 , Question 4
#### Question 5:
Tina had 20 m 5 cm long cloth. She cuts 4 m 50 cm length of cloth from this for making a curtain. How much cloth is left with her?
The length of the cloth left with her will be the difference of these two.
Hence, the length of the cloth left with her is
Therefore, 15.55 m cloth will be remaining.
#### Question 6:
Namita travels 20 km 50 m every day. Out of this, she travels 10 km 200 m by bus and the rest by auto. How much distance does she travel by auto?
#### Question 7:
Aakash bought vegetables weighing 10 kg. Out of this, 3 kg 500 g is onions, 2 kg 75 g is tomatoes and the rest is potatoes. What is the weight of the potatoes?
Total weight of vegetables bought = 10.000 kg
Weight of onions = 3 kg 500 g = 3.500 kg
Weight of tomatoes = 2 kg 75 g = 2.075 kg
Weight of potatoes = Total weight of vegetables bought − (Weight of onions + Weight of tomatoes)
= 10.000 − (3.500 + 2.075)
Hence, the weight of the potatoes was 4.425 kg.
View NCERT Solutions for all chapters of Class 6
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## Elementary Algebra
{$\frac{1-\sqrt {85}}{6},\frac{1+\sqrt {85}}{6}$}
Step 1: We write $7=3x^{2}-x$ as $3x^{2}-x-7=0$. Comparing $3x^{2}-x-7=0$ to the standard form of a quadratic equation, $ax^{2}+bx+c=0$, we find: $a=3$, $b=-1$ and $c=-7$ Step 2: The quadratic formula is: $x=\frac{-b \pm \sqrt {b^{2}-4ac}}{2a}$ Step 3: Substituting the values of a, b and c in the formula: $x=\frac{-(-1) \pm \sqrt {(-1)^{2}-4(3)(-7)}}{2(3)}$ Step 4: $x=\frac{1 \pm \sqrt {1+84}}{6}$ Step 5: $x=\frac{1 \pm \sqrt {85}}{6}$ Step 6: $x=\frac{1-\sqrt {85}}{6}$ or $x=\frac{1+\sqrt {85}}{6}$ Step 7: Therefore, the solution set is {$\frac{1-\sqrt {85}}{6},\frac{1+\sqrt {85}}{6}$}.
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# 5 Easy to Implement Fraction Games
Teaching fractions in the upper elementary classroom can be a daunting task! If you want students to be engaged and learning simultaneously, turning an activity into a game is always a win! I’ve used dominoes for a variety of math games. I just started using them for fractions; of course, the students always want to play more!
## Weigh 'Em up! (Compare Fractions)
1. Students choose two dominoes and turn them over.
2. One side is the numerator; the other side is the denominator.
3. Compare both dominoes.
4. Want to make it a game? Each partner chooses one domino. Then they compare their fraction. The partner with the greatest (or least) fraction wins.
Option for comparing fractions with the same denominator. Have students add both denominators together to create a common denominator.
## Order up! (Ordering Fractions)
1. Students choose five dominoes and turn them over.
2. One side is the numerator; the other side is the denominator.
3. Order the fractions from least to greatest or greatest to least.
4. Want to make it a game? Partners order their dominos and then find the difference between their greatest and least fraction. The partner with the greatest (or least) difference wins.
Option for ordering fractions with the same denominator. Have students pick one domino. The denominator on that one domino represents the denominators for all the other fractions.
## What’s the Value? (Add/Subtract fractions)
1. Students choose two dominoes and turn them over.
2. One side is the numerator; the other side is the denominator.
3. Add or subtract the fractions.
4. Want to make it a game? Partners compare their sum or difference. The partner with the greatest (or least) sum or difference wins.
Option for adding or subtracting fractions with the same denominator. Have students add both denominators together to create a common denominator.
## Fraction Feud (Multiplying Fractions)
1. Students begin with dominoes face down.
2. Each student chooses two dominoes.
3. One side is the numerator; the other side is the denominator.
4. On the count of three, students turn over their dominoes and multiply their fractions together.
5. The student with the highest product wins the dominoes.
Option for multiplying fractions by whole numbers. Have students pick one domino. That domino will represent their whole number. They get the whole number by adding the dots together.
## Divide and Conquer (Dividing Fractions)
1. Students choose two dominoes and turn them over.
3. The other domino is your fraction.
4. One side is the numerator; the other side is the denominator.
5. Want to make it a game? The partner with the greatest (or least) quotient wins.
You can download all five games in print friendly format with one easy step below!
FREE Fractions Games
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# Is the quotient of a rational number and an irrational number irrational?
## Is the quotient of a rational number and an irrational number irrational?
In general, the quotient of two rational numbers (given the denominator is not 0) is rational. Because a rational z contradicts our definition that x is irrational, this must mean z is irrational. Therefore, the product of a rational and irrational number is irrational.
## How do you divide irrational fractions?
0:011:20How to divide an irrational number by another irrational number? - YouTubeYouTubeStart of suggested clipEnd of suggested clipNumbers let's take an example say root 2 divided by 8 root 3 this will be root 2 by root 3 we cannotMoreNumbers let's take an example say root 2 divided by 8 root 3 this will be root 2 by root 3 we cannot further simplify this and this is an irrational. Number.
## What is the quotient of rational and irrational number?
The quotient of a rational number & irrational number is always rational number.
## Is zero rational or irrational number?
Why Is 0 a Rational Number? This rational expression proves that 0 is a rational number because any number can be divided by 0 and equal 0. Fraction r/s shows that when 0 is divided by a whole number, it results in infinity. Infinity is not an integer because it cannot be expressed in fraction form.
## What is the sum of two irrational numbers?
irrationalSum of two irrational numbers is always irrational. Sum of a rational and irrational numbers is always an irrational number.
## How do you divide rational fractions?
To divide two numerical fractions, we multiply the dividend (the first fraction) by the reciprocal of the divisor (the second fraction). For example: We can also use this method to divide rational expressions.
## Is 25.1 a rational number?
The number 25 is a rational number. It is a whole number that can be written as the fraction 25/1. By definition, a rational number is the number...
## Is pi irrational?
Pi is an irrational number, which means that it is a real number that cannot be expressed by a simple fraction. That's because pi is what mathematicians call an "infinite decimal" — after the decimal point, the digits go on forever and ever. (These rational expressions are only accurate to a couple of decimal places.)
## Is an irrational number divided by an irrational number irrational?
The reason of our contradiction is we assumed that 1 divided by irrational is rational. So we proved that 1 divided by irrational is irrational.
## Is 3.666 rational or irrational?
A rational number will contain numbers whose decimal expansion is finite or recurring in nature. For example, 1.67 and 3.666... are rational numbers.
## Is e irrational?
The number e was introduced by Jacob Bernoulli in 1683. More than half a century later, Euler, who had been a student of Jacob's younger brother Johann, proved that e is irrational, that is, that it cannot be expressed as the quotient of two integers.
## Is 1 divided by an irrational number?
Originally Answered: Is 1 divided by an irrational number an irrational number ? Yes. An irrational number is one that can't be represented as a fraction of two integers. The function f(x) = 1/x is a one to one function where the set of all rational numbers is mapped to the set of all rational numbers.
## What is an irrational number divided by an irrational number?
Irrational divided by irrational might be rational (it depends on the two irrational numbers), but will more often be irrational as well. If you pick two irrational numbers at random, their quotient will almost certainly be irrational - you have to choose them very particularly to get a rational quotient.
## What is a rational number divided by?
A rational number is the result of dividing two integers. If the signs of the divisor and dividend are the same, then the quotient will be positive. If the signs of the divisor and dividend are different, then the quotient will be negative.
How did Blackbeard get two powers?
Did Blackbeard eat Whitebeard's Devil Fruit?
Why can Blackbeard eat two fruits?
## How long is supermans life span?
In DC ONE MILLION, he lived on forever. In Kingdom Come, even though he was already cleary in his 50's or 60's. He lived on for another 1000 years. That's when you see him as a old man.
## How long do unopened mayonnaise packets last?
Properly stored, an unopened package of mayonnaise will generally stay at best quality for about 3 to 4 months after the date on the package.
## Do mayonnaise packets need refrigerated?
The small packets of mayo (as well as mustard, salad dressings, etc.) are all shelf-stable and never need to be refrigerated when unopened.
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Explanation of the Logarithm Technique - Deji Adegbite
P.S So I'll be sure you understand the technique, please reply me and explain it to me again with another example.
*********Please get a pen and a piece of paper*********
Let me start from the very beginning.
We know that the logarithm of a number (N) is the power to which another number (the base) must be raised to get that number (N). For example, log 8 (base 2) = 3 i.e. you raise 2 to the power of 3 to get 8. Log 7 (base 49) = 1/2 i.e. you raise 49 to the power of 1/2 to get 7.
Now, lets imagine the logarithm scale as a straight line. Thinking in terms of base 10, we'll have 1 on the left end of the line and 10 on the right end of the line - remember it's in base 10. To get the logarithm of a number, what portion or fraction of the line would we cover to get to that number if we are moving from left to right? For example, if we we're looking for the logarithm of 3.162, i.e. the square root of 10, we would move half way across the line. Lets take another example. If the log scale was in base 16, we would have 1 on the left end of the line and 16 on the right end. Using this base 16 log scale, if we want to get the logarithm of 4, we would only have to move half way across the line to get to 4. This means the logarithm of 4 (base 16) is ½. Using the same base 16 scale, what is the logarithm of 8? ¾ or 0.75. Which means we have covered ¾ of the line by the time we get to 8 (moving from left - i.e. 1 - to right - i.e. 16). Using base 1000, what is the logarithm of 10? 1/3 or 0.333 i.e. we would have covered 1/3 of the line by the time we get to 10 once again moving from left (1) to right (1000 - the base).
Let's now imagine that our base 10 line is divided only into 2 parts. This means we would have 1 on the left end of the line, 3.162 half way across and 10 on the right end of the line. This also means that on this scale, log 1 = 0, log 3.162 = 1/2 and log 10 = 1. Now, let's divide the line into 4 parts. This means we'll have 1 on the left end, 1.778 ¼ way across, 3.162 half way across, 5.622 ¾ way across and 10 on the right end of the line. This means that log 5.622 = ¾ because it is ¾ way across the line and log 1.778 = ¼ because it is ¼ way across.
PS Don't continue if you don't understand. So I'll be sure you understand the technique, please reply me and explain it to me again with another example.
Get a pen and a piece of paper before you continue.
Draw a straight line on your piece of paper and lets imagine that we want to get the log of 5 to the base of 10 i.e. log 5 (base 10). Without dividing the line, we'll have 1 on the left end and 10 on the right end of the line. Let's divide the line into 2 equal parts. Now, we'll have 1 on the left end of the line, 3.162 half way across the line, and 10 on the right end of the line. If we used this scale, we'll be approximating the log of 5 to be ½ which isn't very accurate. We know that 5 is somewhere between 3.162 and 10. Therefore, to make our log scale more accurate, we'll divide the line into 4 equal parts. This means we'll have 1 on the left end of he line, 1.778 ¼ way across the line, 3.162 half way across the line, 5.622 ¾ way across and 10 on the right end of he line. With this scale, we will approximate the log of 5 to be 3.162 since 5.622 is too large i.e. 5.622 is greater than 5. Now we are approximating the log of 5 to be 0.5 or ½. Instead of doing that, we'll do it this way; the number of subdivisions before 3.162 is 2, and the log scale was divided into 4 parts, so, instead of saying ½, we'll say 2/4.
Now, let's further divide our log scale into 8 parts to make it more accurate. Now, we'll have 1.333 1/8 way across, 1.778 2/8 (or ¼) way across, 2.371 3/8 way across, 3.162 4/8 (1/2) way across, 4.216 5/8 way across, 5.622 6/8 (or ¾) way across, 7.498 7/8 way across and finally, we'll have 10. Now, we should have something closer to 5 on our log scale and that is 4.216 (4.216 is closer to 5 than 3.162) which is 5/8 way across. This means that using this scale, we would approximate the log of 5 to be 5/8. Now lets make our log scale more accurate by dividing it into 16 parts, this gives 1 on the left end of the line, 1.154 1/16 across, 1.333 2/16 (or 1/8) across, 1.539 3/16 across, 1.778 4/16 (or 2/8) across, 2.053 5/16 across, 2.371 6/16 (or 3/8) across, 3.651 9/16 across, 4.216 10/16 (or 5/8) across, 4.869 11/16 across, 5.622 12/16 (or 6/8) across, 6.493 13/16 across, 7.498 14/16 (or 7/8) across, 8.659 15/16 across, and 10 on the right end of the line.
Notice now that we have something closer to 5 than the previous result. The previous result gave us 4.216 which was 5/8 way across the line. Now we have 4.869 which is closer. Hence, we would approximate the log of 5 to be 11/16 (i.e. the log of 4.869).
So you see, to make our log scale more accurate, we keep dividing it. So the next time we divide our log scale, we'll divide it in 32 parts. Just continue the trend until you get as close to log 5 as you like but even if we stop here, we'll be approximating our log 5 to 11/16 = 0.687. My calculator tells me that log 5 = 0.698. This gives us a difference of 0.698 - 0.687 = 0.011. which is pretty close already!
PS Read again if you don't understand. For me to be sure you understand the technique, reply me and explain it to me again using another example.
In the technique, we were incrementing the number on the far left of our soroban by 1. This is to help us keep track of how many times we have had to divide our log scale. In our last example, we divided 4 times. This means that there would be 16 subdivisions on our log scale. If we had divided one more time, we would have divided 5 times and we would have 32 subdivisions. This makes our log scale more accurate. Notice that if we divide our log scale t times, then we will have 2^t subdivisions. When we divided only once, we had 2^1 divisions i.e. 1 to 3.162 and 3.162 to 10.
For the rooting and multiplying part, here is the explanation;
Note that when marking our log scale, we don't just mark the line any how. For example, lets say you want to make a base 16 scale, how do you know what is supposed to be half way across the line? You get the square root of 16 first which gives us 4, then, mark half way across the line 4. Now lets divide the log 16 line again. This time, we would have 4 subdivisions with 1 on the left end, 4 half way across, and 16 on the right end of the line. What will be between 1 and 4? That will be ¼ way across the line, i.e. 16 raised to the power of ¼ which is 2. What will be between 4 and 16? Lets see…that's ¾ way across the line so that's 16 raised to the power of ¾ which gives us 8. Now your log scale (base 16) should have 1 on the left end, 2 ¼ way across, 4 ½ way across, 8 ¾ way across and 16 on the right end. Lets have another example. Lets say we have a line with base 10 000. At first, we had 1 on the left end and 10000 on the right end. When we divide the line into 2 parts, what would we have ½ way across? That would be 10000 raised to the power of half which gives 100. So now we would have 1 on the left end, 100 half way across, and 10000 on the right end. Lets say we divide this line again, this time into four equal parts. What would we have between 1 and 100? Notice that's now ¼ way across which means we'll have 10000 raised to the power of ¼ which gives us 10. What would we have between 100 and 10000? Hmm… that's ¾ way across the line and that gives us 10000 raised to the power of ¾ which is 1000. So your new log scale should now have 1 on the left end, 10 ¼ way across and 10000 on the right end.
I hope you're still following me! Let's say we have a base 256 log scale (I hope you still have your pen and paper) with which we want to get the log of 100 i.e. log 100 (base 256); Now let's draw our line and label it. This means we would have 1 on the left end and 256 on the right end. Now, lets divide the line into 2 equal parts. What would be ½ way across the line? That's 256 raised to the power of ½ which is 16. Now, if we stop at this point, we would be approximating the log of 100 to ½ since 16 is the nearest number to 100. We will not say 256 because 256 is greater than 100 and we're not considering greater numbers - only the lesser ones. Now lets divide the line into 4 equal parts. Where do you think the next value will fall on after 16? Don't think too far! The next value after 16 will be 64 i.e. 16 x sqr root(16) = 16 x 4 = 64. You don't believe? Well, lets see… what will be between 1 and 16? That's ¼ way across the line i.e. 256 raised to the power of ¼ which is 4. What will be between 16 and 256? That's 254 raised to the power of ¾ which is 64. At this point, we would approximate the log of 100 to ¾ since 64 is now nearer to 100. Please remember that we're not saying 256 because we're not considering larger numbers. Lets divide our log scale on more time. This means we will have 8 subdivisions. Remember we're sitting on the 64 mark (the number whose log we used as the log of 10). What will be between 64 and 256? That should be 64 x sqr root(4) = 64 x 2 = 128. How did I do that? By getting the square root of the previous unit length and then multiplying. In this case, the previous unit length was 4. The unit length is the next number after 1 on our log scale. Let's continue. We're still sitting on the 64 mark. Why? Because 128 (the number after 64) is greater than 100 and we are not considering higher numbers. So, we're still approximating the log of 100 to ¾? Yes but this time, remember we not have 8 subdivisions so we'll say 6/8 (double the numerator without adding 1 since our new approximation is still equal to our previous approximation).
Now, another example. Let's get log 75 (base 10000). Draw your log line. We have 1 on the left end and 10000 on the right end. Now lets divide the line the first time. This means we have divided only once. How many subdivisions would we have? 2^1 = 2. What do we have ½ way between 1 and 10000? That's 10000 raised to the power of ½ which is 100. Our unit length is 100. That's the next number after 1. Also, at this point, we would approximate log 75 to be 0 i.e. log 1. We're not to approximate it to log 100 which is ½. This is because 100 is greater than 75 and we're not considering higher numbers. Now, lets divide the line a second time. This means we're have 4 subdivisions (since we're dividing the second time, we will have 2^2 subdivisions). What will we have between 1 and 100? Our new unit length is 10 i.e. the square root of the previous unit length which was 100. We're sitting on the 1 mark. So, what will be between 1 and 100, that's 1 x 10 = 10. If we use this scale which is still not very accurate, we would be approximating the logarithm of 75 to be equal to ¼. Why? Because 10 is the closest thing to 75. Why not 100? Because 100 is greater than 75 and we are not considering numbers that are greater. So, to make the scale more accurate, what do we do? We divide the line one more time. Note that this would be the third time we are dividing. So, how many subdivisions will we have on our log scale? That's 2^3 = 8. Why? Because this would be the third time we are dividing.
What would be our new unit length? That would be the square root of our previous unit length i.e. sqr root(10) = 3.162. We are sitting on the 10 mark i.e. in our previous approximation, we approximated the log of 75 to be equal to that of 10. So what will be between 10 and 100? Don't get your calculator yet! The answer will be the new unit length multiplied by the mark we were sitting on. In this case, 3.162 x 10 = 31.62. Now, we've moved a little closer! Remember we will not say 100 because we are not considering higher numbers. We're not sitting on the 31.62 mark. Now, what's the log? How many subdivisions do we have in total? That's 8 or 2 raised to the power of t where t is the number of times we have divided. How many subdivisions do we have before 31.62? 3. How do we get this? Observe that since 31.62 is still less than 75, the numerator in our previous answer is doubled and then 1 is added. In our previous approximation, we approximated the log of 75 to be ¼. The numerator is 1. So our new numerator would be 2 x 1 = 2; 2 + 1 = 3. So, our new approximation would be 3/8.
Let's stop at this point and take another example. I hope you still have your pen and paper.
What is the log of 7 to the base of 256?
Draw your straight line once again. We'll have 1 on the left end of the log scale, and we'll have 256 on the right end of the log scale. Let's divide the log scale the first time. How many subdivisions do we have? We've divided only once so the number of subdivision = 2^1 = 2. Now what is half way between 1 and 256? Hmm… that would be 256 raised to the power of ½ which is 16. At this point, what would be our approximation to log 7? 0/2 = 0. Its not ½ because that's 16 which is greater than 7 and we're not considering higher numbers. What is our unit length? That's 16 i.e. the next number after 1. So on what mark are we sitting? We are sitting on the 1 mark because we are approximating the log of 7 to be equal to that of 1.
Now, let's divide the line the second time. How many subdivisions will we have? That's 2^2 = 4. Remember that the number of subdivisions is 2^t where t represents how many times we have divide. What will be our new unit length? That's the square root of the previous unit length i.e. sqr root(16) = 4. What's the next number after 1? Remember we were sitting on the 1 mark in the first approximation. That's the new unit length multiplied by 1 - 1 x 4 = 4. What is the numerator of our answer? Double the previous numerator and add 1 i.e. 0 x 2 = 0; 0 + 1 = 1. We are adding the 1 because 4 is still less than 7. Remember this time we will be approximating our log of 7 to that of 4 since the next number after 4 (16) is greater than 7 and we are not considering greater numbers. How do we get the denominator? That's the total number of subdivisions which is 2^2 = 4. So the new approximation is ¼. Now, we are sitting on the 4 mark. Let's now divide the line 1 more time. How many subdivisions will we have now? We are dividing for the third time so that would be 2^3 = 8. This will be the new denominator of our answer. What is our new unit length? That's the square root of the previous unit length i.e. sqr root(4) = 2. So, from point 4 where we are sitting on, what will be between 4 and 16? That's 4 (because we're sitting on the 4 mark) multiplied by our new unit length which is 4 x 2 = 8. So, we will have 8 between 4 and 16. What will we approximate log 7 to? We will approximate it to that of log 4. Why? Because 8 is greater than 7 and we are not considering higher numbers. So, how do we approximate? Double the previous numerator to get the new one (don't add 1 after doubling because our approximation has not changed. We are still sitting on the 4 mark) i.e. 2 x 1 = 2. So our approximation will be 2/8 (not ¼).
Let's continue. We'll divide for the last time then I'll stop. Let's divide our log scale 1 more time. How many subdivisions do we now have? We have divided 4 times now so the number of subdivisions will be 2^4 = 16. I hope you're still working with your pen and paper! What will be new unit length? That's the square root of the previous unit length i.e. sqr root(2) = 1.414. Remember we're still sitting on the 4 mark. So what will be between 4 and 8? That's 4 multiplied by the new unit length i.e. 4 x 1.414 = 5.656. This is closer to 7. So how do we approximate? 5.656 is less than 7 so we'll double the previous numerator and add 1 i.e. 2 x 2 = 4; 4 + 1 = 5. So, we'll have 5/16. We're not approximating the log of 7 to that of 8 because 8 is greater than 7 and we're not considering greater numbers. If we stop here, we would be implying that the log of 7 (base 256) is 5/16. So, you'll continue dividing and approximating till you get as close to log 7 as you like. You can try dividing up to 16 times. If you divide up to 16 times, how many subdivisions would we have? That's 2^16 = 65536. You see that would give us something very accurate. The more you divide, the more accurate the log scale becomes.
I hope you understand the whole thing. If you don't read it all over again until you do and mail me. Please use a pen and a piece of paper so you'll understand it better.
If you understand, you'll still have to mail me and explain it to me with another example so that I will know whether you understand it or not.
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# Change the percent to a fraction 64%
0.64. All you have to do is add a 0 and a decimal to it. It's the same number. :)
64/100
Simplest form is 16/25
## Related Questions
Which of the following ordered pairs represents a solution to the linear equation y=6x-2x
The ordered pairs are not given, but they are going to be a solution if they are in the following format:
.
The equation given is:
We can simplify it:
Thus, ordered pairs (x,y) are a solution if they are in the following format:
.
A similar problem, which asks if an ordered pair is a solution of an equation, is given at brainly.com/question/10585472
Since you didn't list the points, I can only give you some example points. Then, you can select the points in your problem.
First, the equation can be simplified to just: y = 4x
Now, just select points for x, then multiply by 4 to get y.
0, 0
1, 4
2, 8
3, 12
4, 16
5, 20
In four years Cranston’s age will be the same as Terrill’s age is now. In two years time, Terrill will be twice as old as Cranston. Find their ages now.
Uuuuh...
c + 4 = t
t + 2 = 2c
c + 4 = 2c – 2
c = 2
t = 6
or
c = t - 4....................................(1)
in 2 years we have:
t + 2 = 2(c + 2)
i.........e......... t + 2 = 2c + 4
=&gt; t - 2c = 2...........................(2)
(1) into (2) for c gives:
t - 2(t - 4) = 2
i.........e......... 8 - t = 2
so, t = 6 and c = 2
therefore, terrill is 6 and cranston is 2
and, in 2 years time they shall be 8 and 4....................................i..... terrill being twice cranston's age
i lost myself the first time i solved it
Cranston's Age: 2 years
Terril's Age: 6 years
Step-by-step explanation:
c = Cranston
t = Terril
Information we have:
c + 4 = t
t + 2 = 2c
So, if Cranston is 2 and Terril is 6, it's true as c + 4 (2 + 4) is 6 and When Cranston and Terril age 2 years, Cranston will be 4 and Terril will be 8, so Terril will be twice of Cranston's age.
*A lot of people forget the part where Cranston will ALSO age, so they get t = 10 and c = 6
Find the max and min values of f(x,y,z)=x+y-z on the sphere x^2+y^2+z^2=81
Using Lagrange multipliers, we have the Lagrangian
with partial derivatives (set equal to 0)
Substituting the first three equations into the fourth allows us to solve for :
For each possible value of , we get two corresponding critical points at .
At these points, respectively, we get a maximum value of and a minimum value of .
Which equation represents a line that is perpendicular to the line y= 1/4x - 1 A) Y=-1/4x - 3
B) Y= 1/4X + 1
C) Y= 4X-5
D) Y= -4X - 2
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Lesson Plan
# Exploring The Coordinate Plane
Students will explore the parts of a Coordinate Plane
Allyson ..
Classroom teacher
Freedom Middle School, Stone Mountain, GA
My Subjects Math, English Language Learning
Objectives
Students will be able to...
• Extend the coordinate system to four quadrants.
• Name points on the plane.
• Graph ordered pairs on the Cartesian Coordinate Plane.
Subjects
Math
#### 1 Opening: Whole Group Exploration -Creating A Cartesian Coordinate Plane
Activity: Creating
Teacher will present the class with an empty grid. Students will follow steps 1-9 as listed in the student instructions. Before step 1 read: In earlier lessons, you graphed points on a coordinate plane where both the x- and y-coordinates were zero or positive numbers. Now let’s include negative numbers in the coordinate plane. After step 4 read: These regions on the coordinate plane are called quadrants. They are numbered with Roman numerals from one to four (I, II, III, IV) starting in the upper right-hand quadrant and moving counterclockwise. After step 5: You can plot points on the coordinate plane using an ordered pair. An ordered pair is a pair of numbers which can be represented as (x, y) that indicate the position of a point on the coordinate plane. For example, the ordered pair for the origin is (0,0).
Student Instructions
1. To begin this process, first draw a horizontal line segment across the width of the grid that splits the grid in half. Draw arrowheads at the ends of your line segment. Label the line x. This horizontal line on the coordinate plane is called the x-axis. 2. Next, draw a line segment perpendicular to your first line segment from the top of the grid that splits the grid in half so that the line segments intersect. Label this line y. This vertical line on the coordinate plane is called the y-axis. 3. Label the point of intersection with 0. This point where the x-axis and y-axis intersect on the coordinate plane is known as the origin. Then, using an interval of 1, label the grid lines to the right and above 0 with positive numbers in numerical order. Finally, label the grid lines to the left and below 0 with negative numbers in numerical order. 4. How many regions are created when the coordinate plane is divided by the perpendicular lines? 5. Label each of the quadrants on your coordinate plane. 6. Plot a point on the coordinate plane anywhere in the first quadrant, and label the point with its ordered pair. 7. Plot a point on the coordinate plane anywhere in the second quadrant, and label the point with its ordered pair. 8. Plot a point on the coordinate plane anywhere in the third quadrant, and label the point with its ordered pair. 9. Plot a point on the coordinate plane anywhere in the fourth quadrant, and label the point with its ordered pair.
#### 2 Independent Practice- Performance Task and Free Response
Each student will have an iPad to complete the interactive tasks on USA Test Prep.
Teacher will assign a Performance Task to give the students practice in matching ordered pairs to the correct coordinate on the grid. Students will match key terms related to the coordinate plane with their meaning.
Lastly, students will complete a constructed response task in which they will look at a coordinate plane with labeled points and write the ordered pair and tell the quadrant in which the point is located.
Teacher will evaluate responses and give students immediate feedback via USA Test Prep.
Student Instructions
• Practice matching ordered pairs to the correct coordinate on the grid.
• Match key terms related to the coordinate plane with their meaning.
• Complete a constructed response task in which you will look at a coordinate plane with labeled points, write the ordered pair, and tell the quadrant in which the point is located.
#### 3 Closing- Reflection - Think Pair Share
Activity: Other — Reflecting
Teacher will provide students with a prompt to explain how they know a point is located in Quadrant I, Quadrant II, Quadrant III, and Quadrant IV.
Student Instructions
Explain using words:
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# Find the Radius of a Circle Like a PRO: Don't be Rookie!
## Basic Definitions and Concepts
Before diving deep into the methods to find the radius of a circle, it's essential to familiarize oneself with some foundational terms and concepts. A circle, despite its simplicity, carries with it a rich vocabulary that forms the basis for more advanced geometric discussions.
• Center of a Circle: The point inside a circle that is equidistant from all points on the circle's edge is termed its center. It serves as the reference point from which various measurements like the radius or diameter are drawn.
• Diameter: This is the straight-line segment that passes through the center of the circle, connecting two points on its boundary. Essentially, it's twice the length of the radius, stretching across the entire circle.
• Circumference: When you trace the outer boundary of a circle, the distance covered is its circumference. Think of it as the "perimeter" for circles. It has a unique relationship with the diameter, often denoted by the famous mathematical constant, Ï€.
### Relationship between Diameter and Radius
The diameter and radius are two fundamental properties of a circle, and they share a straightforward relationship. The diameter is simply twice the length of the radius. This is because the diameter stretches across the entire circle, passing through its center, while the radius only spans from the center to the edge of the circle.
Mathematically, this relationship can be expressed as:
$\text{Diameter (d)} = 2 \times \text{Radius (r)}$ or equivalently, $\text{Radius (r)} = \frac{\text{Diameter (d)}}{2}$
This means that if you know the diameter of a circle, you can immediately determine its radius by halving the diameter's length, and vice versa. Understanding this relationship is pivotal in many geometrical applications and problems related to circles.
To use an analogy: If the circle were a wheel, the radius would be the length of a spoke, while the diameter would be the distance across the entire wheel, from one edge to the opposite edge through the center.
## Visualization of a Circle’s Radius and Diameter
Let us use a visual image to understand Circle Radius and Diameter:
• The circle is represented with a black outline, centered at the coordinate (0.5, 0.5) with a radius of 0.4 units.
• The diameter, shown in blue, extends horizontally across the circle, passing through its center. It is labeled as "Diameter (d)".
• The radius, depicted in red and dashed lines, stretches vertically from the center of the circle to its top edge. It is labeled as "Radius (r)".
## Methods to Find the Radius of a Circle
### 1. Given the Diameter
The most straightforward method to determine the radius when the diameter is known is by simply dividing the diameter by two. This is due to the inherent relationship between the diameter and the radius.
$r = \frac{d}{2}$ Where $$r$$ is the radius and $$d$$ is the diameter.
Example:
If the diameter of a circle is $$16$$ units, the radius would be:
$r = \frac{16}{2} = 8$ Thus, the radius is $$8$$ units.
### 2. Given the Circumference:
The circumference ($$C$$) of a circle relates to the radius using the formula: $C = 2\pi r$ To find the radius from the circumference: $r = \frac{C}{2\pi}$
Example:
For a circle with a circumference of $$31.4$$ units: $r = \frac{31.4}{2\pi} \approx 5$ Hence, the radius is approximately $$5$$ units.
### 3. Using the Area of the Circle:
The area ($$A$$) of a circle is given by: $A = \pi r^2$ To find the radius from the area: $r = \sqrt{\frac{A}{\pi}}$
Example:
For a circle with an area of $$78.5$$ square units: $r = \sqrt{\frac{78.5}{\pi}} \approx 5$ Therefore, the radius is approximately $$5$$ units.
### 4. Through Chord Properties and Intersecting Lines
This method is a bit advanced but can be understood with an example. Given two chords in a circle that intersect, you can use their lengths and the segments they create to find the radius.
Example:
Let's say two chords intersect inside a circle forming segments of lengths $$a$$, $$b$$ on one chord and $$c$$, $$d$$ on the other. The formula relating these segments and the radius ($$r$$) is: $a \times b = c \times d = r^2 - p^2$ where $$p$$ is the perpendicular distance from the center to the point of intersection of the chords.
Given $$a = 3$$, $$b = 4$$, $$c = 2$$, and $$d = 6$$ and the perpendicular distance $$p = 1$$, we can find: $r^2 = a \times b + p^2 = 3 \times 4 + 1 = 13$ $r \approx \sqrt{13}$ Thus, the radius is approximately $$3.6$$ units.
## Hands-On Examples for Beginners
Exercise 1: Given a circle with a diameter of $$10$$ units, determine the radius.
Solution:
Using the formula $$r = \frac{d}{2}$$, we find: $r = \frac{10}{2} = 5$ So, the radius is $$5$$ units.
Exercise 2: A bicycle wheel has a circumference of $$62.8$$ units. What's the radius?
Solution:
Using the formula $$r = \frac{C}{2\pi}$$: $r = \frac{62.8}{2\pi} \approx 10$ Thus, the radius of the bicycle wheel is approximately $$10$$ units.
Example: The area of a park shaped like a circle is $$78.5$$ square units. Find its radius.
Solution:
Step 1: We know $$A = \pi r^2$$.
Step 2: Rearrange to get $$r = \sqrt{\frac{A}{\pi}}$$.
Step 3: Plug in the given area $$A = 78.5$$: $r = \sqrt{\frac{78.5}{\pi}} \approx 5$ Therefore, the radius of the park is approximately $$5$$ units.
1. Using Pythagoras theorem in a Right-Angled Triangle Inscribed in a Circle:
When a right-angled triangle is inscribed in a circle, the hypotenuse of the triangle is the diameter ($$d$$) of the circle. Using the Pythagoras theorem: $a^2 + b^2 = c^2$ where $$c$$ is the hypotenuse (also the diameter), and $$a$$ and $$b$$ are the other two sides, we can determine: $r = \frac{c}{2}$
Example: Let's say in a right-angled triangle, sides $$a = 3$$ units and $$b = 4$$ units. Using the Pythagoras theorem: $c^2 = 3^2 + 4^2$ $c = 5$ Thus, the radius $$r = \frac{5}{2} = 2.5$$ units.
2. Utilizing Segment Properties
For a segment formed by a chord and its arc, the area is the difference between the area of the sector (formed by the arc and the two radii) and the area of the triangle (formed by the two radii and the chord). This relationship can be used to find the radius given the area of the segment and the length of the chord.
Example: The topic can be broad, but for simplicity: if we have a segment area and an angle, we can find the radius using: $\text{Area of Segment} = \frac{\theta}{360} \times \pi r^2 - \frac{1}{2} r^2 \sin(\theta)$ where $$\theta$$ is the angle of the sector in degrees.
3. Finding Radius from Arc Length and Sector Area
For a circle with radius $$r$$, arc length $$L$$, and the corresponding sector area $$A$$, the relationship is: $L = \theta \times \frac{r}{180}$ $A = \frac{\theta}{360} \times \pi r^2$ Given $$L$$ and $$A$$, we can solve for $$r$$.
Example: Given an arc length of $$5$$ units and sector area of $$20$$ square units: From $$L$$, we get $$\theta = \frac{5 \times 180}{r}$$ From $$A$$, we derive $$r = \sqrt{\frac{20 \times 360}{\pi \times \theta}}$$ Solving simultaneously will give us $$r$$.
4. Radius of Inscribed and Circumscribed Circles
Inscribed Circle (Incircle):
The radius (inradius) can be found using the area and semiperimeter (s) of the triangle:
$r_{in} = \frac{\text{Area of triangle}}{s}$
Circumscribed Circle (Circumcircle):
For a triangle with sides $$a$$, $$b$$, and $$c$$, the radius (circumradius) is: $r_{cir} = \frac{abc}{4K}$ where $$K$$ is the area of the triangle.
Example: For a triangle with sides $$3$$, $$4$$, and $$5$$ units: $$r_{in}$$ would use the formula above with semiperimeter and area calculations. $$r_{cir}$$ would be calculated using $$a = 3$$, $$b = 4$$, $$c = 5$$, and $$K$$.
## Mistakes to Avoid
• Confusing Diameter with Radius: Always remember, the diameter is twice the length of the radius.
• Ignoring Units: Whether you're dealing with centimeters, meters, or inches, always ensure you carry the units throughout your calculations.
• Misusing Formulas: For example, using the circumference formula when given the area. Always ensure you understand which formula to apply based on the information given.
## Challenges and Quizzes on Finding the Radius of a Circle
1. Beginner Challenge:
A circle has a diameter of 14 cm. What is its radius?
Solution:
$r = \frac{\text{diameter}}{2} = \frac{14 \text{ cm}}{2} = 7 \text{ cm}$
2. Intermediate Challenge:
The circumference of a circle is $$56\pi$$ cm. Determine the radius of this circle.
Solution:
Using the formula $$C = 2\pi r$$, $r = \frac{C}{2\pi} = \frac{56\pi \text{ cm}}{2\pi} = 28 \text{ cm}$
A sector in a circle has an area of $$25\pi$$ square units and an angle of 90°. Find the radius of the circle.
Solution:
The area of a sector is given by: $A = \frac{\theta}{360} \times \pi r^2$ Plugging in the given values: $25\pi = \frac{90}{360} \times \pi r^2$ Solving for $$r$$, we get: $r^2 = \frac{25 \times 4}{\pi} \implies r = 10 \text{ units}$
4. For Experienced Students - Challenging Quiz
In a triangle with sides measuring 6 cm, 8 cm, and 10 cm, find the radius of the circumscribed circle.
Solution:
The triangle is a right-angled triangle, and for a right-angled triangle, the circumradius $$r_{cir}$$ is:
$r_{cir} = \frac{\text{hypotenuse}}{2}$ Given the hypotenuse (longest side) is 10 cm, $r_{cir} = \frac{10 \text{ cm}}{2} = 5 \text{ cm}$
5. Expert Challenge:
A chord of length 16 cm stands 6 cm away from the center of a circle. Find the radius of the circle.
Solution:
Let's use the Pythagoras theorem. If $$r$$ is the radius and the perpendicular distance from the center to the chord divides the chord into two equal parts of 8 cm each, then: $r^2 = 8^2 + 6^2$ Solving for $$r$$, $r^2 = 64 + 36 = 100 \implies r = 10 \text{ cm}$
What is the radius of a circle?
The radius of a circle is the distance from the center of the circle to any point on its circumference.
How is the radius related to the diameter?
The radius is half the diameter. If you know the diameter, you can find the radius by dividing the diameter by 2.
Is the radius always shorter than the diameter?
Yes, the radius is always half the length of the diameter, so it's always shorter.
What is the significance of the radius in real-world applications?
The radius is crucial in various applications, from calculating the area for land plots to determining the size of gears in machinery.
If a chord's length and its distance from the circle's center are given, can I find the radius?
Yes, using the Pythagoras theorem. If the chord is bisected by the perpendicular from the center, it forms a right-angled triangle.
Does a circle always have a fixed radius, or can it change?
For a given circle, the radius is always fixed. However, if you change the size of the circle, the radius changes accordingly.
How does the radius of a circle affect its area and circumference?
The area and circumference are directly proportional to the square of the radius and the radius, respectively. If the radius increases, both the area and circumference increase, and vice versa.
## Let’s Wrap Up
Understanding the radius of a circle is fundamental to the study of geometry. It not only provides insight into the properties and dimensions of the circle itself but also forms the foundation for many other geometric concepts and formulas. From determining areas to calculating circumferences and delving into more complex geometric properties, the radius serves as a gateway. As we've explored, finding the radius from various given information, whether it's diameter, area, or even arc length, underscores its versatility and significance in the realm of mathematics. Its applicability doesn't stop at the classroom; the radius has real-world implications, from designing wheels to urban planning.
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## Problem on Ages Questions and Answers Part-3
1. The ratio between the parents ages of A and B is 5 : 3 respectively. The ratio between A's age 4 years ago and B's 4 years hence is 1 : 1. What is the ratio between A's age 4 years hence and B's age 4 years ago ?
a) 1 : 3
b) 3 : 1
c) 2 : 1
d) 4 : 1
Explanation: Let A's age be 5x years, then B's age = 3x years
\eqalign{ & \frac{{5x - 4}}{{3x + 4}} = \frac{1}{1} \cr & 5x - 4 = 3x + 4 \cr & 2x = 8 \cr & x = 4 \cr & \frac{{{\text{A's age 4 years hence}}}}{{{\text{B's age 4 years ago }}}} \cr & = \frac{{5x + 4}}{{3x - 4}} \cr & = \frac{{5 \times 4 + 4}}{{3 \times 4 - 4}} \cr & = \frac{{24}}{8} \cr & = \frac{3}{1} \cr & = 3:1 \cr}
2. The ratio between the ages of Neelam and Shiny is 5 : 6 respectively. If the ratio between the one-third age of Neelam and half of Shiny's age is 5 : 9, then what is Shiny's age = ?
a) 25 years
b) 30 years
c) 36 years
d) Cannot be determined
Explanation: Let Neelam's age be 5x years and Shiny's age be 6x years
\eqalign{ & \left( {\frac{1}{3} \times 5x} \right):\left( {\frac{1}{2} \times 6x} \right) = 5:9 \cr & \frac{{5x}}{{3 \times 3x}} = \frac{5}{9} \cr}
Shiny's age cannot be determined
3. 18 years ago, a man was three times as old as his son. Now, the man is twice as old as his son. The sum of the present ages of the man and his son is =
a) 54 year
b) 72 years
c) 105 years
d) 108 years
Explanation: Let the son's age 18 years ago be x years,
Then man's age 18 years ago = 3x years
\eqalign{ & \left( {3x + 18} \right) = 2\left( {x + 18} \right) \cr & 3x + 18 = 2x + 36 \cr & x = 18 \cr}
Sum of their present ages
\eqalign{ & \left( {3x + 18 + x + 18} \right){\text{years}} \cr & \left( {4x + 36} \right){\text{years}} \cr & \left( {4 \times 18 + 36} \right){\text{years}} \cr & {\text{ 108 years}} \cr}
4. The age of a man 10 years ago was thrice the age of his son. 10 years hence, the man's age will be twice the age of his son. The ratio of their present ages is = ?
a) 5 : 2
b) 7 : 3
c) 9 : 2
d) 13 : 4
Explanation: Let son's age 10 years ago be x years,
Then man's age 10 years ago = 3x years
Son's present age = (x + 10) years
Man's present age = (3x + 10) years
\eqalign{ & \left( {3x + 10} \right) + 10 = 2\left( {x + 10 + 10} \right) \cr & 3x + 20 = 2\left( {x + 20} \right) \cr & 3x + 20 = 2x + 40 \cr & x = 20 \cr}
Ratio of present ages of man and the son
\eqalign{ & {\text{ = }}\frac{{3x + 10}}{{x + 10}} \cr & = \frac{{3 \times 20 + 10}}{{20 + 10}} \cr & = \frac{{70}}{{30}} \cr & = 7:3 \cr}
5. Tanya's grandfather was 8 times older to her 16 years ago. He would be 3 times of her age 8 years from now. 8 years ago, what was the ratio of Tanya's age to that of her grandfather ?
a) 1 : 2
b) 1 : 5
c) 3 : 8
d) None of these
Explanation: 16 years ago, let T = x years and G = 8x years
After 8 years from now,
T = (x + 16 + 8) years and G = (8x + 16 + 8) years
\eqalign{ & {\text{8x + 24 = 3}}\left( {x + 24} \right) \cr & 8x - 3x = 72 - 24 \cr & 5x = 48 \cr & 8{\text{years ago,}} \cr & {\text{ }}\frac{{\text{T}}}{{\text{G}}} \cr & = \frac{{x + 8}}{{8x + 8}} \cr & = \frac{{\frac{{48}}{5} + 8}}{{8 \times \frac{{48}}{5} + 8}} \cr & = \frac{{48 + 40}}{{384 + 40}} \cr & = \frac{{88}}{{424}} \cr & = \frac{{11}}{{53}} \cr}
6. Rajan got married 8 years ago. His present age is $$\frac{6}{5}$$ times his age at the time of his marriage. Rajan's sister was 10 years younger to him at the time of his marrige. The age of Rajan's sister is =
a) 32 years
b) 36 years
c) 38 years
d) 40 years
Explanation: Let Rajan's age 8 years ago be x years,
Present age = (x + 8) years
\eqalign{ & x + 8 = \frac{6}{5}x \cr & 5x + 40 = 6x \cr & x = 40 \cr}
Rajan's sister's age 8 years ago
= (40 - 10) years
= 30 years
His sister's age now
= (30+8) years
= 38 years
7. A couple has a son and a daughter. The age of the father is four times that of the son and the age of the daughter is one-third of that of her mother. The wife is 6 years younger to her husband and the sister is 3 years older then her brother. The mother's age is = ?
a) 42 years
b) 48 years
c) 54 years
d) 63 years
Explanation: M → Mother, F → Father, S → Son and D → Daughter
F = 4S, D = $$\frac{1}{3}$$M, M = F - 6 and S = D - 3
\eqalign{ & M = 3D = 3\left( {S + 3} \right) \cr & = 3S + 9 = \frac{3}{4}F + 9 = \frac{3}{4}\left( {M + 6} \right) + 9 \cr & = \frac{3}{4}M + \frac{3}{4} \times 6 + 9 \cr & \Rightarrow \left( {M - \frac{3}{4}M} \right) = \left( {\frac{9}{2} + 9} \right) \cr & \Rightarrow \frac{1}{4}M = \frac{{27}}{2} \cr & \Rightarrow M = \left( {\frac{{27}}{2} \times 4} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 54\,{\text{years}} \cr}
8. Reenu's father was 38 years of age when she was born while her mother was 36 years old when her brother 4 years younger to her was born. What is the difference between the ages of her parents ?
a) 2 Years
b) 4 Years
c) 6 Years
d) 8 years
Explanation: Mother's age when Reenu's brother was born = 36 year
Father's age when Reenu's brother was born = (38 + 4) years = 42 year
Required difference = (42 - 36) years = 6 years
9. A man was asked to state his age in years. His reply was, "Take my age 3 years hence, multiply it by 3 and then subtract 3 times my age 3 years ago and you will know how old I am". What is the age of the man ?
a) 18 years
b) 20 years
c) 24 years
d) 32 years
Explanation: Let the present age of the man be x years
\eqalign{ & {\text{3}}\left( {x + 3} \right) - 3\left( {x - 3} \right) = x \cr & \left( {3x + 9} \right) - \left( {3x - 9} \right) = x \cr & x = 18 \cr}
The present age of the man is 18 years
10. The ratio of a man's age and his son's age is 7 : 3 and the product of their ages is 756. The ratio of their ages after 6 years will be ?
a) 5 : 2
b) 2 : 1
c) 11 : 7
d) 13 : 9
\eqalign{ & {\text{7}}x \times 3x = 756 \cr & 21{x^2} = 756 \cr & {x^2} = 756 \cr & {x^2} = 36 \cr & x = 6 \cr}
\eqalign{ & {\text{ = }}\left( {7x + 6} \right):\left( {3x + 6} \right) \cr & = \left( {7 \times 6 + 6} \right):\left( {3 \times 6 + 6} \right) \cr & = 48:24 \cr & = 2:1 \cr}
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• Application of Normal Random Variables: Approximation to Binomial One of the important applications of the normal distribution is that under certain conditions it can provide a very good approximation to the binomial distribution. We've seen before that sometimes calculating binomial probabilities can be quite tedious, and the solution we suggested before is to use statistical software to do the work for you. In the absence of statistical software, another solution would be to use the normal approximation (when appropriate). Normal calculations never get too complicated; all we have to do is use a table correctly. Let's start with a motivating example. Example: True/False Questions Suppose a student answers 20 true/false questions completely at random. What is the probability of getting no more than 8 correct? Let X be the number of questions the student gets right (successes) out of the 20 questions (trials), when the probability of success is .5. X is therefore a binomial random variable with n = 20 and p = .5, and we are looking for $$P(X \leq 8) = P(X = 0) + P(X = 1) + ... + P(X = 8)$$. Doing this by hand using the binomial distribution formula is very tedious, and requires us to do 9 complex calculations, as shown below: $$\frac{20!}{0!20!} 0.5^0(1-0.5)^{20-0} + \frac{20!}{1!19!} 0.5^1(1-0.5)^{20-1} + ... + \frac{20!}{8!12!} 0.5^8(1-0.5)^{20-8} = 0.2517$$ One option that we have is to use statistical software, which will provide the answer:
• Consider the appearance of the probability histogram for the distribution of X:
• Clearly, the shape of the distribution of X for n = 20, p = 0.5 has a normal appearance: symmetric, bulging at the middle, and tapering at the ends. The following figure should help you visualize this:
• This suggests a method of approximating binomial probabilities:
• Estimate the binomial probability of XB taking a value over a certain interval with the probability that a normal random variable XN takes a value over the same interval, where XN has the same mean and standard deviation as XB, namely $$\mu = np ; \sigma = \sqrt{np(1-p)} • Example Suppose a student answers 20 true/false questions completely at random. Use a normal approximation to estimate the probability of getting no more than 8 correct. The number (X) correct is a binomial random variable that represents the number of successes in 20 trials when the probability of success for each trial is 0.5. X has a mean and standard deviation of: • \( \mu = np = 20*0.5 = 10; \sigma = \sqrt{np(1-p) = \sqrt{20*0.5*(1-0.5)} = 2.24} • and so we approximate the binomial X with a normal random variable having the same mean and standard deviation: Then we solve in the usual way using normal tables: • \( P(X_B \leq 8) \sim P(X_N \leq 8) = P(Z \leq \frac{8-10}{2.24}) = P(Z \leq -0.89) = 0.1867$$
• Unfortunately, the approximated probability, .1867, is quite a bit different from the actual probability, 0.2517.
Rule of Thumb
• Probabilities for a binomial random variable X with n and p may be approximated by those for a normal random variable having the same mean and standard deviation as long as the sample size n is large enough relative to the proportions of successes and failures, p and 1 - p. Our Rule of Thumb will be to require that $$np \geq 10; n(1-p) \geq 10$$
• Example May we use a normal approximation for a binomial X with n = 20 and p = 0.5? In this case, np = 20(.5) = 10 and n(1 - p) = 20(1 - .5) = 10. The criteria are just barely satisfied, and so we should not expect the approximation to be especially good.
• The purpose of the next activity is to give you practice at deciding whether the normal approximation is appropriate for a given binomial random variable. You'll get to practice checking the rule of thumb $$np \geq 10; n(1-p) \geq 10$$ , but also get a visual sense of when the normal approximation is appropriate.
1. Explanation :
Indeed, the rule of thumb is satisfied, since np = 300 * 0.9 = 270 > 10 and n(1 - p) = 300 * 0.1 = 30 > 10. Also, visually, it is quite clear that the normal approximation would be very good in this case. The distribution looks essentially normal.
1. Explanation :
Indeed, np = n(1 - p) = 4 * 0.5 = 2, so the rule of thumb is not satisfied. Visually, although the distribution is symmetric and maybe remotely resembles the normal distribution, it is not "fine" enough for the normal approximation to be appropriate.
1. Explanation :
Indeed, np = 20 * 0.1 = 2 < 10 so the rule of thumb is not satisfied. Visually, it is quite clear that the normal approximation is not appropriate in this case, since the distribution is skewed to the right.
1. Explanation :
Indeed, the rule of thumb is satisfied, since np = 100 * 0.75 = 75 > 10 and n(1 - p) = 100 * 0.25 = 25 > 10. Also, visually, it is quite clear that the normal approximation would be very good in this case. The distribution looks essentially normal.
Q. Recall that when appropriate, a binomial random variable can be approximated by a normal random variable that has the same mean and standard deviation as the binomial random variable. In other words, when appropriate, a binomial random variable with n trials and probability of success p, can be approximated by a normal distribution with mean μ = np and standard deviation σ = sqrt( np (1 - p) ). For those binomial distributions in questions 1-4 above of the previous exercise for which the normal approximation is appropriate, write down which normal distribution you would use to approximate them.
• For example 1:
1. X is binomial with n = 100 and p = 0.75, and would therefore be approximated by a normal random variable having mean μ = 100 * 0.75 = 75 and standard deviation σ = sqrt(100 * 0.75 * 0.25) = sqrt(18.75) = 4.33.
2. Note that if you look at the histogram, this makes sense.
3. The distribution is indeed centered at 75, and extends approximately 3 standard deviations (3 * 4.33 = 13) on each side of the mean (as we know normal distributions do).
• For example 4:
1. X is binomial with n = 300 and p = .9, 0 and would therefore be approximated by a normal random variable having mean μ = 300 * 0.9 = 270 and standard deviation σ = sqrt(300 * 0.9 * 0.1) = sqrt(27) = 5.2.
2. Note that if you look at the histogram, this makes sense.
3. The distribution is indeed centered at 270, and extends approximately 3 standard deviations (3 * 5.2 = 15.6) on each side of the mean (as we know normal distributions do).
• It is possible to improve the normal approximation to the binomial by adjusting for the discrepancy that arises when we make the shift from the areas of histogram rectangles to the area under a smooth curve. For example, if we want to find the binomial probability that X is less than or equal to 8, we are including the area of the entire rectangle over 8, which actually extends to 8.5. Our normal approximation only included the area up to 8. The figure below illustrates this:
• It can be improved upon by making the continuity correction:
• in this case, we would have
• $$P(X_B \leq 8) \sim P(X_N \leq 8.5) = P(Z \leq \frac{8.5 - 10}{2.24}) = P(Z \leq -0.67) = 0.2514$$
• , which is much closer to the actual binomial probability of 0.2517 than our original approximation (0.1867) was.
• Similarly, suppose I wanted to answer: What is the probability that the student gets at least 13 questions right?
• Here, to calculate the exact probability we are including the area of the entire rectangle over 13, which actually starts from 12.5. Our normal approximation only included the area from 13. The continuity correction in this case would be:
• $$P(X_B \geq 13) \sim P(X_N \geq 12.5) = P(Z \leq \frac{12.5 - 10}{2.24}) = P(Z \geq 1.12) = P(Z \leq -1.12) = 0.1314$$
• It turns out that the exact probability in this case (using software) is 0.1316, so the approximation is excellent.
• The purpose of the next activity is to give you guided practice in solving word problems involving a binomial random variable, when the normal approximation is appropriate and is extremely helpful.
Scenario: Left-Handed College Students
• Roughly 10% of all college students in the United States are left-handed. Most academic institutions, therefore, try to have at least a few left-handed chairs in each classroom. 225 students are about to enter a lecture hall that has 30 left-handed chairs for a lecture. What is the probability that this is not going to be enough; in other words, what is the probability that more than 30 (or at least 31) of the 225 students are left-handed?
• Let X be the number of left-handed students (success) out of the 225 students (trials). X is therefore binomial with n = 225 and p = 0.1. We are asked to find P(X > 30) or P(X ≥ 31). Explain why we can use the normal approximation in this case, and state which normal distribution you would use for the approximation.
1. X is binomial with n = 225 and p = 0.1. The normal approximation is appropriate, since the rule of thumb is satisfied: np = 225 * 0.1 = 22.5 > 10, and also n(1 - p) = 225 * 0.9 = 202.5 > 10.
2. We will approximate the binomial random variable X by the random variable Y having a normal distribution with mean μ = 225 * 0.1 = 22.5 and standard deviation σ = sqrt(225 * 0.1 * 0.9) = sqrt(20.25) = 4.5
• Use the normal approximation to find P(X ≥ 31). For the approximation to be better, use the continuity correction as we did in the last example. In other words, rather than approximating P(X ≥ 31) by P(Y ≥ 31), approximate it by P(Y ≥ 30.5).
1. P(X ≥ 31) ≈ (normal approximation + continuity correction) ≈ P(Y ≥ 30.5) = P(Z ≥ (30.5 - 22.5) / 4.5) = P(Z ≥ 1.78) = (symmetry) = P(Z ≤ -1.78) = (table) = 0.0375.
• Already on several occasions we have pointed out the important distinction between a population and a sample.
• In Exploratory Data Analysis, we learned to summarize and display values of a variable for a sample, such as displaying the blood types of 100 randomly chosen adults using a pie chart, or displaying the heights of 150 males using a histogram and supplementing it with the sample mean $$\overline{X}$$ and sample standard deviation (S).
• In our study of Probability and Random Variables, we discussed the long-run behavior of a variable, considering the population of all possible values taken by that variable.
• For example, we talked about the distribution of blood types among all adults and the distribution of the random variable X, representing a male's height.
• In this module, we focus directly on the relationship between the values of a variable for a sample and its values for the entire population from which the sample was taken.
• This module is the bridge between probability and our ultimate goal of the course, statistical inference. In inference, we look at a sample and ask what we can say about the population from which it was drawn.
• In this module, we'll pose the reverse question: If I know what the population looks like, what can I expect the sample to look like?
• Clearly, inference poses the more practical question, since in practice we can look at a sample, but rarely do we know what the whole population looks like. This module will be more theoretical in nature, since it poses a problem which is not really practical, but will present important ideas which are the underpinnings for statistical inference.
• Example: Example #1: Blood Type
1. In the probability section, we presented the distribution of blood types in the entire U.S. population:
2. Assume now that we take a sample of 500 people in the United States, record their blood type, and display the sample results:
3. Note that the percentages (or proportions) that we got in our sample are slightly different than the population percentages. This is really not surprising. Since we took a sample of just 500, we cannot expect that our sample will behave exactly like the population, but if the sample is random (as it was), we expect to get results which are not that far from the population (as we did). If we took yet another sample of size 500:
4. we again get sample results that are slightly different from the population figures, and also different from what we got in the first sample. This very intuitive idea, that sample results change from sample to sample, is called sampling variability.
• Example: Example #2: Heights of Adult Males
1. Heights among the population of all adult males follow a normal distribution with a mean $$\mu = 69$$ inches and a standard deviation $$\sigma = 2.8$$ inches. Here is a probability display of this population distribution:
2. A sample of 200 males was chosen, and their heights were recorded. Here are the sample results:
3. The sample mean is $$\overline{x} = 68.7$$ inches and the sample standard deviation is s = 2.95 inches.
4. Again, note that the sample results are slightly different from the population. The histogram we got resembles the normal distribution, but is not as fine, and also the sample mean and standard deviation are slightly different from the population mean and standard deviation. Let's take another sample of 200 males:
5. The sample mean is $$\overline{x} = 69.065$$ inches and the sample standard deviation is s = 2.659 inches.
• Again, as in Example 1 we see the idea of sampling variability. Again, the sample results are pretty close to the population, and different from the results we got in the first sample.
• In both the examples, we have numbers that describe the population, and numbers that describe the sample. In Example 1, the number 42% is the population proportion of blood type A, and 39.6% is the sample proportion (in sample 1) of blood type A. In Example 2, 69 and 2.8 are the population mean and standard deviation, and (in sample 1) 68.7 and 2.95 are the sample mean and standard deviation.
• parameter and statistic (definition) A parameter is a number that describes the population; a statistic is a number that is computed from the sample.
1. In Example 1: 42% is the parameter and 39.6% is a statistic.
2. In Example 2: 69 and 2.8 are the parameters and 68.7 and 2.95 are the statistics.
• In this course, as in the examples above, we focus on the following parameters and statistics:
• population proportion and sample proportion
• population mean and sample mean
• population standard deviation and sample standard deviation
• The following table summarizes the three pairs, and gives the notation
• (Population) Parameter(Sample) Statistic
Proportion$$p$$$$\hat{p}$$
Mean$$\mu$$$$\overline{x}$$
Standard Deviation$$\sigma$$$$s$$
• The only new notation here is p for population proportion (p = 0.42 for type A in Example 1), and $$\hat{p}$$ for sample proportion
• ( $$\hat{p}$$ = 0.396 for type A in Example 1).
1. Parameters are usually unknown, because it is impractical or impossible to know exactly what values a variable takes for every member of the population.
2. Statistics are computed from the sample, and vary from sample to sample due to sampling variability.
• In the last part of the course, statistical inference, we will learn how to use a statistic to draw conclusions about an unknown parameter, either by estimating it or by deciding whether it is reasonable to conclude that the parameter equals a proposed value. In this module, we'll learn about the behavior of the statistics assuming that we know the parameters. So, for example, if we know that the population proportion of blood type A in the population is 0.42, and we take a random sample of size 500, what do we expect the sample proportion $$\hat{p}$$ to be?
• Example If students picked numbers completely at random from the numbers 1 to 20, the proportion of times that the number 7 would be picked is .05. When 15 students picked a number "at random" from 1 to 20, 3 of them picked the number 7. Identify the parameter and accompanying statistic in this situation.
1. The parameter is the population proportion of random selections resulting in the number 7, which is p = 0.05. The accompanying statistic is the sample proportion of selections resulting in the number 7, which is $$\hat{p} = 3/15 = 0.2$$.
• Example The length of human pregnancies has a mean of 266 days and a standard deviation of 16 days. A random sample of 9 pregnant women was observed to have a mean pregnancy length of 270 days, with a standard deviation of 14 days. Identify the parameters and accompanying statistics in this situation.
1. The parameters are population mean $$\mu = 266$$ and population standard deviation $$\sigma = 16$$. The accompanying statistics are sample mean $$\overline{x} = 270$$ and sample standard deviation s = 14.
• The SAT-Verbal scores of a sample of 300 students at a particular university had a mean of 592 and standard deviation of 73. According to the university's reports, the SAT-Verbal scores of all its students had a mean of 580 and a standard deviation of 110.
1. Explanation :
Indeed, 592 is the sample mean, which is a statistic.
1. Explanation :
Indeed, 110 is the population standard deviation, which is a parameter
1. Explanation :
sample mean which is 592.
1. Explanation :
the population mean which is 580
1. Explanation :
s is the sample standard deviation which is 73.
1. Explanation :
the population standard deviation which is 110.
• The first step to drawing conclusions about parameters based on the accompanying statistics is to understand how sample statistics behave relative to the parameter that summarizes the entire population
• Behavior of Sample Proportion $$\hat{p}$$
1. Approximately 60% of all part-time college students in the United States are female. (In other words, the population proportion of females among part-time college students is p = 0.6.) What would you expect to see in terms of the behavior of a sample proportion of females $$\hat{p}$$ if random samples of size 100 were taken from the population of all part-time college students?
2. As we saw before, due to sampling variability, sample proportion in random samples of size 100 will take numerical values which vary according to the laws of chance: in other words, sample proportion is a random variable. To summarize the behavior of any random variable, we focus on three features of its distribution: the center, the spread, and the shape.
3. Based only on our intuition, we would expect the following:
1. Center: Some sample proportions will be on the low side—say, 0.55 or 0.58—while others will be on the high side—say, 0.61 or 0.66. It is reasonable to expect all the sample proportions in repeated random samples to average out to the underlying population proportion, .6. In other words, the mean of the distribution of $$\hat{p}$$ should be p.
2. Spread: For samples of 100, we would expect sample proportions of females not to stray too far from the population proportion 0.6. Sample proportions lower than 0.5 or higher than 0.7 would be rather surprising. On the other hand, if we were only taking samples of size 10, we would not be at all surprised by a sample proportion of females even as low as 4/10 = 0.4, or as high as 8/10 = 0.8. Thus, sample size plays a role in the spread of the distribution of sample proportion: there should be less spread for larger samples, more spread for smaller samples.
3. Shape: Sample proportions closest to 0.6 would be most common, and sample proportions far from 0.6 in either direction would be progressively less likely. In other words, the shape of the distribution of sample proportion should bulge in the middle and taper at the ends: it should be somewhat normal.
4. Comment
1. The distribution of the values of the sample proportions $$\hat{p}$$ in repeated samples is called the sampling distribution of $$\hat{p}$$.
2. The purpose of the next activity is to check whether our intuition about the center, spread and shape of the sampling distribution of $$\hat{p}$$ was right via simulations.
• We're going to discuss the behavior of sample proportions by investigating these two questions:
1. when we collect random samples what patterns emerge?
2. More specifically, what is the shape, center, and spread of the distribution of sample proportions?
• To investigate these questions we're going to return to the familiar context of the previous example and look at the population of all part-time college students.
• We're assuming that sixty percent of this population is female.
• Now what we're going to be investigating in this movie is what happens as we begin to take random samples from this population.
• I'm going to be collecting random samples of 25 students at a time.
• Each random sample will have a different proportion of females. What we're interested in is what is going to happen when we began to collect many random samples.
• While we run the simulation, Each sample had 25 part-time college students in it and for each sample I calculated the proportion that were female and I recorded that here.
• What looks like happened over the long run is that many of the samples had proportions that were close to the population proportion of 0.6.
• We can also see that as we moved further away from 0.6 we had fewer samples with sample proportions in that range.
• We can get a more accurate sense of the variability in sample proportions by looking at the standard deviation. Here the standard deviation is roughly 10%.
• That tells me that typical samples had proportions that fell between about 0.5 and 0.7. Here in the graph I have marked one standard deviation below and one standard deviation above the mean of 0.6.
• Another thing we notice is that the shape is approximately normal. I have used a mathematical formula here to graph a normal curve on top of the sampling distribution and we can see that the normal distribution models the sample proportions well. This is encouraging.
• It tells me that a normal model will be a good probability model for the sampling distribution of sample proportions.
• When I increased the sample size by a factor of four, the standard deviation decreased to about half of what it was previously. So, I can conclude that larger samples do have less variability.
• I also notice that the mean of the sampling distribution stayed at 0.6. So the mean does not seem to be impacted by sample size. I also see that the distribution is normal for this case.
1. Explanation :
Each sample is represented in the sampling distribution with a dot at its p̂ value.
1. Explanation :
In the simulation, the sample proportions for n = 100 were more tightly grouped about the population proportion.
1. Explanation :
If the sample size is increased, the standard deviation will decrease because larger samples have less variability.
• If repeated random samples of a given size n are taken from a population of values for a categorical variable, where the proportion in the category of interest is p, then the mean of all sample proportions $$\hat{p}$$ is the population proportion (p).
• As for the spread of all sample proportions, theory dictates the behavior much more precisely than saying that there is less spread for larger samples. In fact, the standard deviation of all sample proportions $$\hat{p}$$ is exactly $$\sqrt{\frac{p(1-p)}{n}}$$.
• Since sample size n appears in the denominator of the square root, the standard deviation does decrease as sample size increases. Finally, the shape of the distribution of $$\hat{p}$$ will be approximately normal as long as the sample size n is large enough. The convention is to require both np and n(1 - p) to be at least 10.
• We can summarize all of the above by the following:
• $$\hat{p}$$ has a normal distribution with a mean of $$\mu_{\hat{p}} = p$$ and standard deviation $$\sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}}$$ (and as long as np and n(1 - p) are at least 10).
• Let's apply this result to our example and see how it compares with our simulation.
• In our example, n = 25 (sample size) and p = 0.6. Note that np = 15 ≥ 10 and n(1 - p) = 10 ≥ 10. Therefore we can conclude that $$\hat{p}$$ is approximately a normal distribution with mean p = 0.6 and standard deviation $$\sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.6(1-0.6)}{25}} = 0.097$$ (which is very close to what we saw in our simulation).
Scenario: Student Loans
• According to the National Postsecondary Student Aid Study conducted by the U.S. Department of Education in 2008, 62% of graduates from public universities had student loans.
• We randomly sample college graduates from public universities and determine the proportion in the sample with student loans.
1. Explanation :
Both conditions are met when n = 30. np = (30)(0.62) = 18.6 and n(1 - p) = (30)(0.38) = 11.4. Both are greater than 10. So a normal model is a good fit for the sampling distribution of sample proportions when n = 30.
1. Explanation :
62% of graduates from public universities had student loans. The shape of the distribution of pˆ will be approximately normal as long as the sample size n is large enough. Since our sample size is 50, the mean of the distribution of sample proportions will be equal to the population mean which is 62% or 0.62.
1. Explanation :
The standard deviation of the distribution of sample proportions is equal to sqrt((p * (1 - p)) / n) = sqrt((0.62 * (1 - 0.62)) / 50) = sqrt(0.2356 / 50) = sqrt(0.004712) = 0.0686 = 0.07.
1. Explanation :
The mean of the sampling distribution should be p = 0.62 with standard deviation sqrt( p(1 - p) / n ) = sqrt( 0.62(1 - 0.62 ) / 30) = 0.09. Typical values should fall within one standard deviation of the mean, from about 0.53 to 0.71. This distribution fits this description, as shown in the graph.
1. Explanation :
There is more variability in small samples, so it is more likely to get sample results further from p = 0.24 with a small bag.
1. Explanation :
We expect p̂s within 1 standard deviation of p = 0.20 to be most common. The standard deviation is about 0.06. 4 of the 5 p̂s in this sequence are within 0.06 of p = 0.20.
• The proportion of left-handed people in the general population is about 0.10. To simulate this population, we constructed a collection in which p = 0.10. We then conducted four simulations, drawing random samples of different sizes from this collection. Here you see the resulting sampling distributions and corresponding summary tables:
• Explain how these simulations illustrate the theory discussed above.
• Shape: Theory tells us that if np ≥ 10 and n(1 - p) ≥ 10, then the sampling distribution is approximately normal. When p = 0.10, these conditions are not met for n = 20 or n = 50. We can see that the distributions are skewed to the right for these sample sizes. As the sample size increases, we see the distributions becoming more normal. For n = 100 and for larger samples, the conditions are met, and we see that a normal distribution is a pretty good model for the sampling distribution.
• Note: We only collected 1,005 samples. The theory presumes we have collected all possible samples. So we don't expect the simulations to give perfectly normal distributions.
• Center: All of the sampling distributions are centered at approximately p = 0.10.
• Spread: The standard deviation of each sampling distribution is very close to the value predicted by
• $$\sqrt{\frac{p(1-p)}{n}}$$
• For example, using this formula for n = 20, the standard deviation is predicted to be 0.0671. We see in the simulation that the standard deviation is 0.0675, which is very close to the predicted value.
• A random sample of 100 students is taken from the population of all part-time students in the United States, for which the overall proportion of females is 0.6.
• (a) There is a 95% chance that the sample proportion $$\hat{p}$$ falls between what two values? First note that the distribution of has the mean p = 0.6, standard deviation $$\sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.6(1-0.6)}{100}} = 0.05$$, and a shape that is close to normal, since np = 100(0.6) = 60 and n(1 - p) = 100(0.4) = 40 are both greater than 10. The Standard Deviation Rule applies: the probability is approximately 0.95 that $$\hat{p}$$ falls within 2 standard deviations of the mean, that is, between 0.6 - 2(0.05) and 0.6 + 2(0.05). There is roughly a 95% chance that $$\hat{p}$$ falls in the interval (0.5, 0.7).
• (b) What is the probability that sample proportion $$\hat{p}$$ is less than or equal to 0.56?
• To find P($$\hat{p} \leq 0.56$$ ), we standardize 0.56 to z = (0.56 - 0.60) /0.05 = -0.80:
• P($$\hat{p} \leq 0.56$$ ) = P(Z ≤ -0.8) = 0.2119
• A random sample of 2,500 students is taken from the population of all part-time students in the United States, for which the overall proportion of females is 0.6.
• (a) There is a 95% chance that the sample proportion $$\hat{p}$$ falls between what two values? First note that the distribution of $$\hat{p}$$ has the mean p = 0.6, standard deviation $$\sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.6(1-0.6)}{2500}} = 0.01$$, and a shape that is close to normal, since np = 2500(0.6) = 1500 and n(1 - p) = 2500(0.4) = 1000 are both greater than 10. The standard deviation rule applies: the probability is approximately 0.95 that $$\hat{p}$$ falls within 2 standard deviations of the mean, that is, between 0.6 - 2(0.01) and 0.6 + 2(0.01). There is roughly a 95% chance that $$\hat{p}$$ falls in the interval (0.58, 0.62).
• (b) What is the probability that sample proportion $$\hat{p}$$ is less than 0.56?
• To find P($$\hat{p}$$ ≤ 0.56) , we standardize 0.56 to z = (0.56 - 0.60) / 0.01 = -4.00:
• P($$\hat{p}$$ ≤ 0.56) = P(Z ≤ -4.0) = 0, approximately.
• As long as the sample is truly random, the distribution of $$\hat{p}$$ is centered at p, no matter what size sample has been taken.
• Larger samples have less spread.
• Specifically, when we multiplied the sample size by 25, increasing it from 100 to 2,500, the standard deviation was reduced to 1/5 of the original standard deviation.
• Sample proportion strays less from population proportion 0.6 when the sample is larger: it tends to fall anywhere between 0.5 and 0.7 for samples of size 100, whereas it tends to fall between 0.58 and 0.62 for samples of size 2,500.
• It is not so improbable to take a value as low as 0.56 for samples of 100 (probability is more than 20%) but it is almost impossible to take a value as low as low as 0.56 for samples of 2,500 (probability is virtually zero).
• The purpose of this next activity is to give guided practice in finding the sampling distribution of the sample proportion
• ($$\hat{p}$$), and use it to draw conclusions about what values of $$\hat{p}$$ we are most likely to get.
• The proportion of left-handed people in the general population is about 0.1. Suppose a random sample of 225 people is observed. What is the sampling distribution of the sample proportion (p̂ )? In other words, what can we say about the behavior of the different possible values of the sample proportion that we can get when we take such a sample? (Note: normal approximation is valid because 0.1(225) = 22.5 and 0.9(225) = 202.5 are both more than 10.)
1. The possible values of the sample proportion follow approximately a normal distribution with mean p = 0.1, and standard deviation =$$\sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.1(1-0.1)}{225}} = 0.02$$
• Since the sample proportion has a normal distribution, its values follow the Standard Deviation Rule. What interval is almost certain (probability 0.997) to contain the sample proportion of left-handed people?
1. Since the Standard Deviation Rule applies, the probability is approximately 0.997 that the sample proportion falls within 3 standard deviations of its mean, that is, between 0.1 - 3(0.02) and 0.1 + 3(0.02). There is roughly a 99.7% chance, therefore, that the sample proportion falls in the interval (0.04, 0.16).
• In a sample of 225 people, would it be unusual to find that 40 people in the sample are left-handed?
1. About 18% (40/225) of this sample is left-handed. In the previous problem, we determined that there is roughly a 99.7% chance that a sample proportion will fall between 0.04 and 0.16. So a sample proportion of 0.18 is very unlikely. Note: According to the Standard Deviation Rule, sample proportions greater than 0.16 will occur 0.15% of the time. (100% - 99.7%) / 2 = 0.15%.
• Find the approximate probability of at least 27 in 225 (proportion 0.12) being left-handed. In other words, what is P(p̂ ≥ 0.12)? Guidance: Note that 0.12 is exactly 1 standard deviation (0.02) above the mean (0.1). Now use the Standard Deviation Rule.
1. Note that 0.12 is exactly 1 standard deviation above the mean. The Standard Deviation Rule tells us that there is a 68% chance that the sample proportion falls within 1 standard deviation of its mean, that is, between 0.08 and 0.12. There is therefore a probability of (1 - 0.68) / 2 = 0.16 that the sample proportion falls above 0.12.
• We are now moving on to explore the behavior of the statistic $$\overline{X}$$, the sample mean, relative to the parameter $$\mu$$, the population mean (when the variable of interest is quantitative).
• Example
1. Birth weights are recorded for all babies in a town. The mean birth weight is 3,500 grams, µ = 3,500 g. If we collect many random samples of 9 babies at a time, how do you think sample means will behave?
2. Here again, we are working with a random variable, since random samples will have means that vary unpredictably in the short run but exhibit patterns in the long run.
3. Based on our intuition and what we have learned about the behavior of sample proportions, we might expect the following about the distribution of sample means:
4. Center: Some sample means will be on the low side—say 3,000 grams or so—while others will be on the high side—say 4,000 grams or so. In repeated sampling, we might expect that the random samples will average out to the underlying population mean of 3,500 g. In other words, the mean of the sample means will be µ, just as the mean of sample proportions was p.
5. Spread: For large samples, we might expect that sample means will not stray too far from the population mean of 3,500. Sample means lower than 3,000 or higher than 4,000 might be surprising. For smaller samples, we would be less surprised by sample means that varied quite a bit from 3,500. In others words, we might expect greater variability in sample means for smaller samples. So sample size will again play a role in the spread of the distribution of sample measures, as we observed for sample proportions.
6. Shape: Sample means closest to 3,500 will be the most common, with sample means far from 3,500 in either direction progressively less likely. In other words, the shape of the distribution of sample means should bulge in the middle and taper at the ends with a shape that is somewhat normal. This, again, is what we saw when we looked at the sample proportions.
7. Comment The distribution of the values of the sample mean $$\overline{x}$$ in repeated samples is called the sampling distribution of $$\overline{x}$$.
• If repeated random samples of a given size n are taken from a population of values for a quantitative variable, where the population mean is $$\mu$$ and the population standard deviation is $$\sigma$$, then the mean of all sample means $$\overline{x}$$ is population mean $$\mu$$. As for the spread of all sample means, theory dictates the behavior much more precisely than saying that there is less spread for larger samples. In fact, the standard deviation of all sample means $$\overline{x}$$ is exactly $$\frac{\sigma}{\sqrt{n}}$$. Since the square root of sample size n appears in the denominator, the standard deviation does decrease as sample size increases.
• Scenario: Pell Grant Awards
1. The Federal Pell Grant Program provides need-based grants to low-income undergraduate and certain postbaccalaureate students to promote access to postsecondary education. According to the National Postsecondary Student Aid Study conducted by the U.S. Department of Education in 2008, the average Pell grant award for 2007-2008 was $2,600. Assume that the standard deviation in Pell grant awards was$500.
2. If we randomly sample 36 Pell grant recipients and record the mean Pell grant award for the sample, then repeat the sampling process many, many times, what is the mean and standard deviation of the sample means?
3. The distribution of sample means will have a mean equal to μ = 2,600 and a standard deviation of - $$\frac{\sigma}{\sqrt{n}} = \frac{500}{\sqrt{36} = 83.3$$
4. The population mean for Verbal IQ scores is 100, with a standard deviation of 15. Suppose a researcher takes 50 random samples, with 30 people in each sample. What is mean of the sample means?
5. The mean of the sample means is the population mean; therefore, the mean of the sample means or the sampling distribution of the mean is 100.
6. What is the standard deviation of the sample means? The standard deviation of the sample means is calculated by dividing the population standard deviation by the square root of the sample size; therefore, σ/sqrt(n) = 15/sqrt(30)= 15/5.48= 2.74.
Categorical (example: left-handed or not)p = population proportion$$\hat{p}$$ = sample proportionp$$\sqrt{\frac{p(1-p)}{n}}$$Normal IF np ≥ 10 and n(1 - p) ≥ 10
Quantitative (example: age)μ = population mean, σ = population standard deviation$$\overline{x}$$ = sample mean$$\mu$$$$\frac{\sigma}{\sqrt{n}}$$Normal if n > 30 (always normal if population is normal)
• If a variable is skewed in the population and we draw small samples, the distribution of sample means will be likewise skewed. If we increase the sample size to around 30 (or larger), the distribution of sample means becomes approximately normal.
• To summarize, the distribution of sample means will be approximately normal as long as the sample size is large enough. This discovery is probably the single most important result presented in introductory statistics courses. It is stated formally as the Central Limit Theorem.
• We will depend on the Central Limit Theorem again and again in order to do normal probability calculations when we use sample means to draw conclusions about a population mean. We now know that we can do this even if the population distribution is not normal.
• How large a sample size do we need in order to assume that sample means will be normally distributed? Well, it really depends on the population distribution, as we saw in the simulation. The general rule of thumb is that samples of size 30 or greater will have a fairly normal distribution regardless of the shape of the distribution of the variable in the population.
• Comment: For categorical variables, our claim that sample proportions are approximately normal for large enough n is actually a special case of the Central Limit Theorem.
• Recall our earlier scenario: The Federal Pell Grant Program provides need-based grants to low-income undergraduate and certain postbaccalaureate students to promote access to postsecondary education. According to the National Postsecondary Student Aid Study conducted by the U.S. Department of Education in 2008, the average Pell grant award for 2007-2008 was $2,600. Assume that the standard deviation in Pell grants awards was$500.
1. Explanation :
For n=36, sample means are approximately normal, so we can use the Standard Deviation Rule. Three standard deviations above 2,600 is 2,600 + 3(500/6)) = 2,850). So $2,940 is more than 3 standard deviations above$2,600, thus this sample mean would be surprising
1. Explanation :
The sampling distribution will have a mean of 2,600 and a standard deviation of 70. Using the Standard Deviation Rule, approximately 68% of the values should be between 2,530 and 2,670 (within 1 standard deviation of the mean.). Graph B looks like it could fit this description.
• Household size in the United States has a mean of 2.6 people and standard deviation of 1.4 people.
• (a) What is the probability that a randomly chosen household has more than 3 people?
1. A normal approximation should not be used here, because the distribution of household sizes would be considerably skewed to the right. We do not have enough information to solve this problem.
• (b) What is the probability that the mean size of a random sample of 10 households is more than 3?
1. By anyone's standards, 10 is a small sample size. The Central Limit Theorem does not guarantee sample mean coming from a skewed population to be approximately normal unless the sample size is large.
• (c) What is the probability that the mean size of a random sample of 100 households is more than 3?
1. Note: To review how to determine probabilities for z scores, please refer to the Standard Normal Table section of the Random Variables module.
2. Now we may invoke the Central Limit Theorem: even though the distribution of household size X is skewed, the distribution of sample mean household size $$\overline{X}$$ is approximately normal for a large sample size such as 100. Its mean is the same as the population mean, 2.6, and its standard deviation is the population standard deviation divided by the square root of the sample size: $$\frac{\sigma}{\sqrt{n}} = \frac{1.4}{\sqrt{100}} = 0.14$$ The z-score for 3 is $$\frac{3-2.6}{\frac{1.4}{\sqrt{100}}} = \frac{0.4}{0.14} = 2.86$$ The probability of the mean household size in a sample of 100 being more than 3 is therefore P($$\overline{X}$$ > 3) = P(Z > 2.86) = P(Z < -2.86) = 0.0021.
3. Households of more than 3 people are, of course, quite common, but it would be extremely unusual for the mean size of a sample of 100 households to be more than 3.
• The annual salary of teachers in a certain state X has a mean of $54,000 and standard deviation of σ =$5,000.
• What is the probability that the mean annual salary of a random sample of 5 teachers from this state is more than $60,000? Find this probability or explain why you cannot. 1. Recall from the Exploratory Data Analysis unit that salary distribution is typically skewed to the right. Since 5 is a small sample size, and the Central Limit Theorem does not guarantee that the sample mean coming from a skewed population is approximately normal unless the sample size is larger, we thus do not have enough information to solve the problem. 1. Explanation : According to the Central Limit Theorem, the mean has approximately a normal distribution with the same mean as the population; therefore,$54,000 is the mean of the distribution of the sample means
1. Explanation :
According to the Central Limit Theorem, then, the mean has approximately a normal distribution with the same mean as the population, $54,000, and a standard deviation of: σ/square root(n) = 5000/square root(64) = 625. 1. Explanation : The z-score of 52,000 is: (52,000 - 54,000)/5000/sqrt(64). 1. Explanation : The probability of the z score of -3.21 using the Normal Table is 0.0007 or P(<52,000) = P(z ,-3.2) = (table) = 0.0007. Thus, we find that while it is probably quite common to find teachers in this state with an annual salary that is less than$52,000, it would be extremely unusual for the mean salary of a sample of 64 teachers to be less than \$52,000.
Scores on the math portion of the SAT (SAT-M) in a recent year have followed a normal distribution with mean μ = 507 and standard deviation σ = 111. What is the probability that the mean SAT-M score of a random sample of 4 students who took the test that year is more than 600? Explain why you can solve this problem, even though the sample size (n = 4) is very low.
• Since the scores on the SAT-M in the population follow a normal distribution, the sample mean automatically also follows a normal distribution, for any sample size. Therefore, the mean has a normal distribution with the same mean as the population, 507, and standard deviation
• $$\frac{\sigma}{\sqrt{n}} = \frac{111}{\sqrt{4}} = 55.5$$
• The z-score of 600 is therefore:
• $$\frac{600-507}{111/\sqrt{4}} = \frac{93}{55.5} = 1.68$$
• And therefore,
• $$P(\overline{X} > 600) = P(Z > 1.68) = P(Z < -1.68) = 0.0465$$
• We find that while it is very common to find students who score above 600 on the SAT-M, it would be quite unlikely (4.65% chance) for the mean score of a sample of 4 students to be above 600.
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# Least Common Denominator
... is the Least Common Multiple of the denominators ...
First, let's be clear what a Denominator is:
## Fractions
A Fraction (such as 3/4) has two numbers:
Numerator Denominator
We call the top number the Numerator, it is the number of parts you have.
We call the bottom number the Denominator, it is the number of parts the whole is divided into.
## Fractions with Different Denominators
Sometimes you have two (or more) fractions with different denominators - you may want to add or subtract them - but you need to make the denominators the same before you can do that:
### Example: What is 3/8 + 5/12 ?
Let's try to make the denominators the same ... if you multiply 8 × 3 you get 24, and if you multiply 12 × 2 you also get 24. So, let's try that (important: what you do to the bottom of the fraction, you must also do to the top):
× 3 3 = 9 8 24 × 3
and,
× 2 5 = 10 12 24 × 2
Now we can do the addition: 9/24 + 10/24 = 19/24.
## How to Make the Denominators the Same
The trick is to find the Least Common Multiple of the denominators. (Read about Least Common Multiples now, if you haven't already.)
In the previous example, the Least Common Multiple of 8 and 12 was 24. And so the the Least Common Denominator of 3/8 and 5/12 is 24
So, here are the steps:
• Find the Least Common Multiple of the denominators (this is called the Least Common Denominator).
• Change each fraction (using equivalent fractions) to make their denominators the same as the least common denominator
• Then you can do whatever you want with the fractions (add, subtract, etc) !
### Example: What is 1/6 + 7/15 ?
The Least Common Multiple of 6 and 15 is 30 (try to work that one out yourself!). So, let's do some multiplying to make each denominator equal to 30:
× 5 1 = 5 6 30 × 5
and,
× 2 7 = 14 15 30 × 2
Now we can easily do the addition: 5/30 + 14/30 = 19/30.
## Least Common Multiple Tool
To find the least common denominator automatically, you can use our Least Common Multiple Tool - just put in the denominators, press the button, and the least common denominator is shown for you.
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###### Norm Prokup
Cornell University
PhD. in Mathematics
Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.
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# Exponential Functions - Problem 4
Norm Prokup
###### Norm Prokup
Cornell University
PhD. in Mathematics
Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.
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One of the problems you are going to see a lot of when you are studying exponential functions is, find the exponential function that matches the situation. Well here, the situation is a graph. We want to find an exponential function f of x equals 8 times b to the x that matches this graph.
So the first thing you want to do is you have to know the formula of an exponential function. You have to know this form. And if you know that the graph passes through these points, you can use that to figure out what the coefficients a and b are.
So for example, using the point (1,54) f of x is going to be 54, and x is going to be 1. So 54 equals 8 times b to the 1. That’s one equation. And using a second point, f of x is going to be 24 and x is going to be 3. So you have two equations with 2 unknowns. Easy to solve, all I need to do is take one equation, and divide both sides by the other equation. This is kind of like the method of elimination.
And you see that a lot of cancellation that’s going to happen here. There a’s will cancel, some of the b’s will cancel, and I can also reduce this fraction; 24 over 54. 24 is 2 times 12, this is 2 times 27. So I have 12 over 27, but that reduces even any further. Both of these have a factor of 3. 3 times 4 over 3 times 9, cancel the 3’s. So b² is 4 over 9. That means b is plus or minus 2/3. Now we know that the base of an exponential function has to be positive. So b has to be 2/3. All I have to do is find a.
Let’s use this equation because I'm going to have to raise b to the first power. So it will be a little easier to find a. 54 equals a times b. Equals a times 2/3. Now I just multiply both sides by 3/2. I get 3 times 27, 81 equals a, a equals 81. And so my function is f of x equals 81 times 2/3 to the x.
That’s it. All you have to do when you are finding the equation of an exponential function is, use the points that you are given, (1,54) and (3,24), to set up equations that involve the variables that are the coefficients of your exponential function and solve for them.
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# Polygons0
A polygon is a closed figure whose sides are line segments.
Two consecutive sides of a polygon share a common endpoint. The common endpoints are called the vertices of the polygon. A polygon is named by naming the vertices as you go around the polygon. In Figure 10.4, the polygon has six sides and can be named ABCDEF.
1. A diagonal of a polygon is a line segment that joins two nonconsecutive vertices.
2. An equiangular polygon is a polygon with all angles having the same measure.
3. An equilateral polygon is a polygon with all sides having the same length.
4. A regular polygon is a polygon that is both equiangular and equilateral.
Types of Polygons:
The sum of the interior angles of a polygon with n sides is S = (n − 2)180°.
Example:
Find the sum of the interior angles of a polygon with the given number of sides.
A. 3
B. 4
C. 5
D. 10
E. 15
Solution:
A. n = 3, S = (n − 2)180° = (3 − 2)180° = 1 × 180° = 180°
B. n = 4, S = (n − 2)180° = (4 − 2)180° = 2 × 180° = 360°
C. n = 5, S = (n − 2)180° = (5 − 2)180° = 3 × 180° = 540°
D. n = 10, S = (n − 2)180° = (10 − 2)180° = 8 × 180° = 1,440°
E. n = 15, S = (n − 2)180° = (15 − 2)180° = 13 × 180° = 2,340°
1. Two polygons are congruent if the angles of the first polygon are equal to the corresponding angles of the second polygon and the sides of the first polygon are equal to the corresponding sides of the second polygon.
2. Two polygons are similar if the angles of the first polygon are equal to the corresponding angles of the second polygon and the sides of the first polygon are proportional to the corresponding sides of the second polygon.
3. The altitude or height of a polygon is a line segment from one vertex of a polygon that is perpendicular to the opposite side, or to the opposite side extended.
To show that sides are of the same length, you mark them with the same number of tick marks and then mark angles with arcs.
In that Figure 10.5, the quadrilateral ABCD has the sides marked to show AD = BC and AB = CD. The angles are marked to show that ∠A = ∠C and ∠B = ∠D.
# GRE: Data Interpretation Questions0
Data interpretation questions involve information presented in a table or a graph. The questions may involve simply reading the graph, drawing conclusions from the data shown, making inferences based on the given data, and making comparisons using the data shown. One thing to note about the table or graph provided for a question is that it will always contain a great deal of information not needed for answering your question.
For the GRE, about one-fifth of the 20 questions in a quantitative section will be data interpretation questions.
The data presented in charts and tables are classified by both row and column. The units will always be stated, and the table will frequently involve data that have been rounded. Occasionally, the data will be expressed as percentages. The questions related to a table may ask you to merely read one particular item from the table, to combine values from the table, or to compare values found in the table.
In questions involving graphs, the graph will have a name that indicates what information is being presented. The type of graph used may also aid you in the interpretation of the data.
Line graphs and bar graphs have a scale which may or may not start at zero. The horizontal and vertical axes will each be labeled to show both what is being graphed and the scale along that dimension. Line graphs show the change in a quantity over time. A bar graph can also show the change in a quantity over time, but it can show a comparison of quantities as well. In a bar graph, the width of the bars is generally uniform, and only the height or length varies.
A circle graph or pie chart is used to show a whole divided up into sectors which represent a percentage of the whole. The size of the sector is proportional to the percentage of the whole that it represents. The sum of all of the sectors is always 100%.
When working with graphs and tables, the goal is to use the visual representation to help you find the answer, rather than doing extensive computation. In a line graph that has values for 1980 and 1990, if you are asked to find the value for 1985, the easiest way is to locate the midpoint of the segment between 1980 and 1990 and read the answer from the axis representing the requested value. If you are asked how many times the number of accidents exceeded 40,000, then hold your pencil at the correct level to form a line at 40,000 and count the number of bars going past the line. In circle graphs, it is easier to make comparisons using the percentages given than to use the actual numbers. Whenever possible, use the visual image to answer the question; that is why the data are presented in graphical form.
Example 1:
What is the ratio of the time Jose spent in class to the time he spent at work?
• 3% more
• 33% together
• 6/5
• 3/4
• 5/6
Solution:
The circle graph shows that Jose spent 15% of his time in class and 18% of his time at work. There is no need to change the percentages to hours to get the ratio of time spent in class to time spent at work. The ratio is :
Example 2:
What is the best estimate of the end-of-year bonus for 1965?
• \$3
• \$4
• \$3,000
• \$4,000
• \$5,000
Solution:
On the line graph, if you place an imaginary dot in the middle of the segment between 1960 and 1970, then follow that point across the graph to the scale, it hits at about the \$4 mark. Since the scale is in thousands of dollars, the bonus for 1965 is \$4,000. The answer is D.
Example 3:
How much greater is the preference for Brand Q over Brand S?
• 4%
• 3%
• 2%
• 1%
• 12%
Solution:
The bar graph shows that Brand Q is preferred by 8% and Brand S is preferred by 4%.
Example 4:
What is the least amount of growth for cities that grew between 1950 and 2000?
• 10,000
• 20,000
• 30,000
• 40,000
• 50,000
Solution:
The bar graph shows that the population of city W declined by 40,000, city X increased by 10,000, city Y increased by 20,000, and city Z increased by 30,000. Of the cities that had a population increase, city X had the smallest increase. The answer is A.
Example 5:
What two pieces of equipment together are used the same amount of time as weights and the elliptical machine?
• Glider and rowing machine
• Bicycle and glider
• Rowing machine and bicycle
Solution:
The circle graph shows that the combined percent of time spent on weights and the elliptical machine is 25% + 5% = 30%. Bicycles are used 12% of the time, and gliders are used 18% of the time. Since 12% + 18% = 30%, the correct answer is C.
# Operations with Fractions0
To add two fractions, the fractions must have a common denominator. Add the numerators to get the numerator of the sum, and the denominator of the sum is the common denominator.
If the fractions do not have a common denominator, create one by finding the least common multiple of the denominators, or use the product of the denominators. The product of the denominators is quicker to find, but it can be much larger than the LCM, which could make the computation more difficult.
Example:
Solution:
## Subtracting Fractions
Subtraction of fractions also requires that the fractions have a common denominator. Once the fractions have a common denominator, subtract the numerators, and the result goes over the common denominator.
Order is important in subtraction. If it is reversed, you get the opposite value.
You can write with the fraction being negative rather than the numerator being negative, since the two fractions are equal.
Example:
Solutions:
Addition and subtraction of mixed numbers is similar to the procedures for fractions.
Example:
Solution:
## Multiplying Fractions
Multiplication of fractions does not require the fractions to have common denominators.
You can multiply the numerators together and place the product over the product of the denominators. The next step is to reduce the fraction to lowest terms.
A second way is to divide a numerator and a denominator by a common factor. The second approach is most helpful when the numbers are large or when several fractions are being multiplied.
Exercise:
Multiply these fractions.
Solution:
## Dividing Fractions
When you want to divide fractions, change the problem to multiplication by the reciprocal of the divisor. As in all other division, the divisor fraction cannot be zero.
Example:
Divide these fractions.
Solution:
# What Are Fractions Make Of?0
A fraction is made up of three parts: the numerator, the fraction bar, and the denominator. The numerator tells you how many parts you have, and the denominator tells you how many parts the whole is divided into. The fraction bar is read as “out of.” Otherwise, it is used as a grouping indicator to remind you to simplify the numerator and denominator separately before doing the division.
A fraction can tell you what part of a whole unit that you have, and even that you have more parts than are needed for one whole.
The fraction 3/4 means that you have 3 of the 4 parts that the whole is divided into, while 5/4 means that you have all 4 parts of the one whole, plus 1 part of another whole that is divided into 4 equal parts.
In a fraction a/b where a and b are whole numbers and b is not zero, you say that a/b is a proper fraction when a < b. If a = b or a > b, you say that it is an improper fraction.
You can use fractions to express the ratio of two quantities. If there are units with the numbers, then they need to be the same units. When the units are different, convert to a common unit.
A fraction can be used to indicate division; . This use of fractions is especially helpful when you are dividing a smaller counting number by a larger one.
When you have an improper fraction such as , you can change it so that the number of whole units involved is clear. For 4/4, you have all 4 parts for 1 unit, and 4/4=1. With 7/5, you have 2 more parts than needed to make a whole unit, so . It is understood that With 11/2, you see that 11 ÷ 2 yields 5 with a remainder of 1. The remainder is 1 of 2 parts needed to make the next whole, so it represents 1/2. Thus,
It is very helpful to be able to convert between improper fractions and mixed numbers.
A complex fraction has fractions in the numerator, the denominator, or both.
Examples of complex fractions are
One way to simplify complex fractions is to find the least common multiple of all of the fractions in the numerator and denominator and then multiply the numerator and denominator by this number.
Solution:
1. The only fraction is 1/2, so you need to multiply by
2. The only fraction is 5/6, so you need to multiply by
Equivalent fractions are fractions that are different representations of the same number.
For example, in this set, there are six objects, and two are @s: @@# # # #.
So the fraction that shows the ratio of *s to all objects in the set is 2/6.
In the set @@@@# # # # # # # # there are 4 @s and 12 objects, so the fraction representing *s to objects is 4/12. However in each case, there is one@ out of every three objects.
The ratio of objects in each set is . Thus you say that 1/3, 2/6, 4/12 and are equivalent fractions.
You use equivalent fractions when you put a fraction in lowest terms, and when you want to add or subtract fractions. You can also use equivalent fractions when you compare fractions, but other ways may be faster.
Example:
Write each fraction as an equivalent fraction whose denominator is the number given.
Solution
A. To write ⅔ as an equivalent fraction with a denominator of 30, you need to find what number times 3 equals 30. 30 ÷ 3 = 10, so 3 × 10 = 30.
B. 28 ÷ 7 = 4, so 7 × 4 = 28.
# Other GRE Math Question Formats0
## GRE: Numeric Entry Question
Numeric Entry questions require you to compute the answer to a question and then enter that answer into a box or boxes provided. These questions are problem-solving and reasoning situations similar to GRE multiple choice questions, except that no answer choices are provided. The correct answer to a Numeric Entry question is a decimal or integer with up to eight digits; it may be positive, zero, or negative. The negative sign or decimal point must be included if necessary.
For single box answers, type the answer directly into the box, but take care in your typing. A typo will count as a wrong answer.
When the answer is a fraction, you will be given two boxes, one above a fraction bar for the numerator, and the other below the bar for the denominator. You cannot have a decimal point in the numerator or denominator of a fraction. Fractions do NOT have to be in lowest terms for Numeric Entry questions.
### EXAMPLES
Example 1:
A coat was sold at a 30% discount sale for \$140. What was the original price of the coat?
Solution:
The original price of the coat is 100% of P, and the discount amount is 30% of P, so the sale price is 70% of the original price.
0.70 P = 140
P = 140 ÷ 0.70
P = 200
On the real test, you will answer the question by clicking on the box and entering 200. It would also be correct to enter 200.00, but it is not necessary to do so.
Example 2:
George earns \$15 an hour for each regular-time hour he works; he earns time and one-half for each overtime hour. Overtime hours start after he has worked 32 hours in a week. What is George’s pay this week if he works 45 hours?
Solution:
George works 45 − 32 = 13 overtime hours. Because overtime hours are paid at a rate of time and one-half, the pay rate is 1.5 × \$15 = \$22.50 per hour.
George’s pay is his regular pay plus his overtime pay. 32 × \$15 + 13 × \$22.50 = \$480 + \$292.50 = \$772.50. Enter 772.50 into the box. The decimal point must be entered from the keyboard too. If you enter 772.5, that is also correct.
### SOLUTION STRATEGIES
2. Make sure you enter a complete answer, including any negative signs or decimal points.
3. There are no answer choices to check your solution against, so check to see that it satisfies all of the conditions in the problem.
### EXERCISES
1. Calls made using a particular calling card cost \$0.50 for the first 2 minutes and \$0.15 for each minute after the first 2 minutes. What would be the cost of a 10 minute call?
1. A room is 12 ft wide, 18 ft long, and 9 ft high. How many square yards of carpet would it take to cover the floor of the room?
1. What is the sum of the positive factors of 75?
### SOLUTIONS
1. 1.70
Break the problem into parts: the cost of the first 2 minutes and the cost of the remaining 8 minutes. The cost of the first 2 minutes is \$0.50. The remaining 8 minutes cost \$0.15 each.
\$0.50 + 8(\$0.15) = \$0.50 + \$1.20 = \$1.70.
The call costs \$1.70, so enter 1.70 or 1.7.
1. 24
The floor is 12 ft wide and 18 ft long. (Note that you were also given the height of the room. On the GRE, you may be given more information than you need to solve a problem.) Change the dimensions to yards; 3 ft equals 1 yard, and the length and width are divisible by 3. The room is 4 yards wide and 6 yards long, so the area of the room is 4 × 6 square yards, or 24 square yards. Enter 24.
1. 124
75 = 1 × 75, 75 = 3 × 25, 75 = 5 × 15. Thus, the positive factors of 75 are 1, 3, 5, 15, 25, and 75. 1 + 3 + 5 + 15 + 25 + 75 = 124. The sum of the positive factors of 75 is 124, so enter 124.
# Solving GRE Multiple Choice Questions0
Multiple Choice questions are the typical standardized test questions most test-takers are familiar with. About 50% of the questions on the Quantitative Reasoning sections of the GRE are of this type; on the computer-based test, out of 20 questions in each section, approximately 10 will be Multiple Choice questions. These questions have five answer choices, only one of which is correct. They focus on general problem-solving skills. You are to use the given information and your reasoning skills to select the best answer.
Unless you are told differently, you can assume all numbers are real numbers. Operations on real numbers are also assumed.
Figures provided for these problems show general relationships such as straight lines, collinear points, and adjacent angles. In general, you cannot determine the measures of angles or the lengths of segments from the figure alone. When a figure is NOT drawn to scale, it will be clearly identified as such. From a figure that is drawn to scale, you may estimate the lengths of segments and measurements of angles.
Example 1:
Solution:
Because is a proportion, you can use two properties to transform it. First, use the reciprocal property to get then use the subtraction property to get
So and answer choice C is correct.
Example 2:
In circle P, the two chords intersect at point X, with the lengths as indicated in the figure. Which could NOT be the sum of the lengths a and b, if a and b are integers?
Solution:
When two chords intersect within a circle, the product of the segments on one chord is equal to the product of the segments on the other chord. Because the segments of the first chord are 6 and 8, the product of the lengths is 48. Thus, the product of the lengths a and b must be 48. Possible lengths are 48 and 1, 24 and 2, 16 and 3, 12 and 4, and 8 and 6, so possible values for a + b are 49, 26, 19, 16, and 14. The correct answer is choice A because 30 is not the sum of two integer factors of 48.
Example 3:
What is the sum of the prime numbers between and ?
Solution:
The prime numbers between and are 2, 3, 5, and 7. The sum of these prime numbers is 17, so the answer is choice C.
# GRE Multiple Choice Strategies for the Quantitative Reasoning Section0
Remember the following general strategies when approaching GRE Multiple-choice questions. Note that these strategies might also be helpful in answering the other question types.
## Draw Pictures
Visualize the problem by creating a figure or diagram. This strategy should not take a lot of time, and can prevent careless errors. Sometimes, you are given a figure or a table that you can work with; sometimes, you just have to make your own.
Consider the following examples:
Q1: The greatest number of diagonals that can be drawn from one vertex of a regular 8-sided polygon is:
The correct answer is E. To solve this problem, draw a diagram such as the one that follows:
As you can see, if you draw a regular octagon (8-sided polygon), you can make only five diagonals from one vertex.
Q2: A square is inscribed in a circle. If the area of the inscribed square is 50, then the area of the circle is:
The correct answer is D. To solve this problem, draw a diagram such as the one that follows:
## Apply Logic
Remember, most of the actual calculations for GRE Quantitative problems are fairly simple. In fact, the GRE test writers are just as likely to test your logical reasoning ability or your ability to follow directions as they are to test your ability to plug numbers into an equation.
Consider the following examples:
Q1:
The correct answer is D. To solve this problem, first recognize that . This is true because the c values cancel each other out, leaving you with , which is equivalent to . Therefore, a + b must equal 2 + 16, or 18.
Q2: Which of the following equations correctly translates this statement: “Three less than twice a number is the same as eight more than half the number”?
The correct answer is E. The first step in solving this problem is to recognize that “three less than” means to subtract 3 from some other quantity. Quickly review the answer choices to see that you only need to consider answer choices C and E. Likewise, the statement “eight more than” means to add 8 to some other quantity. Only answer choice E meets both of these criteria. The expression “three less than twice a number” is expressed mathematically as 2x – 3, and “eight more than half the number” is expressed mathematically as 1/2x + 8 .
If the problem requires three steps to reach a solution and you only completed two of the steps, then it is likely that the answer you arrived at will be one of the choices. However, it will not be the correct choice!
Consider the following examples:
Q1: The rectangular garden shown in the figure above has a stone border 2 feet in width on all sides. What is the area, in square feet, of that portion of the garden that excludes the border?
The correct answer is B. This problem asks for the area of the middle portion of the garden. To solve this problem, perform the following calculations, and remember that the border goes around the entire garden. First, subtract the border width from the length of the garden:
12 – 2(2) = 8
Next, subtract the border width from the width of the garden:
6 – 2(2) = 2
The area (length × width) of the portion of the garden that excludes the border is 8 × 2, or 16.
If you only accounted for the border along one length and one width of the garden, you would have gotten answer choice C. Answer choice D is the area of the border around the garden. Answer choice E is the area of the entire garden, including the stone border.
Q2: Which of the following is the slope of a line that is perpendicular to the line ?
The correct answer is E. The first step in solving this problem is to put the equation in the standard form y=mx+b where m is the slope:
The slope of the given line is –3/5. However, you are asked for the slope of a line that is perpendicular to the given line. Perpendicular lines have negative reciprocal slopes, so the correct answer is 5/3.
## Don’t Quit Early
Particularly for questions that ask you to consider selecting more than one answer choice, make sure you examine all the answer choices. Don’t make the mistake of moving on to the next problem as soon as you have found one correct answer choice!
Consider the following examples:
Q1: Which of the following integers are multiples of both 2 and 3? Indicate all such integers.
The correct answer is A and C. Be sure to evaluate each of the answer choices, and make sure to select those choices that meet both criteria in the question stem. Note that some choices are divisible by either 2 or 3, but the question asks for all integers divisible by both 2 and 3.
Q2: For which values of x is the following function undefined?
The correct answer is A and D. Don’t let the term function scare you; this is not really a function question at all. To solve this problem, first recall that a fraction is undefined when the denominator is 0. This is true because you cannot divide a number by 0. Therefore, the correct answer will include the values of x for which . Because both 72 and , answer choices A and D are correct.
## Check the Choices
Take a quick look at the answer choices as you read the problem for the first time. They can provide valuable clues about how to proceed. For example, many answer choices will be in either ascending or descending order. If the question asks you for the least possible value, try the smallest answer choice first. If it does not correctly answer the question, work through the rest of the answer choices from smallest to largest. Remember that one of them is the correct choice.
Consider the following examples:
Q1: If x is an integer and y = 7x + 11, what is the greatest value of x for which y is less than 50?
The correct answer is C. Because the question asks for the greatest value of x, evaluate answer choice A first because it is the greatest value among the answer choices:
• Answer choice A: y=7(7) + 11 = 60. This is not less than 50, so eliminate answer choice A, and look at answer choice B.
• Answer choice By=7(6) + 11 = 53. This is not less than 50, so eliminate answer choice B, and look at answer choice C.
• Answer choice Cy=7(5) + 11 = 46. Because 5 is the greatest of the remaining answer choices and the result is less than 50, answer choice C must be correct.
Q2: Noah’s Ark Fish Store sold 80 guppies and 48 bala sharks on Saturday. Which of the choices below could represent the ratio of bala sharks to guppies sold on Saturday?
The correct answer is A. The first step in solving this problem is to note that the ratio is bala sharks to guppies, which can be expressed mathematically in the following ways: 48 to 80, 48:80, and 48/80. Therefore, the correct answer must include a factor of 48 in the first position. Eliminate answer choices B, D, and E because 5, 25, and 20 are not factors of 48. Answer choice C cannot be correct because, although 6 is a factor of 48 and 5 is a factor of 80, the fraction 48/80 is less than the fraction 6/5 (note that the numerator is larger than the denominator in the second fraction). Therefore, the correct answer must be A; 12/20 = 48/80.
## Pick Numbers for the Variables
You can sometimes simplify your work on a given problem by using actual numbers as “stand-ins” for variables. This strategy works when you have variables in the question and the same variables in the answer choices. You can simplify the answer choices by substituting actual numbers for the variables. Pick numbers that are easy to work with and that meet the parameters of the information given in the question. If you use this strategy, remember that numbers on the GRE can be either positive or negative and are sometimes whole numbers and sometimes fractions. You should also be careful not to use 1 or 0 as your “stand-ins” because they can create “identities,” which can lead to more than one seemingly correct answer choice. The word identity refers to an equality that remains true regardless of the values of any variables that appear within the equality. For example, any number multiplied by 0 is always 0.
In addition, it is sometimes necessary to try more than one number to see if the result always correctly responds to the question. If the numbers that you pick work for more than one answer choice, pick different numbers and try again, focusing on the remaining answer choices.
Consider the following examples:
Q1: If x and y are both positive even integers, which of the following expressions must be even? Select all such expressions.
The correct answer is A and C. The question states that both x and y are positive even integers. Therefore, you can pick any positive even integer and substitute that value for x and y in each of the choices, as follows:
Q2:
The correct answer is E. You are given that both a and b are positive consecutive odd integers, and that b is greater than a. Pick two numbers that fit the criteria: and . Now substitute these numbers into the equation, as follows:
Sometimes the questions include irrelevant data, so be sure you’re working with the correct numbers! Also, it helps to write down key words and phrases in the question. When you are looking at ratio problems, for example, note whether the question is giving a part-to-part ratio or a part-to-whole ratio. The ratio of girls to boys in a class is a part-to-part ratio. The ratio of girls to students in a class is a part-to-whole ratio. Focusing on the right information will help you to quickly answer the question.
Consider the following examples:
Q1: The ratio of two quantities is 4 to 5. If each of the quantities is increased by 3, which of the following could be the ratio of these two new quantities? Indicate all answer choices that apply.
The correct answer is A, B, and C. To understand this problem, realize that, although the ratio of two quantities is 4 to 5, the actual values of the quantities might be very different. For different sets of values, increasing each by 3 will create different ratios. For instance, if the quantities were 4 and 5, increasing each by 3 would result in a ratio of 7 to 8. If the quantities were 8 and 10 (a ratio of 4 to 5), increasing each by 3 would result in a ratio of 11 to 13. If the quantities were 20 and 25 (a ratio of 4 to 5), increasing each by 3 would result in a ratio of 23 to 28. So answer choices A, B, and C are all correct.
Q2: What percent of 5 is 7?
The correct answer is D. To solve this problem, note that 7 is greater than 5, which means that the correct answer must be greater than 100%; eliminate answer choices A, B, and C. To find what percent of 5 the number 7 is, you can simply divide 7 by 5 and multiply by 100%, as follows:
# What Math Should I Know for the GRE?0
The GRE is taken by people with a wide variety of educational backgrounds and undergraduate majors. For that reason, the GRE Quantitative Reasoning section tests mathematical skills and concepts that are assumed to be common for all test-takers. The test questions require you to know arithmetic, algebra, geometry, and basic probability and statistics. You will be expected to apply basic mathematical skills, understand elementary mathematical concepts, reason quantitatively, apply problem-solving skills, recognize what information is relevant to a problem, determine what relationship, if any, exists between two quantities, and interpret tables and graphs.
The GRE does not attempt to assess how much mathematics you know. It seeks to determine whether you can use the mathematics frequently needed by graduate students, and whether you can use quantitative reasoning to solve problems. Specialized or advanced mathematical knowledge is not needed to be successful on the Quantitative Reasoning section of the GRE. You will NOT be expected to know advanced statistics, trigonometry, or calculus, and you will not be required to write a proof.
In general, the mathematical knowledge and skills needed to be successful on the GRE do not extend beyond what is usually covered in the average high school mathematics curriculum. The broad areas of mathematical knowledge needed for success are number properties, arithmetic computation, algebra, and geometry.
Number properties include such concepts as even and odd numbers, prime numbers, divisibility, rounding, and signed (positive and negative) numbers.
In arithmetic computation, order of operations, fractions (including computation with fractions), decimals, and averages will be tested. You may also be asked to solve word problems using arithmetic concepts.
The algebra needed on the GRE includes linear equations, operations with algebraic expressions, powers and roots, standard deviation, inequalities, quadratic equations, systems of equations, and radicals. Again, algebra concepts may be part of a word problem you are asked to solve.
In geometry, concepts tested include the properties of points, lines, planes, and polygons. You may be asked to calculate area, perimeter, and volume, or explore coordinate geometry.
You will be expected to recognize standard symbols for mathematical relationships, such as = (equal), ≠ (not equal), < (less than), > (greater than), || (parallel), and ⊥ (perpendicular). All numbers used will be real numbers. Fractions, decimals, and percentages may be used.
When units of measure are used, they may be in English (or customary) or metric units. If you need to convert between units of measure, the conversion relationship will be given, except for common ones such as converting minutes to hours, inches to feet, or centimeters to meters.
GRE word problems usually focus on doing something or deciding something. The mathematics is only a tool to help you get the necessary result. When answering a question on the GRE, you first need to read the question carefully to see what is being asked. Then, recall the mathematical concepts needed to relate the information you are given in a way that will enable you to solve the problem.
If you have completed the average high school mathematics program, you have been taught the mathematics you need for the GRE. The review of arithmetic, algebra, and geometry provided in this book will help you refresh your memory of the mathematical skills and knowledge you previously learned.
If you are not satisfied with your existing mathematics knowledge in a given area, then review the material provided on that topic in more detail, making sure that you fully understand each section before going on to the next one.
# What’s on the GRE Quantitative Reasoning Section?0
The GRE is given as a computer-based test in the United States. (In some other countries, a paper-based version is used.) On the computer-based GRE General Test there are two 35-minute Quantitative Reasoning sections. The test uses a modified computer-adaptive process in which the computer selects the difficulty of your second section based on how you well you scored on the first section. In other words, if you do well on the first section, you will get a harder second section (and a higher score). If you do poorly on the first section, you will get an easier second section (and a lower score). Since you must answer 20 questions in 35 minutes, you need to answer a question approximately every minute and a half. Within a section, you may skip a question and return to it later in order to maximize your efficiency. You need to finish each section in the allotted time. There is an on-screen calculator that you may use to aid in your calculations.
The questions in the Quantitative Reasoning sections assess your ability to solve problems using mathematical and logical reasoning and basic mathematical concepts and skills. The mathematics content on the GRE General Test does not go beyond what is generally taught in high schools. It includes arithmetic, algebra, geometry, and data analysis. The mathematics content, based on GRE sample tests provided by ETS, comes from the following areas:
• Number properties: approximately 22%
• Arithmetic (often graph-related): approximately 18%
• Algebra: approximately 18%
• Plane and solid geometry: approximately 14%
• Probability and statistics: approximately 8%
• Algebra word problems: approximately 6%
• Arithmetic ratios: approximately 6%
• Coordinate geometry: approximately 4%
• Tables: approximately 4%
There is no guarantee that the questions on each Quantitative Reasoning section on a given GRE test will be divided among the content areas according to these exact percentages, but the total Quantitative Reasoning part of the GRE will be spread among the content areas in approximately this way, based on the sample test materials provided by ETS.
In the revised GRE introduced in 2011, two new question formats have been added. These new question types allow some questions to be asked and answered in more natural and complex ways than the older formats permitted. The types of questions in the Quantitative Reasoning sections of the GRE General Test may now include the following:
• Quantitative Comparisons
• Multiple Choice
• Numeric Entry
• Multiple Response
Quantitative Comparison questions present two mathematical quantities. You must determine whether the first quantity is larger, the second is larger, the two quantities are equal, or if it is impossible to determine the relationship based on the given information.
Multiple Choice questions are questions for which you are to select a single answer from a list of choices. These are the traditional multiple choice questions with five possible answers that most test-takers will be familiar with from other standardized examinations.
In Numeric Entry questions, you are asked to type in the answer to the problem from the keyboard, rather than choosing from answers provided to you. For example, if the answer to the question is 8.2, you click on the answer box and then type in the number 8.2.
Multiple Response questions are similar to multiple choice questions, but you may select more than one of the five choices, if appropriate.
# Top 25 GRE Math Concepts You Absolutely Need to Know0
We know. The GRE Quantitative Reasoning section is hard. But to save you some time and energy, we put together this list of top 25 GRE math concepts you need to know off the top of your head:
1. The area of a circle is A=r2, where r is the radius of the circle.
1. The circumference of a circle is C = 2πr, where r is the radius of the circle. The circumference can also be expressed as πd, because the diameter is always twice the radius.
1. The area of a rectangle is A = lw, where l is the length of the rectangle and w is the width of the rectangle.
1. The area of a triangle is A = ½bh, where b is the base of the triangle and h is the height of the triangle.
1. The volume of a rectangular prism is V = lwh, where l is the length of the rectangular prism, w is the width of the rectangular prism, and h is height of the rectangular prism.
1. The volume of a cylinder is V=r2h, where r is the radius of one of the bases of the cylinder and h is the height of the cylinder.
1. The perimeter is the distance around any object.
1. The Pythagorean Theorem states that c2=a2+b2, where c is the hypotenuse of the triangle and a and b are two sides of the triangle.
1. The following are angle measures and side lengths for Special Right Triangles:
1. In an equilateral triangle, all three sides have the same length, and each of the angles equals 60°.
1. In an isosceles triangle, two sides have the same length, and the angles opposite those sides are congruent.
1. The complete arc of a circle has 360°.
1. A straight line has 180°.
1. A prime number is any number that can only be divided by itself and 1.
1. Squaring a negative number yields a positive result.
1. To change any fraction to a decimal, divide the numerator by the denominator.
1. If two numbers have one or more divisors in common, those are the common factors of the numbers.
1. To calculate the mean, or average, of a list of values, divide the sum of the values by the number of values in the list.
1. The median is the middle value of a list of numbers that is in either ascending or descending order.
1. The mode is the value that appears the greatest number of times in a list.
1. A ratio expresses a mathematical comparison between two quantities. (1/4 or 1:4)
1. A proportion is an equation involving two ratios. (1/4 = x/8 or 1:4 = x:8)
1. When multiplying exponential expressions with the same base, add the exponents.
1. When dividing exponential expressions with the same base, subtract the exponents.
1. When raising one power to another power, multiply the exponents.
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May 20, 2024
A step-by-step guide to dividing decimals by decimals. Learn how to divide, common mistakes to avoid, practical real-life examples, tips and tricks, rounding, misconceptions, and fun practice exercises. Master the art of dividing decimals by decimals today!
## I. Introduction
Dividing decimals by decimals can be a difficult concept for students to master. However, it is an important skill to learn as it is frequently used in real-life situations such as calculating prices at the grocery store or determining rates of change. In this article, we will explore the step-by-step process for dividing decimals by decimals, common mistakes to avoid, practical real-life examples, tips and tricks for easier division, the importance of rounding, common misconceptions, and fun practice exercises to improve skills.
## II. Step-by-Step Guide to Dividing Decimals by Decimals
The process of dividing decimals by decimals is similar to dividing whole numbers, but with the added step of moving the decimal point. To divide decimals by decimals:
1. Write the division problem as a fraction
2. Multiply the numerator and denominator by a power of 10 to make the divisor a whole number
3. Multiply both the numerator and denominator by the same power of 10 again to keep the value the same
4. Divide the resulting numerator by the resulting denominator
5. Place the decimal point in the quotient directly above the decimal point in the dividend
For example, let’s divide 2.5 by 0.5:
1. Write the problem as a fraction: 2.5/0.5
2. Multiply both the numerator and denominator by 10: 25/5
3. Multiply both the numerator and denominator by 10 again: 250/50
4. Divide the numerator by the denominator: 5
5. Place the decimal point in the quotient: 5.0
To check your answer, multiply the quotient by the divisor. In this example, 5.0 x 0.5 = 2.5, which is the original dividend, so the answer is correct.
Some tips to make the division process easier are:
• Clear any extra zeros before and after the numbers in the problem
• Align the decimal points in the division problem when setting it up
• Keep track of the number of decimal places in the divisor and quotient
## III. Top Common Mistakes to Avoid When Dividing Decimals by Decimals
One common mistake students make when dividing decimals by decimals is forgetting to move the decimal point when setting up the problem. This can lead to an incorrect answer. Another mistake is mixing up the divisor and dividend, which can also result in an incorrect answer.
To avoid these common mistakes, students should:
• Double check the placement of the decimal point
• Label the dividend and divisor correctly
• Use scratch paper to keep track of the decimal point when moving it
Let’s look at an example problem of dividing 14.3 by 0.4:
1. Write the problem as a fraction: 14.3/0.4
2. Multiply both the numerator and denominator by 10 to make the divisor a whole number: 143/4
3. Multiply both the numerator and denominator by 10 again: 1430/40
4. Divide the numerator by the denominator: 35.75
5. Place the decimal point in the quotient: 35.75
If the decimal point in the quotient were placed one place to the left, the answer would be 357.5, which is incorrect.
## IV. Practical Real-Life Examples of Dividing Decimals by Decimals
Decimals are frequently used in real-life situations such as determining the price per unit of an item or calculating the miles per gallon of a vehicle. Let’s look at some practical real-life examples involving division of decimals by decimals:
Example 1:
A 25-pound turkey costs \$37.50. What is the cost per pound?
To solve this problem, divide the cost by the weight:
1. Write the problem as a fraction: \$37.50/25 lbs
2. Divide the numerator by the denominator: \$1.50/lb
The cost per pound of the turkey is \$1.50.
Example 2:
A car travels 240 miles on 8 gallons of gasoline. What is the miles per gallon?
To solve this problem, divide the distance by the amount of fuel:
1. Write the problem as a fraction: 240 miles/8 gallons
2. Divide the numerator by the denominator: 30 miles/gallon
The car travels 30 miles on one gallon of gasoline.
## V. Tips and Tricks for Faster and Easier Division of Decimals by Decimals
There are a few tips and tricks that can make dividing decimals by decimals faster and easier. One such trick is to move the decimal point in the divisor to the right until it becomes a whole number. Then, move the decimal point in the dividend the same number of places to the right. Divide and move the decimal point in the quotient the same number of places to the left.
Let’s look at an example problem of dividing 0.06 by 0.2:
1. Move the decimal point in the divisor one place to the right to make it a whole number: 0.2 becomes 2
2. Move the decimal point in the dividend one place to the right: 0.06 becomes 0.6
3. Divide 0.6 by 2: 0.3
4. Move the decimal point in the quotient one place to the left: 0.03
The answer is 0.03.
## VI. The Importance of Rounding When Dividing Decimals by Decimals
Rounding is important when dividing decimals by decimals as it can help simplify the problem and provide a more accurate estimate. When rounding, it is important to keep track of the number of decimal places in the divisor and dividend.
Let’s look at an example problem of dividing 8.563 by 3.72:
1. Round 3.72 to 4
2. Divide 8.563 by 4: 2.14
3. Round 2.14 to the nearest hundredth: 2.14
The answer is approximately 2.14.
## VII. Common Misconceptions About Dividing Decimals by Decimals
One common misconception about dividing decimals by decimals is that it is always necessary to move the decimal point. While it is necessary in most cases, there are some problems where it is not needed.
Another misconception is that the quotient will always have the same number of decimal places as the dividend. In reality, the number of decimal places in the quotient will vary based on the values of the dividend and divisor.
## VIII. Fun Practice Exercises for Mastering Division of Decimals by Decimals
Practicing division of decimals by decimals can be both fun and engaging. Some exercises to try include:
• Divide the distance of a road trip by the number of gallons of gasoline used
• Calculate the cost per ounce of a food item
• Find the time it takes to fill a swimming pool based on gallons per minute
When checking the answers, make sure to multiply the quotient by the divisor to make sure the result is correct.
## IX. Conclusion
Dividing decimals by decimals can be a daunting task, but with the step-by-step process outlined in this article, common mistakes to avoid, practical real-life examples, tips and tricks, and fun practice exercises, anyone can master this skill.
|
Using physics, you can calculate what happens when you swerve. For example, you may be in a car or on a walk when you suddenly accelerate in a particular direction. In this case, just like displacement and velocity, acceleration, a, is a vector.
Assume that you’ve just managed to hit a groundball in a softball game and you’re running to first base. You figure you need the y component of your velocity to be at least 25.0 feet/second and that you can swerve at 90 degrees to your present path with an acceleration of 60.0 feet/second2 in an attempt to dodge the first baseman. Is that acceleration going to be enough to change your velocity to what you need it to be in the tenth of a second that you have before the first baseman touches you with the ball? Sure, you’re up to the challenge!
You can find your change in velocity with the following equation:
Now you can calculate the change in your velocity from your original velocity, as this figure shows.
You can use acceleration and change in time to find a change in velocity.
Finding your new velocity, vf, becomes an issue of vector addition. That means you have to break your original velocity, vi, and your change in velocity,
into components. Here’s what vi equals:
You’re halfway there. Now, what about
the change in your velocity? You know that
and that a = 60.0 feet/second2 at 90 degrees to your present path, as the above figure shows. You can find the magnitude of
because
But what about the angle of
If you look at the figure, you can see that
is at an angle of 90 degrees to your present path, which is itself at an angle of 45 degrees from the positive x-axis; therefore,
is at a total angle of 135 degrees with respect to the positive x-axis. Putting that all together means that you can resolve
into its components:
You now have all you need to perform the vector addition to find your final velocity:
You’ve done it: vf = (17.0, 25.4). The y component of your final velocity is more than you need, which is 25.0 feet/second. Having completed your calculation, you put your calculator away and swerve as planned. And to everyone’s amazement, it works — you evade the startled first baseman and make it to first base safely without going out of the baseline (some tight swerving on your part!). The crowd roars, and you tip your helmet, knowing that it’s all due to your superior knowledge of physics. After the roar dies down, you take a shrewd look at second base. Can you steal it at the next pitch? It’s time to calculate the vectors, so you get out your calculator again (not as pleasing to the crowd).
Notice that total displacement is a combination of where your initial velocity takes you in the given time, added to the displacement you get from constant acceleration.
|
PUMPA - SMART LEARNING
எங்கள் ஆசிரியர்களுடன் 1-ஆன்-1 ஆலோசனை நேரத்தைப் பெறுங்கள். டாப்பர் ஆவதற்கு நாங்கள் பயிற்சி அளிப்போம்
Book Free Demo
Identity $$2$$: ${\left(a+b\right)}^{2}={a}^{2}+\mathit{2ab}+{b}^{2}$
Let us construct a figure of four regions. The two square shaped regions with the dimensions of $$3 × 3$$ (Blue) and $$2 × 2$$ (Yellow). Observe the remaining two rectangle shaped regions. Both are in $$3 × 2$$ (Green) dimension.
By observing the above rectangle, we can notice that:
$$\text{Area of the bigger square = Area of the two small square + Area of the two rectangles}$$
$$3 + 2$$$$2 + 3$$ $$=$$ $$(3 × 3) + (2 × 3) + (3 × 3) + (2 × 2)$$
Now, we simplify the LHS and RHS of the above expression.
LHS $$=$$ $$3 + 2$$$$2 + 3$$ $$=$$ $$5×5 = 25$$
RHS $$=$$ $$(3 × 3) + (2 × 3) + (3 × 3) + (2 × 2)$$
$$=$$ ${3}^{2}+\left(2×3\right)+\left(3×2\right)+{2}^{2}$ $$=$$ $$9+6+6+4 = 25$$
Therefore, LHS $$=$$ RHS
Similarly, if we use the variables in this case instead of number we get:
Assume the square of ABCD of side $$a + b$$. From the above figure, we can get that:
$$\text{The total area of the bigger square = The area of the two small squares × The are of the two rectangles}$$
That is, $$(a + b)^2$$$={a}^{2}+\mathit{ab}+\mathit{ba}+{b}^{2}$
Since, $$ba=ab$$; $$(a + b)^2$$ $={a}^{2}+\mathit{ab}+\mathit{ab}+{b}^{2}={a}^{2}+2\mathit{ab}+{b}^{2}$.
Therefore, ${\left(a+b\right)}^{2}={a}^{2}+\mathit{2ab}+{b}^{2}$ is a identity.
Example:
Simplify the expression $\left(a+6\right)\left(a+6\right)$ using the identity ${\left(a+b\right)}^{2}={a}^{2}+\mathit{2ab}+{b}^{2}$.
Now write the given expression $\left(a+6\right)\left(a+6\right)$ with respect to the given identity ${\left(a+b\right)}^{2}={a}^{2}+\mathit{2ab}+{b}^{2}$.
$\begin{array}{l}\left(a+b\right)\left(a+b\right)=\left(a+6\right)\left(a+6\right)\\ \\ {\left(a+6\right)}^{2}=\left({a}^{2}+2\left(6×6\right)a+{6}^{2}\right)\end{array}$
Now simplify the expression.
$=\left({a}^{2}+72a+36\right)$.
Therefore, $\left(a+6\right)\left(a+6\right)$$=\left({a}^{2}+72a+36\right)$.
|
# Math Snap
## A cylindrical steam boiler is to be constructed having a capacity of 1000 cubic $\mathrm{m}$. The material for the side costs Php 20 per sq. meter and for the ends Php 30 per sq. meter. Find the radius when teh cost is least.
#### STEP 1
Assumptions 1. The boiler is cylindrical with a volume of 1000 cubic meters. 2. The cost of material for the cylindrical side is Php 20 per square meter. 3. The cost of material for the ends (circular bases) of the cylinder is Php 30 per square meter. 4. The goal is to minimize the total cost of the material. 5. The formula for the volume of a cylinder is $V = \pi r^2 h$, where $r$ is the radius and $h$ is the height. 6. The formula for the surface area of the sides of the cylinder is $A_{side} = 2\pi rh$. 7. The formula for the surface area of the ends of the cylinder is $A_{ends} = 2\pi r^2$.
#### STEP 2
Express the height $h$ in terms of the volume $V$ and radius $r$. $h = \frac{V}{\pi r^2}$
#### STEP 3
Substitute $h$ in the surface area formulas. For the cylindrical side: $A_{side} = 2\pi rh = 2\pi r \left(\frac{V}{\pi r^2}\right) = \frac{2V}{r}$ For the ends: $A_{ends} = 2\pi r^2$
#### STEP 4
Calculate the cost of the material for the cylindrical side and the ends. Cost of the side: $C_{side} = 20 \times A_{side} = 20 \times \frac{2V}{r} = \frac{40V}{r}$ Cost of the ends: $C_{ends} = 30 \times A_{ends} = 30 \times 2\pi r^2 = 60\pi r^2$
#### STEP 5
Write the total cost $C$ as a function of $r$. $C(r) = C_{side} + C_{ends} = \frac{40V}{r} + 60\pi r^2$
#### STEP 6
Differentiate $C(r)$ with respect to $r$ to find the critical points. $C'(r) = -\frac{40V}{r^2} + 120\pi r$
#### STEP 7
Set $C'(r) = 0$ to solve for $r$. $-\frac{40V}{r^2} + 120\pi r = 0$ $120\pi r^3 = 40V$ $r^3 = \frac{40V}{120\pi}$ $r^3 = \frac{V}{3\pi}$
#### STEP 8
Substitute $V = 1000$ cubic meters into the equation. $r^3 = \frac{1000}{3\pi}$ $r = \left(\frac{1000}{3\pi}\right)^{\frac{1}{3}}$
##### SOLUTION
Calculate $r$. $r = \left(\frac{1000}{3\pi}\right)^{\frac{1}{3}} \approx 6.8 \text{ meters}$ The radius that minimizes the cost is approximately 6.8 meters.
|
# Pascal s Law & Surface Area of a Piston. Lessons 2 and 3
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1 Pascal s Law & Surface Area of a Piston Lessons 2 and 3 Remember: Pretty Please My Dear Aunt Sally (rom left to right; Parentheses; Power; Multiply; Divide; Add, Subtract) We have learned how to measure a brake drum to determine if we can resurface it, there charging only a labor fee, or, if the drum is worn or scored beyond it s discard limit. Today, we re going to discuss how a 100-pound person can stop a 4000-pound car. Or how a 1 servo piston inside an automatic transmission can increase fluid pressure enough to stop an input drum connected to the torque converter from spinning. By using the brake system s ability to multiply force, we can use a math formula to calculate the applied force on the brake pad pressing on the rotor. You can also calculate the surface area of a piston using the formula for finding the area of a circle. Some of the math and math terms we ll be using to today are: Algebraic Expressions - variables and numbers that can be combined with the operations of arithmetic. ormulas - Equations with several variables. Variables - letters or symbols that can represent real numbers Piston - container that transfers pressure into motion. Pascal - scientist who worked with atmospheric pressure. Pressure decreases with height and a vacuum existed above the atmosphere. The formula we ll be using to determine Pressure, orce and Area: P A = = A = P * A P Area = r 2 (r = radius) Diameter = 2r Pressure - (pounds per square inch [psi]) A = area (square inches) = force (pounds) Surface area - the area of the face of the object. In this instance, surface area would be the face of the top of the piston, (Pi) = 3.14 X r X r = area of a circle 1
2 (Pi) approx. Radius - the distance from the center of the circle to any point on the circle. (Half of the diameter) Diameter - the line that passes through the center point and ends at two points on the circumference. Center - point in a circle that is equidistant from the all points around the circumference of the circle. Define Pascal s Law. Use the formulas to calculate pressures. Use the formula to calculate surface area. Area = r2 (r = radius), Diameter = 2r Example: 1. If a force of 100 pounds is applied to a piston in a hydraulic cylinder of 4 square inches in area, what is the pressure? P = A 100 pounds P = 2 4in pounds P = 25psi( ) 2 in 2. If 80 pounds per square inch of pressure is applied to a 2 square inch area. What is the total force? 3. If total force is 120 pounds per square inch and pressure is 30 pounds per square inch, what is the area? 4. If we have 50 pounds of force applied to an area of.5 square inches, what would be the pressure? 2
3 5. A master cylinder piston has 100 pounds per square inch of force applied to a 1.5 square inch piston. What is the pressure? 5A. If the force stays the same, but the diameter of the piston is increased to 4.50 inches, what is the pressure? 5B. What is the surface area of the 4.50 diameter piston in square inches? 5C. How much how force is being applied to the 4.50 piston if the pressure is 1000 psi? 6. If the diameter of a wheel cylinder cup is 23/32 inches, determine the force applied to the piston if 1700psi is produced. Remember to convert! 7. ind the area of a piston with a 7-inch diameter. 8. ind the area of a circle with a 20 mm (millimeters) diameter. 9. ind the diameter of a piston with an area of square inches. 10. ind the radius of a circle with an area of 25 square centimeters. 3
4 Larry s Truck Worksheet Name: AM-1: PM Date: This can be a team project. You ve just seen that Larry put a 6 lift kit on his ord -150 and a set of 33 s Super Swamper Thornbird Tires. The Suspension System has been modified to handle the oversized tires, however, Larry cheaped out and didn t modify the Brake System. Larry now has a problem stopping his Big oot whannabe. Congratulations! You ve just been promoted to head braking system engineer. Based on our lessons on Pascal s Law and your knowledge of automotive braking systems, you ve been assigned the task of developing a plan to fix Larry s truck. Remember: Larry doubled the size of the original tires. The truck s stock brake system parts sizes are: 1/4 in brake lines; 1/2 in bore master cylinder; 7/8 in bore wheel cylinders and 2 in caliper pistons. Your task: Now, you must: 1. Determine what parts should be replaced on Larry s truck that will make it stop better. 2. We know it is possible to replace all the parts that could affect the trucks stopping ability, but is it physically possible to bolt them to the truck? or example, can we purchase a replacement master cylinder that has a different size bore and is the same physical size as the stock master cylinder that came standard on the truck, and still be bolted to the truck without modifications? 3. List the parts you would replace, and why. You must show all calculations and justify your solution in a word essay. NOTE: There is no one right answer. The written paper will account for one of your required marking period essays. If this was a team effort, each team member must turn in a separate essay on his or her portion of the project. Good luck and have fun! 4
5 North Montco Technical Career Center Math-In-CTE Lessons 2 and 3 Worksheet Pascal s Law Name: AM-1: PM Date: Work through traditional math examples. 1. ind the time traveled for a distance of 150 miles at a rate of 55miles per hour. Remember: T = d/r 2. ind the diameter of a circle that has a circumference of 36 in. Remember: C =πd 3. If a force of 120 pounds is applied to a piston in a hydraulic cylinder of 8 square inches in area, what is the pressure? 4. If 95 pounds per square-inch of pressure is applied to a 3 square in area, what is the total force? 5. Solve for the value of x. 15 =x /3 6. ind the value of r when 3 =57/r 7. ind the value of R in terms of E and I from the formula, E =IR 5
6 North Montco Technical Career Center Math-In-CTE Lessons 2 and 3 Homework Pascal s Law Name: AM-1: PM Date: 1. If we have 50 pounds of force applied to an area of.75 square inches, what would be the pressure? 2. If total force is 110 pounds per square inch and pressure is 45 pounds per square inch, what is the area? 3. A master cylinder has 200 pounds per square inch of force applied to a 2 square inch piston. If the diameter of the caliper piston is 4 inches, find the area of the caliper piston in square inches. Determine the force of the caliper piston. 4. ind the area of a piston with a 5-inch diameter. 5. ind the radius of a circle with an area of 18 square centimeters. 6. ind the diameter of a circle with an area of 25 square centimeters. 7. ind the average rate traveled when the distance from point A to point B was 92 miles and the time of travel was 2 hours and 15 minutes. 8. ind the value of x when 36 =4 /x 2 6
### Cylinder Balance and Percent Changes Lesson 11
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# Factors of 701: Prime Factorization, Methods, and Examples
In division, if the resultant is not a whole number we cannot say it is the factor of that number that is used as a divisor or in the numerator like in this solution 701.
Only 1 and 701 satisfy the factor of 701.
### Factors of 701
Here are the factors of number 701.
Factors of 701: 1 and 701
### Negative Factors of 701
The negative factors of 701 are similar to its positive aspects, just with a negative sign.
Negative Factors of 701: -1 and -701
### Prime Factorization of 701
The prime factorization of 701 is the way of expressing its prime factors in the product form.
Prime Factorization: 1 x 701
In this article, we will learn about the factors of 701 and how to find them using various techniques such as upside-down division, prime factorization, and factor tree.
## What Are the Factors of 701?
The factors of 701 are 1 and 701. These numbers are the factors as they do not leave any remainder when divided by 701.
The factors of 701 are classified as prime numbers and composite numbers. The prime factors of the number 701 can be determined using the prime factorization technique.
## How To Find the Factors of 701?
You can find the factors of 701 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 701, create a list containing the numbers that are exactly divisible by 701 with zero remainders. One important thing to note is that 1 and 701 are the 701’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 701 are determined as follows:
$\dfrac{701}{1} =701$
$\dfrac{701}{701} = 1$
Therefore, 1 and 701 are the factors of 701.
### Total Number of Factors of 701
For 701, there are two positive factors and two negative ones. So in total, there are four factors of 701.
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 701 is given as:
Factorization of 701 is 1 x 701.
The exponent of 1 and 701 is 1.
Adding 1 to each and multiplying them together results in two.
Therefore, the total number of factors of 701 is four whereas two are positive, and two 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 701 by Prime Factorization
The number 701 is a prime 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 701 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 701, 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 701 can be expressed as:
701 = 1 x 701
## Factors of 701 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 701, the factor pairs can be found as:
1 x 701 = 701
The possible factor pairs of 701 are given as (1, 701).
All these numbers in pairs, when multiplied, give 701 as the product.
The negative factor pairs of 701 are given as:
-1 x -701 = 701
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 and -701 are called negative factors of 701.
The list of all the factors of 701, including positive as well as negative numbers, is given below.
Factor list of 701: 1, -1, 701, and -701
## Factors of 701 Solved Examples
To better understand the concept of factors, let’s solve some examples.
### Example 1
How many factors of 701 are there?
### Solution
The total number of Factors of 701 is two.
Factors of 701 are 1 and 701.
### Example 2
Find the factors of 701 using prime factorization.
### Solution
The prime factorization of 701 is given as:
701 $\div$ 1 = 701
701 $\div$ 701 = 1
So the prime factorization of 701 can be written as:
1 x 701 = 701
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# How do you solve Log(x+10) - log(x+4) = log x?
Dec 9, 2015
Apply properties of logarithms and solve the resulting quadratic equation to find
$x = 2$ or $x = - 5$
#### Explanation:
We will use the following properties:
• $\log \left(x\right) - \log \left(y\right) = \log \left(\frac{x}{y}\right)$
• ${e}^{\log} \left(x\right) = x$
$\log \left(x + 10\right) - \log \left(x + 4\right) = \log \left(x\right)$
$\implies \log \left(\frac{x + 10}{x + 4}\right) = \log \left(x\right)$
$\implies {e}^{\log} \left(\frac{x + 10}{x + 4}\right) = {e}^{\log} \left(x\right)$
$\implies \frac{x + 10}{x + 4} = x$
$\implies x + 10 = x \left(x + 4\right) = {x}^{2} + 4 x$
$\implies {x}^{2} + 3 x - 10 = 0$
$\implies \left(x + 5\right) \left(x - 2\right) = 0$
$\implies x = 2$ or $x = - 5$
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# Solve the following system of equations: $x\ +\ 2y\ =\ \frac{3}{2}$ $2x\ +\ y\ =\ \frac{3}{2}$
Given:
The given system of equations is:
$x\ +\ 2y\ =\ \frac{3}{2}$
$2x\ +\ y\ =\ \frac{3}{2}$
To do:
We have to solve the given system of equations.
Solution:
The given system of equations can be written as,
$x+2y=\frac{3}{2}$
$\Rightarrow 2(x+2y)=3$ (On cross multiplication)
$\Rightarrow 2x+4y=3$---(i)
$2x+y=\frac{3}{2}$
$\Rightarrow 2(2x+y)=3$ (On cross multiplication)
$\Rightarrow 4x+2y=3$
$\Rightarrow x=\frac{3-2y}{4}$----(ii)
Substitute $x=\frac{3-2y}{4}$ in equation (i), we get,
$2(\frac{3-2y}{4})+4y=3$
$\frac{3-2y}{2}+4y=3$
Multiplying by $2$ on both sides, we get,
$2(\frac{3-2y}{2})+2(4y)=2(3)$
$3-2y+8y=6$
$6y=6-3$
$y=\frac{3}{6}$
$y=\frac{1}{2}$
Substituting the value of $y=\frac{1}{2}$ in equation (ii), we get,
$x=\frac{3-2(\frac{1}{2})}{4}$
$x=\frac{3-1}{4}$
$x=\frac{2}{4}$
$x=\frac{1}{2}$
Therefore, the solution of the given system of equations is $x=\frac{1}{2}$ and $y=\frac{1}{2}$.
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Updated on: 10-Oct-2022
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|
# 4th Class Mental Ability Algebra
Algebra
Category : 4th Class
Algebra
Learning Objectives
• Fraction
• Unitary Method
Fraction
Fraction represents a part of whole or any number of equal parts. A line separates the two terms. Number above the line is called numerator and the number below the line is called denominator. Denominator of a fraction tells us how many parts the whole is divided into and the numerator tells us how many parts are taken out the whole. Let 4 kg of flour is divided into five equal parts. The amount of each part is represented as $\frac{4}{5}$ kg. Here, the number four-fifth is known as fractional number and its symbol $\frac{4}{5}$ is called fraction.
Understanding Fractions
Fractions are formed when we have a whole that is divided into so many equal parts. The shaded portion of a figure can also be represented by a fraction as shown in the table given below.
Shaded Part Fraction Fractional Number Figure 1 part out of 6 equal parts $\frac{1}{4}$ One-fourth 3 part out of 6 equal parts $\frac{3}{6}$ Three-sixth 5 part out of 8 equal parts $\frac{5}{8}$ Five-eighth 2 part out of 6 equal parts $\frac{2}{6}$ Two-sixth 4 parts out of 7 equal parts $\frac{4}{7}$ Four-seventh 5 part out of 6 equal parts $\frac{5}{6}$ Five-sixth
Types of Fractions
There are several types of fractions. Let us study the fractions and their types.
Proper and Improper Fractions
A proper fraction is one whose numerator is smaller than the denominator. For example, $\frac{3}{4},\frac{17}{21},\frac{27}{29}$ are proper fractions. An improper fraction is one whose numerator is bigger than the denominator. For example, $\frac{5}{4},\frac{8}{5},\frac{23}{21}$ are improper fractions.
Mixed Fractions
A mixed fraction is the ratio of a whole number and a fraction combined into one mixed number.
For example, $1\frac{3}{4},2\frac{7}{9},3\frac{4}{5}$ are mixed fractions.
Equivalent Fractions
Fractions which have the same value though they may look different are called equivalent fractions. For example, $\frac{1}{3},\frac{2}{6},\frac{3}{9},\frac{4}{12}$ are equivalent fractions because all are one-third.
• To find equivalent fractions with terms bigger than the given fraction, we simply multiply the numerator and denominator of the given fraction by a common number.
• To find equivalent fractions with terms smaller than the given fraction, we simply divide the numerator and denominator of the given fraction by a common number.
Decimals
A decimal or decimal number is a fraction written in special form. A decimal is a fraction which has 10, 100, 1000, etc. for its denominator and it is expressed in the decimal system of notation. For example, $\frac{9}{10}$ and $1\frac{9}{100}$ can be written in decimal form as 0.9 and 1.09, respectively. A decimal number is broadly divided into two parts: Whole number part and Decimal part. The two parts are separated by a dot (.) called the decimal point. From the decimal point, as we move on the left, the place value is multiplied by 10 and as we move on the right, it is divided by 10.
Decimal Fraction Whole Part Decimal Part 5.3 5 3 147.81 147 81 15.679 15 679 0.8 0 8 1.004 1 004
Decimal Place Value Chart
The following chart shows the value of each place in a decimal number.
Hundreds Tens Ones Decimal Point Tenths Hundredths Thousandths Ten Thousandths 100 10 1 . $\frac{1}{10}$ $\frac{1}{100}$ $\frac{1}{1000}$ $\frac{1}{10000}$
Here, the values of left of the decimal point are whole numbers and numbers written after the decimal point or right of the decimal point is always less than one. Putting a zero before the decimal point indicates that there are no whole numbers.
What is Unitary Method?
The term ‘unitary method’ is a technique which has evolved from the concept of 'unit' which means one. Therefore, the unitary method is also known as the method of ones.
In unitary method:
• To get more value we multiply.
• To get less value we divide.
Example:
1. If the price of one bicycle is Rs.1200, then what is the price of such 5 bicycles?
Solution: Here, price of 1 bicycle is given and we have to find the price of 5 such bicycles. Therefore, we multiply the price of 1 bicycle by 5.
Hence, price of 5 bicycles = Rs. ($1200\times 5$) = Rs.6000
2. If the price of 8 bicycles is Rs. 10200, then what is the price of 1 such bicycle?
Solution: Here, price of 8 bicycles is given and we have to find the price of 1 such bicycle. Therefore, we divide the price of 8 bicycles by 8.
Hence, price of 1 bicycle = Rs. ($10200-8$) = Rs.1275
Rules Applied in Unitary Method
Rule 1: If we know the value of one then to find the value of many, we do multiplication.
Example:
1. If the cost of 1 chair is Rs. 345, then find the cost of 7 such chairs.
Solution: Given that the cost of 1 chair = Rs. 345
Therefore, the cost of 7 chairs = Rs. ($345\times 7$) = Rs.2415
Rule 2: If we know the value of many then to find the value of one, we do division.
2. If the cost of 8 pencils is Rs. 68, then find the cost of 1 such pencil.
Solution: Given that the cost of 8 pencils = Rs.68
Therefore, the cost of 1 pencil = Rs. ($68\div 8$) = Rs.8.50
Rule 3: If we know the value of many then to find the value of more than given or less than given, we first need to calculate the value of one.
3. If the cost of 4 apples is Rs. 38, then find the cost of 6 such apples.
Solution: Given that the cost of 4 apples = Rs.38
Therefore, the cost of 1 apple = Rs. ($38\div 4$) = Rs.9.50
So, the cost of 6 apples =Rs. ($9.50\times 6$) = Rs.57
4. A train is running at a uniform speed. It covers 840 km in 7 hours. How much distance will it cover in 5 hours?
Solution: Clearly, in more hours, the train will cover more distance and in less hours, the train will cover less distance.
Here, distance covered in 7 hours = 840 km
So, the distance covered in 1 hour = ($840\div 7$) km = 120 km
Therefore, the distance covered in 5 hours = ($120\times 5$) km = 600 km
Problems Based on Unitary Method
Some examples of problems based on unitary method are given below. Let us study the following examples.
Example:
1. There are 1250 toffees in a packet. How many toffees are there in 9 such packets?
Solution: Number of toffees in 1 packet = 1250
Therefore, the number of toffees in 9 packets = $1250\times 9$ = 11250
2. If the cost of half dozen bananas is Rs. 15, then find the cost of one banana.
Solution: Given that the cost of half dozen bananas = Rs.15
Or, the cost of 6 bananas = Rs.15
Therefore, the cost of 1 banana = T ($15-6$) = Rs.2.50
3. If 8 litres of petrol is consumed by a car in covering a distance of 324 km, then how many kilometres will it cover in 10 litres of petrol?
Solution: In 8 litres of petrol, car covered the distance = 324 km
Therefore, in 1 litre of petrol, the car covered the distance
= ($324\div 8$) km = 40.5 km
So, in 10 litres of petrol, the car will cover the distance
= ($40.5\times 10$) km = 405 km
1. Which one of the following is a pair of like fractions?
(a) $\frac{3}{17},\frac{17}{3}$
(b) $\frac{12}{18},\frac{16}{10}$
(c) $\frac{4}{17},\frac{14}{17}$
(d) $\frac{14}{21},\frac{14}{22}$
(e) None of these
Answer (c) is correct.
Explanation: Fractions in which the denominators are the same, are called like fractions. Therefore, $\frac{4}{17},\frac{14}{17}$ are like fractions.
2. In the decimal number 24.9785, what is the place value of 8?
(a) Tenths (b) Hundredths
(c) Thousandths (d) Ten thousandths
(e) None of these
Answer (c) is correct.
Explanation: Here, 8 is at third place right from the decimal point.
Therefore, it is at $\frac{1}{1000}$ place in place value chart.
So, the place value of 8 is thousandths.
3. In which one of the following figures, the fraction represented by shaded parts is equal to$\frac{1}{3}$ ?
(a) (b)
(c) (d)
(e) None of these
Answer (d) is correct.
Explanation: Here, we have:
In option (a); fraction represented by the shaded parts = $\frac{3}{5}$
In option (b); fraction represented by the shaded parts = $\frac{3}{4}$
In option (c); fraction represented by the shaded parts = $\frac{2}{7}$
In option (d); fraction represented by the shaded parts = $\frac{2}{6}$ = $\frac{1}{3}$
Therefore, the required figure is given in option (d).
4. The fraction represented by shaded parts in the figure given below is equal to:
(a) 0.2 (b) 0.4
(c) 0.5 (d) 0.6
(e) None of these
Answer (c) is correct.
Explanation: Number of shaded parts = 5
Total number of parts = 10
Fraction represented by shaded parts = $\frac{5}{10}$
Therefore, the required decimal number = $5\div 10$ = 0.5
5. Which one of the following represents the sum of fractions represented by shaded parts in the figures shown below?
(a) $\frac{7}{8}+\frac{7}{8}$
(b) $\frac{7}{8}+\frac{1}{8}$
(c) $\frac{1}{8}+\frac{1}{8}$
(d) $\frac{1}{8}+\frac{1}{4}$
(e) None of these
Answer (b) is correct.
Explanation: Here, we observe that:
Fraction represented by the shaded parts in left side figure = $\frac{7}{8}$
and, fraction represented by the shaded parts in right side figure = $\frac{1}{8}$
Thus, the required sum of fractions = $\frac{7}{8}+\frac{1}{8}$
6. If we arrange the digits of decimal number 35.564 in the decimal place value chart, place value of 4 will be:
(a) thousands (b) tenths
(c) hundredths (d) thousandths
(e) None of these
Answer (d) is correct.
Explanation:
Tens Ones Decimal Point Tenths Hundredths Thousandths 3 5 5 6 4
7. Arrange the decimals given below in descending order:
0.392, 3.092, 0.0392, 3.0092
(a) 0.0392, 0.392, 3.0092, 3.092
(b) 3.092, 0.392, 0.0392, 3.0092
(c) 3.0092, 3.092, 0.392, 0.0392
(d) 3.092, 3.0092, 0.392, 0.0392
(e) None of these
Answer (d) is correct.
Explanation: Here, we have:
$0.392=\frac{392}{1000}=\frac{392\times 10}{1000\times 10}=\frac{3920}{10000}$
$3.092=\frac{3092}{1000}=\frac{3092\times 10}{1000\times 10}=\frac{30920}{10000}$
$0.0392=\frac{392}{10000}$ and $3.0092=\frac{30092}{10000}$
Clearly, 30920 > 30092 > 3920 > 392; or, 3.092 > 3.0092 > 0.392 > 0.0392
Hence, the decimals in descending order are: 3.092, 3.0092, 0.392, 0.0392
8. In the magic square given below, the sum of fractions in each row, each column and each diagonal is same. Find the sum of missing fractions P and Q.
$\frac{4}{13}$ $\frac{3}{13}$ $\frac{8}{13}$ P $\frac{5}{13}$ $\frac{1}{13}$ $\frac{2}{13}$ Q $\frac{6}{13}$
(a) $\frac{6}{13}$ (b) $\frac{17}{13}$
(c) $\frac{19}{13}$ (d) $\frac{21}{13}$
(e) None of these
Answer (a) is correct.
Explanation: Here,
Sum of fractions in one diagonal = $\frac{4}{13}+\frac{5}{13}+\frac{6}{13}=\frac{4+5+6}{13}=\frac{15}{13}$
Hence, P = $\frac{15}{13}-\left( \frac{4}{13}+\frac{2}{13} \right)=\frac{15}{13}-\frac{6}{13}=\frac{15-6}{13}=\frac{9}{13}$
And, Q = $\frac{15}{13}-\left( \frac{3}{13}+\frac{5}{13} \right)=\frac{15}{13}-\frac{8}{13}=\frac{15-8}{13}=\frac{7}{13}$
Also, numerators of all fractions are: 1, 2, 3, 4, 5, 6, 7, 8, 9 where, 7 and 9 are missing. So, the required sum (P + Q.) = $\frac{9}{13}+\frac{7}{13}+\frac{16}{13}$
9. I bought $2\frac{1}{4}$ kg of apples, $1\frac{1}{4}$ kg of oranges and $\frac{1}{2}$ kg of grapes. How much weight did I carry home?
(a) 16 kg (b) 8 kg
(c) 6 kg (d) 4 kg
(e) None of these
Answer (d) is correct.
Explanation: Here,
Sum of all weights = $\left( 2\frac{1}{4}+1\frac{1}{4}+\frac{1}{2} \right)$ kg
= $\left( \frac{2\times 4+1}{4}+\frac{1\times 4+1}{4}+\frac{1\times 2}{2\times 2} \right)$ kg
=$\left( \frac{8+1}{4}+\frac{4+1}{4}+\frac{2}{4} \right)$ kg
= $\left( \frac{9}{4}+\frac{5}{4}+\frac{2}{4} \right)$ kg = $\left( \frac{9+5+2}{4} \right)$ kg
= $\left( \frac{16}{4} \right)$ kg = $\left( \frac{16\div 4}{4\div 4} \right)$ kg = $\frac{4}{1}$ kg = 4 kg
10. Match the following and choose the correct option.
(A) Rs.10 (1) Five 50 paise coins (B) Rs.1 (2) Twenty 50 paise coins (C) Rs.2.50 (3) Nine 50 paise coins (D) Rs.4.50 (4) Two 50 paise coins
(a) A - 4, B - 2, C - 1, D - 3
(b) A - 2, B - 3, C - 4, D - 1
(c) A - 1, B - 3, C - 2, D - 4
(d A - 2, B - 4, C - 1, D - 3
(e) None of these
Answer (d) is correct.
Explanation: (i) Rs. 10 is equal to twenty 50 paise coins.
(ii) Rs. 1 is equal to two 50 paise coins.
(iii) Rs. 2.50 is equal to five 50 paise coins.
(iv) Rs. 4.50 is equal to nine 50 paise coins.
11. Paul has thirty four 50 paise coins. How much amount in rupees does he have?
(a) Rs. 34 (b) Rs. 17
(c) Rs. 68 (d) Rs. 8.50
(e) None of these
Answer (b) is correct.
Explanation: Two 50 paise coins make Rs.1
Therefore, thirty four 50 paise coins make Rs. $\left( \frac{34}{2} \right)$ = Rs.17
12. If the cost of 6 ball pens is Rs. 63, then the cost 2 such ball pens is:
(a) Rs.28 (b) Rs.94.50
(c) Rs.21 (d) Rs.31.50
(e) None of these
Answer (c) is correct.
Explanation: Here, the cost of 6 ball pens = Rs.63
Now, the cost of 1 ball pen = Rs. ($63\div 6$) = Rs.10.50
Hence, the cost of 2 ball pens = Rs. ($10.50\times 2$) = Rs.21
13. How many paise should be added to Rs. 2.10 to make it 385 paise?
(a) 165 paise (b) 170 paise
(c) 185 paise (d) 175 paise
(e) None of these
Answer (d) is correct.
Explanation: As we know, subtraction is the inverse process of addition.
Here, Rs. 2.10 = ($2.10\times 100$) paise = 210 paise
Difference of 385 paise and 210 paise = ($385-210$) paise = 175 paise
Hence, 175 paise should be added to Rs. 2.10 to get 385 paise.
14. If a cow gives 13 litres of milk per day, then how much milk does the cow gives in a week?
(a) 84 litres (b) 77 litres
(c) 104 litres (d) 91 litres
(e) None of these
Answer (d) is correct.
Explanation: As we know, the number of days in a week = 7
Now, in 1 day, the cow gives 13 litres of milk.
So, in 7 days, the cow will give ($13\times 7$) litres = 91 litres of milk.
Hence, the cow gives 91 litres of milk in a week.
15. If the cost of eight baskets of fruits is Rs.2568, then the cost of 1 basket of fruits is:
(a) Rs.321 (b) Rs.214
(c) Rs.341 (d) Rs.361
(e) None of these
Answer (a) is correct.
Explanation: Given that 8 baskets of fruits cost Rs.2568.
Therefore, 1 basket of fruits cost Rs. ($2568\div 8$) = Rs.321
16. A man works for 6 days and earns Rs.5118. How much does he earn in a day?
(a) Rs.848 (b) Rs.798
(c) Rs.853 (d) Rs.913
(e) None of these
Answer (c) is correct.
Explanation: Given that the man works 6 days and earn Rs.5118.
Therefore, the man works 1 day and earns Rs. ($5118\div 6$) = Rs.853
17. Mohit types 900 words in half an hour. How many words would he type in 6 minutes?
(a) 120 words (b) 210 words
(c) 150 words (d) 180 words
(e) None of these
Answer (d) is correct.
Explanation: Since, half an hour = 30 minutes
Now, in 30 minutes, Mohit types 900 words
So, in 1 minute, Mohit types ($900\div 30$) words = 30 words
Therefore, in 6 minutes, Mohit would type = ($30\times 6$) words = 180 words.
18. If the cost of half dozen guavas is Rs.15, then find the cost of two and half dozen guavas.
(a) Rs.120 (b) Rs.90
(c) Rs.75 (d) Rs.60
(e) None of these
Answer (c) is correct.
Explanation: Since, half dozen = 6. Now, the cost of 6 guavas = Rs.15
Therefore, the cost of 1 guava = Rs. ($15-6$) = Rs.2.50
Now, two and half dozen = $12+12+6=30$
Hence, the cost of 30 guavas = Rs. ($2.50\times 30$) = Rs.75
19. A local farmer sells onions in 5 kg bags at Rs.87.50 per bag and 9 kg bags at Rs.148.50 per bag. Is there any monetary advantage gained if you buy the 9 kg bag?
(a) Maybe (b) Yes
(c) No (d) Can't say
(e) None of these
Answer (b) is correct.
Explanation: Here, if you buy 5 kg bag, the cost of 1 kg of onions = Rs. ($87.50\div 5$) = Rs.17.50
And, if you buy 9 kg bag, the cost of 1 kg of onions = Rs. ($148.50\div 9$) = Rs.16.50
Hence, 9 kg bag is comparatively cheaper than that of 5 kg bag.
So, yes there is a monetary advantage gain in buying the 9 kg bag.
20. I can buy 6 burgers for Rs.90. How much money would I need if I wanted to buy 90 such burgers?
(a) Rs.1350 (b) Rs.1290
(c) Rs.1250 (d) Rs.1130
(e) None of these
Answer (a) is correct.
Explanation: Given that the cost of 6 burgers = Rs.90
Therefore, the cost of 1 burger = Rs. ($90\div 6$) = Rs.15
Hence, the cost of 90 burgers =Rs. ($15\times 90$) =Rs.135
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# CBSE Class 12th Math 2 – Inverse Trigonometric Functions MCQs
#### If cosec-1x = y, for a ≤ y ≤ b and y ≠ {0}, the values of a and b are _______ and ________.
Correct! Wrong!
Given that cosec-1x = y then a ≤ y ≤ b and y ≠ {0} ---------- (1) and the principal value of cosec-1 (x) is –π/2, π/2 – {0} ---------- (2) By comparing equations (1) and (2), a = -π/2 b = π/2 Hence, if cosec-1x = y, for a ≤ y ≤ b and y ≠ {0}, the values of a and b are –π/2 and π/2.
#### The principal value along with range of cot-1x is ___________.
Correct! Wrong!
The domain of the cot function (cotangent function) is the set {x: x ∈ R and x ≠ nΠ, n ∈ Z} and range is R. It means that cotangent function is not defined for integral multiples of Π. If we restrict the domain of cotangent function to (0, Π), then it is bijective with its range as R. In fact, cotangent function restricted to any of the intervals (–Π, 0), (0, Π), (Π, 2Π) etc. is bijective and its range is R. Thus, cot-1can be defined as a function whose domain is R and range as any of the intervals (–Π, 0), (0, Π), (Π, 2Π), etc. These intervals give different branches of the cot-1function. The function with range (0, Π) is called the principal value branch of the function cot-1. We thus have cot-1: R → (0, Π) Hence, the principal value along with range of cot-1x is (0, Π).
#### The principal value of sin-1 (-1) is _____________.
Correct! Wrong!
Let sin-1 (-1) = y. Then, sin y = -1 = -sin(π/2) = sin(-π/2) We know that the range of the principal value branch of sin-1is[ –π/2, π/2] and sin(-π/2) = -1 Hence, the principal value of sin-1 (-1) is –π/2.
#### The principal value of sec-1 (2) is _________.
Correct! Wrong!
Let sec-1 (2) = y. Then, sec y = 2 = sec (π/3) We know that the range of the principal value branch of sin-1is [0, Π] – {π/2} and sec(π/3) = 2. Therefore, principal value of sec-1 (2) is π/3.
#### The principal value of sin-1 (0) is ____________.
Correct! Wrong!
Let sin-1 (0) = y. Then, sin y = 0 We know that the range of the principal value branch of sin-1is –π/2, π/2 and sin 0 = 0 and hence, sin y = sin 0 ⇒ y = 0 ⇒ sin-1 (0) = 0 Therefore, principal value of sin-1 (0) is 0.
Date
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CBSE Class 12th Math 2 - Inverse Trigonometric Functions MCQs
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# How do you construct polynomial equations with the given roots?
## 1. $2$, $4$ and $- 7$. 2. $5$ and $\sqrt{3}$.
Aug 11, 2017
1. ${x}^{3} + {x}^{2} - 34 x + 56 = 0$
2. ${x}^{3} - 5 {x}^{2} - 3 x + 15 = 0$
#### Explanation:
Note that if a polynomial in $x$ has a zero $a$ then it has a factor $\left(x - a\right)$ and vice versa.
$\textcolor{w h i t e}{}$
For question 1 we can construct a polynomial:
$f \left(x\right) = \left(x - 2\right) \left(x - 4\right) \left(x + 7\right) = {x}^{3} + {x}^{2} - 34 x + 56$
Any polynomial with these zeros will be a multiple (scalar or polynomial) of this $f \left(x\right)$.
So the polynomial equation:
${x}^{3} + {x}^{2} - 34 x + 56 = 0$
has roots $2$, $4$ and $- 7$.
$\textcolor{w h i t e}{}$
For question 2 we can multiply out $\left(x - 5\right) \left(x - \sqrt{3}\right)$ but this will result in a polynomial with irrational coefficients:
$\left(x - 5\right) \left(x - \sqrt{3}\right) = {x}^{2} - \left(5 + \sqrt{3}\right) x + 5 \sqrt{3}$
If - as is probably the case - we want a polynomial with integer coefficients, then we also need the rational conjugate $- \sqrt{3}$ to be a zero and $\left(x + \sqrt{3}\right)$ a factor.
Then we can define:
$g \left(x\right) = \left(x - 5\right) \left(x - \sqrt{3}\right) \left(x + \sqrt{3}\right) = \left(x - 5\right) \left({x}^{2} - 3\right) = {x}^{3} - 5 {x}^{2} - 3 x + 15$
Any polynomial with these zeros will be a multiple (scalar or polynomial) of this $g \left(x\right)$.
So the polynomial equation:
${x}^{3} - 5 {x}^{2} - 3 x + 15 = 0$
has roots $5$, $\sqrt{3}$ and $- \sqrt{3}$.
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# Isosceles Triangles
## Properties of triangles with two equal sides.
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Isosceles Triangles
### Isosceles Triangles
An isosceles triangle is a triangle that has at least two congruent sides. The congruent sides of the isosceles triangle are called the legs. The other side is called the base. The angles between the base and the legs are called base angles. The angle made by the two legs is called the vertex angle. One of the important properties of isosceles triangles is that their base angles are always congruent. This is called the Base Angles Theorem.
For DEF\begin{align*}\triangle DEF\end{align*}, if DE¯¯¯¯¯¯¯¯EF¯¯¯¯¯¯¯¯\begin{align*}\overline{DE} \cong \overline{EF}\end{align*}, then DF\begin{align*}\angle D \cong \angle F\end{align*}.
Another important property of isosceles triangles is that the angle bisector of the vertex angle is also the perpendicular bisector of the base. This is called the Isosceles Triangle Theorem. (Note this is ONLY true of the vertex angle.) The converses of the Base Angles Theorem and the Isosceles Triangle Theorem are both true as well.
Base Angles Theorem Converse: If two angles in a triangle are congruent, then the sides opposite those angles are also congruent. So for DEF\begin{align*}\triangle DEF\end{align*}, if DF\begin{align*}\angle D \cong \angle F\end{align*}, then DE¯¯¯¯¯¯¯¯EF¯¯¯¯¯¯¯¯\begin{align*}\overline{DE} \cong \overline{EF}\end{align*}.
Isosceles Triangle Theorem Converse: The perpendicular bisector of the base of an isosceles triangle is also the angle bisector of the vertex angle. So for isosceles DEF\begin{align*}\triangle DEF\end{align*}, if EG¯¯¯¯¯¯¯¯DF¯¯¯¯¯¯¯¯\begin{align*}\overline{EG} \perp \overline{DF}\end{align*} and DG¯¯¯¯¯¯¯¯GF¯¯¯¯¯¯¯¯\begin{align*}\overline{DG} \cong \overline{GF}\end{align*}, then DEGFEG\begin{align*}\angle DEG \cong \angle FEG\end{align*}.
What if you were presented with an isosceles triangle and told that its base angles measure x\begin{align*}x^\circ\end{align*} and y\begin{align*}y^\circ\end{align*}? What could you conclude about x and y?
### Examples
#### Example 1
Find the value of x\begin{align*}x\end{align*} and the measure of each angle.
The two angles are equal, so set them equal to each other and solve for x\begin{align*}x\end{align*}.
(4x+12)15=x=(5x3)\begin{align*}(4x+12)^\circ & = (5x-3)^\circ\\ 15 = x\end{align*}
Substitute x=15\begin{align*}x = 15\end{align*}; the base angles are [4(15)+12]\begin{align*}[4(15) +12]^\circ\end{align*}, or 72\begin{align*}72^\circ\end{align*}. The vertex angle is 1807272=36\begin{align*}180^\circ -72^\circ-72^\circ =36^\circ\end{align*}.
#### Example 2
True or false: Base angles of an isosceles triangle can be right angles.
This statement is false. Because the base angles of an isosceles triangle are congruent, if one base angle is a right angle then both base angles must be right angles. It is impossible to have a triangle with two right (90\begin{align*}90^\circ\end{align*}) angles. The Triangle Sum Theorem states that the sum of the three angles in a triangle is 180\begin{align*}180^\circ\end{align*}. If two of the angles in a triangle are right angles, then the third angle must be 0\begin{align*}0^\circ\end{align*} and the shape is no longer a triangle.
#### Example 3
Which two angles are congruent?
This is an isosceles triangle. The congruent angles are opposite the congruent sides. From the arrows we see that SU\begin{align*}\angle S \cong \angle U\end{align*}.
#### Example 4
If an isosceles triangle has base angles with measures of 47\begin{align*}47^\circ\end{align*}, what is the measure of the vertex angle?
Draw a picture and set up an equation to solve for the vertex angle, v\begin{align*}v\end{align*}. Remember that the three angles in a triangle always add up to 180\begin{align*}180^\circ\end{align*}.
47+47+vvv=180=1804747=86\begin{align*}47^\circ + 47^\circ + v & = 180^\circ\\ v & = 180^\circ - 47^\circ - 47^\circ\\ v & = 86^\circ\end{align*}
#### Example 5
If an isosceles triangle has a vertex angle with a measure of 116\begin{align*}116^\circ\end{align*}, what is the measure of each base angle?
Draw a picture and set up and equation to solve for the base angles, b\begin{align*}b\end{align*}.
116+b+b2bb=180=64=32\begin{align*}116^\circ + b + b & = 180^\circ\\ 2b & = 64^\circ\\ b & = 32^\circ\end{align*}
### Review
Find the measures of x\begin{align*}x\end{align*} and/or y\begin{align*}y\end{align*}.
Determine if the following statements are true or false.
1. Base angles of an isosceles triangle are congruent.
2. Base angles of an isosceles triangle are complementary.
3. Base angles of an isosceles triangle can be equal to the vertex angle.
4. Base angles of an isosceles triangle are acute.
Fill in the proofs below.
1. Given: Isosceles CIS\begin{align*}\triangle CIS\end{align*}, with base angles C\begin{align*}\angle C\end{align*} and S\begin{align*}\angle S\end{align*} IO¯¯¯¯¯¯\begin{align*}\overline{IO}\end{align*} is the angle bisector of CIS\begin{align*}\angle CIS\end{align*} Prove: IO¯¯¯¯¯¯\begin{align*}\overline{IO}\end{align*} is the perpendicular bisector of CS¯¯¯¯¯¯¯\begin{align*}\overline{CS}\end{align*}
Statement Reason
1. 1. Given
2. 2. Base Angles Theorem
3. CIOSIO\begin{align*}\angle CIO \cong \angle SIO\end{align*} 3.
4. 4. Reflexive PoC
5. CIOSIO\begin{align*}\triangle CIO \cong \triangle SIO\end{align*} 5.
6. CO¯¯¯¯¯¯¯¯OS¯¯¯¯¯¯¯\begin{align*}\overline{CO} \cong \overline{OS}\end{align*} 6.
7. 7. CPCTC
8. IOC\begin{align*}\angle IOC\end{align*} and IOS\begin{align*}\angle IOS\end{align*} are supplementary 8.
9. 9. Congruent Supplements Theorem
10. IO¯¯¯¯¯¯\begin{align*}\overline{IO}\end{align*} is the perpendicular bisector of CS¯¯¯¯¯¯¯\begin{align*}\overline{CS}\end{align*} 10.
1. Given: Isosceles ICS\begin{align*}\triangle ICS\end{align*} with C\begin{align*}\angle C\end{align*} and S\begin{align*}\angle S\end{align*} IO¯¯¯¯¯¯\begin{align*}\overline{IO}\end{align*} is the perpendicular bisector of \begin{align*}\overline{CS}\end{align*} Prove: \begin{align*}\overline{IO}\end{align*} is the angle bisector of \begin{align*}\angle CIS\end{align*}
Statement Reason
1. 1.
2. \begin{align*}\angle C \cong \angle S\end{align*} 2.
3. \begin{align*}\overline{CO} \cong \overline{OS}\end{align*} 3.
4. \begin{align*}m\angle IOC = m\angle IOS = 90^\circ\end{align*} 4.
5. 5.
6. 6. CPCTC
7. \begin{align*}\overline{IO}\end{align*} is the angle bisector of \begin{align*}\angle CIS\end{align*} 7.
On the \begin{align*}x-y\end{align*} plane, plot the coordinates and determine if the given three points make a scalene or isosceles triangle.
1. (-2, 1), (1, -2), (-5, -2)
2. (-2, 5), (2, 4), (0, -1)
3. (6, 9), (12, 3), (3, -6)
4. (-10, -5), (-8, 5), (2, 3)
5. (-1, 2), (7, 2), (3, 9)
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
Base
The side of a triangle parallel with the bottom edge of the paper or screen is commonly called the base. The base of an isosceles triangle is the non-congruent side in the triangle.
Base Angles
The base angles of an isosceles triangle are the angles formed by the base and one leg of the triangle.
Base Angles Theorem Converse
The base angles theorem converse states if two angles in a triangle are congruent, then the sides opposite those angles are also congruent.
Isosceles Triangle Theorem
The Isosceles Triangle Theorem states that the perpendicular bisector of the base of an isosceles triangle is also the angle bisector of the vertex angle.
Triangle Sum Theorem
The Triangle Sum Theorem states that the three interior angles of any triangle add up to 180 degrees.
Vertical Angles
Vertical angles are a pair of opposite angles created by intersecting lines.
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# 1TTO §8.2 Powers
Now comes a powerful paragraph!
1 / 26
Slide 1: Slide
WiskundeMiddelbare schoolhavo, vwoLeerjaar 1
This lesson contains 26 slides, with interactive quizzes and text slides.
Lesson duration is: 30 min
## Items in this lesson
Now comes a powerful paragraph!
§8.2 Powers
#### Slide 2 -Slide
When using your calculator:
5 kwadraat kun je op twee manieren doen:
tik 5
tik =.
Of: tik 5 en dan 'dakje' [dus ^] en dan 2 en dan =
x2
#### Slide 3 -Slide
When using your calculator:
tik 3
tik dakje [ = ^-knop)
tik 4
en tik dan =
34=
#### Slide 4 -Slide
Here is a table with length = 2 m and also width = 2 m.
The Area (=oppervlakte) = 2 x 2 = 4
Multiplying a number by itself -in this case 2 muliplied by 2- is called:
SQUARING a number.
The result is the SQUARE of a number. Let's look at a few more examples!
m2
Square = vierkant!
#### Slide 5 -Slide
Square = vierkant = 'kwadraat'
#### Slide 6 -Slide
In the next slide
give the answers to these 5 exercises.
First copy in your notebook:
12,22,32,42,52
#### Slide 7 -Slide
The 4 answers to the SQUARES of 1,2,3,4, and 5 are:
A
1, 2, 3, 4 and 5
B
1, 4, 9, 16 and 25
C
2, 4, 6, 8 and 10
#### Slide 8 -Quiz
Solutions:
So, for example:
the SQUARE of 5 is
5 x 5 = 25.
#### Slide 9 -Slide
However ...
sometimes you want to multiply a number by itself more than once. Then we can use a similar Notation as for the Square:
+ For example: 3 x 3 x 3 x 3 =
+ See the next slide for more! New examples and the new Words that come with this.
+ Like the word: POWER.
34
#### Slide 12 -Slide
is the same as ........
43
A
4 x 4 x 4
B
4 + 4 +4
C
4 x 3
D
64
#### Slide 13 -Quiz
Explanation:
means the same as:
4 x 4 x 4,
but it also means: 64, because that is the outcome.
It doesn't mean 4 + 4 + 4, which you could write as 3 x 4 = 12.
Once more: = 4 x 4 x 4 = 64
Remember: a POWER is about multiplying, not adding!
43
43
#### Slide 14 -Slide
is the same as .....
24
A
8
B
2 x 2 x 2 x 2
C
16
D
2 + 2 + 2 + 2
#### Slide 15 -Quiz
Explanation:
is the same as:
+ 2 x 2 x 2 x 2 , because a power is a multiplication, that is repeated.
+ 16, because that's the outcome.
+ however, it's not about adding, so not 2 + 2 + 2 +2.
24
#### Slide 16 -Slide
REMEMBER THIS? IT'S FROM FROM §8.1!
HOWEVER, THE POWERS HAVE TO HAVE A PLACE IN THIS NOW!
SEE NEXT SLIDE.
#### Slide 17 -Slide
Watch below: between the 'brackets' and the 'multiplying/dividing' the POWERS are placed!
So it comes 2nd in place.
#### Slide 18 -Slide
Hoe Haakje
Moeten Machten
Wij wortel
Van Deze vermenigvuldigen delen
------------>
Onvoldoendes Afkomen? optellen aftrekken
------------------------------>
#### Slide 19 -Slide
THINK about this one for a while ... (You don't have to actually work it out, yet!)
How to work it out and what is the final answer?
#### Slide 20 -Slide
Here is the Solution:
(Study it carefully, because you're going to make one yourself!)
H
M
W
VD ->
OA ->
#### Slide 21 -Slide
1. Work this one out on paper.
2. Take a pic and
3. Send it in in next slide!
#### Slide 22 -Slide
My work in a Photo:
#### Slide 23 -Open question
Solution:
The POWERS between the BRACKETS have to be worked out first!
Following the PRIORITY rules we get:
7 x ( 36 - 32) = 7 x 4 = 28
A common mistake would be:
7 x ( 12 - 10 ) = 7 x 2 = 14 In this case you've done 6 x 2 and 2 x 5 instead of
calculating POWERS!
#### Slide 24 -Slide
Homework time.
+ §8.2 make 11, 14, 15, 16.
#### Slide 25 -Slide
As for 18, above, use the remark :
'Calculate as in the example above' !
So don't just give the answers -they are in the Solutionsbook too- but show the process of Calculating!
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# What is the vertex, axis of symmetry, the maximum or minimum value, and the range of parabola y=-x^2+2x-5?
May 25, 2018
$y = - {\left(x - 1\right)}^{2} - 4$
#### Explanation:
All you Need can you see in the following formula
$y = - \left({x}^{2} - 2 x + 1 - 1\right) - 5 = - {\left(x - 1\right)}^{2} - 4$
See below:
#### Explanation:
To find the answers in a more straightforward manner, let's first convert from standard form to vertex form. To do that, we're changing from
$y = A {x}^{2} + B x + C$
into
$y = {\left(x - h\right)}^{2} + k$
To do that, we complete the square:
$y = - {x}^{2} + 2 x - 5$
$y = - \left({x}^{2} - 2 x\right) - 5$
$y = - \left({x}^{2} - 2 x + 1\right) - 5 + 1$
$y = - {\left(x - 1\right)}^{2} - 4$
Vertex
In this form, the vertex (or, in other words, the "pointy bit") is given to us in the $\left(h , k\right)$ term. Here we have $\left(1 , - 4\right)$
Axis of Symmetry
The axis of symmetry splits the parabola in 2 equal parts. It runs through the vertex straight up and down (for a parabola that either opens up or opens down). And so we can write an equation for the line that is vertical and runs through the vertex. In this question, it's $x = 1$.
Maximum/minimum value
Here we're dealing with looking at the vertex and seeing if it is as high as the parabola goes (i.e. maximum and is when there is a negative sign sitting in front of the ${\left(x - h\right)}^{2}$ term) or as low as it goes (i.e. minimum and is when there is a positive sign sitting there).
In our case, there's a negative sign and so this will be a maximum. The $y$ coordinate of the vertex shows the value.
$y = - 4$
Range
Since we know the maximum value is $y = - 4$, all $y$ values less than and including $- 4$ will be part of the parabola graph.
$y \le - 4$
To see all of this in graph form, here's the graph:
graph{-x^2+2x-5[-10,10,-10,0]}
For more on parabolas, you may find this helpful:
http://jwilson.coe.uga.edu/emt725/class/sarfaty/emt669/instructionalunit/parabolas/parabolas.html
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Tutorial of Orthogonal Matching Pursuit
A tutorial for Beginners and Dummies
Koredianto Usman
Sekolah Teknik Elektro dan Informatika
8 Mei 2014
Koredianto Usman
Tutorial of Orthogonal Matching Pursuit
Outline 1 Problem Statement 2 The Orthogonal Matching Pursuit 3 Koherensi Basis 2 / 13 .
65 −0.1.8 0 Now.6 −1 Therefore y = A · x gives: −1.65 −0. Given that : y = and A = 0. Problem Statement Consider the following very simple example: Given the following sparse signals −1.8 0 A= 0.707 0.25 0.707 0.6 −1 y= How to find original x? 3 / 13 .707 0.25 −1.2 x = 1 0 The following is the measurement matrix A (2 x 3): −0.707 0.
8 0 b2 = b3 = 0.8 0 A= 0.6 −1 Columns in matrix A are called BASIS (CHEN and DONOHO : ATOMS). Orthogonal Matching Pursuit −1.707 0.6 −1 4 / 13 .707 0. In the example.65 BASIS Previous example: Given : y = and 0.2. we have the following atoms: b1 = −0.25 −0.707 0.707 0.
Orthogonal Matching Pursuit Interpretation of equation Ax = y Since A = [b1 b2 b3 ]. and if we let : x = [a b c].707 0.65 = −1.8 0 −1.707 −0. then A · x = a · b1 + b · b2 + c · b3 A · x is the linear combination of b1 .707 0.2.2 −0.8 0 A·x = · 1 0.707 0. b3 .6 −1 0.6 −1 0 −0. b2 .25 5 / 13 .2 · +1· +0· =y = 0. −1.
2 −0.6 −1 0. b3 that will influence the STRONGEST to y .707 0.707 −0. dan last is b3 .25 From the equation above.707 0.707 0. 6 / 13 .2. ORTHOGONAL MATCHING PURSUIT works reversely: we start finding which of b1 .8 0 −1.8 0 A·x = · 0 0. Orthogonal Matching Pursuit −1. it is clear that atom b1 contribute the biggest influence in y.65 = −1.2 · +1· +0· =y = 0. and so on.6 −1 1 −0. next is b2 . Then the SECOND STRONGEST from the residual. b2 .
y > calculate the residue ri = pi − pi · < pi .707) b2 (0.707 .25) X b3 (0 .2. Orthogonal Matching Pursuit STRONGEST influence is measured using DOT PRODUCT / INNER PRODUCT OMP Algorithm: 1 find atom with has the biggest inner product with y pi = maxj < bj . -1) 7 / 13 . 0. 0. 0.8 .65 .6) y (-1. y > find atom with has the biggest inner product with ri 4 repeat step 2 and 3 until residue achieve a certain threshold Geometrically: 2 3 Y b1 (-0.
17 and < y . Then. We next count the residue: −1. we see b1 gives the biggest inner product.34 < y .34) · 0. b3: < y .34. Orthogonal Matching Pursuit Here the dot product of y to any of b1.25 Taking the absolute value. b1 > = − (−1. b1 >= −1.25 0. b2 >= 1.707 8 / 13 . b1 is chosen as the atom in first step.65 0.707 r1 = y − b1 · < y .2. b2. b3 >= 0. DOT PRODUCT −1.
-1) Next we count the DOT PRODUCT of this residue to b2 and b3 (no need to count with b1 . 0. b3 >= −0.2. we get b2 as the next strongest influence.25) b2 (0. Orthogonal Matching Pursuit Y Residue 1 b1 (-0.6) X b3 (0 . 9 / 13 . < r 1. 0.8 .707 . b1 >= 1 < r 1.7 Taking the absolute value.65 .707) y (-1. since this residue must perpendicular to b1 ). 0.
b3 >= −0.7 0.099 10 / 13 .7 0.6 −0. we finally count the final DOT PRODUCT.2.8 r 2 = r 1− < r 1. between r2 with the last b3 : < r 2. Orthogonal Matching Pursuit Next we count again the residue: −0.099 From residue r2.099 = 0. b2 > b2 = − (1) · 0.
34. Orthogonal Matching Pursuit The following base is in STRONGEST influence in order: b1 . with influence measure from dot product is : −1.34 Therefore. 1. dan −0.2. b3 . the reconstructed x is 1 −0.099 −1. b2 .099 −1.2 Original was : 1 0 11 / 13 .
(C). Koherensi Rendah 12 / 13 . Koherensi Sedang. b2 b2 A3 b2 A3 A3 A2 A2 A2 A1 b1 b1 b1 A1 (A) (B) A1 (C) Figure : Ilustrasi Koherensi Basis. (B).3. Koherensi tinggi. (A). Base Coherency Base Coherency indicate how close one base to the others.
Base Coherency A simple measure for base coherency is INNER PRODUCT. Aj >) Range of Coherency Value µ is : 0 ≤ |µ| ≤ 1 0 : least coherency 1 : full coherent We expect least coherency in our base in order to success in MP. 13 / 13 .j). µ = max(i.i6=j (< Ai .3.
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# FINDING NATURE OF QUADRATIC EQUATION BY GRAPHING
Finding Nature of Quadratic Equation by Graphing :
Here we are going to see some example problems of finding nature of solution of quadratic equation with graph.
To obtain the roots of the quadratic equation ax2 + bx + c = 0 graphically, we first draw the graph of y = ax2 +bx +c .
The solutions of the quadratic equation are the x coordinates of the points of intersection of the curve with X axis.
## Finding the Nature of Solution of Quadratic Equations Graphically
To determine the nature of solutions of a quadratic equation, we can use the following procedure.
(i) If the graph of the given quadratic equation intersect the X axis at two distinct points, then the given equation has two real and unequal roots.
(ii) If the graph of the given quadratic equation touch the X axis at only one point, then the given equation has only one root which is same as saying two real and equal roots.
(iii) If the graph of the given equation does not intersect the X axis at any point then the given equation has no real root.
## Finding Nature of Quadratic Equation by Graphing - Questions
Question 1 :
Graph the following quadratic equations and state their nature of solutions.
(i) x2 + x + 7 = 0
Solution :
Draw the graph for the function y = x2 + x + 7
Let us give some random values of x and find the values of y.
x-4-3-2-101234 x216941014916 x-4-3-2-101234 +7777777777 y19139779131927
Points to be plotted :
(-4, 19) (-3, 13) (-2, 9) (-1, 7) (0, 7) (1, 9) (2, 13) (3, 19) (4, 27)
To find the x-coordinate of the vertex of the parabola, we may use the formula x = -b/2a
x = -1/2(1) = -1/2
By applying x = -1/2, we get the value of y.
y = (-1/2)2 + (-1/2) + 7
y = (1/4) - (1/2) + 7
y = (1 - 2 + 28)/4 = 27/4
Vertex (1/2, 27/4)
The graph of the given parabola does not intersect the x-axis at any point. Hence it has no real roots.
(iv) x2 − 9 = 0
Solution :
Let us give some random values of x and find the values of y.
x-4-3-2-101234 x216941014916 -9-9-9-9-9-9-9-9-9-9 y70-5-8-9-8-507
Points to be plotted :
(-4, 7) (-3, 0) (-2, -5) (-1, -8) (0, -9) (1, -8) (2, -5) (3, 0) (4, 7)
To find the x-coordinate of the vertex of the parabola, we may use the formula x = -b/2a
x = 0/2(1) = 0
By applying x = 0, we get the value of y.
y = 02 - 9
y = -9
Vertex (0, -9)
The graph of the given parabola intersects the x-axis at two points. Hence it has two real and unequal roots.
After having gone through the stuff given above, we hope that the students would have understood, "Finding Nature of Quadratic Equation by Graphing".
Apart from the stuff given in this section "Finding Nature of Quadratic Equation by Graphing"if you need any other stuff in math, please use our google custom search here.
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# Chapter 3 - Section 3.2 - Corresponding Parts of Congruent Triangles - Exercises: 25
First, we need to prove $\triangle FED\cong\triangle GED$ by SSS. Then we can deduce $\angle DEF\cong\angle DEG$. Then prove $\angle DEF$ and $\angle DEG$ must be right $\angle$s. That means $\overline{DE}\bot\overline{FG}$
#### Work Step by Step
*PLANNING: First, we need to prove $\triangle FED\cong\triangle GED$. Then we can deduce $\angle DEF\cong\angle DEG$. So $\angle DEF$ and $\angle DEG$ must be right $\angle$s. That means $\overline{DE}\bot\overline{FG}$ 1. $E$ is the midpoint of $\overline{FG}$ (Given) 2. $\overline{EF}\cong\overline{EG}$ (The midpoint of a line divides it into 2 congruent lines) 3. $\overline{DF}\cong\overline{DG}$ (Given) 4. $\overline{DE}\cong\overline{DE}$ (Identity) So now we have all 3 sides of $\triangle FED$ are congruent with 3 corresponding sides of $\triangle GED$. Therefore, 5. $\triangle FED\cong\triangle GED$ (SSS) 6. $\angle DEF\cong\angle DEG$ (CPCTC) However, we see that $\angle DEF+\angle DEG=\angle FEG=180^o$ (since $\overline{FG}$ is a line) Therefore, the value of each angle must be $90^o$. So, 7. $\angle DEF$ and $\angle DEG$ are both right $\angle$s. 8. $\overline{DE}\bot\overline{FG}$ (if a line intersects another one and creates 2 right angles, then those 2 lines are perpendicular with each other)
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# What is GCF of 27 and 36?
GCF of 27 and 36 is 9
#### How to find GCF of two numbers
1. Steps to find GCF of 27 and 36 2. What is GCF of two numbers? 3. What are Factors? 4. Examples of GCF
### Example: Find gcf of 27 and 36
• Factors for 27: 1, 3, 9, 27
• Factors for 36: 1, 2, 3, 4, 6, 9, 12, 18, 36
Hence, GCf of 27 and 36 is 9
#### How do we define GCF?
In mathematics we use GCF or greatest common method to find out the greatest possible positive integer which can completely divide the given numbers. It is written as GCF (27, 36).
#### Properties of GCF
• The GCF of two given numbers where one of them is a prime number is either 1 or the number itself.
• GCF of two consecutive numbers is always 1.
• Given two numbers 27 and 36, such that GCF is 9 where 9 will always be less than 27 and 36.
• Product of two numbers is always equal to the product of their GCF and LCM.
#### How do you explain factors?
In mathematics, a factor is a number or also it can be an algebraic expression that divides another number or any expression completely and that too without leaving any remainder. A factor of a number can be positive or negative.
#### Properties of Factors
• Every number is a factor of zero (0), since 27 x 0 = 0 and 36 x 0 = 0.
• Every number other than 1 has at least two factors, namely the number itself and 1.
• Every factor of a number is an exact divisor of that number, example 1, 3, 9, 27 are exact divisors of 27 and 1, 2, 3, 4, 6, 9, 12, 18, 36 are exact divisors of 36.
• Factors of 27 are 1, 3, 9, 27. Each factor divides 27 without leaving a remainder.
Simlarly, factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, 36. Each factor divides 36 without leaving a remainder.
#### Steps to find Factors of 27 and 36
• Step 1. Find all the numbers that would divide 27 and 36 without leaving any remainder. Starting with the number 1 upto 13 (half of 27) and 1 upto 18 (half of 36). The number 1 and the number itself are always factors of the given number.
27 ÷ 1 : Remainder = 0
36 ÷ 1 : Remainder = 0
27 ÷ 3 : Remainder = 0
36 ÷ 2 : Remainder = 0
27 ÷ 9 : Remainder = 0
36 ÷ 3 : Remainder = 0
27 ÷ 27 : Remainder = 0
36 ÷ 4 : Remainder = 0
36 ÷ 6 : Remainder = 0
36 ÷ 9 : Remainder = 0
36 ÷ 12 : Remainder = 0
36 ÷ 18 : Remainder = 0
36 ÷ 36 : Remainder = 0
Hence, Factors of 27 are 1, 3, 9, and 27
And, Factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, and 36
#### Examples of GCF
Sammy baked 27 chocolate cookies and 36 fruit and nut cookies to package in plastic containers for her friends at college. She wants to divide the cookies into identical boxes so that each box has the same number of each kind of cookies. She wishes that each box should have greatest number of cookies possible, how many plastic boxes does she need?
Since Sammy wants to pack greatest number of cookies possible. So for calculating total number of boxes required we need to calculate the GCF of 27 and 36.
GCF of 27 and 36 is 9.
What is the difference between GCF and LCM?
Major and simple difference betwen GCF and LCM is that GCF gives you the greatest common factor while LCM finds out the least common factor possible for the given numbers.
What is the relation between LCM and GCF (Greatest Common Factor)?
GCF and LCM of two numbers can be related as GCF(27, 36) = ( 27 * 36 ) / LCM(27, 36) = 9.
What is the GCF of 27 and 36?
GCF of 27 and 36 is 9.
Ram has 27 cans of Pepsi and 36 cans of Coca Cola. He wants to create identical refreshment tables that will be organized in his house warming party. He also doesn't want to have any can left over. What is the greatest number of tables that Ram can arrange?
To find the greatest number of tables that Ram can stock we need to find the GCF of 27 and 36. Hence GCF of 27 and 36 is 9. So the number of tables that can be arranged is 9.
Rubel is creating individual servings of starters for her birthday party. He has 27 pizzas and 36 hamburgers. He wants each serving to be identical, with no left overs. Can you help Rubel in arranging the same in greatest possible way?
The greatest number of servings Rubel can create would be equal to the GCF of 27 and 36. Thus GCF of 27 and 36 is 9.
Ariel is making ready to eat meals to share with friends. She has 27 bottles of water and 36 cans of food, which she would like to distribute equally, with no left overs. What is the greatest number of boxes Ariel can make?
The greatest number of boxes Ariel can make would be equal to GCF of 27 and 36. So the GCF of 27 and 36 is 9.
Mary has 27 blue buttons and 36 white buttons. She wants to place them in identical groups without any buttons left, in the greatest way possible. Can you help Mary arranging them in groups?
Greatest possible way in which Mary can arrange them in groups would be GCF of 27 and 36. Hence, the GCF of 27 and 36 or the greatest arrangement is 9.
Kamal is making identical balloon arrangements for a party. He has 27 maroon balloons, and 36 orange balloons. He wants each arrangement tohave the same number of each color. What is the greatest number of arrangements that he can make if every balloon is used?
The greatest number of arrangements that he can make if every balloon is used would be equal to GCF of 27 and 36. So the GCF of 27 and 36 is 9.
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# Section 3.3. Differentiation of Polynomials and Rational Functions. Difference Equations to Differential Equations
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1 Difference Equations to Differential Equations Section 3.3 Differentiation of Polynomials an Rational Functions In tis section we begin te task of iscovering rules for ifferentiating various classes of functions. By te en of Section 3.5 we will be able to ifferentiate any algebraic or trigonometric function as a matter of routine witout reference to te limits use in Section 3.2. Differentiation of polynomials We first note tat if f is a first egree polynomial, say, f(x) = ax + b for some constants a an b, ten f is an affine function an ence its own best affine approximation. Tus f (x) = a for all x. In particular, if f is a constant function, say, f(x) = b for all x, ten f (x) = 0 for all x. Next we consier te case of a monomial f(x) = x n, were n is a positive integer greater tan 1. Ten Now f (x) 0 f(x + ) f(x) (x + ) n x n. (3.3.1) 0 (x + ) n = x n + nx n 1 + R(), (3.3.2) were R() represents te remaining terms in te expansion. Since every term in R() as a factor of raise to a power greater tan or equal to 2, it follows tat R() is o(). Hence we ave f x n + nx n 1 + R() x n (x) 0 nx n 1 + R() 0 ( nx n 1 + R() ) 0 = nx n 1 R() + lim 0 = nx n 1. Since from our previous result f (x) = 1 wen f(x) = x, tis formula also works in te case n = 1. Hence we ave te following proposition. Proposition For any positive integer n, x xn = nx n 1. (3.3.3) 1 Copyrigt c by Dan Slougter 2000
2 2 Differentiation of Polynomials an Rational Functions Section 3.3 If f(x) = x 3, ten f (x) = 3x 2, as we saw in an example in Section 3.2. Similarly, t t5 = 5t 4. Hence, for example, te equation of te line tangent to te curve x = t 5 at ( 1, 1) is or x = 5(t + 1) 1, x = 5t + 4. Once we establis results for te erivative of a constant times a function an for te erivative of te sum of two functions, similar to te results we ave for limits, we will be able to easily ifferentiate any polynomial. So suppose f is a ifferentiable function an let k(x) = cf(x), were c is any constant. Ten k (x) 0 k(x + ) k(x) 0 cf(x + ) cf(x) 0 c(f(x + ) f(x)) = c lim 0 f(x + ) f(x) = cf (x). Tat is, te erivative of a constant times a function is te constant times te erivative of te function. Proposition If f is ifferentiable an c is any constant, ten x (cf(x)) = c f(x). (3.3.4) x If f(x) = 14x 3, ten f (x) = (14)(3x 2 ) = 42x 2. Now suppose f an g are bot ifferentiable functions an let k(x) = f(x) +. Ten k k(x + ) k(x) (x) 0 (f(x + ) + g(x + )) (f(x) + ) 0 0 ( f(x + ) f(x) + f(x + ) f(x) 0 = f (x) + g (x). g(x + ) ) + lim 0 g(x + )
3 Section 3.3 Differentiation of Polynomials an Rational Functions 3 Hence te erivative of te sum of two functions is te sum of teir erivatives. A similar argument woul sow tat te erivative of te ifference of two functions is te ifference of teir erivatives. Proposition If f an g are bot ifferentiable, ten (f(x) + ) = x x f(x) + (3.3.5) x an (f(x) ) = x x f(x). (3.3.6) x Putting te preceing results togeter, we are now in a position to easily ifferentiate any polynomial, as te next examples will illustrate. Suppose f(x) = 3x 5 6x 2 + 2x 16. Ten f (x) = x (3x5 6x 2 + 2x 16) = x (3x5 ) x (6x2 ) + x (2x) x (16) = 3 x x5 6 x x2 + 2 x x 0 = (3)(5x 4 ) (6)(2x) + (2)(1) = 15x 4 12x + 2. Of course, it is not necessary to write out in etail all te steps in ifferentiating a polynomial as we i in te preceing example. For example, if g(t) = 3t 12 6t 2 + t, ten g (t) = (3)(12t 11 ) (6)(2t) + 1 = 36t 11 12t + 1. In particular, since g(1) = 2 an g (1) = 25, te best affine approximation to g at t = 1 is T (t) = 25(t 1) 2 = 25t 27. Differentiation of rational functions We next consier te problem of ifferentiating te quotient of two functions wose erivatives are alreay known. In particular, combining tis result wit our result for polynomials will enable us to easily ifferentiate any rational function. We migt ope tat, analogous to te last two results an te relate results for limits, te erivative of te quotient of two functions woul be equal to te quotient of teir erivatives. Tis turns out not to be true; neverteless, tere is a nice rule for ifferentiating quotients.
4 4 Differentiation of Polynomials an Rational Functions Section 3.3 Suppose f an g are bot ifferentiable functions an let k(x) = f(x). Ten, at all points were 0, k k(x + ) k(x) (x) f(x + ) g(x + ) f(x) f(x + ) g(x + )f(x) g(x + ) f(x + ) g(x + )f(x) 0 g(x + ) It turns out tat by aing an subtracting te term f(x) (a stanar matematical trick of aing 0) in te numerator, we can simplify tis limit into a form tat we can evaluate. Tat is, Now an k (x) 0 f(x + ) g(x + )f(x) g(x + ) f(x + ) f(x) + f(x) g(x + )f(x) 0 g(x + ) (f(x + ) f(x)) f(x)(g(x + ) ) 0 g(x + ) f(x + ) f(x) g(x + ) f(x). 0 g(x + ) f(x + ) f(x) lim 0 g(x + ) lim f(x) 0 = lim 0 f(x + ) f(x) = f(x) lim 0 g(x + ). = f (x), (3.3.7) = f(x)g (x), (3.3.8) lim g(x + ) = lim g(x + ) = = 0 0 ()2, (3.3.9) were te limits in (3.3.7) an (3.3.8) follow from te ifferentiability of f an g, wile te limit in (3.3.9) follows from te continuity of g (wic is a consequence of te ifferentiability of g). Putting everyting togeter, we ave a result known as te quotient rule. k (x) = f (x) f(x)g (x) () 2, (3.3.10)
5 Section 3.3 Differentiation of Polynomials an Rational Functions 5 Quotient Rule If f an g are bot ifferentiable, ten x at all points were 0. f(x) = Suppose f(x) = 2x + 1 x 2. Ten f(x) f(x) x x () 2 (3.3.11) (x 2) (2x + 1) (2x + 1) (x 2) f (x) = x x (x 2) 2 (x 2)(2) (2x + 1)(1) = (x 2) 2 2x 4 2x 1 = (x 2) 2 5 = (x 2) 2. Hence, for example, f(3) = 7 an f (3) = 5, so te equation of te line tangent to te grap of f at (3, 7) is y = 5(x 3) + 7, or y = 5x Suppose g(z) = 1 z 2. Ten g (z) = z 2 z (1) (1) z (z2 ) z 4 = (z2 )(0) 2z z 4 = 2 z 3. Note tat we may write tis result in te form wic is consistent wit our previous result z z 2 = 2z 3, z zn = nz n 1. However, we erive te latter uner te assumption tat n was a positive integer. We will now sow tat we can exten tis result to te case of negative integer exponents.
6 6 Differentiation of Polynomials an Rational Functions Section 3.3 Suppose f(x) = x n, were n is a negative integer. Ten, using te quotient rule an te fact tat n > 0, f (x) = x xn = 1 x x n x n (1) (1) = x x (x n ) x 2n = (x n )(0) ( nx n 1 ) x 2n = nx n 1 x 2n = nx n 1+2n = nx n 1. We can now state te more general result. Proposition For any integer n 0, x xn = nx n 1. (3.3.12) If f(x) = 1 x, ten f (x) = x x 1 = x 2 = 1 x 2. Similarly, 5 x x 3 = x (5x 3 ) = 15x 4 = 15 x 4. We will eventually see tat (3.3.12) ols for rational an irrational exponents as well. We will consier te rational case in Section 3.4, but we will not ave te tools for anling te irrational case until we iscuss exponential an logaritm functions in Capter 6. Differentiation of proucts We will close tis section wit a iscussion of a rule for ifferentiating te prouct of two functions. Since te prouct of two rational functions is again a rational function, tis will not exten te class of functions tat we know ow to ifferentiate routinely. However, tis rule will be very useful in te future an, even at te present point, can elp simplify some problems.
7 Section 3.3 Differentiation of Polynomials an Rational Functions 7 Suppose f an g are bot ifferentiable an k(x) = f(x). Ten k (x) 0 k(x + ) k(x) f(x + )g(x + ) f(x). (3.3.13) 0 Aing an subtracting f(x + ) in te numerator (again, te matematical trick of aing 0 in a useful manner) will elp simplify tis limit. Namely, k f(x + )g(x + ) f(x + ) + f(x + ) f(x) (x) 0 f(x + )(g(x + ) ) + (f(x + ) f(x)) 0 0 ( f(x + ) Now g(x + ) lim f(x+) 0 ( g(x + ) f(x + ) f(x) + 0 f(x+) lim 0 g(x + ) )). = f(x)g (x) (3.3.14) an f(x + ) f(x) lim 0 = lim 0 f(x + ) f(x) = f (x), (3.3.15) were, as wit te erivation of te quotient rule, we ave use te ifferentiability of f an g as well as te continuity of f in evaluating te limits. Putting everyting togeter, we ave k (x) = f(x)g (x) + f (x). (3.3.16) a result known as te prouct rule. Prouct Rule ten If If f an g are bot ifferentiable, ten f(x) = f(x) + f(x). (3.3.17) x x x f(x) = (x 4 3x 2 + 6x 3)(6x 3 + 2x + 5), f (x) = (x 4 3x 2 + 6x 3) x (6x3 + 2x + 5) + (6x 3 + 2x + 5) x (x4 3x 2 + 6x 3) = (x 4 3x 2 + 6x 3)(18x 2 + 2) + (6x 3 + 2x + 5)(4x 3 6x + 6). Of course, in tis example, f is just a polynomial so we coul also fin f by multiplying out te two factors of f an ifferentiating te polynomial term by term as usual. However,
8 8 Differentiation of Polynomials an Rational Functions Section 3.3 te prouct rule gives us a quicker route to te erivative. Altoug te result is not simplifie into te stanar form of a polynomial, for most applications tis form is just as useful as any oter. It is wort noting tat altoug we can now ifferentiate any rational function in teory, in practice our metos may not be te most useful. For example, te function f(x) = (x 2 + 1) 567 is a polynomial, an so we know ow to ifferentiate it. However, at tis point te only way we coul perform te ifferentiation woul be to expan f(x) into stanar polynomial form an ten ifferentiate term by term. In Section 3.4 we will learn ow to anle tis problem more irectly. At te same time we will exten te class of functions tat we can ifferentiate routinely to inclue all algebraic functions. Problems 1. Fin te erivative of eac of te following functions. (a) f(x) = x 3 + 6x (b) = 13x 5 6x (c) g(t) = 3t 6t 2 () y(t) = 4t 3 18t + 3 (e) f(t) = (3t 6) 2 (f) f(x) = (4x + 5)(6x 2 1) 2. Fin te erivative of eac of te following functions. (a) f(x) = (2x + 1) 2 (b) g(t) = (t 2 3) 3 (c) = x 3 2x + 5 (e) f(t) = 3t4 8t + 1 2t (g) (t) = 3 t (i) (z) = 8z 3 1 2z () (s) = 2s s2 s (f) x(t) = 3 t 3 16t2 () f(x) = 41 3x 7 (j) f(s) = 31 s s 2 16s 3. For eac of te following, make use of te prouct rule in fining te erivative of te epenent variable wit respect to te inepenent variable. (a) s = (t 2 6t + 3)(8t 4 + 6t 2 7) (b) q = (13t 4 + 5t)(3t 5 + 4t t 31) (c) y = (x 2 2x + 3)(2x x 6)(3x 2 4x + 1) () z = (x2 3x + 6)(8x 2 + 3x 2) x 2 6
9 Section 3.3 Differentiation of Polynomials an Rational Functions 9 4. Suppose f(2) = 2, f (2) = 6, g(2) = 3, an g (2) = 4. Fin k (2) for eac of te following. (a) k(x) = f(x) (b) k(x) = f(x) (c) k(x) = f(x)() 2 f(x) f(x) () k(x) = 5. Suppose an object moves along te x-axis so tat its position at time t is x = t + t3 6. (a) Fin te velocity, v(t) = ẋ(t), of te object. (b) Wat is v(0)? Wat oes tis say about te irection of motion of te object at time t = 0? (c) Wen is te object at te origin? Wat is te velocity of te object wen it is at te origin? () For wat values of t is te object moving towar te rigt? (e) For wat values of t is te object moving towar te left? (f) Wat is appening at te points were v(t) = 0? (g) Fin te acceleration of te object, a(t) = v(t). () Wen is te acceleration positive? Wen is it negative? (i) Notice tat v(1) < 0 an a(1) > 0. Wat oes tis say about te motion at time t = 1? 6. (a) Using only te prouct rule an te fact tat x = 1, sow tat x x x2 = 2x. (b) Now use te prouct rule to sow tat x x3 = 3x 2. (c) Let n > 1 an suppose we know tat x xm = mx m 1 for all m < n. Use te prouct rule to sow tat x xn = nx n 1.
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### Math 230.01, Fall 2012: HW 1 Solutions
Math 3., Fall : HW Solutions Problem (p.9 #). Suppose a wor is picke at ranom from this sentence. Fin: a) the chance the wor has at least letters; SOLUTION: All wors are equally likely to be chosen. The
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Sheets Algebra Cheat Sheets provide you with a tool for teaching your students note-taking, problem-solving, and organizational skills in the context of algebra lessons. These sheets teach the concepts
### f(x) f(a) x a Our intuition tells us that the slope of the tangent line to the curve at the point P is m P Q =
Lecture 6 : Derivatives and Rates of Cange In tis section we return to te problem of finding te equation of a tangent line to a curve, y f(x) If P (a, f(a)) is a point on te curve y f(x) and Q(x, f(x))
### Option Pricing Using the Binomial Model
Finance 400 A. Penati - G. Pennacci Option Pricing Using te Binomial Moel Te Cox-Ross-Rubinstein (CRR) tecnique is useful for valuing relatively complicate options, suc as tose aving American (early exercise)
### The Derivative. Philippe B. Laval Kennesaw State University
The Derivative Philippe B. Laval Kennesaw State University Abstract This handout is a summary of the material students should know regarding the definition and computation of the derivative 1 Definition
### List the elements of the given set that are natural numbers, integers, rational numbers, and irrational numbers. (Enter your answers as commaseparated
MATH 142 Review #1 (4717995) Question 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 Description This is the review for Exam #1. Please work as many problems as possible
### Solution Derivations for Capa #7
Solution Derivations for Capa #7 1) Consider te beavior of te circuit, wen various values increase or decrease. (Select I-increases, D-decreases, If te first is I and te rest D, enter IDDDD). A) If R1
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Pressure Pressure is force per unit area: F P = A Pressure Te direction of te force exerted on an object by a fluid is toward te object and perpendicular to its surface. At a microscopic level, te force
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# How many solution sets do systems of linear inequalities have?
yamaguchityler | Certified Educator
Linear inequalities can either have no solution, one specific solution, or an infinite amount of solutions. Thus, the total possible would equal three.
For instance, say we have a variable x. Although we do not know what x is, we can determine it's value depending on what inequalities it has been placed next to.
For instance, say we have X>10. This would indicate that X is obviously a number greater than 10. It also cannot equal 10 because it is not a "greater than or equal to" notation. This simply says that X is larger than 10. Similarly, if we also have X> 15, we now also know that X is larger than 15 as well. However, if this is all we're given, then we will have no idea what X is. There is an infinite number of solutions because X could be any number larger than 15.
When there is no answer, say we have Y>9. We know that Y is not definitely above 9. However, say there is also Y<7. This is impossible and has no answer because Y cannot be both greater than 9, and above 7. Thus, Y is undefined and unknown.
Finally, say we have this 5<X<7. Assuming we are not taking decimals and fractions into place, this would leave us with an answer that X=6. Obviously this changes when taking into account decimals and proportions, however this is for example only. In this case, there is only one answer and X has to equal 6.
A system of linear inequalities can have none, one, or an infinite number of solutions; therefore, there are three.
In order for a number to be a solution to a linear inequality it must satisfy all linear inequalities. For example, in a system with two linear inequalities:
This has no solutions: x>7 and x<4 because a number cannot be greater than 7 AND less than 4
This has one solution: x>=7 and x<=7 because the only value that satisfies both inequalities is 7
This has an infinite number of solutions: x>1 and x>2 because all numbers greater than 2 satisfy both inequalities
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```M203E Practice Quiz 4 Solutions
[1] Say you roll a fair die twice.
(a) What's the chance that the first time the die comes up 2 or more?
There are 36 outcomes, 6 possibilities for the first roll, and for
each of those 6, there are 6 for the second roll. Five of the 6
outcomes of the first roll consist of the die coming up 2 or more,
so there are 5x6 outcomes for two rolls where the first roll is a 2
or more. Thus the chance of the first roll being a 2 or more is
5x6/(6x6) = 5/6.
(b) What's the chance that the first time the die comes up 2 and the
second time the die comes up 5?
The chance is 1/36.
(c) What's the chance that either the first time the die comes up 2
or the second time the die comes up 5?
Here are the 6 outcomes where you get a 2 on the first roll:
(2,1), (2,2), (2,3), (2,4), (2,5), (2,6).
Here are the 6 outcomes where you get a 5 on the second roll:
(1,5), (2,5), (3,5), (4,5), (5,5), (6,5)
Thus there are 11 outcomes where you get either a 2 on the first roll
or a 5 on the second roll (not 12, since you can't count (2,5) twice).
Thus the chance is 11/36.
[2] A bad of marbles has 3 red marbles and 5 blue marbles.
(a) If you reach into the bag and pick a marble at random,
what's the chance it is red?
The chance is 3/8.
(b) If you reach into the bag twice and pick a marble each time
(without replacing the marbles you pick), what's the chance
the marbles have different colors?
There are 8x7 ways to pick two marbles. Of these, there are 3x5 ways
for the first marble to be red and the second marble to be blue,
and there are 5x3 ways for the first marble to be blue and the
second marble to be red. Thus the chance is (3x5 + 5x3)/(8x7) = 30/56.
(c) If you reach into the bag and pick a marble, record the color,
and replace it back in the bag, and then reach into the bag and pick a
marble again, what's the chance now that the marbles have different colors?
There are now 8x8 ways to pick two marbles, because now you could pick
the same marble twice, but there are still just 30 ways, as before, to pick
marbles of different colors. So the chance is 30/64.
[3] A fisheries official tags and releases 100 fish into a pond.
After a short period of time (to allow the tagged fish to mix in
with the whole fish population in the pond), she catches a sample
of 20 fish, of which 4 are tagged. What does she estimate the
pond's fish population to be?
Experimentally the chance of a random fish having a tag is 4/20.
But if N is the total fish population, the mathematical chance
of a random fish having a tag is 100/N. Thus 4/20 = 100/N,
so N = 100x20/4 = 500.
[4] In February 2006 a group of eight people from Lincoln won a jackpot valued at
177 million dollars. In an Omaha World Herald column
(Feb 26, 2006), Harold Andersen (whom the journalism building here at
UNL is named after) wrote: ``The Lottery Commission's intent seems pretty obvious:
Encourage more gambling by Nebraskans, in spite of the fact that the law of
probabilities makes it less likely that a record-breaking Powerball payoff
will strike in Nebraska again.'' Is there such a law
of probabilities? If you flip a fair coin three times and it comes up
heads each time, is it more likely to come up heads or tails the
next time you flip it? Justify your answer to the second question
and then indicate whether you agree or disagree with Andersen's
assertion that a second payoff is less likely as a result of there
having been a payoff recently.
The probability of a fair coin coming up heads doesn't change, even if
it's come up heads several times recently. It's true that in the long run
we expect the average rate of heads to be about 50% (this is the law
of probabilities that Andersen is referring to, sometimes called the
Law of Large Numbers), but this is what naturally happens if heads
and tails are equally likely. There's nothing that forces some tails
to happen to make up for a few heads. For example, if half the population
has brown hair, and if one day you see several brown haired people
on the street, that doesn't mean that all of a sudden it becomes more
likely to see a person with blonde hair. Regarding the lottery payoff,
the chance of a win in Lincoln does not change just because Lincoln
has had a winner recently. In fact, after a big payoff in Lincoln,
the chance of another win in Lincoln is probably more than before,
because more people in Lincoln may buy tickets from the store that
sold the winner, on the false premise of that store being "lucky".
The more Lincolnites that buy tickets, the more likely it will be
for someone from Lincoln to win!
```
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## Using Graphs and Equations to Identify Proportional Relationships
We venture into the realm of proportional relationships, specifically focusing on how we can use graphs and equations to identify them. Grasping this concept is integral to expanding your understanding of mathematical relationships. So, let's get started!
### Step 1: Understanding Proportional Relationships
Before we start identifying proportional relationships, we need to understand what they are. A proportional relationship between two variables is one where the ratio of the two variables is constant. In other words, as one variable increases, the other increases at a consistent rate. In equation terms, we often express this as $$y=kx$$, where $$k$$ represents the constant of proportionality.
### Step 2: Identifying Proportional Relationships in Equations
When examining an equation, you can identify a proportional relationship if it takes the form $$y=kx$$, where $$k$$ is the constant of proportionality. For example, in the equation $$y=3x$$, the constant of proportionality is 3, indicating that $$y$$ is always three times the value of $$x$$.
### Step 3: Identifying Proportional Relationships in Graphs
When graphed, proportional relationships create a straight line that passes through the origin (0,0). This occurs because the constant ratio results in a constant slope, creating a linear graph. Therefore, if you see a graph where the line passes through the origin and is straight, you can conclude that it represents a proportional relationship.
### Step 4: Analyzing Real-World Scenarios
Applying this knowledge, you can analyze real-world scenarios and determine if they represent proportional relationships. For instance, if you're looking at a graph or equation describing the relationship between the price of an item and its quantity, and you notice the traits of a proportional relationship, you can conclude that the cost per unit is consistent.
### Step 5: Practice!
As with any math concept, practice is key. Try to identify proportional relationships in different equations and graphs. Also, create your own proportional relationships, graph them, and identify the constant of proportionality.
Remember, mathematics is a language, and each new concept you learn is like adding words to your mathematical vocabulary. Each new word increases your ability to understand, interpret, and describe the world around you. Happy learning!
### Example
Suppose you're working a part-time job that pays you $15 per hour. We can express this as a proportional relationship, where your total earnings (E) are proportional to the number of hours (h) you work. The constant of proportionality is your wage rate,$15/hour.
We can write this as an equation: $$E = 15h$$.
Now, let's graph this relationship. On the x-axis, we put the hours worked, and on the y-axis, we put the total earnings.
If you work for 0 hours, you earn $0. This gives us the point (0,0) on the graph. If you work for 1 hour, you earn$15. This gives us the point (1,15) on the graph. If you work for 2 hours, you earn $30. This gives us the point (2,30) on the graph. As you continue plotting these points and draw a line through them, you'll see that it's a straight line that passes through the origin (0,0). This is a key characteristic of proportional relationships. The slope of this line is the wage rate,$15/hour, which is the constant of proportionality in this relationship.
So, through both the equation and the graph, we've identified a proportional relationship between the hours worked and the total earnings.
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# Integration by Parts in Calculus
Examples with detailed solutions and exercises with answers on how to use the technique of integration by parts to find integrals are presented.
## Review Integration by Parts
The method of integration by parts may be used to easily integrate products of functions. The main idea of integration by parts starts the derivative of the product of two function u and v as given by Rewrite the above as Take the integral of both side of the above equation follows Noting that
the above is simplified to obtain the rule of integration by parts. Note that any choice of which function in the integral on the left is chosen as u and which is chosen as dv/dx must simplify the integral on the right of the above formula.
## Examples with Detailed Solutions
In what follows c is a constant of integration.
### Example 1
Evaluate the integral Solution to Example 1
3 is a constant and may therefore be taken out of the integrand outside the inetgral sign.
Let $$u = x$$ and $$\dfrac{dv}{dx} = e^x$$ , hence $$\dfrac{du}{dx} = 1$$ and $$v = \displaystyle \int e^x \; dx = e^x$$ and using the method of integration by parts, we have \begin{aligned} & \color{red}{\text{given integral}} \\[8pt] & \int 3 \; x \; e^x \; dx \\[15pt] & \color{red}{\text{constant $$3$$ out of integral}} \\[8pt] & = 3 \int x \; e^x \; dx \\[15pt] &\color{red}{\text{apply integration by parts}}\\[8pt] & = 3 \left( x \; e^x - \int 1 \cdot e^x \; dx \right) \\[15pt] &\color{red}{\text{evaluate integral and simplify to obtain the final answer}} \\[8pt] & \int 3 x e^x \; dx = 3 \; x \; e^x - 3 \; e^x + c \end{aligned}
### Example 2
Evaluate the integral $\int x \; \sin(x) \; dx$ Solution to Example 2
Let $$u = x$$ and $$\dfrac{dv}{dx} = \sin(x)$$, hence $$\dfrac{du}{dx} = 1$$ and $$v = - \cos(x)$$. Hence \begin{aligned} & \color{red}{\text{given integral}} \\[8pt] & \int x \; \sin(x) \; dx \\[15pt] &\color{red}{\text{integration by parts}}\\[8pt] & = x (-\cos(x)) - \int 1 \cdot (-cos(x)) \; dx \\[15pt] &\color{red}{\text{evaluate integral and simplify obtain the final answer}} \\[8pt] & \int x \; \sin(x) \; dx = - x \cos(x) + \sin(x) + c \end{aligned}
### Example 3
Calculate the integral $\int x^2 \; \cos x \; dx$ Solution to Example 3
Let $$u = x^2$$ and $$\dfrac{dv}{dx} = \cos(x)$$, hence $$\dfrac{du}{dx} = 2x$$ and $$v = sin(x)$$ and apply the integration by parts. $\int \; x^2 \cos(x) \; dx = x^2 \sin (x) - 2 \int x\; \sin (x) \; dx \qquad (I)$ We now need to apply the method of integration by parts to the integral $$\displaystyle \int x\; \sin (x) \; dx$$ . In example 2 we evaluated the integral $$\displaystyle \int x\; \sin (x) \; dx = - x \cos(x) + \sin(x)$$ and we may substitute this in the above integral. Hence the final result is given by \begin{aligned} & \color{red}{\text{right side of (I) above}} \\[8pt] & x^2 \sin (x) - 2 \int x\; \sin (x) \; dx \\[15pt] &\color{red}{\text{substitute integral on the right of (I)}} \\[8pt] & = x^2 \sin(x) - 2 (- x \cos(x) + \sin(x)) + c \\[15pt] &\color{red}{\text{simplify to obtain the final answer}} \\[8pt] & \int x^2 \; \cos x \; dx = x^2 \sin(x) + 2 x \cos(x) - 2 \sin(x) + c \end{aligned}
### Example 4
Evaluate the integral $\int x \; \ln x \; dx$ Solution to Example 4:
Let $$u = \ln(x)$$ and $$\dfrac{dv}{dx} = x$$ , hence $$\dfrac{du}{dx} = \dfrac{1}{x}$$and $$v = \dfrac{x^2}{2}$$ and apply the integration by parts. \begin{aligned} &\color{red}{\text{Given the integral}} \\[8pt] & \int x \; \ln x \; dx \\[15pt] & \color{red}{\text{integration by parts}} \\[8pt] & = \dfrac{x^2}{2} \; \ln x - \int \dfrac{x^2}{2} \; \dfrac{1}{x} \; dx \\[15pt] &\color{red}{\text{simplify integrand}} \\[8pt] & = \dfrac{x^2}{2} \; \ln x - \int \dfrac{x}{2} \; dx \\[15pt] &\color{red}{\text{evaluate integral to obtain the final answer }} \\[8pt] & \int x^2 \; \ln x \; dx = \dfrac{x^2}{2} \; \ln x - \dfrac{1}{4} \; x^2 + c \end{aligned}
### Example 5
Calculate the integral $\int x \; \cos \left(\dfrac{x}{3} \right) \; dx$ Solution to Example 5:
Let $$u = x$$ and $$\dfrac{dv}{dx} = \cos \left(\dfrac{x}{3} \right)$$, hence $$\dfrac{du}{dx} = 1$$ and $$v = 3 \sin \left(\frac{x}{3}\right)$$ \begin{aligned} &\color{red}{\text{Given the integral}} \\[8pt] & \int x \; \cos \left(\dfrac{x}{3} \right) \; dx \\[15pt] &\color{red}{\text{integration by parts}} \\[8pt] & = x \cdot 3 \sin \left(\dfrac{x}{3}\right) - \int 1 \cdot 3 \sin \left(\dfrac{x}{3}\right) dx \\[15pt] &\color{red}{\text{evaluate integral and simplify to obtain the final answer}} \\[8pt] & \int x \; \cos \left(\dfrac{x}{3} \right) \; dx = 3 x \sin\left(\dfrac{x}{3}\right) + 9 \cos\left(\dfrac{x}{3}\right) + c \end{aligned}
### Example 6
Use integration by parts to evaluate the integral $\int \ln x \; dx$ Solution to Example 6:
We first rewrite the integrand $$\ln(x)$$ as $$1 \cdot ln(x)$$
$\int \ln(x) dx = \int 1 \cdot ln(x) \;dx$ Let $$u = \ln(x)$$ and $$\dfrac{dv}{dx} = 1$$ , hence $$\dfrac{du}{dx} = \dfrac {1}{x}$$ and $$v = x$$. Using integration by parts, we obtain \begin{aligned} & \int 1 \cdot \ln(x) \;dx = x \ln(x) - \int x \cdot (1/x) dx \\[15pt] &\color{red}{\text{simplify the integrand on the right side}} \\[8pt] & = x \ln(x) - \int 1 \cdot dx \\[15pt] &\color{red}{\text{evaluate integral and simplify to obtain the final answer}} \\[8pt] & \int \ln(x) \; dx = x \ln(x) - x + c \end{aligned}
### Example 7
Use integration by parts to evaluate the integral $\int x^2 \; (\ln x)^2 \; dx$ Solution to Example 7:
Let $$\dfrac{dv}{dx} = x^2$$ and $$u = (ln(x))^2$$ hence $$v = \dfrac{x^3}{3}$$ and $$\dfrac{du}{dx} = 2 \dfrac{\ln x}{x}$$ and use integration by parts to write \begin{aligned} & \int x^2 \; (\ln x)^2 \; dx = \dfrac{x^3}{3} \; (\ln(x))^2 - \int \dfrac{x^3}{3} \left( 2 \dfrac{\ln x}{x} \right) dx \\[15pt] &\color{red}{\text{simplify the integrand on the right}} \\[8pt] & = \dfrac{x^3}{3} \; (\ln(x))^2 - \dfrac{2}{3} \int x^2 \; \ln x \; dx \\[15pt] & \color{red}{\text{Let $$\dfrac{dw}{dx} = x^2$$ and $$z = \ln(x)$$ hence $$w = \dfrac{x^3}{3}$$ and $$z' = \dfrac{1}{x}$$ and use the }} \\[8pt] & \color{red}{\text{ integration by parts one more time on the integral on the right}} \\[8pt] &= \dfrac{x^3}{3} (\ln(x))^2 - \dfrac{2}{3} \left(\dfrac{x^3}{3} \ln(x) - \int \dfrac{x^3}{3} \dfrac{1}{x} \; dx \right) \\[15pt] & \color{red}{\text{simplify the integrand on the right}} \\[8pt] &= \dfrac{x^3}{3} (\ln(x))^2 - \dfrac{2}{3} \left(\dfrac{x^3}{3} \ln(x) - \dfrac{1}{3} \int x^2 \; dx \right) \\[15pt] & \color{red}{\text{Expand and simplify}} \\[8pt] &= \dfrac{x^3}{3} (\ln(x))^2 - \dfrac{2}{9} x^3 \ln(x) + \dfrac{2}{9} \int x^2 \; dx \\[15pt] & \color{red}{\text{Integrate on the right to obtain the final answer}} \\[8pt] & \int x^2 \; (\ln x)^2 \; dx = (\ln(x))^2 \dfrac{x^3}{3} - \dfrac{2}{9}x^3 \ln(x) + \dfrac{2}{27} x^3 + c \\[15pt] \end{aligned}
### Example 8
Use integration by parts to evaluate the integral $\int e^x \sin (2x) dx$ Solution to Example 8:
Let $$I = \int e^x \sin (2x) \; dx$$ which is the integral to evaluate.
Let $$v = \sin(2x)$$ and $$\dfrac{du}{dx} = e^x$$ hence $$\dfrac{dv}{dx} = 2 \cos(2x)$$ and $$u = e^x$$.
Use integration by parts as follows \begin{aligned} & \int e^x \sin(2x) dx = e^x \sin(2x) - \int e^x \cdot 2 \cos(2x) dx \\[15pt] & \color{red}{\text{Take the constant $$2$$ out of the integral sign and rewrite as}} \\[8pt] & = e^x \sin(2x) - 2 \int e^x \cdot \cos(2x) dx \\[15pt] & \color{red}{\text{We now let $$w = \cos(2x)$$ and $$\dfrac{dz}{dx} = e^x$$ hence $$w'= -2 \sin(2x)$$ and $$z = e^x$$ and }} \\[8pt] & \color{red}{\text{use the integration by parts one more time to the integral on the right}}\\[8pt] & = e^x \sin(2x) - 2 \left(e^x \cdot \cos(2x) - \left(\int e^x \cdot (-2 \sin(2x)) dx \right)\right) \\[15pt] & \color{red}{\text{Expand and simplify}} \\[8pt] & = e^x \sin(2x) - 2 e^x \cdot \cos(2x) - 4 \left(\int e^x \cdot \sin (2x) \right) dx \\[15pt] & \color{red}{\text{Note that the integral on the right is the intergal $$I = \int e^x \cdot \sin (2x) dx$$}}\\[8pt] & \color{red}{\text{we are trying to evaluate , hence the above may be written as}}\\[8pt] & I = e^x \sin(2x) - 2 e^x \cdot \cos(2x) - 4 I \\[15pt] & \color{red}{\text{Solve the last equation for $$I$$ to obtain the final answer}} \\[8pt] & I = \int e^x \cdot \sin (2x) dx = \dfrac{e^x \sin(2x) - 2e^x \cos(2x)}{5} + c \end{aligned}
## Exercises
Use the table of integrals and the method of integration by parts to find the integrals below. [Note that you may need to use the method of integration by parts more than once].
1. $$\displaystyle \int x \; \cos(x) \; dx$$
2. $$\displaystyle \int x \; e^{2x} \; dx$$
3. $$\displaystyle \int x^{1/3} \; \ln x \; dx$$
4. $$\displaystyle \int \dfrac{\ln x}{x^2} \; dx$$
5. $$\displaystyle \int x^3 \; \cos x \; dx$$
6. $$\displaystyle \int x^2 \; e^{-3x} \; dx$$
7. $$\displaystyle \int e^x \; \cos(2 x) \; dx$$
1. $$x \sin(x) + \cos(x) + c$$
2. $$\dfrac{x}{2} e^{2x} - \dfrac{1}{4} e^{2x} + c$$
3. $$\dfrac{3}{4} x^{\frac{4}{3}} \ln(x) - \dfrac{9}{16} x^{\frac{4}{3}} + c$$
4. $$- \dfrac{\ln x}{x} - \dfrac{1}{x} + c$$
5. $$3(x^2 - 2) \cos(x) + (x^3 - 6 x) \sin(x) + c$$
6. $$- \dfrac{1}{27} ( 9 x^2 + 6 x + 2 ) e^{-3x} + c$$
7. $$\dfrac{1}{5} ( e^x \cos(2x) + 2 e^x \sin(2x) ) + c$$
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# Chess Board: How to find Number of Squares and Rectangles
Views:4104
###### This concept is important for CAT and MBA entrance exams.
We all have encountered a frequently asked question, "How many squares are there in a 8*8 chessboard?"
We usually think the answer is 8*8=64, right?
But in this case we have just counted 1*1 squares. What about the 2*2 squares, 3*3 squares, 4*4 squares and so on?
Number of 1*1 squares= 8*8=64
Number of 2*2 squares= 7*7=49
Number of 3*3 squares= 6*6=36
Number of 4*4 squares= 5*5=25
Number of 5*5 squares= 4*4=16
Number of 6*6 squares= 3*3=9
Number of 7*7 squares= 2*2=4
Number of 8*8 squares= 1*1=1
Total number of Squares= 82+72+62+...+22+12= 204
Can you see a pattern?
In a 8*8 chessboard, the total number of squares is ∑82
We can generalize this in the following way:
###### Total number of squares in a n*n chessboard will be = ∑n2; n varying from 1 to n.
Now let us calculate the number of rectangles in 8*8 chessboard.
A rectangle can have the following dimensions: 1*1, 1*2, 1*3, 1*4… 1*8, 2*2, 2*3, 2*4, …2*8, 3*3, 3*4, ….7*8, 8*8.
We can approach the problem in the following way:
To form a rectangle, we need 4 lines. (2 sets of parallel lines). We can select these lines by = 9C2 * 9C2 = 36*36=1296.
We can generalize this in the following way:
Total number of rectangles in a n*n chessboard will be
= n+1C2*n+1C2
Note: The squares will also be included in counting rectangles.
Calculating it every time is cumbersome. Also, we can be asked to calculate the number of squares and rectangles in a m*n board. In this case, the above-derived formulas won't work.
Having a standard formula to calculate the number of squares and rectangles in an m*n chessboard will simplify the problem.
Let us see how we can get a generalized formula to calculate the number of squares and rectangles in an m*n chessboard.
In a m*n board,
Total number of rectangles in a m*n board
= m+1C2*n+1C2
(A rectangle can be formed by selecting 2 lines from m+1 lines and 2 lines from n+1 lines) For example, number of rectangles in a 2*3 board will be =18
Total number of squares in a m*n board= ∑ (m*n); m, n varying from 1 to m,n respectively.
For example, number of squares in 2*3 board = 2*3+1*2=8
For your practice, you can calculate the number of squares and rectangles in a 6*7 board.
Let's solve some examples based on this concept.
Suggested Video :
Learn the Concept of Painted Sides of a Cube Agreement
###### Solved Examples
Example 1: In how many ways can you place 2 rooks on 8*8 chessboard such that they are not in attacking positions?
Solution:
Number of ways of selecting 1st rook= 64C1
Number of ways of selecting 2nd rook (it should not be in the same row or column) =7*7/2
Therefore, total number of ways= 64*7*7/2= 32*49=1568.
Example 2: If two squares are chosen on a 8*8 chessboard, what is the probability that they have one side in common?
Solution:
Total number of ways of selecting 2 squares= 64C2=2016
To count the favourable cases, we will consider 3 cases.
Case 1: The corner squares.
There are 4 corner squares. For each corner square, we can select the other square in 2 ways.
Therefore, for this cases, favourable cases= 4*2=8
Case 2: The squares on the edges.
There are 24 squares on the edges. For each such square, we can select the other square in 3 ways.
Therefore, for this cases, favourable cases= 24*3=72
Case 3: The inner squares.
There are 36 inner squares. For each such square, we can select the other square in 4 ways.
Therefore, for this case, favourable cases= 36*4=144
On adding these three cases we get 8+72+144=224
But this is not the final answer.
We have counted every case twice.
So total favourable cases= 224/2=112
The probability= 112/2016
Example 3: If two squares are chosen on a 8*8 chessboard, what is the probability that they have one vertex in common?
Solution:
Total number of ways of selecting 2 squares= 64C2= 2016
To count the favourable cases, we will consider 3 cases.
Case 1: The corner squares.
There are 4 corner squares. For each corner square, we can select the other square in 1 ways.
Therefore, for this cases, favourable cases= 4*1=4
Case 2: The squares on the edges.
There are 24 squares on the edges. For each such square, we can select the other square in 2 ways.
Therefore, for this cases, favourable cases= 24*2=48
Case 3: The inner squares
There are 36 inner squares. For each such square, we can select the other square in 4 ways.
Therefore, for this cases, favourable cases= 36*4=144
On adding these three cases we get 4+48+144=196
But this is not the final answer.
We have counted every case twice.
So total favourable cases= 196/2=98
The probability= 98/2016.
###### Key Learning:
Remember the formulas to calculate the number of squares and rectangles in a m*n board. It can be directly applied to questions on chessboard.
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# Writing recurring decimals as a fraction
When writing recurring decimal to fraction why do we move the decimal place equal to the number of digits in the repeating string? For example, to write $$0.10\overline{4357}$$ as a fraction, we let $$x= 0.10\overline{4357}$$, we then multiply both sides by $$10^4$$ because that's the number of digits in the repeating string ($$4357$$) to get $$10^4x=1043.574357...$$, we then subtract $$x$$ from $$10^{4}x$$ and $$0.1043574357....$$ to get $$9999x = 1043.47$$, we then multiply both sides by $$100$$ to get $$999900x=104347$$, then divide to get $$x = 104347/999900.$$ So, why does this work?
• Simply because using that multiplier lines up one of the decimals to the other so that all decimals beyond some place cancel out on subtraction. Then what's left is a terminating decimal which is easy to make into a rational. Sep 27, 2022 at 4:18
• If it's easier for you, notice that the decimal repeats after 6 digits, so multiply by $10^6$, we get $104,357.4357...$ now you can subtract off $10^2 \cdot 0.104357... = 10.4357...$ and we have $999,900x = 104,347$
– Igor
Sep 27, 2022 at 5:56
• Why not multpily with $100$ first (the number of decimal digits after the comma and before the period) ? Then , you have $100x=10+\frac{4357}{9999}=\frac{104347}{9999}$ Sep 27, 2022 at 7:21
$$0.4357435743574357\cdots = \frac{4357}{9999}. \tag1$$
Everything in the posted question can be inferred from (1) above. So, the question reduces to determining why (1) above is true.
(1) above may be re-stated as
$$\frac{4357}{10000} \times \left[1 + \frac{1}{(10000)^1} + \frac{1}{(10000)^2} + \frac{1}{(10000)^3} + \cdots \right]$$
$$= \frac{4357}{10000} \times \frac{1}{1 - \frac{1}{10000}} = \frac{4357}{10000} \times \frac{10000}{9999} = \frac{4357}{9999}. \tag2$$
In order to explain why (2) above is true, you have to understand that it is part of the following general formula for geometric series:
To simplify the situation, if $$0 < t < 1$$, then
$$1 + t + t^2 + t^3 + \cdots = \frac{1}{1 - t}. \tag3$$
That is, (2) is an immediate consequence of $$(3)$$, with $$t$$ set to the value $$~\dfrac{1}{10000}.$$
So, the posted question reduces to explaining why (3) is true, for any value of $$t$$ such that $$0 < t < 1.$$
This particular question has been beaten to death, many times on MathSE. I presume that you can find pertinent MathSE articles if you search on the string : "geometric series".
Alternatively, see the Wikipedia article on Geometric Series.
The Wikipedia article may be glossing over the fact that for any fixed constant $$t$$ such that $$0 < t < 1,$$ you have that
$$\lim_{n\to\infty} t^n = 0. \tag4$$
This is easily proven by realizing that since $$0 < t < 1,$$ you must have that $$\log(t)$$ is some fixed negative number. Therefore, $$\displaystyle \log\left[t^n\right] = n \times$$ some fixed negative number.
Therefore, the limit as $$n \to \infty$$ of $$\log\left[t^n\right]$$ must be $$-\infty$$.
This demonstrates that $$\lim_{n\to \infty} t^n = 0.$$
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# Pairs of Lines - Angles Made by a Transversal
## Notes
### 1. Pairs Of Angles:
#### a. Corresponding angles:
• Corresponding angles have distinct vertex points,
• Corresponding angles lie on the same side of the transversal and
• Corresponding angles one angle is interior and the other is exterior.
• Pairs of Corresponding angles:
### (i) Alternate interior angles:
• Alternate interior angles have different vertices
• Alternate interior angles are on opposite sides of the transversal and
• Alternate interior angles lie ‘between’ the two lines i.e., both angles are interior.
### (ii) Alternate exterior angles:
• Alternate exterior angles have different vertices
• Alternate exterior angles are on opposite sides of the transversal
• Alternate exterior angles are not lain ‘between’ the two lines i.e., both angles are exterior.
The Z-shape stands for alternate angles:
#### C. Consecutive Interior Angles:
• The pairs of angles on one side of the transversal but inside the two lines are called Consecutive Interior Angles.
• If the transversal cuts across parallel lines then the interior angles are supplementary (add to 180°) than you can conclude the lines are parallel.
### 2. Pairs of Lines - Angles Made by a Transversal:
• A transversal gives rise to several types of angles.
• Line l intersects lines m and n at points P and Q respectively. Therefore, line l is a transversal for lines m and n.
• ∠ 1, ∠ 2, ∠7, and ∠8 are called exterior angles, while ∠3, ∠4, ∠5, and ∠6 are called interior angles.
• Interior angles on the same side of the transversal are also referred to as consecutive interior angles or allied angles or co-interior angles.
(a) Corresponding angles:
(i) ∠ 1 and ∠ 5 (ii) ∠ 2 and ∠ 6
(iii) ∠ 4 and ∠ 8 (iv) ∠ 3 and ∠ 7
(b) Alternate interior angles:
(i) ∠ 4 and ∠ 6 (ii) ∠ 3 and ∠ 5
(c) Alternate exterior angles:
(i) ∠ 1 and ∠ 7 (ii) ∠ 2 and ∠ 8
(d) Interior angles on the same side of the transversal:
(i) ∠ 4 and ∠ 5 (ii) ∠ 3 and ∠ 6.
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# Mathematics Class 7 Chapter 2- Exercise 2.5 – Fractions and Decimals
1. Which is greater?
(i) 0.5 or 0.05
Solution:
We compare the tenth place digit:
So, 5 > 0
Hence, 0.5 > 0.05
(ii) 0.7 or 0.5
Solution:
We compare the tenths place digit:
so, 7 > 5
Hence, 0.7 > 0.5
(iii) 7 or 0.7
Solution:
We compare the one’s place digit:
so, 7 > 0
Hence, 7 > 0.7
(iv) 1.37 or 1.49
Solution:
we compare the tenths place digit:
So, 3 < 4
Hence, 1.37 < 1.49
(v) 2.03 or 2.30
Solution:
We compare tenths place digit:
So, 0 < 3
Hence, 2.03 < 2.30
(vi) 0.8 or 0.88
Solution:
we write it as 0.80 or 0.88
since here tenth place is the same
so, we compare the hundredth place:
we get 0 < 8
Hence, 0.80 < 0.88
2. Express as rupees as decimals:
(i) 7 paise
Solution:
(ii) 7 rupees 7 paise
Solution:
(iii) 77 rupees 77 paise
Solution:
(iv) 50 paise
Solution:
(v) 235 paise
Solution:
3. (i) Express 5 cm in meter and kilometer
Solution:
(ii) Express 35 mm in cm, m and km
Solution:
4. Express in kg:
(i) 200 g
Solution:
(ii) 3470 g
Solution:
(iii) 4 kg 8 g
Solution:
5. Write the following decimal numbers in the expanded form:
(i) 20.03
Solution:
(ii) 2.03
Solution:
(iii) 200.03
Solution:
(iv) 2.034
Solution:
6. Write the place value of 2 in the following decimal numbers:
(i) 2.56
Solution:
Place value of 2 in 2.56 is ones
(ii) 21.37
Solution:
Place value of 2 in 21.37 is tens
(iii) 10.25
Solution:
Place value of 2 in 10.25
2/10 = 0.2
So, place value of 2 is tenths
(iv) 9.42
Solution:
Place value of 2 in 9.42
2/100 = 0.02
So, place value of 2 is hundredths
(v) 63.352
Solution:
Place value of 2 in 63.352
2/1000 = 0.002
So, place value of 2 is thousandths
7. Dinesh went from place A to place B and from there to place C. A is 7.5 km from B and B is 12.7 km from C. Ayub went from place A to place D and from there to place C. D is 9.3 km from A and C is 11.8 km from D. Who travelled more and by how much?
Solution:
Given,
Distance from A to B = 7.5 km
Distance from B to C = 12.7 km
Distance travelled by Dinesh = AB + BC
= 7.5 km + 12.7 km
= 20.2 km
Now, Given
Distance from A to D = 9.3 km
Distance from D to C = 11.8 km
Dstance travelled by Ayub from A to C = AD + DC
= 9.3 km + 11.8 km
= 21.1 km
So, 21.1 > 20.2
= 21.1km – 20.2km
=0.9 km
Hence Ayub travelled more distance than Dinesh by 0.9km
8. Shyama bought 5 kg 300 g apples and 3 kg 250 g mangoes. Sarala bought 4 kg 800 g oranges and 4 kg 150 g bananas. Who bought more fruits?
Solution:
Given,
Shyama bought
Apple = 5 kg 300 g
Mangoes = 3 kg 250 g
Now, Total fruits bought by Shyama = 5 kg 300 g apple + 3 kg 250 g mangoes
= 5.300 kg apple + 3.250 kg mangoes (convert gram into kg)
=8.550 kg of fruits
Given,
Sarala bought
Oranges = 4 kg 800 g
Bananas = 4 kg 150 g
Now, Total fruits bought by Sarala = 4 kg 800 g oranges + 4 kg 150 g bananas
= 4.800 kg oranges + 4.150 kg bananas (convert gram into kg)
= 8.950 kg of fruits
So,
8.950 kg > 8.550 kg
Hence, Sarala bought more fruits than Shyama.
9. How much less is 28 km than 42.6 km?
Solution:
Given, 28 km < 42.6
Now,
= 42.6 km – 28 km
= 14.6 km
Hence, 28 km is les than 42.6 km by 14.6 km
error:
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# Math Snap
## $\begin{array}{l}\frac{d}{100}=\frac{l}{\omega} \text { is } \\ \frac{36}{100}=\frac{x}{90}\end{array}$
#### STEP 1
Assumptions1. We have two proportions given, $\frac{d}{100}=\frac{l}{\omega}$ and $\frac{36}{100}=\frac{x}{90}$. . We are assuming that these proportions are equal to each other.
3. We are looking to solve for the variable $x$.
#### STEP 2
First, we can set the two proportions equal to each other.
$\frac{d}{100}=\frac{l}{\omega} = \frac{36}{100}=\frac{x}{90}$
#### STEP 3
We can simplify this equation by focusing on the proportion we are interested in, which is $\frac{36}{100}=\frac{x}{90}$.
#### STEP 4
To solve for $x$, we can cross-multiply.
$x = \frac{36}{100} \times90$
##### SOLUTION
Now, we can calculate the value of $x$.
$x = \frac{36}{100} \times90 =32.4$So, $x =32.4$.
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How to Create a Figure for an Analytic Proof - dummies
# How to Create a Figure for an Analytic Proof
When you do an analytic proof, your first step is to draw a figure in the coordinate system and label its vertices. This figure will make the algebra part easier, when you have to prove something about the figure.
Here’s an example. Say you’re given the following proof: First, prove analytically that the midpoint of the hypotenuse of a right triangle is equidistant from the triangle’s three vertices, and then show analytically that the median to this midpoint divides the triangle into two triangles of equal area.
Your first step in an analytic proof is to draw a figure in the x-y coordinate system and give its vertices coordinates. You want to put the figure in a convenient position that makes the math work out easily. For example, sometimes putting one of the vertices of your figure at the origin, (0, 0), makes the math easy because adding and subtracting with zeros is so simple. Quadrant I is also a good choice because all coordinates are positive there.
The figure you draw has to represent a general class of shapes, so you make the coordinates letters that can take on any values. You can’t label the figure with numbers (except for using zero when you place a vertex at the origin or on the x– or y-axis) because that’d give the figure an exact size and shape—and then anything you proved would only apply to that particular shape rather than to an entire class of shapes.
Here’s how you create your figure for the triangle proof in this example:
• Choose a convenient position and orientation for the figure in the x-y coordinate system. Because the x– and y-axes form a right angle at the origin (0, 0), that’s the natural choice for the position of the right angle of the right triangle, with the legs of the triangle lying on the two axes. Then you have to decide which quadrant the triangle should go in. Unless you have some reason to pick a different quadrant, quadrant I is the best way to go.
• Choose suitable coordinates for the two vertices on the x– and y-axes. You’d often go with something like (a, 0) and (0, b) but here, because you’re going to end up dividing these coordinates by 2 when you use the midpoint formula, the math will be easier if you use (2a, 0) and (0, 2b). Otherwise, you have to deal with fractions. Egad! The figure shows the final diagram.
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<meta http-equiv="refresh" content="1; url=/nojavascript/"> Interpretation of Circle Graphs ( Read ) | Statistics | CK-12 Foundation
# Interpretation of Circle Graphs
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Best Score
Practice Interpretation of Circle Graphs
Best Score
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Interpretation of Circle Graphs
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The chart shows the Patrick family’s monthly budget. Make a circle graph to display their budget. Then find how much the family budgets for in each category if their monthly income is $2,500. Rent and utilities - 35% Food - 25% Savings - 10% Transportation - 15% Clothing - 10% Other - 5% What does the family budget look like? Can you interpret where and how they spend their money? This Concept is all about interpreting circle graphs. Use it to answer these questions at the end of the Concept. ### Guidance In a circle graph , the circle represents the whole. A circle graph can be used to compare the parts with the whole. It is a useful way to visually display data. Each of the parts is called a sector. Let’s look at how data is expressed in a circle graph. The circle graph shows how the students at Grandville Middle School voted for a school mascot. Which mascot got the least number of votes? Since there are no values shown on the circle graph, you need to visually determine which sector is the smallest. The smallest sector is the one for the Panther. The Panther got the least number of votes for mascot. The circle graph shows the method of transportation used by students to get to Grandville Middle School. What percent of the students use the school bus to get to school? The sectors of a circle graph must add up to 100%. Find the sum of the three given percents and then subtract from 100%. $15 \% + 20 \% + 25 \% &= 60 \%\\100 \% - 60 \% &= 40 \%$ 40% of the students use the school bus to get to school. This circle graph shows the results of a music survey. What fraction of the people surveyed said country was their favorite type of music? The fractions in the sectors of a circle graph must add up to 1. Find the sum of the three given fractions and then subtract from 1. $\frac{1}{4} + \frac{1}{2} + \frac{1}{10} & = \frac{5}{20} + \frac{10}{20} + \frac{2}{20} = \frac{17}{20}\\1 - \frac{17}{20} & = \frac{20}{20} - \frac{17}{20} = \frac{3}{20}$ $\frac{3}{20}$ of the people surveyed said country was their favorite type of music. We can show the same circle graph with its percents or with the measures of its central angles. The sum of the measures of the central angles in a circle graph is $360^ \circ$ . Notice that you can take the percent, change it to a decimal and multiply it by 360 to find the number of degrees. Check this out with the following examples. Use the two graphs above to answer the following questions. #### Example A Which number of degrees is equal to 40%? Solution: $144^\circ$ #### Example B True or false. 25% is the same as a $90^\circ$ angle? Solution: True #### Example C How many degrees is equal to 15%? Solution: $54^\circ$ Here is the original problem once again. The chart shows the Patrick family’s monthly budget. Make a circle graph to display their budget. Then find how much the family budgets for in each category if their monthly income is$2,500.
Rent and utilities - 35%
Food - 25%
Savings - 10%
Transportation - 15%
Clothing - 10%
Other - 5%
What does the family budget look like? Can you interpret where and how they spend their money?
Patrick's family spends most of their money on rent, utilities and food. They spend the least amount of money on other things and clothing. These are a couple of statements that we can make about their monthly budget.
### Vocabulary
Here are the vocabulary words in this Concept.
Circle Graph
a visual display of data in a circle. A circle graph is created from percentages with the entire circle representing the whole. The sectors of the circle graph are divided according to degrees which are created out of $360^\circ$ .
Sector
the section of a circle graph. Each section is known as a sector. Each sector can be measured in degrees and given a percentage.
### Guided Practice
Here is one for you to try on your own.
The table shows the results of the favorite school lunch of students in the seventh grade at Grandville Middle School. Make a circle graph for the results of the survey.
Favorite Food % of Students Surveyed
Pizza 30%
Grilled Cheese 35%
Hamburger 10%
Chicken fingers 25%
Step 1: Find the measure of the central angle by multiplying $360^\circ$ by the percent.
Favorite Food % of Students Surveyed Degrees in Central Angle
Pizza 30% 30% of $360^\circ = 0.30 \times 360^\circ = 108^\circ$
Grilled Cheese 35% 30% of $360^\circ = 0.35 \times 360^\circ = 126^\circ$
Hamburger 10% 10% of $360^\circ = 0.10 \times 360^\circ = 36^\circ$
Chicken fingers 25% 25% of $360^\circ = 0.25 \times 360^\circ = 90^\circ$
Step 2: Draw a circle with a compass. Draw one radius. Use that radius as a side of one central angle. Measure and draw the other central angles using a protractor.
Step 3: Label each sector with a title and percent and give a title to the entire circle graph.
What about if we have actual data?
If we have actual data, we first need to find the percent for each sector of the circle graph. Then we can find the measures of the central angles of the circle and make the circle graph.
### Video Review
Here is a video for review.
### Practice
Directions: Use the survey to answer each question.
A survey of 300 people asked them to name their favorite spectator sport. The results are shown in the circle graph below.
1. What was the most favorite spectator sport of the people surveyed?
2. What was the least favorite spectator sport of the people surveyed?
3. What percent of the people surveyed said that football was their favorite spectator sport?
4. How many people said that basketball was their favorite spectator sport?
5. How many more people said that soccer was their favorite sport than ice hockey?
6. The table shows the how much money the students in the seventh grade have raised so far for a class trip. Make a circle graph that shows the data.
Fundraiser Amount
Car wash $150 Book sale$175
Bake sale $100 Plant sale$75
7. Make a list of 5 popular ice cream flavors. Then survey your classmates asking them which of the 5 flavors is their favorite ice cream flavor. Use the data to make a circle graph.
8. Use a newspaper to locate a circle graph of some data. Then write five questions about the data.
This circle graph shows the results of a survey taken among students about their favorite school lunches. Use the graph to answer the following questions.
9. What percent of the students enjoy soup as a lunch?
10. What is the favorite choice of students for school lunch?
11. What is the least favorite choice?
12. What percent of the students enjoy salad?
13. What percent of the students did not choose salad as a favorite choice?
14. What percent of the students chose either pizza or tacos as their favorite choice?
15. What percent of the students chose chicken sandwich and pizza as their favorite choice?
16. What percent of the students did not choose chicken or pizza?
17. What is your favorite choice for lunch?
18. If you could add a food choice to this survey, what would it be?
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# 3.1 – The Pythagorean Theorem
## Objectives
• Derive the Pythagorean theorem $a^2+b^2=c^2$ from the relationship between right triangles and squares.
• Use the Pythagorean theorem to find the values of missing sides in a right triangle.
• Prove that triangles are right triangles using the converse of the Pythagorean theorem.
• Define and give examples of Pythagorean triples.
• Use the Pythagorean theorem to solve real-world problems.
## Key Terms
• Hypotenuse – The side opposite from the right angle in a right triangle.
• It is the triangle’s longest side.
• Leg – Either of the two shorter sides of a right triangle.
• Always touches the right angle.
• Pythagorean Theorem – The theorem that relates the side lengths of a right triangle.
• The theorem states that the square of the hypotenuse equals the sum of the squares of the legs: $a^2+b^2=c^2$
• Pythagorean Triple – A set of three whole numbers, a, b, and c, that satisfies the equation $a^2+b^2=c^2$.
• If the side lengths of a triangle form a Pythagorean triple, it is a right triangle.
## Notes
Right Triangles
• Right Triangles are:
• Triangles with one right angle (an angle measuring 90°).
• The other two angles in a right triangle are always acute (less than 90°).
• Remember: the area of a square is $(side)^2$
Pythagorean Theorem
• Pythagorean Theorem
• Pythagorean theorem: If a triangle is a right triangle, then the square of its longest side equals the sum of the squares of its other two sides.
• Converse: If the square of the longest side of a triangle equals the sum of the squares of its other two sides, then the triangle is a right triangle
• Solve for a missing side of a right triangle
• 2. Plug in the side lengths you know.
• 3. Solve the equation for the unknown side length
• Examples
• Tests for Right Triangles
• Common Pythagorean Triples
• How to Find a Pythagorean Triple:
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# What is the common ratio of the geometric sequence: 128, -16, 2, -1/4, …?
Nov 16, 2015
$r = \frac{1}{8}$
Explanation also solves the rest of the sequence structure!
#### Explanation:
where a_1=128 ;color(white)(xx) a_2 = (-16) ; color(white)(xx)a_3=2 color(white)(xx)and a_4=(-1/4)...
Let any term in this sequence be ${a}_{i}$
Let a constant be k
let the ratio be r
Two points to note:
Point 1:
it is reducing so $r < 1$
Point 2:
The sequence alternates positive to negative so involves ${\left(- 1\right)}^{f \left(i\right)}$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Consider the alternating +ve and -ve
If ${a}_{i} = {a}_{1}$ then we have${\left(- 1\right)}^{i + 1} k {r}^{i} = {\left(- 1\right)}^{2} k {r}^{1} = 128$
if${a}_{i} = {a}_{2}$ then we have ${\left(- 1\right)}^{i + 1} k {r}^{i} = {\left(- 1\right)}^{3} k {r}^{2} = \left(- 16\right)$
To find r ignore the ${\left(- 1\right)}^{i + 1}$ part remembering that this alternates the +ve and -ve. Thus the solution for r will be +ve
$\frac{k {r}^{2}}{k {r}^{1}} = r = \frac{16}{128} = \frac{1}{8}$
You are not asked to find the value of k.
But you could by
${\left(- 1\right)}^{2} k {\left(\frac{1}{8}\right)}^{1} = 128$
so $k = 128 \times \frac{8}{1} \div i \mathrm{de} 1 = 1024$
for a quick check I select ${a}_{3}$
a_3 = 2 -> ? [ (-1)^(3+1) (1024) times (1/8)^3 ] = 2
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A matrix is an ordered array of numbers. The order of a matrix is stated as the number of rows by the number of columns. For example, a matrix with 3 rows and 4 columns has an order of 3 x 4. If we have a matrix A, then a1,3 would be the element in A that is in the first row and third column. Two matrices are equal if and only if they have the same order or size and each corresponding element is equal.
Test Objectives
• Demonstrate the ability to find the order of a matrix
• Demonstrate the ability to find a specific entry in a matrix
• Demonstrate the ability to solve a matrix equation
Introduction to Matrices Practice Test:
#1:
Instructions: state the order.
$$a)\hspace{.2em}$$ $$\left[ \begin{array}{ccc}5 & -1 & 3\\ 3 & 7 & 0\end{array}\right]$$
$$b)\hspace{.2em}$$ $$\left[ \begin{array}{ccc}-9 & 19 & 17\\ 2 & 3 & 0 \\ 1 & -1 & 5\end{array}\right]$$
#2:
Instructions: state the order.
$$a)\hspace{.2em}$$ $$A=\left[ \begin{array}{ccccc}5 & 2 & 19 & 13 & 22 \\ -17 & -4 & 8 & 29 & 1 \\ -9 & -7 & -3 & 44 & 50\end{array}\right]$$
Instructions: what is element: a1,2
$$b)\hspace{.2em}$$ $$A=\left[ \begin{array}{cc}5 & 6 \\ 1 & -1 \\ 7 & 3\end{array}\right]$$
#3:
Instructions: what is element: a2,3
$$a)\hspace{.2em}$$ $$A=\left[ \begin{array}{cc}5 & 6 \\ 1 & -1 \\ 7 & 3\end{array}\right]$$
b) What type of matrix has the same number of rows as columns?
#4:
Instructions: if A = B, what is the value of x?
$$a)\hspace{.2em}$$ $$A=\left[ \begin{array}{cc}1 & 5 \\ 6 & x\end{array}\right]$$ $$B=\left[ \begin{array}{cc}1 & 5 \\ 6 & 7\end{array}\right]$$
Instructions: if A = B, what is the value of x and y?
$$b)\hspace{.2em}$$
$$A=\left[ \begin{array}{cc}-3 & 1 \\ 6 & x + 5\end{array}\right]$$ $$B=\left[ \begin{array}{cc}y + 2 & 1 \\ 6 & -2\end{array}\right]$$
#5:
Instructions: find x, y, and z such that A = B.
$$a)\hspace{.2em}$$
$$A=\left[ \begin{array}{ccc}-1 & 5 & 9 \\ x - 1 & y + 2 & 4\end{array}\right]$$ $$B=\left[ \begin{array}{ccc}-1 & 5 & 9 \\ 13 & -21 & z + 4\end{array}\right]$$
$$b)\hspace{.2em}$$
$$A=\left[ \begin{array}{ccc}3 & 5 & 11 \\ x + 1 & y - 7 & 7\end{array}\right]$$ $$B=\left[ \begin{array}{ccc}3 & 11 & 5 \\ -14 & 12 & z - 9\end{array}\right]$$
Written Solutions:
#1:
Solutions:
a) 2 x 3
b) 3 x 3
#2:
Solutions:
a) 3 x 5
b) 6
#3:
Solutions:
a) Doesn't exist
b) Square Matrix
#4:
Solutions:
$$a)\hspace{.2em}7$$
$$b)\hspace{.2em}x=-7, y=-5$$
#5:
Solutions:
$$a)\hspace{.2em}x=14, y=-23, z=0$$
$$b)\hspace{.2em}A ≠ B$$
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# Lesson 9: The Distributive Property, Part 1
Let's use the distributive property to make calculating easier.
## 9.1: Number Talk: Ways to Multiply
Find each product mentally.
$5 \boldcdot 102$
$5 \boldcdot 98$
$5 \boldcdot 999$
## 9.2: Ways to Represent Area of a Rectangle
1. Select all the expressions that represent the area of the large, outer rectangle in figure A. Explain your reasoning.
• $6 + 3 + 2$
• $6 \boldcdot 3 + 6 \boldcdot 2$
• $6 \boldcdot 3 + 2$
• $6 \boldcdot 5$
• $6 (3+2)$
• $6 \boldcdot 3 \boldcdot 2$
2. Select all the expressions that represent the area of the shaded rectangle on the left side of figure B. Explain your reasoning.
• $4 \boldcdot 7 + 4 \boldcdot 2$
• $4 \boldcdot 7 \boldcdot 2$
• $4 \boldcdot 5$
• $4 \boldcdot 7 - 4 \boldcdot 2$
• $4(7-2)$
• $4(7+2)$
• $4 \boldcdot 2 - 4 \boldcdot 7$
## 9.3: Distributive Practice
Complete the table. If you get stuck, skip an entry and come back to it, or consider drawing a diagram of two rectangles that share a side.
column 1 column 2 column 3 column 4 value
row 1 $5 \boldcdot 98$ $5 (100-2)$ $5 \boldcdot 100 - 5 \boldcdot 2$ $500 - 10$ 490
row 2 $33 \boldcdot 12$ $33 (10 + 2)$
row 3 $3 \boldcdot 10 - 3 \boldcdot 4$ $30-12$
row 4 $100 (0.04 + 0.06)$
row 5 $8 \boldcdot \frac 1 2 + 8 \boldcdot \frac 1 4$
row 6 $9 + 12$
row 7 $24 - 16$
## Summary
When we need to do mental calculations, we often come up with ways to make the calculation easier to do mentally.
Suppose we are grocery shopping and need to know how much it will cost to buy 5 cans of beans at 79 cents a can. We may calculate mentally in this way: $$5\boldcdot {79}$$ $$5\boldcdot {70}+5\boldcdot {9}$$ $$350+45$$ $$395$$
In general, when we multiply two numbers (or factors), we can break up one of the factors into parts, multiply each part by the other factor, and then add the products. The result will be the same as the product of the two original factors.
When we break up one of the factors and multiply the parts we are using the distributive property.
The distributive property also works with subtraction. Here is another way to find $5\boldcdot 79$: $$5\boldcdot 79$$ $$5\boldcdot {(80-1)}$$ $$400-5$$ $$395$$
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Lesson 8
Relating Area to Circumference
8.1: Irrigating a Field (5 minutes)
Warm-up
The purpose of this activity is for students to estimate the area of a circle by comparing it to a surrounding square.
Launch
Explain that some farms have circular fields because they use center-pivot irrigation. If desired, display these images to familiarize students with the context.
Provide quiet think time followed by whole-group discussion.
Student Facing
A circular field is set into a square with an 800 m side length. Estimate the field's area.
Anticipated Misconceptions
Students might think the answer should be 640,000 m2 because that is the area of the square, not realizing that they are being asked to find the area of a circle. Ask them what shape is the region where the plants are growing.
Some students might incorrectly calculate the area of the square to be 6,400 m2 and therefore estimate that the circle would be about 5,000 m2.
Some students might try to use what they learned in the previous lessons about the relationship between the area of a circle and the area of a square with side length equal to the circle's radius. Point out that the question is asking for an estimate and answer choices all differ by a factor of 10.
Activity Synthesis
Discuss the estimation strategies students used to answer the question. Ask students what the area of the square is in square meters ($$800 \boldcdot 800$$, or 640,000). Ask them if the circle's area is greater than or less than the square's area (less). Then ask them to use the picture to determine the best estimate (500,000 since the circle is close in area to the square).
8.2: Making a Polygon out of a Circle (20 minutes)
Activity
The purpose of this activity is for students to use what they know about finding the area of a parallelogram to develop the formula for the area of a circle. This activity builds on the work students did in grade 6 when cutting and rearranging shapes in order to calculate their areas. In this activity, students cut and rearrange parts of a circle to approximate a parallelogram. They see that the area of the parallelogram would be calculated by multiplying half of the circle’s circumference times its radius. Since students are not familiar with the process of writing proofs, it is necessary to walk them through writing the justification that uses the formula for the area of the parallelogram to develop the formula for the area of the circle.
The construction in this activity shows that the constructed parallelogram has a height at most the radius of the circle and a base at most half the circumference of the circle. Establishing equality is beyond grade level and will be addressed again in high school.
The process used to decompose the circle and recompose it into a shape resembling a parallelogram is a good example of MP8. The pie shaped wedges are successively cut in half and rearranged. Each time, the sides of the shape look more like line segments.
Watch for students who identify that the rearranged circle pieces resemble a parallelogram as the pieces get smaller. Also watch for how they estimate the width and height of this parallelogram and invite them to share during the discussion.
Launch
Arrange students in groups of 2. Each group needs a circular object, with a diameter between 3 and 5 inches, and a thick marker with which to trace it. Also provide each group a sheet of white paper, a sheet of colored paper, a pair of scissors, and glue or tape. Remind students that in the past they decomposed and rearranged a shape to figure out its area. Demonstrate how to do the first 4 steps of the activity, and invite students to follow along with your example. Ask how the area of the new shape differs from that of the circle. Solicit some ideas on what the new shape resembles and how the area of such a shape could be approximated. Without resolving this, ask students to continue the process.
Engagement: Develop Effort and Persistence. Provide prompts, reminders, guides, rubrics, or checklists that focus on increasing the length of on-task orientation in the face of distractions. For example, provide students with a task checklist which makes all the required components of the visual display explicit.
Supports accessibility for: Attention; Social-emotional skills
Conversing, Reading: MLR2 Collect and Display. As students work in groups to make sense of the shapes glued on the paper, circulate and listen to the language students use as they compare the shapes and discuss how to find the area of the shape that resembles a parallelogram. Write down the words and phrases students use to explain why the areas of both shapes are equal and why the area of the shape is half of the circumference multiplied by the radius. As students review the language collected in the visual display, encourage students to clarify the meaning of a word or phrase. For example, a phrase such as “the base is half of the circle” can be clarified with the phrase, “the base of the parallelogram is half of the circumference of the circle.” A phrase such as, “the height is the radius” can be clarified with the phrase, “the height of the parallelogram is equal to the radius of the circle.” This routine will provide feedback to students in a way that supports sense-making while simultaneously increasing meta-awareness of language.
Design Principle(s): Support sense-making; Maximize meta-awareness
Student Facing
Your teacher will give you a circular object, a marker, and two pieces of paper of different colors.
Follow these instructions to create a visual display:
1. Using a thick marker, trace your circle in two separate places on the same piece of paper.
2. Cut out both circles, cutting around the marker line.
3. Fold and cut one of the circles into fourths.
4. Arrange the fourths so that straight sides are next to each other, but the curved edges are alternately on top and on bottom. Pause here so your teacher can review your work.
5. Fold and cut the fourths in half to make eighths. Arrange the eighths next to each other, like you did with the fourths.
6. If your pieces are still large enough, repeat the previous step to make sixteenths.
7. Glue the remaining circle and the new shape onto a piece of paper that is a different color.
1. How do the areas of the two shapes compare?
2. What polygon does the shape made of the circle pieces most resemble?
3. How could you find the area of this polygon?
Anticipated Misconceptions
Students might not fold the wedges accurately or make a straight cut. Remind them that the halves must be equal.
Activity Synthesis
Ask students which polygon resembles the shape they constructed of circle pieces and how they know. (A parallelogram or rectangle)
For a dynamic visualization, see http://ggbm.at/RUqSMrjn, created in GeoGebra by Malin Christersson, or display this image.
Ask, “If we could continue cutting the wedges in half, how would that affect the new shape?” (It would look even more like a parallelogram or rectangle. The bumpy top and bottom straighten out, and the slanted height becomes more vertical.)
Note that we are going to refer to the bumpy parallelogram-ish shape as the “parallelogram” (in quotes). Ask students to describe comparisons we can make between the measurements in the circle and the “parallelogram.” Record and display these ideas for all to see. It may be helpful to write over the actual images themselves. Students should notice the following measurements, but if they do not, prompt them to look for them:
• The base of the “parallelogram” is approximately equal to half of the circle’s circumference.
• The height of the “parallelogram” is approximately equal to the radius of the circle.
• The areas of the 2 shapes are equal.
Tell students to label these measurements on their visual display:
• $$\text{Circumference} = \pi d$$” around the circle
• $$\frac12 \text{Circumference} = \pi r$$” at the base of the “parallelogram”
• “Radius” on the radius of the circle (needs to be drawn in)
• “Radius” on the height of the “parallelogram” (needs to be drawn in)
If students struggle to understand these relationships through the abstract variables, consider measuring your example circle and using the numerical measurements to talk about these relationships.
Ask students to discuss the different ways we can calculate the area of the “parallelogram.” Students should share the following ways:
• $$\text{Area} = \text{Base} \boldcdot \text{Height}$$
• $$\text{Area} = \frac12 \text{Circumference} \boldcdot \text{Radius}$$
• $$\text{A} = \pi r \boldcdot r$$
• $$\text{A} = \pi r^2$$
Tell students to record these ideas on their visual display. Display student work for all to see. If students do not bring up one of these ideas, make it explicit in the discussion.
8.3: Making Another Polygon out of a Circle (10 minutes)
Optional activity
The purpose of this activity is for students to consider a different way to cut and reassemble a circle into something resembling a polygon in order to calculate its area. This time the polygon is a triangle, but the area of the circle can still be found by multiplying $$\frac12$$ times the circumference times the radius.
In the previous activity, students had experience following along as the teacher developed the justification. This time give students the opportunity to write their own justification for the area of a circle. As students work, monitor and select students who have clear, but different, explanations to share during the whole-group discussion. In particular, select students who use the following steps:
• $$\text{Area} = \frac12 \boldcdot \text{base} \boldcdot \text{height}$$
• $$\text{Area} = \frac12 \boldcdot \text{circumference} \boldcdot \text{radius}$$
• $$\text{Area} = \frac12 \boldcdot (\pi d) \boldcdot r$$
• $$\text{Area} = \pi r \boldcdot r$$
• $$\text{Area} = \pi r^2$$
If the bands making up the circle really did not stretch, then they would not form rectangles when they are unwound because the circumference of the inner circle is not the same as the circumference of the outer circle in each band. A rectangle is an appropriate approximation for the shape in terms of calculating its area.
Launch
Give students quiet work time followed by partner and whole-class discussion. Applet created in GeoGebra by timteachesmath.
Representation: Internalize Comprehension. Activate or supply background knowledge about finding the area of triangles and circles. Allow students to use calculators to ensure inclusive participation in the activity.
Supports accessibility for: Memory; Conceptual processing
Student Facing
Imagine a circle made of rings that can bend but not stretch.
Watch the animation.
1. What polygon does the new shape resemble?
2. How does the area of the polygon compare to the area of the circle?
3. How can you find the area of the polygon?
4. Show, in detailed steps, how you could find the polygon’s area in terms of the circle’s measurements. Show your thinking. Organize it so it can be followed by others.
5. After you finish, trade papers with a partner and check each other’s work. If you disagree, work to reach an agreement. Discuss:
• Do you agree or disagree with each step?
• Is there a way they can make the explanation clearer?
Launch
Give students quiet work time followed by partner and whole-class discussion.
Representation: Internalize Comprehension. Activate or supply background knowledge about finding the area of triangles and circles. Allow students to use calculators to ensure inclusive participation in the activity.
Supports accessibility for: Memory; Conceptual processing
Student Facing
Imagine a circle made of rings that can bend, but not stretch.
1. What polygon does the new shape resemble?
2. How does the area of the polygon compare to the area of the circle?
3. How can you find the area of the polygon?
4. Show, in detailed steps, how you could find the polygon’s area in terms of the circle’s measurements. Show your thinking. Organize it so it can be followed by others.
5. After you finish, trade papers with a partner and check each other’s work. If you disagree, work to reach an agreement. Discuss:
• Do you agree or disagree with each step?
• Is there a way to make the explanation clearer?
Anticipated Misconceptions
If students struggle to imagine the circle and how it is cut and rearranged, suggest a familiar material for the rings that bends but does not stretch (for example, a cord or chain).
Activity Synthesis
Ask selected students to explain their steps for finding the area in terms of the circle’s measures. Ask the class whether they agree, disagree, or have questions after each student shares their reasoning.
Writing, Speaking, Listening: MLR1 Stronger and Clearer Each Time. After students have the opportunity to think about how to find the area of the triangle in terms of the circle’s measurements, ask students to write a brief explanation of their process. As students prepare their explanation, look for students who state that the area of the shape that resembles a triangle is half of the circumference multiplied by the radius of the circle. Ask each student to meet with 2–3 other partners for feedback. Provide students with prompts for feedback that will help them strengthen their ideas and clarify their language (e.g., “Can you explain how…,” “You should expand on...,” etc.). Students can borrow ideas and language from each partner to refine and clarify their original explanation. This will help students revise and refine both their ideas and their verbal and written output.
Design Principles(s): Optimize output (for explanation); Maximize meta-awareness
8.4: Tiling a Table (5 minutes)
Activity
The purpose of this activity is for students to apply the formula for area of a circle to solve a problem in context. The diameter of the circle is given, so students must first determine the radius.
Launch
Display this image of table top for all to see. Ask students, “What do you notice? What do you wonder?”
Quiet work time followed by a whole-class discussion.
Student Facing
Elena wants to tile the top of a circular table. The diameter of the table top is 28 inches. What is its area?
Student Facing
A box contains 20 square tiles that are 2 inches on each side. How many boxes of tiles will Elena need to tile the table?
Anticipated Misconceptions
Students may square the diameter, forgetting that they need to determine the radius first.
Activity Synthesis
Invite students to share their strategies for finding the area of the tabletop. After 3 students have shared their strategies, ask the class what formula all of the students used for finding the area of a circle. Record and display this formula, $$A = \pi r^2$$, for all to see. Here, $$A$$ is the area of the circle, and $$r$$ is the radius of the circle.
Engagement: Develop Effort and Persistence. Encourage and support opportunities for peer interactions. Prior to the whole-class discussion, invite students to share their work with a partner. Display sentence frames to support student conversation such as: “First, I _____ because….”, “I noticed _____ so I….”, “Why did you...?”, and “I agree/disagree because….”
Supports accessibility for: Language; Social-emotional skills
Lesson Synthesis
Lesson Synthesis
The main ideas are:
• We can find the area of a circle if we know the radius or the diameter.
• We know that the radius is half the length of the diameter.
• The formula for finding area of a circle is $$A=\pi r^2$$.
Discussion Questions:
• How would you find the area of a circle with a radius of 10? (Multiply $$\pi$$ times 100, because $$10^2 = 100$$.)
• How would you find the area of a circle with a diameter of 10? (Multiply $$\pi$$ times 25, because $$10 \div 2 = 5$$ and $$5^2 = 25$$.)
Student Lesson Summary
Student Facing
If $$C$$ is a circle’s circumference and $$r$$ is its radius, then $$C=2\pi r$$. The area of a circle can be found by taking the product of half the circumference and the radius.
If $$A$$ is the area of the circle, this gives the equation:
$$A = \frac12 (2\pi r) \boldcdot r$$
This equation can be rewritten as:
$$A=\pi r^2$$
(Remember that when we have $$r \boldcdot r$$ we can write $$r^2$$ and we can say “$$r$$ squared.”)
This means that if we know the radius, we can find the area. For example, if a circle has radius 10 cm, then the area is about $$(3.14) \boldcdot 100$$ which is 314 cm2.
If we know the diameter, we can figure out the radius, and then we can find the area. For example, if a circle has a diameter of 30 ft, then the radius is 15 ft, and the area is about $$(3.14) \boldcdot 225$$ which is approximately 707 ft2.
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This Homework answers app provides step-by-step instructions for solving all math problems. We will also look at some example problems and how to approach them.
## The Best Homework answers app
One instrument that can be used is Homework answers app. First, it is important to read the problem carefully and identify the key information. Second, students should consider what type of operation they need to use to solve the problem. Third, they should work through the problem step-by-step, using each piece of information only once. By following these steps, students will be better prepared to tackle even the most challenging math problems.
However, more often than not, we need to solve a system of equations in order to find all of the unknown values. Fortunately, there are a variety of methods that we can use to solve systems of equations, including elimination and substitution. With a little practice, solving algebra problems can be easy and even fun!
Algebra is the branch of mathematics that deals with the solution of equations. In an equation, the unknown quantity is represented by a letter, usually x. The object of algebra is to find the value of x that will make the equation true. For example, in the equation 2x + 3 = 7, the value of x that makes the equation true is 2. To solve an equation, one must first understand what each term in the equation represents. In the equation 2x + 3 = 7, the term 2x represents twice the value of x; in other words, it represents two times whatever number is assigned to x. The term 3 represents three units, nothing more and nothing less. The equal sign (=) means that what follows on the left-hand side of the sign is equal to what follows on the right-hand side. Therefore, in this equation, 2x + 3 is equal to 7. To solve for x, one must determine what value of x will make 2x + 3 equal to 7. In this case, the answer is 2; therefore, x = 2.
In other words, x would be equal to two (2). However, if x represented one third of a cup of coffee, then solving for x would mean finding the value of the whole cup. In this case, x would be equal to three (3). The key is to remember that, no matter what the size of the fraction, solving for x always means finding the value of the whole. With a little practice, solving for x with fractions can become second nature.
A composition of functions solver can be a useful tool for solving mathematical problems. In mathematics, function composition is the operation of combining two functions to produce a third function. For example, if f(x) = 2x + 1 and g(x) = 3x - 5, then the composition of these two functions, denoted by g o f, is the function defined by (g o f)(x) = g(f(x)) = 3(2x + 1) - 5 = 6x + 8. The composition of functions is a fundamental operation in mathematics and has many applications in science and engineering. A composition of functions solver can be used to quickly find the composition of any two given functions. This can be a valuable tool for students studying mathematics or for anyone who needs to solve mathematical problems on a regular basis. Thanks to the composition of functions solver, finding the composition of any two given functions is now quick and easy.
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# Inertial Frame of Reference – Understand its Concept with Examples
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## Frame of Reference
JEE is closer than we think. The JEE Main application form is here and this paves a way for students to boost their preparation for the exam. Taking this into consideration, we’re here to bring you the most important topics that will make a difference to your last-minute efforts. Today, we’ll be covering the ‘Inertial Frame of Reference’, which is a top-scoring topic from physics, considering the Engineering Entrance Exams study material. We are going to cover the concepts of inertial frame of reference, non-inertial frame of reference, its examples, etc.
Let us begin with the concept!
In Physics, a frame of reference is defined to easily calculate the position of an object at rest or in motion at a given specific time.
To do this, we take the help of a reference point of three mutually perpendicular axes X, Y, and Z.
Let us consider the system given below where the three axis X, Y, and Z are mutually perpendicular to each other.
Suppose we want to find out the position of the square box in the picture given above. Here the reference point is ‘O’ because the three axes X, Y, and Z intersect each other at this point. At a given specific time, the coordinates of X, Y and Z give the accurate position of the object at a given time. This entire setup of a reference point and a specific time is known as Frame of Reference.
## Type of Frame of Reference
Based on the motion of an object, the frame of reference is classified into two types:
(a) Inertial frame of reference.
(b) Non- Inertial frame of reference.
### Inertial Frame of Reference
Let us take an example of two persons A and B sitting inside a moving bus. A is sitting behind B as shown in the picture given below-
At any given time as the bus is moving forward, if we ask A to measure the distance between B and A or to measure the velocity of B, we will find that from A’s point of reference at any time, there is no change either in the distance between A and B neither there is any change in the velocity of B unless and until there is no other external force applied on B. Thus B’s velocity from A’s frame of reference remains constant even if the bus is moving forward. This frame of reference where an object does not change its speed or velocity be it at rest or in motion is defined as an inertial frame of reference.
### Non-Inertial frame of reference
Coming to this, our experts feel that this is a very important topic considering the JEE Main Exam Pattern. So, let’s dive into it!
Let us again take the example of two persons A and B where A is sitting on a station and B is sitting on a moving train.
From the point when the train proceeds towards the station to when the train leaves the station, from B’s point of view or B’s reference point, the velocity of A is not constant at any point of time as A is stationary but B is moving. This frame of reference which is accelerating with respect to inertial frame of reference (Here inertial frame of reference is A) is known as non- inertial frame of reference.
## Summarised Notes
• A frame of reference which is constant or uniform when in rest or in motion is known as inertial frame of reference.
• A frame of reference which is moving or non-uniform is known as non-inertial frame of reference.
As we all know, practice is the key to success. Therefore, boost your preparation by starting your practice now.
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