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# Base Area of a Cone – Definition, Formula, Examples, Facts, FAQs
Home » Math Vocabulary » Base Area of a Cone – Definition, Formula, Examples, Facts, FAQs
## What Is the Base Area of a Cone?
The surface area of a cone’s circular base is referred to as its base area. It is simply the area of the base of a cone.
A cone is a three-dimensional geometric shape with a circular base and a curved surface that narrows down to a single point known as the apex or vertex. The base area is crucial in defining the cone’s total surface area and volume.
Before moving on to calculating the base area of a cone, it is important to recall the basics.
Parts of a Cone:
There are three parts of a cone. They are:
• Base: The flat circular surface at a cone’s bottom face.
• Apex (Vertex): The apex is the pointed tip of a cone where all of its slanted sides converge.
• Lateral Surface: The curved surface that joins the base with the apex.
Dimensions of a Cone:
• Height of a Cone: The vertical distance from the apex to the base.
• Slant Height of a Cone: The distance from the apex to any point on the circular edge.
• Radius of the Base of a Cone: The distance from the center of the circular base to its edge.
## Base Area of a Cone Formula
The formula for calculating the base area of a cone is obtained from the area of a circle. Since a cone’s base is a circle, we can calculate its base area using the method for calculating the area of a circle:
Base Area (A) $= \pi \times r^{2}$
Here
A denotes the base area of the cone
π (pi) is a mathematical constant approximately equal to 3.14159
r denotes the radius of the circular base of the cone
## How to Find the Base Area of a Cone
Follow the steps given below to find the base area of a cone:
• Note down the radius (r) of the circular base, which is the distance from the circle’s center to any point on its edge.
• Put the values in the formula for the base area.
$A = \pi \times r^{2}$
• Simplify and assign the unit “square units” depending on the unit of the radius.
Example: If a cone has a radius of its circular base to be 5 units, then what will be the base area of the cone?
$A = \pi \times (5^{2}) = 25\pi$ square units.
## Facts about the Base Area of a Cone
• The base area of a cone increases as the radius of the circular base increases.
• The cone has the largest base area among all three-dimensional shapes with a fixed volume.
• Total Surface Area of a Cone = Base Area + Curved Surface Area
• Volume of a Cone $= \frac{1}{3} \times$Base Area $\times$ Height $= \frac{1}{3}\pi r^{2}h$
## Conclusion
In this article, we learned about the base area of a cone, which is represented by πr2, where r is the radius of the circular base of the cone. Understanding this fundamental concept is essential for calculating the cone’s volume and surface area. Let’s solve a few examples and practice MCQs for better comprehension.
## Solved Examples on the Base Area of a Cone
Example 1: Calculate the base area of a 4-unit-radius cone.
Solution:
Use the formula for the base area of a cone.
$A = \pi \times r^{2}$
$A = \pi \times (4 \;units)^{2}$
$A = \pi \times 16\; units^{2}$
$A = 50.27\; units^{2}$
Therefore, the base area of the cone is 50.27 square units.
Example 2: A traffic cone has a radius of the base 7 inches. Find the base area of the cone.
Solution:
Use the formula for the base area of a cone to find the base area of the traffic cone.
$A = \pi \times r^{2}$
On substituting the value of the radius into the formula, we get
$A = \pi (7\; inches)^{2}$
$A = \pi \times 49$ square inches
$A = 3.14 \times 49$ square inches
$A = 153.86$ square inches
Therefore, the base area of the traffic cone is 153.86 square inches.
Example 3: A cone’s circular base has to be covered with marble, and its base area is 36 square feet. What is the total cost to cover the entire base if the cost of marble is $5 per square foot? Solution: Total Cost = Base Area$\times$Cost per square foot Total Cost$= 36 \times \$5$
Total Cost $= \$180$Example 4: Find the radius of a cone that has a base area of 100 square units. Solution: Base Area (A) = 100 square units Base Area$(A) = \pi \times r^{2}$Substitute the value of the base area in the formula, we get$100 = \pi \times r^{2}$Let us divide both sides of the equation by π.$r^{2} = \frac{100}{\pi}$Take the square root of both sides.$r = \sqrt{\frac{100}{\pi}}r = \sqrt{31.83098862}r = 5.64$units (rounded to two decimal places) Therefore, the cone radius is about 5.64 units. Example 5: Find the radius of the cone whose base area is 36π square meters. Solution: Base Area$(A) = 36\pi$square units On applying the formula, we get Base Area$(A) = \pi r^{2}36\pi = \pi r^{2}$Divide both sides of the equation by π$r^{2} = 36$Take the square root of both sides.$r = \sqrt{36}r = 6$units Therefore, the radius of the cone is 6 units. ## Practice Problems on the Base Area of a Cone 1 ### What is the formula for computing the base area of a cone?$A = \pi r^{2}A = 2\pi rA = 4\pi r^{2}A = 0.5\pi r^{2}$CorrectIncorrect Correct answer is:$A = \pi r^{2}$Base area of a cone$= \pi r^{2}$2 ### What is the base area of a cone whose base has the radius of 10 units? 100 square units 50π square units 100π square units 314 square units CorrectIncorrect Correct answer is: 100 square units$A = \pi r^{2}A = \pi \times (10)^{2} = 100\pi \;unit^{2}$3 ### Volume of a cone =$(Base\; Area)^{2}Base \;Area \times Height\frac{1}{2} \times Base \;Area \times Height\frac{1}{3} \times Base \;Area \times Height$CorrectIncorrect Correct answer is:$\frac{1}{3} \times Base \;Area \times Height$Volume of a cone$= \frac{1}{3} \times Base \;Area \times Height$## Frequently Asked Questions about the Base Area of a Cone When the radius of a cone’s base is doubled, the base area of the cone grows four times. Base Area (A)$= \pi r^{2}$Let the new radius be “2r.” So, the new base area$= \pi \times (2r)^{2} = \pi 4r^{2} = 4(\pi \times r^{2})\$
Hence, the new base area grows four times the original base area by doubling the radius.
The base area is not directly related to the slant height; it only depends on the radius.
No, the base area is separate from the lateral surface area of a cone. Lateral surface area is simply the curved surface area of the cone.
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# Fractions for Parents part 1
Published on June 3, 2015
Fractions for Parents part 1
If your child is starting fractions at school and you realise that you have forgotten everything you once knew on fractions then this is for you.
A very quick terminology recap: the number on top of the fraction is the numerator and the number below the line is the denominator.
If the number on top is bigger than the one below then we have an improper fraction. (7/4). And if it’s a number written next to a fraction it’s called a mixed numeral.
Here is six tenths. This fraction can be simplified because both the numerator and the denominator can be divided by the same number – in this case 2. Six divided by 2 is 3 and 10 divided by 2 is 5: Three fifths can’t be simplified any further because apart from 1 you can’t divide both 3 and 5 by the same number and arrive at a whole number.
If your child struggles with this concept, you might like to use pizza to help explain.
This is a quarter (or a fourth) of a pizza. Now we have 2 quarters of a pizza. But it’s clear from just looking at it that 2/4 of a pizza is the same as half a pizza.
If the denominators are the same then addition is very easy – you just add the numerators.
For example, what is three sevenths plus two sevenths? The denominators are the same, so we just add the numerators together to get 5: the answer is five sevenths.
If the denominators aren’t the same you need to find a common denominator.
For example what is one sixth plus one quarter.
Now, the lowest common denominator is the lowest number that is a multiple of both denominators. Many of you will instantly know that 12 is the lowest common denominator here: as 12 is divisible by both 4 and 6, but no lower number is.
However, if you’re not sure – or if your child is not sure – a foolproof way is to simply multiply the denominators as this guarantees that the new denominator will be a multiple of both. So in this case, it’s six times four which is 24.
I’ll do this sum using both methods. First, with 12 as the common denominator:
We had to multiply six by 2 to get 12, so we also multiply the numerator by 2 and one sixth becomes two twelfths. We had to multiply 4 by 3 to get to 12, so we multiply this numerator by 3: 3 quarters becomes 9 twelfths.
Now we have the same denominator so we can add the numerators. 3 plus 9 = 11. The answer is 11 twelfths.
Now using the other method.
We multiplied 6 by 4 to get 24, so we also multiply the numerator by 4, giving us 4. And for this one we multiplied 4 by 6 to get to 24 so we multiply the numerator by 6, giving us 18.
Now we can add the numerators to give us 22 over 24.
Both 22 and 24 are divisible by 2, so we can simplify: 22 divided by 2 is 11and 24 divided by 2 is 12 – giving us the same answer as above: 11/12.
4. Improper fractions and mixed numerals
Let’s do a simple sum: 4/5 plus 3/5. The denominator is the same so we can just add the numerators giving us an answer of seven fifths.
This is an improper fraction; there is nothing wrong with this, but usually it is peferable to express it as a mixed number.
Conceptually, once we reach five fifths we have a whole number. That leaves us with 2 fifths left over, so we can write it as 1 and 2 fifths.
Here’s a more complex example: Convert 25/7 to a mixed numeral.
The process is to divide 25 by 7. Given that 3 times 7 = 21, the answer is 3, with 4 left over. So of our original 25 sevenths, 21 of the sevenths equals 3, and there are 4 sevenths left over. So the mixed numeral is 3 and 4 sevenths.
Again, if your child is having difficulty with the concept of mixed numerals, pizza can help.
Imagine 3 halves of a pizza: 1 half is here, another half is here and another half is here. Thus we have 3 over 2, 3 halves of pizza, but by looking at the picture it is clear that another way of saying this is that we have one and a half pizzas.
Okay, that’s enough fractions for the moment – I’m going to have a pizza!
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## Search This Blog
### Hard Conditional Probability Problem
Hard Conditional Probability Problem - 22 April
Four friends - Anna, Brian, Christy and Drake are asked to choose any number between 1 and 5.
Can you calculate the probability that any of them chose the same number ?
1. Let us take this one step at a time.
Let us calculate the probability that Anna and Brian have the same number in their mind.
101/125
Now, there's a 1/5 chance that the numbers will be same and 4/5 chance that the numbers are different.
Let us now include Christy in this data. There can be two cases.
1) Anna and Brian have the same number. In that case, Christy will have only one number to compare.
2) Anna and Brian did not have the same number. In that case, Christy will have two numbers to compare to.
For the first case, the probability will be 5/25. This is if Anna and Brian did have the same numbers.
But if Anna and Brian did not have the same numbers, there is a 2/5 probability that Christy is having the same number (this is because Christy gets to match her number with both Anna and Brian). In that case, we can simply multiply the probabilities.
4/5 * 2/5 = 8/25
Otherwise, if Christy is not having the same number, the probability is 3/5. Now multiplying with the previous chain:
4/5 * 3/5 = 12/25
Now, we can include Drake in our calculations. If we follow the path where Drake's number matches with Anna and Brian, the probability will be 25/125.
Now let us join that with Christy's probability. If Christy's number matches with Anna and Brian and Drakes' number also matches, then the probability will be:
4/5 * 2/5 = 40/125
If Christy's number does not match with Anna and Brian but Drake's number matches with Christy's, the probability will be:
4/5 * 3/5 * 3/5 = 36/125
But if Christy's number does not matches with Anna and Brian and even Drake's does not matches with Christy, then the probability will be:
4/5 * 3/5 * 2/5 = 24/125
Now, we have to tell the probability when all the four friends have same numbers, so we will just add up the probability where all the numbers matches:
25/125 + 40/125 + 36/125 = 101/125.
2. Why can't it be this way ??
same concept of 1- p(event)
p(event) = no person chooses the same number
1st person has 5 choices
2nd person 4
3rd person 3
4th person 2
5th person 1
so total = 5*4*3*2*1= 120
total no of all possible cases = 5^5 = 3125
so prob = 1-(120/3125) = 601/625... just asking.. sorry if i am wrong.
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Study Materials
# NCERT Solutions for Class 9th Mathematics
Page 3 of 5
## Chapter 2. Polynomials
### Exercise 2.3
Chapter 2. Polynomials
Exercise 2.3
Solution:
(i)By remainder theorem, the required remainder is equal to p(x) = (-1)
P (-1) = x3+ 3x2+ 3x+1
= (-1)3 + 3 (-1)2 + 3(-1) + 1
= -1 + 3 – 3 + 1 = 0
Required remainder is p (-1) = 0
Required remainder is p (1/2) = 0
(iv) By remainder theorem, the required remainder is equal to p(x) = - π
P(x) = x3+ 3x2+ 3x+1
P(π) = (- π)3 + 3(-π)2 + 3(-π) + 1
= (-π)3 + 3π2 + 3π + 1
Required remainder is p (π) = 0
Required remainder is p (- 5/2) = 0
Q.2. find the remainder when x3- ax+ 6x - a is divided by x- a.
Solution: Let P(x) = x3- ax2+ 6x- a . P(x) = a.
By remainder theorem
P (a) = (a)3- a(a)2+ 6(a) - a
=a3-a3 + 6a - a
Remainder = 5a
Q3. Check whether 7 + 3x is a factor of 3x3 + 7x.
Solution:
g(x) = 7 + 3x = 0
⇒ 3x = - 7
P(x) 0
Therefore, 7 + 3x is not a factor of P(x).
Page 3 of 5
Chapter Contents:
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Page 8 / 15
$f\left(x\right)=-2{\left(x+3\right)}^{2}-6$
$f\left(x\right)={x}^{2}+6x+4$
Domain is $\text{\hspace{0.17em}}\left(-\infty ,\infty \right).\text{\hspace{0.17em}}$ Range is $\text{\hspace{0.17em}}\left[-5,\infty \right).$
$f\left(x\right)=2{x}^{2}-4x+2$
$k\left(x\right)=3{x}^{2}-6x-9$
Domain is $\text{\hspace{0.17em}}\left(-\infty ,\infty \right).\text{\hspace{0.17em}}$ Range is $\text{\hspace{0.17em}}\left[-12,\infty \right).$
For the following exercises, use the vertex $\text{\hspace{0.17em}}\left(h,k\right)\text{\hspace{0.17em}}$ and a point on the graph $\text{\hspace{0.17em}}\left(x,y\right)\text{\hspace{0.17em}}$ to find the general form of the equation of the quadratic function.
$\left(h,k\right)=\left(2,0\right),\left(x,y\right)=\left(4,4\right)$
$f\left(x\right)={x}^{2}-4x+4$
$\left(h,k\right)=\left(-2,-1\right),\left(x,y\right)=\left(-4,3\right)$
$\left(h,k\right)=\left(0,1\right),\left(x,y\right)=\left(2,5\right)$
$f\left(x\right)={x}^{2}+1$
$\left(h,k\right)=\left(2,3\right),\left(x,y\right)=\left(5,12\right)$
$\left(h,k\right)=\left(-5,3\right),\left(x,y\right)=\left(2,9\right)$
$f\left(x\right)=\frac{6}{49}{x}^{2}+\frac{60}{49}x+\frac{297}{49}$
$\left(h,k\right)=\left(3,2\right),\left(x,y\right)=\left(10,1\right)$
$\left(h,k\right)=\left(0,1\right),\left(x,y\right)=\left(1,0\right)$
$f\left(x\right)=-{x}^{2}+1$
$\left(h,k\right)=\left(1,0\right),\left(x,y\right)=\left(0,1\right)$
Graphical
For the following exercises, sketch a graph of the quadratic function and give the vertex, axis of symmetry, and intercepts.
$f\left(x\right)={x}^{2}-2x$
Vertex Axis of symmetry is $\text{\hspace{0.17em}}x=1.\text{\hspace{0.17em}}$ Intercepts are
$f\left(x\right)={x}^{2}-6x-1$
$f\left(x\right)={x}^{2}-5x-6$
Vertex $\text{\hspace{0.17em}}\left(\frac{5}{2},\frac{-49}{4}\right),\text{\hspace{0.17em}}$ Axis of symmetry is $\text{\hspace{0.17em}}\left(0,-6\right),\left(-1,0\right),\left(6,0\right).$
$f\left(x\right)={x}^{2}-7x+3$
$f\left(x\right)=-2{x}^{2}+5x-8$
Vertex Axis of symmetry is $\text{\hspace{0.17em}}x=\frac{5}{4}.\text{\hspace{0.17em}}$ Intercepts are
$f\left(x\right)=4{x}^{2}-12x-3$
For the following exercises, write the equation for the graphed quadratic function.
$f\left(x\right)={x}^{2}-4x+1$
$f\left(x\right)=-2{x}^{2}+8x-1$
$f\left(x\right)=\frac{1}{2}{x}^{2}-3x+\frac{7}{2}$
Numeric
For the following exercises, use the table of values that represent points on the graph of a quadratic function. By determining the vertex and axis of symmetry, find the general form of the equation of the quadratic function.
$x$ –2 –1 0 1 2 $y$ 5 2 1 2 5
$f\left(x\right)={x}^{2}+1$
$x$ –2 –1 0 1 2 $y$ 1 0 1 4 9
$x$ –2 –1 0 1 2 $y$ –2 1 2 1 –2
$f\left(x\right)=2-{x}^{2}$
$x$ –2 –1 0 1 2 $y$ –8 –3 0 1 0
$x$ –2 –1 0 1 2 $y$ 8 2 0 2 8
$f\left(x\right)=2{x}^{2}$
Technology
For the following exercises, use a calculator to find the answer.
Graph on the same set of axes the functions
What appears to be the effect of changing the coefficient?
Graph on the same set of axes $\text{\hspace{0.17em}}f\left(x\right)={x}^{2},f\left(x\right)={x}^{2}+2\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}f\left(x\right)={x}^{2},f\left(x\right)={x}^{2}+5\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}f\left(x\right)={x}^{2}-3.\text{\hspace{0.17em}}$ What appears to be the effect of adding a constant?
The graph is shifted up or down (a vertical shift).
Graph on the same set of axes
What appears to be the effect of adding or subtracting those numbers?
The path of an object projected at a 45 degree angle with initial velocity of 80 feet per second is given by the function $\text{\hspace{0.17em}}h\left(x\right)=\frac{-32}{{\left(80\right)}^{2}}{x}^{2}+x\text{\hspace{0.17em}}$ where $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ is the horizontal distance traveled and $\text{\hspace{0.17em}}h\left(x\right)\text{\hspace{0.17em}}$ is the height in feet. Use the TRACE feature of your calculator to determine the height of the object when it has traveled 100 feet away horizontally.
50 feet
A suspension bridge can be modeled by the quadratic function $\text{\hspace{0.17em}}h\left(x\right)=.0001{x}^{2}\text{\hspace{0.17em}}$ with $\text{\hspace{0.17em}}-2000\le x\le 2000\text{\hspace{0.17em}}$ where $\text{\hspace{0.17em}}|x|\text{\hspace{0.17em}}$ is the number of feet from the center and $\text{\hspace{0.17em}}h\left(x\right)\text{\hspace{0.17em}}$ is height in feet. Use the TRACE feature of your calculator to estimate how far from the center does the bridge have a height of 100 feet.
Extensions
For the following exercises, use the vertex of the graph of the quadratic function and the direction the graph opens to find the domain and range of the function.
Vertex $\text{\hspace{0.17em}}\left(1,-2\right),\text{\hspace{0.17em}}$ opens up.
Domain is $\text{\hspace{0.17em}}\left(-\infty ,\infty \right).\text{\hspace{0.17em}}$ Range is $\text{\hspace{0.17em}}\left[-2,\infty \right).$
Vertex $\text{\hspace{0.17em}}\left(-1,2\right)\text{\hspace{0.17em}}$ opens down.
Vertex $\text{\hspace{0.17em}}\left(-5,11\right),\text{\hspace{0.17em}}$ opens down.
Domain is $\text{\hspace{0.17em}}\left(-\infty ,\infty \right)\text{\hspace{0.17em}}$ Range is $\text{\hspace{0.17em}}\left(-\infty ,11\right].$
Vertex $\text{\hspace{0.17em}}\left(-100,100\right),\text{\hspace{0.17em}}$ opens up.
For the following exercises, write the equation of the quadratic function that contains the given point and has the same shape as the given function.
Contains $\text{\hspace{0.17em}}\left(1,1\right)\text{\hspace{0.17em}}$ and has shape of $\text{\hspace{0.17em}}f\left(x\right)=2{x}^{2}.\text{\hspace{0.17em}}$ Vertex is on the $\text{\hspace{0.17em}}y\text{-}$ axis.
$f\left(x\right)=2{x}^{2}-1$
Contains $\text{\hspace{0.17em}}\left(-1,4\right)\text{\hspace{0.17em}}$ and has the shape of $\text{\hspace{0.17em}}f\left(x\right)=2{x}^{2}.\text{\hspace{0.17em}}$ Vertex is on the $\text{\hspace{0.17em}}y\text{-}$ axis.
Contains $\text{\hspace{0.17em}}\left(2,3\right)\text{\hspace{0.17em}}$ and has the shape of $\text{\hspace{0.17em}}f\left(x\right)=3{x}^{2}.\text{\hspace{0.17em}}$ Vertex is on the $\text{\hspace{0.17em}}y\text{-}$ axis.
$f\left(x\right)=3{x}^{2}-9$
Contains $\text{\hspace{0.17em}}\left(1,-3\right)\text{\hspace{0.17em}}$ and has the shape of $\text{\hspace{0.17em}}f\left(x\right)=-{x}^{2}.\text{\hspace{0.17em}}$ Vertex is on the $\text{\hspace{0.17em}}y\text{-}$ axis.
Contains $\text{\hspace{0.17em}}\left(4,3\right)\text{\hspace{0.17em}}$ and has the shape of $\text{\hspace{0.17em}}f\left(x\right)=5{x}^{2}.\text{\hspace{0.17em}}$ Vertex is on the $\text{\hspace{0.17em}}y\text{-}$ axis.
$f\left(x\right)=5{x}^{2}-77$
Contains $\text{\hspace{0.17em}}\left(1,-6\right)\text{\hspace{0.17em}}$ has the shape of $\text{\hspace{0.17em}}f\left(x\right)=3{x}^{2}.\text{\hspace{0.17em}}$ Vertex has x-coordinate of $\text{\hspace{0.17em}}-1.$
Real-world applications
Find the dimensions of the rectangular corral producing the greatest enclosed area given 200 feet of fencing.
50 feet by 50 feet. Maximize $\text{\hspace{0.17em}}f\left(x\right)=-{x}^{2}+100x.$
Find the dimensions of the rectangular corral split into 2 pens of the same size producing the greatest possible enclosed area given 300 feet of fencing.
Find the dimensions of the rectangular corral producing the greatest enclosed area split into 3 pens of the same size given 500 feet of fencing.
125 feet by 62.5 feet. Maximize $\text{\hspace{0.17em}}f\left(x\right)=-2{x}^{2}+250x.$
Among all of the pairs of numbers whose sum is 6, find the pair with the largest product. What is the product?
Among all of the pairs of numbers whose difference is 12, find the pair with the smallest product. What is the product?
$6\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}-6;\text{\hspace{0.17em}}$ product is –36; maximize $\text{\hspace{0.17em}}f\left(x\right)={x}^{2}+12x.$
Suppose that the price per unit in dollars of a cell phone production is modeled by $\text{\hspace{0.17em}}p=\text{}45-0.0125x,\text{\hspace{0.17em}}$ where $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ is in thousands of phones produced, and the revenue represented by thousands of dollars is $\text{\hspace{0.17em}}R=x\cdot p.\text{\hspace{0.17em}}$ Find the production level that will maximize revenue.
A rocket is launched in the air. Its height, in meters above sea level, as a function of time, in seconds, is given by $\text{\hspace{0.17em}}h\left(t\right)=-4.9{t}^{2}+229t+234.\text{\hspace{0.17em}}$ Find the maximum height the rocket attains.
2909.56 meters
A ball is thrown in the air from the top of a building. Its height, in meters above ground, as a function of time, in seconds, is given by $\text{\hspace{0.17em}}h\left(t\right)=-4.9{t}^{2}+24t+8.\text{\hspace{0.17em}}$ How long does it take to reach maximum height?
A soccer stadium holds 62,000 spectators. With a ticket price of $11, the average attendance has been 26,000. When the price dropped to$9, the average attendance rose to 31,000. Assuming that attendance is linearly related to ticket price, what ticket price would maximize revenue?
\$10.70
A farmer finds that if she plants 75 trees per acre, each tree will yield 20 bushels of fruit. She estimates that for each additional tree planted per acre, the yield of each tree will decrease by 3 bushels. How many trees should she plant per acre to maximize her harvest?
what is the VA Ha D R X int Y int of f(x) =x²+4x+4/x+2 f(x) =x³-1/x-1
can I get help with this?
Wayne
Are they two separate problems or are the two functions a system?
Wilson
Also, is the first x squared in "x+4x+4"
Wilson
x^2+4x+4?
Wilson
thank you
Wilson
Wilson
f(x)=x square-root 2 +2x+1 how to solve this value
Wilson
what is algebra
The product of two is 32. Find a function that represents the sum of their squares.
Paul
if theta =30degree so COS2 theta = 1- 10 square theta upon 1 + tan squared theta
how to compute this 1. g(1-x) 2. f(x-2) 3. g (-x-/5) 4. f (x)- g (x)
hi
John
hi
Grace
what sup friend
John
not much For functions, there are two conditions for a function to be the inverse function: 1--- g(f(x)) = x for all x in the domain of f 2---f(g(x)) = x for all x in the domain of g Notice in both cases you will get back to the element that you started with, namely, x.
Grace
sin theta=3/4.prove that sec square theta barabar 1 + tan square theta by cosec square theta minus cos square theta
acha se dhek ke bata sin theta ke value
Ajay
sin theta ke ja gha sin square theta hoga
Ajay
I want to know trigonometry but I can't understand it anyone who can help
Yh
Idowu
which part of trig?
Nyemba
functions
Siyabonga
trigonometry
Ganapathi
differentiation doubhts
Ganapathi
hi
Ganapathi
hello
Brittany
Prove that 4sin50-3tan 50=1
False statement so you cannot prove it
Wilson
f(x)= 1 x f(x)=1x is shifted down 4 units and to the right 3 units.
f (x) = −3x + 5 and g (x) = x − 5 /−3
Sebit
what are real numbers
I want to know partial fraction Decomposition.
classes of function in mathematics
divide y2_8y2+5y2/y2
wish i knew calculus to understand what's going on 🙂
@dashawn ... in simple terms, a derivative is the tangent line of the function. which gives the rate of change at that instant. to calculate. given f(x)==ax^n. then f'(x)=n*ax^n-1 . hope that help.
Christopher
thanks bro
Dashawn
maybe when i start calculus in a few months i won't be that lost 😎
Dashawn
what's the derivative of 4x^6
24x^5
James
10x
Axmed
24X^5
Taieb
comment écrire les symboles de math par un clavier normal
SLIMANE
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# What is the equation of the line that is normal to the polar curve f(theta)=2sin(3theta+pi/3) - 2thetacostheta at theta = pi/4?
Jul 27, 2018
$y = \frac{- 8 - 8 \sqrt{3}}{- 12 - 4 \sqrt{3} + 2 \pi} \left(x - \left(\frac{2 - 2 \sqrt{3} - \pi}{4}\right)\right) + \left(\frac{2 - 2 \sqrt{3} - \pi}{4}\right)$
#### Explanation:
To find the equation of the line, we need a $\textcolor{red}{p o i n t}$, and we need a $\textcolor{b l u e}{s l o p e}$.
First, let's find the $\textcolor{red}{p o i n t}$.
Plug in $\frac{\pi}{4}$ into $f \left(\theta\right)$.
$f \left(\frac{\pi}{4}\right) = 2 \sin \left(3 \left(\frac{\pi}{4}\right) + \frac{\pi}{3}\right) - 2 \left(\frac{\pi}{4}\right) \left(\cos \left(\frac{\pi}{4}\right)\right)$
$f \left(\frac{\pi}{4}\right) = \frac{\sqrt{2} \left(2 - 2 \sqrt{3} - \pi\right)}{4}$
We have the polar coordinate $\left(\frac{\pi}{4} , \frac{\sqrt{2} \left(2 - 2 \sqrt{3} - \pi\right)}{4}\right)$.
We have to convert this to Cartesian form. Using the formulas:
$x = r \cos \theta$
$y = r \cos \theta$
We find that the Cartesian coordinate is color(red)(((2-2sqrt3-pi)/4,(2-2sqrt3-pi)/4).
Good! We have found the $\textcolor{red}{p o i n t}$.
Now, we need to find the $\textcolor{b l u e}{s l o p e}$.
The formula for the tangent line of a polar function is:
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{\mathrm{dr}}{d \theta} \sin \theta + r \cos \theta}{\frac{\mathrm{dr}}{d \theta} \cos \theta - r \sin \theta}$
If you notice, $\sin \theta$ and $\cos \theta$ will be the same in this case, because $\sin \left(\frac{\pi}{4}\right)$ and $\cos \left(\frac{\pi}{4}\right)$ are the same, so for this case, our $\frac{\mathrm{dy}}{\mathrm{dx}}$ is simply:
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{\mathrm{dr}}{d \theta} + r}{\frac{\mathrm{dr}}{d \theta} - r}$
We solved $r$ already previously, which was $\frac{\sqrt{2} \left(2 - 2 \sqrt{3} - \pi\right)}{4}$.
Now, we need to find $\frac{\mathrm{dr}}{d \theta}$.
$f ' \left(\theta\right) = \frac{\mathrm{dr}}{d \theta} = 6 \cos \left(3 \theta + \frac{\pi}{3}\right) + 2 \theta \sin \theta - 2 \cos \theta$
Now, we plug in $\frac{\pi}{4}$ and get:
$f ' \left(\frac{\pi}{4}\right) = \frac{\mathrm{dr}}{d \theta} = \frac{\sqrt{2} \left(- 10 - 6 \sqrt{3} + \pi\right)}{4}$
Going back to the tangent line formula:
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{\sqrt{2} \left(- 10 - 6 \sqrt{3} + \pi\right)}{4} + \frac{\sqrt{2} \left(2 - 2 \sqrt{3} - \pi\right)}{4}}{\frac{\sqrt{2} \left(- 10 - 6 \sqrt{3} + \pi\right)}{4} - \frac{\sqrt{2} \left(2 - 2 \sqrt{3} - \pi\right)}{4}}$
After some tedious simplification, we find this reduces to color(blue)(dy/dx = (-8-8sqrt3)/(-12-4sqrt3+2pi)
We have our $\textcolor{red}{p o i n t}$:
color(red)(((2-2sqrt3-pi)/4,(2-2sqrt3-pi)/4)
We have our $\textcolor{b l u e}{s l o p e}$:
color(blue)(dy/dx = (-8-8sqrt3)/(-12-4sqrt3+2pi)
All that is left is to plug in these values into the point slope formula:
$\left(y - \textcolor{red}{{y}_{1}}\right) = \textcolor{b l u e}{m} \left(x - \textcolor{red}{{x}_{1}}\right)$
We get:
$y - \left(\frac{2 - 2 \sqrt{3} - \pi}{4}\right) = \frac{- 8 - 8 \sqrt{3}}{- 12 - 4 \sqrt{3} + 2 \pi} \left(x - \left(\frac{2 - 2 \sqrt{3} - \pi}{4}\right)\right)$
So the equation is:
$y = \frac{- 8 - 8 \sqrt{3}}{- 12 - 4 \sqrt{3} + 2 \pi} \left(x - \left(\frac{2 - 2 \sqrt{3} - \pi}{4}\right)\right) + \left(\frac{2 - 2 \sqrt{3} - \pi}{4}\right)$
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Zeckendorf's theorem, named after Belgian mathematician Edouard Zeckendorf, is a theorem about the representation of integers as sums of Fibonacci numbers.
Zeckendorf's theorem states that every positive integer can be represented uniquely as the sum of one or more distinct Fibonacci numbers in such a way that the sum does not include any two consecutive Fibonacci numbers. More precisely, if N is any positive integer, there exist positive integers ci ≥ 2, with ci + 1 > ci + 1, such that
$$N = \sum_{i = 0}^k F_{c_i},$$
where Fn is the nth Fibonacci number. Such a sum is called the Zeckendorf representation of N. The Fibonacci coding of N can be derived from its Zeckendorf representation.
For example, the Zeckendorf representation of 100 is
100 = 89 + 8 + 3.
There are other ways of representing 100 as the sum of Fibonacci numbers – for example
100 = 89 + 8 + 2 + 1
100 = 55 + 34 + 8 + 3
but these are not Zeckendorf representations because 1 and 2 are consecutive Fibonacci numbers, as are 34 and 55.
For any given positive integer, a representation that satisfies the conditions of Zeckendorf's theorem can be found by using a greedy algorithm, choosing the largest possible Fibonacci number at each stage.
istory
While the theorem is named after the eponymous author who published his paper in 1972, the same result had been published 20 years earlier by Lekkerkerker.[1] As such, the theorem is an example of Stigler's Law of Eponymy.
Proof
Zeckendorf's theorem has two parts:
1. Existence: every positive integer n has a Zeckendorf representation.
2. Uniqueness: no positive integer n has two different Zeckendorf representations.
The first part of Zeckendorf's theorem (existence) can be proven by induction. For n = 1, 2, 3 it is clearly true (as these are Fibonacci numbers), for n = 4 we have 4 = 3 + 1. Now suppose each nk has a Zeckendorf representation. If k + 1 is a Fibonacci number then we're done, else there exists j such that Fj < k + 1 < Fj + 1 . Now consider a = k + 1 − Fj . Since ak, a has a Zeckendorf representation (by the induction hypothesis). At the same time, Fj + a < Fj + 1 and since Fj + 1 = Fj + Fj − 1 (by definition of Fibonacci numbers), a < Fj − 1, so the Zeckendorf representation of a does not contain Fj − 1 . As a result, k + 1 can be represented as the sum of Fj and the Zeckendorf representation of a.
The second part of Zeckendorf's theorem (uniqueness) requires the following lemma:
Lemma: The sum of any non-empty set of distinct, non-consecutive Fibonacci numbers whose largest member is Fj is strictly less than the next largest Fibonacci number Fj + 1 .
The lemma can be proven by induction on j.
Now take two non-empty sets of distinct non-consecutive Fibonacci numbers S and T which have the same sum. Consider sets S and T which are equal to S and T from which the common elements have been removed (i.e. S = S\T and T = T\S). Since S and T had equal sum, and we have removed exactly the elements from S $$\cap$$ T from both sets, S and T must have the same sum as well.
Now we will show by contradiction that at least one of S and T is empty. Assume the contrary, i.e. that S and T are both non-empty and let the largest member of S be Fs and the largest member of T be Ft. Because S and T contain no common elements, FsFt. Without loss of generality, suppose Fs < Ft. Then by the lemma, the sum of S is strictly less than Fs + 1 and so is strictly less than Ft, whereas the sum of T is clearly at least Ft. This contradicts the fact that S and T have the same sum, and we can conclude that either S or T must be empty.
Now assume (again without loss of generality) that S is empty. Then S has sum 0, and so must T. But since T can only contain positive integers, it must be empty too. To conclude: S = T = ∅ which implies S = T, proving that each Zeckendorf representation is unique.
Fibonacci multiplication
One can define the following operation a\circ b on natural numbers a, b: given the Zeckendorf representations $$a=\sum_{i=0}^kF_{c_i}\;(c_i\ge2) and b=\sum_{j=0}^lF_{d_j}\;(d_j\ge2)$$ we define the Fibonacci product $$a\circ b=\sum_{i=0}^k\sum_{j=0}^lF_{c_i+d_j}$$.
For example, the Zeckendorf representation of 2 is F_3, and the Zeckendorf representation of 4 is $$F_4 + F_2$$ ($$F_1$$ is disallowed from representations), so $$2 \circ 4 = F_{3+4} + F_{3+2} = 13 + 5 = 18.$$
A simple rearrangement of sums shows that this is a commutative operation; however, Donald Knuth proved the surprising fact that this operation is also associative.
Representation with negafibonacci numbers
The Fibonacci sequence can be extended to negative index n using the re-arranged recurrence relation
$$F_{n-2} = F_n - F_{n-1}, \,$$
which yields the sequence of "negafibonacci" numbers satisfying
$$F_{-n} = (-1)^{n+1} F_n. \,$$
Any integer can be uniquely represented[2] as a sum of negafibonacci numbers in which no two consecutive negafibonacci numbers are used. For example:
−11 = F−4 + F−6 = (−3) + (−8)
12 = F−2 + F−7 = (−1) + 13
24 = F−1 + F−4 + F−6 + F−9 = 1 + (−3) + (−8) + 34
−43 = F−2 + F−7 + F−10 = (−1) + 13 + (−55)
0 is represented by the empty sum.This gives a system of coding integers, similar to the representation of Zeckendorf's theorem. In the string representing the integer x, the nth digit is 1 if Fn appears in the sum that represents x; that digit is 0 otherwise. For example, 24 may be represented by the string 100101001, which has the digit 1 in places 9, 6, 4, and 1, because 24 = F−1 + F−4 + F−6 + F−9 . The integer x is represented by a string of odd length if and only if x > 0.
Note that 0 = F−1 + F−2 , for example, so the uniqueness of the representation does depend on the condition that no two consecutive negafibonacci numbers are used.
This gives a system of coding integers, similar to the representation of Zeckendorf's theorem. In the string representing the integer x, the nth digit is 1 if Fn appears in the sum that represents x; that digit is 0 otherwise. For example, 24 may be represented by the string 100101001, which has the digit 1 in places 9, 6, 4, and 1, because 24 = F−1 + F−4 + F−6 + F−9 . The integer x is represented by a string of odd length if and only if x > 0.
See also
Fibonacci coding
Ostrowski numeration
References
Historical note on the name Zeckendorf Representation by R Knott, University of Surrey
Knuth, Donald. "Negafibonacci Numbers and the Hyperbolic Plane" Paper presented at the annual meeting of the Mathematical Association of America, The Fairmont Hotel, San Jose, CA. 2008-12-11 <http://www.allacademic.com/meta/p206842_index.html>
Knuth, Donald E. (1988). "Fibonacci multiplication". Applied Mathematics Letters 1 (1): 57–60. doi:10.1016/0893-9659(88)90176-0. ISSN 0893-9659. Zbl 0633.10011.
Zeckendorf, E. (1972). "Représentation des nombres naturels par une somme de nombres de Fibonacci ou de nombres de Lucas". Bull. Soc. R. Sci. Liège (in French) 41: 179–182. ISSN 0037-9565. Zbl 0252.10011.
External links
Weisstein, Eric W., "Zeckendorf's Theorem", MathWorld.
Weisstein, Eric W., "Zeckendorf Representation", MathWorld.
Zeckendorf's theorem at cut-the-knot
G.M. Phillips (2001), "Zeckendorf representation", in Hazewinkel, Michiel, Encyclopedia of Mathematics, Springer, ISBN 978-1-55608-010-4
"Sloane's A101330 : Knuth's Fibonacci (or circle) product", The On-Line Encyclopedia of Integer Sequences. OEIS Foundation.
This article incorporates material from proof that the Zeckendorf representation of a positive integer is unique on PlanetMath, which is licensed under the Creative Commons Attribution/Share-Alike License.
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# Opposite Sides of a Parallelogram are Equal
Here we will discuss about the opposite sides of a parallelogram are equal in length.
In a parallelogram, each pair of opposite sides are of equal length.
Given: PQRS is a parallelogram in which PQ ∥ SR and QR ∥ PS.
To prove: PQ = SR and PS = QR
Construction: Join PR
Proof:
Statement In ∆PQR and ∆RSP;1. ∠QPR = ∠SRP2. ∠QRP = ∠RPS3. PR = PR4. ∆PQR ≅ ∆RSP 5. PQ = SR and PS = QR. (Proved) Reason 1. PQ ∥ RS and RP is a transversal.2. PS ∥ QR and RP is a transversal.3. Common side4. By ASA criterion of congruency. 5. CPCTC
Converse of the above given theorem
A quadrilateral is a parallelogram if each pair of opposite sides are equal.
Given: PQRS is a quadrilateral in which PQ = SR and PS = QR
To prove: PQRS is a parallelogram
Proof: In ∆PQR and ∆RSP, PQ = SR, QR = SP (given) and PR is the common side.
Therefore, by SSS criterion of congruency, ∆PQR ≅ ∆RSP
Therefore, ∠QPR = ∠PRS, ∠QRP = ∠RPS (CPCTC)
Therefore, PQ ∥ SR, QR ∥ PS
Hence, PQRS is a parallelogram.
Solved examples based on the theorem of opposite sides of a parallelogram are equal in length:
1. In the parallelogram PQRS, Pq = 6 cm and SR : RQ = 2 : 1. Find the perimeter of the parallelogram.
Solution:
In the parallelogram PQRS, PQ ∥ SR and SP ∥ RQ.
The opposite sides of a parallelogram are equal. So, SR + PQ = 6 cm.
AS SR : RQ = 23 : 1, $$\frac{6 cm}{RQ}$$ = $$\frac{2}{1}$$
⟹ RQ = 3 cm
Also, RQ = SP = 3 cm.
Therefore, perimeter = PQ + QR + RS + SP
= 6 cm + 3 cm + 6 cm + 3 cm
= 18 cm.
2. In the parallelogram ABCD, ∠ABC = 50°. Find the measures of ∠BCD, ∠CBA and ∠DAB.
Solution:
AS AB ∥ DC, ∠ABC + ∠BCD = 180°
Therefore, ∠BCD = 180° - ∠ABC
= 180° - 50°
= 130°
As opposite angles in a parallelogram are equal,
∠CDA = ∠ABC = 50° and
∠DAB = ∠BCD = 130°
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Geometry
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# Geometry - PowerPoint PPT Presentation
## Geometry
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##### Presentation Transcript
1. Geometry 12.2 – 12.5 Prep for the STAR Test
2. Geometry 12.2 Pyramids
3. 10 6 Let’s solve #7 as we learn the formula for lateral area!! 7) Find the lateral area and total area of this regular pyramid. LA = ½ pl A = ½ b(h) LA = ½ 36(10) A = 3(10) A = 30 LA = 180 square units We have 6 triangles! 10 LA = nF LA = 6(30) 6 LA = 180 square units The lateral area of a regular pyramid with n lateral faces is n times the area of one lateral face. The best way to know the formulas is to understand them rather than memorize them. L.A. = nF OR… The lateral area of a regular pyramid equals half the perimeter of the base times the slant height. Why? Imagine a curtain on the ground around the pyramid(perimeter) and you are pulling the curtain up the slant height. Since the curtain will wrap around triangles we need the ½ . L.A. = ½ pl
4. 10 6 7) Find the lateral area and total area of this regular pyramid. A = ½ a(p) TA = LA + B A = ½ 3√3(36) TA = 180 + 54√3 sq. units A = 3√3(18) 30 3√3 A = 54√3 3 6 The total area of a pyramid is its lateral area plus the area of its base. T.A. = L.A. + B That makes sense!
5. h l Let’s solve #9 as we learn the formula volume!! 9. Find the volume of a regular square pyramid with base edge 24 and lateral edge 24. Draw a square pyramid with the given dimensions. Must be a 30-60-90. V = 1/3 B(h) 1/3 the volume of its prism. V = 1/3 24(24)(h) 122 + x2 = (12√3)2 12√3 12√3 12√2 V = 8(24)(h) 24 V = 192(h) 12 12 V = 192(12√2) 24 V = 2304√2 sq. units The volume of a pyramid equals one third the area of the base times the height of the pyramid. Why? Because a rectangular prism with the same base and same height would be V = bh and the pyramid, believe it or not, would hold 1/3 the amount of water.
6. 12.3 Cylinders and Cones
7. To find lateral area (L.A.): Take a can of soup Peel off the label Multiply circumference by height height circumference
8. To find total area (T.A.): Take the label (LA) Pop the top and find the area of the base Add 2 base areas to the LA A = πr² Top Bottom
9. To find volume (V): Start with the area of the base Multiply it by height That’s how much soup is in the can ! H r A = πr²
10. Lateral Area of a Cylinder: L.A. The lateral area of a cylinder equals the circumference of a base times the height of the cylinder. L.A = 2πrH 8 6 L.A = dπH which is • L.A = CH which is LA = 12π • 8 = 96π square units
11. Total Area of a Cylinder: T.A. The total area of a cylinder is the lateral area plus twice the area of a base. T.A = L.A. + 2B 8 6 • which is TA = 96π + 2(π • 6²) = 96π + 2(36π) = 96π + 72π = 168π square units T.A. = 2πrH + 2πr²
12. Volume of a Cylinder: V The volume of a cylinder equals the area of a base times the height of the cylinder. V = πr²H 8 6 • V = (π • 6²) • 8 = 36π • 8 = 288π cubic units
13. Lateral Area of a Cone: L.A. 10 8 • L.A = πrl 6 LA = π • 6 •10 = 60π square units
14. Total Area of a Cone: T.A. 10 T.A = L.A. + B 8 • 6 which is TA = 60π + (π • 6²) = 60π + 36π = 96π square units T.A. = πrl + πr²
15. Volume of a Cone: V The volume of a cone equals one third the area of the base times the height of the cone. 10 V = πr²h 8 • 6 V = 1/3 • π • 6² • 8 = 96π cubic units
16. 12.4 Spheres
17. Surface Area Formula Surface Area = What does this represent? It represents the area of a circle with radius = r. r
18. Volume Formula Volume = A way to remember which formula is cubed… The volume formula has two 3’s and volume is measured in units3!!
19. 1. radius = 10 Area: _____ Volume: _____ Surface Area = Volume = = 4π(10)2 = 4π(100) = 400πunits2
20. Geometry 12.5 Areas and Volumes of Similar Solids
21. Scale Factor If the scale factor of two solids is a:b, then (1) the ratio of corresponding perimeters is a:b (2) the ratio of base areas, of lateral areas, and of the total area is a²:b² (3) the ratio of volumes is a³:b³ SCALE FACTOR: 1:2 Base circumference: 6π:12π 1:2 Lateral areas: 15π:60π 1:4 Volumes: 12π:96π 1:8 10 8 5 4 • 3 6
22. Exercises Find the missing information. 2:7 5:6 3:10 2:5 2:5 1:9 25:36 9:100 4:25 4:25 1:27 8:343 8:125
23. HW • Midpoint WS
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# Addition Strategies: Explanation & Word Problems
Jul 24, 2023
#### Introduction:
• In this session, you will understand which addition strategy can be used and learn to solve word problems by using different addition strategies that were learned in previous sessions.
1. Solve.
2. Solve.
3. Ashley has 5 blue toys and 9 red toys. How many toys are there with Ashley in all?
## Explain Addition Strategies:
Choose a strategy to solve the problem.
We can explain our work and the strategy we used.
Example:
Milan says that he wants to find 8 + 5.
Solution:
Make 10 to add 8 + 5
Let us find this.
#### Doubles Plus:
Let us solve 8 + 5 by using doubles plus.
Solution:
8 + 5
= 3 + 5 + 5
= 3 + 10
= 13
#### Number Line:
We can also solve 8 + 5 by using a number line.
Here we use a number line as well as visual learning.
#### Visual Learning:
Example:
Jake has 9 pink balls and some red balls. He has 17 balls in all. How many red balls does Jake have in all?
9 + 8 = 17.
Activity:
Make 10 to Add:
Solve.
#### Addition and Multiplication Using Counters & Bar-Diagrams
Introduction: We can find the solution to the word problem by solving it. Here, in this topic, we can use 3 methods to find the solution. 1. Add using counters 2. Use factors to get the product 3. Write equations to find the unknown. Addition Equation: 8+8+8 =? Multiplication equation: 3×8=? Example 1: Andrew has […]
#### Dilation: Definitions, Characteristics, and Similarities
Understanding Dilation A dilation is a transformation that produces an image that is of the same shape and different sizes. Dilation that creates a larger image is called enlargement. Describing Dilation Dilation of Scale Factor 2 The following figure undergoes a dilation with a scale factor of 2 giving an image A’ (2, 4), B’ […]
#### How to Write and Interpret Numerical Expressions?
Write numerical expressions What is the Meaning of Numerical Expression? A numerical expression is a combination of numbers and integers using basic operations such as addition, subtraction, multiplication, or division. The word PEMDAS stands for: P → Parentheses E → Exponents M → Multiplication D → Division A → Addition S → Subtraction Some examples […]
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# Distributive Property to Evaluate Formulas with Decimals
## Multiplying values inside the brackets by value outside
Estimated6 minsto complete
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Practice Distributive Property to Evaluate Formulas with Decimals
Progress
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Distributive Property to Evaluate Formulas with Decimals
Have you ever had to use a formula to figure something out? Well sometimes when you work with properties, it is necessary.
Let's say that you had a rectangle that was half as large as this one. That would mean that the side lengths of the rectangle would be 6 inches and 3.5 inches.
Now what if you had two of them? What would be the area of the two rectangles?
2(6 x 3.5)
Do you know how to figure this out?
This Concept is about the distributive property and formulas. By the end of it, you will know how to tackle this problem.
### Guidance
We can also use and apply the Distributive Property when working with a formula. Let’s think about the formula for finding the area of a rectangle.
We know that the area of a rectangle can be found by using the formula:
For this example, we would multiply 12 times 4 and get an area of 48 square inches.
How can we find the area of both of these rectangles?
You can see that they have the same width. The width is four and a half inches. However, there are two lengths.
We need to find the product of a number and a sum.
Here is our expression.
Now we can use the Distributive Property to find the area of these two rectangles.
Notice that we used what we have already learned about multiplying decimals and whole numbers with the Distributive Property. When we distributed 4.5 with each length, we were able to find the sum of the products. This gives us the area of the two rectangles.
Use what you have learned to answer these questions about formulas, area and the distributive property.
#### Example A
What is the formula for finding the area of a square?
Solution: A = s^2
#### Example B
Which property is being illustrated: 4(a + b) = 4a + 4b
Solution: The Distributive Property
#### Example C
What is the formula for finding the area of a rectangle?
Solution: A = length x width
Remember the rectangle from the beginning of the Concept? Now you are ready to work on that problem. Take a look.
Let's say that you had a rectangle that was half as large as this one. That would mean that the side lengths of the rectangle would be 6 inches and 3.5 inches.
Now what if you had two of them? What would be the area of the two rectangles?
2(6 x 3.5)
Do you know how to figure this out?
To figure this out, we have to multiply the value outside the parentheses by both values inside the parentheses.
The area of the two rectangles is 84 square inches.
### Vocabulary
Numerical expression
a number sentence that has at least two different operations in it.
Product
the answer in a multiplication problem
Sum
The Distributive Property
the property that involves taking the product of the sum of two numbers. Take the number outside the parentheses and multiply it by each term in the parentheses.
Area
the space inside a figure.
### Guided Practice
Here is one for you to try on your own.
Use the Distributive Property to find the area of the rectangles.
First, we can write an expression to solve it.
Next, we can solve it.
The area of the two rectangles is .
### Practice
Directions: Practice using the Distributive Property to solve each problem.
1. 3.2(4 + 7)
2. 2.5(6 + 8)
3. 1.5(2 + 3)
4. 3.1(4 + 15)
5. 6.5(2 + 9)
6. 7.5(2 + 3)
7. 8.2(9 + 3)
8. 4(5.5 + 9)
9. 5(3.5 + 7)
10. 2(4.5 + 5)
11. 3.5(2.5 + 3)
12. 2.5(9 + 1.5)
13. 3.2(7 + 8.3)
14. 1.5(8.9 + 2.5)
15. 3.5(2.5 + 8.2)
### Vocabulary Language: English
Area
Area
Area is the space within the perimeter of a two-dimensional figure.
distributive property
distributive property
The distributive property states that the product of an expression and a sum is equal to the sum of the products of the expression and each term in the sum. For example, $a(b + c) = ab + ac$.
Numerical expression
Numerical expression
A numerical expression is a group of numbers and operations used to represent a quantity.
Product
Product
The product is the result after two amounts have been multiplied.
Sum
Sum
The sum is the result after two or more amounts have been added together.
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# Combinations and Permutations
What’s the Difference?
In English we use the word “combination” loosely, without thinking if the order of things is important. In other words:
“My fruit salad is a combination of apples, grapes and bananas” We don’t care what order the fruits are in, they could also be “bananas, grapes and apples” or “grapes, apples and bananas”, it’s the same fruit salad.
“The combination to the safe is 472”. Now we do care about the order. “724” won’t work, nor will “247”. It has to be exactly 4-7-2.
So, in Mathematics we use more precise language:
• When the order doesn’t matter, it is a Combination.
• When the order does matter it is a Permutation.
So, we should really call this a “Permutation Lock”!
In other words:
A Permutation is an ordered Combination.
To help you to remember, think “Permutation … Position”
## Permutations
There are basically two types of permutation:
1. Repetition is Allowed: such as the lock above. It could be “333”.
2. No Repetition: for example, the first three people in a running race. You can’t be first and second.
### 1. Permutations with Repetition
These are the easiest to calculate.
When a thing has n different types … we have n choices each time!
For example: choosing 3 of those things, the permutations are:
n × n × n
(n multiplied 3 times)
More generally: choosing r of something that has n different types, the permutations are:
n × n × … (r times)
(In other words, there are n possibilities for the first choice, THEN there are n possibilities for the second choice, and so on, multiplying each time.)
Which is easier to write down using an exponent of r:
n × n × … (r times) = nr
Example: in the lock above, there are 10 numbers to choose from (0,1,2,3,4,5,6,7,8,9) and we choose 3 of them:
10 × 10 × … (3 times) = 103 = 1,000 permutations
So, the formula is simply:
nr where n is the number of things to choose from, and we choose r of them, repetition is allowed, and order matters.
### 2. Permutations without Repetition
In this case, we have to reduce the number of available choices each time.
### Example: what order could 16 pool balls be in?
After choosing, say, number “14” we can’t choose it again.
So, our first choice has 16 possibilities, and our next choice has 15 possibilities, then 14, 13, 12, 11, … etc. And the total permutations are:
16 × 15 × 14 × 13 × … = 20,922,789,888,000
But maybe we don’t want to choose them all, just 3 of them, and that is then:
16 × 15 × 14 = 3,360
In other words, there are 3,360 different ways that 3 pool balls could be arranged out of 16 balls.
Without repetition our choices get reduced each time.
But how do we write that mathematically? Answer: we use the “factorial function”
The factorial function (symbol:) just means to multiply a series of descending natural numbers. Examples: 4! = 4 × 3 × 2 × 1 = 24 7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5,040 1! = 1 Note: it is generally agreed that 0! = 1. It may seem funny that multiplying no numbers together gets us 1, but it helps simplify a lot of equations.
So, when we want to select all of the billiard balls the permutations are:
16! = 20,922,789,888,000
But when we want to select just 3 we don’t want to multiply after 14. How do we do that? There is a neat trick: we divide by 13!
16 × 15 × 14 × 13 × 12 …13 × 12 … = 16 × 15 × 14
That was neat. The 13 × 12 × … etc. gets “cancelled out”, leaving only 16 × 15 × 14.
The formula is written:
n!(n − r)! where n is the number of things to choose from, and we choose r of them, no repetitions, order matters.
### Example Our “order of 3 out of 16 pool balls example” is:
16! = 16! = 20,922,789,888,000 = 3,360 (16-3)! 13! 6,227,020,800
(which is just the same as: 16 × 15 × 14 = 3,360)
### Example: How many ways can first and second place be awarded to 10 people?
10! = 10! = 3,628,800 = 90 (10-2)! 8! 40,320
(which is just the same as: 10 × 9 = 90)
### Notation
Instead of writing the whole formula, people use different notations such as these:
## Combinations
There are also two types of combinations (remember the order does not matter now):
1. Repetition is Allowed: such as coins in your pocket (5,5,5,10,10)
2. No Repetition: such as lottery numbers (2,14,15,27,30,33)
### 1. Combinations with Repetition
Actually, these are the hardest to explain, so we will come back to this later.
### 2. Combinations without Repetition
This is how lotteries work. The numbers are drawn one at a time, and if we have the lucky numbers (no matter what order) we win!
The easiest way to explain it is to:
• assume that the order does matter (i.e. permutations),
• then alter it so the order does not matter.
Going back to our pool ball example, let’s say we just want to know which 3 pool balls are chosen, not the order.
We already know that 3 out of 16 gave us 3,360 permutations.
But many of those are the same to us now, because we don’t care what order!
For example, let us say balls 1, 2 and 3 are chosen. These are the possibilities:
Order does matter Order doesn’t matter 1 2 3 1 3 2 2 1 3 2 3 1 3 1 2 3 2 1 1 2 3
So, the permutations have 6 times as many possibilities.
In fact, there is an easy way to work out how many ways “1 2 3” could be placed in order, and we have already talked about it. The answer is:
3! = 3 × 2 × 1 = 6
(Another example: 4 things can be placed in 4! = 4 × 3 × 2 × 1 = 24 different ways, try it for yourself!)
So, we adjust our permutations formula to reduce it by how many ways the objects could be in order (because we aren’t interested in their order any more):
That formula is so important it is often just written in big parentheses like this:
where n is the number of things to choose from, and we choose r of them, no repetition, order doesn’t matter.
It is often called “n choose r” (such as “16 choose 3”)
And is also known as the Binomial Coefficient.
### Notation
As well as the “big parentheses”, people also use these notations:
Just remember the formula:
n!r!(n − r)!
### Example: Pool Balls (without order)
So, our pool ball example (now without order) is:
16!3!(16−3)! = 16!3! × 13!
= 20,922,789,888,0006 × 6,227,020,800
= 560
Or we could do it this way:
16×15×143×2×1 = 33606 = 560
It is interesting to also note how this formula is nice and symmetrical:
In other words, choosing 3 balls out of 16, or choosing 13 balls out of 16 have the same number of combinations.
16!3!(16−3)! = 16!13!(16−13)! = 16!3! × 13! = 560
### Pascal’s Triangle
We can also use Pascal’s Triangle to find the values. Go down to row “n” (the top row is 0), and then along “r” places and the value there is our answer. Here is an extract showing row 16:
1 14 91 364 …
1 15 105 455 1365 …
1 16 120 560 1820 4368 …
### 1. Combinations with Repetition
OK, now we can tackle this one …
Let us say there are five flavors of ice-cream: banana, chocolate, lemon, strawberry and vanilla.
We can have three scoops. How many variations will there be?
Let’s use letters for the flavors: {b, c, l, s, v}. Example selections include
• {c, c, c} (3 scoops of chocolate)
• {b, l, v} (one each of banana, lemon and vanilla)
• {b, v, v} (one of banana, two of vanilla)
(And just to be clear: There are n=5 things to choose from, and we choose r=3 of them.
Order does not matter, and we can repeat!)
Now, I can’t describe directly to you how to calculate this, but I can show you a special technique that lets you work it out.
Think about the ice cream being in boxes, we could say “move past the first box, then take 3 scoops, then move along 3 more boxes to the end” and we will have 3 scoops of chocolate!
So, it is like we are ordering a robot to get our ice cream, but it doesn’t change anything, we still get what we want.
We can write this down as (arrow means move, circle means scoop).
In fact, the three examples above can be written like this:
{c, c, c} (3 scoops of chocolate): {b, l, v} (one each of banana, lemon and vanilla): {b, v, v} (one of banana, two of vanilla):
OK, so instead of worrying about different flavors, we have a simpler question: “how many different ways can we arrange arrows and circles?”
Notice that there are always 3 circles (3 scoops of ice cream) and 4 arrows (we need to move 4 times to go from the 1st to 5th container).
So (being general here) there are r + (n−1) positions, and we want to choose r of them to have circles.
This is like saying “we have r + (n−1) pool balls and want to choose r of them”. In other words, it is now like the pool balls question, but with slightly changed numbers. And we can write it like this:
where n is the number of things to choose from, and we choose r of them repetition allowed, order doesn’t matter.
Interestingly, we can look at the arrows instead of the circles, and say “we have r + (n−1) positions and want to choose (n−1) of them to have arrows”, and the answer is the same:
So, what about our example, what is the answer?
(3+5−1)! = 7! = 5040 = 35 3!(5−1)! 3!×4! 6×24
There are 35 ways of having 3 scoops from five flavors of ice-cream.
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## Finding the Area of Parallelograms
6-8
1
Students will use their knowledge of rectangles to discover the area formula for parallelograms.
In this lesson, students will use the area formula for rectangles to discover the area formula for parallelograms. As necessary, review the concept of area as well as the area formula, A = bh.
It may also be beneficial to review the properties of parallelograms as they relate to other quadrilaterals. You might ask students questions such as, "Is every square a parallelogram?" or "Is any parallelogram also a rhombus?" To help students answer these questions, you may want to use the diagram below; or, you may want to take some time to create the diagram with your students.
To begin the lesson, have students look at a U.S. map. Ask students, "What state is in the shape of a parallelogram?" Although not exactly, Tennessee is roughly a parallelogram. Working in groups of three, students should discuss methods that could be used to determine the area of Tennessee. (At the end of the lesson, students will return to this problem and discuss their findings with the class.)
Distribute the Rectangles and Parallelograms and the Quadrilateral Area Record Activity Sheets. Give students time to determine the area of shapes A-E, and have them record the information on the record sheet.
For each rectangle, students can simply count the number of squares, or they can multiply the length by the width. Alternatively, they could practice either metric or customary measurements by measuring the length and width using a ruler, and then multiplying to find the area. For each parallelogram, students will likely need to count the squares to determine the area; they will need to combine partial squares to form full squares when making their estimates.
Then, have student cut out shapes A, B, and C.
With rectangle A, have students cut from the lower left corner to a point on the top edge that is three units in from the upper left vertex; this cut will form a 45-degree angle, which divides each of the squares through which it passes exactly in half. Then, for rectangles B and C on the activity sheet, have students remove a triangle by cutting from the lower left corner diagonally to any point along the top edge. One such cut is shown below. (It might be helpful for students to first draw a straight line with a ruler. You should also encourage them to choose a point along the top where the edge and one of the grid lines meet.) As shown below, students should then place the removed triangle at the other end of the rectangle. Encourage students to make a different cut than other members of their group.
Students should then determine the area of the resulting parallelograms and record the results on the record sheet.
Working in the other direction, have students cut out shapes D and E. From these shapes, have students remove the right triangle on either the right or left side, and move it to the other side. Students should realize that this modification changed the parallelogram to a rectangle with the same area. As above, students should determine the area of the resulting rectangle and record the information on the record sheet.
As students complete the record sheet, verify the accuracy of their measurements and calculations. Without correct values, students will not discover that the formulas for rectangles and parallelograms are identical.
To explore other shapes, students can use the Shape Cutter tool.
Students can create either a rectangle or parallelogram, make an appropriate cut, and then rearrange the pieces. Further, students could even make other cuts to show that two non-rectangular parallelograms with the same base and height have the same area, as shown below. (If the red piece were moved to the other side, notice that a different parallelogram would be formed.)
After the explorations, discuss what happened as a class, and pose the following questions to students:
• How was the shape changed? [The rectangle was made into a parallelogram. Or, the parallelogram was made into a rectangle.]
• Have the dimensions changed? [The side lengths have changed, but the base and height have not changed.]
• Has the area changed? [No.]
• Did other students in your group make a different cut? If so, did they get a different area than you did? [The area should be the same, regardless of the cut that was made.]
At the conclusion of this discussion, students should realize that the area formula for a rectangle, A = bh, is also the area formula for parallelograms.
As a final piece, students should also recognize the relationship between the area formulas for trapezoids and parallelograms. The formula for a trapezoid is A = ½(b1 + b2) × h. For a parallelogram, the bases are equal, so b1 = b2. Therefore, using the trapezoid formula to calculate the area of a parallelogram results in the following:
A = ½(b1 + b2) × h = ½(b1 + b1) × h = b1h
As in previous lessons in this unit, students can also use the Area Tool for Parallelograms to investigate the relationship of the height and the length of the base to the area of a parallelogram.
At the end of the lesson, return to the motivating problem: What is the area of the state of Tennessee? Students should now use measurements from the map to determine the height and base, and then they should use the formula to find the area.
Assessment Options
1. Have students determine the area of the state of Tennessee.
2. Allow students to work in pairs. Each student should create a parallelogram, measure the dimensions, and calculate the area. Then, they should tell each other the area of their parallelogram and one of the dimensions (either height or base). Their partner should determine the unknown dimension. To ensure that students have experience using various sizes, distribute blank pieces of 8½" × 11" paper, or use geoboards.
3. Provide problems for students to complete individually. For example, a backyard shaped like a parallelogram with a base of 7.9 yards and a height of 2.3 yards. What is the area of this backyard? [18.17 yards2.] In addition, provide problems for which the area and either the base or height are known, and have students find the missing factor.
Extensions
1. Students can create parallelograms by giving coordinates on the coordinate plane; another student can draw the parallelogram and determine its area. Students can use the distance formula to calculate the base and height if the parallelogram is not in a typical orientation.
2. Students can create parallelograms of varying sizes using geoboards. In pairs, one student can create a parallelogram, and the other student can determine its area using the area formula. To verify the result, students can use Pick’s Theorem: I + ½P - 1, where I is the number of points in the interior of the polygon and P is the number of points on the perimeter of the polygon.
3. Annabella Milbanke (aka, Anne Isabelle Milbanke, Lady Byron) was known as the "Princess of Parallelograms." Students can research her on the Internet. Students can report to the class on why Milbanke was known as the Princess of Parallelograms.
4. Move on to the last lesson, Finding the Area of Irregular Figures.
Questions for Students
1. What relationship is shared by parallelograms and rectangles that allow the same formula to be used to find the area of each?
[A rectangle and parallelogram with the same base and height have the same area. When a triangle is removed from a rectangle and reattached to form a parallelogram, the base, height, and area remain the same.]
2. Other than using a formula, what methods could you use to determine the area of a parallelogram? Give an example and show step-by-step how you found the area. Which is easier — your method, or using the formula?
[A parallelogram could be divided into several parts. One possible division is dividing it into a rectangle surrounded by two triangles, as shown below. The area of each piece could then be calculated, and the areas could be added together.
Although this method will yield the correct answer, it requires using an area formula three times instead of just once. Therefore, it is probably easier to use the parallelogram formula.]
3. When finding the area of a parallelogram, why multiply base times height, instead of base times side?
[The height is the perpendicular distance from the base to the top. The length of the side can change, depending on the orientation of the parallelogram, but the height never changes.]
4. Can the area formula for parallelograms be extended to rhombuses? Why or why not?
[Yes, because a rhombus is a parallelogram with four congruent sides.]
Teacher Reflection
• What alternative methods did students use to calculate the area of parallelograms? Will they always work? Did students clearly explain these methods?
• Did you find the online assessments worthwhile? Did they relate to the lesson? Did they challenge students? If not, how could you change the online assessments so that they were more challenging?
• Were students involved throughout the lesson?
• How does this lesson address the needs of the diverse learner?
### Discovering the Area Formula for Triangles
6-8
In this lesson, students develop the area formula for a triangle. Students find the area of rectangles and squares, and compare them to the areas of triangles derived from the original shape.
### Finding the Area of Parallelograms
6-8
Students will use their knowledge of rectangles to discover the area formula for parallelograms.
### Finding the Area of Trapezoids
6-8
Students discover the area formula for trapezoids, as well as explore alternative methods for calculating the area of a trapezoid.
### Finding the Area of Irregular Figures
6-8
Students will estimate the areas of highly irregular shapes and will use a process of decomposition to calculate the areas of irregular polygons.
### Learning Objectives
Students will:
• Measure rectangles, using appropriate units of measure, and determine their areas.
• Use their knowledge of rectangle formulas to derive an area formula for parallelograms.
• Explore alternative methods for determining the area of parallelograms.
• Use the formula they have derived to calculate the area of a parallelogram (given the base and the height).
### NCTM Standards and Expectations
• Understand, select, and use units of appropriate size and type to measure angles, perimeter, area, surface area, and volume.
• Select and apply techniques and tools to accurately find length, area, volume, and angle measures to appropriate levels of precision.
• Develop and use formulas to determine the circumference of circles and the area of triangles, parallelograms, trapezoids, and circles and develop strategies to find the area of more-complex shapes.
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### Transcription
21DecimalsIn most everyday applications, one encounters numbers written in decimalnotation such as the price of a comodity, the Gross National Product, thediameter of an atom, etc. In this section, we introduce the concept of decimalnumbers.The word decimal comes from the Latin decem, meaning ”ten”. Thus, decimal numbers can be expressed as powers of ten. To be more precise, adecimal number such as 128.294 can be written in expanded form as128.294 1 102 2 101 8 100 2 111 9 2 4 3.101010The dot in 128.294 is called the decimal point.Decimal place value is an extension of whole number place value, and it hasa symmetry as shown in the table of Figure 21.1Figure 21.1Remark 21.1Note that every whole number is also a decimal number. For example, 15 15.0.Pictorial Representations of DecimalsFor beginning grade school students, it is helpful to introduce decimals usingpictorial representations. We can use base ten blocks and decide that 1 flat11represents a unit, 1 long represents 10and 1 cube represents 100as shown inFigure 21.2.1
Figure 21.2In this model, the number 1.23 is represented in Figure 21.3.Figure 21.3To represent a decimal such as 3.235, we can think of the block shown in11Figure 21.4(a) as a unit. Then a flat represents 10, a long represents 100,1and a cube represents 1000 . Using these objects, we show a representation of3.235 in Figure 21.4(b).2
Figure 21.4The number 3.235 is read ”three and two hundred thirty five thousandths”.Example 21.1Write the number 30.0012 in expanded form.Solution.The expanded form is1111 0 2 1 3 2 41010101030.0012 3 101 0 100 0 Multiplying and Dividing Decimals by Powers of TenLet’s see the effect of multiplying a decimal number by a power of ten. Consider the number 25.723. writing this number in expanded form we find25.723 2 10 5 7 111 2 3 .1010010003
If we multiply this number by 100 102 then using the distributive propertywe obtain111(100) (25.723) (100) (2 10 5 7 10 2 100 3 1000)11 100 2 10 100 5 100 7 10 100 2 1001 100 3 10001 2000 500 70 2 3 10 2572.3and the notational effect is to move the decimal point two places to the right.Note that 2 is the exponent of the power of 10 we are multiplying by andalso the number of zeros in 100.Now, let’s try and divide 25.723 by 100. In this case, we have the followingcomputation.25.723 100 111(2 10 5 7 10 2 100 3 1000) 100111111 2 10 100 5 100 7 10 100 2 10010011 3 1000100111112 10 5 100 7 1000 2 10000 3 1000000.25723so that the effect is to move the decimal point two places to the left. Summarizing we have the following rules:(1) Multiplying a decimal number by 10n , where n W, is the same as moving the decimal point n places to the right.(2) Dividing a decimal number by 10n , where n W, is the same as movingthe decimal point n places to the left.Example 21.2Compute each of the following:(a) (103 )·(253.26) (b) (253.26) 103 (c) (1000)·(34.764) (d) 34.764 1000Solution.(a) (103 ) · (253.26) 253, 260.(b) (253.26) 103 0.25326(c) (1000) · (34.764) 34, 764(d) 34.764 1000 0.034764Representing a Terminating Decimal as a FractionA number such as 0.33 . . . where the ellipsis dots indicate that the string4
of 3s continues without end is called a nonterminating decimal. On thecontrary, a decimal like 24.357, which has finitely many digits to the right ofthe decimal point, is an example of a terminating decimal. Such a numbercan be written as a fraction. To see this, write first the expanded form24.357 2 10 4 357 .10 100 1000Now, find the common denominator and obtain24.357 30050724, 35724000 .10001000 1000 10001000Note that the number of zeros at the bottom is just the number of digits tothe right of the decimal point.Now, what about converting a fraction into a terminating decimal number.Not all fractions have terminating decimal expansion. For example, 13 0.33 · · · . However, a fraction ab , where the prime factorizations of b consistsonly of powers of 2 and 5, has a terminating decimal expansion. We illustratethis in the following example.Example 21.343as a decimal number.Express 1250Solution.4343 2·54 125043·2324 ·54 34410,000 0.0344.Practice ProblemsProblem 21.1Write the following decimals in expanded form.(a) 273.412 (b) 0.000723 (c) 0.020305Problem 21.2Write the following decimals as fractions in simplified form and determinethe prime factorization of the denominator in each case.(a) 0.324 (b) 0.028 (c) 4.25Problem 21.3Write these fractions as terminating decimals.7(b) 2218·54(a) 205
Problem 21.4If you move the decimal point in a number two places to the left, the valueof the number is divided byor multiplied by.Problem 21.5Write each of the following as a decimal number.(a) Forty-one and sixteen hundredths(b) Seven and five thousandthsProblem 21.6You ask a fourth grader to add 4.21 18. The child asks,”Where is thedecimal point in 18?” How would you respond?Problem 21.7A sign in a store mistakenly says that apples are sold for .89 cents a pound.What should the sign say?Problem 21.8Write each of the following numbers in expanded form.(a) 0.023 (b) 206.06 (c) 0.000132 (d) 312.0103Problem 21.9Rewrite the following sums as decimals.7 (a) 4 103 3 102 5 10 6 10683(b) 4 10 10 102(c) 4 104 10322(d) 10 1044 10778102Problem 21.10Write each of the following as numerals.(a) Five hundred thirty six and seventy-six ten thousandths(b) Three and eight thousandths(c) Four hundred thirty-six millionths(d) Five million and two tenthsProblem 21.11Write each of the following terminating decimals in common fractions.(a) 0.436 (b) 25.16 (c) 28.19026
Problem 21.12Determine which of the following represent terminating decimals.26(b) 133(c) 65(a) 2612 ·5625Problem 21.13Explain how to use base-ten blocks to represent ”two and three hundredfourty-five thousandths”Problem 21.14Write the following numbers in words.(a) 0.013 (b) 68,485.532 (c) 0.0082 (d) 859.080509Problem 21.15Determine, without converting to decimals, which of the following fractionshas a terminating decimal representation.625421(b) 326(c) 125(d) 130(a) 45400Problem 21.16A student reads the number 3147 as ”three thousand one hundred and fortyseven” What’s wrong with this reading?Problem 21.17It is possible to write a decimal number in the form M 10n where 1 M 10 and n {0, 1, 2, 3, · · ·}. This is known as the scientific notation. Such a notation is useful in expressing large numbers. For example,760, 000, 000, 000 7.6 109 . Write each of the following in scientific notation.(a) 4326 (b) 1,000,000 (c) 64,020,000 (d) 71,000,000,000 (e) 0.0001236Problem 21.18Find each of the following products and quotients.(a) (6.75)(1, 000, 000) (b) 19.514 100, 000 (c) (2.96 1016 )(1012 )(d)2.96 10161012Ordering Terminating DecimalsWe use two methods for comparing two terminating decimal numbers. Fraction MethodTwo decimals can be ordered by converting each to fractions in the forma, where a and b are whole numbers, and determine which is greater. Webillustrate this method in the next example.7
Example 21.4Compare the numbers 0.9 and 0.36.Solution.Writing each number as a fraction we find 0.9 Thus,3690 0.360.9 100100910 90100and 0.36 36.100 Place Value MethodOrdering decimals with this method is much like ordering whole numbers.For example, to determine the larger of 247,761 and 2,326,447 write bothnumerals as if they had the same number of digits (by adding zeros whennecessary); that is, write0, 247, 761 and 2, 326, 447.Next, start at the left and find the first place value where the face values aredifferent and compare these two digits. The number containing the greaterface value in this place is the greater of the two original numbers. In ourexample, the first place value from the left where the face values are differentis in the ”million” position. Since 0 2, we have 247, 761 2, 326, 447. Thesame process applies when comparing decimal numbers as shown in the nextexample.Example 21.5Compare 2.35714 and 2.35709Solution.The first digits from the left that differ are 1 and 0. Since 0 1 we concludethat 2.35709 2.35714Practice ProblemsProblem 21.19Order the following decimals from greatest to lowest: 13.4919, 13.492, 13.49183,13.49199.8
Problem 21.20If the numbers 0.804, 0.84 and 0.8399 are arranged on a number line, whichis furthest to the right?Problem 21.21Which of the following numbers is the greatest: 100, 0003 , 10005 , 100, 0002 ?Problem 21.22The five top swimmers in an event had the following times.Emily64.54 secondsMolly64.46 secondsMartha 63.59 secondsKathy64.02 secondsRhonda 63.54 secondsList them in the order they placed.Problem 21.23Write the following numbers from smallest to largest: 25.412, 25.312, 24.999,25.412412412.Problem 21.24Order the following from smallest to largest by changing each fraction to a11 17, 29 .decimal: 53 , 18Mental Computation and EstimationSome of the tools used for mental computations with whole numbers can beused to perform mental computations with decimals, as seen in the following: Using Compatible Numbers7.91 3.85 4.09 0.15 (7.91 4.09) (3.85 0.15) 12 4 16. Using Properties17 0.25 23 0.25 (17 23) 0.25 40 0.25 10where we used the distributive property of multiplication over addition. Compensation3.76 1.98 3.74 2 5.749
using additive compensation.Computational estimations of operations on whole numbers and fractionscan also be applied to estimate the results of decimal operations.Example 21.6Estimate 1.57 4.36 8.78 using (i) range estimation, (ii) front-end adjustment, and (iii) rounding.Solution.Range: A low estimate is 1 4 8 13 and a high estimate is 2 5 9 16.Front-end: Since 0.57 0.36 0.78 1.50, 1.57 4.36 8.78 13 1.50 14.50.Rounding: Rounding to the nearest whole number we obtain 1.57 4.36 8.78 2 4 9 15.Practice ProblemsProblem 21.25Round 0.3678(a) up to the next hundredth(b) down to the preceding hundredth(c) to the nearest hundredth.Problem 21.26Suppose that labels are sold in packs of 100.(a) If you need 640 labels, how many labels would you have to buy?(b) Does this application require rounding up, down, or to the ”nearest”?Problem 21.27Mount Everest has an altitude of 8847.6 m and Mount Api has an altitudeof 7132.1 m. How much higher is Mount Everest than Mount Api?(a) Estimate using rounding.(b) Estimate using the front-end strategy.Problem 21.28A 46-oz can of apple juice costs 1.29. How can you estimate the cost perounce?10
Problem 21.29Determine by estimating which of the following answers could not be correct.(a) 2.13 0.625 1.505(b) 374 1.1 41.14(c) 43.74 2.2 19.88181818.Problem 21.30Calculate mentally. Describe your method.(a) 18.43 9.96(b) 1.3 5.9 1.3 64.1(c) 4.6 (5.8 2.4)(d) 51.24 103(e) 0.15 105Problem 21.31Estimate using the indicated techniques.(a) 4.75 5.91 7.36 using range and rounding to the nearest whole number.(b) 74.5 6.1; range and rounding.(c) 3.18 4.39 2.73 front-end with adjustment.(d) 4.3 9.7 rounding to the nearest whole number.Problem 21.32Round the following.(a) 97.26 to the nearest tenth(b) 345.51 to the nearest ten(c) 345.00 to the nearest ten(d) 0.01826 to the nearest thousandth(e) 0.498 to the nearest tenth11
Figure 21.4(a) as a unit. Then a at represents 1 10, a long represents 1 100, and a cube represents 1 1000: Using these objects, we show a representation of 3.235 in Figure 21.4(b). 2. Figure 21.4 The number 3.235 is read "three and two hundred thirty ve thousandths". Example 21.1 Write the number 30.0012 in expanded form. Solution. The expanded form is 30:0012 3 101 0 100 0 1 10 0 1 .
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# Video: MATH-DIFF-INT-2018-S1-Q05
Let π₯ = 2π‘Β³ β 15π‘Β² + 36π‘ β 1 and π¦ = π‘Β² β 8π‘ + 11. For which values of π‘ does this curve have a vertical tangent?
04:49
### Video Transcript
Let π₯ equal two π‘ cubed two minus 15π‘ squared plus 36π‘ minus one and π¦ equal π‘ squared minus eight π‘ plus 11. For which values of π‘ does this curve have a vertical tangent?
Letβs start by working out what the question is asking us to find, mathematically, when it asks us to find the values of π‘ where this curve has a vertical tangent. Now, if the question hadβve asked us to find a horizontal tangent, what weβd have been looking for is points on the curve where the tangent to the curve is horizontal. Now, these will be the points where the gradient of the line is zero. And so, we can say that at these points, dπ¦ by dπ₯ will be equal to zero. Or, the change in π¦ with respect to π₯ is zero.
Now, the question asks us to find the values of π‘ for which the curve has a vertical tangent and not a horizontal tangent. So what this means is the points on the line where the tangent to that point is vertical. And so, we can say that at these points, the change in π₯ with respect to π¦ is zero. In mathematical terms, this means that dπ₯ by dπ¦ is equal to zero, since dπ₯ by dπ¦ represents the change in π₯ with respect to π¦. In order to find the values of π‘ for which the curve has vertical tangent, we can simply find the values of π‘ so that dπ₯ by dπ¦ is equal to zero.
Now, the equation for our curve is in parametric equations. And in order to find dπ¦ by dπ₯ or dπ₯ by dπ¦ in terms of parametrics, we will use the following equation. We have that dπ¦ by dπ₯ is equal to dπ¦ by dπ‘ over dπ₯ by dπ‘. From this, we obtain that dπ₯ by dπ¦ is equal to dπ₯ by dπ‘ over dπ¦ by dπ‘. In our case, we have that dπ₯ by dπ¦ is equal to zero. And since dπ₯ by dπ¦ is also equal to dπ₯ by dπ‘ over dπ¦ by dπ‘, we obtain that dπ₯ by dπ‘ over dπ¦ by dπ‘ is equal to zero.
Next, we will use the fact that if any fraction is equal to zero, then this means that the numerator must be equal to zero. And so, therefore, this tells us that dπ₯ by dπ‘ is equal to zero. The next step is to find dπ₯ by dπ‘. Weβre given in the question that π₯ is equal to two π‘ cubed minus 15π‘ squared plus 36π‘ minus one. And this is simply a polynomial. When differentiating polynomial terms, we simply multiply by the power and decrease the power by one.
For the first term here, we have two π‘ cubed. So when we differentiate this with respect to π‘, we multiply by the power, so thatβs multiplying by three. And a two times three gives us a six. Then, we decrease the power by one, so π‘ cubed goes to π‘ squared, leaving us with six π‘ squared. For the next term, we have negative 15π‘ squared. So we multiply it by the power to give us negative 30. Then we decrease the power by one. So we end up with π‘ to the power of one, which is simply π‘. And so, this term becomes negative 30π‘. Then, the next term is 36π‘. And we multiply it by the power of π‘, which is just one, to give us 36 and decrease the power of π‘ by one. So that becomes π‘ to the power of zero or just one.
Then, the last term here is negative one, which is simply a constant. And when we differentiate a constant by anything, it always becomes zero. And so, this term differentiates to zero. However, adding zero does absolutely nothing here. So we can just ignore it. Now, we have found dπ₯ by dπ‘. We can set it equal to zero to give us that six π‘ squared minus 30π‘ plus 36 is equal to zero. We spot that each term here is a multiple of six. So we can divide the whole equation by six, which gives us π‘ squared minus five π‘ plus six is equal to zero.
Factorising this gives us π‘ minus two times π‘ minus three is equal to zero. So we obtain that π‘ minus two equals zero, which gives us that π‘ is equal to two. Or, π‘ minus three is equal to zero, which gives us that π‘ is equal to three. And so, we have found the values of π‘ for which dπ₯ by dπ‘ is equal to zero. And, therefore, these values of π‘ also give dπ₯ by dπ¦ is equal to zero. And so, these must be the values of π‘ for which the curve has a vertical tangent, giving us a solution of π‘ is equal to two or π‘ is equal to three.
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In mathematics, a partial derivative of a function of several variables is its derivative with respect to one of those variables with the others held constant (as opposed to the total derivative, in which all variables are allowed to vary). Partial derivatives are useful in vector calculus and differential geometry.
The partial derivative of a function ${\displaystyle f}$ with respect to the variable ${\displaystyle x}$ is written as ${\displaystyle f_{x},\partial _{x}f,{\frac {\partial f}{\partial x}}}$ . The partial-derivative symbol ${\displaystyle \partial }$ is a rounded letter, distinguished from the straight d of total-derivative notation. The notation was introduced by Adrien-Marie Legendre and gained general acceptance after its reintroduction by Carl Gustav Jacob Jacobi.[citation needed]
## Introduction
Suppose that ${\displaystyle f}$ is a function of more than one variable. For instance,
${\displaystyle f(x,y)=x^{2}+xy+y^{2}}$
A graph of ${\displaystyle z=x^{2}+xy+y^{2}}$ . We want to find the partial derivative at ${\displaystyle (1,1,3)}$ that leaves ${\displaystyle y}$ constant; the corresponding tangent line is parallel to the ${\displaystyle x}$-axis.
It is difficult to describe the derivative of such a function, as there are an infinite number of tangent lines to every point on this surface. Partial differentiation is the act of choosing one of these lines and finding its slope. Usually, the lines of most interest are those that are parallel to the ${\displaystyle x}$-axis, and those that are parallel to the ${\displaystyle y}$-axis.
This is a slice of the graph at the right at ${\displaystyle y=1}$
A good way to find these parallel lines is to treat the other variable as a constant. For example, to find the tangent line of the above function at ${\displaystyle (1,1,3)}$ that is parallel to the ${\displaystyle x}$-axis, we treat ${\displaystyle y}$ as a constant one. The graph and this plane are shown on the right. On the left, we see the way the function looks on the plane ${\displaystyle y=1}$ . By finding the tangent line on this graph, we discover that the slope of the tangent line of ${\displaystyle f}$ at ${\displaystyle (1,1,3)}$ that is parallel to the ${\displaystyle x}$-axis is three. We write this in notation as
${\displaystyle {\frac {\partial z}{\partial x}}(1,1,3)=3}$
or as "The partial derivative of ${\displaystyle z}$ with respect to ${\displaystyle x}$ at ${\displaystyle (1,1,3)}$ is 3."
## Definition
The function ${\displaystyle f}$ can be reinterpreted as a family of functions of one variable indexed by the other variables:
${\displaystyle f(x,y)=f_{x}(y)=x^{2}+xy+y^{2}}$
In other words, every value of ${\displaystyle x}$ defines a function, denoted ${\displaystyle f_{x}}$ , which is a function of one real number.[1] That is,
${\displaystyle f_{x}(y)=x^{2}+xy+y^{2}}$
Once a value of ${\displaystyle x=a}$ is chosen, then ${\displaystyle f(x,y)}$ determines a function ${\displaystyle f_{a}}$ which sends ${\displaystyle y}$ to ${\displaystyle a^{2}+ay+y^{2}}$ :
${\displaystyle f_{a}(y)=a^{2}+ay+y^{2}}$
In this expression, ${\displaystyle a}$ is a constant, not a variable, so ${\displaystyle f_{a}}$ is a function of only one real variable, that being ${\displaystyle y}$ . Consequently the definition of the derivative for a function of one variable applies:
${\displaystyle f_{a}'(y)=a+2y}$
The above procedure can be performed for any choice of ${\displaystyle a}$ . Assembling the derivatives together into a function gives a function which describes the variation of ${\displaystyle f}$ in the ${\displaystyle y}$ direction:
${\displaystyle {\frac {\partial f}{\partial y}}(x,y)=x+2y}$
This is the partial derivative of ${\displaystyle f}$ with respect to ${\displaystyle y}$ . Here ${\displaystyle \partial }$ is a rounded ${\displaystyle d}$ called the partial derivative symbol. To distinguish it from the letter ${\displaystyle d}$ , ${\displaystyle \partial }$ is sometimes pronounced "der", "del", "dah", or "partial" instead of "dee".
In general, the partial derivative of a function ${\displaystyle f(x_{1},\ldots ,x_{n})}$ in the direction ${\displaystyle x_{i}}$ at the point ${\displaystyle (a_{1},\ldots ,a_{n})}$ is defined to be:
${\displaystyle {\frac {\partial f}{\partial x_{i}}}(a_{1},\ldots ,a_{n})=\lim _{h\to 0}{\frac {f(a_{1},\ldots ,a_{i}+h,\ldots ,a_{n})-f(a_{1},\ldots ,a_{n})}{h}}}$
In the above difference quotient, all the variables except ${\displaystyle x_{i}}$ are held fixed. That choice of fixed values determines a function of one variable ${\displaystyle f_{a_{1},\ldots ,a_{i-1},a_{i+1},\ldots ,a_{n}}(x_{i})=f(a_{1},\ldots ,a_{i-1},x_{i},a_{i+1},\ldots ,a_{n})}$ , and by definition,
${\displaystyle {\frac {df_{a_{1},\ldots ,a_{i-1},a_{i+1},\ldots ,a_{n}}}{dx_{i}}}(a_{1},\ldots ,a_{n})={\frac {\partial f}{\partial x_{i}}}(a_{1},\ldots ,a_{n})}$
In other words, the different choices of ${\displaystyle a}$ index a family of one-variable functions just as in the example above. This expression also shows that the computation of partial derivatives reduces to the computation of one-variable derivatives.
An important example of a function of several variables is the case of a scalar-valued function ${\displaystyle f(x_{1},\ldots ,x_{n})}$ on a domain in Euclidean space ${\displaystyle \mathbb {R} ^{n}}$ (e.g. on ${\displaystyle \mathbb {R} ^{2}}$ or ${\displaystyle \mathbb {R} ^{3}}$). In this case ${\displaystyle f}$ has a partial derivative ${\displaystyle {\frac {\partial f}{\partial x_{j}}}}$ with respect to each variable ${\displaystyle x_{j}}$ . At the point ${\displaystyle a}$ , these partial derivatives define the vector
${\displaystyle \nabla f(a)=\left({\frac {\partial f}{\partial x_{1}}}(a),\ldots ,{\frac {\partial f}{\partial x_{n}}}(a)\right)}$
This vector is called the gradient of ${\displaystyle f}$ at ${\displaystyle a}$ . If ${\displaystyle f}$ is differentiable at every point in some domain, then the gradient is a vector-valued function ${\displaystyle \nabla f}$ which takes the point ${\displaystyle a}$ to the vector ${\displaystyle \nabla f(a)}$ . Consequently the gradient determines a vector field.
## Examples
The volume of a cone depends on height and radius
Consider the volume ${\displaystyle V}$ of a cone; it depends on the cone's height ${\displaystyle h}$ and its radius ${\displaystyle r}$ according to the formula
${\displaystyle V(r,h)={\frac {\pi }{3}}r^{2}h}$
The partial derivative of ${\displaystyle V}$ with respect to ${\displaystyle r}$ is
${\displaystyle {\frac {\partial V}{\partial r}}={\frac {2\pi }{3}}rh}$
It describes the rate with which a cone's volume changes if its radius is varied and its height is kept constant. The partial derivative with respect to ${\displaystyle h}$ is
${\displaystyle {\frac {\partial V}{\partial h}}={\frac {\pi }{3}}r^{2}}$
and represents the rate with which the volume changes if its height is varied and its radius is kept constant.
Now consider by contrast the total derivative of ${\displaystyle V}$ with respect to ${\displaystyle r}$ and ${\displaystyle h}$ . They are, respectively
${\displaystyle {\frac {dV}{dr}}=\overbrace {{\frac {2\pi }{3}}rh} ^{\frac {\partial V}{\partial r}}+\overbrace {{\frac {\pi }{3}}r^{2}} ^{\frac {\partial V}{\partial h}}\cdot {\frac {\partial h}{\partial r}}}$
and
${\displaystyle {\frac {dV}{dh}}=\overbrace {{\frac {\pi }{3}}r^{2}} ^{\frac {\partial V}{\partial h}}+\overbrace {{\frac {2\pi }{3}}rh} ^{\frac {\partial V}{\partial r}}\cdot {\frac {\partial r}{\partial h}}}$
We see that the difference between the total and partial derivative is the elimination of indirect dependencies between variables in the latter.
Now suppose that, for some reason, the cone's proportions have to stay the same, and the height and radius are in a fixed ratio ${\displaystyle k}$ :
${\displaystyle k={\frac {h}{r}}={\frac {\partial h}{\partial r}}}$
This gives the total derivative:
${\displaystyle {\frac {dV}{dr}}={\frac {2\pi }{3}}rh+k{\frac {\pi }{3}}r^{2}}$
Equations involving an unknown function's partial derivatives are called partial differential equations and are common in physics, engineering, and other sciences and applied disciplines.
## Notation
For the following examples, let ${\displaystyle f}$ be a function in ${\displaystyle x,y,z}$ .
First-order partial derivatives:
${\displaystyle {\frac {\partial f}{\partial x}}=f_{x}=\partial _{x}f}$
Second-order partial derivatives:
${\displaystyle {\frac {\partial ^{2}f}{\partial x^{2}}}=f_{xx}=\partial _{xx}f}$
Second-order mixed derivatives:
${\displaystyle {\frac {\partial ^{2}f}{\partial y\,\partial x}}={\frac {\partial }{\partial y}}\left({\frac {\partial f}{\partial x}}\right)=f_{xy}=\partial _{yx}f}$
Higher-order partial and mixed derivatives:
${\displaystyle {\frac {\partial ^{i+j+k}f}{\partial x^{i}\,\partial y^{j}\,\partial z^{k}}}=f^{(i,j,k)}}$
When dealing with functions of multiple variables, some of these variables may be related to each other, and it may be necessary to specify explicitly which variables are being held constant. In fields such as statistical mechanics, the partial derivative of ${\displaystyle f}$ with respect to ${\displaystyle x}$ , holding ${\displaystyle y}$ and ${\displaystyle z}$ constant, is often expressed as
${\displaystyle \left({\frac {\partial f}{\partial x}}\right)_{y,z}}$
## Formal definition and properties
Like ordinary derivatives, the partial derivative is defined as a limit. Let ${\displaystyle U}$ be an open subset of ${\displaystyle \mathbb {R} ^{n}}$ and ${\displaystyle f:U\to \mathbb {R} }$ a function. We define the partial derivative of ${\displaystyle f}$ at the point ${\displaystyle {\vec {a}}=(a_{1},\ldots ,a_{n})\in U}$ with respect to the ${\displaystyle i}$-th variable ${\displaystyle x_{i}}$ as
${\displaystyle {\frac {\partial }{\partial x_{i}}}f({\vec {a}})=\lim _{h\to 0}{\frac {f(a_{1},\ldots ,a_{i-1},a_{i}+h,a_{i+1},\ldots ,a_{n})-f(a_{1},\ldots ,a_{n})}{h}}}$
Even if all partial derivatives ${\displaystyle {\frac {\partial f}{\partial x_{i}}}(a)}$ exist at a given point ${\displaystyle a}$ , the function need not be continuous there. However, if all partial derivatives exist in a neighborhood of ${\displaystyle a}$ and are continuous there, then ${\displaystyle f}$ is totally differentiable in that neighborhood and the total derivative is continuous. In this case, we say that ${\displaystyle f}$ is a ${\displaystyle C^{1}}$ function. We can use this fact to generalize for vector valued functions (${\displaystyle f:U\to \mathbb {R} ^{m}}$) by carefully using a componentwise argument.
The partial derivative ${\displaystyle {\frac {\partial f}{\partial x}}}$ can be seen as another function defined on ${\displaystyle U}$ and can again be partially differentiated. If all mixed second order partial derivatives are continuous at a point (or on a set), we call ${\displaystyle f}$ a ${\displaystyle C^{>}2}$ function at that point (or on that set); in this case, the partial derivatives can be exchanged by Clairaut's theorem:
${\displaystyle {\frac {\partial ^{2}f}{\partial x_{i}\,\partial x_{j}}}={\frac {\partial ^{2}f}{\partial x_{j}\,\partial x_{i}}}}$
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Simplifying Algebraic Expressions Quiz 1
1. Which expression is not equivalent to the other three?A) 2n-3+4n-5B) -8-4n+10nC) 2(3n-4)D) 6(n-5)2. Jake says the expressions 3(x+2) and 3x+2 are equivalent. Marie says they are not. Who is correct?A) Jake,because they are both equal to 3x+2B) Marie, because the first expression is equal to 3x+5C) Marie, because the first expression is equal to 3x+63. Use the distributive property to rewrite the expression in simplest form:3(x+5)A) 3x+15B) 3x+8C) 3x+35D) 3x+54. Use the distributive property to rewrite the expression in simplest form:-4(m-7n)A) -4m-28nB) -4m+28nC) -4m+7nD) -4m-7n5. Use the distributive property to rewrite the expression in simplest form:6(3a+4b+2c)A) 18a+24b+12cB) 18a+10b+12cC) 18a+24b+2cD) 9a+10b+8c6. Use the distributive property to rewrite the expression in simplest form:-2(3p+7r-9)A) -6p+14r+18B) -6p-7r-18C) -6p-14r-18D) -6p-14r+187. How many terms are in the following expression?2x+4-5y-6A) 4B) 6C) 9D) 58. What are the Constants in the following expression?2x+4-5y-6A) 2 and 5B) 4 and 6C) 2 and -5D) 4 and -69. What are the coefficients in the following expression?2x+4-5y-6A) 4 and 6B) 2 and -5C) 2 and 5D) 4 and -610. Harry says (7+8)+9=(8+7)+9 is an example of the associative property. Hannah says it is an example of the commutative property. Who is correct and why?A) Hannah, because the order of numbers changed:B) Hannah, because the grouping changed.C) Harry, because the order of numbers changedD) Harry, because the grouping changed.11. Which expression is equivalent to 2.5(x - 1) + 4.5( -x - 2)?A) 7x+11.5B) -2x+11.5C) 7x-11.5D) -2x-11.512. Nell writes the expression (3.6x + 6) - 4. She rewrites the expression using the associative property. Which expression could Nell have written using the associative property?A) 3.6(x+6)-4B) 9.6x-4C) 5.6xD) 3.6x(6-4)13. Which expression is equivalent to 6x + 7.5?A) -3(-2x+2.5)B) -3(-2x-7.5)C) -3(-2x+7.5)D) -3(-2x-2.5)14. Mary solved the problem below and came up with the answer listed. Is * 5 Mary correct? If not, state what Mary did wrong and give the correct answer. 9y- 3 + 5y 9y- 5y - 3 Mary's answer: 4y - 3A) No, Mary should have added 5y to 9y and gotten 14y + 3.B) Yes, Mary's answer is correct.C) No, Mary should have added 5y to 9y and gotten 14y - 3.D) No, Mary should have subtracted 3 from 9y and then added 5y to get 11 y15. Simplify the expression: 9a+5-7a-2A) 2a-3B) 16a+3C) 16a-7D) 2a+316. Simplify the expression: -8m+3-6m-9A) -2m-6B) -14m+6C) -14m-6D) 14m-617. Simplify the expression: 4x-12-9x-5A) -5x-17B) -13x-17C) 5x-17D) -5x-718. Simplify the expression: 2(2x+10)+3x+4A) 7x+14B) 7x+17C) 5x+14D) 7x+2419. Simplify the expression: -3(9y+5)+ 7y-22A) -20y-27B) -14y-37C) -20y-37D) -20y-1720. A) -7x-2B) -9x+2C) -7x+2D) -7x+18
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# My Family's Lady Bugs (day 1 of 3)
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## Objective
SWBAT use ratios, percent, and graph to test for and express proportionality.
#### Big Idea
Students become comfortable using a variety of mathematical tools to show equivalence.
## Intro & Rationale
This lesson is the first of three in which students use ratio, percent, and graphing to test for proportionality. In this first one students are representing a single population with ratios and on a graph. The focus is on using the tools and emphasizing how they show equivalence. The focus in on multiple methods and representations and the teachers role is to highlight and share student ideas and allow them to lead the learning.
## Warm up
15 minutes
The warm up warm up percent ratio equivalence.docx asks tells students that 60% of the lady bugs in my yard have spots and asks them to figure out how many we would expect to find with spots with different given totals. The first total is 100, which is easy for them because they know that 60% means that 60 out of the 100 total will be spotted. They then need to scale up and simplify to find the number out of a total of 300 and then out of 10. The last one asks how many out of a total of 15 would be spotted.
As always, students work in small math family groups of 3 or 4 and I circulate to highlight multiple methods and good ideas that students come up with. In one of the videos (need for organization) one student recognizes the difficulty in sharing his work when it is disorganized. This is an idea I might share with the class. I may show his "disorganized" work under the document camera and say that "Zavier thinks his work looks a little 'all over the place', so he is going to try to organize it in a table to make it easier to share...what a great idea!"
I expect students to have a little struggle with the last one, because none of the equivalent ratios that we have so far, 60/100, 180/300, & 6/10, can be scaled directly to a denominator of 15. I expect students to realize that they need to simplify to 3/5 before scaling it up.
## Exploration
20 minutes
After they have finished the warm up I point out how they have actually found 60% of 300, 10, and 15 and then ask them to show what the information looks like in a graph. If they have not used a table for their data and are having trouble figuring out how to graph the information, I suggest organizing their work in the table first.
I provide them with the graph with axes already labeled and numbered, because I didn't want them to be able to fit all the data on the graph. I intend for them to struggle with 180/300 so they have to figure out another way to represent it on the graph. This helps to reinforce the concept of equivalence in proportions. As I circulate I listen for this question. I would expect them to ask if they can simplify, but if they don't I ask what they think they could try. If they still don't suggest it I tell them I will check in with some other groups and see if they have any ideas. If they do suggest simplifying I tell them to give it a try and see where that point ends up on the graph and if it makes sense to them. Some students may ask if it has to be simplified all the way to which I respond "I wonder if that matters? Try it and see where it ends up on the graph". As several students begin asking I draw it to the attention of the class, "several of you are wondering if you can simplify 180/300 to make it fit on the graph, what an interesting idea, I can't wait to see what that looks like!"
I also look to see who might be connecting the points and noticing again that it is a straight line through zero, which is not a new idea to them. When I see students doing this I will show their work to the class and ask if anyone else is noticing this pattern. This helps them to connect to the graph as a tool for showing equivalence and proportionality. Having a variety of tools for communicating their thinking and their solutions is really helpful for my ELL students.
## Whole class discussion
19 minutes
I ask a few follow up questions for whole class discussion that focus in on the tools that can be used to test for proportionality. After they have explained what each point represents or "says" and noticed that all the points they put on their graph lie in a straight line that passes through zero, I ask them to explain what this tells us about the ratios at each point. I expect them to conclude that they are all proportional or equivalent to each other and that they all show the same thing, 60% of the lady bugs are spotted. If students are a little unsure I might ask at each point "does this point say that 60% are spotted?", "how about this point?".
This is the perfect segue into the next questions that I might use to highlight the ratios and percents as tools. I might ask "How did we already know they are proportional before we graphed them?", or "How does this proportionality show in the ratios?" I want to include in the discussion the idea that no matter whether we simplify partially, fully, scale up to the percent or any other total the ratios are equivalent.
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Paul's Online Notes
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### Section 3.3 : Differentiation Formulas
18. Determine where the function $$h\left( z \right) = 6 + 40{z^3} - 5{z^4} - 4{z^5}$$ is increasing and decreasing.
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Hint : Recall the various interpretations of the derivative. One of them is exactly what we need to get the problem started.
Start Solution
Recall that one of the interpretations of the derivative is that it gives the rate of change of the function. Since we are talking about where the function is increasing and decreasing we are clearly talking about the rate of change of the function.
So, we’ll need the derivative.
$\require{bbox} \bbox[2pt,border:1px solid black]{{h'\left( z \right) = 120{z^2} - 20{z^3} - 20{z^4} = - 20{z^2}\left( {z + 3} \right)\left( {z - 2} \right)}}$
Note that the derivative was factored for later steps and doesn’t really need to be done in general.
Hint : Where is the function not changing?
Show Step 2
Next, we need to know where the function is not changing and so all we need to do is set the derivative equal to zero and solve.
$h'\left( z \right) = 0\hspace{0.5in} \Rightarrow \hspace{0.5in} - 20{z^2}\left( {z + 3} \right)\left( {z - 2} \right) = 0$
From this it is pretty easy to see that the derivative will be zero, and hence the function will not be moving, at,
$\require{bbox} \bbox[2pt,border:1px solid black]{{z = 0 \hspace{0.25in} z = - 3 \hspace{0.25in} z = 2}}$
Hint : How does the increasing/decreasing behavior of the function relate to the sign (positive or negative) of the derivative?
Show Step 3
To get the answer to this problem all we need to know is where the derivative is positive (and hence the function is increasing) or negative (and hence the function is decreasing). Because the derivative is continuous we know that the only place it can change sign is where the derivative is zero. So, as we did in this section a quick number line will give us the sign of the derivative for the various intervals.
Here is the number line for this problem.
From this we get the following increasing/decreasing information.
\require{bbox} \bbox[2pt,border:1px solid black]{\begin{align*}{\mbox{Increasing :}} & \,\, - 3 < z < 0,\,\,\,\,\,0 < z < 2\\ {\mbox{Decreasing :}} & \,\, - \infty < z < - 3,\,\,\,\,2 < z < \infty \end{align*}}
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# HOW TO FIND THE MISSING COORDINATE USING SLOPE
How to Find the Missing Coordinate Using Slope ?
Here we are going to see, how to find the missing coordinate using slope.
If the given points are collinear, then their slopes will be equal.
By suing this concept, we may find the missing coordinate. Let us look into some problems based on this concept.
## How to Find the Missing Coordinate Using Slope - Questions
Question 1 :
Show that the given points are collinear: (-3,-4) , (7,2) and (12,5)
Solution :
Let A (-3, -4) B (7, 2) and C (12, 5) the given points.
Slope of AB = (y2 - y1)/(x2 - x1)
= [2 - (-4)]/[7 - (-3)]
= (2 + 4)/(7 + 3)
= 6/10
= 3/5
Slope of BC = (y2 - y1)/(x2 - x1)
= (5 - 2)/(12 - 7)
= 3/5
Hence the given points are collinear.
Question 2 :
If the three points (3,-1) , (a, 3) and (1,-3) are collinear, find the value of a.
Solution :
Let the given points be A (3,-1) B (a, 3) and C (1,-3)
Slope of AB = (y2 - y1)/(x2 - x1)
= (3-(-1))/(a - 3)
= (3 + 1)/(a - 3)
= 4/(a-3) ----(1)
Slope of BC = (y2 - y1)/(x2 - x1)
= (-3-3)/(1 - a)
= -6/(1 - a) ----(2)
(1) = (2)
4/(a - 3) = -6/(1 - a)
4(1 - a) = -6(a - 3)
4 - 4a = -6a + 18
6a - 4a = 18 - 4
2a = 14
a = 7
Hence the value of a is 7.
Question 3 :
The line through the points (-2, a) and (9, 3) has slope -1/2. Find the value of a.
Solution :
Slope of the line passing through the given points = -1/2
(-2, a) (9, 3)
m (y2 - y1)/(x2 - x1)
m = (3 - a)/(9 - (-2))
(3 - a)/(9 + 2) = -1/2
(3 - a)/11 = -1/2
2(3 - a) = -11
6 - 2a = -11
-2a = -11 -6
a = 17/2
Hence the value of a is 17/2.
Question 4 :
The line through the points (-2,6) and (4,8) is perpendicular to the line through the points (8,12) and (x,24) . Find the value of x.
Solution :
Slope of the line passing through the points (-2, 6) and (4, 8).
m = (8 - 6)/(4 - (-2))
= 2/(4+2)
= 2/6
= 1/3 -----(1)
Slope of the line passing through the points (8, 12) and (x, 24) .
m = (24 - 12)/(x - 8)
= 12/(x - 8) -----(2)
Since these lines are perpendicular to each other,
(1/3) ⋅ 12/(x - 8) = -1
4/(x - 8) = -1
4 = -(x - 8)
4 = -x + 8
x = 8 - 4
x = 4
After having gone through the stuff given above, we hope that the students would have understood, "How to Find the Missing Coordinate Using Slope".
Apart from the stuff given in this section "How to Find the Missing Coordinate Using Slope"if you need any other stuff in math, please use our google custom search here.
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# LCM of 100 and 90
The lcm of 100 and 90 is the smallest positive integer that divides the numbers 100 and 90 without a remainder. Spelled out, it is the least common multiple of 100 and 90. Here you can find the lcm of 100 and 90, along with a total of three methods for computing it. In addition, we have a calculator you should check out. Not only can it determine the lcm of 100 and 90, but also that of three or more integers including hundred and ninety for example. Keep reading to learn everything about the lcm (100,90) and the terms related to it.
## What is the LCM of 100 and 90
If you just want to know what is the least common multiple of 100 and 90, it is 900. Usually, this is written as
lcm(100,90) = 900
The lcm of 100 and 90 can be obtained like this:
• The multiples of 100 are …, 800, 900, 1000, ….
• The multiples of 90 are …, 810, 900, 990, …
• The common multiples of 100 and 90 are n x 900, intersecting the two sets above, $\hspace{3px}n \hspace{3px}\epsilon\hspace{3px}\mathbb{Z}$.
• In the intersection multiples of 100 ∩ multiples of 90 the least positive element is 900.
• Therefore, the least common multiple of 100 and 90 is 900.
Taking the above into account you also know how to find all the common multiples of 100 and 90, not just the smallest. In the next section we show you how to calculate the lcm of hundred and ninety by means of two more methods.
## How to find the LCM of 100 and 90
The least common multiple of 100 and 90 can be computed by using the greatest common factor aka gcf of 100 and 90. This is the easiest approach:
lcm (100,90) = $\frac{100 \times 90}{gcf(100,90)} = \frac{9000}{10}$ = 900
Alternatively, the lcm of 100 and 90 can be found using the prime factorization of 100 and 90:
• The prime factorization of 100 is: 2 x 2 x 5 x 5
• The prime factorization of 90 is: 2 x 3 x 3 x 5
• Eliminate the duplicate factors of the two lists, then multiply them once with the remaining factors of the lists to get lcm(100,100) = 900
In any case, the easiest way to compute the lcm of two numbers like 100 and 90 is by using our calculator below. Note that it can also compute the lcm of more than two numbers, separated by a comma. For example, enter 100,90. Push the button only to start over.
The lcm is...
Similar searched terms on our site also include:
## Use of LCM of 100 and 90
What is the least common multiple of 100 and 90 used for? Answer: It is helpful for adding and subtracting fractions like 1/100 and 1/90. Just multiply the dividends and divisors by 9 and 10, respectively, such that the divisors have the value of 900, the lcm of 100 and 90.
$\frac{1}{100} + \frac{1}{90} = \frac{9}{900} + \frac{10}{900} = \frac{19}{900}$. $\hspace{30px}\frac{1}{100} – \frac{1}{90} = \frac{9}{900} – \frac{10}{900} = \frac{-1}{900}$.
## Properties of LCM of 100 and 90
The most important properties of the lcm(100,90) are:
• Commutative property: lcm(100,90) = lcm(90,100)
• Associative property: lcm(100,90,n) = lcm(lcm(90,100),n) $\hspace{10px}n\neq 0 \hspace{3px}\epsilon\hspace{3px}\mathbb{Z}$
The associativity is particularly useful to get the lcm of three or more numbers; our calculator makes use of it.
To sum up, the lcm of 100 and 90 is 900. In common notation: lcm (100,90) = 900.
If you have been searching for lcm 100 and 90 or lcm 100 90 then you have come to the correct page, too. The same is the true if you typed lcm for 100 and 90 in your favorite search engine.
Note that you can find the least common multiple of many integer pairs including hundred / ninety by using the search form in the sidebar of this page.
Questions and comments related to the lcm of 100 and 90 are really appreciated. Use the form below or send us a mail to get in touch.
Please hit the sharing buttons if our article about the least common multiple of 100 and 90 has been useful to you, and make sure to bookmark our site.
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Posted in Least Common Multiple
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Video: Identifying Missing Values in Additions Represented in Words and Numerals
Complete the following: Two hundreds + ＿ hundreds = five hundreds, 200 + ＿ = ＿.
01:55
Video Transcript
Complete the following, two hundreds plus what hundreds equals five hundreds. 200 plus what equals what.
If we look carefully at this problem, we can see that we’ve got two additions written in different ways. The first calculation uses words, two hundreds plus what hundreds equals five hundreds. And although we have two empty spaces in our second calculation, we can see that we’re using numerals, 200 plus what equals what? Let’s look at the first addition to begin with.
Our addition begins with two hundreds. On a place value grid, two hundreds look like this. Now, what do we add to two hundreds to give us five hundreds? Let’s count on from two until we reach five, two, three, four, five. We’ve added three more hundreds. Two hundreds plus three hundreds equal five hundreds.
Now, let’s perform the same addition, this time we’re going to represent it using numerals. 200 plus one, two, 300 equals 500. We’ve represented the same calculation in two different ways. Two hundreds plus three hundreds equals five hundreds. 200 plus 300 equals 500. The missing word and numbers are three, 300 and 500.
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# 265L13 - LESSON 13 Lagrange Multipliers READ Section 15.8...
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LESSON 13 Lagrange Multipliers READ: Section 15.8. NOTES: At the end of the last lesson, we looked at the problem of finding the maximum value of f ( x, y ) = x 2 + y for points ( x, y ) on the circle x 2 + y 2 = 1. The idea used there was to parameterize the circle as x = cos t , y = sin t , 0 t 2 π . Then those component functions were plugged into f for x and y . The result was a function of just one variable ( t ). Now we can determine the max and min of f along the boundary in the old-fashioned calculus I way. In this lesson, we will look at another way to determine the max and min of f along the boundary. To help visualize the problem, imagine we are walking about on the surface given by equation f ( x, y ) = x 2 + y , but we can’t walk just anywhere on the surface: we are constrained to the part of the surface directly above the circle x 2 + y 2 = 1 in the x , y -plane. The question we are trying to answer is: under that constraint, what is the highest point we will reach and what is the lowest point we will reach as we go all the way around above the circle. The question is usually posed as: what are the maximum and minimum values of x 2 + y subject to the constraint x 2 + y 2 = 1? Lagrange devised a method for solving such problems. The method can be explained geometrically. Suppose the maximum value on the surface z = f ( x, y ) directly above the constraint curve g ( x, y ) = 0 is M . (In our example above, the constraint curve would be given by equation g ( x, y ) = x 2 + y 2 - 1 = 0.) Imagine the level curve for the surface produced by slicing through the surface at z = M . Now down in the x , y -plane, the constraint curve and the level curve will intersect, since there is a spot on the surface with z = M somewhere above the constraint curve. Call the point of intersection ( x 0 , y 0 ). If there is any justice in life, the level curve and the constraint curve ought to be tangent to each other at the point where they intersect, since if the constraint curve crossed over the level curve, then that would mean there would be points above the constraint curve that were higher than M , which we assumed was the maximum value above the constraint curve. So it seems reasonable that the level curve and the constraint curve are tangent to each other at the
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# How to Find Rate of Change: A Comprehensive Guide with Examples
## Introduction
Rate of change is a fundamental concept in calculus and related fields that measures how one quantity changes in relation to another quantity over a period of time. It is used to analyze trends, predict future outcomes, and solve practical problems in fields like physics, economics, and engineering. Understanding rate of change is crucial for anyone studying calculus or pursuing a career in a related field.
In this article, we will provide a comprehensive guide on how to find rate of change. We will cover everything from the basics of finding rate of change to more advanced topics like derivatives and graphing techniques. By the end of this article, you will have a solid understanding of what rate of change is, how to calculate it, and how it is used in real-world problems.
## A step-by-step guide on how to find the rate of change
The rate of change is simply the rate at which a quantity changes over a period of time. It is calculated by dividing the change in the quantity by the change in time. The formula for finding the rate of change is:
Rate of change = (change in quantity) / (change in time)
To find the rate of change, you need to identify the quantity you are measuring and the period of time over which you are measuring it. Let’s walk through a few examples.
Example 1: Jane drives 60 miles in 2 hours. What is her average speed?
To find Jane’s average speed, we need to calculate her rate of change, which is the change in distance (60 miles) divided by the change in time (2 hours).
Rate of change = (60 miles) / (2 hours) = 30 miles per hour
Therefore, Jane’s average speed is 30 miles per hour.
Example 2: The temperature outside drops from 80 degrees Fahrenheit to 60 degrees Fahrenheit over the course of 4 hours. What is the rate of change of the temperature?
In this example, the quantity we are measuring is temperature, and the period of time we are measuring it over is 4 hours. To find the rate of change, we need to calculate the change in temperature (which is 20 degrees Fahrenheit) and divide it by the change in time (which is 4 hours).
Rate of change = (20 degrees Fahrenheit) / (4 hours) = 5 degrees Fahrenheit per hour
Therefore, the rate of change of the temperature is 5 degrees Fahrenheit per hour.
Tips and Tricks: When identifying which variable is the independent variable (the variable that is changing over time) and which variable is the dependent variable (the variable that is being affected by the change in the independent variable), look for keywords like “per” or “for every.” These keywords indicate a rate of change problem. You can also use context to determine which variable is which. For example, in the two examples above, the independent variable was time in the first example and temperature in the second example.
## Applications of rate of change in real-world problems
Rate of change has a wide range of applications in fields like physics, economics, and engineering. It is used to analyze trends, predict future outcomes, and solve practical problems. Here are a few examples:
Speed and Acceleration: In physics, rate of change is used to calculate speed (distance over time) and acceleration (change in velocity over time).
Growth Rates: In economics, rate of change is used to calculate growth rates of populations, inflation rates, and stock prices over time.
Engineering: In engineering, rate of change is used to calculate stress and strain on materials over time.
These are just a few examples of how rate of change is used to solve problems in real-world settings. It’s clear that understanding rate of change is crucial for solving practical problems in a variety of fields.
## Understanding rate of change through graphs
Another way to understand rate of change is through graphing. The slope of a graph represents the rate of change of the function. The steeper the slope, the greater the rate of change.
Example 3: The graph below shows the distance traveled by a car over time. What is the rate of change of the car’s speed between 0 and 2 seconds?
To find the rate of change of the car’s speed, we need to find the slope of the graph between 0 and 2 seconds. To do this, we need to identify two points on the graph: (0,0) and (2,16). The change in distance is 16 – 0 = 16, and the change in time is 2 – 0 = 2. Therefore, the slope (rate of change) of the graph between 0 and 2 seconds is:
Rate of change = (16 – 0) / (2 – 0) = 8 units per second
Therefore, the rate of change of the car’s speed between 0 and 2 seconds is 8 units per second.
Understanding graphing techniques is important for visualizing and communicating rate of change. When you’re given a problem to solve, try graphing the data to see if you can visualize the rate of change more clearly.
## Finding the rate of change of a function
There are several methods for finding the rate of change of a function, including the use of derivatives and difference quotients.
The Derivative: The derivative of a function is a way to find the rate of change of the function at a specific point. It is a mathematical tool that can be used to find the slope (rate of change) of a curve at any point along the curve. The basic idea behind the derivative is to find the rate at which a function is changing at a specific point by taking the limit of the difference quotient as the change in x approaches 0.
The Difference Quotient: The difference quotient is another way to find the rate of change of a function at a specific point. It is defined as the slope of the secant line between two points on the curve that are very close together.
Example 4: Let’s say we have the function f(x) = x^2. We want to find the rate of change of the function at x = 3.
To find the rate of change of the function at x = 3 using the derivative, we first need to find the derivative of the function:
f'(x) = 2x
Now we can plug in x = 3 to get:
f'(3) = 2(3) = 6
Therefore, the rate of change of the function f(x) = x^2 at x = 3 is 6.
There are many mathematical concepts and formulas related to finding the rate of change of a function, including limits, integrals, and derivatives. These concepts are beyond the scope of this article, but it’s important to know that they exist and are used extensively in calculus and related fields.
## How to find the average rate of change
The average rate of change is similar to the instantaneous rate of change, but it is calculated over a specific interval of time instead of at a specific point in time. The formula for finding the average rate of change is:
Average rate of change = (final value – initial value) / (final time – initial time)
Example 5: Let’s say we have the function f(x) = x^3. We want to find the average rate of change of the function between x = 1 and x = 3.
To find the average rate of change of the function between x = 1 and x = 3, we need to find the change in the function (f(3) – f(1)) and the change in x (3 – 1). Therefore:
Average rate of change = (f(3) – f(1)) / (3 – 1) = (27 – 1) / 2 = 13
Therefore, the average rate of change of the function f(x) = x^3 between x = 1 and x = 3 is 13.
Note that the average rate of change is different from the instantaneous rate of change, which is the rate of change at a specific point in time. The average rate of change is calculated over a specific interval of time, while the instantaneous rate of change is calculated at a single point in time.
## Conclusion
In summary, rate of change is a fundamental concept in calculus and related fields that measures how one quantity changes in relation to another quantity over a period of time. It is used to analyze trends, predict future outcomes, and solve practical problems in fields like physics, economics, and engineering. To find rate of change, you need to identify the quantity you are measuring and the period of time over which you are measuring it. There are several methods for finding the rate of change of a function, including the use of derivatives and difference quotients. Understanding rate of change is crucial for practical problem solving and future studies in calculus and related fields.
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Trigonometric functions are functions that model periodic phenomena. They are based on a repeating event, therefore we use a circle, angles and trigonometric ratios to define and represent a trigonometric function. Trigonometric functions can model relationships between different quantities that follow a periodic nature: height over time, distance over time, temperature over time and so on.
The main types of trigonometric functions are based on the primary trigonometric ratios of sine, cosine and tangent. The sine and cosine functions are most frequently used to model real life periodic events.
Sine Function has the following characteristics and graph:
Domain: {x ∈ R}, Range: {y ∈ R| -1≤ y ≤ 1}
Cosine Function has the following graph and characteristics:
Domain: {x ∈ R}, Range: {y ∈ R|-1 ≤ y ≤ 1}
Tangent Function has the following graph and characteristics
Domain: {x ∈ R| x ≠ 90 + 180n}, Range: {y ∈ R}
## Transformations of Trigonometric Functions
Above you have observed the parent trigonometric functions. But every function can be transformed.
Periodic functions have the following general equation that describes possible transformations (sin function could be replaced by cos or tan functions) :
##### f(x) = asin(k(x – d))+c
• a – amplitude and reflection in the x-axis if negative
• k – determines the period of a function: P = Standard Period/k
• d – phase shift left or right
• c – vertical shift, determine the equation of the axis of the curve: y = c
Example: f(x) = -2sin(12(x – 1)) + 11
• a = 2, reflection in the x-axis
• k = 12, so Period = 360/12 = 30
• d = 1, so phase shift 1 unit to the right
• c = 11, so AoC is y = 11
• max value is 13, min value is 9
## Applications of Trigonometric Functions
Trigonometric functions can be used to model a variety of situations that are periodic.
For example, seasons change periodically, heart beats periodically, a wheel rotates or a wave moves also periodically.
Periodic behaviour is when something repeats over and over and over again over the same period and resulting in the same outputs each time.
Check out the video lesson below explaining how certain events can be described as periodic functions. Learn how to graph and find equations of the trigonometric functions based on the information provided.
Take a quiz
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ELEC 321
# Tutorial 12
Updated 2019-04-09
Suppose we have the Markov chains $X_n$. The Markov property is $\mathbb P(X_{n+1}\vert X_n, X_{n+1},\dotsc)=\mathbb P(X_{n+1}\vert X_n)$.
For homogeneous MC, the initial probability distribution is $p_i(0)=\mathbb P(X_0=i)$, and that $\sum_i p_i=1$.
Chapman-Kolmogrov Equation
Example
There exists 2 white balls and 2 black balls. At each time, one ball is drawn and the color is flipped with probability $a$, and not flipped with probability 1-$a$. Ball is placed back into the urn.
Let $X_n$ be the number of black balls in the urn.
a. Is $X_n$ a Markov chain?
There are five possible outcomes for $X_n$: $X_n\in\{0,1,2,3,4\}$. Let $X_n=k$, so $k$ is the number of black balls. Then
Since we have shown that $X_{n+1}$ only depends on the last value, $k=X_n$. Then $X_n$ is a Markov chain.
b. Find the transition matrix
Let the rows be the number of black balls the transition is ‘from’ and the column be the number of black balls the transition is going ‘to’.
If there are 0 black balls to begin with. Then there are two cases:
• a white ball is picked and no color changes
• a white ball is picked and color changes (black balls + 1)
If there are 1 black ball in the urn. There are these outcomes:
• a black ball is picked and color changes
Thus, $P_{10}=\mathbb P(\text{pick black ball and changes color})$. Since picking and changing color are independent, $P_{10}=\mathbb P(\text{pick black})\mathbb P(\text{change color})$. Which is $\frac{1}{4}\times a$.
• Similarly, using the steps for the black ball above, the probability of picking the white ball and changing color is just $\frac{3}{4}\times a$.
• Picking any colored ball and not change color contributes to the same transition.
• For picking a black ball and not change color, the probability is ...
• Picking a white ball and not change color, the probability is ...
The sum of the two above adds up to $1-a$.
Using the same arguments for the other states, we come to the final matrix:
## Classes of State
Accessibility: State $j$ is accessible from $i$ if and only if $\exists n \in \mathbb N, p_{ij}>0$
Communicating State: States $i$ and $j$ communicate if and only if $i$ is accessible from $j$ and $j$ is accessible from $i$.
Communicating Class: The set of all communicating states.
Irreducible: A state is irreducible if and only if all states in the state space belongs to a single communicating class.
Recurrence: The state $i$ is recurring if and only if the probability of visiting state $i$ is 1.
To check, consider: $\sum_{n=1}^\infty p_{ii}(n)=\infty$
Transient: The state $i$ is transient if and only if the probability of visiting state $i$ is less than 1.
To check, consider: $% $
### Properties
1. State machine is irreducible if and only if all states are recurrent
2. All states in a communicating class are either transient or recurring (if one state in a communicating class is transient, all of the states in the same class are)
Example
Given transition matrix
Let’s draw the state diagram.
Since 1 is accessible from 0 and 0 is accessible from 1, then $1,0\in \mathcal C$. Since 2 is also accessible from any of the states and any state is accessible from 2 given some time, we see that $2\in\mathcal C$ also.
Thus $\mathcal C=\{0,1,2\}$.
Example
Given transition matrix
We see that 0 is not communicating with any state but itself, and that 1 and 2 are communicating amongst each other. Thus we have two communicating classes: $\mathcal C_1=\{0\},\mathcal C_2=\{1,2\}$.
We see that both communicating classes are recurrent.
Example
Given transition matrix
We see that none of the states are communicating except for themselves:
To check if state 0 is transient, we check $% $:
If we take the sum, it converges to $% $. Therefore state 0 is transient.
Similarity, we could argue that state 2 is also transient.
For state 1, if we end up in state 1, the only outcome after is state 1. Therefore state 1 is recurring. We can check this by taking the summation: $\sum_{n=1}^{\infty}p_{11}(n)=\sum_{n=1}^\infty(1)=\infty$.
## Periodicity of a State
Given a Markov chain, the period $d_i$ of a state is defined as
### Properties
1. All states in a communicating class has the same period
2. A state is aperiodic if there exists a class, where all states in the class have a period of 1. From property 1, we see that if one state in the class has a period of 1, then all of the other states in the same class also has a period of 1. Thus it will make the class aperiodic.
3. Regular: a Markov chain is regular if and only if it is irreducible and aperiodic
Example
copy diagram
All states belong to the same communicating class: $\mathcal C=\{0,1,2,3\}$ and we only need to find the period of one state to determine the period of the entire class.
Observe the period of from state 0 to state 0. We see that $p_{00}(1)=0,p_{00}(2)=0.5,p_{00}(3)=0,p_{00}(4)=0.75$. Thus the period is $d_0=\gcd(n=2,n=4)=2$.
Example
Given transition matrix
We see that this is not regular because its one and only communicating class is periodic.
## Long-Run Behavior
This only applies to regular Markov chains.
Doesn’t matter what the starting state is, after a long time, the probability approaches to a certain probability - the steady state probability.
Suppose we start the chain at state $i$, then $\mathbb P(X_n=j \vert X_o=i)=\pi_j>0$
$\pi$ can be obtained as a solution of $\pi=\pi P$
Example
Given a 2 state MC with the transition matrix
We first check that this MC is regular, and indeed it is.
To compute the steady state probability, $\pi=\pi P$:
We obtain the equations and solve for $\pi_1,\pi_2$. Alas, we get $\pi_1=\frac{b}{a+b}$ and $\pi_2=\frac{a}{a+b}$.
Suppose that $% $ and $a=0.3, b=0.8$. Compute the probability at $n=2$.
Because $p(n)=p(0)p^n$ we have $% $.
Multiplying them out using preferred methods, we see that $% $.
Example
Given a buffer
At $t=0$, the buffer contains 3 packets. Assume no more packet arrives, the packets in the buffer is transmitted. Transmission is successful with probability $p$.
Let $Y_n$ denote the number of packets in the buffer.
a. Is $Y_n$ a MC?
There are four outcomes: $Y_n\in\{0,1,2,3\}$
Let $Y_n=k$, we see that $Y_{n+1}\in\{k, k-1\}$, thus $Y_{n+1}$ only depends on $k$ and therefore $Y_n$ is a Markov chain.
The transition matrix is
It is not regular because all states don’t belong to a single communicating class.
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# Symmetry and Transformations
## Presentation on theme: "Symmetry and Transformations"— Presentation transcript:
Symmetry and Transformations
G.3cd Symmetry and Transformations Transformations
Lines of Symmetry If a line can be drawn through a figure so the one side of the figure is a reflection of the other side, the line is called a “line of symmetry.” Some figures have 1 or more lines of symmetry. Some have no lines of symmetry. Four lines of symmetry One line of symmetry Two lines of symmetry Infinite lines of symmetry No lines of symmetry Transformations
Point of Symmetry A figure with point of symmetry can be turned about a center point less than 360° and coincide with the original figure. For example: For each figure, state the angle that the figure can be turned and coincide with the original figure. Transformations
Be careful!! Transformations
Types of Transformations
Reflections: These are like mirror images as seen across a line or a point. Translations ( or slides): This moves the figure to a new location with no change to the looks of the figure. Rotations: This turns the figure clockwise or counter-clockwise but doesn’t change the figure. Dilations: This reduces or enlarges the figure to a similar figure. Transformations
Reflections You can reflect a figure using a line or a point. All measures (lines and angles) are preserved but in a mirror image. Example: The figure is reflected across line l . l You could fold the picture along line l and the left figure would coincide with the corresponding parts of right figure. Transformations
Reflections – continued…
Reflection across the x-axis: the x values stay the same and the y values change sign (x , y) (x, -y) Reflection across the y-axis: the y values stay the same and the x values change sign (x , y) (-x, y) Example: In this figure, line l : n l reflects across the y axis to line n (2, 1) (-2, 1) & (5, 4) (-5, 4) reflects across the x axis to line m. (2, 1) (2, -1) & (5, 4) (5, -4) m Transformations
Reflections across specific lines:
To reflect a figure across the line y = a or x = a, mark the corresponding points equidistant from the line. i.e. If a point is 2 units above the line its corresponding image point must be 2 points below the line. Example: Reflect the fig. across the line y = 1. (2, 3) (2, -1). (-3, 6) (-3, -4) (-6, 2) (-6, 0) Transformations
Translations (slides)
If a figure is simply moved to another location without change to its shape or direction, it is called a translation (or slide). If a point is moved “a” units to the right and “b” units up, then the translated point will be at (x + a, y + b). If a point is moved “a” units to the left and “b” units down, then the translated point will be at (x - a, y - b). Example: A Image A translates to image B by moving to the right 3 units and down 8 units. B A (2, 5) B (2+3, 5-8) B (5, -3) Transformations
Composite Reflections
If an image is reflected over a line and then that image is reflected over a parallel line (called a composite reflection), it results in a translation. Example: C B A Image A reflects to image B, which then reflects to image C. Image C is a translation of image A Transformations
Rotations An image can be rotated about a fixed point.
The blades of a fan rotate about a fixed point. An image can be rotated over two intersecting lines by using composite reflections. Image A reflects over line m to B, image B reflects over line n to C. Image C is a rotation of image A. A B C m n Transformations
Rotation is simply turning about a fixed point.
For our purposes, the fixed point will be the origin Rotate 90 counterclockwise about the origin Rotate 180 about the origin Rotate 90clockwise about the origin
CLOCKWISE is like a right turn.
Hands in the air on the wheel.
Left hand is x Right hand is y
Which hand is at 12 o’clock first?
Make a clockwise turn. Which hand is at 12 o’clock first? X
Rotate 90 degrees clockwise.
Change the sign of x & switch the order of x and y.
Example: Rotate 90 degrees clockwise.
Rotate 90° clockwise
Rotate 90° clockwise
COUNTERCLOCKWISE is like a left turn.
Hands in the air on the wheel.
Left hand is x Right hand is y
Make a counterclockwise turn.
Which hand is at 12 o’clock first? Y
Rotate 90 degrees counterclockwise.
Change the sign of y & Switch the order of x and y
Example: Rotate 90 degrees counterclockwise.
Rotate 90° counterclockwise
Rotate 90° counterclockwise
Rotating 180 degrees changes the sign of the x and the sign of the y
Rotating 180 degrees changes the sign of the x and the sign of the y. (or rotate 90° twice) (for 270° rotate 90° 3x)
Rotate 180 degrees. Keep the order & change the sign of both x & y.
Example: Rotate 180 degrees.
Rotate 180°
Rotate 180°
Angles of rotation In a given rotation, where A is the figure and B is the resulting figure after rotation, and X is the center of the rotation, the measure of the angle of rotation AXB is twice the measure of the angle formed by the intersecting lines of reflection. Example: Given segment AB to be rotated over lines l and m, which intersect to form a 35° angle. Find the rotation image segment KR. A B 35 ° Transformations
Angles of Rotation . . Since the angle formed by the lines is 35°, the angle of rotation is 70°. 1. Draw AXK so that its measure is 70° and AX = XK. 2. Draw BXR to measure 70° and BX = XR. 3. Connect K to R to form the rotation image of segment AB. A B 35 ° X K R Transformations
Dilations A dilation is a transformation which changes the size of a figure but not its shape. This is called a similarity transformation. Since a dilation changes figures proportionately, it has a scale factor k. If the absolute value of k is greater than 1, the dilation is an enlargement. If the absolute value of k is between 0 and 1, the dilation is a reduction. If the absolute value of k is equal to 0, the dilation is congruence transformation. (No size change occurs.) Transformations
Dilations – continued…
In the figure, the center is C. The distance from C to E is three times the distance from C to A. The distance from C to F is three times the distance from C to B. This shows a transformation of segment AB with center C and a scale factor of 3 to the enlarged segment EF. In this figure, the distance from C to R is ½ the distance from C to A. The distance from C to W is ½ the distance from C to B. This is a transformation of segment AB with center C and a scale factor of ½ to the reduced segment RW. C E A F B C R A B W Transformations
Dilations – examples… Find the measure of the dilation image of segment AB, 6 units long, with a scale factor of S.F. = -4: the dilation image will be an enlargment since the absolute value of the scale factor is greater than 1. The image will be 24 units long. S.F. = 2/3: since the scale factor is between 0 and 1, the image will be a reduction. The image will be 2/3 times 6 or 4 units long. S.F. = 1: since the scale factor is 1, this will be a congruence transformation. The image will be the same length as the original segment, 1 unit long. Transformations
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There are 366 different Starters of The Day, many to choose from. You will find in the left column below some starters on the topic of Geometry. In the right column below are links to related online activities, videos and teacher resources.
A lesson starter does not have to be on the same topic as the main part of the lesson or the topic of the previous lesson. It is often very useful to revise or explore other concepts by using a starter based on a totally different area of Mathematics.
Main Page
### Geometry Starters:
Angle Estimates: Estimate the sizes of each of the angles then add your estimates together.
Area Two: How many different shapes with an area of 2 square units can you make by joining dots on this grid with straight lines?
Christmas Tables: Which of the two shapes has the largest area? You will be surprised!
Cross Perimeter: Calculate the distance around the given shape
Dice Reflections: A dice is reflected in two mirrors. What numbers can be seen?
Hexagon: On a full page in the back of your exercise book draw a perfectly regular hexagon.
Mathterpiece: Memorise a picture made up of geometrical shapes
Oblongs: Find the dimensions of a rectangle given the perimeter and area.
Pentagon: On a full page in the back of your exercise book draw a perfectly regular pentagon.
Polygon Riddle 1: Solve the riddle to find the name of the polygon then sum the interior angles.
Quad Areas: Calculate the areas of all the possible quadrilaterals that can be constructed by joining together dots on this grid.
Reflective Cat: On squared paper copy the drawing of the face then reflect it in three different lines.
Rotational Symmetry: Draw a pattern with rotational symmetry of order 6 but no line symmetry.
Sectors: Work out which sectors fit together to make complete circles. Knowledge of the sum of the angles at a point will help find more than one correct solution to this puzzle.
Square Angles: Find a trapezium, a triangle and a quadrilateral where all of the angles are square numbers.
Square Circle Kite: Write down the names of all the mathematical shapes you know.
Stair Perimeter: Use the information implied in the diagram to calculate the perimeter of this shape.
Step Perimeter: Is it possible to work out the perimeter of this shape if not all the side lengths are given?
Find The Radius: Find the radius of the circle from the small amount of information provided.
Geometry Snack: Find the value of the marked angle in this diagram from the book Geometry Snacks
Quad Midpoints: What shape is created when the midpoints of the sides of a quadrilateral are joined together?
Three Right Triangles: Calculate the lengths of the unlabelled sides of these right-angled triangles.
#### Volume
Use formulae to solve problems involving the volumes of cuboids, cones, pyramids, prisms and composite solids.
Transum.org/go/?to=volume
### Curriculum for Geometry:
#### Year 5
Pupils should be taught to use the properties of rectangles to deduce related facts and find missing lengths and angles more...
Pupils should be taught to distinguish between regular and irregular polygons based on reasoning about equal sides and angles more...
#### Year 6
Pupils should be taught to draw 2-D shapes using given dimensions and angles more...
Pupils should be taught to compare and classify geometric shapes based on their properties and sizes and find unknown angles in any triangles, quadrilaterals, and regular polygons more...
Pupils should be taught to recognise angles where they meet at a point, are on a straight line, or are vertically opposite, and find missing angles. more...
#### Years 7 to 9
Pupils should be taught to draw and measure line segments and angles in geometric figures, including interpreting scale drawings more...
Pupils should be taught to describe, sketch and draw using conventional terms and notations: points, lines, parallel lines, perpendicular lines, right angles, regular polygons, and other polygons that are reflectively and rotationally symmetric more...
Pupils should be taught to use the standard conventions for labelling the sides and angles of triangle ABC, and know and use the criteria for congruence of triangles more...
Pupils should be taught to derive and illustrate properties of triangles, quadrilaterals, circles, and other plane figures [for example, equal lengths and angles] using appropriate language and technologies more...
Pupils should be taught to identify and construct congruent triangles, and construct similar shapes by enlargement, with and without coordinate grids more...
Pupils should be taught to apply the properties of angles at a point, angles at a point on a straight line, vertically opposite angles more...
Pupils should be taught to derive and use the sum of angles in a triangle and use it to deduce the angle sum in any polygon, and to derive properties of regular polygons more...
Pupils should be taught to apply angle facts, triangle congruence, similarity and properties of quadrilaterals to derive results about angles and sides, including Pythagoras’ Theorem, and use known results to obtain simple proofs more...
Pupils should be taught to interpret mathematical relationships both algebraically and geometrically. more...
#### Years 10 and 11
Pupils should be taught to identify and apply circle definitions and properties, including: centre, radius, chord, diameter, circumference, tangent, arc, sector and segment more...
Pupils should be taught to {apply and prove the standard circle theorems concerning angles, radii, tangents and chords, and use them to prove related results} more...
Pupils should be taught to construct and interpret plans and elevations of 3D shapes more...
Pupils should be taught to calculate arc lengths, angles and areas of sectors of circles more...
Pupils should be taught to apply the concepts of congruence and similarity, including the relationships between lengths, {areas and volumes} in similar figures more...
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### Notes:
Geometry is a branch of mathematics concerned with questions of shape, size, relative position of figures, and the properties of space. Geometry arose independently in a number of early cultures as a body of practical knowledge concerning lengths, areas, and volumes, with elements of a formal mathematical science emerging in the West as early 6th Century BC.
See also the topics of Angles, Area, Bearings, Circles, Enlargements, Mensuration, Pythagoras, Shape, Shape (3D), Symmetry, Transformations and Trigonometry.
### Geometry Teacher Resources:
Angle Theorems: Diagrams of the angle theorems with interactive examples.
Circle Parts Kim's Game: A memory game to be projected to help the whole class revise the names for the parts of a circle.
Circle Theorems: Diagrams of the circle theorems to be projected onto a white board as an effective visual aid.
Geometry Toolbox: Create your own dynamic geometrical diagrams using this truly amazing tool from GeoGebra.
Kite Maths: Can you make a kite shape from a single A4 size sheet of paper using only three folds?
Pin Board: Rows and columns of dots that can be joined using straight lines to create shapes.
Polygons: Name the polygons and show the number of lines and order of rotational symmetry.
### Geometry Activities:
Angle Chase: Find all of the angles on the geometrical diagrams.
Angle Parallels: Understand and use the relationship between parallel lines and alternate and corresponding angles.
Angle Points: Apply the properties of angles at a point, angles on a straight line and vertically opposite angles.
Angles in a Triangle: A self marking exercise involving calculating the unknown angle in a triangle.
Area Maze: Use your knowledge of rectangle areas to calculate the missing measurement of these composite diagrams.
Area of a Triangle: Calculate the areas of the given triangles in this self marking quiz.
Area Two: How many different shapes with an area of 2 square units can you make by joining dots on this grid with straight lines?
Areas of Composite Shapes: Find the areas of combined (composite) shapes made up of one or more simple polygons and circles.
Circle Theorems Exercise: Show that you understand and can apply the circle theorems with this self marking exercise.
Congruent Triangles: Test your understanding of the criteria for congruence of triangles with this self-marking quiz.
Estimating Angles: Estimate the size of the given acute angles in degrees.
Formulae Pairs: Find the matching pairs of diagrams and formulae for basic geometrical shapes.
Geometry Toolbox: Create your own dynamic geometrical diagrams using this truly amazing tool from GeoGebra.
Measuring Units: Check your knowledge of the units used for measuring with this multiple choice quiz about metric and imperial units.
Polybragging: The Transum version of the Top Trumps game played online with the properties of polygons.
Polygon Angles: A mixture of problems related to calculating the interior and exterior angles of polygons.
Polygon Pieces: Arrange the nine pieces of the puzzle on the grid to make different polygons.
Polygon Properties: Connect the names of the polygons with the descriptions of their properties.
Polygons: Name the polygons and show the number of lines and order of rotational symmetry.
Proof of Circle Theorems: Arrange the stages of the proofs for the standard circle theorems in the correct order.
Quad Areas: Calculate the areas of all the possible quadrilaterals that can be constructed by joining together dots on this grid.
Similar Shapes: Questions about the scale factors of lengths, areas and volumes of similar shapes.
Surface Area: Work out the surface areas of the given solid shapes.
Tantrum: A game, a puzzle and a challenge involving counters being placed at the corners of a square on a grid.
Transformations: Draw transformations online and have them instantly checked. Includes reflections, translations, rotations and enlargements.
Vectors: An online exercise on addition and subtraction of vectors, multiplication of vectors by a scalar, and diagrammatic representations of vectors.
Volume: Use formulae to solve problems involving the volumes of cuboids, cones, pyramids, prisms and composite solids.
Finally there is Topic Test, a set of 10 randomly chosen, multiple choice questions suggested by people from around the world.
### Geometry Investigations:
Area shapes: Investigate polygons with an area of 4 square units. This is your starting point, you can decide how to proceed.
How Many Rectangles?: Investigate the number of rectangles on a grid of squares. What strategies will be useful in coming up with the answer?
Maxvoltray: Find the maximum volume of a tray made from an A4 sheet of paper. A practical mathematical investigation.
Pin Board: Rows and columns of dots that can be joined using straight lines to create shapes.
Polygon Areas: Investigate polygons with an area of 4 sq. units. Investigate polygons with other areas.
Rectangle Perimeters: The perimeter of a rectangle is 28cm. What could its area be?
Tantrum: A game, a puzzle and a challenge involving counters being placed at the corners of a square on a grid.
Tessellations: Drag the shapes onto the canvas to create tessellating patterns and investigate the laws of tessellations.
### Geometry Videos:
Construct a congruent triangle: Construction (with compass and straight edge) of a triangle congruent to a given triangle.
Different types of Triangle: Euclid and his friends explain how many different kinds of triangle there are.
Parallelogram: Instructional video showing how the area of a parallelogram can be determined.
### Geometry Worksheets/Printables:
Angle Chase Worksheets: A set of printable Angle Chase sheets on which pupils fill in the missing angles.
Polybragging Cards: Printable cards for the Polybragging game. Use the properties of polygons to win.
Links to other websites containing resources for Geometry are provided for those logged into 'Transum Mathematics'. Subscribing also opens up the opportunity for you to add your own links to this panel. You can sign up using one of the buttons below:
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#### Angle Chase
Find all of the angles on the geometrical diagrams.
Transum.org/go/?to=chase
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Add Three 1-Digit Numbers Year 2 Addition and Subtraction Learning Video Clip | Classroom Secrets
Maths Resources & WorksheetsYear 2 Maths LessonsAutumn Block 2 (Addition and Subtraction)16 Add Three 1-Digit Numbers › Add Three 1-Digit Numbers Year 2 Addition and Subtraction Learning Video Clip
# Add Three 1-Digit Numbers Year 2 Addition and Subtraction Learning Video Clip
## Step 16: Add Three 1-Digit Numbers Year 2 Addition and Subtraction Learning Video Clip
Cleo and Blake are back in the prehistoric times to solve more maths problems. Help them to add three 1-digit numbers as they continue to explore.
More resources for Autumn Block 2 Step 16.
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Discussion points for teachers
1. Fill in the boxes below to show the number of each type of egg. What is the total of all of the eggs?
Discuss how many dinosaur eggs can be seen in each colour. Discuss putting this into the calculation. Discuss which numbers would be easiest to add first and why.
7 + 3 + 6 = 16
2. Help Cleo work out the total number of items she has in her backpack. Discuss how many of each item Cleo has.
Discuss how we could write this into a calculation. Would we use numerals or words? Discuss which numbers would be easiest to add first and why.
seven + seven + six = twenty
3. Help Blake to count how many pterodactyls can be seen over head by writing the numbers into the boxes below and then find the total.
Discuss how many pterodactyls can be seen in each group. Discuss writing this into the calculation. Discuss which pair of numbers would be the easiest to add first and why.
6 + 4 + 9 = 19
4. Help Cleo and Blake to find the route across the river which has the most steppingstones. Write a calculation for each route and use the comparison symbols to compare the three answers.
Discuss how many stones can be seen in each group and how we would write this into the calculation. Discuss which groups of stones would be easiest to add together first for each route. Discuss which comparison symbols we could use to order the totals.
A. 9 + 8 + 5 = 22; B. 9 + 9 + 5 = 23; C. 8 + 8 + 4 = 20 23 > 22 > 20 or 20 < 22 < 23
Optional discussion points:
Discuss how we know which number must come first when using the comparison symbols.
National Curriculum Objectives
Mathematics Year 2: (2C2a) Add and subtract numbers mentally, including: adding three one-digit numbers
This resource is available to play with a Premium subscription.
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Rs Aggrawal 2020 2021 Solutions for Class 6 Maths Chapter 16 Triangles are provided here with simple step-by-step explanations. These solutions for Triangles are extremely popular among Class 6 students for Maths Triangles Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggrawal 2020 2021 Book of Class 6 Maths Chapter 16 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggrawal 2020 2021 Solutions. All Rs Aggrawal 2020 2021 Solutions for class Class 6 Maths are prepared by experts and are 100% accurate.
#### Page No 196:
We get a triangle by joining the three non-collinear points A, B, and C.
(i) The side opposite to ∠C is AB.
(ii) The angle opposite to the side BC is ∠A.
(iii) The vertex opposite to the side CA is B.
(iv) The side opposite to the vertex B is AC.
#### Page No 196:
The measures of two angles of a triangle are 72° and 58°.
Let the third angle be x
Now, the sum of the measures of all the angles of a triangle is 180o.
$\therefore$ + 72+ 58o = 180o
⇒ x + 130= 180o
⇒ x = 180o$-$ 130o
⇒ x = 50o
The measure of the third angle of the triangle is 50o.
#### Page No 196:
The angles of a triangle are in the ratio 1:3:5.
Let the measures of the angles of the triangle be (1x), (3x) and (5x)
Sum of the measures of the angles of the triangle = 180o
∴ 1x + 3x + 5x = 180o
⇒ 9x = 180o
⇒ x = 20o
1x = 20o
3x = 60o
5x = 100o
The measures of the angles are 20o, 60o and 100o
#### Page No 196:
In a right angle triangle, one of the angles is 90o.
It is given that one of the acute angled of the right angled triangle is 50o.
We know that the sum of the measures of all the angles of a triangle is 180o.
Now, let the third angle be x.
Therefore, we have:
90o + 50o= 180o
⇒ 140o = 180o
⇒ x = 180o $-$ 140o
⇒ x = 40o
The third acute angle is 40o.
#### Page No 196:
Given:
∠A = 110o and ∠B = ∠C
Now, the sum of the measures of all the angles of a traingle is 180o .
∠A + ∠B + ∠C = 180o
⇒ 110o + ∠B + ∠B = 180o
⇒ 110o + 2∠B = 180o
⇒ 2∠B = 180$-$ 110o
⇒ 2∠B = 70o
⇒ ∠B = 70/ 2
⇒ ∠B = 35o
∴ ∠C = 35o
The measures of the three angles:
∠A = 110o, ∠B = 35o, ∠C = 35o
#### Page No 196:
Given:
∠A = ∠B + ∠C
We know:
∠A + ∠B + ∠C = 180o
⇒ ∠B +∠C + ∠B + ∠C = 180o
⇒ 2∠B + 2∠C = 180o
⇒ 2(∠B +∠C) = 180o
⇒ ∠B + ∠C = 180/2
⇒ ∠B + ∠C = 90o
$\therefore$ ∠A = 90o
This shows that the triangle is a right angled triangle.
#### Page No 196:
Let 3∠A = 4 ∠B = 6 ∠C = x
Then, we have:
#### Page No 196:
(i) It is an obtuse angle triangle as one angle is 130o, which is greater than 90o.
(ii) It is an acute angle triangle as all the angles in it are less than 90o.
(iii) It is a right angle triangle as one angle is 90o.
(iv) It is an obtuse angle triangle as one angle is 92o, which is greater than 90o.
#### Page No 197:
Equilateral Triangle: A triangle whose all three sides are equal in length and each of the three angles measures 60o.
Isosceles Triangle: A triangle whose two sides are equal in length and the angles opposite them are equal to each other.
Scalene Triangle: A triangle whose all three sides and angles are unequal in measure.
(i) Isosceles
AC = CB = 2 cm
(ii) Isosceles
DE = EF = 2.4 cm
(iii) Scalene
All the sides are unequal.
(iv) Equilateral
XY = YZ = ZX = 3 cm
(v) Equilateral
All three angles are 60o.
(vi) Isosceles
Two angles are equal in measure.
(vii) Scalene
All the angles are unequal.
#### Page No 197:
In ∆ABC, if we take a point D on BC, then we get three triangles, namely ∆ADB, ∆ADC and ∆ABC.
#### Page No 197:
(i) No
If the two angles are 90o each, then the sum of two angles of a triangle will be 180o, which is not possible.
(ii) No
For example, let the two angles be 120o and 150o. Then, their sum will be 270o, which cannot form a triangle.
(iii) Yes
For example, let the two angles be 50o and 60o, which on adding, gives 110o. They can easily form a triangle whose third angle is 180o $-$ 110o = 70o.
(iv) No
For example, let the two angles be 70o and 80o, which on adding, gives 150o. They cannot form a triangle whose third angle is 180o$-$ 150= 30o, which is less than 60o.
(v) No
For example, let the two angles be 50o and 40o, which on adding, gives 90o . Thus, they cannot form a triangle whose third angle is 180o $-$ 90o = 90o, which is greater than 60o.
(vi) Yes
Sum of all angles = 60o + 60o + 60o = 180o
#### Page No 197:
(i) A triangle has 3 sides 3 angles and 3 vertices.
(ii) The sum of the angles of a triangle is 180o
(iii) The sides of a scalene triangle are of different lengths.
(iv) Each angle of an equilateral triangle measures 60o.
(v) The angles opposite to equal sides of an isosceles triangle are equal.
(vi) The sum of the lengths of the sides of a triangle is called its perimeter.
#### Page No 197:
Correct option: (c)
A triangle has 6 parts: three sides and three angles.
#### Page No 197:
Correct option: (b)
(a) Sum = 30° + 60° + 70° = 160o
This is not equal to the sum of all the angles of a triangle.
(b) Sum = 50° + 70° + 60° = 180o
Hence, it is possible to construct a triangle with these angles.
(c) Sum = 40° + 80° + 65° = 185o
This is not equal to the sum of all the angles of a triangle.
(d) Sum = 72° + 28° + 90° = 190o
This is not equal to the sum of all the angles of a triangle.
#### Page No 197:
(b) 80o
Let the measures of the given angles be (2x)o, (3x)o and (4x)o.
$\therefore$ (2x)o + (3x)+ (4x)o = 180o
⇒ (9x)o = 180o
⇒ x = 180 / 9
⇒ x = 20o
$\therefore$ 2x = 40o, 3x = 60o, 4x = 80o
Hence, the measures of the angles of the triangle are 40o, 60o, 80o.
Thus, the largest angle is 80o.
#### Page No 198:
Correct option: (d)
The measure of two angles are complimentary if their sum is 90o degrees.
Let the two angles be x and y, such that x + y = 90o .
Let the third angle be z.
Now, we know that the sum of all the angles of a triangle is 180o.
x + y + z = 180o
⇒ 90o + z = 180o
⇒ = 180o $-$ 90
= 90o
The third angle is 90o.
#### Page No 198:
Correct option: (c)
Let ∠A = 70o
The triangle is an isosceles triangle.
We know that the angles opposite to the equal sides of an isosceles triangle are equal.
$\therefore$ ∠B = 70o
We need to find the vertical angle ∠C.
Now, sum of all the angles of a triangle is 180o.
∠A + ∠B + ∠C = 180o
⇒ 70o + 70o + ∠C = 180o
⇒ 140+ ∠C = 180o
⇒ ∠C = 180o $-$ 140o
⇒ ∠C = 40o
#### Page No 198:
Correct option: (c)
A triangle having sides of different lengths is called a scalene triangle.
#### Page No 198:
Correct option: (a)
In the isosceles ABC, the bisectors of ∠B and ∠C meet at point O.
Since the triangle is isosceles, the angles opposite to the equal sides are equal.
∠B = ∠C
$\therefore$ ∠A + ∠B + ∠C = 180o
⇒ 40o + 2∠B = 180o
⇒ 2∠B = 140o
⇒ ∠B = 70o
Bisectors of an angle divide the angle into two equal angles.
So, in ∆BOC:
∠OBC = 35o and ∠OCB = 35o
∠BOC + ∠OBC + ∠OCB = 180o
⇒ ∠BOC + 35o + 35o = 180o
⇒ ∠BOC = 180o - 70o
⇒ ∠BOC = 110o
#### Page No 198:
Correct option: (b)
The sides of a triangle are in the ratio 3:2:5.
Let the lengths of the sides of the triangle be (3x), (2x), (5x).
We know:
Sum of the lengths of the sides of a triangle = Perimeter
(3x) + (2x) + (5x) = 30
⇒ 10x = 30
⇒ x = 30
10
⇒ x = 3
First side = 3x = 9 cm
Second side = 2x = 6 cm
Third side = 5x = 15 cm
The length of the longest side is 15 cm.
#### Page No 198:
Correct option: (d)
Two angles of a triangle measure 30° and 25°, respectively.
Let the third angle be x.
x + 30o + 25o = 180o
x = 180o $-$ 55o
x = 125o
#### Page No 198:
Correct option: (c)
Each angle of an equilateral triangle measures 60o.
|
Warm Up Given: ∠ 1 ≅ ∠ 2 m ∠ 2 = 60° m ∠ 3 = 60° Prove: ∠ 1 ≅ ∠ 3 1 2 3 1 1 2 2 3 3.
Presentation on theme: "Warm Up Given: ∠ 1 ≅ ∠ 2 m ∠ 2 = 60° m ∠ 3 = 60° Prove: ∠ 1 ≅ ∠ 3 1 2 3 1 1 2 2 3 3."— Presentation transcript:
Warm Up Given: ∠ 1 ≅ ∠ 2 m ∠ 2 = 60° m ∠ 3 = 60° Prove: ∠ 1 ≅ ∠ 3 1 2 3 1 1 2 2 3 3
Angle Congruence Theorems Students will use the angle congruence theorems and their other theorems, postulates, and definitions to construct 2-column proofs.
What Do We Know So Far? Our definitions: congruence, midpoint, angle bisector Our postulates: segment addition, angle addition Our algebraic properties (reflexivity, symmetry, transitivity, and addition, subtraction, multiplication, division, substitution) Our segment congruence and angle congruence theorems (reflexivity, symmetry, transitivity)
Right Angle Congruence Theorem If two angles are right angles, then they are congruent. StatementsReasons 1. ∠ A and ∠ B are right angles Given 2. m ∠ A = 90° Definition of a right angle 3. m ∠ B = 90° Definition of a right angle 4. m ∠ A = m ∠ B Substitution property of equality 5. ∠ A ≅ ∠ B Definition of congruent angles
Linear Pair Postulate If two angles form a linear pair, then they are supplementary. A BC D Question: Why is this a postulate?
Congruent Supplements Theorem If two angles are supplementary to the same angle (or two congruent angles) then they are congruent. If m 1 + m 2 = 180 0 and m 2 + m 3 = 180 0, then 1 3.
1 2 3 If m 1 + m 2 = 180 0 and m 2 + m 3 = 180 0, then 1 3.
StatementsReasons 1. ∠ 1 and ∠ 2 are a linear pair. Given 2. ∠ 1 and ∠ 2 are supplementary Linear Pair Postulate 3. m ∠ 1 + m ∠ 2 = 180 Definition of supplementary angles 4. ∠ 3 and ∠ 2 are a linear pair. Given 5. ∠ 3 and ∠ 2 are supplementary Linear Pair Postulate 6. m ∠ 3 + m ∠ 2 = 180 Definition of supplementary angles 7. m ∠ 3 = 180 - m ∠ 2 Subtraction POE 8. m ∠ 1 = 180 - m ∠ 2 Subtraction POE 9. m ∠ 1 = m ∠ 3 Substitution 10. ∠ 1 ≅ ∠ 3 Definition of Congruence PP Proof of the Congruent Supplements Theorem
Vertical Angles Theorem: Vertical angles are congruent. (Angle A ≅ Angle B) A B
Congruent Complements Theorem If two angles are complementary to the same angle (or two congruent angles) then the two angles are congruent. If m 4 + m 5 = 90 0 and m 5 + m 6 = 90 0, then 4 6.
If m 4 + m 5 = 90 0 and m 5 + m 6 = 90 0, then 4 6. 6 6 5 5 4 4
Jigsaw Activity Step 1: Each group will complete one problem from worksheet 2.6. Each member of the group will be an expert on their particular problem. Step 2: One member from each group will move to a second group, so that each of the new groups has (at least) one expert on each problem. Step 3: Each member will present his or her problem and how they solved it.
Exit Ticket Complete the following proof on a piece of loose-leaf paper. Given: ∠ A and ∠ B are complementary. m ∠ C + m ∠ B = 90° Prove: ∠ A ≅ ∠ C
What did we talk about? Properties of Angle Congruence 1.Reflexive 2. Symmetric 3. Transitive Right Angle Congruence Theorem Congruent Supplements Theorem Congruent Complements Theorem Linear Pair Postulate Vertical Angles Theorem
Practice Problems 1.Find the values of x and y. 1.What conclusions can you draw about the angles in the following diagram? Justify your answer.
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# What is the limit of x^n?
Jul 13, 2015
${\lim}_{n \to \infty} {x}^{n}$ behaves in seven different ways according to the value of $x$
#### Explanation:
If $x \in \left(- \infty , - 1\right)$ then as $n \to \infty$, $\left\mid {x}^{n} \right\mid \to \infty$ monotonically, but alternates between positive and negative values. ${x}^{n}$ does not have a limit as $n \to \infty$.
If $x = - 1$ then as $n \to \infty$, ${x}^{n}$ alternates between $\pm 1$. So again, ${x}^{n}$ does not have a limit as $n \to \infty$.
If $x \in \left(- 1 , 0\right)$ then ${\lim}_{n \to \infty} {x}^{n} = 0$. The value of ${x}^{n}$ alternates between positive and negative values but $\left\mid {x}^{n} \right\mid \to 0$ is monotonically decreasing.
If $x = 0$ then ${\lim}_{n \to \infty} {x}^{n} = 0$. The value of ${x}^{n}$ is constant $0$ (at least for $n > 0$).
If $x \in \left(0 , 1\right)$ then ${\lim}_{n \to \infty} {x}^{n} = 0$ The value of ${x}^{n}$ is positive and ${x}^{n} \to 0$ monotonically as $n \to \infty$.
If $x = 1$ then ${\lim}_{n \to \infty} {x}^{n} = 1$. The value of ${x}^{n}$ is constant $1$.
If $x \in \left(1 , \infty\right)$ then as $n \to \infty$, then ${x}^{n}$ is positive and ${x}^{n} \to \infty$ monotonically. ${x}^{n}$ does not have a limit as $n \to \infty$.
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# Solving Probability Part two
Updated on October 31, 2008
Solving PROBABILITY Part Two
Problem Number One:
An experiment consists of getting 6 cards from a standard 52-card deck. What is the probability of getting (a) 6 black cards? (b) 6 face cards ? (c) 6 hearts ? (d) 6 non-face cards ?
First find the Sample space( S) : We use combination 52C6 since we are taking 6 cards out of 52. 52C6 = n!/r!(n-r)! = 52 !/6! (52 - 6) ! =20, 358, 520
(a) Find the event A of getting 6 black cards. Since there are 26 black cards we get 26C6. 26C6 = 230,230 Probability of getting 6 black cards = A/S = 230,230/20,358,520 =.011308779
(b) Find the event of getting 6 face cards. Since there are 12 face cards we get 12C6 12C6 = 924 Probability of getting 6 face cards = 924/20,358,520 = .000045386
(c ) Find the event of getting 6 hearts. Since there are 13 hearts we use 13C6 = 1716 Probability of getting 6 hearts = 1716/20,358,520 = .00008428
( d) Find the event of getting 6 nonface cards. We use 40C6 since there are 40 nonface cards. 40C6 =3838380 Probabilty of getting nonface cards = 3838380/20,358,520 = .18854
Sample Problem Two
: If a number consisting of four different digits is randomly formed from the digits 1, 3, 5, 7 and 9, what is the probability that it is (a) divisible by 5 ? (b) less than 3000.?
First find the Sample Space (S) . We use "Multiplication Principle" in finding the event of forming 4-digit number without repetition. Since there are 5 digits to be selected from: A = (5) (4) (3) (2) = 120
(a) Find probability that it is divisible by 5. To be divisible by 5, it must end in 5. (4) (3) (2) (1) = 24 So here is the arrangement. You fill in first the last digit . (1) is placed because there is only one possible number which is 5. P = A/S = 24/120 =1/5 = 0.20
(b) Find the probability that it is less than 3000 In order to be less than 3000 it should never start with 3, 5, 7, 9. So there is only one number qualified for the first position which is one. So the arrangement will be (1) (4) (3) (2) = 24 P = 24/120 = 1/5 = 0.20
Problem Number Three:
What is the probability that a in batch which consists of 100 boxes of flashlight batteries, 5 boxes which contains 8 batteries each contains one that is defective?
First find the Sample Space using 100C5 = 75,287,520
Find the event A that one battery is defective . Use 40C5 since there are 5 boxes = 658,008
Probability = A/S = 658,008/75,287,520 = 0.0087399
Problem Number Four :
Three couples are seated in a circular table. What is the probability that each husband is seated beside his wife?
Sample Space : Number of ways of arranging six persons in a circular table. This is circular permutation. So the formula is (n - 1) ! (6 - 1) ! = 5 * 4 * 3 * 2 = 120.
The event that each husband is seated with his wife. Since there are three couples : 2 * (3 - 1) ! = 2 * 2 = 4 Probability + A/S = 4/ 120 = .033
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If you find any mistakes, please make a comment! Thank you.
## Compute limits using L’Hospital’s Rule III
Solution:
### Part a
Since $|\sin x|\leq 1$, we have
\label{30-3-1}
\lim_{x\to \infty}\frac{\sin x}{x}=0.
Therefore, by \eqref{30-3-1},
$$\lim_{x\to \infty}\frac{x-\sin x}{x}=\lim_{x\to \infty}\frac{1-\frac{\sin x}{x}}{1}=\frac{1-0}{1}=1.$$
In this case, we cannot apply L’Hospital’s Rule because the limit $\lim_{x\to \infty}(1-\cos x)$ does not exists.
### Part b
We consider
$$\ln x^{\sin(1/x)}=\sin(1/x)\ln x=-\sin (1/x)\ln(1/x).$$ Using substitution $u=1/x$, we have
\begin{align*}
\lim_{x\to \infty }\ln x^{\sin(1/x)}
=&\ \lim_{u\to 0^+ }-\sin u\ln u\\
=&\ -\lim_{u\to 0^+ }\frac{\ln u}{\csc u}\\
\text{L’Hospital’s Rule}\ =&\ \lim_{u\to 0^+ }\frac{\frac{1}{u}}{\csc u\cot u}\\
=&\ \lim_{u\to 0^+ }\frac{\sin^2 u}{u\cos u}\\
=&\ \lim_{u\to 0^+ }\frac{\sin^2 u}{u^2}\lim_{u\to 0^+ }\frac{u}{\cos u}\\
\text{Use Example 1}\ =&\ 1\cdot \frac{0}{1}=0.
\end{align*} Here we used Example 1,
$$\lim_{x\to 0}\frac{\sin x}{x}=1.$$ Therefore by Theorem 20.5, we have
$$\lim_{x\to \infty} x^{\sin(1/x)}=e^{0}=1.$$
### Part c
Clearly, we have
$$\lim_{x\to 0^+}(1+\cos x)=2$$ and
$$\lim_{x\to 0^+}(e^x-1)=(1^+-1)=0^+.$$ Therefore,
$$\lim_{x\to 0^+}\frac{1+\cos x}{e^x-1}=\frac{2}{0^+}=\infty.$$
### Part d
Repeatedly applying L’Hospital’s Rule, we have
\begin{align*}
\lim_{x\to 0}\frac{1-\cos 2x -2x^2}{x^4}
=&\ \lim_{x\to 0}\frac{2\sin 2x-4x}{4x^3} \\
=&\ \lim_{x\to 0}\frac{4\cos 2x-4}{12x^2} \\
=&\ \lim_{x\to 0}\frac{-8\sin 2x}{24x}\\
=&\ \lim_{x\to 0}\frac{-16\cos 2x}{24}\\
=&\ \frac{-16}{24}=-\frac{2}{3}.
\end{align*}
Another solution:
We can also do it using Exercise 30.2 (a). Namely,
$$\lim_{u\to 0}\frac{u^3}{\sin u-u}=-6\quad \text{ or }\quad \lim_{u\to 0}\frac{\sin u-u}{u^3}=-\frac{1}{6}.$$
Apply L’Hospital’s Rule and set $u=2x$. We have
\begin{align*}
\lim_{x\to 0}\frac{1-\cos 2x -2x^2}{x^4}
=&\ \lim_{x\to 0}\frac{2\sin 2x-4x}{4x^3} \\
=&\ \lim_{u\to 0}\frac{2\sin u-2 u}{\frac{1}{2}u^3} \\
=&\ 4\lim_{u\to 0}\frac{\sin u-u}{u^3}\\
=&\ -\frac{4}{6}=-\frac{2}{3}.
\end{align*}
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# Angle Sum Property of a Triangle & Exterior Angle Theorem
Angle Sum Property of a Triangle
Triangle is the smallest polygon which has three sides and three interior angles.
In the given triangle, ∆ABC, AB, BC, and CA represent three sides. A, B and C are the three vertices and ∠ABC, ∠BCA and ∠CAB are three interior angles of ∆ABC.
Figure 1 Triangle ABC
Theorem 1: Angle sum property of triangle states that the sum of interior angles of a triangle is 180°.
Proof: Consider a ∆ABC, as shown in the figure below. To prove the above property of triangles, draw a line $\overleftrightarrow {PQ}$ parallel to the side BC of the given triangle.
Since PQ is a straight line, it can be concluded that:
∠PAB + ∠BAC + ∠QAC = 180° ………(1)
SincePQ||BC and AB, AC are transversals,
Therefore, ∠QAC = ∠ACB (a pair of alternate angle)
Also, ∠PAB = ∠CBA (a pair of alternate angle)
Substituting the value of ∠QAC and∠PAB in equation (1),
∠ACB + ∠BAC + ∠CBA= 180°
Thus, the sum of the interior angles of a triangle is 180°.
Exterior Angle Property of a Triangle:
Theorem 2: If any side of a triangle is extended, then the exterior angle so formed is the sum of the two opposite interior angles of the triangle.
In the given figure, the side BC of ∆ABC is extended. The exterior angle ∠ACD so formed is the sum of measures of ∠ABC and ∠CAB.
Proof: From figure 3, ∠ACB and ∠ACD forms a linear pair since they represent the adjacent angles on a straight line.
Thus, ∠ACB + ∠ACD = 180° ……….(2)
Also, from the angle sum property it follows that:
∠ACB + ∠BAC + ∠CBA = 180° ……….(3)
From equation (2) and (3) it follows that:
∠ACD = ∠BAC + ∠CBA
This property can also be proved using concept of parallel lines as follows:
In the given figure, side BCof ∆ABC is extended. A line $\overleftrightarrow {CE}$ parallel to the side AB is drawn, then: Since $\overline {BA} ~||~\overline{CE}$ and $\overline{AC}$ is the transversal,
∠CAB = ∠ACE ………(4) (Pair of alternate angles)
Also, $\overline {BA} ~||~\overline{CE}$ and $\overline{BD}$iis the transversal
Therefore, ∠ABC = ∠ECD ……….(5) (Corresponding angles)
We have, ∠ACB + ∠BAC + ∠CBA = 180° ………(6)
Since,the sum of angles on a straight line is 180°
Therefore, ∠ACB + ∠ACE + ∠ECD = 180° ………(7)
Since, ∠ACE + ∠ECD = ∠ACD(From figure 4)
Substituting this value in equation (7);
∠ACB + ∠ACD = 180° ………(8)
From the equations (6) and (8) it follows that,
∠ACD = ∠BAC + ∠CBA
Hence it can be seen that the exterior angle of a triangle is equals to the sum of its opposite interior angles.
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# Mastering Binomial Expansion
Mastering Binomial Expansion
1 / 25
Slide 1: Slide
This lesson contains 25 slides, with interactive quizzes and text slides.
## Items in this lesson
Mastering Binomial Expansion
#### Slide 1 -Slide
This item has no instructions
Learning Objective
Understand and master the technique of expanding products of two or more binomials.
#### Slide 2 -Slide
This item has no instructions
What do you already know about expanding binomials?
#### Slide 3 -Mind map
This item has no instructions
What are Binomials?
A binomial is a polynomial with two terms. For example, (x + 3) or (2y - 5).
#### Slide 4 -Slide
This item has no instructions
Expanding Binomials
Expanding a binomial means to multiply it out. For example, (a + b)(c + d) = ac + ad + bc + bd.
#### Slide 5 -Slide
This item has no instructions
Product of Two Binomials
Expanding the product of two binomials involves using the distributive property to multiply each term of the first binomial by each term of the second binomial.
#### Slide 6 -Slide
This item has no instructions
Interactive Example 1
Expand (x + 2)(x - 3) = x^2 - x - 6.
#### Slide 7 -Slide
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Expanding Three Binomials
Expanding three binomials involves systematically multiplying each term of each binomial by each term of the other binomials.
#### Slide 8 -Slide
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Interactive Example 2
Expand (a + b)(c - d)(e + f) = ace + acf - ade - adf + bce + bcf - bde - bdf.
#### Slide 9 -Slide
This item has no instructions
Expanding Four Binomials
Expanding four binomials follows the same pattern of systematically multiplying each term of each binomial by each term of the other binomials.
#### Slide 10 -Slide
This item has no instructions
Interactive Example 3
Expand (p + q)(r - s)(t + u)(v - w).
#### Slide 11 -Slide
This item has no instructions
Quiz Time!
Test your understanding with a quiz on expanding binomials.
#### Slide 12 -Slide
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Common Errors
Avoid common errors such as forgetting to multiply every term in each binomial or misapplying the distributive property.
#### Slide 13 -Slide
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Practice Makes Perfect
Practice expanding various binomial products to strengthen your skills.
#### Slide 14 -Slide
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Real-life Applications
Understanding binomial expansion is important in fields such as engineering, physics, and economics.
#### Slide 15 -Slide
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Review and Recap
Recap the key concepts and techniques learned in this lesson.
#### Slide 16 -Slide
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Assessment
Assess students' understanding through a short assignment on expanding binomials.
#### Slide 17 -Slide
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Q&A Session
Address any questions or concerns students may have regarding binomial expansion.
#### Slide 18 -Slide
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Further Exploration
Explore advanced techniques in binomial expansion for those interested in delving deeper.
#### Slide 19 -Slide
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Lesson Conclusion
Congratulations on mastering the art of expanding binomial products!
#### Slide 20 -Slide
This item has no instructions
Homework
Practice expanding various binomial products as homework.
#### Slide 21 -Slide
This item has no instructions
Feedback
Provide feedback on the lesson and express any concerns or suggestions.
#### Slide 22 -Slide
This item has no instructions
Write down 3 things you learned in this lesson.
#### Slide 23 -Open question
Have students enter three things they learned in this lesson. With this they can indicate their own learning efficiency of this lesson.
Write down 2 things you want to know more about.
#### Slide 24 -Open question
Here, students enter two things they would like to know more about. This not only increases involvement, but also gives them more ownership.
Ask 1 question about something you haven't quite understood yet.
#### Slide 25 -Open question
The students indicate here (in question form) with which part of the material they still have difficulty. For the teacher, this not only provides insight into the extent to which the students understand/master the material, but also a good starting point for the next lesson.
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# Geometry
## Lines & Angles
According to the Curve Wikipedia article, lines are required to be straight and so-called curved lines should actually be referred to simply as curves [1].
A horizontal line goes side-to-side or forwards-and-backwards, while a vertical line goes up-and-down. A diagonal line can either refer to a slanted line or a line that goes between two nonadjacent corners of a shape or object [2].
Lines intersect if they come together.
When lines intersect, they create one or more angles depending on the length of the lines and the number of lines that are intersecting.
Angles are typically measured in degrees (°). An entire circle is 360°. A right angle is 90°, which is equivalent to a quarter of a circle. A right angle can be symbolized with either a small square or a dot (this is demonstrated in the Right angle Wikipedia article) [3].
Two lines are perpendicular if they form a right angle (this is the case for a horizontal line and a vertical line). Two lines are parallel if they are positioned so that they'd never intersect if continued indefinitely (this is the case for two vertical lines).
## Dimensions, Faces, & Bases
A 0 dimensional (0D) object is a point, a 1D object is a line, a 2D object is a flat shape, and a 3D object is non-flat shape.
The different 2D shapes that are on the surface of a 3D shape are called faces.
The term base can be used to refer to the bottom face of an object [4]. The bottom face of an object is typically only considered a base if it can, on its own, keep the object in a steady and stable position. This is why it can be said, for example, that a poor argument is "baseless" and "falls flat on its face".
## Polytopes
A polytope is a shape with all flat sides [5]. Some examples of different polytopes are discussed below:
• polygon - a 2D polytope [5]
• quadrilateral - a polygon with 4 sides [5]
• parallelogram - a quadrilateral with two pairs of parallel sides [6]
• rectangle - a parallelogram with all right angles [6]
• rhombus (plural: rhombi or rhombuses [7]) - a parallelogram in which all sides are of equal length [6]
• square - a parallelogram with all right angles and having all sides of equal length [6]
• polyhedron (plural: polyhedra or polyhedrons) - a 3D polytope [8]
• hexahedron (plural: hexahedra) - a polyhedron with 6 faces [9]
• parallelepiped - a hexahedron with faces that are parallelograms [9]
• cuboid - a parallelepiped with faces that are rectangles [9]
• rhombohedron - a parallelepiped with faces that are rhombi [9]
• cube - a parallelepiped with faces that are squares [9]
## Shape & Object Measurements
The diagonal length of a polytope refers to the length of its longest diagonal or diagonals.
The diameter of a circle refers to its longest length across [10].
An equilateral polygon has all of its sides as the same length, an equiangular polygon has all of its angles as the same number of degrees, and a regular polygon is both equilateral and equiangular [11].
Two objects are similar if they have the same shape and congruent if they have the same shape and size [12].
The term form factor can refer to an object's standardized shape and size [13]. Some examples of objects with a form factor are mentioned in the Form factor Wikipedia article.
Length can refer to the longer measurement of a rectangle, the longest measurement, or the longer base measurement. Width can refer to the shorter measurement of a rectangle, the side-to-side measurement, or the shorter base measurement. Depth can refer to front-to-back or vertical measurement [14] [15] [16] [17] [18 pages 11+60].
## Landscape vs. Portrait Orientation
An item that is in landscape orientation is wider-than-taller, whereas an item in portrait orientation is taller-than-wider [19].
## Display Pixels & Resolution
Modern-day displays show a picture by creating a grid of congruent rectangular cells. The cells in this grid are known as pixels. When the word resolution is applied to these displays, it's referring to the dimensions, rows, or columns of pixels [20].
## Dimension Measurements
Dimension measurements measure the sides of a shape or item. As is mentioned in the Writing article, this website will use an x to separate numbers in dimension measurements, with the x serving as a substitute for the word "by" [21]. Some important examples of dimension measurement are discussed below:
• Paper and photo print sizes typically give the smaller dimension followed by the bigger dimension [22] [23].
• Images and video typically use width followed by height [20].
• Display resolution will typically fit in one of the three categories:
• Like images and videos, display resolution with a definitive orientation typically uses width followed by height [20].
• Flexible-orientation displays typically use the larger number followed by the smaller number for resolution.
• Flexible orientation displays on devices measured in portrait orientation sometimes have the smaller number first for resolution [24] but other times have the larger number first [25].
• Boxes and other containers are typically measured as length-by-width-by-height/depth, with length being the longer base measurement and width being the shorter base measurement [17].
• The GS1 organization, which is responsible for barcodes, also specifies a variety of other supply chain standards [26]. One such standard is the GDSN Package Measurement Rules Standard, which specifies two ways of positioning and measuring items:
• For "consumer (end-user) trade items" [18 page 9], GS1 specifies to locate the "default front" [18 page 9] of an item and to then assign width, height, and depth based on this front [18 page 11]. Locating the front seems to be largely based on surface area and product markings [18 page 9]. When it comes to orienting the front, the ideal seems to be to do so via product markings; however, if this option isn't available, GS1 specifies giving the item a portrait rather than landscape orientation [18 page 10].
• For a "non-consumer trade item" [18 page 59], GS1 specifies to locate the "natural base" and to then assign depth/length, width, and height based on this base. In this context, the depth/length is the longer side of the base and the width is the shorter side of the base. For items that don't have a specific height, the height is assigned to the shortest length [18 page 60].
As is demonstrated by these and other examples [14] [15] [16], there's not a universal consensus regarding how to measure shapes, objects, and items. Therefore, in order to measure shapes, objects, and items, the following system will be used for dimension measurements:
• Paper and photo print sizes will give the smaller dimension followed by the larger dimension.
• Images and videos will take the width followed by the height.
• Display resolution will use the larger dimension followed by the smaller dimension unless the display is only meant to be used in portrait orientation.
• For 3D items that have specific side-to-side, height, and front-to-back measurements, the dimensions given will be side-to-side, then height, and finally front-to-back. This will be indicated by placing a W for width after the 1st number, an H for height after the 2nd number, and a D for depth after the 3rd number.
• For 3D items that lack a specific side-to-side measurement or front but have a specific depth, the dimensions given will be longer non-depth measurement, then shorter non-depth measurement, and finally the depth measurement. This will be indicated by only placing a D for depth after the 3rd number.
• For 3D items measured without specific side-to-side, height, and front-to-back measurements, the dimensions given will simply be largest to smallest. This will be indicated by not using any letters for any of three dimensions.
## Axes
3D space is typically measured with three axes: x, y, and z [27]. This is shown in the image below [28]:
These axes can also apply to the measurements of objects. For example, HDD, SSD, and SSHD manufacturers will sometimes describe the height of these devices as the z-height to indicate that the height of these devices is the shortest dimension [29] [30] [31] [32] [33] [34].
## Sources
1. Curve. Wikipedia.
2. Diagonal. Wikipedia.
3. Right angle. Wikipedia.
4. Base (geometry). Wikipedia.
5. Polygon. Wikipedia.
7. Rhombus. Wikipedia.
8. Polyhedron. Wikipedia.
9. Hexahedron. Wikipedia.
10. Diameter. Wikipedia.
11. Regular polygon. Wikipedia.
12. Congruence (geometry). Wikipedia.
13. Form factor. Wikipedia.
14. Measurement: Length, width, height, depth. Education Development Center, Inc. (C) 1994-2014.
15. Thread started by Herb on May 31, 1999 5:22:02 PM. What is Length in a Rectangle?. Ask Dr. Math. (C) 1994-2015 The Math Forum.
16. Thread started by Kelsey on Dec. 2, 2003 9:08:48 PM. Length or Width?. Ask Dr. Math. (C) 1994-2015 The Math Forum.
17. uspacking. Box Size & How To Measure. Oct. 28, 2011. U.S. Packaging & Wrapping LLC.
18. GDSN Package Measurement Rules Standard. Release 2.1, Oct. 2015. (C) 2015 GS1 AISBL.
19. Page orientation. Wikipedia.
20. Display resolution. Wikipedia.
21. Multiplication sign. Wikipedia.
22. Paper size. Wikipedia.
23. Photo print sizes. Wikipedia.
24. Moto X Pure Edition (2015) - Unlocked Smartphone - Motorola. (C) 2015 Motorola Mobility LLC.
25. DROID Turbo 2 - Shatterproof Android Smartphone - Motorola. (C) 2015 Motorola Mobility LLC.
26. What we do. GS1.
27. Three-dimensional space. Wikipedia.
28. Falcorian. File:Coord planes color.svg. Feb. 14, 2006 7:29 AM. Wikipedia.
29. Patrick Schmid and Achim Roos. 9.5 Versus 12.5 mm: Which Notebook HDD Is Right For You?. May 22, 2010 12:00 AM. Tom's Hardware.
30. The World’s First 7mm, 2.5-Inch Form Factor Drive. Mar. 2012. (C) 2012 Seagate Technology LLC.
31. 600 SSD. May 2013. (C) 2013 Seagate Technology LLC.
32. Making the Move Toward Slim. Aug. 2013. (C) 2013 Seagate Technology LLC.
33. Design Flexibility with 7.0mm Z-Series Drives. Oct. 2013. (C) 2013 HGST, Inc.
34. Kent Smith. Z-height: The shifting landscape of SSDs. Apr. 11, 2014 7:00 AM. (C) 2015 TechSpot, Inc.
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## Pages
### Trigonometry identities - 1
Trigonometric identity: sin2(θ) + cos2(θ) = 1
In order to prove this trigonometric identity, you will apply the Pythagorean Theorem. Let us consider a right angled triangle ABC as shown below:
Image 1: Proof of sin2(x) + cos2(x) =1
In the above right triangle, with respect to angle theta, the trigonometric ratios of sine and cosine are defined as follows:
Image 2: Proof of sin2(x) + cos2(x) =1
By applying Pythagorean Theorem in the above triangle, we get the following result:
(AB)2 + (BC)2 = (AC)2
Notice that in the trigonometric ratios (1) and (2) defined above, we have AC in the denominator of both fractions. Thus, in order to include the trigonometric ratios in this equation, we will divide the equation by (AC)2. The equation then becomes,
Image 3: Proof of sin2(x) + cos2(x) =1
In this equation, (AC)2 over itself (on the right side) is equal to 1. From result (1) obtained above, we know that sin(θ) is AB over AC and cos(θ) is BC over AC. Applying this , we get
Image 4: Proof of sin2(x) + cos2(x) =1
Thus we arrive at a result, which is one of the most important trigonometric identities in the study of Trigonometry.
#### 1 comment:
1. I am here to share some simple information about precalculus that is,It is a part of mathematics which prepares a student to learn and clear the basics of calculus and by learning this its simple for students to deal with calculus,It is just like a first step towards calculus.
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# How do you factor x^3+x^2-14x-24?
##### 1 Answer
$\textcolor{red}{{x}^{3} + {x}^{2} - 14 x - 24 = \left(x + 2\right) \left(x + 3\right) \left(x - 4\right)}$
#### Explanation:
We start from the given 3rd degree polynomial
${x}^{3} + {x}^{2} - 14 x - 24$
Use the monomial $- 14 x$
It is equal to $- 4 x - 10 x$
${x}^{3} + {x}^{2} - 4 x - 10 x - 24$
Rearrange
${x}^{3} - 4 x + {x}^{2} - 10 x - 24$
Regroup
$\left({x}^{3} - 4 x\right) + \left({x}^{2} - 10 x - 24\right)$
Factoring
$x \left({x}^{2} - 4\right) + \left(x + 2\right) \left(x - 12\right)$
$x \left(x + 2\right) \left(x - 2\right) + \left(x + 2\right) \left(x - 12\right)$
Factor out the common binomial factor $\left(x + 2\right)$
$\left(x + 2\right) \left[x \left(x - 2\right) + \left(x - 12\right)\right]$
Simplify the expression inside the grouping symbol [ ]
$\left(x + 2\right) \left[{x}^{2} - 2 x + x - 12\right]$
$\left(x + 2\right) \left({x}^{2} - x - 12\right)$
Factoring the trinomial ${x}^{2} - x - 12 = \left(x + 3\right) \left(x - 4\right)$
We now have the factors
$\left(x + 2\right) \left(x + 3\right) \left(x - 4\right)$
Final answer
$\textcolor{red}{{x}^{3} + {x}^{2} - 14 x - 24 = \left(x + 2\right) \left(x + 3\right) \left(x - 4\right)}$
God bless ....I hope the explanation is useful.
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# Average Value of a Function
## Definition of Average Value
One of the main applications of definite integrals is to find the average value of a function $$y = f\left( x \right)$$ over a specific interval $$\left[ {a,b} \right].$$
In order to find this average value, one must integrate the function by using the Fundamental Theorem of Calculus and divide the answer by the length of the interval.
So, the average (or the mean) value of $$f\left( x \right)$$ on $$\left[ {a,b} \right]$$ is defined by
$\bar f = \frac{1}{{b - a}}\int\limits_a^b {f\left( x \right)dx} .$
## The Mean Value Theorem for Definite Integrals
Let $$y = f\left( x \right)$$ be a continuous function on the closed interval $$\left[ {a,b} \right].$$ The mean value theorem for integrals states that there exists a point $$c$$ in that interval such that
$f\left( c \right) = \frac{1}{{b - a}}\int\limits_a^b {f\left( x \right)dx} .$
In other words, the mean value theorem for integrals states that there is at least one point $$c$$ in the interval $$\left[ {a,b} \right]$$ where $$f\left( x \right)$$ attains its average value $$\bar f:$$
$f\left( c \right) = \bar f = \frac{1}{{b - a}}\int\limits_a^b {f\left( x \right)dx} .$
Geometrically, this means that there is a rectangle whose area exactly represents the area of the region under the curve $$y = f\left( x \right).$$ The value of $$f\left( c \right)$$ represents the height of the rectangle and the difference $$\left( {b - a} \right)$$ represents the width.
## Root Mean Square Value of a Function
The root mean square value $$\left( {RMS} \right)$$ is defined as the square root of the average (mean) value of the squared function $${{{\left[ {f\left( x \right)} \right]}^2}}$$ over an interval $$\left[ {a,b} \right].$$ The corresponding integration formula is written in the form
$RMS = \sqrt {\frac{1}{{b - a}}\int\limits_a^b {{{\left[ {f\left( x \right)} \right]}^2}dx} } .$
The $${RMS}$$ value has many applications in mathematics, physics and engineering. For example, in physics, the RMS value of an alternating current $$\left( {AC} \right)$$ is equal to the value of the direct current $$\left( {DC} \right)$$ that dissipates the same power in a resistor.
## Solved Problems
Click or tap a problem to see the solution.
### Example 1
The average value of a function $$y = f\left( x \right)$$ over the interval $$x \in \left[ {1,5} \right]$$ is $$2.$$ What is the value of $\int\limits_1^5 {f\left( x \right)dx}?$
### Example 2
Find the average value of the cubic function $f\left( x \right) = {x^3}$ on the interval $$\left[ {0, 1} \right].$$
### Example 3
Find the average value of the square root function $f\left( x \right) = \sqrt{x}$ on the interval $$\left[ {0, 25} \right].$$
### Example 4
Find the average value of the cosine function $f\left( x \right) = \cos{x}$ on the interval $$\left[ {0, \frac{\pi }{2}} \right].$$
### Example 1.
The average value of a function $$y = f\left( x \right)$$ over the interval $$x \in \left[ {1,5} \right]$$ is $$2.$$ What is the value of $\int\limits_1^5 {f\left( x \right)dx}?$
Solution.
By definition,
$\bar f = \frac{1}{{b - a}}\int\limits_a^b {f\left( x \right)dx} ,\;\; \Rightarrow \int\limits_a^b {f\left( x \right)dx} = \bar f\left( {b - a} \right).$
Substituting the given values, we obtain
$\int\limits_1^5 {f\left( x \right)dx} = 2\left( {5 - 1} \right) = 8.$
### Example 2.
Find the average value of the cubic function $f\left( x \right) = {x^3}$ on the interval $$\left[ {0, 1} \right].$$
Solution.
We use the integration formula
$\bar f = \frac{1}{{b - a}}\int\limits_a^b {f\left( x \right)dx} .$
Hence,
$\bar f = \frac{1}{{1 - 0}}\int\limits_0^1 {{x^3}dx} = \int\limits_0^1 {{x^3}dx} = \left. {\frac{{{x^4}}}{4}} \right|_0^1 = \frac{1}{4}.$
### Example 3.
Find the average value of the square root function $f\left( x \right) = \sqrt{x}$ on the interval $$\left[ {0, 25} \right].$$
Solution.
Using the definition of the average value, we can write
$\bar f = \frac{1}{{b - a}}\int\limits_a^b {f\left( x \right)dx} = \frac{1}{{25}}\int\limits_0^{25} {\sqrt x dx} = \frac{1}{{25}}\int\limits_0^{25} {{x^{\frac{1}{2}}}dx} = \frac{1}{{25}} \cdot \left. {\frac{{2{x^{\frac{3}{2}}}}}{3}} \right|_0^{25} = \frac{1}{{25}} \cdot \frac{{2{{\left( {\sqrt {25} } \right)}^3}}}{3} = \frac{{250}}{{75}} = \frac{{10}}{3}.$
### Example 4.
Find the average value of the cosine function $f\left( x \right) = \cos{x}$ on the interval $$\left[ {0, \frac{\pi }{2}} \right].$$
Solution.
We use the formula
$\bar f = \frac{1}{{b - a}}\int\limits_a^b {f\left( x \right)dx} .$
So we set up and evaluate the following integral
$\bar f = \frac{1}{{\frac{\pi }{2} - 0}}\int\limits_0^{\frac{\pi }{2}} {\cos xdx} = \frac{2}{\pi }\int\limits_0^{\frac{\pi }{2}} {\cos xdx} = \frac{2}{\pi }\left. {\sin x} \right|_0^{\frac{\pi }{2}} = \frac{2}{\pi }.$
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# How do you find vertical, horizontal and oblique asymptotes for f(x) =(x^2+4x-7)/(x-7)?
Nov 2, 2016
The VA Is $x = 7$, the OA is $y = x + 11$, and there is no HA.
#### Explanation:
$\frac{{x}^{\textcolor{red}{2}} + 4 x - 7}{{x}^{\textcolor{b l u e}{1}} - 7}$
To find the vertical asymptote (VA), set the denominator equal to zero and solve for $x$. At this value of $x$, $f \left(x\right)$ is undefined.
$x - 7 = 0$
$x = 7$ is the VA
For horizontal/oblique asymptotes, compare the degree of the numerator to the degree of the denominator. If the degree of the numerator is one more than the degree of the denominator, there is an oblique asymptote (OA), but no horizontal asymptote (HA).
In this example, the degree of the numerator is $\textcolor{red}{2}$ and the degree of the denominator is $\textcolor{b l u e}{1}$. There is an OA but no HA.
To find the OA, use the first two coefficients produced by synthetic division. You do not need the remainder.
$7 | 1 \textcolor{w h i t e}{a a a} 4 \textcolor{w h i t e}{a a} - 7$
$\textcolor{w h i t e}{a a} \downarrow \textcolor{w h i t e}{{a}^{22}} 7$
$\textcolor{w h i t e}{a a a}$--------------------
color(white)(a^2a)color(magenta)1color(white)(a^2a)color(magenta)(11
The OA is $y = \textcolor{m a \ge n t a}{1} x + \textcolor{m a \ge n t a}{11} = x + 11$
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# Using the generalized pigeonhole principle we can
• Notes
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Using the generalized pigeonhole principle we can conclude that among 100 people, there are at least 100 / 12 = 9 who are born in the same month. Example. Suppose each point in the plane is colored either red or blue. Show that there always exist two points of the same color that are exactly one feet apart. Solution. Consider an equilateral triangle with the length of each side being one feet. The three corners of the triangle are colored red or blue. By pigeonhole principle, two of these three points must have the same color. Example. Given a sequence of n integers, show that there exists a subsequence of con- secutive integers whose sum is a multiple of n . Solution. Let x 1 , x 2 , . . . , x n be the sequence of n integers. Consider the following n sums. x 1 , x 1 + x 2 , x 1 + x 2 + x 3 , . . . , x 1 + x 2 + · · · + x n If any of these n sums is divisible by n , then we are done. Otherwise, each of the n sums have a non-zero remainder when divided by n . There are at most n 1 different possible remainders: 1 , 2 , . . . n 1. Since there are n sums, by the pigeonhole principle, at least two of the n sums have the same remainder when divided by n . Let p and q , p < q , be integers such that for some integers c 1 and c 2 , x 1 + x 2 + · · · + x p = c 1 n + r and x 1 + x 2 + · · · + x q = c 2 n + r Subtracting the two sums, we get x p +1 + · · · + x q = ( c 2 c 1 ) n Hence, x p +1 + · · · + x q is divisible by n . Example. Show that in any group of six people there are either three mutual friends or three mutual strangers. Solution. Consider one of the six people, say A . The remaining five people are either friends of A or they do not know A . By the pigeonhole principle, at least 5 / 2 = 3 of the five people are either friends of A or are unacquainted with A . In the former case, if any two of the three people are friends then these two along with A would be mutual friends, otherwise the three people would be strangers to each other. The proof for the latter case, when three or more people are unacquainted with A , proceeds in the same manner.
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January 10, 2011 Lecture Outline 5 Example. A chess master who has 11 weeks to prepare for a tournament decides to play at least one game every day but, in order not to tire himself, he decides not to play more than 12 games during any calendar week. Show that there exists consecutive days during which the chess master will have played exactly 21 games. Solution. Let a i , 1 i 77, be the total number of games that the chess master has played during the first i days. Note that the sequence of numbers a 1 , a 2 , . . . , a 77 is a strictly increasing sequence. We have 1 a 1 < a 2 < . . . < a 77 11 × 12 = 132 Now consider the sequence a 1 + 21 , a 2 + 21 , . . . , a 77 + 21. We have 22 a 1 + 21 < a 2 + 21 < . . . < a 77 + 21 153 Clearly, this sequence is also a strictly increasing sequence. The numbers a 1 , a 2 , . . . , a 77 , a 1 + 21 , a 2 + 21 , . . . , a 77 + 21 (154 in all) belong to the set { 1 , 2 , . . . , 153 } . By the pigeonhole principle there must be two numbers out of the 154 numbers that must be the same. Since no two numbers in a 1 , a 2 , . . . , a 77 are equal and no two numbers in a 1 +21 , a 2 +21 , . . . , a 77 +21 are equal there must exist i and j such that a i = a j + 21. Hence during the days j + 1 , j + 2 , . . . , i , exactly 21 games must have been played.
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• Winter '10
• RajivSir
• Algorithms, Prime number, Rational number, Generalized Pigeonhole Principle
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# Numerical semigroup
In mathematics, a numerical semigroup is a special kind of a semigroup. Its underlying set is the set of all nonnegative integers except a finite number and the binary operation is the operation of addition of integers. Also, the integer 0 must be an element of the semigroup. For example, while the set {0, 2, 3, 4, 5, 6, ...} is a numerical semigroup, the set {0, 1, 3, 5, 6, ...} is not because 1 is in the set and 1 + 1 = 2 is not in the set. Numerical semigroups are commutative monoids and are also known as numerical monoids.[1][2]
The definition of numerical semigroup is intimately related to the problem of determining nonnegative integers that can be expressed in the form x1n1 + x2 n2 + ... + xr nr for a given set {n1, n2, ..., nr} of positive integers and for arbitrary nonnegative integers x1, x2, ..., xr. This problem had been considered by several mathematicians like Frobenius (1849 – 1917) and Sylvester (1814 – 1897) at the end of the 19th century.[3] During the second half of the twentieth century, interest in the study of numerical semigroups resurfaced because of their applications in algebraic geometry.[4][5]
## Definition and examples
### Definition
Let N be the set of nonnegative integers. A subset S of N is called a numerical semigroup if the following conditions are satisfied.
1. 0 is an element of S
2. NS, the complement of S in N, is finite.
3. If x and y are in S then x + y is also in S.
There is a simple method to construct numerical semigroups. Let A = {n1, n2, ..., nr} be a nonempty set of positive integers. The set of all integers of the form x1 n1 + x2 n2 + ... + xr nr is the subset of N generated by A and is denoted by 〈 A 〉. The following theorem fully characterizes numerical semigroups.
### Theorem
Let S be the subsemigroup of N generated by A. Then S is a numerical semigroup if and only if gcd (A) = 1. Moreover, every numerical semigroup arises in this way.[6]
### Examples
The following subsets of N are numerical semigroups.
1. 〈 1 〉 = {0, 1, 2, 3, ...}
2. 〈 1, 2 〉 = {0, 1, 2, 3, ...}
3. 〈 2, 3 〉 = {0, 2, 3, 4, 5, 6, ...}
4. Let a be a positive integer. 〈 a, a + 1, a + 2, ... , 2a - 1 〉 = {0, a, a + 1, a + 2, a + 3, ...}.
5. Let b be an odd integer greater than 1. Then 〈 2, b 〉 = {0, 2, 4, . . . , b − 3 , b − 1, b, b + 1, b + 2, b + 3 , ...}.
## Embedding dimension, multiplicity
The set A is a set of generators of the numerical semigroup 〈 A 〉. A set of generators of a numerical semigroup is a minimal system of generators if none of its proper subsets generates the numerical semigroup. It is known that every numerical semigroup S has a unique minimal system of generators and also that this minimal system of generators is finite. The cardinality of the minimal set of generators is called the embedding dimension of the numerical semigroup S and is denoted by e(S). The smallest member in the minimal system of generators is called the multiplicity of the numerical semigroup S and is denoted by m(S).
## Frobenius number and genus
There are several notable numbers associated with a numerical semigroup S.
1. The set NS is called the set of gaps in S and is denoted by G(S).
2. The number of elements in the set of gaps G(S) is called the genus of S (or, the degree of singularity of S) and is denoted by g(S).
3. The greatest element in G(S) is called the Frobenius number of S and is denoted by F(S).
### Examples
Let S = 〈 5, 7, 9 〉. Then we have:
• The set of elements in S : S = {0, 5, 7, 9, 10, 12, 14, ...}.
• The minimal set of generators of S : {5, 7, 9}.
• The embedding dimension of S : e(S) = 3.
• The multiplicity of S : m(S) = 5.
• The set of gaps in S : G(S) = {1, 2, 3, 4, 6, 8, 11, 13}.
• The Frobenius number of S : F(S) = 13.
• The genus of S : g(S) = 8.
Numerical semigroups with small Frobenius number or genus
n Semigroup S
with F(S) = n
Semigroup S
with g(S) = n
1 〈 2, 3 〉 〈 2, 3 〉
2 〈 3, 4, 5 〉 〈 3, 4, 5 〉
〈 2, 5 〉
3 〈 4, 5, 6, 7 〉
〈 2, 5 〉
〈 4, 5, 6, 7, 〉
〈 3, 5, 7 〉
〈 3, 4 〉
〈 2, 7 〉
4 〈 5, 6, 7, 8, 9 〉
〈 3, 5, 7 〉
〈 5, 6, 7, 8, 9 〉
〈 4, 6, 7, 9 〉
〈 3, 7, 8 〉
〈 4, 5, 7 〉
〈 4, 5, 6 〉
〈 3, 5, 〉
〈 2, 9 〉
## Computation of Frobenius number
### Numerical semigroups with embedding dimension two
The following general results were known to Sylvester.[7][not in citation given] Let a and b be positive integers such that gcd (a, b) = 1. Then
• F(〈 a, b 〉) = (a − 1) (b − 1) − 1 = ab − (a + b).
• g(〈 a, b 〉) = (a − 1)(b − 1) / 2.
### Numerical semigroups with embedding dimension three
There is no known general formula to compute the Frobenius number of numerical semigroups having embedding dimension three or more. It is also known that no polynomial formula can be found to compute the Frobenius number or genus of a numerical semigroup with embedding dimension three.[8] Interestingly, it is known that every positive integer is the Frobenius number of some numerical semigroup with embedding dimension three.[9]
### Rödseth's algorithm
The following algorithm, known as Rödseth's algorithm,[10] [11] can be used to compute the Frobenius number of a numerical semigroup S generated by {a1, a2, a3} where a1 < a2 < a3 and gcd ( a1, a2, a3) = 1. Its worst-case complexity is not as good as Greenberg's algorithm [12] but it is much simpler to describe.
• Let s0 be the unique integer such that a2s0a3 mod a1, 0 ≤ s0 < a1.
• The continued fraction algorithm is applied to the ratio a1/s0:
• a1 = q1s0s1, 0 ≤ s1 < s0,
• s0 = q2s1s2, 0 ≤ s2 < s1,
• s1 = q3s2s3, 0 ≤ s3 < s2,
• ...
• sm−1 = qm+1sm,
• sm+1 = 0,
where qi ≥ 2, si ≥ 0 for all i.
• Let p−1 = 0, p0 = 1, pi+1 = qi+1pipi−1 and ri = (sia2pia3)/a1.
• Let v be the unique integer number such that rv+1 ≤ 0 < rv, or equivalently, the unique integer such
• sv+1/pv+1a3/a2 < sv/pv·
• Then, F(S) = −a1 + a2(sv − 1) + a3(pv+1 − 1) − min{a2sv+1, a3pv}.
## Special classes of numerical semigroups
An irreducible numerical semigroup is a numerical semigroup such that it cannot be written as the intersection of two numerical semigroups properly containing it. A numerical semigroup S is irreducible if and only if S is maximal, with respect to set inclusion, in the collection of all numerical semigroups with Frobenius number F(S).
A numerical semigroup S is symmetric if it is irreducible and its Frobenius number F(S) is odd. We say that S is pseudo-symmetric provided that S is irreducible and F(S) is even. Such numerical semigroups have simple characterizations in terms of Frobenius number and genus:
• A numerical semigroup S is symmetric if and only if g(S) = (F(S) + 1)/2.
• A numerical semigroup S is pseudo-symmetric if and only if g(S) = (F(S) + 2)/2.
## References
1. ^ Garcia-Sanchez, P.A. "Numerical semigroups minicourse". Retrieved 6 April 2011.
2. ^ Finch, Steven. "Monoids of Natural Numbers" (PDF). INRIA Algorithms Project. Retrieved 7 April 2011.
3. ^ J.C. Rosales and P.A. Garcia-Sanchez (2009). Numerical Semigroups. Springer. ISBN 978-1-4419-0159-0.
4. ^ V. Barucci, et. al. (1997). "Maximality properties in numerical semigroups and applications to one-dimensional analytically irreducible local domains". Memoirs of the Amer. Math. Soc. 598.
5. ^ Martino, Ivan; Martino, Luca (2013-11-14). "On the variety of linear recurrences and numerical semigroups". Semigroup Forum. 88 (3): 569–574. doi:10.1007/s00233-013-9551-2. ISSN 0037-1912.
6. ^ García-Sánchez, J.C. Rosales, P.A. (2009). Numerical semigroups (First. ed.). New York: Springer. p. 7. ISBN 978-1-4419-0160-6.
7. ^ J. J. Sylvester (1884). "Mathematical questions with their solutions". Educational Times. 41 (21).
8. ^ F. Curtis (1990). "On formulas for the Frobenius number of a numerical semigroup". Mathematica Scandinavica. 67 (2): 190–192. Retrieved 8 April 2011.
9. ^ J. C. Rosales, et. al. (2004). "Every positive integer is the Frobenius number of a numerical semigroup with three generators". Mathematica Scandinavica. 94: 5–12. Retrieved 14 March 2015.
10. ^ J.L. Ramírez Alfonsín (2005). The Diophantine Frobenius Problem. Oxford University Press. pp. 4–6. ISBN 978-0-19-856820-9.
11. ^ Ö.J. Rödseth (1978). "On a linear Diophantine problem of Frobenius". J. Reine Angew. Math. 301: 171–178.
12. ^ Harold Greenberg (1988). "Solution to a linear Diophantine equation for non-negative integers". Journal of Algorithms. 9: 343–353. doi:10.1016/0196-6774(88)90025-9.
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# Multiplying fractions
When multiplying fractions in order to find the product of two or more fractions, you just need to follow these three simple steps basically.
Step 1: Multiply the numerators together. The numerators are also called top numbers.
Step 2: Multiply the denominators together. The denominators are also called bottom numbers.
Step 3: Finally, try to simplify the product if needed to get the final answer.
For example, notice what we do when we multiply the following fractions: 3/4 × 4/6.
Step 1: Multiply 3 and 4 to get 12 and 12 is the numerator of the product
Step 2: Multiply 4 and 6 to get 24 and 24 is the denominator of the product
3/4 × 4/6 = (3 × 4)/(4 × 6) = 12/24
Step 3: Divide both the numerator and the denominator by 12 to simplify the fraction. 12 is the greatest common factor (GCF) of 12 and 24.
3/4 × 4/6 = 1/2
The example above is straightforward. However, when multiplying fractions, you may wonder about the following cases.
• Multiplying fractions with different denominators
• Multiplying fractions with the same denominator
• Multiplying fractions with whole numbers
• Multiplying fractions with mixed numbers
• Multiplying improper fractions
Depending on which situation(s) you encounter, there are rules to follow when you multiply fractions with different types of fractions .
## Rules of multiplying fractions
Rule 1: The most important rule is to multiply straight across. In other words, multiply the numerators to get the new numerator or the numerator of the product. Multiply the denominators to get the new denominator or the denominator of the product.
Rule 2: Another important rule is to always convert mixed fractions, also called mixed numbers into improper fractions before multiplying.
Rule 3: Convert whole numbers into fractions before doing multiplication.
Rule 4: Multiplying fractions is not the same as adding fractions. Therefore, you must not look for the least common denominator!
Rule 5: Simplify the product or write the fraction you end with after performing multiplication in lowest terms if needed.
## Multiplying fractions with different denominators
When you multiply fractions with different denominators, just keep in mind rule 4 stated above. Do not look for a common denominator! The rule for adding fractions and multiplying fractions are not the same.
For example, notice that we do not look for a common denominator when we multiply the following fractions: 1/5 × 2/3.
Step 1: Multiply 1 and 2 to get 2
Step 2: Multiply 5 and 3 to get 15
1/5 × 2/3 = (1 × 2)/(5 × 3) = 2/15
Step 3: 2/15 is already written in lowest terms since the greatest common factor of 2 and 15 is 1.
1/5 × 2/3 = 1/2
## Multiplying fractions with the same denominator
When you multiply fractions with the same denominator, just do the same thing you do when the fractions have unlike denominators.
Example: Multiply 3/4 and 1/4
3/4 × 1/4 = (3 × 1)/(4 × 4) = 3/16
## Multiplying fractions with whole numbers
When you multiply fractions with whole numbers, just keep in mind rule 3 stated above. Convert the whole number into a fraction before doing multiplication.
Notice that any whole number x can be written as a fraction x/1 since any number divided by 1 will return the same number.
For example if you multiply the whole number 5 by another fraction, write 5 as 5/1 before you multiply.
Example: Multiply 5 and 2/3
5 × 2/3 = 5/1 × 2/3
5 × 2/3 = (5 × 2)/(1 × 3) = 10/3
## Multiplying fractions with mixed numbers
When multiplying fractions with mixed numbers, it is important to remember rule 2. You must first convert any mixed number into a fraction before you multiply.
Suppose you are multiplying a fraction by 2 1/3. Since 2 1/3 is a mixed number, you must convert it into a fraction.
2 1/3 = (2 × 3 + 1)/3 = (6 + 1) / 3 = 7/3
Example: Multiply 1/6 and 2 1/3
1/6 × 2 1/3 = 1/6 × 7/3
1/6 × 7/3 = (1 × 7)/(6 × 3) = 7/18
## Multiplying improper fractions
The multiplication of improper fractions is performed by following rule 1. Just multiply straight across. One thing you definitely do not want to do here is to convert the improper fractions to mixed numbers.
This will be very counterproductive as you will have to convert them right back into improper fractions.
Example: Multiply 9/2 and 3/5
9/2 × 3/5 = (9 × 3)/(2 × 5) = 27/10
## A couple of tips and trick to follow when multiplying fractions
1. I recommend that you become familiar with the multiplication table. You will be able to perform the multiplication of fractions much quicker.
2. Sometimes, it is a good idea to simplify the fractions before multiplying to make calculations easier.
Take a look at the following example:
10 / 20 × 3 / 15
10 / 20 can be simplified as 1 / 2
Divide the numerator and the denominator by 10
3 / 15 can be simplified as 1 / 5
Divide the numerator and the denominator by 3
This way, it is easier to do the multiplication to get
1 / 10
2. Sometimes, it is a good idea to simplify the fractions before multiplying.
Take a look at the following example:
10 / 20 × 3 / 15
10 / 20 can be simplified as 1 / 2
After we divide the numerator and the denominator by 10
3 / 15 can be simplified as 1 / 5
After we divide the numerator and the denominator by 3
This way, it is easier to do the multiplication to get
1 / 10
3. If you have three or more fractions, just multiply all numerators and all denominators
## Going a little deeper! Why do we multiply fractions straight across?
I would like to introduce the topic with an interesting example about pizza.
Suppose that you bought a medium pizza and the pizza has 8 slices.
If someone eats half of your pizza, or 4 slices, you are left with
4 / 8
From the illustration below, you can also see that the leftover is the same as
1 / 2
If you decide that you are only going to eat 1 slice out of the 4 slices remaining, you are eating
1 / 4
of the leftover.
Remember that the leftover is
1 / 2
You can also argue that you only ate 1 slice out of 8 slices or
1 / 8
Thus, we can see that eating 1/4 of 1/2 is the same as eating 1/8.
Another way to get
1 / 8
is to perform the following multiplication:
1 / 4 × 1 / 2 = 1 / 8
We get this answer by multiplying the numbers on top (numerators): 1 × 1 = 1
and by multiplying the numbers at the bottom (denominators): 4 × 2 = 8
This is an interesting result but all you need to remember is the following:
When you multiply fractions, you must multiply straight across.
When the word 'of' is placed between two fractions, it means multiplication.
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# I have been asked to investigate further into the stair totals and other step stairs on different number grids. Firstly, I will change the sizes of the stairs from 3 steps to 4, then 5, and so on to 8 step stairs.
Extracts from this document...
Introduction
Maths coursework - part two In part two, I have been asked to investigate further into the stair totals and other step stairs on different number grids. Firstly, I will change the sizes of the stairs from 3 steps to 4, then 5, and so on to 8 step stairs. When I have finished investigating that, I will change the sizes of the grids. I will also try to find the nth term for the nth term. So I can use it to find it the nth terms of any sized step stairs, or any three step stair on any grid. 4 step stairs Stair number = 1 Stair total = 120 31 32 33 34 21 22 23 24 11 12 13 14 1 2 3 4 Stair number = 2 Stair total = 130 32 33 34 354 22 23 24 25 12 13 14 15 2 3 4 5 Stair number = 3 Stair total = 140 33 34 35 36 23 24 25 26 13 14 15 16 3 4 5 6 A table to show the stair numbers And stair totals Stair no 1 2 3 Stair total 120 130 140 Finding the nth term 10 20 30 120 130 140 +10 +10 +10 10n + 110 = stair total Checking the nth term for stair 4 10n + 110 (10 x 4) ...read more.
Middle
4 5 6 7 8 Stair number = 3 Stair total = 700 63 646566 65 66 67 68 69 53 54 55 56 57 58 59 43 44 45 46 47 48 49 33 34 35 36 37 38 39 23 24 25 26 27 28 29 13 14 15 16 17 18 19 3 4 5 6 7 8 9 A table to show the stair numbers and Their stair totals Stair number 1 2 3 Stair total 644 672 700 Finding the nth term 28 56 94 644 672 700 +28 +28 +28 28n + 616 Checking the nth term for stair 4 28n + 616 (28 x 4)+ 616 = 728 = stair total for stair 4 8 step stairs Stair number = 1 Stair total = 960 71 72 73 74 75 76 78 79 61 62 63 64 65 66 67 68 51 52 53 54 55 56 57 58 41 42 43 44 45 46 47 48 31 32 33 34 35 36 377 38 21 22 23 24 25 26 27 28 11 12 13 14 15 16 17 18 1 2 3 4 5 6 7 8 Stair number = 2 Stair total = 996 72 73 74 75 76 77 78 79 62 63 64 65 66 67 68 69 52 ...read more.
Conclusion
+ 32 56 = stair total for stair 4 8 x 8 grid Stair number = 1 Stair total = 42 17 18 19 9 10 11 1 2 3 Stair number = 2 Stair total = 48 18 19 20 10 11 12 2 3 4 Stair number = 3 Stair total = 54 19 20 21 11 12 13 3 4 5 Stair no 1 2 3 Stair total 42 48 54 Nth term 6 12 18 42 48 54 +6 +6 +6 6n + 36 Check nth term for stair 4 6n + 36 (6 x 4) + 36 60 = stair total for stair 4 Nth term of nth term 6n + 28 +4 6n + 32 +4 6n + 36 +4 4 8 12 28 32 36 +4 +4 +4 4n + 24 Then with the answer, put 6n + in front of it. Check nth term for fourth nth term (9 x 9) grid 4n + 24 (4 x 4) +24 40 = 6n + 40 = nth term for step stair on 9 x 9 grid. After all of my investigating, I have come up with a formula to find out the nth term of any three step stair on any grid. I have also found out how to find the stair totals of any three step stair on a 10 x 10 grid. ?? ?? ?? ?? ...read more.
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# Related GCSE Number Stairs, Grids and Sequences essays
1. ## Maths coursework- stair totals. I shall be investigating the total and difference in sets ...
3 star(s)
Therefore due to the above diagram it is evident that the formula must be, 6x + 44. It is easy to work out because there is six 'x's then you have to times x by 6 then you just add the other numbers together to get 44.
2. ## Investigate the number of winning lines in the game Connect 4.
board, the number of winning lines = h - 4. I then decided to alter the width of the grid, instead of constantly keeping it at 1 square I decide to increase it by 1 square on each separate grid. However I still only drew on the winning vertical lines.
1. ## Number stairs
65 66 45 46 47 48 49 50 51 52 53 54 55 34 35 36 37 38 39 40 41 42 43 44 23 24 25 26 27 28 29 30 31 32 33 12 13 14 15 16 17 18 19 20 21 22 1 2 3 4
2. ## Number Grids Investigation Coursework
= w ((n - 1) (n - 1)) = w (n - 1)2 As the formula I worked out was D = w (n - 1)2 and the expression for the difference between
1. ## For other 3-step stairs, investigate the relationship between the stair total and the position ...
The results are conclusive and consistent, proving the theory to be accurate and reliable. Using any of the algebra equation from the above table, e.g. 10x-90 or 10x-110 we can prove the results for any 4-step grid: 21 Formula: 10x- 90 31 32 1: 10 x 51 - 90= 420
2. ## Maths Grids Totals
59 62 63 64 65 66 67 68 69 72 73 74 75 76 77 78 79 82 83 84 85 86 87 88 89 12 x 89 = 1068 19 x 82 = 1558 1558 - 1068 = 490.
1. ## number stairs
+ 4 + 12 + 13 + 22 = 56 If I move the 3-step stair another unit to the right, the 3-step stair would be made up of 3 + 4 + 5 + 13 + 14 + 23 = 62 I have created a table of a few
2. ## Number Grids
= 672 832 - 672 = 160 I predict that my next 5 x 5 grid will result in an answer of 160. 6 7 8 9 10 16 17 18 19 20 26 27 28 29 30 36 37 38 39 40 46 47 48 49 50 10 x
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You are viewing an older version of this Concept. Go to the latest version.
Multi-Step Equations
Maintain balance of an equation throughout all steps needed to solve.
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Practice Multi-Step Equations
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Solve Equations with the Distributive Property and Combining Like Terms
Do you like candy? Take a look at this yummy dilemma.
Eight children were given some candy. Then six different children were given the same unknown amount of candy. Next, two children were that same unknown amount of candy plus three additional pieces of candy. The total number of pieces of candy given out was thirty -eight.
If this is the case, what is the unknown amount of candy?
Do you know how to solve this problem? Write an equation and then solve it for the unknown amount of candy. Pay attention and you will see this dilemma at the end of the Concept.
Guidance
To solve some multi-step equations you will need to use the distributive property and combine like terms. When this happens, you will see that there is more than one term with the same variable or there is more than one number in the equation. You always want to combine everything that you can before moving on to solving the equation.
Let's apply this to the following situation.
Solve for \begin{align*}m\end{align*}: \begin{align*}6(1 +2m) -3m = 24\end{align*}
Apply the distributive property to the left side of the equation. Multiply each of the two numbers inside the parentheses by 6 and then add those products.
Next, subtract the like terms—\begin{align*}12m\end{align*} and \begin{align*}3m\end{align*}—on the left side of the equation.
Finally, solve as you would solve any two-step equation. Subtract 6 from both sides of the equation.
Now, divide both sides of the equation by 9.
The value of \begin{align*}m\end{align*} is 2.
Here is another one.
Solve for \begin{align*}b\end{align*}: \begin{align*}-4(2 + 3b) + 5b = 13\end{align*}
Apply the distributive property to the left side of the equation. Multiply each of the two numbers inside the parentheses by -4 and then add those products.
Next, add the like terms on the left side of the equation. To add those like terms, \begin{align*}-12b\end{align*} and \begin{align*}5b\end{align*}, you will need to use what you know about adding integers. To review how to add and subtract integers, look back at Lesson 2.5.
Finally, solve as you would solve any two-step equation. Since -8 is added to \begin{align*}(-7b)\end{align*}, you can subtract -8 from both sides of the equation to solve it.
Now, divide both sides of the equation by -7.
The value of \begin{align*}b\end{align*} is -3.
Example A
\begin{align*}6(x+4)+3x - 2 = 54\end{align*}
Solution: \begin{align*}x = 4\end{align*}
Example B
\begin{align*}6y + 3(y-4) = 33\end{align*}
Solution: \begin{align*}y = 5\end{align*}
Example C
\begin{align*}5(a+3)+6(a+1)+8a=40\end{align*}
Solution: \begin{align*}a = 1\end{align*}
Now let's go back to the dilemma from the beginning of the Concept.
Eight children were given some candy. Then six different children were given the same unknown amount of candy. Next, two children were that same unknown amount of candy plus three additional pieces of candy. The total number of pieces of candy given out was thirty -eight.
First, write an equation. Break down the problem a piece at a time. We will use \begin{align*}c\end{align*} for the unknown amount of candy given.
\begin{align*}8c + 6c + 2(c + 3) = 38\end{align*}
This is our equation.
Now solve the equation by first getting rid of the parentheses.
\begin{align*}8c + 6c + 2c + 6 = 38\end{align*}
Next, combine like terms.
\begin{align*}16c + 6 = 38\end{align*}
Now subtract six from both sides of the equation.
\begin{align*}16c + 6 - 6 = 38 - 6\end{align*}
\begin{align*}16c = 32\end{align*}
\begin{align*}c = 2\end{align*}
The unknown amount of candy was two pieces.
Vocabulary
Like Terms
terms that include a common variable.
Commutative Property of Addition
states that the order that you add different numbers does not change the sum.
Associative Property of Addition
states that you can change the groupings of numbers being added without changing the sum.
Distributive Property
states that you can multiply a term outside of a set of parentheses with the terms inside the parentheses to simplify the set of parentheses.
Guided Practice
Here is one for you to try on your own.
Solve for \begin{align*}x\end{align*}.
\begin{align*}-5x+3(x+1)-4x=45\end{align*}
Solution
First, distribute the three and get rid of the parentheses.
\begin{align*}-5x+3x+3-4x=45\end{align*}
Now combine like terms.
\begin{align*}-2x+3-4x=45\end{align*}
Combine again because there are many terms to combine in this problem.
\begin{align*}-6x+3=45\end{align*}
Next subtract three from both sides of the equation.
\begin{align*}-6x+3-3=45-3\end{align*}
\begin{align*}-6x=42\end{align*}
\begin{align*}x = -7\end{align*}
This is our answer.
Practice
Directions: Distribute and combine like terms and then solve each equation.
1. \begin{align*}x+8(x+2)=52\end{align*}
2. \begin{align*}2y+6(y+3)=34\end{align*}
3. \begin{align*}4y+2(y-2)=8\end{align*}
4. \begin{align*}9y+3(y-6)=30\end{align*}
5. \begin{align*}6(x+2)-4x=30\end{align*}
6. \begin{align*}3(y-1)+2(y+3)=13\end{align*}
7. \begin{align*}4(a+3)-2(a+6)=20\end{align*}
8. \begin{align*}6(x+2)-4x+6=36\end{align*}
9. \begin{align*}-9(x+3)+4x=-2\end{align*}
10. \begin{align*}-4(y+3)-2y=24\end{align*}
11. \begin{align*}4(a+2)-9=11\end{align*}
12. \begin{align*}-8(y+2)-16=16\end{align*}
13. \begin{align*}5(a+4)-6a+1=12\end{align*}
14. \begin{align*}x+3x+2x+3(x+1)=30\end{align*}
15. \begin{align*}2x+4x+6x-2(x+3)=34\end{align*}
Vocabulary Language: English
Associative Property
Associative Property
The associative property states that you can change the groupings of numbers being added or multiplied without changing the sum. For example: (2+3) + 4 = 2 + (3+4), and (2 X 3) X 4 = 2 X (3 X 4).
Commutative Property
Commutative Property
The commutative property states that the order in which two numbers are added or multiplied does not affect the sum or product. For example $a+b=b+a \text{ and\,} (a)(b)=(b)(a)$.
distributive property
distributive property
The distributive property states that the product of an expression and a sum is equal to the sum of the products of the expression and each term in the sum. For example, $a(b + c) = ab + ac$.
like terms
like terms
Terms are considered like terms if they are composed of the same variables with the same exponents on each variable.
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## GeoGebra Tutorial Video 2 – Constructing a Square
This is the second video tutorial on using GeoGebra. This is the video of the article GeoGebra Tutorial 3 – Constructing a Square. Sorry for some grammar lapses (I am not very good in impromptu), and I am really having a hard time imitating American accent to make most of the viewers understand this video. Sometimes my tongue gets twisted — so twisted that it automatically adds “s” to singular nouns (chuckles) like squares instead of square.
Anyway, this is a math blog, so I suppose it is appropriate for us to leave the grammar stuff to language teachers, although you are very much welcome to comment on errors, grammar may it be.
My next video is on Graphs and Sliders, so keep posted.
## Derivative and the Maximum Area Problem
Note: This is the third part of the Derivative Concept Series. The first part is The Algebraic and Geometric Meaning of Derivative and the second part is Derivative in Real Life Context.
Introduction
The computation of derivative is often seen in maximum and minimum problems. In this article, we will discuss why do we get the derivative of a function and equate it to 0 when we want to get its maximum or minimum. To give you a concrete example, let us consider the problem below.
Find the maximum area a rectangle with perimeter 10 units.
Without using calculus, we can substitute values for the rectangle’s length, compute for its width and its corresponding area. If we set the interval to 0.5, then we can come up with the table shown in Figure 1.
Figure 1 - Table showing the length, width, and area of a rectangle with perimeter 10.
Looking at the table above, we can observe that a rectangle of length of 2.5, a square, has the maximum area. If we have prior calculus knowledge, however, we know that whatever the value of our perimeter, a square having the given perimeter will always have the maximum area.
Using elementary algebra, if we let $x$ be the width of our rectangle, it follows that the length is $5-x$. Let $f(x)$ be the area of the rectangle. In effect, the area of the rectangle is described by the equation $f(x) = 5x - x^2$. We want to maximize the area, which implies that we want to find the maximum value of $f(x)$.
Figure 2 – A rectangle with Perimeter 10 and width x units.
In elementary calculus, to compute for the maximum value of $f(x)$, we get its derivative, which is equal to $5 - 2x$, which we will denote $f'(x)$. We then equate the $f'(x)$ to $0$ resulting to the equation $5-2x=0 \Rightarrow x = 2.5$ which is exactly the maximum value in the table above.
Derivative and Equation to 0
In the article the Algebraic and Geometric Meaning of Derivative, we have learned that the derivative of a function is the slope of the line tangent to that function at a particular point. From elementary algebra, we also have learned the properties of slopes. If a line is rising to the right, the slope is greater than 0; if the line is rising to the to the left, then the slope is less than 0. We have also learned that a horizontal line has slope 0 and the vertical line has an undefined slope.
Figure 3 – Properties of slope of a straight line.
In the problem above, we calculated by getting the derivative (the slope of the line tangent to a function at a particular point) and equate it to $0$. But a line with slope $0$ is a horizontal line. In effect, we are looking for a horizontal tangent of $f(x) = 5x-x^2$. To give a clearer picture let us look at the graph of $f(x) = 5x - x^2$.
Figure 4 – Tangent lines of 5x – x2.
From the graph it is clear that the maximum point of the function is where the tangent line (red line) horizontal. In fact, there are only three possible cases that tangent line could be horizontal as shown in Figure 5: first, the minimum of a function (blue graph); second, the inflection point (red graph); and the third is the maximum of the function (green graph).
It should also be noteworthy to say that all the ordered pairs (length, area) or(width, area) in Figure 1 will be on the blue curve in Figure 4.
Figure 5 – Cases of a graph where the tangent is horizontal.
The derivative has many applications and it is seen in many topics in calculus. In the next Derivative Tutorial, we are going to discuss how the derivative is used in other context.
Summary
• The derivative of a function is the slope of the line tangent to a function at a particular point.
• The horizontal line has slope zero.
• In solving maxima and minima problems, we get the derivative of a function and equate to zero to get the minimum or maximum. We do this because geometrically, we want to get the line tangent to a function at a particular point that is horizontal.
## CaR Tutorial 1- Constructing an Isosceles Triangle
CaR or Compass and Ruler is a free dynamic geometry software written in Java by Rene Grothman. The CaR window is shown below.
Figure 1 - The Compass and Ruler window.
The upper part of the window contains the menu bar and toolbar. The toolbar contain tools in constructing and editing mathematical objects. The left window below the toolbar is the Objects window and the right pane is the Drawing pad where we construct drawings.
Tutorial 1 – Constructing an Isosceles Triangle
In the first tutorial, we are going to create an isosceles triangle by using the center of the circle and two points on its circumference.
1.) We will not need the Coordinate axes so click the Show grid icon until the Show the Grid icon until the grid or axes is not shown. 2.)Click the Circle tool, then click the drawing pad to determine the center of the circle, and click another location to determine its radius. 3.) Click the Point tool and click another location on the circumference circle. After step 3, your drawing should look like the figure below. Figure 2 - Circle with 2 points on its circumference. 4.) Click the Segment tool and two points to construct a side of the triangle. Continue until the triangle is formed. 5.) Click the Move button to drag the points and observe what happens to the triangle. Explain why the triangle is always isosceles. 6.) Click the Hide object button and click the circle to hide it. 7.) Next, we will change the name of the points and display their names. To do this, right click a point to display the Edit Point dialog box. In the Name text box, type A, then click the Show Object Names icon. Figure 1 - The Edit Point Dialog box. Use the same process to change the name of the other points.
Congratulations, you have finished the first Compass and Ruler tutorial.
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# Lesson 16
Multi-step Experiments
Let’s look at probabilities of experiments that have multiple steps.
### 16.1: True or False?
Is each equation true or false? Explain your reasoning.
$$8=(8+8+8+8)\div3$$
$$(10+10+10+10+10)\div5=10$$
$$(6+4+6+4+6+4)\div6=5$$
### 16.2: Spinning a Color and Number
The other day, you wrote the sample space for spinning each of these spinners once.
What is the probability of getting:
1. Green and 3?
2. Blue and any odd number?
3. Any color other than red and any number other than 2?
### 16.3: Cubes and Coins
The other day you looked at a list, a table, and a tree that showed the sample space for rolling a number cube and flipping a coin.
1. Your teacher will assign you one of these three structures to use to answer these questions. Be prepared to explain your reasoning.
1. What is the probability of getting tails and a 6?
2. What is the probability of getting heads and an odd number?
2. Suppose you roll two number cubes. What is the probability of getting:
1. Both cubes showing the same number?
2. Exactly one cube showing an even number?
3. At least one cube showing an even number?
4. Two values that have a sum of 8?
5. Two values that have a sum of 13?
3. Jada flips three quarters. What is the probability that all three will land showing the same side?
### 16.4: Pick a Card
Imagine there are 5 cards. They are colored red, yellow, green, white, and black. You mix up the cards and select one of them without looking. Then, without putting that card back, you mix up the remaining cards and select another one.
1. Write the sample space and tell how many possible outcomes there are.
2. What structure did you use to write all of the outcomes (list, table, tree, something else)? Explain why you chose that structure.
3. What is the probability that:
1. You get a white card and a red card (in either order)?
2. You get a black card (either time)?
3. You do not get a black card (either time)?
4. You get a blue card?
5. You get 2 cards of the same color?
6. You get 2 cards of different colors?
In a game using five cards numbered 1, 2, 3, 4, and 5, you take two cards and add the values together. If the sum is 8, you win. Would you rather pick a card and put it back before picking the second card, or keep the card in your hand while you pick the second card? Explain your reasoning.
### Summary
Suppose we have two bags. One contains 1 star block and 4 moon blocks. The other contains 3 star blocks and 1 moon block.
If we select one block at random from each, what is the probability that we will get two star blocks or two moon blocks?
To answer this question, we can draw a tree diagram to see all of the possible outcomes.
There are $$5 \boldcdot 4 = 20$$ possible outcomes. Of these, 3 of them are both stars, and 4 are both moons. So the probability of getting 2 star blocks or 2 moon blocks is $$\frac{7}{20}$$.
In general, if all outcomes in an experiment are equally likely, then the probability of an event is the fraction of outcomes in the sample space for which the event occurs.
### Glossary Entries
• probability
The probability of an event is a number that tells how likely it is to happen. A probability of 1 means the event will always happen. A probability of 0 means the event will never happen.
For example, the probability of selecting a moon block at random from this bag is $$\frac45$$.
• random
Outcomes of a chance experiment are random if they are all equally likely to happen.
• sample space
The sample space is the list of every possible outcome for a chance experiment.
For example, the sample space for tossing two coins is:
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Home » Maths » integration of 1/1+cosx , 1/1 -cos x
# integration of 1/1+cosx , 1/1 -cos x
We can easily find the integration of $\frac {1}{1+ cos x}$ and $\frac {1}{1 -cos x}$ using trigonometric identities and fundamental integration techniques, The formula of these integral is given as
$\int \frac {1}{1+ cos x} \; dx = \tan \frac {x}{2} + C$
or
$\int \frac {1}{1+ cos x} \; dx = -\cot x + \csx x + C$
$\int \frac {1}{1- cos x} \; dx == -\cot {x}{2} + C$
or
$\int \frac {1}{1- cos x} \; dx = -\cot x -\csc x + C$
## Proof of integration of 1/1+cosx
Method I
$\int \frac {1}{1+ cos x} \; dx = \int \frac {1}{2\cos^2 \frac {x}{2} }$
$=\frac {1}{2} \int \sec^2 {x}{2} \; dx$
Now let ${x}{2} =t$
Then $dx= 2dt$
Therefore
$=\frac {2 }{2} \int \sec^2 t \; dt$
$=\tan t + C$
Substituting back
$= \tan {x}{2} + C$
Method II
$\int \frac {1}{1+ cos x} \; dx = \int \frac {1}{1+ cos x} \times \frac {1- cos x}{1- cos x}$
$= \int \frac {1- cos x}{1- cos^2 x } \; dx = \int \frac {1-cos x}{\sin^2 x} \; dx$
$=\int ( \csc^2 x – \cot x \csc x) \; dx$
$= -\cot x + \csx x + C$
## Proof of integration of 1/1-cosx
Method I
$\int \frac {1}{1- cos x} \; dx = \int \frac {1}{2\sin^2 \frac {x}{2} }$
$=\frac {1}{2} \int \csc^2 {x}{2} \; dx$
Now let ${x}{2} =t$
Then $dx= 2dt$
Therefore
$=\frac {2 }{2} \int \csc^2 t \; dt$
$=-\cot t + C$
Substituting back
$= -\cot {x}{2} + C$
Method II
$\int \frac {1}{1- cos x} \; dx = \int \frac {1}{1- cos x} \times \frac {1+ cos x}{1+ cos x}$
$= \int \frac {1+ cos x}{1- cos^2 x } \; dx = \int \frac {1+cos x}{\sin^2 x} \; dx$
$=\int ( \csc^2 x + \cot x \csc x) \; dx$
$= -\cot x – \csx x + C$
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# Point of Intersection Formula
Point of intersection means the point at which two lines intersect. These two lines are represented by the equation a1x+ b1x + c1= 0 and a2x+ b2x + c2 = 0 respectively. Given figure illustrate the point of intersection of two lines.
We can find the point of intersection of three or more lines also. By solving the two equations, we can find the solution for point of intersection of two lines.
### Solved Examples
Question 1: Find out the point of intersection of two lines $x^{2}$ + 2x + 1 = 0 and 2$x^{2}$ + 3x + 5 = 0 ?
Solution:
Given straight line equations are: $x^{2}$ + 2x + 1 = 0 and 2$x^{2}$ + 3x + 5 = 0
Here,
a1 = 1
b1 = 2
c1 = 1
a2 = 2, b2 = 3, c2 = 5
Intersection point can be calculated using this formula,
x = $\frac{b_{1}c_{2}-b_{2}c_{1}}{a_{1}b_{2}-a_{2}b_{1}}$; y = $\frac{a_{2}c_{1}-a_{1}c_{2}}{a_{1}b_{2}-a_{2}b_{1}}$
(x,y) = ($\frac{2\times5-3\times1}{1\times3-2\times2}$,$\frac{2\times1-1\times5}{1\times3-2\times2}$)
(x,y) = ($\frac{10-3}{3-4}$,$\frac{2-5}{3-4}$)
(x,y) = (-7,3)
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# What Does “Squared” Mean in Math? Explain It to a 10-Year-Old
Jul 10, 2024 | Ankeny
This beginner-friendly overview of square numbers in math is for everyone, from seven- to seventy-year-olds, looking to learn new mathematical concepts or refresh their memory. Read on to find simple definitions, and fun applications and exercises.
## What Is a Square Number?
A square number is the result of multiplying a number by itself.
For instance, if you take the number 3 and multiply it by 3, you get 9.
9 is a square number because it’s the product of multiplying 3 by itself.
In math, we write this as 3² (we read it as “three to the second power” or “three squared”). We call the “3” in 3² the base and we call “²” the exponent.
Let’s look at some examples to understand square numbers better:
• 2² which can also be expressed as 2 x 2 = 4
• 4² which can also be expressed as 4 x 4 = 16
• 5² which can also be expressed as 5 x 5 = 25
Refresh Your Memory: What is a Factor in Math?
### What’s the Difference Between a Square Number and a Square Root?
In addition to “square numbers,” you might have also heard about “square roots.”
Square numbers and square roots are like two sides of the same coin — they represent opposite actions.
How?
Since square roots are a whole new topic, let’s look at a brief example:
As we said, a square number is a result of multiplying a number by itself. For example, 3 multiplied by 3 equals 9 (3² = 9), so the number 9 is a square number.
A square root is the opposite of a square number. A square root is the number you multiply by itself to get another number. We show it with this symbol: √ (radical symbol or surd). For example, the square root of 9 is 3 (√9 = 3), and the square root of 25 is 5 (√25 = 5).
## Why Are They Called Square Numbers?
If you thought square numbers must have something to do with squares, you would have been correct!
Square numbers are called “squared” because they make the shape of a square.
Squares have sides of equal length. To find the area of a square, you just need to multiply one side by itself, or “square” it.
## How Do We Use Square Numbers in Math?
Squares are like puzzle pieces in math. They fit into various math classes and fields.
Pre-Algebra and Algebra: In algebra, we use square numbers in equations. Let’s say that we have the expression . In algebra, this means “x squared” or “x multiplied by itself.” For example, if , then would be or which equals .
Geometry: In geometry we use square numbers to measure the area inside of a square based on its side lengths. Let’s look at a square with sides of length that are 4 centimeters long for this example. To find the area of this square, we square the length of one of its sides. In this case, the length of one side is 4 centimeters. So, the area of the square would be 16 square centimeters.
Number Theory: In number theory, we use square numbers to study patterns and relationships between numbers. For example, the sequence 1, 4, 9, 16, 25, 36 is a list of square numbers. How do we know this? Each number in the sequence is made by multiplying a natural number by itself.
Here’s proof:
• 1 = 1 × 1
• 4 = 2 × 2
• 9 = 3 × 3
• 16 = 4 × 4
• 25 = 5 × 5
• 36 = 6 × 6
Check out this video demonstrating a cool trick for squaring numbers ending in 0 or 5.
## Examples of Square Numbers
Let’s see how numbers from 1 to 12 are squared.
• 1² = 1 which can also be expressed 1 × 1 = 1
• 2² = 4 which can also be expressed 2 × 2 = 4
• 3² = 9 which can also be expressed 3 × 3 = 9
• 4² = 16 which can also be expressed 4 × 4 = 16
• 5² = 25 which can also be expressed 5 × 5 = 25
• 6² = 36 which can also be expressed 6 × 6 = 36
• 7² = 49 which can also be expressed 7 × 7 = 49
• 8² = 64 which can also be expressed 8 × 8 = 64
• 9² = 81 which can also be expressed 9 x 9 = 81
• 10² = 100 which can also be expressed 10 x 10 = 100
• 11² = 121 which can also be expressed 11 × 11 = 121
• 12² = 144 which can also be expressed 12 × 12 = 144
Check out this short guide to squaring any number (no matter how big!) using mental math and Number Sense.
## How Can We Use Square Numbers Outside of Classroom?
We can easily use (and practice!) square numbers outside the classroom, especially when we want to calculate the surfaces.
Here are a couple of examples of how you can put square numbers to practice, plus how we use them to plan and build the spaces you are using every day:
### Practice with Toy Bricks
Take your toy bricks with studs and a studded baseplate to place them on. The rule of the game is: count the number of studs on a piece and put that many bricks one next to another on the baseplate.
• Start with the smallest brick which has 1 stud. Since the brick only has 1 stud, leave it alone on your baseplate.
• Next, take the brick with 2 studs. Since we have 2 studs, we’ll put 2 bricks on the baseplate, one next to another.
• Now let’s take a brick with 3 studs. Since we have 3 studs, we’ll put 3 bricks next to one another on the baseplate.
And so on!
As you start to count the number of studs in each group of bricks, you will get square numbers of studs on single bricks.
• 1(1 brick only) equals 1 stud
• 2(or 2 studs x 2 studs) equals 4 studs
• 3(or 3 studs x 3 studs) equals 9 studs
### Refresh Your Memory of Square Numbers While Building Virtual Houses
Kids over the age of 12 might like this one!
If you are familiar with virtual house-building games, you’ll know that these games often provide you with “plots” overlaid with square patterns
Next time you outline your square living room or bedroom, count the number of squares and you’ll get a square number. Simple as that!
### Learn How Engineers and Architects Use Square Numbers toPlan & Build Spaces
From planning playgrounds to rooms and buildings, you can see square numbers in action in any square-shaped space.
Take the measuring tape and measure one side of your room. Multiply that number by itself and you’ll get the size of your floor if the room is perfectly square. For example, if each side of your room is 10ft wide, the surface of your square floor would be 10x10ft = 100ft2 (read as “square feet”).
Is 100 a square number?
Yes! Yes, it is.
Ask your parents, engineers, or architects in your family to see how they use square numbers to plan and build the spaces you use every day.
## Practice Square Numbers in Math
Your time to shine! Let’s review what we’ve learned with these simple exercises.
Exercise 1: Square Number Multiplication
Let’s calculate these:
72=___________
92=___________
52=___________
112=___________
Exercise 2: Missing Square Numbers
Fill in the missing square numbers in the sequence:
1, __, 9, __, 25, __, 49, 64, __, 100
Exercise 3: Square Number Word Problems
Try to solve these 3 word problems:
• If a square garden has an area of 64 square feet, what is the length of each side? (Hint: What number can we multiply by itself to get 64?)
• Sarah wants to build a square picture frame with an area of 100 square inches. What should be the length of each side? (Hint: What number can we multiply by itself to get 100?)
• A square rug has an area of 49 square meters. What is the length of one side of the rug? (Hint: What number can we multiply by itself to get 49?)
Completed the exercises?
A Neat Trick: Learn How to Square Any Number
Find answers to common queries regarding the properties and applications of square numbers.
### 1.When do students learn about square numbers at school?
Students usually encounter the basic square numbers when learning addition in early elementary school. They then formally begin to learn about square numbers later in elementary school, typically around grades 4 to 6.
Mathnasium works with elementary school students of all ages and skill levels to help them master math, including square numbers.
### 2. Is zero a square number?
Yes, zero is a square number because 0 x 0 = 0.
### 3. Is every positive number a square number?
Not all positive numbers can be square numbers. For example, 7 is a positive number, but it’s not a square number because you can’t make 7 by multiplying a number by itself.
### 4. Can you square a negative number?
Yes, you can square a negative number. Squaring a negative number also means multiplying the number by itself. When you square a negative number, the result is always positive because multiplying a negative number by a negative number always makes a positive number.
Let’s look at these examples:
-2 squared is (-2) x (-2) = 4
-3 squared is (-3) x (-3) = 9
-4 squared is (-4) x (-4) = 16
## Learn & Master Square Numbers with Math Tutors Near You
Mathnasium’s specially trained tutors work with students of all skill levels to help them learn and master any K-12 math topic, including squared numbers.
Our tutors assess each student’s skills to create personalized learning plans that will put them on the best path to math mastery.
Find a Mathnasium Learning Center near you, schedule an assessment, and enroll today!
Find a Math Tutor Near You
## Answers to Square Number Practice Exercises
Exercise 1: Square Number Multiplication
5= 25
7= 49
9= 81
112 = 121
Exercise 2: Missing Square Numbers
1, _4_, 9, _16_, 25, _36_, 49, 64, _81_, 100
Exercise 3: Square Number Word Problems
• If a square garden has an area of 64 square feet, what is the length of each side? (Hint: What number can we multiply by itself to get 64?) The length of each side is 8 ft, because 64 = 82.
• Sarah wants to build a square picture frame with an area of 100 square inches. What should be the length of each side? (Hint: What number can we multiply by itself to get 100?) The length of each side should be 10 inches, because 100 = 102.
• A square rug has an area of 49 square meters. What is the length of one side of the rug? (Hint: What number can we multiply by itself to get 49?) The length of one side of the rug is 7 meters, because 49=72.
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# Vectors
## The basic ideaEdit
A vector is a mathematical concept that has both magnitude and direction. Detailed explanation of vectors may be found at the Wikibooks module Linear Algebra/Vectors in Space. In physics, vectors are used to describe things happening in space by giving a series of quantities which relate to the problem's coordinate system.
A vector is often expressed as a series of numbers. For example, in the two-dimensional space of real numbers, the notation (1, 1) represents a vector that is pointed 45 degrees from the x-axis towards the y-axis with a magnitude of $\sqrt 2$.
Commonly in physics, we use position vectors to describe where something is in the space we are considering, or how its position is changing at that moment in time. Position vectors are written as summations of scalars multiplied by unit vectors. For example:
$a \hat{i} + b \hat{j} + c \hat{k}$
where a, b and c are scalars and $\hat{i}, \hat{j}$ and $\hat{k}$ are unit vectors of the Cartesian (René Descartes) coordinate system. A unit vector is a special vector which has magnitude 1 and points along one of the coordinate frame's axes. This is better illustrated by a diagram.
A vector itself is typically indicated by either an arrow: $\vec{v}$, or just by boldface type: v, so the vector above as a complete equation would be denoted as:
$\vec{v} = a \hat{i} + b \hat{j} + c \hat{k}$
The magnitude of a vector is computed by $|\vec{v}| = \sqrt{\sum_i(x_i^2)}$. For example, in two-dimensional space, this equation reduces to:
$|\vec{v}| = \sqrt{x^2+y^2}$.
For three-dimensional space, this equation becomes:
$|\vec{v}| = \sqrt{x^2+y^2+z^2}$.
# ExercisesEdit
Find the magnitude of the following vectors. Answers below.
$\vec{v} = (4, 3)$
$|\vec{v}| = \sqrt{4^2 + 3^2} = 5$
$\vec{v} = (5, 3)$
$|\vec{v}| = \sqrt{5^2 + 3^2} = \sqrt{34}$
$\vec{v} = (1, 0)$
$|\vec{v}| = \sqrt{1^2 + 0^2} = 1$
$\vec{v} = (4, 4)$
$|\vec{v}| = \sqrt{4^2 + 4^2 } = \sqrt{32}$
$\vec{v} = (5, 0, 0)$
$|\vec{v}| = \sqrt{5^2 + 0^2 + 0^2} = 5$
## Using vectors in physicsEdit
Many problems, particularly in mechanics, involve the use of two- or three-dimensional space to describe where objects are and what they are doing. Vectors can be used to condense this information into a precise and easily understandable form that is easy to manipulate with mathematics.
Position - or where something is, can be shown using a position vector. Position vectors measure how far something is from the origin of the reference frame and in what direction, and are usually, though not always, given the symbol $\vec{r}$. It is usually good practice to use $\vec{r}$ for position vectors when describing your solution to a problem as most physicists use this notation.
Velocity is defined as the rate of change of position with respect to time. You may be used to writing velocity, v, as a scalar because it was assumed in your solution that v referred to speed in the direction of travel. However, if we take the strict definition and apply it to the position vector - which we have already established is the proper way of representing position - we get:
$\frac{d\vec{r}}{dt} = \frac{da}{dt}\hat{i} + \frac{db}{dt}\hat{j} + \frac{dc}{dt}\hat{k}$
However, we note that the unit vectors are merely notation rather than terms themselves and are in fact not differentated, only the scalars which represent the vector's components in each direction differentiate.
Assuming that each component is not a constant and thus has a non-zero derivative, we get:
$\vec{v} = a' \hat{i} + b' \hat{j} + c' \hat{k}$ where a', b' and c' are simply the first derivatives with respect to time of each original position vector component.
Here it is clear that velocity is also a vector. In the real world this means that each component of the velocity vector indicates how quickly each component of the position vector is changing - that is, how fast the object is moving in each direction.
# Vectors in MechanicsEdit
Vector notation is ubiquitous in the modern literature on solid mechanics, fluid mechanics, biomechanics, nonlinear finite elements and a host of other subjects in mechanics. A student has to be familiar with the notation in order to be able to read the literature. In this section we introduce the notation that is used, common operations in vector algebra, and some ideas from vector calculus.
## VectorsEdit
A vector is an object that has certain properties. What are these properties? We usually say that these properties are:
• a vector has a magnitude (or length)
• a vector has a direction.
To make the definition of the vector object more precise we may also say that vectors are objects that satisfy the properties of a vector space.
The standard notation for a vector is lower case bold type (for example $\mathbf{a}\,$).
In Figure 1(a) you can see a vector $\mathbf{a}$ in red. This vector can be represented in component form with respect to the basis ($\mathbf{e}_1, \mathbf{e}_2\,$) as
$\mathbf{a} = a_1 \mathbf{e}_1 + a_2 \mathbf{e}_2 \,$
where $\mathbf{e}_1\,$ and $\mathbf{e}_2\,$ are orthonormal unit vectors. Orthonormal means they are at right angles to each other (orthogonal) and are unit vectors. Recall that unit vectors are vectors of length 1. These vectors are also called basis vectors.
You could also represent the same vector $\mathbf{a}\,$ in terms of another set of basis vectors ($\mathbf{g}_1,\mathbf{g}_2\,$) as shown in Figure 1(b). In that case, the components of the vector are $(b_1,b_2)\,$ and we can write
$\mathbf{a} = b_1 \mathbf{g}_1 + b_2 \mathbf{g}_2 \, ~.$
Note that the basis vectors $\mathbf{g}_1\,$ and $\mathbf{g}_2\,$ do not necessarily have to be unit vectors. All we need is that they be linearly independent, that is, it should not be possible for us to represent one solely in terms of the others.
In three dimensions, using an orthonormal basis, we can write the vector $\mathbf{a}\,$ as
$\mathbf{a} = a_1 \mathbf{e}_1 + a_2 \mathbf{e}_2 + a_2 \mathbf{e}_3 \,$
where $\mathbf{e}_3\,$ is perpendicular to both $\mathbf{e}_1\,$ and $\mathbf{e}_2\,$. This is the usual basis in which we express arbitrary vectors.
## Vector Algebra OperationsEdit
Some vector operations are shown in Figure 2.
If $\mathbf{a}\,$ and $\mathbf{b}\,$ are vectors, then the sum $\mathbf{c} = \mathbf{a} + \mathbf{b}\,$ is also a vector (see Figure 2(a)).
The two vectors can also be subtracted from one another to give another vector $\mathbf{d} = \mathbf{a} - \mathbf{b}\,$.
### Multiplication by a scalarEdit
Multiplication of a vector $\mathbf{b}\,$ by a scalar $\lambda\,$ has the effect of stretching or shrinking the vector (see Figure 2(b)).
You can form a unit vector $\hat\mathbf{b}\,$ that is parallel to $\mathbf{b}\,$ by dividing by the length of the vector $|\mathbf{b}|\,$. Thus,
$\hat\mathbf{b} = \frac{\mathbf{b}}{|\mathbf{b}|} ~.$
### Scalar product of two vectorsEdit
The scalar product or inner product or dot product of two vectors is defined as
$\mathbf{a}\cdot\mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos(\theta)$
where $\theta\,$ is the angle between the two vectors (see Figure 2(b)).
If $\mathbf{a}\,$ and $\mathbf{b}\,$ are perpendicular to each other, $\theta = \pi/2\,$ and $\cos(\theta) = 0\,$. Therefore, ${\mathbf{a}}\cdot{\mathbf{b}} = 0$.
The dot product therefore has the geometric interpretation as the length of the projection of $\mathbf{a}\,$ onto the unit vector $\hat\mathbf{b}\,$ when the two vectors are placed so that they start from the same point.
The scalar product leads to a scalar quantity and can also be written in component form (with respect to a given basis) as
${\mathbf{a}}\cdot{\mathbf{b}} = a_1 b_1 + a_2 b_2 + a_3 b_3 = \sum_{i=1..3} a_i b_i~.$
If the vector is $n$ dimensional, the dot product is written as
${\mathbf{a}}\cdot{\mathbf{b}} = \sum_{i=1..n} a_i b_i~.$
Using the Einstein summation convention, we can also write the scalar product as
${\mathbf{a}}\cdot{\mathbf{b}} = a_i b_i~.$
Also notice that the following also hold for the scalar product
1. ${\mathbf{a}}\cdot{\mathbf{b}} = {\mathbf{b}}\cdot{\mathbf{a}}$ (commutative law).
2. ${\mathbf{a}}\cdot{(\mathbf{b}+\mathbf{c})} = {\mathbf{a}}\cdot{\mathbf{b}} + {\mathbf{a}}\cdot{\mathbf{c}}$ (distributive law).
### Vector product of two vectorsEdit
The vector product (or cross product) of two vectors $\mathbf{a}\,$ and $\mathbf{b}\,$ is another vector $\mathbf{c}\,$ defined as
$\mathbf{c} = {\mathbf{a}}\times{\mathbf{b}} = |\mathbf{a}||\mathbf{b}|\sin(\theta) \hat{\mathbf{c}}$
where $\theta\,$ is the angle between $\mathbf{a}\,$ and $\mathbf{b}\,$, and $\hat{\mathbf{c}}\,$ is a unit vector perpendicular to the plane containing $\mathbf{a}\,$ and $\mathbf{b}\,$ in the right-handed sense (see Figure 3 for a geometric interpretation)
In terms of the orthonormal basis $(\mathbf{e}_1,\mathbf{e}_2,\mathbf{e}_3)\,$, the cross product can be written in the form of a determinant
${\mathbf{a}}\times{\mathbf{b}} = \begin{vmatrix} \mathbf{e}_1 & \mathbf{e}_2 & \mathbf{e}_3 \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix}~.$
In index notation, the cross product can be written as
${\mathbf{a}}\times{\mathbf{b}} \equiv e_{ijk} a_j b_k ~.$
where $e_{ijk}$ is the Levi-Civita symbol (also called the permutation symbol, alternating tensor).
## Identities from Vector AlgebraEdit
Some useful vector identities are given below.
1. ${\mathbf{a}}\times{\mathbf{b}} = - {\mathbf{b}}\times{\mathbf{a}}$.
2. ${\mathbf{a}}\times{\mathbf{b}+\mathbf{c}} = {\mathbf{a}}\times{\mathbf{b}} + {\mathbf{a}}\times{\mathbf{c}}$.
3. ${\mathbf{a}}\times{({\mathbf{b}}\times{\mathbf{c}})} = \mathbf{b}({\mathbf{a}}\cdot{\mathbf{c}}) - \mathbf{c}({\mathbf{a}}\cdot{\mathbf{b}})$ ~.
4. ${({\mathbf{a}}\times{\mathbf{b}})}\times{\mathbf{c}} = \mathbf{b}({\mathbf{a}}\cdot{\mathbf{c}}) - \mathbf{a}({\mathbf{b}}\cdot{\mathbf{c}})$ ~.
5. ${\mathbf{a}}\times{\mathbf{a}} = \mathbf{0}$~.
6. ${\mathbf{a}}\cdot{({\mathbf{a}}\times{\mathbf{b}})} = {\mathbf{b}}\cdot{({\mathbf{a}}\times{\mathbf{b}})} = \mathbf{0}$~.
7. ${({\mathbf{a}}\times{\mathbf{b}})}\cdot{\mathbf{c}} = {\mathbf{a}}\cdot{({\mathbf{b}}\times{\mathbf{c}})}$~.
## Vector CalculusEdit
So far we have dealt with constant vectors. It also helps if the vectors are allowed to vary in space. Then we can define derivatives and integrals and deal with vector fields. Some basic ideas of vector calculus are discussed below.
## Derivative of a vector valued functionEdit
Let $\mathbf{a}(x)\,$ be a vector function that can be represented as
$\mathbf{a}(x) = a_1(x) \mathbf{e}_1 + a_2(x) \mathbf{e}_2 + a_3(x) \mathbf{e}_3\,$
where $x\,$ is a scalar.
Then the derivative of $\mathbf{a}(x)\,$ with respect to $x\,$ is
$\cfrac{d\mathbf{a}(x)}{dx} = \lim_{\Delta x\rightarrow 0} \cfrac{\mathbf{a}(x+\Delta x) - \mathbf{a}(x)}{\Delta x} = \cfrac{da_1(x)}{dx} \mathbf{e}_1 + \cfrac{da_2(x)}{dx} \mathbf{e}_2 + \cfrac{da_3(x)}{dx} \mathbf{e}_3~.$
Note: In the above equation, the unit vectors $\mathbf{e}_i$ (i=1,2,3) are assumed constant.
If $\mathbf{a}(x)\,$ and $\mathbf{b}(x)\,$ are two vector functions, then from the chain rule we get
\begin{align} \cfrac{d({\mathbf{a}}\cdot{\mathbf{b}})}{x} & = {\mathbf{a}}\cdot{\cfrac{d\mathbf{b}}{dx}} + {\cfrac{d\mathbf{a}}{dx}}\cdot{\mathbf{b}} \\ \cfrac{d({\mathbf{a}}\times{\mathbf{b}})}{dx} & = {\mathbf{a}}\times{\cfrac{d\mathbf{b}}{dx}} + {\cfrac{d\mathbf{a}}{dx}}\times{\mathbf{b}} \\ \cfrac{d[{\mathbf{a}}\cdot{({\mathbf{a}}\times{\mathbf{b}})}]}{dt} & = {\cfrac{d\mathbf{a}}{dt}}\cdot{({\mathbf{b}}\times{\mathbf{c}})} + {\mathbf{a}}\cdot{\left({\cfrac{d\mathbf{b}}{dt}}\times{\mathbf{c}}\right)} + {\mathbf{a}}\cdot{\left({\mathbf{b}}\times{\cfrac{d\mathbf{c}}{dt}}\right)} \end{align}
## Scalar and vector fieldsEdit
Let $\mathbf{x}\,$ be the position vector of any point in space. Suppose that there is a scalar function ($g\,$) that assigns a value to each point in space. Then
$g = g(\mathbf{x})\,$
represents a scalar field. An example of a scalar field is the temperature. See Figure4(a).
If there is a vector function ($\mathbf{a}\,$) that assigns a vector to each point in space, then
$\mathbf{a} = \mathbf{a}(\mathbf{x})\,$
represents a vector field. An example is the displacement field. See Figure 4(b).
## Gradient of a scalar fieldEdit
Let $\varphi(\mathbf{x})\,$ be a scalar function. Assume that the partial derivatives of the function are continuous in some region of space. If the point $\mathbf{x}\,$ has coordinates ($x_1, x_2, x_3\,$) with respect to the basis ($\mathbf{e}_1, \mathbf{e}_2, \mathbf{e}_3\,$), the gradient of $\varphi\,$ is defined as
$\boldsymbol{\nabla}{\varphi} = \frac{\partial \varphi}{\partial x_1}~\mathbf{e}_1 + \frac{\partial \varphi}{\partial x_2}~\mathbf{e}_2 + \frac{\partial \varphi}{\partial x_3}~\mathbf{e}_3 ~.$
In index notation,
$\boldsymbol{\nabla}{\varphi} \equiv \varphi_{,i}~\mathbf{e}_i ~.$
The gradient is obviously a vector and has a direction. We can think of the gradient at a point being the vector perpendicular to the level contour at that point.
It is often useful to think of the symbol $\boldsymbol{\nabla}{}$ as an operator of the form
$\boldsymbol{\nabla}{} = \frac{\partial }{\partial x_1} ~\mathbf{e}_1 + \frac{\partial }{\partial x_2} ~\mathbf{e}_2 + \frac{\partial }{\partial x_3} ~\mathbf{e}_3 ~.$
## Divergence of a vector fieldEdit
If we form a scalar product of a vector field $\mathbf{u}(\mathbf{x})\,$ with the $\boldsymbol{\nabla}{}$ operator, we get a scalar quantity called the divergence of the vector field. Thus,
$\boldsymbol{\nabla}\cdot{\mathbf{u}} = \frac{\partial u_1}{\partial x_1} + \frac{\partial u_2}{\partial x_2} + \frac{\partial u_3}{\partial x_3} ~.$
In index notation,
$\boldsymbol{\nabla}\cdot{\mathbf{u}} \equiv u_{i,i} ~.$
If $\boldsymbol{\nabla}\cdot{\mathbf{u}} = 0$, then $\mathbf{u}\,$ is called a divergence-free field.
The physical significance of the divergence of a vector field is the rate at which some density exits a given region of space. In the absence of the creation or destruction of matter, the density within a region of space can change only by having it flow into or out of the region.
## Curl of a vector fieldEdit
The curl of a vector field $\mathbf{u}(\mathbf{x})\,$ is a vector defined as
$\boldsymbol{\nabla}\times{\mathbf{u}} = \det \begin{vmatrix} \mathbf{e}_1 & \mathbf{e}_2 & \mathbf{e}_3 \\ \frac{\partial }{\partial x_1} & \frac{\partial }{\partial x_2} & \frac{\partial }{\partial x_3} \\ u_1 & u_2 & u_3 \\ \end{vmatrix}$
The physical significance of the curl of a vector field is the amount of rotation or angular momentum of the contents of a region of space.
## Laplacian of a scalar or vector fieldEdit
The Laplacian of a scalar field $\varphi(\mathbf{x})\,$ is a scalar defined as
$\nabla^2{\varphi} := \boldsymbol{\nabla}\cdot{\boldsymbol{\nabla}{\varphi}} = \frac{\partial^2 \varphi}{\partial x_1} + \frac{\partial^2 \varphi}{\partial x_2} + \frac{\partial^2 \varphi}{\partial x_3}~.$
The Laplacian of a vector field $\mathbf{u}(\mathbf{x})\,$ is a vector defined as
$\nabla^2{\mathbf{u}} := (\nabla^2{u_1}) \mathbf{e}_1 + (\nabla^2{u_2}) \mathbf{e}_2 + (\nabla^2{u_3}) \mathbf{e}_3 ~.$
## Identities in vector calculusEdit
Some frequently used identities from vector calculus are listed below.
1. $\boldsymbol{\nabla}\cdot{(\mathbf{a} + \mathbf{b})} = \boldsymbol{\nabla}\cdot{\mathbf{a}} + \boldsymbol{\nabla}\cdot{\mathbf{b}}$~.
2. $\boldsymbol{\nabla}\times{(\mathbf{a} + \mathbf{b})} = \boldsymbol{\nabla}\times{\mathbf{a}} + \boldsymbol{\nabla}\times{\mathbf{b}}$~.
3. $\boldsymbol{\nabla}\cdot{(\varphi\mathbf{a})} = {(\boldsymbol{\nabla}{\varphi})}\cdot{\mathbf{a}} + \varphi (\boldsymbol{\nabla}\cdot{\mathbf{a}})$~.
4. $\boldsymbol{\nabla}\times{(\varphi\mathbf{a})} = {(\boldsymbol{\nabla}{\varphi})}\times{\mathbf{a}} + \varphi (\boldsymbol{\nabla}\times{\mathbf{a}})$~.
5. $\boldsymbol{\nabla}\cdot{({\mathbf{a}}\times{\mathbf{b}})} = {\mathbf{b}}\cdot{(\boldsymbol{\nabla}\times{\mathbf{a}})} - {\mathbf{a}}\cdot{(\boldsymbol{\nabla}\times{\mathbf{b}})}$~.
## Green-Gauss Divergence TheoremEdit
Let $\mathbf{u}(\mathbf{x})\,$ be a continuous and differentiable vector field on a body $\Omega\,$ with boundary $\Gamma\,$. The divergence theorem states that
${ \int_{\Omega} \boldsymbol{\nabla}\cdot{\mathbf{u}}~dV = \int_{\Gamma} {\mathbf{n}}\cdot{\mathbf{u}}~dA }$
where $\mathbf{n}\,$ is the outward unit normal to the surface (see Figure 5).
In index notation,
$\int_{\Omega} u_{i,i}~dV = \int_{\Gamma} n_i u_i~dA$
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# Decimals to Fractions
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We know that decimals and fractions both are ways of showing parts of a whole. With decimals, the parts are always broken into tenths, hundredths, thousandths, and so on. The digits to the left of the decimal point represent a whole number. The digits to the right of the number represent the parts of a whole.
When we use decimals, we show all the digits horizontally or in a line.
Suppose there is one whole pan of brownies. It would be the same as the digit 1 to the left of the decimal point. Since there are no parts of a whole brownie, there would be a zero to the right of the decimal point. One whole is the same as 1.0
A second pan of brownies that had been cut into equal parts, but some pieces had been eaten, is like the digits to the right of the decimal point.
If 4 pieces out of 10 had been eaten (and since 10 - 4 = 6), then we would say 0.6 or 6 tenths of the brownie remains. We just count the tenths!
Now, if we want to convert this to a fraction, we just show the parts of the whole.
So, 0.6 =
6
10
6 out of 10 pieces of brownie
Everything is not always cut into 10 equal parts like the brownie. What if something was cut into one hundred parts?
Well, then we would move two digits to the right of the decimal point.
So, if we had a really large pan of brownies cut into 100 parts and 18 pieces were eaten (100 - 18 = 82), then 0.82 or eighty-two hundredths are the parts of the whole brownie that remain.
We can write 0.82 in fraction form as
82
100
.
Here's a strategy to help in converting decimals to fractions.
• If the decimal is in the tenths, just write the digits in decimal as the numerator over a denominator of 10.
• If the decimal is in the hundredths, write the digits in decimal as the numerator over a denominator of 100.
• If the decimal is in the thousandths, put the digits in decimal as the numerator over a denominator of 1000.
Remember to reduce the obtained fraction into its lowest form.
Let's take a look at some examples.
## Example 1
Write 0.7 as a fraction.
0.7 =
7
10
## Example 2
Write 0.75 as a fraction.
0.75 =
75
100
Now, reduce
75
100
into its lowest form.
75
100
=
75 ÷ 25
100 ÷ 25
=
3
4
So, 0.75 =
3
4
## Example 3
Write 0.126 as a fraction.
0.126 =
126
1000
Now, reduce
126
1000
into its lowest form.
126
1000
=
126 ÷ 2
1000 ÷ 2
=
63
500
So, 0.126 =
63
500
## Decimals to Fractions
• Decimals and fractions both are ways of showing parts of a whole.
• To convert a decimal to a fraction, follow these steps:
• Write the digits in the decimal as the numerator.
• Write as many zeroes after 1 in the denominator, as there were decimal places (digits to the right of the decimal point) in the decimal number.
• Reduce the fraction to its lowest form.
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# How Do I Graph Equations of the form Ax + By = C?
## Presentation on theme: "How Do I Graph Equations of the form Ax + By = C?"— Presentation transcript:
How Do I Graph Equations of the form Ax + By = C?
A4.d How Do I Graph Equations of the form Ax + By = C? Course 3 Warm Up Problem of the Day Lesson Presentation
Find the slope of the line that passes through each pair of points.
Course 3 A4.d How Do I Graph Equations of the form Ax + By = C? Warm Up Find the slope of the line that passes through each pair of points. 1. (3, 6) and (–1, 4) 2. (1, 2) and (6, 1) 3. (4, 6) and (2, –1) 4. (–3, 0) and (–1, 1) 1 2 1 5 7 2 1 2
Using Slopes and Intercepts
Course 3 12-3 Using Slopes and Intercepts Problem of the Day Write the equation of a straight line that passes through fewer than two quadrants on a coordinate plane. x = 0 or y = 0
Using Slopes and Intercepts
Course 3 12-3 Using Slopes and Intercepts Learn to use slopes and intercepts to graph linear equations.
Insert Lesson Title Here
Course 3 12-3 Using Slopes and Intercepts Insert Lesson Title Here Vocabulary x-intercept y-intercept slope-intercept form
Using Slopes and Intercepts
Course 3 12-3 Using Slopes and Intercepts You can graph a linear equation easily by finding the x-intercept and the y-intercept. The x-intercept of a line is the value of x where the line crosses the x-axis (where y = 0). The y-intercept of a line is the value of y where the line crosses the y-axis (where x = 0).
Using Slopes and Intercepts
Course 3 12-3 Using Slopes and Intercepts Example 1: Finding x-intercepts and y-intercepts to Graph Linear Equations Find the x-intercept and y-intercept of the line 4x – 3y = 12. Use the intercepts to graph the equation. Find the x-intercept (y = 0). 4x – 3y = 12 4x – 3(0) = 12 4x = 12 4x 4 12 = x = 3 The x-intercept is 3.
Using Slopes and Intercepts
Course 3 12-3 Using Slopes and Intercepts Example 1 Continued Find the y-intercept (x = 0). 4x – 3y = 12 4(0) – 3y = 12 –3y = 12 –3y –3 12 = y = –4 The y-intercept is –4.
Course 3 12-3 Using Slopes and Intercepts Additional Example 1 Continued The graph of 4x – 3y = 12 is the line that crosses the x-axis at the point (3, 0) and the y-axis at the point (0, –4).
Using Slopes and Intercepts
Course 3 12-3 Using Slopes and Intercepts The form Ax + By = C, where A, B, C are real numbers, is called the Standard Form of a Linear Equation. Helpful Hint
Using Slopes and Intercepts
Course 3 12-3 Using Slopes and Intercepts Check It Out: Example 1 Find the x-intercept and y-intercept of the line 8x – 6y = 24. Use the intercepts to graph the equation. Find the x-intercept (y = 0). 8x – 6y = 24 8x – 6(0) = 24 8x = 24 8x 8 24 = x = 3 The x-intercept is 3.
Check It Out: Example 1 Continued
Course 3 12-3 Using Slopes and Intercepts Check It Out: Example 1 Continued Find the y-intercept (x = 0). 8x – 6y = 24 8(0) – 6y = 24 –6y = 24 –6y –6 24 = y = –4 The y-intercept is –4.
Check It Out: Example 1 Continued
Course 3 12-3 Using Slopes and Intercepts Check It Out: Example 1 Continued The graph of 8x – 6y = 24 is the line that crosses the x-axis at the point (3, 0) and the y-axis at the point (0, –4).
y = mx + b Using Slopes and Intercepts 12-3
Course 3 12-3 Using Slopes and Intercepts In an equation written in slope-intercept form, y = mx + b, m is the slope and b is the y-intercept. y = mx + b Slope y-intercept
Example 2A: Using Slope-Intercept Form to Find Slopes and y-intercepts
Course 3 12-3 Using Slopes and Intercepts Example 2A: Using Slope-Intercept Form to Find Slopes and y-intercepts Write each equation in slope-intercept form, and then find the slope and y-intercept. 2x + y = 3 2x + y = 3 –2x –2x Subtract 2x from both sides. y = 3 – 2x Rewrite to match slope-intercept form. y = –2x + 3 The equation is in slope-intercept form. m = –2 b = 3 The slope of the line 2x + y = 3 is –2, and the y-intercept is 3.
Using Slopes and Intercepts
Course 3 12-3 Using Slopes and Intercepts Check It Out: Example 2A Write each equation in slope-intercept form, and then find the slope and y-intercept. 4x + y = 4 –4x –4x Subtract 4x from both sides. y = 4 – 4x Rewrite to match slope-intercept form. y = –4x + 4 The equation is in slope-intercept form. m = –4 b = 4 The slope of the line 4x + y = 4 is –4, and the y-intercept is 4.
Example 2B: Using Slope-Intercept Form to Find Slopes and y-intercepts
Course 3 12-3 Using Slopes and Intercepts Example 2B: Using Slope-Intercept Form to Find Slopes and y-intercepts 5y = 3x 5y = 3x = x 3 5 5y Divide both sides by 5 to solve for y. y = x + 0 3 5 The equation is in slope-intercept form. m = 3 5 b = 0 The slope of the line 5y = 3x is , and the y-intercept is 0. 3 5
Find the slope and y-intercept.
Course 3 12-3 Using Slopes and Intercepts Check It Out: Example 2B Find the slope and y-intercept. 7y = 2x 7y = 2x = x 2 7 7y Divide both sides by 7 to solve for y. y = x + 0 2 7 The equation is in slope-intercept form. m = 2 7 b = 0 The slope of the line 7y = 2x is , and the y-intercept is 0. 2 7
Example 2C: Using Slope-Intercept Form to Find Slopes and y-intercepts
Course 3 12-3 Using Slopes and Intercepts Example 2C: Using Slope-Intercept Form to Find Slopes and y-intercepts 4x + 3y = 9 4x + 3y = 9 –4x –4x Subtract 4x from both sides. 3y = 9 – 4x Rewrite to match slope-intercept form. 3y = –4x + 9 = –4x 3 3y 9 Divide both sides by 3. y =- x + 3 4 3 The equation is in slope-intercept form. The slope of the line 4x+ 3y = 9 is – , and the y-intercept is 3. 4 3 m =- 4 3 b = 3
Using Slopes and Intercepts
Course 3 12-3 Using Slopes and Intercepts Check It Out: Example 2C 5x + 4y = 8 5x + 4y = 8 –5x –5x Subtract 5x from both sides. 4y = 8 – 5x Rewrite to match slope-intercept form. 4y = 8 – 5x = –5x 4 4y 8 Divide both sides by 4. y =- x + 2 5 4 The equation is in slope-intercept form. The slope of the line 5x + 4y = 8 is – , and the y-intercept is 2. 5 4 m =- 5 4 b = 2
Example 3: Writing Slope-Intercept Form
Course 3 12-3 Using Slopes and Intercepts Example 3: Writing Slope-Intercept Form Write the equation of the line that passes through (3, –4) and (–1, 4) in slope-intercept form. Find the slope. = y2 – y1 x2 – x1 4 – (–4) –1 – 3 8 –4 = = –2 The slope is –2. Substitute either point and the slope into the slope-intercept form. y = mx + b Substitute –1 for x, 4 for y, and –2 for m. 4 = –2(–1) + b 4 = 2 + b Simplify.
Using Slopes and Intercepts
Course 3 12-3 Using Slopes and Intercepts Example 3 Continued Solve for b. 4 = 2 + b –2 –2 Subtract 2 from both sides. 2 = b Write the equation of the line, using –2 for m and 2 for b. y = –2x + 2
Using Slopes and Intercepts
Course 3 12-3 Using Slopes and Intercepts Check It Out: Example 3 Write the equation of the line that passes through (1, 2) and (2, 6) in slope-intercept form. Find the slope. = y2 – y1 x2 – x1 6 – 2 2 – 1 4 1 = = 4 The slope is 4. Substitute either point and the slope into the slope-intercept form. y = mx + b Substitute 1 for x, 2 for y, and 4 for m. 2 = 4(1) + b 2 = 4 + b Simplify.
Check It Out: Example 3 Continued
Course 3 12-3 Using Slopes and Intercepts Check It Out: Example 3 Continued Solve for b. 2 = 4 + b –4 –4 Subtract 4 from both sides. –2 = b Write the equation of the line, using 4 for m and –2 for b. y = 4x – 2
Example 4: Entertainment Application
Course 3 12-3 Using Slopes and Intercepts Example 4: Entertainment Application A video club charges \$8 to join, and \$1.25 for each DVD that is rented. The linear equation y = 1.25x + 8 represents the amount of money y spent after renting x DVDs. Graph the equation by first identifying the slope and y-intercept. The equation is in slope-intercept form. y = 1.25x + 8 m =1.25 b = 8
Using Slopes and Intercepts
Course 3 12-3 Using Slopes and Intercepts Example 4 Continued Cost of DVDs Cost The slope of the line is 1.25, and the y-intercept is 8. The line crosses the y-axis at the point (0, 8) and moves up 1.25 units for every 1 unit it moves to the right. Number of DVDs
Using Slopes and Intercepts
Course 3 12-3 Using Slopes and Intercepts Check It Out: Example 4 A salesperson receives a weekly salary of \$500 plus a commission of 5% for each sale. Total weekly pay is given by the equation y = 0.05x Graph the equation using the slope and y-intercept. The equation is in slope-intercept form. y = 0.05x + 500 m =0.05 b = 500
Check It Out: Example 4 Continued
Course 3 12-3 Using Slopes and Intercepts Check It Out: Example 4 Continued Weekly Salary x y 500 1000 1500 2000 10,000 5000 15,000 Salary The slope of the line is 0.05, and the y-intercept is 500. The line crosses the y-axis at the point (0, 500) and moves up 0.05 units for every 1 unit it moves to the right. Sales
Using Slopes and Intercepts Insert Lesson Title Here
Course 3 12-3 Using Slopes and Intercepts Insert Lesson Title Here Lesson Quiz Write each equation in slope-intercept form, and then find the slope and y-intercept. 1. 2y – 6x = –10 2. –5y – 15x = 30 Write the equation of the line that passes through each pair of points in slope-intercept form. 3. (0, 2) and (4, –1) 4. (–2, 2) and (4, –4) y = 3x – 5; m = 3; b = –5 y = –3x – 6; m = –3; b = –6 y = – x + 2 3 4 y = –x
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# `5x - 2y - 1 = 0` Solve each equation for y
mathace | Certified Educator
5x-2y-1=0
5x-2y-1+1=0+1
5x-2y=1
5x-5x-2y=-5x+1
-2y=-5x+1
y=(-5/-2)x+1
y=(5/2)x+1
kspcr111 | Student
`5x-2y-1=0`
`5x-2y=1`
`-2y=1-5x`
`2y=5x-1`
`Y=(5/2)x-1/2`
llovejoy | Student
To solve this equation for ‘y’ you need to rearrange the equation so that ‘y’ becomes the subject of the equation – you will get an equation that states ‘y=(the rest of the equation)’. To do this you need to step by step rearange the terms that are currently on the ‘y’ side of the equation so that ‘y’ can be isolated.
Remember that when rearranging algebra equations the first rule is that ‘what you do on one side of the equation, you must also do on the other side of the equation’ this is because the equation must remain balanced.
The equation is:
`5x-2y-1=0`
Step 1: the first (and easiest) term of the equation that can be rearranged is the ‘-1’. We want to ‘get rid’ of the -1 from the ‘y’ side of the equation, but still keep the equation balanced. Adding +1 to the ‘y’ side of the equation will cancel the -1 out (because -1+1=0). Then because of the rule ‘what you do on one side of the equation, you must also do on the other side of the equation’ we also need to add +1 to the other side.
`5x-2y-1+1=0+1`
`5x-2y=1`
Step 2: the next term to rearrange is the ‘5x’. We want to ‘get rid’ of the 5x from the ‘y’ side of the equation, but still keep the equation balanced. Subtracting -5x from the ‘y’ side of the equation will cancel the 5x out (because 5x-5x=0 regardless of the actual value of x). Then because of the rule ‘what you do on one side of the equation, you must also do on the other side of the equation’ we also need to place a -5x on the other side.
`+5x-5x-2y=-5x+1`
`-2y=-5x+1`
Step 3: now we just need to isolate the ‘y’ by ‘getting rid’ of the -2 from the ‘y’ side of the equation. The –2 is not a term of the equation, it is a coefficient of y so is treated differently. The full term ‘-2y’ means that -2 is being multiplied with ‘y’ so to ‘get rid’ of the -2 we need to divided by -2. (Because -2/-2=1 and any number multiplied by 1 is still the same number 1y=y). Then because of the rule ‘what you do on one side of the equation, you must also do on the other side of the equation’ we also need to divide the other side of the equation by -2. (NOTE: all terms on the other side of the equation need to be divided by -2)
`(-2y)/(-2)=(-5x+1)/(-2)`
`(-2y)/(-2)=(-5x)/(-2)+(1)/(-2)`
`y=-(5x)/(-2)+(1)/(-2)`
Step 4: now to tidy up the equation. `(-5x)/(-2)` is a negative divided by a negative so the negatives cancel each other out. Then `(5x)/(2)` can be more correctly written as the fraction `(5)/(2)x` ; and `(1)/(-2)` can be more correctly written as -`–(1)/(2)` .
`y=(5)/(2)x-(1)/(2)`
So the quick version, solving for y, would look like this:
`5x-2y-1=0`
`5x-2y=1`
`-2y=-5x+1`
`y=(-5x)/(-2)+(1)/(-2)`
`y=(5)/(2)x-(1)/(2)`
becamymica | Student
5x - 2y -1 =0
( add 2y to both sides)
5x - 1 = 2y
( divide both sides by 2)
(5/2)x - 1/2 = y
or y = (5/2)x - 1/2
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# 1988 IMO Problems/Problem 5
## Problem
In a right-angled triangle $ABC$ let $AD$ be the altitude drawn to the hypotenuse and let the straight line joining the incentres of the triangles $ABD, ACD$ intersect the sides $AB, AC$ at the points $K,L$ respectively. If $E$ and $E_1$ dnote the areas of triangles $ABC$ and $AKL$ respectively, show that $$\frac {E}{E_1} \geq 2.$$
## Solution
Lemma: Through the incenter $I$ of $\triangle{ABC}$ draw a line that meets the sides $AB$ and $AC$ at $P$ and $Q$, then: $$\frac{AB}{AP} \cdot AC + \frac{AC}{AQ} \cdot AB = AB+BC+AC$$ Proof of the lemma: Consider the general case: $M$ is any point on side $BC$ and $PQ$ is a line cutting AB, AM, AC at P, N, Q. Then:
$\frac{AM}{AN}=\frac{S_{APMQ}}{\triangle{APQ}}=\frac{\triangle{APM}+\triangle{AQM}}{\triangle{PQA}}=\frac{\frac{AP}{AB}\triangle{ABM}+\frac{AQ}{AC}\triangle{ACM}}{\frac{AP\cdot AQ}{AB \cdot AC}}=$
$=\frac{AC}{AQ}\cdot \frac{BM}{BC}+\frac{AB}{AP}\cdot \frac{CM}{BC}$
If $N$ is the incentre then $\frac{AM}{AN}=\frac{AB+BC+CA}{AB+AC}$, $\frac{BM}{BC}=\frac{AB}{AB+AC}$ and $\frac{CM}{BC}=\frac{AC}{AC+AB}$. Plug them in we get: $$\frac{AB}{AP} \cdot AC + \frac{AC}{AQ} \cdot AB = AB+BC+AC$$
Back to the problem Let $I_1$ and $I_2$ be the areas of $\triangle{ABD}$ and $\triangle{ACD}$ and $E$ be the intersection of $KL$ and $AD$. Thus apply our formula in the two triangles we get: $$\frac{AD}{AE} \cdot AB + \frac{AB}{AK} \cdot AD = AB+BD+AD$$ and $$\frac{AD}{AE} \cdot AC + \frac{AC}{AL} \cdot AD = AC+CD+AD$$ Cancel out the term $\frac{AD}{AE}$, we get: $$\frac{AB+BD+AD-\frac{AB}{AK} \cdot AD }{AC+CD+AD- \frac{AC}{AL} \cdot AD }=\frac{AB}{AC}$$ $$AB \cdot CD + AB \cdot AD - \frac{AB \cdot AC \cdot AD}{AL}=AC \cdot BD+ AC \cdot AD -\frac{AB \cdot AC \cdot AD}{AK}$$ $$AB+AB \cdot \frac{CD}{AD}-\frac{AB \cdot AC}{AL}=AC+ AC \cdot \frac{BD}{AD} - \frac{AB \cdot AC}{AK}$$ $$AB+AC - \frac{AB \cdot AC}{AL}=AB+AC - \frac{AB \cdot AC}{AK}$$ $$\frac{AB \cdot AC}{AK} = \frac{AB \cdot AC}{AL}$$ So we conclude $AK=AL$.
Hence $\angle{AKI_1}=45^o=\angle{ADI_1}$ and $\angle{ALI_2}=45^o=\angle{ADI_2}$, thus $\triangle{AK_1} \cong \triangle{ADI_1}$ and $\triangle{ALI_2} \cong \triangle{ADI_2}$. Thus $AK=AD=AL$. So the area ratio is: $$\frac{E}{E_1}=\frac{AB \cdot AC}{AD^2} = \frac{BC}{AD} =\frac{BD+CD}{\sqrt{BD \cdot CD}}\geq 2$$
This solution was posted and copyrighted by shobber. The original thread for this problem can be found here: [1]
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# If f’’(x) be continuous at x=0.If, $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{2f(x)-3af(2x)+bf(8x)}{{{\sin }^{2}}x}$ exists and $f(0)\ne 0,$ $f'(0)\ne 0,$Then the value of $\dfrac{3a}{b}$___________________.
Last updated date: 28th May 2024
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Hint: To solve the problem we can directly put the limits and see that with which indeterminate form it matches and then we can solve it accordingly by equating it with that form.
Firstly we will write the given values,
$f(0)\ne 0,$$f'(0)\ne 0,$……………………………………….. (1)
Now we will write the given expression and name it as L,
$L=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{2f(x)-3af(2x)+bf(8x)}{{{\sin }^{2}}x}$………………………………… (2)
As we don’t know what to do next, we will put the limits directly to see what happens with the above equation,
$\therefore L=\dfrac{2f(0)-3af(2\times 0)+bf(8\times 0)}{{{\sin }^{2}}\left( 0 \right)}$
$\therefore L=\dfrac{2f(0)-3af(0)+bf(0)}{{{\sin }^{2}}0}$
As we all know the value of sin 0 is zero, therefore we will get,
$\therefore L=\dfrac{2f(0)-3af(0)+bf(0)}{0}$
As it is mentioned that the limit ‘exists’ and if the limit exists then the value of the above equation should match with one of the indeterminate forms so that we can find the limit using L-Hospitals Rule.
As the denominator of the given limit is becoming zero therefore we can say that it is of the $\dfrac{0}{0}$ form which is a type of indeterminate form.
Therefore we can conclude that both the numerator and denominator are zero. Therefore we will get,
$\therefore 2f(0)-3af(0)+bf(0)=0$
Taking f(0) we will get,
$\therefore f(0)\times \left( 2-3a+b \right)=0$
As it is given in (1) that f(0) is not equal to zero, therefore we will get,
$\left( 2-3a+b \right)=0$
$\therefore 2+b=3a$………………………………………………. (3)
Now as we know that the L is giving the indeterminate form therefore we can use the L-Hospital’s Rule which is given below,
L-Hospital’s Rule:
$\underset{x\to a}{\mathop{\lim }}\,\dfrac{f(x)}{g(x)}=\underset{x\to a}{\mathop{\lim }}\,\dfrac{\dfrac{d}{dx}f(x)}{\dfrac{d}{dx}g(x)}$
By using above rule we will write the equation (2) as shown below,
$L=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\dfrac{d}{dx}\left[ 2f(x)-3af(2x)+bf(8x) \right]}{\dfrac{d}{dx}\left( {{\sin }^{2}}x \right)}$
To solve further we should know the formulae given below,
Formulae:
\begin{align} & 1.\dfrac{d}{dx}f(x)=f'(x) \\ & 2.\dfrac{d}{dx}\sin x=\cos x \\ & 3.\dfrac{d}{dx}f{{(x)}^{2}}=2f(x)\times f'(x) \\ \end{align}
By using Formulae (1) in numerator of ‘L’ we will get,
$L=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\left[ 2f'(x)-3af'(2x)\dfrac{d}{dx}2x+bf'(8x)\dfrac{d}{dx}8x \right]}{\dfrac{d}{dx}\left( {{\sin }^{2}}x \right)}$
By using formula (3) in denominator we will get,
$L=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\left[ 2f'(x)-3af'(2x)\times 2+bf'(8x)\times 8 \right]}{\left( 2\sin x \right)\dfrac{d}{dx}\left( \sin x \right)}$
By using the formula (2) in the denominator we can get,
$L=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\left[ 2f'(x)-6af'(2x)+8bf'(8x) \right]}{2\sin x\times \cos x}$
To proceed further we should the formula given below,
Formula:
$\sin 2x=2\times \sin x\times \cos x$
By using above formula we can write L as,
$L=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\left[ 2f'(x)-6af'(2x)+8bf'(8x) \right]}{\sin 2x}$
Now if we put the limits directly we will get,
$L=\dfrac{\left[ 2f'(0)-6af'(2\times 0)+8bf'(8\times 0) \right]}{\sin \left( 2\times 0 \right)}$
$L=\dfrac{\left[ 2f'(0)-6af'(0)+8bf'(0) \right]}{\sin \left( 0 \right)}$
As we all know the value of sin 0 is zero,
$\therefore L=\dfrac{\left[ 2f'(0)-6af'(0)+8bf'(0) \right]}{0}$
Again we are getting the condition of indeterminate form and we can easily say that it’s a $\dfrac{0}{0}$ form and to make the above equation in this form, the Numerator should have to be Zero.
$\therefore 2f'(0)-6af'(0)+8bf'(0)=0$
Taking f’(0) common we will get,
$\therefore f'(x)\left( 2-6a+8b \right)=0$
If we see the equation (1) we will come to know that the value of f’(x) is not equal to Zero.
$\therefore \left( 2-2\left( 3a \right)+8b \right)=0$………………………………….. (4)
Now we will put the value of equation (3) in above equation we will get,
$\therefore \left[ 2-2\left( b+2 \right)+8b \right]=0$
$\therefore \left[ 2-2b-4+8b \right]=0$
$\therefore 6b-2=0$
$\therefore b=\dfrac{1}{3}$………………………………………………………….. (5)
Now put the value of ‘b’ in equation (4),
$\therefore \left( 2-2\left( 3a \right)+8\left( \dfrac{1}{3} \right) \right)=0$
$\therefore 2-6a+\dfrac{8}{3}=0$
$\therefore 2+\dfrac{8}{3}=6a$
$\therefore \dfrac{6+8}{3}=6a$
$\therefore \dfrac{14}{3\times 6}=a$
$\therefore \dfrac{7}{3\times 3}=a$
$\therefore a=\dfrac{7}{9}$………………………………………………………….. (6)
Now consider, $\dfrac{3a}{b}$ and put the values of (5) and (6) in it, then we will get,
$\therefore \dfrac{3a}{b}=\dfrac{3\times \dfrac{7}{9}}{\dfrac{1}{3}}$
$\therefore \dfrac{3a}{b}=\dfrac{\dfrac{7}{3}}{\dfrac{1}{3}}$
$\therefore \dfrac{3a}{b}=\dfrac{7}{3}\times \dfrac{3}{1}$
$\therefore \dfrac{3a}{b}=7$
Therefore the value of $\dfrac{3a}{b}$ is 7.
Note: While taking the derivative of a given expression do remember to use chain rule in f(x) also, If we won’t use the chain rule then we will not be able to get any answer.
Chain Rule:
$.\dfrac{d}{dx}f{{(x)}^{2}}=2f(x)\times \dfrac{d}{dx}f(x)$
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# 1.8 The real numbers (Page 3/13)
Page 3 / 13
$\text{These decimals either stop or repeat.}$
What do these examples tell us?
Every rational number can be written both as a ratio of integers , $\left(\frac{p}{q},$ where p and q are integers and $q\ne 0\right),$ and as a decimal that either stops or repeats.
Here are the numbers we looked at above expressed as a ratio of integers and as a decimal:
Fractions Integers
Number $\frac{4}{5}$ $-\phantom{\rule{0.2em}{0ex}}\frac{7}{8}$ $\frac{13}{4}$ $-\phantom{\rule{0.2em}{0ex}}\frac{20}{3}$ $-2$ $-1$ $0$ $1$ $2$ $3$
Ratio of Integers $\frac{4}{5}$ $-\phantom{\rule{0.2em}{0ex}}\frac{7}{8}$ $\frac{13}{4}$ $-\phantom{\rule{0.2em}{0ex}}\frac{20}{3}$ $-\phantom{\rule{0.2em}{0ex}}\frac{2}{1}$ $-\phantom{\rule{0.2em}{0ex}}\frac{1}{1}$ $\frac{0}{1}$ $\frac{1}{1}$ $\frac{2}{1}$ $\frac{3}{1}$
Decimal Form $0.8$ $-0.875$ $3.25$ $-6.\stackrel{\text{–}}{6}$ $-2.0$ $-1.0$ $0.0$ $1.0$ $2.0$ $3.0$
## Rational number
A rational number is a number of the form $\frac{p}{q},$ where p and q are integers and $q\ne 0.$
Its decimal form stops or repeats.
Are there any decimals that do not stop or repeat? Yes!
The number $\pi$ (the Greek letter pi , pronounced “pie”), which is very important in describing circles, has a decimal form that does not stop or repeat.
$\pi =3.141592654...$
We can even create a decimal pattern that does not stop or repeat, such as
$2.01001000100001\dots$
Numbers whose decimal form does not stop or repeat cannot be written as a fraction of integers. We call these numbers irrational.
## Irrational number
An irrational number is a number that cannot be written as the ratio of two integers.
Its decimal form does not stop and does not repeat.
Let’s summarize a method we can use to determine whether a number is rational or irrational.
## Rational or irrational?
If the decimal form of a number
• repeats or stops , the number is rational .
• does not repeat and does not stop , the number is irrational .
Given the numbers $0.58\stackrel{\text{–}}{3},0.47,3.605551275...$ list the rational numbers irrational numbers.
## Solution
$\begin{array}{cccccc}\text{Look for decimals that repeat or stop.}\hfill & & & & & \text{The}\phantom{\rule{0.2em}{0ex}}3\phantom{\rule{0.2em}{0ex}}\text{repeats in}\phantom{\rule{0.2em}{0ex}}0.58\stackrel{\text{–}}{3}.\hfill \\ & & & & & \text{The decimal}\phantom{\rule{0.2em}{0ex}}0.47\phantom{\rule{0.2em}{0ex}}\text{stops after the}\phantom{\rule{0.2em}{0ex}}7.\hfill \\ & & & & & \text{So}\phantom{\rule{0.2em}{0ex}}0.58\stackrel{\text{–}}{3}\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}0.47\phantom{\rule{0.2em}{0ex}}\text{are rational.}\hfill \end{array}$
$\begin{array}{cccccc}\text{Look for decimals that neither stop nor repeat.}\hfill & & & & & 3.605551275\text{…}\phantom{\rule{0.2em}{0ex}}\text{has no repeating block of}\hfill \\ & & & & & \text{digits and it does not stop.}\hfill \\ & & & & & \text{So}\phantom{\rule{0.2em}{0ex}}3.605551275\text{…}\phantom{\rule{0.2em}{0ex}}\text{is irrational.}\hfill \end{array}$
For the given numbers list the rational numbers irrational numbers: $0.29,0.81\stackrel{\text{–}}{6},2.515115111\text{…}.$
$0.29,0.81\stackrel{\text{–}}{6}$ $2.515115111\text{…}$
For the given numbers list the rational numbers irrational numbers: $2.6\stackrel{\text{–}}{3},0.125,0.418302\text{…}$
$2.6\stackrel{\text{–}}{3},0.125$ $0.418302\text{…}$
For each number given, identify whether it is rational or irrational: $\sqrt{36}$ $\sqrt{44}.$
1. Recognize that 36 is a perfect square, since ${6}^{2}=36.$ So $\sqrt{36}=6,$ therefore $\sqrt{36}$ is rational.
2. Remember that ${6}^{2}=36$ and ${7}^{2}=49,$ so 44 is not a perfect square. Therefore, the decimal form of $\sqrt{44}$ will never repeat and never stop, so $\sqrt{44}$ is irrational.
For each number given, identify whether it is rational or irrational: $\sqrt{81}$ $\sqrt{17}.$
rational irrational
For each number given, identify whether it is rational or irrational: $\sqrt{116}$ $\sqrt{121}.$
irrational rational
We have seen that all counting numbers are whole numbers, all whole numbers are integers, and all integers are rational numbers. The irrational numbers are numbers whose decimal form does not stop and does not repeat. When we put together the rational numbers and the irrational numbers, we get the set of real number s .
## Real number
A real number is a number that is either rational or irrational.
All the numbers we use in elementary algebra are real numbers. [link] illustrates how the number sets we’ve discussed in this section fit together.
find the solution to the following functions, check your solutions by substitution. f(x)=x^2-17x+72
Aziza is solving this equation-2(1+x)=4x+10
No. 3^32 -1 has exactly two divisors greater than 75 and less than 85 what is their product?
x^2+7x-19=0 has Two solutions A and B give your answer to 3 decimal places
3. When Jenna spent 10 minutes on the elliptical trainer and then did circuit training for20 minutes, her fitness app says she burned 278 calories. When she spent 20 minutes onthe elliptical trainer and 30 minutes circuit training she burned 473 calories. How manycalories does she burn for each minute on the elliptical trainer? How many calories doesshe burn for each minute of circuit training?
.473
Angelita
?
Angelita
John left his house in Irvine at 8:35 am to drive to a meeting in Los Angeles, 45 miles away. He arrived at the meeting at 9:50. At 3:30 pm, he left the meeting and drove home. He arrived home at 5:18.
p-2/3=5/6 how do I solve it with explanation pls
P=3/2
Vanarith
1/2p2-2/3p=5p/6
James
Cindy
4.5
Ruth
is y=7/5 a solution of 5y+3=10y-4
yes
James
Cindy
Lucinda has a pocketful of dimes and quarters with a value of $6.20. The number of dimes is 18 more than 3 times the number of quarters. How many dimes and how many quarters does Lucinda have? Rhonda Reply Find an equation for the line that passes through the point P ( 0 , − 4 ) and has a slope 8/9 . Gabriel Reply is that a negative 4 or positive 4? Felix y = mx + b Felix if negative -4, then -4=8/9(0) + b Felix -4=b Felix if positive 4, then 4=b Felix then plug in y=8/9x - 4 or y=8/9x+4 Felix Macario is making 12 pounds of nut mixture with macadamia nuts and almonds. macadamia nuts cost$9 per pound and almonds cost $5.25 per pound. how many pounds of macadamia nuts and how many pounds of almonds should macario use for the mixture to cost$6.50 per pound to make?
Nga and Lauren bought a chest at a flea market for $50. They re-finished it and then added a 350 % mark - up Makaila Reply$1750
Cindy
the sum of two Numbers is 19 and their difference is 15
2, 17
Jose
interesting
saw
4,2
Cindy
Felecia left her home to visit her daughter, driving 45mph. Her husband waited for the dog sitter to arrive and left home 20 minutes, or 13 hour later. He drove 55mph to catch up to Felecia. How long before he reaches her?
hola saben como aser un valor de la expresión
NAILEA
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How to Do Division in Math
The process of dividing is a very simple procedure you just have to remember the tables and there will be no trick left behind. Usually for children division is considered to be the most hectic and difficult part of Math especially in grade 5 when the long division is involved along with the complicated numbers, but if children memorize the tables till 12 at least then their job can easily be solved and instead of getting into a hassle of division they will start enjoying this portion of Math. Try learning the division from a small level, do not try to jump on a high level. Simple numbers are used in the process of division mentioned in this article so that you can perfectly understand the exact process.
Instructions
• 1
We will start with an easy process of division, by taking small and simple numbers. We will be dividing 25 by 2.
• 2
In the above expression 25 is known as the dividend where as 2 is called the divisor.
• 3
As you know 2 x 1 = 2 so there is no remainder left behind, because when you minus 2 from 2 the answer is zero and 1 is the quotient.
• 4
Now come to the next number of the dividend which is 5. Start remembering the table of 2 and see whether 5 come in 2’s table or there is any number which is smaller than and near the digit 5, yes there is 2 x 2 = 4.
• 5
Minus 5 from 4, now you will receive a remainder which is 1.
• 6
There is no other number left in the dividend with the help of which you can vanish the remainder.
• 7
Now from the division of dividend with the divisor the result is shown as 12 is the quotient and 1 is the remainder.
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Review question
Ref: R6514
Solution
Find the value of $k$ for which the simultaneous equations \begin{align*} 2x - ky &= 1 \\ (k+3)x - 9y &= k \end{align*}
have
1. an infinite number of solutions,
2. no solutions.
Approach 1: using algebra
From the first equation, we can write $$$\label{eq:eq-for-x-in-terms-of-k-and-y} x = \frac{1+ky}{2}$$$
and substitute this into the second equation, which yields the equation $(k+3) \frac{1+ky}{2} - 9y = k.$ After multiplying this through by $2$ and rearranging, we have that $\left( (k+3)k - 18 \right) y + (k+3) - 2k = 0.$ Expanding this leads us to the equation $\left( k^2 + 3k - 18 \right) y + 3 - k = 0.$ The quadratic coefficient for $y$ can be factored, giving the equation $(k+6)(k-3)y + 3-k = 0.$
So $(k+6)(k-3)y-(k-3)=0$, and factorising, we have
$$$\label{eq:linear-equation-in-y-and-k} (k-3)((k+6)y-1)=0.$$$
Thus $k = 3$, or $(k+6)y-1=0$.
There are three cases to consider:
Case 1: $k \ne 3$ and $k \ne -6$
In this case, we have a unique solution, namely, $y = \frac{1}{k+6} \quad\text{and}\quad x = \frac{(k+6) + k}{2(k+6)} = \frac{k+3}{k+6}.$
Case 2: $k = 3$
If $k = 3$, equation $\eqref{eq:linear-equation-in-y-and-k}$ is true for any choice of $y$, and any choice of $y$ gives a unique value of $x$ (namely, that given obtained from equation $\eqref{eq:eq-for-x-in-terms-of-k-and-y}$). Thus there are infinitely many solutions $(x,y)$.
Case 3: $k = -6$
In this case, equation $\eqref{eq:linear-equation-in-y-and-k}$ can never hold, as it leads to the contradictory statement that $9 = 0$. There are therefore no solutions.
Approach 2: Using coordinate geometry
Each of the two equations can be considered as the equation of a straight line.
In general, these two lines will meet in one point.
But if their gradients are equal, then they either have no solutions (if the lines are distinct and parallel) or infinitely many (if the lines coincide).
What are their gradients? The line $2x - ky = 1$ has gradient $\dfrac{2}{k}$ (if $k \neq 0$, and is vertical if $k=0$) and $(k+3)x - 9y = k$ has gradient $\dfrac{k+3}{9}$.
Their gradients are equal if $\dfrac{2}{k} = \dfrac{k+3}{9}$, where $k\ne0$, which rearranges to $k^2 + 3k -18 = 0$. (If $k=0$, the second line has gradient $\frac13$, which is not vertical, so the lines are not parallel.)
This factorises to $(k+6)(k-3)=0$, so their gradients are equal when $k = 3$ and when $k=-6$.
If $k = 3,$ then the original equations are $2x-3y=1$, and $6x-9y=3$, which coincide, and so there are infinitely many solutions $(x,y)$.
If $k = -6$, then the original equations become $2x+6y=1$, or $x + 3y = 0.5$, and $-3x-9y=-6$, or $x+3y =2$. In this case the lines are parallel and distinct, so there are no solutions.
This GeoGebra applet helps us to see what is happening.
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# Find the length of the perpendicular from the point $\left( {3,2,1} \right)$to the line$\dfrac{{x - 7}}{{ - 2}} = \dfrac{{y - 7}}{2} = \dfrac{{z - 6}}{3}$.
Last updated date: 20th Jun 2024
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Hint: A line is said to be perpendicular to the other line if the angle between them is${90^ \circ }$, a straight line is defined as a line that does not have any curve or the angle made between any two points on the line is always 180 degrees.
In this question, the length of the perpendicular is needed to be determined from the $\left( {3,2,1} \right)$to the line$\dfrac{{x - 7}}{{ - 2}} = \dfrac{{y - 7}}{2} = \dfrac{{z - 6}}{3}$ for which consider a point M on the line which makes an angle of 90 degrees with the given point and use the concept that the sum of the product of the direction ratio of both the line will be equal to 0.
Complete step by step answer:
Let M be a point on the given line where the line from point $P\left( {3,2,1} \right)$is perpendicular
Consider
$\dfrac{{x - 7}}{{ - 2}} = \dfrac{{y - 7}}{2} = \dfrac{{z - 6}}{3} = h$
Hence we can write
$\dfrac{{x - 7}}{{ - 2}} = h \\ x - 7 = - 2h \\ x = 7 - 2h - - - (i) \\$
$\dfrac{{y - 7}}{2} = h \\ y - 7 = 2h \\ y = 7 + 2h - - - (ii) \\$
$\dfrac{{z - 6}}{3} = h \\ z - 6 = 3h \\ z = 6 + 3h - - - (iii) \\$
So the coordinate of point
$M\left( {x,y,z} \right) = M\left( {7 - 2h,7 + 2h,6 + 3h} \right)$
Now the direction ratio of the line PM will be equal to:
$\left( {7 - 2h - 3,7 + 2h - 2,6 + 3h - 1} \right) \\ \left( {4 - 2h,5 + 2h,5 + 3h} \right) \\$
The direction ratio of the given line on which PM is perpendicular is given as$\left( { - 2,2,3} \right)$,
Since line PM is perpendicular to the given line so the sum of the product of the direction ratio of both the line will be equal to 0, hence we can write
$\left( {4 - 2h} \right)\left( { - 2} \right) + \left( {5 + 2h} \right)\left( 2 \right) + \left( {5 + 3h} \right)\left( 3 \right) = 0 \\ - 8 + 4h + 10 + 4h + 15 + 9h = 0 \\ 17h = - 17 \\ h = - \dfrac{{17}}{{17}} \\ h = - 1 \\$
Now put the value of $h$ in coordinate point of M
$M\left( {7 - 2h,7 + 2h,6 + 3h} \right) = \left( {9,5,3} \right)$
Hence the length of the perpendicular line PM will be
$PM = \sqrt {{{\left( {9 - 3} \right)}^2} + {{\left( {5 - 2} \right)}^2} + {{\left( {3 - 1} \right)}^2}} \\ = \sqrt {{6^2} + {3^2} + {2^2}} \\ = \sqrt {36 + 9 + 4} \\ = \sqrt {49} \\ = 7 \\$
Note:
In two dimensional plane, a line is represented by the general equation$y = mx + c$, where m is the slope of the line in that 2-d plane but in the case of a three-dimensional plane a straight line is defined as the intersection of two planes which is the equation of two planes together given as ${a_1}x + {b_1}y + {c_1}z + {d_1} = 0,$${a_2}x + {b_2}y + {c_2}z + {d_2} = 0$.
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# If mid-points of the sides of $\Delta ABC$ are $(1,2),(0,1)$ and $(0,1),$ then find the coordinates of the three vertices of $\Delta ABC.$
Last updated date: 20th Jun 2024
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Hint: Here we will find the midpoint of given points using related formulas. Here we will find the midpoint for all points that means $AB$ and $BC$ and $CA$ then finally we will find the coordinate points using some rules and some formula. Then finally we will get the vertices of the given points.
Formula used:
Finding the midpoint of given point $= \dfrac{{\;{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2}$
Let $A({x_1},{y_1}),$ $B({x_2},{y_2}),$ and $C({x_3},{y_3}),$ be the vertices of $\Delta ABC,$ let $P(1,2),Q(0, - 1)$ and $R(2, - 1)$ be the mid-points of sides $BC,AC$ and $AB$
Since, $P$ is the mid-point of $AB.$
$A({x_1},{y_1}),$ $B({x_2},{y_2}),$ and $C({x_3},{y_3}),$
$\therefore \dfrac{{{x_1} + {x_2}}}{2} = 1$ and $\dfrac{{{y_1} + {y_2}}}{2} = 2$
$\Rightarrow {x_1} + {x_2} = 2 - - - - - - (1)$
$\Rightarrow {y_2} + {y_3} = 4 - - - - - - (2)$
$Q(0,1)$ is the mid-point of $BC$
$\therefore \dfrac{{{x_2} + {x_3}}}{2} = 0$ and $\dfrac{{{y_2} + {y_3}}}{2} = 1$
$\Rightarrow {x_2} + {x_3} = 0 - - - - - - (3)$
$\Rightarrow {y_2} + {y_3} = 2 - - - - - - (4)$
$R(1,0)$ is the mid-point of $AC$
$\therefore \dfrac{{{x_3} + {x_1}}}{2} = 1$ and $\dfrac{{({y_3} + {y_1})}}{2} = 0$
$\Rightarrow {x_3} + {x_1} = 2 - - - - - - - (5)$
$\Rightarrow {y_3} + {y_1} = 0 - - - - - - (6)$
On adding equation $1,3\& 5$
$2({x_1} + {x_2} + {x_3}) = 4 \\ {x_1} + {x_2} + {x_3} = \dfrac{4}{2} \\ {x_1} + {x_2} + {x_3} = 2 - - - - - - (7) \\$
On adding equation $2,4\& 6$
$2({y_1} + {y_2} + {y_3}) = 6 \\ {y_1} + {y_2} + {y_3} = 3 - - - - - - (6) \\$
Subtracting equation $(1)$ from equation $(7)$
${x_1} + {x_2} + {x_3} = 2 \\ {x_1} + {x_2}{\text{ = 2}} \\ {\text{( - ) ( - ) ( - )}} \\ {\text{ - - - - - - - - - - - - - - - - - - - - }} \\ \Rightarrow {{\text{x}}_3} = 0 \\$
Subtracting equation $(3)$ from equation $(7)$
${x_1} + {x_2} + {x_3} = 2 \\ {\text{ + }}{{\text{x}}_2}{\text{ + }}{{\text{x}}_3}{\text{ = 2}} \\ {\text{( - ) ( - ) ( - )}} \\ {\text{ - - - - - - - - - - - - - - - - - - - - }} \\ \Rightarrow {{\text{x}}_1} = 2 \\$
Subtracting equation $(5)$ from equation $(7)$
${x_1} + {x_2} + {x_3} = 2 \\ {{\text{x}}_1}{\text{ + + }}{{\text{x}}_3}{\text{ = 2}} \\ {\text{( - ) ( - ) ( - )}} \\ {\text{ - - - - - - - - - - - - - - - - - - - - }} \\ \Rightarrow {x_2} = 0 \\$
Subtracting equation $(2)$ from equation $(8)$
${y_1} + {y_2} + {y_3} = 3 \\ {{\text{y}}_1}{\text{ + }}{{\text{y}}_2}{\text{ = 4}} \\ {\text{( - ) ( - ) ( - )}} \\ {\text{ - - - - - - - - - - - - - - - - - - - - }} \\ \Rightarrow {y_3} = - 1 \\$
Subtracting equation $(4)$ from equation $(8)$
${y_1} + {y_2} + {y_3} = 3 \\ {\text{ + }}{{\text{y}}_2}{\text{ + }}{{\text{y}}_3}{\text{ = 2}} \\ {\text{ ( - ) ( - ) ( - )}} \\ {\text{ - - - - - - - - - - - - - - - - - - - - }} \\ \Rightarrow {y_1} = 1 \\$
Subtracting equation $(6)$ from equation $(8)$
${y_1} + {y_2} + {y_3} = 3 \\ {{\text{y}}_1}{\text{ + + }}{{\text{y}}_3}{\text{ = 0}} \\ {\text{( - ) ( - ) ( - )}} \\ {\text{ - - - - - - - - - - - - - - - - - - - - }} \\ \Rightarrow {y_2} = 3 \\$
The value of ${x_1} = 2,{x_2} = 0$ and ${x_3} = 0$
The value of ${y_1} = 1,{y_2} = 3$ and ${y_3} = - 1$
Hence, the coordinates of the vertices of the $\Delta ABC$ are $A(2,1),B(0,3),C(0, - 1).$
The medial triangle or midpoint triangle of a triangle $ABC$ is the triangle with vertices at the midpoints of the triangle's sides $AB,{\text{ }}AC$ and $BC.$ If you draw lines from each corner (or vertex) of a triangle to the midpoint of the opposite sides, then those three lines meet at a center, or centroid, of the triangle. The centroid is the triangle's center of gravity, where the triangle balances evenly.
Here Measure the distance between the two end points, and divide the result by $2.$ This distance from either end is the midpoint of that line. Alternatively, add the two $x$ coordinates of the endpoints and divide by $2.$Do the same for the $y$ coordinates.
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# The maximum Youth Allowance is reduced by $1 for every$4 that the youth's parents income is over $31 400. By how much is Hannah's Youth Allowance reduced if her parents earn a combined income of$35 624?
Hannah's reduction in Youth Allowance is $2112 #### Explanation: Let the total income be $t$Then the amount over $31400" " is " "$[t-31400] The count of every every$4 in $[t-31400] -> ($[t-31400])/($4) Notice the way in which the units of measurement ($) cancels out leaving just a count: color(white)("d")-> color(white)("d") (cancel($)[t-31400])/(cancel($)4)
We are told that the actual joint income is $35624.00 This is substituted for $t : \textcolor{w h i t e}{\text{d}} \to \frac{35624 - 31400}{4} = 2112$For each of these counts the youth allowance is reduced by$1
Total reduction in Hannah's allowance is: $1 xx 2112 =$2112
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How to Perform Arithmetic Operations on Radicals
Key Terms
o Associative
o Distributive
Objectives
o Identify additional properties of multiplication and addition (associative and distributive properties)
o Simplify radical expressions when appropriate
Primer for Arithmetic Operations on Radicals
We will look carefully at how to work with radicals by way of arithmetic operations (addition, subtraction, division, and multiplication). First, we introduce a mathematical concept that may or may not be familiar to you, but that is crucial to properly understanding this topic. When we multiply two numbers, for instance, we can generally write the product as another number. For example, . But what if one of the factors is unknown or cannot be written in an exact form? Let's consider multiplication of 4 by some unknown number c.
We cannot simplify this expression further or write it as a single known number (because c is unknown). Likewise, consider multiplication of 4 by an irrational number . Because we cannot write as an exact decimal or as a fraction with integers in the numerator and denominator, we cannot write the product (in this case) as an exact decimal or a fraction.
Thus, the way we will treat radicals is very similar to the way we must treat unknown numbers. We will leave them (in many cases) in radical form, knowing that they correspond to a specific, known number, but in the interest of exactness, we will not attempt to evaluate the radical. The results of any arithmetic operations will, in such cases, be expressions that include radicals.
Recall that we identified addition and multiplication as commutative operations. Thus, for any two numbers a and b,
To aid in our look at radical operations, we can also consider some additional properties of addition and multiplication (and, by implication, subtraction and division, since these operations can be rewritten as addition and multiplication, respectively). First, we note that addition and multiplication are both associative, meaning that we can group terms or factors in any way we wish. Given three numbers a, b, and c,
This property simply says that we can perform a series of additions or multiplications in any order. (Recall from our study of the order of operations that expressions in parentheses must be evaluated first-hence we express the associative property as we have above.)
Additionally, we can show that multiplication is distributive, meaning that if we multiply a number a by the sum b + c, then
We can illustrate this property in a more concrete manner by considering that the product of two factors x and y can be viewed as x sets of y objects. For instance, 7 sets of 8 objects is equal to 56 objects. But 7 sets of 8 is the same as 3 sets of 8 and 4 sets of 8, or 2 sets of 8 and 5 sets of 8. Consider a graphical illustration of multiplication for this example.
Note above that and that . Furthermore, according to the representation of multiplication that we have used above, not only is , but . Thus, we can see from this example that multiplication is indeed distributive.
Interested in learning more? Why not take an online Pre-Algebra course?
We can now apply these concepts and properties to our understanding of radicals.
Let's say we wanted to perform the following addition:
As we have discussed, the square root of a number that is not a perfect square is irrational; thus, we cannot express it exactly as either a finite decimal or a fraction containing an integer numerator and integer denominator. Note, however, that we can use the fact that multiplication is distributive: . We'll start by rewriting the radical expressions slightly-this does not change the value of the expression, however.
Note that we cannot simplify this expression any further and still keep it exact. Also, note that we can omit the multiplication symbol () whenever doing so does not affect the clarity of the expression. Thus, we will, for the most part, omit this symbol for the remainder of this article. Let's look at a couple other examples.
Note that we cannot perform these operations if the numbers under the radicals are different. Thus, for instance, the expressions below cannot be further simplified.
We can also multiply and divide radicals.
Practice Problem: Evaluate each expression, where possible.
a. b. c. d.
e. f.
Solution: In each case, simply follow the pattern demonstrated earlier. Until you get more familiar with the operations, you may need to very carefully follow the distributive property to find sums or differences of radical expressions. Note that the expression in part a cannot be evaluated any further; the expression in part f can be only partially evaluated.
a. b.
c. d.
e.
f.
To aid in performing the operations above, we must often simplify radicals. To illustrate, consider the following expression.
A cursory look at this expression might seem to indicate that the expression cannot be simplified any further. Nevertheless, if we simplify the radical in the second term, we will find that we can actually perform the addition. The manipulations below rely only on the rules we have discussed so far.
Thus,
Above, we simplified the radical expression ; this process is in some ways similar to reducing a fraction to lowest terms. In simplifying a radical, we are writing an equivalent expression that is easier to work with and that often leaves less room for error. The goal of this process is to write the radical expression such that the number under the radical is not divisible by a perfect square. For instance, in the radical expression , 8 is divisible by 4, which is a perfect square. As a result, we can simplify this expression by factoring out the 4, as we did above. We then calculate the square root of 4 (which is 2), leaving only a 2 under the radical. Since 2 is not divisible by any perfect squares, the radical form is in simplest form.
Practice Problem: Evaluate each radical expression, where possible.
a. b. c.
d. e. f.
Solution: Each expression requires some degree of simplification of the radical. In some cases, you may take a slightly different approach, but the answer should be the same nonetheless. Not every step of the process is shown; if you are unsure of how a particular answer is achieved, consult the examples discussed above.
a.
b.
c.
d.
e.
f.
Another case where simplification is beneficial is a radical occurring in the denominator of a fraction. Consider the quotient expression we encountered above:
One (perfectly legitimate) approach to evaluating this expression is to take the square root of 16 (which is 4) and then simplify the radical in the denominator.
We are left, however, with a square root in the denominator. Although this is not an incorrect result, preventing occurrences of radicals in the denominator is often beneficial. Thus, we must find a way to remove the radical from the denominator. First, let's take note of the following, where y is any number:
Of course, multiplying any number by 1 yields the original number; furthermore, if the numerator and denominator of a fraction are the same, then that fraction is equal to 1. Thus, we can remove a radical from the denominator of a fraction by calculating an equivalent fraction in the manner shown below.
This is the same result we obtained earlier using a different approach to evaluating this expression. Below is another example of simplifying a fraction with a radical in the denominator.
Finally, note that in such cases, we can write the expression with the radical in the numerator or with the radical as a factor multiplied by a fraction.
Practice Problem: Simplify each of the following fractions.
a. b. c. d.
Solution: In each case, multiply the numerator and denominator by the radical in the denominator and evaluate. Before doing so, however, performing other simplifications may make the process easier. In part c, the result can be written either as a single fraction or as a fraction minus an integer.
a.
b.
c.
d.
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# 12.17: Probability of Independent Events
Difficulty Level: At Grade Created by: CK-12
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Practice Probability of Independent Events
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Have you ever wondered if two things can happen at once?
Jana has two decks of cards. Each deck has ten cards in it. There are three face cards in the first deck and four in the second. What are the chances that Jana will draw a face card from both decks?
This is an independent event. In this Concept, you will learn how to figure out probabilities like this one.
### Guidance
You now know the difference between independent and dependent events. Think about independent events. If one event does not impact the result of a second event, then the two events are independent. For example, there are two different spinners A\begin{align*}A\end{align*} and B\begin{align*}B\end{align*}. The result of spinning spinner A\begin{align*}A\end{align*} does not affect the result of spinning spinner B\begin{align*}B\end{align*}.
But now we ask a new question. What is the probability of two completely independent events both occurring? For example, what is the probability of spinner A\begin{align*}A\end{align*} landing on red and spinner B\begin{align*}B\end{align*} landing on blue?
We could create a tree diagram to show all of the possible options and figure out the probability, but that is very complicated. There is a simpler way.
Notice that this probability equals the product of the two independent probabilities.
P(red-blue)=P(red)P(blue)=1413=112\begin{align*}P(\text{red-blue}) & = P(\text{red}) \cdot P(\text{blue})\\ & = \frac{1}{4} \cdot \frac{1}{3}\\ & = \frac{1}{12}\end{align*}
Where did these fractions come from?
They came from the probability of the sample space of each spinner. The first spinner has four possible options, so the probability is 14\begin{align*}\frac{1}{4}\end{align*}. The second spinner has three possible options, so the probability is 13\begin{align*}\frac{1}{3}\end{align*}. The Probability Rule takes care of the rest.
In fact, this method works for any independent events as summarized in this rule.
Probability Rule: The probability that two independent events, A\begin{align*}A\end{align*} and B\begin{align*}B\end{align*}, will both occur is:
P(A and B)=P(A)P(B)\begin{align*}P(A \ \text{and} \ B) = P (A) \cdot P (B)\end{align*}
Write the Probability Rule in your notebook.
What is the probability that if you spin the spinner two times, it will land on yellow on the first spin and red on the second spin?
To find the solution, use the rule.
P(yellow and red)P(yellow)P(red)=P(yellow)P(red)=35=25\begin{align*}P(\text{yellow and red}) & = P(\text{yellow}) \cdot P(\text{red})\\ P (\text{yellow}) & = \frac{3}{5}\\ P (\text{red}) & = \frac{2}{5}\end{align*}
So:
P(yellow and red)=3525=625\begin{align*}P(\text{yellow and red}) & = \frac{3}{5} \cdot \frac{2}{5}\\ & = \frac{6}{25}\end{align*}
The probability of both of these events occurring is 625\begin{align*}\frac{6}{25}\end{align*}.
Now it's time for you to try a few on your own.
#### Example A
What is the probability of spinner A landing on red and spinner B landing on red?
Solution:112\begin{align*}\frac{1}{12}\end{align*}
#### Example B
What is the probability of spinner A landing on blue or yellow and spinner B landing on blue?
Solution:16\begin{align*}\frac{1}{6}\end{align*}
#### Example C
What is the probability of spinner A landing on yellow and spinner B landing on red or green?
Solution:16\begin{align*}\frac{1}{6}\end{align*}
Here is the original problem once again.
Jana has two decks of cards. Each deck has ten cards in it. There are three face cards in the first deck and four in the second. What are the chances that Jana will draw a face card from both decks?
To figure this out, let's write the probability of picking a face card from the first deck.
310\begin{align*}\frac{3}{10}\end{align*}
Now let's write the probability of picking a face card from the second deck.
410=25\begin{align*}\frac{4}{10} = \frac{2}{5}\end{align*}
Now we can multiply and simplify.
310×25\begin{align*}\frac{3}{10} \times \frac{2}{5}\end{align*}
325\begin{align*}\frac{3}{25}\end{align*}
### Vocabulary
Here are the vocabulary words in this Concept.
Independent Events
events where one event does not impact the result of another.
Probability Rule
P(A)P(B)=Probability of A and B\begin{align*}P(A) \cdot P(B) = \text{Probability of} \ A \ \text{and} \ B\end{align*}
### Guided Practice
Here is one for you to try on your own.
The probability of rain tomorrow is 40 percent. The probability that Jeff’s car will break down tomorrow is 3 percent. What is the probability that Jeff’s car will break down in the rain tomorrow?
To find the solution, use the rule.
P(rain and break)P(rain)P(break)=P(rain)P(break)=40%=40100=25=3%=3100\begin{align*}P(\text{rain and break}) & = P (\text{rain}) \cdot P (\text{break})\\ P (\text{rain}) & = 40\% = \frac{40}{100} = \frac{2}{5}\\ P (\text{break}) & = 3\% = \frac{3}{100}\end{align*}
So:
P(rain and break)=253100=3250\begin{align*}P (\text{rain and break}) & = \frac{2}{5} \cdot \frac{3}{100}\\ & = \frac{3}{250}\end{align*}
You can see that the probability is very small.
### Video Review
Here is a video for review.
### Practice
Directions: Solve each problem.
1. Mia spins the spinner two times. What is the probability that the arrow will land on 2 both times?
2. Mia spins the spinner two times. What is the probability that the arrow will land on 2 on the first spin and 3 on the second spin?
3. Mia spins the spinner two times. What is the probability that the arrow will land on an even number on the first spin and and an odd number on the second spin?
4. Mia spins the spinner two times. What is the probability that the arrow will land on an odd number on the first spin and a number less than 4 on the second spin?
5. A laundry bag has 8 black socks and 2 white socks. If you pull out a sock, then put it back in the bag and pull out a second sock, what is the probability that both socks will be black? Write your answer as a decimal.
6. A laundry bag has 8 black socks and 2 white socks. If you pull out a sock, then put it back in the bag and pull out a second sock, what is the probability that both socks will be white? Write your answer as a decimal.
7. A laundry bag has 8 black socks and 2 white socks. If you pull out a sock, then put it back in the bag and pull out a second sock, what is the probability that the first sock will be black and the second sock will be white? Write your answer as a decimal.
8. Dirk has two 52-card decks. Each deck has 4 Aces, 4 Kings, 4 Queens, and so on. What is the probability that Dirk will pick an Ace out of each deck?
9. Dirk has two 52-card decks. Each deck has 4 Aces, 4 Kings, 4 Queens, and so on. What is the probability that Dirk will pick a face card (Jack, Queen, King) out of each deck?
10. Dirk has two 52-card decks. Each deck has 4 Aces, 4 Kings, 4 Queens, and so on. What is the probability that Dirk will pick a card lower than a Jack out of each deck?
11. Karina flips a coin 3 times. What is the probability that she will flip heads 3 times in a row?
12. Karina flips a coin 4 times. What is the probability that she will flip heads 4 times in a row?
13. Karina flips a coin 4 times. What is the probability that she will NOT flip heads 4 times in a row?
14. Karina flips a coin 5 times. What is the probability that she will flip heads twice in a row?
15. Karina flips a coin 5 times. What is the probability that she will flip tails four times in a row?
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
TermDefinition
Independent Events Two events are independent if the occurrence of one event does not impact the probability of the other event.
Probability Rule The probability of two independent events A and B both occurring is P( A and B) = P(A) $\cdot$ P(B).
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# The Distributive Property
Fun and awesome way to learn!!! :)
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## Karina and Michelle Plascencia and Banegas
on 4 February 2013
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#### Transcript of The Distributive Property
VOCAB Term- A number, variable, or the product or quotient of a number and one or more variables. Assessment Questions Assessment Questions Assessment Questions The Distributive Property Constant- A term that has no variable factor Coefficient- The numerical factor when a term has a variable. Like Terms- Terms with exactly the same variable factors in a variable expression. The Distributive Property An algebra property which is used to multiply a single term and two or more terms inside a set of parentheses. Simplifying an Expression
EXAMPLE 2(5x + 3)
2(5x + 3) = 2(5x) + 2(3)
= 10x + 6 (3b - 2)(1/3)
(3b - 2)(1/3) = 3b(1/3) - 2(1/3)
= b - 2/3 Explanation Using the Distributive Property of -1
EXAMPLE Simplify: -(6x + 4)
-(6x + 4)= -1(6x + 4)
= -1(6x) +(-1)(4)
= -6x - 4
-(2x + 1)
-(2x + 1)= -1(2x + 1)
= -1(2x) + (-1)(1)
= -2x - 1 Explanation When you see a distributive property problem, it usually has a number outside the parenthesis, but in this part of the distributive property it is a -1 or just a negative sign that represents -1. You still do the same thing; you multiply -1 by everything inside the parenthesis. Then solve. Combining Like Terms
EXAMPLE Explanation When you are combining like terms, there is usually more than one variable and number. First you look for the variables.
12r+5+3r-5
12r+5+3r-5
15r
Then you look for the numbers.
12r+5+3r-5
0
Now whatever you have left is the answer. Make sure to double check your work. In a problem that is solved by the Distributive property, there are numbers inside of the parenthesis and a number on the outside. What you do is you take the number on the outside and you multiply it by every number on the inside. Once you multiply, you are left with a simple equation that you solve and the answer is the answer overall. (example shown above) n-10+9n-3 10n -13 -5n+3(6+7n) +18+21n 16n+18 Question #1 Question #2 Assessment Question Match each vocab. word with its definition ____. Term
____. Constant
____. Coefficient
____. Like Terms A.The numerical factor when a
term has a variable.
B. Terms with exactly the same
variable factors in a variable
expression.
C. A term that has a variable
factor.
D. A number or variable, or the
product of a number & one or
more variables.
Full transcript
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Review question
Which point on the first circle is closest to the second? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource
Ref: R9452
Solution
The point on the circle $(x-5)^2+(y-4)^2=4$ which is closest to the circle $(x-1)^2+(y-1)^2=1$ is
1. $(3.4,2.8)$,
2. $(3,4)$,
3. $(5,2)$,
4. $(3.8,2.4)$.
Here is a diagram of the circles:
Approach 1
The right-angle triangle with the circle centres at two of its vertices has sides of length $3$ and $4$. Its hypotenuse is therefore of length $5$.
Using similar triangles, we have $\dfrac{3}{5}=\dfrac{a}{4}=\dfrac{b}{3}$. Thus $a = 2.4, b = 1.8$, and the point on the first circle closest to the second is $(3.4,2.8)$. The answer is (a).
Approach 2
The vector joining $(5,4)$ and $(1,1)$, the centres of the circles, is $\begin{pmatrix}-4\\-3\end{pmatrix}$. The length of this vector is $\sqrt{(-4)^2+(-3)^2}=5$.
The first circle has radius $2$, and so the point on this closest to the second is $\dfrac{2}{5}$ths of the way along this vector. Hence we can calculate the point: $\begin{pmatrix}5\\4\end{pmatrix} + \frac{2}{5}\begin{pmatrix}-4\\-3\end{pmatrix} = \begin{pmatrix}5\\4\end{pmatrix} + \begin{pmatrix}-1.6\\-1.2\end{pmatrix} = \begin{pmatrix}3.4\\2.8\end{pmatrix}.$
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## Tuesday, November 22, 2005
### Vectors... wow! Topic: Vector Geometry. Level: AMC/AIME.
Problem #1: Find the area of the triangle formed by the heads of two vectors, $A = (5, 67^{\circ})$ and $B = (4, 37^{\circ})$, and the origin.
Solution: So, we have two vectors with pretty ugly angles, but we do notice that they differ by exactly $30^{\circ}$! Hmm, then we know two sides and the angle between them...
And consequently we remember the formula $[ABC] = \frac{1}{2}ab\sin{\angle C}$, which makes us happy.
Hence the area of the triangle is $\frac{1}{2}(5)(4)(\sin{30^{\circ}}) = 5$. QED.
--------------------
Comment: Notice that the formula $\frac{|A||B|\sin{\theta}}{2} = \frac{|A \times B|}{2}$, where $A \times B$ is the cross product of $A$ and $B$.
--------------------
Problem #2: Show that the vectors $A = (3, 4, 5)$ and $B = (2, -4, 2)$ are perpendicular.
Solution: Consider the dot product of $A$ and $B$, $A \cdot B$. The general definition of this is $A \cdot B = |A||B|\cos{\theta}$, where $\theta$ is the angle between the two vectors. Notice that if two vectors are perpendicular to each other, the angle $\theta = 90^{\circ} \Rightarrow \cos{\theta} = 0 \Rightarrow A \cdot B = 0$.
We can easily show that the dot product is distributive (left as an exercise to the reader), that is, $A \cdot (B+C) = A \cdot B + A \cdot C$.
Thus $A \cdot B = [(3, 0, 0)+(0, 4, 0)+(0, 0, 5)] \cdot [(2, 0, 0)+(0, -4, 0)+(0, 0, 2)]$. Distributing this, we see that only the terms that are parallel remain because all the axes are perpendicular to each other ($\theta = 90^{\circ}$ between any two axes).
So $A \cdot B = (3, 0, 0) \cdot (2, 0, 0)+(0, 4, 0) \cdot (0, -4, 0)+(0, 0, 5) \cdot (0, 0, 2) = 6 +(-16)+10 = 0$. So $A$ is perpendicular to $B$. QED.
--------------------
Comment: In effect, we have just shown that given any two vectors $X = (x_1, x_2, \ldots, x_n)$ and $Y = (y_1, y_2, \ldots, y_n)$, we have $X \cdot Y = x_1y_1+x_2y_2+\cdots+x_ny_n$, an interesting and useful result (these are all in $n$-spaces, where there actually exist $n$ dimensions).
--------------------
Practice Problem #1: Show that the dot product is distributive: $A \cdot (B+C) = A \cdot B+ A \cdot C$.
Practice Problem #2: Find the area of the triangle formed by the points $(5,2)$, $(7,8)$, and $(2,1)$.
Practice Problem #3: Show that, given three vectors of equal magnitude $A$, $B$, and $C$, the orthocenter of the triangle formed by the heads of each of the vectors is $A+B+C$.
Practice Problem #4: Using Practice Problem #3, show that the centroid ($G$), circumcenter ($O$), and orthocenter ($H$) of a triangle are collinear and that $OG:GH = 1:2$.
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# Least Common Denominator Calculator
Find the least common denominator for a set of fractions by providing a list of denominators below.
## Solution:
The least common denominator of 22, 88, & 132 is 264
### Steps to Solve
264 is the smallest number that is evenly divisible by 22, 88, & 132 without a remainder
264 ÷ 22 = 12
264 ÷ 88 = 3
264 ÷ 132 = 2
Learn how we calculated this below
## How to Find the Least Common Denominator
A denominator is the bottom number of a fraction, or the number below the fraction line. For a fraction 1/3, the denominator is 3.
A common denominator is a denominator that is common to the fractions being operated on. For the denominator to be common it must be the same in all fractions.
For instance, 1 / 3 and 2 / 3 have common denominators – the denominators are the same.
1 / 3 and 2 / 5 do not have common denominators – the denominators are different.
The least common denominator is the smallest common denominator. It is the smallest whole number that is evenly divisible by all uncommon denominators. The least common denominator is also referred to as the lowest common denominator or the least common multiple.
The least common denominator for the fractions 1 / 3 and 2 / 5 is 15.
15 ÷ 3 = 5
15 ÷ 5 = 3
Note that there can be no remainder when dividing by one of the denominators by the least common denominator.
There are a few methods that can be used to find the least common denominator. The easiest is to use the calculator above. If you want to do the work yourself, follow along to learn two methods to solve it.
### Method One: Find the Least Common Denominator Using Factorization
One way to find the least common denominator is to use prime factorization. Find the prime factors of each denominator. Then multiply all of the prime factors, multiplying the factors that are common to both only once, to find the least common denominator.
You’ll probably find our factors calculator helpful to find the factors of your denominators, including the greatest common factor.
#### Example 1: Find the Least Common Denominator of 90 and 36
Step one: find the prime factors of 36.
Divide 36 by 2, which equals 18. 18 and 2 are factors. 18 can be factored again into 6 and 3. 6 can be factored into 2 and 3. The prime factors are thus [3, 3, 2, 2].
Step two: find the prime factors of 90.
Divide 90 by 10, which equals 9. 9 and 10 are factors. 9 can be factored again into 3 and 3. 10 can be factored into 5 and 2. The prime factors are thus [5, 3, 3, 2].
Step three: find the prime factors that are common to both 36 and 90.
The prime factors of 36 and 90 are [5, 3, 3, 2, 2].
Step four: multiply the factors together to find the least common denominator. Note that since 3, 3, and 2 are common between 36 and 90 they are only used once.
LCD = 5 × 3 × 3 × 3 × 2 × 2
LCD = 180
#### Example 2: Find the Least Common Denominator of 105 and 165
Step one: find the prime factors of 105.
Divide 105 by 7, which equals 15. 15 and 7 are factors. 15 can be factored again into 5 and 3. The prime factors are thus [7, 5, 3].
Step two: find the prime factors of 165.
Divide 165 by 11 which equals 15. 15 and 11 are factors. 15 can be factored again into 5 and 3. The prime factors are thus [11, 5, 3].
Step three: find the prime factors that are common to both 105 and 165.
The prime factors of 105 and 165 are [11, 7, 5, 3].
Step four: multiply the factors together to find the least common denominator. Note that 5 and 3 are common between 105 and 165 so they are only used once.
LCD = 11 × 7 × 5 × 3
LCD = 1,155
#### Example 3: Find the Least Common Denominator of 24 and 42
Step one: find the prime factors of 24.
Divide 24 by 2, which equals 12. 12 and 2 are factors. 12 can be factored again into 6 and 2. 6 can be factored again into 2 and 3. The prime factors are thus [3, 2, 2, 2].
Step two: find the prime factors of 42.
Divide 42 by 7 which equals 6. 7 and 6 are factors. 6 can be factored again into 2 and 3. The prime factors are thus [7, 3, 2].
Step three: find the prime factors that are common to both 24 and 42.
The prime factors for 24 and 42 are [7, 3, 2, 2, 2].
Step four: multiply the factors together to find the least common denominator. Note that 3 and 2 are common between 24 and 42 so they are only used once.
LCD = 7 × 3 × 2 × 2 × 2
LCD = 168
### Method Two: Find the Least Common Denominator by Finding all Multiples
You can also find the least common denominator by finding all of the multiples of each denominator and finding the smallest multiple that is common to both.
To find multiples of a number, multiply it by 2 to find the first multiple. Then multiply it by 3 to find the second. Continue by multiplying it by 4, and so on to find all of the multiples.
For example, let’s find the least common denominator of 3 and 5 by finding all multiples.
The multiples of 3 are [3,6,9,12,15,18,21,24,27,30,…]
The multiples of 5 are [5,10,15,20,25,30,…]
Observe that the multiples that are common to both 3 and 5 are 15 and 30. The smallest of these is 15, which makes it the least common denominator.
Thus, the least common denominator of 3 and 5 is 15.
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# How do you solve compound inequalities 5a-4>16 or 3a + 2 <17?
Jul 21, 2018
$a > 4$ or $a < 5$
#### Explanation:
We have the following:
$\textcolor{\lim e g r e e n}{5 a - 4 > 16}$ and $\textcolor{b l u e}{3 a + 2 < 17}$
Let's start with our green inequality. We can add $4$ to both sides to get
$\textcolor{\lim e g r e e n}{5 a > 20}$
Next, divide both sides by $5$ to get
$\textcolor{\lim e g r e e n}{a > 4}$
Next, let's look at our blue inequality. Let's subtract $2$ from both sides to get
$\textcolor{b l u e}{3 a < 15}$
Lastly, let's divide bot hsides by $3$ to get
$\textcolor{b l u e}{a < 5}$
Our solutions are
$a > 4$ or $a < 5$
Hope this helps!
$a > 4$ or $a < 5$
#### Explanation:
1) Solving first inequality:
$5 a - 4 > 16$
$5 a - 4 + 4 > 16 + 4$
$5 a > 20$
$\frac{5 a}{5} > \frac{20}{5}$
$a > 4$
2) Solving second inequality:
$3 a + 2 < 17$
$3 a + 2 - 2 < 17 - 2$
$3 a < 15$
$\frac{3 a}{3} < \frac{15}{3}$
$a < 5$
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# Whole Numbers Basics
Top
Sub Topics Whole number is one of the classification of number systems. Whole numbers are numbers without fractions, percentages or decimals. Zero is neither a fraction nor a decimal, so zero is a whole number. A whole number is denoted by W. Whole numbers can be finite or infinite. Finite defines the numbers in the set are countable and infinite means the numbers in the set are uncountable. They include only the rounded value. The term whole number is one you’ll find often in mathematics. Whole numbers can never be negative. Whole numbers stop decreasing at Zero.
## Difference Between Whole Numbers and Natural Numbers
The following differences between whole numbers and natural numbers are given below:
• A whole number is a positive integer including zero. The set of natural numbers is the set of positive integers beginning at one.
• Natural numbers are whole numbers, however whole numbers are not natural because zero is not a natural number. Infinite number of zeroes can be attached at the end to whole numbers.
• Smallest whole number is 0 greatest
• Smallest natural number is 1greatest
The difference between natural and whole numbers is where they start. A whole number is any positive number including a zero. A natural number is a positive integers number which starts at one.
## Operations on Whole Numbers
There are some basic operations performed by whole numbers. They are:
When two whole numbers are added we get a Whole Number. Therefore, whole numbers are closed under addition.
Examples : 4 + 5 = 9, 1 + 2 = 3, 9 + 7 = 16
Whole number Subtraction
When two whole numbers are subtracted we get a whole number. In some cases, subtraction of whole number does not always give a whole number. Therefore, whole numbers are not closed under subtraction.
e.g. 9 - 2 = 7, 8 - 11 = - 3
From the above, in some problems we can notice that the the difference of whole numbers is not a whole number, the result obtained is an integer.
Whole number Multiplication
When two whole numbers are multiplied we get a whole number. Therefore, whole numbers are closed under multiplication
Examples : 7 * 3 = 21, 5 * 4 = 20, 6 * 6 = 36, 7 * 6 = 42
Whole Number Division
Dividing a whole number by another does not always give a whole number. Whole numbers are not closed under division.
Examples : $\frac{12}{6}$ = 2, $\frac{5}{15}$ = 0.33, $\frac{36}{128}$ = 0.2812
From the above, we see that in some problems the result obtained is a fraction. We can say that division of two whole numbers is not always a whole number.
## Properties of Whole Numbers
Important properties of whole numbers are explained below.
Closure property
Addition of two whole numbers will always be a whole number.
Examples : 56 + 13 = 69, 2 + 8 = 10, 35 + 10 = 45
Commutative property
If a and b are two whole numbers, then
a + b = b + a
Examples : 5 + 6 = 6 + 5, 2 + 9 = 9 + 2, 89 + 96 = 96 + 89
Associative property
Consider a, b and c to be whole numbers then,
a + ( b + c) = (a + b) + c
Examples : 4 + (5 + 10) = (4 + 5) + 10 = 19, 5 + ( 2 + 3 ) = (5 + 2) + 3 = 10
If we add Zero with any whole result would be same whole number
Suppose a is a whole number, then
a + 0 = 0 + a = a
Examples: 1 + 0 = 0 + 1 = 1, 63 + 0 = 0 + 63 = 63
Multiplicative Property
If we multiply 1 with any whole number result would be number itself.
Suppose a is a whole number, then
a * 1 = 1 * a = a
Distributive Property
Let a, b and c be three whole numbers then,
a * ( b + c) = (a * b) + (a * c)
Examples : 2 * (6 + 3) = ( 2 * 6) + (2 * 3) = 18, 5 * ( 6 + 3 ) = (5 * 6) + (5 * 3 ) = 45
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# 2021 AMC 12A Problems/Problem 23
## Problem
Frieda the frog begins a sequence of hops on a $3 \times 3$ grid of squares, moving one square on each hop and choosing at random the direction of each hop-up, down, left, or right. She does not hop diagonally. When the direction of a hop would take Frieda off the grid, she "wraps around" and jumps to the opposite edge. For example if Frieda begins in the center square and makes two hops "up", the first hop would place her in the top row middle square, and the second hop would cause Frieda to jump to the opposite edge, landing in the bottom row middle square. Suppose Frieda starts from the center square, makes at most four hops at random, and stops hopping if she lands on a corner square. What is the probability that she reaches a corner square on one of the four hops?
$\textbf{(A)} ~\frac{9}{16}\qquad\textbf{(B)} ~\frac{5}{8}\qquad\textbf{(C)} ~\frac{3}{4}\qquad\textbf{(D)} ~\frac{25}{32}\qquad\textbf{(E)} ~\frac{13}{16}$
## Solution
We can draw a state diagram with three states: center, edge, and corner. Denote center by M, edge by E, and corner by C. There are a few ways Frieda can reach a corner in four or less moves: EC, EEC, EEEC, EMEC. Then, calculating the probabilities of each of these cases happening, we have $1\cdot\tfrac{1}{2}+1\cdot\tfrac{1}{4}\cdot\tfrac{1}{2}+1\cdot\tfrac{1}{4}\cdot\tfrac{1}{4}\cdot\tfrac{1}{2}+1\cdot\tfrac{1}{4}\cdot1\cdot\tfrac{1}{2}=\tfrac{25}{32}$, so the answer is $\boxed{D}$. ~IceWolf10
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# 13.1.1: Factorial Notations and Square Tables
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## 2x2 Designs
We’ve just started talking about a 2x2 Factorial design, which means that we have two IVs (the number of numbers indicates how many IVs we have) and each IV has two levels (the numbers represent the number of level for each IV). We said this means the IVs are crossed. To illustrate this, take a look at the following tables. Table $$\PageIndex{1}$$ is a conceptual version. Although not exactly accurate, many call these types of tables a Punnett Square because it shows the combination of different levels of two categories.
IV Levels IV2 Level 1 IV1 Level I IV1 Level 2 DV DV DV DV
Our study on distraction is a 2x2 design, so what would that look like in this type of table?
Example $$\PageIndex{1}$$
Create a "Punnett's Square" for the IVs and DV of the Distraction scenario.
Solution
IV Levels IV2 (Reward): Yes IV1 (Distraction): Yes IV1 (Distraction): No DV = Number of differences spotted DV = Number of differences spotted DV = Number of differences spotted DV = Number of differences spotts
You could have just as easily made IV1 the reward and IV2 the Distraction, and the table would still be correct.
Let’s talk about this crossing business. Here’s what it means for the design. For the first level of Distraction (Yes), we measure the number of differences spotted performance for the people who were rewarded, as well as for the people who were not rewarded. So, for the people who were distracted we also manipulated whether or not they earned a reward. In the second level of the Distraction IV (No), we also manipulate reward, with some people earning a reward and some people not. We collect how many differences were spotted in all conditions.
We could say the same thing, but talk from the point of view of the second IV. For example, for participants who were rewarded, some are distracted and some are not. Similarly, for participants who were not rewarded, we distract some of the participants and don't distract some of them.
Each of the four squares representing a DV, is called a condition. So, we have 2 IVs, each with 2 levels, for a total of 4 conditions. This is why we call it a 2x2 design. 2x2 = 4. The notation tells us how to calculate the total number of conditions.
## Factorial Notation
Anytime all of the levels of each IV in a design are fully crossed, so that they all occur for each level of every other IV, we can say the design is a fully factorial design. We use a notation system to refer to these designs. The rules for notation are as follows. Each IV get’s it’s own number. The number of levels in the IV is the number we use for the IV. Let’s look at some examples:
2x2 = There are two IVS, the first IV has two levels, the second IV has 2 levels. There are a total of 4 conditions, 2x2 = 4.
2x3 = There are two IVs, the first IV has two levels, the second IV has three levels. There are a total of 6 conditions, 2x3 = 6
3x2 = There are two IVs, the first IV has three levels, the second IV has two levels. There are a total of 6 conditions, 3x2=6.
4x4 = There are two IVs, the first IV has 4 levels, the second IV has 4 levels. There are a total of 16 condition, 4x4=16
2x3x2 = There are a total of three IVs. The first IV has 2 levels. The second IV has 3 levels. The third IV has 2 levels. There are a total of 12 condition. 2x3x2 = 12.
Let’s practice a little with this notation.
Exercise $$\PageIndex{1}$$
What is the factorial design notation for a study with two IVs, one has 2 levels and the other has 3 levels?
2x3
There are two IVs, so there are two numbers. Each number represents the number of levels for each IV.
Exercise $$\PageIndex{2}$$
What is the factorial design notation with a study with the following IVs:
2 (task presentation: computer or paper) by
2 (task difficulty: easy or hard) by
2 (student: high school or college)
2x2x2
There are three IVs, so there are three numbers. Each IV only has two levels, so there are three two’s! Notice that there are no threes. The number of IVs is represented in the number of numbers.
Okay, let’s try something a little more challenging.
Let’s do a couple more to make sure that we have this notation business down.
Exercise $$\PageIndex{5}$$
For one of Dr. MO’s dissertation studies, participants read about a character, then rated that character on several personality traits (DV). The race and gender of the character were varied systematically. Here are the IVs:
• Race of participant: White or Black
• Gender of participant: Woman or man
• Race of character: White or Black or none/neutral (the character’s race was not mentioned)
• Gender of character: Woman or man or none/neutral (the character’s gender was not mention)
What kind of factorial design was this study? 2x2? 2x3? Something else (what?)?
This was a 2x2x3x3 study because:
• Race of participant: White or Black = 2
• Gender of participant: Woman or man = 2
• Race of character: White or Black or none/neutral = 3
• Gender of character: Woman or man or none/neutral -= 3
Dr. MO wanted 30 participants in each cell, so she had to have 1,080 participants!
Last one! This one has a few more questions to better understand the scenario.
Exercise $$\PageIndex{6}$$
Dr. MO has more Star Wars collectibles than can fit in her office, and she’d like to sell some. Her research question is whether she’d get a better price through CraigsList or eBay? Also, should she take the picture or use a stock photo?
1. What is the DV? (It is not explicitly labeled.)
2. For each IV, what are the levels?
3. Is this a 2x2 factorial design? If not, what kind of design is it?
4. List out each of the combinations of the levels of the IVs.
1. The DV is the price, or how much Dr. MO could earn for selling each collectible.
2. IV1’s levels are CraigsList or eBay, so the IV name could be something like “website” or “platform”. IVs’ levels are personal photo or stock photo, so the IV name could be something like “Photo Type.”
3. This is a 2x2 factorial design: 2 (Platform: CraigsList or eBay) by 2 (Photo Type: Personal or Stock)
4. List out each of the combinations of the levels of the IVs:
• Posted on CraigsList with a personal photo.
• Posted on Craigslists with a stock photo.
• Posted on eBay with a personal photo.
• Posted on eBay with a stock photo.
Just for fun, let’s illustrate a 2x3 design using the same kinds of tables we looked at before for the 2x2 design.
IV Levels IV2 Level 1 IV1 Level I IV1 Level 2 DV DV DV DV DV DV
Our very first example of tiem spent studying is a 2x3 design, so what would that look like in this type of table?
Example $$\PageIndex{2}$$
Create a "Punnett's Square" for the IV of Mindset (Growth or Fixed) and the IV of Job (Full-Time, Part-Time, or None) for time spend studying.
Solution
IV Levels IV2 (Job): Full-Time IV1 (Mindset): Growth IV1 (Mindset): Fixed DV = Minutes spent studying DV = Minutes spent studying DV = Minutes spent studying DV = Minutes spent studying DV = Minutes spent studying DV = Minutes spent studying
You could have just as easily made IV1 the Job and IV2 the Mindset, and the table would still be correct.
All we did was add another row for the second IV. It’s a 2x3 design, so it should have 6 conditions. As you can see there are now 6 cells to measure the DV.
You might have noticed in the list of notation for different factorial designs that you can have three IVs (that's the 2x3x2 design). In fact, you can have as many IVs with as many levels as you'd like, but the Central Limit Theorem shows (through complicated math that we aren't going to go into) that each condtion (or cell) should have at least 30-50 participants, that can get expensive quickly! If a 2x2 has 4 conditions, and you want at least 30 participants in each condition, then you'd need 120 participants. If you have a 2x3, then you'd need at least 180 participants (2 * 3 * 30 = 180). So, for a 2x3x2, how many participants would you want?
\begin{aligned} \text{Participants} &=2 * 3 * 2 = 12 \\ \text{Participants}&=12 * 30 = 360 \end{aligned} \nonumber
Students always want to know how we would represent more than two IVs in a Punnett's Square, and the answer is that we don't. We create two Punnett Squares.
Let's say that we were looking at time spend studying for those with different mindsets and who have different jobs for different kinds of schools (community colleges or universities). That could look like:
IV Levels IV2 (Job): Full-Time IV1 (Mindset): Growth IV1 (Mindset): Fixed DV = Minutes spent studying MATH DV = Minutes spent studying MATH DV = Minutes spent studying DV = Minutes spent studying DV = Minutes spent studying DV = Minutes spent studying
AND:
IV Levels IV2 (Job): Full-Time IV1 (Mindset): Growth IV1 (Mindset): Fixed DV = Minutes spent studying MATH DV = Minutes spent studying MATH DV = Minutes spent studying DV = Minutes spent studying DV = Minutes spent studying DV = Minutes spent studying
You could have just as easily made IV1 the Job and IV2 the Mindset, or even made a table for only students with Growth Mindset (and had IV1 be the type of school) and another table for only students with Fixed Mindset and the table would still be correct. It doesn't matter statistically which IV is placed where, it's more about interpreting and understanding what is besting tested.
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# Carson Daly Under The Microscope (01/16/2020)
How will Carson Daly perform on 01/16/2020 and the days ahead? Let’s use astrology to perform a simple analysis. Note this is of questionable accuracy – don’t get too worked up about the result. I will first calculate the destiny number for Carson Daly, and then something similar to the life path number, which we will calculate for today (01/16/2020). By comparing the difference of these two numbers, we may have an indication of how smoothly their day will go, at least according to some astrology enthusiasts.
PATH NUMBER FOR 01/16/2020: We will consider the month (01), the day (16) and the year (2020), turn each of these 3 numbers into 1 number, and add them together. Here’s how it works. First, for the month, we take the current month of 01 and add the digits together: 0 + 1 = 1 (super simple). Then do the day: from 16 we do 1 + 6 = 7. Now finally, the year of 2020: 2 + 0 + 2 + 0 = 4. Now we have our three numbers, which we can add together: 1 + 7 + 4 = 12. This still isn’t a single-digit number, so we will add its digits together again: 1 + 2 = 3. Now we have a single-digit number: 3 is the path number for 01/16/2020.
DESTINY NUMBER FOR Carson Daly: The destiny number will consider the sum of all the letters in a name. Each letter is assigned a number per the below chart:
So for Carson Daly we have the letters C (3), a (1), r (9), s (1), o (6), n (5), D (4), a (1), l (3) and y (7). Adding all of that up (yes, this can get tedious) gives 40. This still isn’t a single-digit number, so we will add its digits together again: 4 + 0 = 4. Now we have a single-digit number: 4 is the destiny number for Carson Daly.
CONCLUSION: The difference between the path number for today (3) and destiny number for Carson Daly (4) is 1. That is smaller than the average difference between path numbers and destiny numbers (2.667), indicating that THIS IS A GOOD RESULT. But don’t get too excited yet! As mentioned earlier, this is not scientifically verified. If you want to see something that people really do vouch for, check out your cosmic energy profile here. Go ahead and see what it says for you – you’ll be glad you did.
### Abigale Lormen
Abigale is a Masters in Business Administration by education. After completing her post-graduation, Abigale jumped the journalism bandwagon as a freelance journalist. Soon after that she landed a job of reporter and has been climbing the news industry ladder ever since to reach the post of editor at Tallahasseescene.
#### Latest posts by Abigale Lormen (see all)
Abigale Lormen
Abigale is a Masters in Business Administration by education. After completing her post-graduation, Abigale jumped the journalism bandwagon as a freelance journalist. Soon after that she landed a job of reporter and has been climbing the news industry ladder ever since to reach the post of editor at Tallahasseescene.
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Difference between revisions of "1990 AIME Problems/Problem 11"
Problem
Someone observed that $6! = 8 \cdot 9 \cdot 10$. Find the largest positive integer $n^{}_{}$ for which $n^{}_{}!$ can be expressed as the product of $n - 3_{}^{}$ consecutive positive integers.
Solution 1
The product of $n - 3$ consecutive integers can be written as $\frac{(n - 3 + a)!}{a!}$ for some integer $a$. Thus, $n! = \frac{(n - 3 + a)!}{a!}$, from which it becomes evident that $a \ge 3$. Since $(n - 3 + a)! > n!$, we can rewrite this as $\frac{n!(n+1)(n+2) \ldots (n-3+a)}{a!} = n! \Longrightarrow (n+1)(n+2) \ldots (n-3+a) = a!$. For $a = 4$, we get $n + 1 = 4!$ so $n = 23$. For greater values of $a$, we need to find the product of $a-3$ consecutive integers that equals $a!$. $n$ can be approximated as $^{a-3}\sqrt{a!}$, which decreases as $a$ increases. Thus, $n = 23$ is the greatest possible value to satisfy the given conditions.
Solution 2
Let the largest of the $n-3$ consecutive positive integers be $k$. Clearly $k$ cannot be less than or equal to $n$, else the product of $n-3$ consecutive positive integers will be less than $n!$.
Key observation: Now for $n$ to be maximum the smallest number (or starting number) of the $n-3$ consecutive positive integers must be minimum, implying that $k$ needs to be minimum. But the least $k > n$ is $n+1$.
So the $n-3$ consecutive positive integers are $5, 6, 7…, n+1$
So we have $\frac{(n+1)!}{4!} = n!$ $\Longrightarrow n+1 = 24$ $\Longrightarrow n = 23$
Kris17
Generalization:
Largest positive integer $n$ for which $n!$ can be expressed as the product of $n-a$ consecutive positive integers is $(a+1)! - 1$
For ex. largest $n$ such that product of $n-6$ consecutive positive integers is equal to $n!$ is $7!-1 = 5039$
Proof: Reasoning the same way as above, let the largest of the $n-a$ consecutive positive integers be $k$. Clearly $k$ cannot be less than or equal to $n$, else the product of $n-a$ consecutive positive integers will be less than $n!$.
Now, observe that for $n$ to be maximum the smallest number (or starting number) of the $n-a$ consecutive positive integers must be minimum, implying that $k$ needs to be minimum. But the least $k > n$ is $n+1$.
So the $n-a$ consecutive positive integers are $a+2, a+3, … n+1$
So we have $\frac{(n+1)!}{(a+1)!} = n!$ $\Longrightarrow n+1 = (a+1)!$ $\Longrightarrow n = (a+1)! -1$
Kris17
See also
1990 AIME (Problems • Answer Key • Resources) Preceded byProblem 10 Followed byProblem 12 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
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Jump to a New ChapterIntroduction to the SAT IIContent and Format of the SAT II Math IICStrategies for SAT II Math IICMath IIC FundamentalsAlgebraPlane GeometrySolid GeometryCoordinate GeometryTrigonometryFunctionsStatisticsMiscellaneous MathPractice Tests Are Your Best Friends
10.1 Characteristics of a Function 10.2 Evaluating Functions 10.3 Compound Functions 10.4 Inverse Functions 10.5 Domain and Range
10.6 Graphing Functions 10.7 Identifying the Graphs of Polynomial Functions 10.8 Review Questions 10.9 Explanations
Evaluating Functions
Evaluating a function simply means finding f(x) at some specific value x. The Math IIC will likely ask you to evaluate a function at some particular constant. Take a look at the following example:
If f(x) = x2 – 3, what is f(5)?
Evaluating a function at a constant involves nothing more than substituting the constant into the definition of the function. In this case, substitute 5 for x:
It’s as simple as that.
The Math IIC may also ask questions in which you are asked to evaluate a function at a variable rather than a constant. For example:
If f(x) = , what is f(x + 1)?
To solve problems of this sort, follow the same method you did for evaluating a function at a constant: substitute the variable into the equation. To solve the sample question, substitute (x + 1) for x in the definition of the function:
Operations on Functions
Functions can be added, subtracted, multiplied, and divided just like any other quantities. There are a few rules that help make these operations easier. For any two functions f(x) and g(x):
Rule Example Addition If f(x) = sin x, and g(x) = cos x:(f + g)(x) = sin x + cos x Subtraction If f(x) = x2 + 5 and g(x) = x2 + 2x + 1:(f – g)(x) = x2 + 5 – x2 – 2x – 1 = –2x + 4 Multiplication If f(x) = x and g(x) = x3 + 8:(f g)(x) = x (x3 + 8) = x4 + 8x Division If f(x) = 2 cos x, and g(x) = 2 sin2 x:
As usual, when dividing, you have to be aware of possible situations where you inadvertently divide by zero. Since division by zero is not allowed, you should just remember that any time you are dividing functions, like f(x) /g(x), the resulting function is undefined wherever the function in the denominator equals zero.
Jump to a New ChapterIntroduction to the SAT IIContent and Format of the SAT II Math IICStrategies for SAT II Math IICMath IIC FundamentalsAlgebraPlane GeometrySolid GeometryCoordinate GeometryTrigonometryFunctionsStatisticsMiscellaneous MathPractice Tests Are Your Best Friends
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MAT 221 Lecture_Unit-1
# MAT 221 Lecture_Unit-1 - Chapter 1 Introduction to...
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Chapter 1 Introduction to Probability Definition 1.0.1. An experiment is a procedure we perform (quite often hypothetical) that produces some result. Definition 1.0.2. An outcome is a possible result of an experiment. Definition 1.0.3. An event is a certain set of outcomes of an experiment. Definition 1.0.4. The sample space is the collection or set of all possible outcomes of an experiment. The letter S is used to designate the sample space, which is the universal set of outcomes of an experiment. A sample space is called discrete if it is a finite or a countably infinite set. It is called continuous or a continuum otherwise. Example 1.0.5. Experiment: Tossing two coins. Sample Space: S = { HH, HT, TH, TT } . Event=at least one Head; E = { HH, HT, TH } . Definition 1.0.6. Two events are mutually exclusive if they cannot occur at the same time. Example 1.0.7. S = { H, T } , E 1 = { H } , E 2 = { T } . Then events E 1 and E 2 are mutually exclusive. In this case, E 1 E 2 = . Definition 1.0.8 ( Classical definition of Probability ) . The probability of an event equals the ratio of its favorable outcomes to the total number of outcomes provided that all outcomes are equally likely. Definition 1.0.9. The probability of event E is given by P ( E ) = n ( E ) n ( S ) , where n ( S ) < and n ( · ) denotes the number of elements. Example 1.0.10. Find the probability of obtaining at least one head when two coins are tossed. 1
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CHAPTER 1. INTRODUCTION TO PROBABILITY Definition 1.0.11 ( Axioms of Probability ) . Let S be the sample space, E be an event. We assign to each event E a number P ( E ), which we call the probability of the event E . This number is so chosen as to satisfy the following three conditions: Axiom 1: 0 P ( E ) 1. Axiom 2: P ( S ) = 1. Axiom 3: P ( E F ) = P ( E ) + P ( F ), if events E and F are mutually exclusive. Example 1.0.12. Show that P ( n i =1 E i ) = n summationdisplay i =1 P ( E i ), where E i E j = when i negationslash = j . Example 1.0.13. Find the probability of rolling an even number when a dice is rolled. Proposition 1.0.14. Let E and F be events and E c be the compliment of E . Then 1. P ( E c ) = 1 P ( E ) . 2. If E F , then P ( E ) P ( F ) . 3. P ( E F ) = P ( E ) + P ( F ) P ( E F ) . 4. P ( i =1 E i ) = summationdisplay i =1 P ( E i ) , where E 1 , E 2 , . . . are any sequence of mutually exclusive events (i.e., events for which E i E j = when i negationslash = j ).
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Volume of a Cone
The volume of a right circular cone is one-third of the volume of a right circular cylinder of the same base and same height.
$\therefore$ If $h$ is the height of the cone and $r$ is the radius of the base, then
${\text{Volume }} = {\text{ }}\frac{1}{3}{\text{ }} \times {\text{ area of the base }} \times {\text{ altitude}}$
$\therefore$ $V = \frac{1}{3}\pi {r^2}h$ (as the area of the base $= \pi {r^2}$)
Rule: The volume of a cone equals the area of the base times one-third the altitude.
Example:
The circumference of the base of a $9$m high conical tent is $44$m. Find the volume of the air contained in it.
Solution:
Circumference of the base $= 2\pi r = 44$m
$\therefore$ $2 \times \frac{{22}}{7} \times r = 44$
$r = \frac{{44 \times 7}}{{44}} = 7$m
$\because$ height of the conical tent $= 9$m
$\therefore$ volume of air $= \frac{1}{3}\pi {r^2}h$
$= \frac{1}{3} \times \frac{{22}}{7} \times 7 \times 9 = 462$ cubic m
Example:
The vertical height of a conical tent is $42$ dm and the diameter of its base is $5.4$ m. Find the number of people it can accommodate if each person is to be allowed $2916$ cubic dm of space.
Solution:
Here height $h = 42$ dm
Diameter $= 5.4$ m $= 54$ dm
Radius $r = 27$ dm
Volume $= \frac{1}{3}\pi {r^2}h$
$= \frac{1}{3} \times \frac{{22}}{7} \times 27 \times 27 \times 42$
$= 32076$ Cubic dm
Space allowed for $1$ person $= 2916$ cubic dm
$\therefore$ the required number of people $= \frac{{32076}}{{2916}} = 11$ people
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StatlectThe Digital Textbook
# Covariance
Covariance is a measure of association between two random variables. It is positive if the deviations of the two variables from their respective means tend to have the same sign and negative if the deviations tend to have opposite signs.
The covariance between two random variables and , denoted by , is defined as follows:provided the above expected values exist and are well-defined.
## Interpretation
To understand the meaning of covariance, let us analyze how it is constructed. It is the expected value of the product , where and are defined as follows:i.e. and are the deviations of and from their respective means.
When is positive, it means that:
• either and are both above their respective means;
• or and are both below their respective means.
On the contrary, when is negative, it means that:
• either is above its mean and is below its mean;
• or is below its mean and is above its mean.
In other words, when is positive, and are concordant (their deviations from the mean have the same sign); when is negative, and are discordant (their deviations from the mean have opposite signs). Sincea positive covariance means that on average and are concordant; on the contrary, a negative covariance means that on average and are discordant.
Thus, the covariance of and provides a measure of the degree to which and tend to "move together": a positive covariance indicates that the deviations of and from their respective means tend to have the same sign; a negative covariance indicates that deviations of and from their respective means tend to have opposite signs. Intuitively, we could express the concept as follows:
When , and do not display any of the above two tendencies.
## A covariance formula
The following covariance formula is often used to compute the covariance between two random variables:
Proof
First expand the product:Then, by linearity of the expected value:
This formula also makes clear that the covariance exists and is well-defined only as long as , and exist and are well-defined.
## Example
The following example shows how to compute the covariance between two discrete random variables.
Example Let be a random vector and denote its components by and . Let the support of be: and its joint probability mass function be:The support of is:and its marginal probability mass function is:The expected value of is:The support of is:and its marginal probability mass function is:The expected value of is:Using the transformation theorem, we can compute the expected value of :Hence, the covariance between and is:
## More details
The following subsections contain more details on covariance.
### Covariance of a random variable with itself
Let be a random variable, then:
Proof
It descends from the definition of variance:
### Symmetry
The covariance operator is symmetric:
Proof
Using the definition of covariance:
### Variance of the sum of two random variables
Let and be two random variables. Then the variance of their sum is:
Proof
The above formula is derived as follows:
Thus, to compute the variance of the sum of two random variables we need to know their covariance.
Obviously then, the formula:holds only when and have zero covariance.
The formula for the variance of a sum of two random variables can be generalized to sums of more than two random variables (see variance of the sum of n random variables).
### Bilinearity of the covariance operator
The covariance operator is linear in both of its arguments. Let , and be three random variables and let and be two constants. Then, the first argument is linear:
Proof
This is proved using the linearity of the expected value:
By symmetry, also the second argument is linear:
Linearity in both the first and second argument is called bilinearity.
By iteratively applying the above arguments, one can prove that bilinearity holds also for linear combinations of more than two variables:
### Variance of the sum of n random variables
The variance of the sum of random variables is:
Proof
This is demonstrated using the bilinearity of the covariance operator (see above):
This formula implies that when all the random variables in the sum have zero covariance with each other, then the variance of the sum is just the sum of the variances:This is true, for example, when the random variables in the sum are mutually independent (because independence implies zero covariance).
## Solved exercises
Below you can find some exercises with explained solutions:
1. Exercise set 1 (covariance between discrete random variables).
2. Exercise set 2 (covariance between absolutely continuous random variables).
The book
Most learning materials found on this website are now available in a traditional textbook format.
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## How do you find the angles in a quadrilateral triangle?
Subtract the sum of the angles from 180 degrees to get the missing angle. For example if a triangle in a quadrilateral had the angles of 30 and 50 degrees, you would have a third angle equal to 100 degrees (180 – 80 = 100).
## How many angles does a quadrilateral triangle have?
four angles
A quadrilateral is a polygon that has exactly four sides. (This also means that a quadrilateral has exactly four vertices, and exactly four angles.) Discussions of 2-D shapes sometimes refer only to the boundary (the line segments that form the edges of the figure) or to the interior as well.
What is quadrilateral triangle formula?
Area of General Quadrilateral Formula = 1/2 x diagonals length x ( sum of the height of two triangles ).
Do all quadrilaterals angles add up to 360?
The sum of the interior angles of any quadrilateral is 360°. Consider the two examples below. You could draw many quadrilaterals such as these and carefully measure the four angles. You would find that for every quadrilateral, the sum of the interior angles will always be 360°.
### What do angles in a quadrilateral add up to?
360°
Quadrilaterals are composed of two triangles. Seeing as we know the sum of the interior angles of a triangle is 180°, it follows that the sum of the interior angles of a quadrilateral is 360°.
### What is the sum of all angles of a quadrilateral?
According to the angle sum property of a Quadrilateral, the sum of all the four interior angles is 360 degrees.
Do quadrilaterals have 5 angles?
Defining a Quadrilateral Since it is a polygon, you know that it is a two-dimensional figure made up of straight sides. A quadrilateral also has four angles formed by its four sides.
Do all quadrilaterals have 3 sides and 3 angles?
One, two, three, four. These are all quadrilaterals. They all have four sides, four vertices, and, clearly, four angles. One angle, two angles, three angles, and four angles.
#### What are triangles and quadrilaterals?
A triangle is a closed figure with three straight sides and three angles. A quadrilateral has four straight sides and four angles.
#### Do quadrilaterals equal to 360?
A quadrilateral is a polygon which has 4 vertices and 4 sides enclosing 4 angles and the sum of all the angles is 360°. When we draw a draw the diagonals to the quadrilateral, it forms two triangles. Both these triangles have an angle sum of 180°. Therefore, the total angle sum of the quadrilateral is 360°.
What is the sum of three angles of a quadrilateral?
The sum of all angle of quadrilateral is 360° . The fourth angle is greater than 90°. Then, it is an obtuse angle. ∴ The measure of its fourth angle is an obtuse angle.
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All the solutions provided in McGraw Hill My Math Grade 4 Answer Key PDF Chapter 3 Review will give you a clear idea of the concepts.
Vocabulary Check
Write the letter of each definition on the line next to the correct vocabulary word.
Concept Check
Write a fact family for each set of numbers.
Question 14.
3, 7, 21 ___ ____ ___ ____
3 Ă— 7 = 21
7 Ă— 3 = 21
21 Ă· 3 = 7
21 Ă· 7 = 3
They are all divisible by 3.
Question 15.
9, 5, 45 ___ ____ ____ _____
9 Ă— 5 = 45
5 Ă— 9 = 45
45 Ă· 5 = 9
45 Ă· 9 = 5
Use repeated subtraction to divide.
Question 16.
42 Ă· 7 = ____
42 Ă· 7 = 6
42 – 7 = 35 – 7 = 28 – 7 = 21 – 7 = 14 – 7 = 7 – 7 = 0
Question 17.
56 Ă· 8 = ____
56 Ă· 8 = 7
56 – 8 = 48 – 8 = 40 – 8 = 32 – 8 = 24 – 8 = 16 – 8 = 8 – 8 = 0
Question 18.
36 Ă· 9 = ____
36 Ă· 9 = 4
36 – 9 = 27 – 9 = 18 – 9 = 9 – 9 = 0
Question 19.
Use multiplication to complete the number sentence.
5 times as many
5 times 3 is equal to 15.
5 Ă— 3 = 15
Identify the property or rule shown by each equation.
Question 20.
6 Ă— 8 = 8 Ă— 6
The commutative property is a math rule that says that the order in which we multiply numbers does not change the product.
So, the property for 6 Ă— 8 = 8 Ă— 6 is commutative property.
Question 21.
(3 Ă— 2) Ă— 6 = 3 Ă— (2 Ă— 6)
The associative property is a math rule that says that the way in which factors are grouped in a multiplication problem does not change the product.
So, the property for (3 Ă— 2) Ă— 6 = 3 Ă— (2 Ă— 6) is associative property.
Find the factors of each number.
Question 22.
16
1 Ă— 16 = 16
2 Ă— 8 = 16
4 Ă— 4 = 16
8 Ă— 2 = 16
16 Ă— 1 = 16
The factors of 16 are 1, 2, 4, 8, 16.
Question 23.
18
1 Ă— 18 = 18
2 Ă— 9 = 18
3 Ă— 6 = 18
6 Ă— 3 = 18
9 Ă— 2 = 18
18 Ă— 1 = 18
The factors of 18 are 1, 2, 3, 6, 9 and 18.
Question 24.
15
1 Ă— 15 = 15
3 Ă— 5 = 15
The factors of 15 are 1, 3, 5, and 15.
List the first five multiples.
Question 25.
2
The first five multiples of 2 are 2, 4, 6, 8 and 10.
Question 26.
10
The first five multiples of 10 are 10, 20, 30, 40 and 50.
Question 27.
12
The first five multiples of 12 are 12, 24, 36, 48 and 60.
Problem Solving
Question 28.
There are 8 cans of soup. There are 3 times as many cans of vegetables. How many cans of vegetables are there? Write an equation to find the unknown. Use a variable for the unknown.
There are 8 cans of soup. There are 3 times as many cans of vegetables.
8 Ă— 3 = 24
So, there will be 24 cans of vegetables.
Question 29.
Claire has 15 green beads, 8 blue beads, and 4 yellow beads. If she puts them onto 3 strings equally, how many beads are on each string?
15 + 8 + 4 = 27
27/3 = 9
Thus there are 9 beads on each string.
Question 30.
Stefanie and Eva want to share the shells that they collected on their trip to the beach. They have 18 shells in all. Use related facts and draw an array that will help them decide how they can evenly divide their shells.
They can each get ___ shells.
18 Ă· 2 = 9
Thus they can get 9 shells each.
Question 31.
If 7 people each ride a roller coaster 5 times, and it is $2 per person per ride, what is the total price they paid for the rides? Answer: 7 people each ride a roller coaster 5 times, and it is$2 per person per ride
7 Ă— 5 = 35
35 Ă— $2 =$70
Thus the total price they paid for the rides is \$70.
Test Practice
Question 32.
Marina has scored 9 points on each of 11 quizzes. How many points has she scored in all?
A. 9 points
B. 20 points
C. 90 points
D. 99 points
Given,
Marina has scored 9 points on each of 11 quizzes.
9 Ă— 11 = 99 points
Option D is the correct answer.
Reflect
Use what you learned about multiplication and division to complete the graphic organizer.
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## Word Problems with Fractions
Can't listen to audio right now? Turn on captions.
### Transcript
In this video, we're going to talk about word problems with fractions. The very first rule that we'll talk about is as a general rule when translating words to math, the word is means equals and the word of means multiply. This is a very important guide and this will help us translate many word problems involving fractions into math that we can do. For example, very simple question.
This is a little simpler then you'd see on the test. What is three fifths of 400? So this might actually be something that would be part of a larger problem on the test, but we'll just treat it as its own thing right now. Three fifths of 400. The of means multiply, so this just mean three fifths times 400.
We write 400 as a fraction, we notice that we can cancel. Once we've canceled them we can multiply and then we get the answer. Another question this is starting to get a little more test like. Bills monthly cable bill is two sevenths of his monthly rent if he pays \$300 on cable each month what is his monthly rent? So this is interesting.
The first thing I'll do is I'll introduce some letters to stand for these things. C equals cable, R equals rent. Now that first sentence, Cable is two-sevenths of rent, that means cable equals two-sevenths times R. Now once I have this equation, I'll just substitute the value I have. The cable is actually \$300.
Now, we want to get the R by itself, so I have to multiply by the reciprocal of that fraction. The reciprocal of two-sevenths would be seven-halves. So I multiply both sides by seven-halves. Cancel, and then I just complete the multiplication. For this one, now this one is starting to get to be like something that actually might be on the test.
Cathy's salary is 3/7 of Nora's salary. And is five-fourths of Teresa's salary. Nora's salary is what fraction of Teresa's salary? So, this is a very confusing question. So, first thing I'm gonna say is, we'll represent each one of the salaries by the first letter of the female's name.
So C is three-sevenths of N. Also, C is five-fourths of T. So we can translate that first sentence like this. Now notice that the question is asking us to relate Nora and Teresa's salary. So really, C is irrelevant. Really that part of the equation drops out.
And we just want to compare Teresa and Nora. And we want Nora's salary is what fraction of. So we want Nora by herself, and we want a fraction times Teresa. So to get Nora by herself, we have to multiply by the reciprocal of three sevenths that would be multiplying by seven thirds. That would get N by itself.
We just multiply and we get the fraction of thirty five over twelve. In most word problems, of means multiply and is means equals.
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# Lesson video
In progress...
Hi, my name's Mrs. Dunnett.
In this lesson, we're going to be solving equations involving a function of X.
Here we're given a function of X is equal to five X, and we're going to use this to solve the fine equations.
So I'm part A.
We've got a function of X is equal to 35.
We know that our function of X is five X, so we can write five X equals 35, and we then use, inverse operations to solve this equation to find the value of X.
Just like you would do with any normal linear equation.
So we're going to divide both sides by five and we get X is equal to seven.
Let's have a look at part B.
This time we've got function of X is equal to negative 10.
So again, we got five X equals negative 10.
So our function of X is equal to negative 10.
We divided by five and we get X equals negative two.
Part C we've got our function of X is equal to 1.
6 and probably getting the hang of this now.
So we write five X equals 1.
6.
Use inverse operations to solve.
So we're going to divide by five and we get X is equal to 0.
32.
And finally, we've got function of X equal to three.
So we've got five X is equal to three, divide both sides by five and we get X is equal to three fifths.
Or you can write that answer as a decimal if you want.
So the answer is X is equal to three fifths.
Here's some equations for you to solve.
Pause the video, to complete the questions and restart when you have finished.
All we had to do here was divided by three to get the value for X in each question.
We've now got a different function of X.
This time, our function of X is equal to five X plus two.
And we're going to solve the following equation.
So for our first equation, we've got function of X is equal to 32.
So we're at five X plus two equals 32.
And we use inverse operations to solve this.
So the first thing we need to do, is subtract two from both sides.
We get five X equals 30, and then we divide both sides by five to find X is equal to six.
In the next question we've got function of X is equal to 18.
So we're at five X plus two equals 18.
Again, use inverse operation, subtract in two and divided by five, to get X is equal to 16 over five.
Now we can leave this answer as a fraction.
We could change it into a mixed number, if you wanted, or even a decimal, it's entirely up to you.
Just make sure you do it very accurately, if you're going to change it, into a decimal.
And next up we've got function of X is equal to negative eight.
So we write five X plus two equals negative eight.
We're going to subtract two from both sides.
Negative eight, take away two is negative ten.
So we're moving further away from zero.
And then we divide by five and we get X is equal to negative two.
And finally, we've got a function of X is equal to five.
So for this question, we do the same again.
We write our function of X equal to five, and we solve it, using inverse operations.
So we subtract two, divided by five and we get an answer of X is equal to 0.
6, or we could write that as a fraction if we wanted to.
So that will be X is equal to three fifths.
Here's some equations for you to solve.
Pause the video, complete the questions and restart when you have finished.
This time you should have been using the inverse operations, add three and divide by five to solve.
Now we're going to look at a quadratic equation.
So we've got our function of X this time has got a squared term in it, and we're going to solve the following equations.
So for the first one, we want our function of X to equal 12.
So the function of X is three X squared, and we put that equal to 12 and we then solve it using inverse operations.
You have to be very careful here to remember, that if we multiply X by itself, X squared, we then times it my three.
So we're doing the inverse of that.
So the first thing we're going to do is we're going to divide by three.
And we get X squared is equal to four, and then we square root four and we get X is equal to positive or negative two.
So we're getting two solutions there because we've got a quadratic equation.
For the next question, we've got F of X is equal to 1.
92.
So we write our function of X, three X squared equal to 1.
92, and we use inverse operations to solve.
So I divide by three, to get X squared is equal to 0.
64.
Square root it.
And we get X is equal to positive or negative 0.
8.
And we could leave that as a fraction if we wanted to.
But as the question is given in decimals, it makes more sense to leave our answer as X equals a plus or minus 0.
8.
Here are some questions for you to try.
Pause the video to complete the task and restart when you've finished.
Notice that there are two solutions for each question as our equations are quadratic.
So we get a positive and a negative solution.
Our next function of X involves a fraction.
So we've got our function of X is equal to three X take away nine, all divided by two.
And we want to solve this equation when function of X is equal to 18.
So we write our function of X equal to 18.
Just as we've been doing with all the previous examples.
And then we need to think about, how we're going to solve this equation.
So hopefully you've got some experience of solving fractional equations prior to this lesson.
And you'll see, the first thing you want to do really is to get rid of that, dividing by two.
So we're going to multiply both sides by two.
And that gets rid of that fraction on the left hand side of the equation, and we get 36 on the right hand side of the equation.
And then we add nine to both sides and divide by three.
And we get X is equal to 15.
Let's have a look at these two functions now.
So this time we've got a function of X and a function G of X.
So you can see we've got two different functions here, and we want to solve the equation, F of X equals G of X.
So we have to equate these two functions.
So we put 10X minus four is equal to eight minus five X, and then we start to solve this equation to find the value of X.
So the first thing I'm going to do is add five X to both sides.
You could of course, add four to both sides, if you wish, it doesn't really matter here.
But we'll start off by adding five X to both sides.
And we get 15X tech where four is eight.
Add four to each side of the equation and we get 15X equals 12, and then we divide by 15 and we get X is equal to four fifths when simplified.
Here are two final questions for you to try.
Pause the video to complete each one and restart when you have finished.
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# What is a Parallelogram – Definition, Properties & Examples
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The word ‘parallelogram’ is derived from the Greek word ‘parallelogrammon’ which means “bounded by parallel lines”. A parallelogram is a quadrilateral that is bounded by parallel lines. It is a 2D shape in which the opposite sides are parallel and equal.
There are three types of parallelograms – square, rectangle, and rhombus, and each of them has its own unique properties.
Let’s understand what is a parallelogram and the properties of parallelogram.
## What is a Parallelogram?
A parallelogram is a special type of quadrilateral that is formed by parallel lines. In a parallelogram, both pairs of opposite sides are parallel and equal. Hence, a parallelogram is defined as a quadrilateral in which both pairs of opposite sides are parallel and equal.
The above figure shows three types of parallelograms.
• Rectangle
• Square
• Rhombus
## Properties of a Parallelogram
The following are the basic properties of parallelograms that help you identify them.
• The opposite sides of a parallelogram are parallel. Here, $\text{AB} ‖ \text{CD}$ and $\text{BC} ‖ \text{DA}$.
• The opposite sides of a parallelogram are equal. In the above figure, $\text{AB} = \text{CD}$ and $\text{BC} = \text{DA}$.
• The opposite angles of a parallelogram are equal. Here, $\angle \text{A} = \angle \text{C}$ and $\angle \text{B} = \angle \text{D}$
• The diagonals of a parallelogram bisect each other. Here, $\text{AC} = \text{BD}$
• Adjacent angles are supplementary. In the above figure, $\angle \text{A} + \angle \text{B} = 180^{\circ}$, $\angle \text{B} + \angle \text{C} = 180^{\circ}$, $\angle \text{C} + \angle \text{D} = 180^{\circ}$ and $\angle \text{D} + \angle \text{A} = 180^{\circ}$.
• The diagonals divide the parallelogram into two congruent triangles. Here, $\triangle \text{ABD} \cong \triangle \text{BCD}$, and $\triangle \text{ABC} \cong \triangle \text{ACD}$.
### Diagonals of a Parallelogram Are Equal
The diagonals of a parallelogram divide it into two congruent triangles, i.e., in a parallelogram $\text{ABCD}$, $\triangle \text{ABD} \cong \triangle \text{BCD}$, and $\triangle \text{ABC} \cong \triangle \text{ACD}$.
Let’s see how to prove the above statement.
Since $\text{ABCD}$ is a parallelogram, the opposite sides are equal.
Therefore, $\text{AB} = \text{CD}$ and $\text{BC} = \text{DA}$.
Now, in $\triangle \text{ABD}$ and $\triangle \text{CBD}$
$\text{AB} = \text{CD}$ (Opposite sides of a parallelogram)
$\text{DA} = \text{BC}$ (Opposite sides of a parallelogram)
$\text{BD} = \text{BD}$ ( common)
Thus, $\triangle \text{ABD} \cong \triangle \text{CBD}$ (SSS congruency criterion).
Therefore, we can say that the diagonal of a parallelogram divides it into two congruent triangles.
### Opposite Sides of a Parallelogram Are Equal
The opposite sides of a parallelogram are equal, i.e., in a parallelogram $text{ABCD}$, $\text{AB} = \text{CD}$, and $\text{BC} = \text{DA}$.
Let’s see how to prove the above statement.
In the above figure, $\text{ABCD}$ is a parallelogram, and $\text{AC}$ is one of the diagonals.
The diagonal $\text{AC}$ divides parallelogram $\text{ABCD}$ into two triangles, namely, $\triangle \text{ABC}$ and $\triangle \text{ABC}$.
In order to prove that opposite sides are equal i.e., $\text{AB} = \text{CD}$, and $\text{BC} = \text{DA}$, we need to first prove that $\triangle \text{ABC} \cong \triangle \text{ABC}$
In $\triangle \text{ABC}$ and $\triangle \text{CDA}$, $\text{BC} || \text{DA}$ and $\text{AC}$ is a transversal.
So, $\angle \text{BCA} = \angle \text{DAC}$ (Pair of alternate angles)
And $\text{AC} = \text{CA}$ (common)
Thus, $\triangle \text{ABC} \cong \triangle \text{CDA}$ (ASA rule).
Therefore, the corresponding parts $\text{AB} = \text{CD}$ and $\text{DA}= \text{BC}$.
### Opposite Angles of a Parallelogram Are Equal
In a parallelogram the opposite angles are equal.
Let’s see how to prove the above statement, i.e, in a parallelogram $\text{ABCD}$, $\angle \text{A} = \angle \text{C}$, and $\angle \text{B} = \angle \text{D}$.
In the above figure, in a parallelogram $\text{ABCD}$, $\text{AB} || \text{CD}$ and $\text{AD} || \text{BC}$.
Consider $\triangle \text{ABC}$ and $\triangle \text{ADC}$
$\text{AC} = \text{AC}$ (common side)
We know that alternate interior angles are equal.
$\angle \text{BAC} = \angle \text{ACD}$
$\angle \text{BCA} = \angle \text{CAD}$
Therefore, $\triangle \text{ABC} \cong \triangle \text{ADC}$
Hence, $\angle \text{A} = \angle \text{C}$, and $\angle \text{B} = \angle \text{D}$. (Corresponding Parts of Congruent Triangles).
### Adjacent Angles of a Parallelogram Are Supplementary
In a parallelogram the adjacent angles are supplementary.
Let’s see how to prove the above statement, i.e, in a parallelogram $\text{ABCD}$, $\angle \text{A} + \angle \text{B} = 180^{\circ}$, and $\angle \text{C} + \angle \text{D} = 180^{\circ}$.
In the above figure, $\text{AB} ∥ \text{CD}$ and $\text{AD}$ is a transversal.
We know that interior angles on the same side of a transversal are supplementary.
Therefore, $\angle \text{A} + \angle \text{D} = 180^{\circ}$
Similarly, $\angle \text{B} + \angle \text{C} = 180^{\circ}$, $\angle \text{C} + \angle \text{D} = 180^{\circ}$ and $\angle \text{A} + \angle \text{B} = 180^{\circ}$.
Therefore, the sum of any two adjacent angles of a parallelogram is equal to $180^{\circ}$.
## Practice Problems
1. Define parallelogram.
2. State True or False
• Adjacent sides of a parallelogram are equal
• Adjacent sides of a parallelogram are parallel
• Opposite sides of a parallelogram are equal
• Opposite sides of a parallelogram are parallel
• Adjacent angles of a parallelogram are equal
• Adjacent angles of a parallelogram are supplementary
• Opposite angles of a parallelogram are equal
• Opposite angles of a parallelogram are supplementary
## FAQs
### What is a parallelogram in geometry?
In geometry, a parallelogram is a quadrilateral (4-sided 2D shape) that has its opposite sides parallel and equal in length.
### Are all the angles of a parallelogram equal?
No, all the angles of a parallelogram are not equal. Only the opposite angles of a parallelogram are equal, whereas the adjacent angles of a parallelogram are supplementary, i.e., their sum is $180^{\circ}$.
### What is the difference between a parallelogram and a quadrilateral?
All parallelograms are quadrilaterals but all quadrilaterals are not necessarily parallelograms. For example, a trapezium is a quadrilateral, but not a parallelogram. For a quadrilateral to be a parallelogram, all the opposite sides must be parallel and equal to each other.
### Is a rhombus a parallelogram?
Yes, a rhombus is a parallelogram in which the opposite sides are parallel and the opposite angles are congruent. Apart from this, all the sides of a rhombus are equal and the diagonals bisect each other at right angles.
### Is a trapezium a parallelogram?
No, a trapezium is not a parallelogram, since all opposite sides of the trapezium are not parallel to each other. A trapezium has only one pair of opposite sides parallel to each other. Also, a trapezium doesn’t have opposite sides equal to each other. Hence, it is a quadrilateral but not a parallelogram.
## Conclusion
A parallelogram is a special type of quadrilateral that is formed by parallel lines. In a parallelogram, both pairs of opposite sides are parallel and equal. A parallelogram has certain unique properties that distinguish it from the other quadrilaterals. The three types of parallelograms are square, rectangle, and rhombus.
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# Exponential Properties Involving Products
## Add exponents to multiply exponents by other exponents
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Practice Exponential Properties Involving Products
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Recognize and Apply the Power of a Product Property
Have you ever tried to square a monomial? Do you know how to do it? Take a look at this dilemma.
A square platform has a side length of 6a2\begin{align*}6a^2\end{align*}.
How can we find the area of the platform?
This Concept will show you how to use the Power of a Product with monomials. Then you will be able to find the area of the square platform.
### Guidance
When multiplying monomials, an exponent is applied to the constant, variable, or quantity that is directly to its left. However, we only applied exponents to single variables.
Exponents can also be applied to products using parentheses.
Look at this one.
(5x)4\begin{align*}(5x)^4\end{align*}
If we apply the exponent 4 to whatever is directly to its left, we would apply it to the parentheses, not just the x\begin{align*}x\end{align*}. The parentheses are directly to the left of the 4. This indicates that the entire product in the parentheses is taken to the 4th\begin{align*}4^{th}\end{align*} power. We can also write this in expanded form.
(5x)4=(5x)(5x)(5x)(5x)
Now we multiply the monomials as we have already learned—by placing like factors next to each other, multiplying the coefficients, and simplifying using exponents.
=5555xxxx=625x4
This is the Power of a Product Property which says, for any nonzero numbers a\begin{align*}a\end{align*} and b\begin{align*}b\end{align*} and any integer n\begin{align*}n\end{align*}
(ab)n=anbn\begin{align*}(ab)^n=a^n b^n\end{align*}
Here is another one.
(7h)3=(7h)(7h)(7h)=777hhh=343h3
You can see that whether we have positive or negative integers or both, we can still use the Power of a Product Property. You may have already noticed a pattern with the exponents and the final product. When you multiply like bases, there is another shortcut—you can add the exponents of like bases. Another way of saying it is:
aman=am+n
Take a look at this one.
(2x4)5=(2x4)(2x4)(2x4)(2x4)(2x4)=22222x4x4x4x4x4=22222x4+4+4+4+4=32x20
Write the definition of this property and one problem down in your notebook.
Simplify each monomial.
#### Example A
(6x3)2\begin{align*}(6x^3)^2\end{align*}
Solution: 36x6\begin{align*}36x^6\end{align*}
#### Example B
(2x3y3)3\begin{align*}(2x^3y^3)^3\end{align*}
Solution: 8x9y9\begin{align*}8x^9y^9\end{align*}
#### Example C
(3x2y2z)4\begin{align*}(-3x^2y^2z)^4\end{align*}
Solution: 81x8y8z4\begin{align*}81x^8y^8z^4\end{align*}
Now let's go back to the dilemma from the beginning of the Concept.
Here is the side length of the square platform.
6a2\begin{align*}6a^2\end{align*}
We want to find the area of the platform. To figure out the area, we will use the following formula.
A=s2\begin{align*}A = s^2\end{align*}
Now we substitute the side length into the formula.
A=(6a2)2\begin{align*}A = (6a^2)^2\end{align*}
Next, we can square the monomial.
36a4\begin{align*}36a^4\end{align*}
### Vocabulary
Monomial
a single term of variables, coefficients and powers.
Coefficient
the number part of a monomial or term.
Variable
the letter part of a term
Exponent
the little number, the power, that tells you how many times to multiply the base by itself.
Base
the number that is impacted by the exponent.
Expanded Form
write out all of the multiplication without an exponent.
Power of a Product Property
(ab)n=an(bn)\begin{align*}(ab)^n=a^n(b^n)\end{align*}
### Guided Practice
Here is one for you to try on your own.
(2x4)5\begin{align*}(-2x^4)^5\end{align*}
Solution
(2x4)5=(2x4)(2x4)(2x4)(2x4)(2x4)=(2xxxx)(2xxxx)(2xxxx)(2xxxx)=22222xxxxxxxxxxxxxxxxxxxx=32x20
### Practice
Directions: Simplify.
1. (6x5)2\begin{align*}(6x^5)^2\end{align*}
2. (13d5)2\begin{align*}(-13d^5)^2\end{align*}
3. (3p3q4)3\begin{align*}(-3p^3 q^4)^3\end{align*}
4. (10xy2)4\begin{align*}(10xy^2)^4\end{align*}
5. (4t3)5\begin{align*}(-4t^3)^5\end{align*}
6. (18r2s3)2\begin{align*}(18 r^2 s^3)^2\end{align*}
7. (2r11s3t2)3\begin{align*}(2r^{11}s^3 t^2)^3\end{align*}
8. (7x2)2\begin{align*}(7x^2)^2\end{align*}
9. (2y2)3\begin{align*}(2y^2)^3\end{align*}
10. (5x2)3\begin{align*}(5x^2)^3\end{align*}
11. (12y3)2\begin{align*}(12y^3)^2\end{align*}
12. (5x5)5\begin{align*}(5x^5)^5\end{align*}
13. (2x2y2z)3\begin{align*}(2x^2y^2z)^3\end{align*}
14. (3x4y3z2)3\begin{align*}(3x^4y^3z^2)^3\end{align*}
15. (5x4y3z3)3\begin{align*}(-5x^4y^3z^3)^3\end{align*}
### Vocabulary Language: English
Base
Base
When a value is raised to a power, the value is referred to as the base, and the power is called the exponent. In the expression $32^4$, 32 is the base, and 4 is the exponent.
Coefficient
Coefficient
A coefficient is the number in front of a variable.
Expanded Form
Expanded Form
Expanded form refers to a base and an exponent written as repeated multiplication.
Exponent
Exponent
Exponents are used to describe the number of times that a term is multiplied by itself.
Monomial
Monomial
A monomial is an expression made up of only one term.
Power
Power
The "power" refers to the value of the exponent. For example, $3^4$ is "three to the fourth power".
Product of Powers Property
Product of Powers Property
The product of powers property states that $a^m \cdot a^n = a^{m+n}$.
Variable
Variable
A variable is a symbol used to represent an unknown or changing quantity. The most common variables are a, b, x, y, m, and n.
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# Compute $\lim_{n\to\infty} \left(\int_0^{\pi} \frac{\sin^2 n x}{\sin x} \ dx-\sum_{k=1}^n \frac{1}{k}\right)$
Compute the limit $$\lim_{n\to\infty} \left(\int_0^{\pi} \frac{\sin^2 n x}{\sin x} \ dx-\sum_{k=1}^n \frac{1}{k}\right)$$
-
Lemma: $\int_{0}^{\pi} e^{(2n+1)ix} dx = \frac{-2}{i}(2n+1)$.
Now let $t=e^{ix}$. We'll simplify the integrand: $$\frac{\sin^2(nx)}{\sin x} = (\frac{t^{n}-t^{-n}}{2i})^2 (\frac{t-t^{-1}}{2i})^{-1}$$ $$=\frac{i}{-2} \frac{t^{2n+1}+t^{1-2n}-2t}{t^2-1} = \frac{i}{-2} t \frac{t^{2n}-1}{t^2-1}(1-t^{-2n})$$ $$=\frac{i}{-2} \sum_{i=0}^{n-1} (t^{2i+1} - t^{2(i-n)+1})$$
Now, by applying the lemma, we find that the integral is $$\frac{i}{-2}\frac{-2}{i} \sum_{i=0}^{n-1} \frac{1}{2i+1} - \frac{1}{2(i-n)+1}=$$ $$=2 \sum_{i=0}^{n-1} \frac{1}{2i+1} =2(H_{2n}-\frac{1}{2}H_n)$$
So the limit in question is $\lim_{n \to \infty} 2(H_{2n}-\frac{1}{2}H_n) - H_{n} = \lim 2H_{2n}-2H_{n}$. Now we can either use $H_{n} \sim \log n + \gamma + O(n^{-1})$ or compare $H_{2n}-H_{n}$ to the integral $\int_{n}^{2n} \frac{dt}{t} = \ln 2$ to conclude that the limit is $\ln 4$.
-
By relating your last relation $$\lim_{n \to \infty} 2H_{2n}-2H_{n}$$ to Catalan-Botez identity and using the Taylor series of $\ln(x+1)$, the answer is clearly $2 \ln 2$. Thanks! (+1) – OFFSHARING Dec 22 '12 at 17:27
@Chris'ssister There's also the integral approach, which is elaborated here: math.stackexchange.com/questions/155190/… – Ofir Dec 22 '12 at 17:31
yeah. That series is a well-known series and one may find lots of approaches. – OFFSHARING Dec 22 '12 at 17:33
Let's suppose that $\ \displaystyle f(n):=\int_0^{\pi} \frac{\sin^2 n x}{\sin x} \ dx\$ then : \begin{align} f(n+1)-f(n)&=\int_0^{\pi} \frac{\sin^2((n+1) x)-\sin^2(n x)}{\sin x} \ dx\\ &=\int_0^{\pi} \frac{\cos(2n x)-\cos(2(n+1) x)}{2\sin x} \ dx\\ &=\int_0^{\pi} \frac{\cos(2n x)(1-\cos(2x))+\sin(2nx)\sin(2x)}{2\sin x}dx\\ &=\int_0^{\pi} \frac{\cos(2n x)2\sin(x)^2)+\sin(2nx)2\sin(x)\cos(x)}{2\sin x} dx\\ &=\int_0^{\pi} \cos(2n x)\sin(x)+\sin(2nx)\cos(x)\;dx\\ &=\int_0^{\pi} \sin((2n+1) x)\;dx\\ &=\frac 2{2n+1} \\ \end{align}
So that your limit (as $\ (n+1)\to +\infty$) is the series : $$f(0)+\sum_{n=0}^\infty \left(\frac 2{2n+1}-\frac 1{n+1}\right)=2\sum_{n=0}^\infty \left(\frac 1{2n+1}-\frac 1{2n+2}\right)=2\,\log(1+1)=\log(4)$$ (using the expansion of $\;\log(1+x)\,$ at $\,x=1$)
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thank you for your solution! (+1) – OFFSHARING Dec 22 '12 at 17:22
Sorry for the errancies @Chris'ssister ... :-) – Raymond Manzoni Dec 22 '12 at 17:23
it's OK, and I'm always glad to receive your nice solutions! :) – OFFSHARING Dec 22 '12 at 17:58
Your questions are welcome too @Chris'ssister ! – Raymond Manzoni Dec 22 '12 at 18:03
The point is that you can write $\sum_{k=1}^nsin(2k-1)x=\frac{cos(2nx)-1}{-2sinx}=\frac{sin^2(nx)}{sinx}$, so the integral is equal to $I=lim_{n\to\infty}\sum_{k=1}^n(\int_{0}^\pi(sin(2k-1)xdx-\frac{1}{k})=lim_{n\to\infty}\sum_{k=1}^n(\frac{2}{2k-1}-\frac{1}{k})$, then using $\sum_{k=1}^{n}\frac{1}{k}\sim log(n)+c$, $c$ is the Euler constant, so $I=lim_{n\to\infty}2[\sum_{k=1}^{2n}\frac{1}{k}-\sum_{k=1}^n\frac{1}{2k}]-\sum_{k=1}^n\frac{1}{k}=lim_{n\to\infty}(2log2n+2c-logn-c-logn-c)=2log2$.
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I think that the answer is precisely log(4) since this is the answer of the series on the second line. A very good start! (+1) – OFFSHARING Dec 22 '12 at 17:12
@Chris'ssister, thanks fixed. – ougao Dec 22 '12 at 18:10
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# Area of a Triangle
If ∆ be the area of a triangle ABC, Proved that, ∆ = ½ bc sin A = ½ ca sin B = ½ ab sin C
That is,
(i) ∆ = ½ bc sin A
(ii) ∆ = ½ ca sin B
(iii) ∆ = ½ ab sin C
Proof:
(i) ∆ = ½ bc sin A
Let ABC is a triangle. Then the following three cases arise:
Case I: When the triangle ABC is acute-angled:
Now form the above diagram we have, sin C = AD/AC sin C = AD/b, [Since, AC = b] AD = b sin C ……………………….. (1) Therefore, ∆ = area of triangle ABC = 1/2 base × altitude
= ½ ∙ BC ∙ AD
= ½ ∙ a ∙ b sin C, [From (1)]
= ½ ab sin C
Case II: When the triangle ABC is obtuse-angled:
Now form the above diagram we have, sin (180° - C) = AD/AC sin C = AD/AC, [Since, sin (π - θ) = sin θ] sin C = AD/b, [Since, AC = b]AD = b sin C ……………………….. (2)Therefore, ∆ = area of the triangle ABC
= ½ base x altitude
= ½ ∙ BC ∙ AD
= ½ ∙ a ∙ b sin C, [From (1)]
= ½ ab sin C
Case III: When the triangle ABC is right-angled
Now form the above diagram we have, ∆ = area of triangle ABC = ½ base x altitude = ½ ∙ BC ∙ AD = ½ ∙ BC ∙ AC = ½ ∙ a ∙ b
= ½ ∙ a ∙ b ∙ 1, [Since, ∠C = 90°. Therefore, sin C = sin 90° = 1]
= ½ ab sin C
Therefore, in all three cases, we have ∆ = ½ ab sin C
In a similar manner we can prove the other results, (ii) ∆ = ½ ca sin B and (iii) ∆ = ½ ab sin C.
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# Texas Go Math Grade 4 Lesson 2.2 Answer Key Explore Decimal Place Value
Refer to our Texas Go Math Grade 4 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 4 Lesson 2.2 Answer Key Explore Decimal Place Value.
## Texas Go Math Grade 4 Lesson 2.2 Answer Key Explore Decimal Place Value
Essential Question
How can you find the value of a digit using its place-value position?
Place value is the value of a digit according to its position in the number such as ones, tens, hundreds, and so on.
Example: 5 in 3,458 represents 5 tens, or 50
Unlock the Problem
Connect Decimals, like whole numbers, can be written in standard form, word form, and expanded form.
How can you write the value of each digit in 5.76 using decimal expander strips?
Activity Explore expanded form.
Materials: decimal expander strips
STEP 1. Place the decimal expander strip in front of you, with the largest rectangle on the left-hand side.
STEP 2. Fold along the first two dashed lines as shown, keeping the largest rectangle toward the back.
STEP 3. Continue folding along the dashed lines. Then turn the paper around so that the largest rectangle is on the right-hand side.
STEP 4. Write 5, 7, 6, placing one digit in each section, as shown. Insert a decimal point before the 7.
STEP 5. Unfold the decimal expander, and use numbers and symbols to write 5 + 0.7 + 0.06.
STEP 6. Use the second strip of paper to make another decimal expander for 5.76. This time, write the word that names the value of each digit. The strip should read 5 ones 7 tenths 6 hundredths.
Math Talk
Mathematical Processes:
Explain how you can write 3.5 as 35 tenths.
The given number is: 3.5
Now,
3.5 = 35 × 0.1
= 35 × $$\frac{1}{10}$$
= 35 tenths
Hence, from the above,
We can conclude that
We can write 3.5 as 35 tenths using the fractions and place-value positions
Example Use a place-value chart.
Shortfin mako shark teeth range in length from 0.64 centimeters to 5.08 centimeters. Eli found a Shortfin mako shark tooth measuring 2.54 centimeters on the beach.
you can use a place-value chart to help you understand decimal place values.
Write the decimal above in the place-value chart.
What is the value of digit 4 in 2.54?
The value of the digit 4 is 4Â hundredths, or 0.04.
Math Talk
Mathematical Processes:
In the whole number 277, the value of each digit is 10 times as great as the place-value position to its right. Explain why, in the decimal 2.77, the value of the digit 7 in the tenth place is 10 times as great as the value of the digit 7 in the hundredth place.
It is given that
In the whole number 277, the value of each digit is 10 times as great as the place-value position to its right
Now,
We know that,
The place-value positions for the given number is:
Hence, from the above,
We can conclude that
In the decimal 2.77, the value of the digit 7 in the tenth place is 10 times as great as the value of the digit 7 in the hundredth place due to the rules present in the decimal number system
Share and Show
Question 1.
What is the place-value position of the digit 8 in 0.98? ____________
The given number is: 0.98
Now,
The given place-value chart for 0.98 is:
Now,
From the above place-value chart,
We can observe that
The digit 8 in 0.98 present in “Hundredths” place-value
Hence, from the above,
We can conclude that
The place-value position of the digit 8 in 0.98 is: 0.08
Write the value of the underlined digit.
Go Math 4th Grade Lesson 2.2 Explore Decimal Place Value Question 2.
2.1
The given number is: 2.1
Now,
The representation of the place-value chart for the given number is:
Hence, from the above,
We can conclude that
The value of the underlined digit in the given number is: 0.1
Question 3.
0.09
The given number is: 0.09
Now,
The representation of the place-value chart for the given number is:
Hence, from the above,
We can conclude that
The value of the underlined digit in the given number is: 0.09
Question 4.
6.54
The given number is: 6.54
Now,
The representation of the place-value chart for the given number is:
Hence, from the above,
We can conclude that
The value of the underlined digit in the given number is: 0.04
Question 5.
0.3
The given number is: 0.3
Now,
The representation of the place-value chart for the given number is:
Hence, from the above,
We can conclude that
The value of the underlined digit in the given number is: 0.3
Write the number in two other forms.
Question 6.
3.0 + 0.9 + 0.02
The given Expanded form is:
3.0 + 0.9 + 0.02
Now,
We know that,
The “Word Form” is a form that contains only words but not numbers
The “Expanded form” is a way to write a number by showing the value of each digit
The “Standard Form” is a form that contains only numbers but not any words
So,
The representation of the Standard form for the given expanded form is: 3.92
The representation of the Word form for the given expanded form is:
Three and Ninety-two hundredths
Hence, from the above,
We can conclude that
The representation of the Standard form for the given expanded form is: 3.92
The representation of the Word form for the given expanded form is:
Three and Ninety-two hundredths
Tenth Decimal Place Go Math 4th Grade Lesson 2.2 Question 7.
seventeen hundredths
The given Word form is: Seventeen hundredths
Now,
We know that,
The “Word Form” is a form that contains only words but not numbers
The “Expanded form” is a way to write a number by showing the value of each digit
The “Standard Form” is a form that contains only numbers but not any words
So,
The representation of the Standard form for the given Word form is: 0.17
The representation of the Expanded form for the given Word form is:
0 + 0.1 + 0.07
Hence, from the above,
We can conclude that
The representation of the Standard form for the given Word form is: 0.17
The representation of the Expanded form for the given Word form is:
0 + 0.1 + 0.07
Problem Solving
Practice: Copy and Solve
Write the number in two other forms.
Question 8.
8.26
The given Standard form is: 8.26
Now,
We know that,
The “Word Form” is a form that contains only words but not numbers
The “Expanded form” is a way to write a number by showing the value of each digit
The “Standard Form” is a form that contains only numbers but not any words
So,
The representation of the Expanded form for the given Standard form is:
8 + 0.2 + 0.06
The representation of the Word form for the given Standard form is:
Eight and Twenty-six hundredths
Hence, from the above,
We can conclude that
The representation of the Expanded form for the given Standard form is:
8 + 0.2 + 0.06
The representation of the Word form for the given Standard form is:
Eight and Twenty-six hundredths
Question 9.
one and two tenths
The given Word form is:
One and two tenths
Now,
We know that,
The “Word Form” is a form that contains only words but not numbers
The “Expanded form” is a way to write a number by showing the value of each digit
The “Standard Form” is a form that contains only numbers but not any words
So,
The representation of the Expanded form for the given Word form is:
1 + 0.2
The representation of the Standard form for the given Word form is: 1.2
Hence, from the above,
We can conclude that
The representation of the Expanded form for the given Word form is:
1 + 0.2
The representation of the Standard form for the given Word form is: 1.2
Question 10.
10 + 6 + 0.7 + 0.02
The given Expanded form is:
10 + 6 + 0.7 + 0.02
Now,
We know that,
The “Word Form” is a form that contains only words but not numbers
The “Expanded form” is a way to write a number by showing the value of each digit
The “Standard Form” is a form that contains only numbers but not any words
So,
The representation of the Word form for the given Expanded form is:
Sixteen and Seventy-two hundredths
The representation of the Standard form for the given Expanded form is: 16.72
Hence, from the above,
We can conclude that
The representation of the Word form for the given Expanded form is:
Sixteen and Seventy-two hundredths
The representation of the Standard form for the given Expanded form is: 16.72
Lesson 2.2 4th Grade Go Math Decimal Place Value Question 12.
Explain how you write the number 7.04 in expanded form.
The given number is: 7.04
Now,
We know that,
The “Word Form” is a form that contains only words but not numbers
The “Expanded form” is a way to write a number by showing the value of each digit
The “Standard Form” is a form that contains only numbers but not any words
Now,
From the given number,
We can observe that
The given number is in Standard form
So,
The representation of the Expanded form for the given Standard form is:
7 + 0 + 0.04
Hence, from the above,
We can conclude that
The representation of the Expanded form for 7.04 is:
7 + 0 + 0.04
Use the chart for 13-14.
Question 13.
A marine researcher recorded the lengths of some sharks he observed offshore. The length of which shark has the digit 7 in the tenths place?
It is given that
A marine researcher recorded the lengths of some sharks he observed offshore
Now,
The given chart is:
Now,
The representation of the place-value table is:
Now,
From the above chart,
According to the positions of the above place-value table,
We can observe that
The length of “Blackfin” has the digit 7 in the tenth place
Hence, from the above,
We can conclude that
The length of “Blackfin shark” has the digit 7 in the tenth place
Question 14.
Multi-Step What’s the Error?
Randy said that the black nose shark is one and twenty-seven tenths meters in length. Describe Randy’s error and write the decimal in word form.
It is given that
Randy said that the black nose shark is one and twenty-seven tenths meters in length
Now,
The given chart is:
Now,
The representation of the place-value table is:
Now,
From the above chart,
We can observe that
The length of a Blacknose Shark is: 1.27 meters
Now,
We know that,
The “Word Form” is a form that contains only words but not numbers
The “Expanded form” is a way to write a number by showing the value of each digit
The “Standard Form” is a form that contains only numbers but not any words
So,
1.27 is in the Standard form
Now,
The representation of the Word form for the given Standard form is:
One and Twenty-seven hundredths
But,
According to Randy,
The representation of the Word form for the given Standard form is:
One and Twenty-seven-tenths
So,
The error is:
Since after the decimal point there are 2 numbers, the last digit must have the place-value of hundredths and after the decimal point, if there is only 1 number, then the last digit must have the place-value of a tenth
Hence, from the above,
We can conclude that
The error of Randy is:
Since after the decimal point there are 2 numbers, the last digit must have the place-value of hundredths and after the decimal point, if there is only 1 number, then the last digit must have the place-value of a tenth
The representation of the Word form for 1.27 is:
One and Twenty-seven tenths
Go Math Lesson 2.2 4th Grade Place Value Question 15.
H.O.T.
What’s the Question? The answer is 275 hundredths.
Question 16.
An adult kangaroo jumps 7.6 meters in a single leap. What is 7.6 written in word form?
(A) seven and one-sixth
(B) seven and six tenths
(C) seven and six hundredths
(D) seventy-six hundredths
It is given that
An adult kangaroo jumps 7.6 meters in a single leap
Now,
We know that,
The “Word Form” is a form that contains only words but not numbers
The “Expanded form” is a way to write a number by showing the value of each digit
The “Standard Form” is a form that contains only numbers but not any words
So,
The representation of 7.6 in the word form is:
Seven and Six tenths
Hence, from the above,
We can conclude that
The representation of 7.6 in the Word form is:
Question 17.
During a blizzard, a city received 13.6 inches of snow. What is 13.6 written in expanded form?
(A) 10 + 0.3 + 0.06
(B) 10 + 3 + 0.06
(C) 10 + 3 + 0.6
(D) 10 + 3 + 6
It is given that
During a blizzard, a city received 13.6 inches of snow
Now,
We know that,
The “Word Form” is a form that contains only words but not numbers
The “Expanded form” is a way to write a number by showing the value of each digit
The “Standard Form” is a form that contains only numbers but not any words
So,
The representation of 13.6 in the Expanded form is:
10 + 3 + 0.6
Hence, from the above,
We can conclude that
The representation of 13.6 in the Expanded form is:
Question 18.
A runner finished a race in 9.84 seconds. What is the value of digit 4 in 9.84?
(A) 4 hundredths
(B) 4 ones
(C) 4 tenths
(D) 4 tens
It is given that
A runner finished a race in 9.84 seconds
Now,
The representation of the place-value chart for the given number is:
Hence, from the above,
We can conclude that
The value of digit 4 in 9.84 is:
TEXAS Test Prep
Question 19.
The average annual precipitation in Philadelphia, Pennsylvania, is 42.05 inches. What is the value of the digit 5 in 42.05?
(A) five tens
(B) five ones
(C) five-tenths
(D) five-hundredths
It is given that
The average annual precipitation in Philadelphia, Pennsylvania, is 42.05 inches
Now,
The representation of 42.05 in the place-value chart is:
Hence, from the above,
We can conclude that
The value of digit 5 in 42.05 is:
### Texas Go Math Grade 4 Lesson 2.1 Homework and Practice Answer Key
Write the number in two other forms.
Question 1.
6.7
The given Standard form is: 6.7
Now,
We know that,
The “Word Form” is a form that contains only words but not numbers
The “Expanded form” is a way to write a number by showing the value of each digit
The “Standard Form” is a form that contains only numbers but not any words
So,
The representation of the Word form for the given standard form is:
Six and Seven tenths
The representation of the Expanded form for the given Standard form is:
6 + 0.7
Hence, from the above,
We can conclude that
The representation of the Word form for the given standard form is:
Six and Seven tenths
The representation of the Expanded form for the given Standard form is:
6 + 0.7
Question 2.
one and twelve hundredths
The given Word form is:
One and twelve hundredths
Now,
We know that,
The “Word Form” is a form that contains only words but not numbers
The “Expanded form” is a way to write a number by showing the value of each digit
The “Standard Form” is a form that contains only numbers but not any words
So,
The representation of the Standard form for the given Word form is: 1.12
The representation of the Expanded form for the given Word form is:
1 + 0.1 + 0.02
Hence, from the above,
We can conclude that
The representation of the Standard form for the given Word form is: 1.12
The representation of the Expanded form for the given Word form is:
1 + 0.1 + 0.02
Question 3.
80 + 7 + 0.9 + 0.03
The given Expanded form is:
80 + 7 + 0.9 + 0.03
Now,
We know that,
The “Word Form” is a form that contains only words but not numbers
The “Expanded form” is a way to write a number by showing the value of each digit
The “Standard Form” is a form that contains only numbers but not any words
So,
The representation of the Standard form for the given Expanded form is: 87.93
The representation of the Word form for the given Expanded form is:
Eighty-seven and Ninety-three hundredths
Hence, from the above,
We can conclude that
The representation of the Standard form for the given Expanded form is: 87.93
The representation of the Word form for the given Expanded form is:
Eighty-seven and Ninety-three hundredths
Question 4.
53.02
The given Standard form is: 53.02
Now,
We know that,
The “Word Form” is a form that contains only words but not numbers
The “Expanded form” is a way to write a number by showing the value of each digit
The “Standard Form” is a form that contains only numbers but not any words
So,
The representation of the Expanded form for the given Standard form is:
50 + 3 + 0 + 0.02
The representation of the Word form for the given Standard form is:
Fifty-three and Two hundredths
Hence, from the above,
We can conclude that
The representation of the Expanded form for the given Standard form is:
50 + 3 + 0 + 0.02
The representation of the Word form for the given Standard form is:
Fifty-three and Two hundredths
Write the value of the underlined digit.
Question 5.
3.6
The given number is: 3.6
Now,
The representation of the place-value chart for the given number is:
Hence, from the above,
We can conclude that
The value of digit 6 in the given number is: 0.6
Question 6.
0.04
The given number is: 0.04
Now,
The representation of the place-value chart for the given number is:
Hence, from the above,
We can conclude that
The value of digit 4 in the given number is: 0.04
4th Grade Decimal Place Value Lesson 2.2 Question 7.
9.05
The given number is: 9.05
Now,
The representation of the place-value chart for the given number is:
Hence, from the above,
We can conclude that
The value of digit 0 in the given number is: 0
Question 8.
2.81
The given number is: 2.81
Now,
The representation of the place-value chart for the given number is:
Hence, from the above,
We can conclude that
The value of digit 1 in the given number is: 0.01
Problem Solving
Use the chart for 9-10.
Question 9.
Jordan recorded the distances he walked this week. On which day did he walk two and twenty-five hundredths kilometers?
It is given that
Jordan recorded the distances he walked this week
Now,
The given chart is:
Now,
The representation of the place-value table is:
Now,
The representation of the given Word form in the standard form is: 2.25
Now,
From the above chart,
According to the place-value table,
We can observe that
On Friday,
Jordan walked 2.25 kilometers
Hence, from the above,
We can conclude that
On Friday, Jordan walked two and twenty-five hundredth kilometers
Question 10.
Leah said Jordan walked 3 and 5 tenths kilometers on Wednesday. What error did she make?
It is given that
Leah said Jordan walked 3 and 5 tenths kilometers on Wednesday
Now,
The given chart is:
Now,
From the given chart,
We can observe that
Jordan walked 3.05 kilometers on Wednesday
But,
According to Leah,
Jordan walked 3 and 5 tenths kilometers on Wednesday
So,
The representation of three and five-tenths in the standard form is: 3.5
So,
The error is:
Since after the decimal point there are 2 numbers, the last digit must have the place-value of hundredths and after the decimal point, if there is only 1 number, then the last digit must have the place-value of a tenth
Hence, from the above,
We can conclude that
The error made by Leah is:
Since after the decimal point there are 2 numbers, the last digit must have the place-value of hundredths and after the decimal point, if there is only 1 number, then the last digit must have the place-value of a tenth
Lesson Check
Question 11.
What is 8.07 written in word form?
(A) eight and seven-tenths
(B) eighty-seven tenths
(C) eight and seven hundredths
(D) eighty-seven hundredths
The given Standard form is: 8.07
Now,
We know that,
The “Word Form” is a form that contains only words but not numbers
The “Expanded form” is a way to write a number by showing the value of each digit
The “Standard Form” is a form that contains only numbers but not any words
So,
The representation of 8.07 in the Word form is:
Eight and Seven hundredths
Hence, from the above,
We can conclude that
The representation of 8.07 in the Word form is:
Question 12.
What is the value of the underlined digit in 37.08?
(A) 8 tenths
(B) 8 ones
(C) 8 tens
(D) 8 hundredths.
The given number is: 37.08
Now,
The representation of the given number in the place-value chart is:
Hence, from the above,
We can conclude that
The value of digit 8 in 37.08 is:
Question 13.
What is the expanded form for 70.26?
(A) 7 + 2 + 0.6
(B) 70 + 2 + 0.6
(C) 7 + 0.2 + 0.06
(D) 70 + 0.2 + 0.06
The given Standard form is: 70.26
Now,
We know that,
The “Word Form” is a form that contains only words but not numbers
The “Expanded form” is a way to write a number by showing the value of each digit
The “Standard Form” is a form that contains only numbers but not any words
So,
The representation of 70.26 in the Expanded form is:
70 + 0 + 0.2 + 0.06
Hence, from the above,
We can conclude that
The Expanded form for 70.26 is:
Question 14.
What is thirty-seven and nine-tenths written in expanded form?
(A) 30 + 7 + 0.9
(B) 3 + 0.7 + 0.09
(C) 30 + 7 + 0.09
(D) 37 + 0.9
The given Word form is:
Thirty-seven and nine-tenths
Now,
We know that,
The “Word Form” is a form that contains only words but not numbers
The “Expanded form” is a way to write a number by showing the value of each digit
The “Standard Form” is a form that contains only numbers but not any words
So,
The representation of the given Word form in the Expanded form is:
30 + 7 + 0.9
Hence, from the above,
We can conclude that
The representation of Thirty-seven and nine-tenths in the Expanded form is:
Question 15.
What is the place value of 9 in 42.96?
(A) tenths
(B) ones
(C) tens
(D) hundredths
The given number is: 42.96
Now,
The representation of 42.96 in the place-value chart is:
Hence, from the above,
We can conclude that
The place-value of 9 in 42.96 is:
Question 16.
Matthias walked around the track for 9.28 minutes. What is the value of digit 2 in 9.28?
(A) 2 ones
(B) 2 tenths
(C) 2 hundred
(D) 2 hundredths
It is given that
Matthias walked around the track for 9.28 minutes
Now,
The representation of 9.28 in the place-value chart is:
Hence, from the above,
We can conclude that
The value of digit 2 in 9.28 is:
Question 17.
Multi-Step During a rainstorm, Dallas received 1.25 inches of rain and Arlington received 3.25 inches. which statement is true about 1.25 and 3.25?
(A) The value of 2 is 2 tenths.
(B) The value of 1 is 1 hundred.
(C) The value of 5 is 5 tenths.
(D) The value of 2 is 2 tens.
It is given that
During a rainstorm, Dallas received 1.25 inches of rain and Arlington received 3.25 inches.
Now,
The representation of 1.25 in the place-value chart is:
Now,
The representation of 3.25 in the place-value chart is:
Hence, from the above,
We can conclude that
The statement that is true about 1.25 and 3.25 is:
Question 18.
Multi-Step What is the value of the digit to the right of the underlined digit in 76.56?
(A) 6 hundred
(B) 5 tenths
(C) 6 hundredths
(D) 6 ones
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## Tag Archives: application of basic calculus
### What is a catenary or a chain curve? A nice little application of basic calculus
Let us continue our exploration of basic calculus and its application. As you will discover, the invention of calculus is a triumph of the human intellect, it is a fountain head of many ideas in pure mathematics as well as mine of applications to sciences and engineering and even economics and humanities!
Hanging cables
Imagine a cable, like a telephone line or TV cable, strung from one support to another and hanging freely. The cable’s weight per unit length is w and horizontal tension at its lowest part is a vector of length H. If we choose a coordinate system for the plane of the cable in which the x-axis is horizontal, the force of gravity is straight down, the positive y-axis points straight up, and the lowest point of the cable lies at the point $y=H/w$ on the y-axis. (Fig 1), then it can be shown that the cable lies along the graph of the hyperbolic cosine
$y=(H/w)\cosh (wx/H)$.
Such a curve is sometimes called a chain curve or a catenary, the latter deriving from the Latin catena meaning ‘chain’.
a) Let $P(x,y)$ denote an arbitrary point on the cable. Fig 2 displays the tension at P as a vector of length (magnitude) T, as well as the tension at H at the lowest point A. Show that the cable’s slope at P is
$\tan \phi =dy/dx=\sinh (wx/H)$
b) Using the result from part (a) and the fact that the tension at P must equal H (the cable is not moving or swinging), show that $T=wy$. This means that the magnitude of the tension at $P(x,y)$ is exactly equal to the weight of y units of cable.
2) (Continuation of above problem). The length of arc AP in Fig 2 is
$s=(1/a)\sinh (ax)$ where $a=w/H$. Show that the coordinates of P may be expressed in terms of s as
$x=(1/a) (\sinh)^{-1} (as)$ and $y=\sqrt (s^{2}+1/a^{2})$.
3) The sag and horizontal tension in a cable. The ends of a cable 32 feet long and weighing 2 pounds per foot are fastened at the same level in posts 30 feet apart.
i) Model the cable with the equation
$y=(1/a)\cosh (ax)$ for $-15 \leq x \leq 15$
Use information from problem 2 above to show that a satisfies the equation
$16a=\sinh 15a$
ii) Estimate the horizontal tension in the cable at the cable’s lowest point.
I hope you enjoy a lot 🙂
More later…
Nalin Pithwa
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# Activities to Teach Students About Fractions of a Group
Teaching fractions to students can be a challenging task, especially when dealing with fractions of a group. However, it is critical to ensure that learners have a solid foundation in this area as fractions are a fundamental part of mathematics that students need to learn. Fortunately, with the right activities, teachers can make learning fractions of a group an enjoyable experience for learners. In this article, we explore some activities to teach students about fractions of a group.
1. Fraction Pizza
Fraction pizza is an exciting game that teachers can use to introduce students to basic fractions. Assemble a set of pizza slice cards, each with various fractions and a corresponding numerator and denominator. Divide students into groups of two or three and have them work together to create a whole pizza from the slice cards. After forming the pizza, students can discuss how they divided the pizza into fractions and figure out how many slices each person will get.
2. Wagon Wheels
Wagon wheels is another engaging activity that will teach students about fractions of a group. Ask students to work in pairs or small groups to create a container of wagon wheels, such as crackers or cookies. After, they should count the total number of wagon wheels contained within the container, then work together to divide the cookies into equal groups. After the dividing process, they will understand how fractions work.
3. Fraction Action
Fraction action is a fun, interactive game that encourages teamwork and helps students learn about fractions of a group. Divide the students into teams of two or three and give each team a basket of manipulatives, such as buttons, blocks, or pompoms. Set a timer for two minutes and ask the students to come up with as many different ways to group their manipulatives as possible, such as three-quarters, four-fifths, or six-eighths. Award points for each correct answer, and the team with the most points at the end of the game wins.
4. Fraction Frenzy
Fraction frenzy is a fast-paced activity that will get students up and moving as they learn fractions of a group. First, students need to arrange themselves in groups of four. You need to set a timer for one minute and ask students to brainstorm as many picture examples of fractions as they can. Once the time is up, each team should line up and present their pictures to the group. The other teams should then choose the pictures that demonstrate the most significant understanding of fractions of a group.
In conclusion, fractions of a group can be challenging for students to learn, but with the right activities, students will learn and retain the concept. Teachers need to remember that hands-on activities and games will help motivate students and improve their understanding of fractions. By incorporating fun, interactive activities into your lesson plans, you can ensure that your students gain a solid foundation in fractions of a group and ultimately develop a love for math!
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# constant function equation
By on Dec 30, 2020 in Uncategorized | 0 comments
x). If f(x) = 0 for all x, then what must be true? The differential equation of the function y = ax? . Create an account to start this course today. y) is not dependent on the input variable (e.g. We've learned that a constant function is a function that always has the same value no matter what our input is. For functions between preordered sets, constant functions are both order-preserving and order-reversing; conversely, if f is both order-preserving and order-reversing, and if the domain of f is a lattice, then f must be constant. The independent variable x does not appear on the right side of the function expression and so its value is "vacuously substituted". If the answer is yes, then you don't have a constant function. Main Concept. How Long Does IT Take To Get A PhD IN Nursing? This is a constant function. (a) What are the constant solutions of the equation? No matter what we plug in for x, we get y = 0. Definition of constant function in the Definitions.net dictionary. Services. Namely y(0) = 2, y(���2.7) = 2, y(��) = 2, and so on. The constant functions cut through the vertical axis in the value of the constant and they are parallel to the horizontal axis (and therefore they do not cut through it). You may like to read some of the things you can do with lines: [7] This is often written: We see this because it is not a horizontal line. Meaning of constant function. Second-order linear equations with constant coefficients are very important, especially for applications in mechanical and electrical engineering (as we will see). 8 + b where 'a' and 'b' are constant is: a) y' = 2x y" b) y' = a y" c) y' = 2 y" d) Ex y" | 2 Now consider the function y = 7. What the various symbols mean. In the equation "7x2 - 3x + 6," the number 6 is a constant, whereas 7 is the multiplying coefficient and. credit-by-exam regardless of age or education level. Alternatively, it may be calculated using the Arrhenius equation. Enrolling in a course lets you earn progress by passing quizzes and exams. Identity function, also called an identity relation, is a function that always returns the same value that was used as its argument. {{courseNav.course.mDynamicIntFields.lessonCount}} lessons This means there is no change in the y value, so the graph stays constantly on y = b, forming a horizontal line. x On many pages, both formats are shown. ( [5], In the context of a polynomial in one variable x, the non-zero constant function is a polynomial of degree 0 and its general form is f(x) = c where c is nonzero. A. Did you know… We have over 220 college - Definition & Examples, GED Math: Quantitative, Arithmetic & Algebraic Problem Solving, Introduction to Statistics: Help and Review, Introduction to Statistics: Tutoring Solution, Holt McDougal Larson Geometry: Online Textbook Help, Glencoe Math Connects: Online Textbook Help, Cambridge Pre-U Mathematics - Short Course: Practice & Study Guide, Cambridge Pre-U Mathematics: Practice & Study Guide, NY Regents Exam - Earth Science: Help and Review, High School Physics: Homework Help Resource, NY Regents Exam - Integrated Algebra: Test Prep & Practice, NY Regents Exam - Geometry: Test Prep & Practice, TExES Mathematics 7-12 (235): Practice & Study Guide, CLEP College Algebra: Study Guide & Test Prep. Find constants \displaystyle A and \displaystyle B such that the function \displaystyle y=A\sin x+B\cos x satisfies the differential equation \displaystyle y''+y'-2y=\sin x. In today���s lecture, we begin our study of the types of real-valued functions commonly used by scientists and This function, also denoted as ��� (), is called the "natural exponential function", or simply "the exponential function". Earn Transferable Credit & Get your Degree, Cubic Function: Definition, Formula & Examples, Greatest Integer Function: Definition & Examples, What is a Linear Function? If you have, lucky you for finding an awesome sale, but not only that, you actually came across a relationship between the items in that bin and their price that is an example of a constant function. We will become comfortable identifying constant functions through real-world and mathematical examples. Starting with the easy ones . ↦ Graphically speaking, a constant function, y = b, has a y-value of b everywhere. If x = 3, y = 7 or if x = 5, y = 7 ; y is always 7 no matter what our input is. An constant function is a function that always returns the same constant value. ���Inverse��� of Constant Function. For a power series, the constant term is what results from substituting x = 0 into the formula. Let's think about why this is the case. Another special type of linear function is the Constant Function ... it is a horizontal line: f(x) = C. No matter what value of "x", f(x) is always equal to some constant value. To unlock this lesson you must be a Study.com Member. Function: A function is a relation between a set of inputs and a set of permissible outputs. Sociology 110: Cultural Studies & Diversity in the U.S. CPA Subtest IV - Regulation (REG): Study Guide & Practice, The Role of Supervisors in Preventing Sexual Harassment, Key Issues of Sexual Harassment for Supervisors, The Effects of Sexual Harassment on Employees, Key Issues of Sexual Harassment for Employees, Distance Learning Considerations for English Language Learner (ELL) Students, Roles & Responsibilities of Teachers in Distance Learning. See original article A constant function is an even function, i.e. Substituting x = 0, we get: f(x) = 3 + 5(0) + 7(0)2 + 9(0)3 + 11(0)4 + ���. Efimov (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. and career path that can help you find the school that's right for you. . Mathematically speaking, a constant function is a function that has the same output value no matter what your input value is. This is the general second���order homogeneous linear equation with constant coefficients.. Theorem A above says that the general solution of this equation is the general linear combination of any two linearly independent solutions. To determine if something represents a constant function, ask yourself if you can get different outputs by varying your inputs. 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Given that constants a, b, and c are real numbers and f(x) = a, if x is less than 0, b, if x = 0, c, if x is greater than 0. {\displaystyle (x\mapsto c)'=0} [1][2][3] For example, the function y(x) = 4 is a constant function because the value of y(x) is 4 regardless of the input value x (see image). Have you ever been shopping and you see a bargain bin, like the one pictured, where everything in the bin is a set price? the graph of a constant function is symmetric with respect to the y-axis. ) Consider our example of y = 7. Being comfortable with this definition, and keeping this question in mind, we now have the ability to recognize constant functions and even create our own. To decide if a function is a constant function, ask yourself, is it possible to get different outputs by using different inputs? - Definition, Types & Examples, Life Skills and Guidance Resources for High School Students, Illinois Common Core Social Studies Standards. What does constant function mean? A 12.5- mu F capacitor is connected to a power supply that keeps a constant potential difference of 24.0 V across the plates. Therefore, this is not a constant function. To learn more, visit our Earning Credit Page. Confusion may arise however, as not all scientific calculators show the equation being equal to the answer in the formula display. C) The function representing the price of items in a dollar store (meaning everything costs a dollar). ′ Note that the value of f(x) is always k, independently of the value of x. The rate constant may be found experimentally, using the molar concentrations of the reactants and the order of reaction. The total amount of energy radiated per second by a perfectblackbody depends only on its temperature T andarea A: This relationship is called theStefan-Boltzmann Law. That is, the output value of the function at any input value in its domain is the same, independent of the input. Already registered? 0 = 3. © copyright 2003-2020 Study.com. Viewed 31 times 0 $\begingroup$ The ... so you could maybe try to find out the number of solutions to the equation by plotting the two sides of the equation. succeed. In both cases ��� classical format and wave format ��� all equations can be reduced to+ Read More This is because zero times any number is zero. Constant solutions In general, a solution to a di詮�erential equation is a function. Well, if y = 0 (that is, when y is the constant function zero), then y��� = 0 and the equation is reduced to 0 ��� 0 = 0 which is an expression that is always true. In this lesson, we will learn about what a constant function is and what it looks like on a graph. The input is any item in the store, and the output is the price of the item. This way, for instance, if we wanted to represent a quantity that stays constant over the course of time t, we would use a constant function f(t)=k, in which the variable tdoes not appear. c In this example, the answer is yes, because if I input x = 1, I get y = 1 + 4 or y = 5, and if I input x = 2, then I get y = 2 + 4 or y = 6. For example, think about the function y = x + 4. For example, all solutions to the equation y0 = 0 are constant. 100 C. 105 D. The answer is undefined, Working Scholars® Bringing Tuition-Free College to the Community. We first learn how to solve the homogeneous equation. Laura received her Master's degree in Pure Mathematics from Michigan State University. Select a subject to preview related courses: Furthermore, if we let B be our variable representing the books in the bin, and we let P be the variable representing the price, our function would be P = 3.99, and graphically, we would have a horizontal line through P = 3.99. Because a constant function does not change, its derivative is 0. V7. This represents a constant function. Since we can get different outputs by varying our inputs, this is not a constant function. The converse is also true. Since our output is always $1.00 no matter what our input is, this represents a constant function. Laplace���s Equation and Harmonic Functions In this section, we will show how Green���s theorem is closely connected with solutions to Laplace���s partial di詮�erential equation in two dimensions: (1) ���2w ���x2 ���2w ���y2 = 0, where w(x,y) is some unknown function of two variables, assumed to be twice di詮�erentiable. Can You Take The Accuplacer Test Online At Home? Common Core State Standards in New Mexico, Study.com's Workforce College Accelerator for Employees. No matter what book we take out of that bin, the corresponding cost is$3.99. For example, in the equation "6x - 4 = 8," both 4 and 8 are constants because their values are fixed. Tech and Engineering - Questions & Answers, Health and Medicine - Questions & Answers, True or false (give a reason for your answer) if y = ln 10 .then y'= fraction {1}{10}, A function y(t) satisfies the differential equation \frac{dy}{dt} = y^{4} - 7y^{3} + 10y^{2}. Over 83,000 lessons in all major subjects, {{courseNav.course.mDynamicIntFields.lessonCount}}, Compounding Functions and Graphing Functions of Functions, Understanding and Graphing the Inverse Function, Polynomial Functions: Properties and Factoring, Polynomial Functions: Exponentials and Simplifying, Exponentials, Logarithms & the Natural Log, Equation of a Line Using Point-Slope Formula, High School Algebra II: Homework Help Resource, Biological and Biomedical Can we get different outputs by varying our inputs? What is Planck���s Equation? You can test out of the As a real-valued function of a real-valued argument, a constant function has the general form y(x) = c or just y = c.[4], The graph of the constant function y = c is a horizontal line in the plane that passes through the point (0, c). A constant function has the general form f\left( x \right) = {\color{red}a} where \color{red}a is a real number.. From the general formula, the output of a constant function regardless of its input value (usually denoted by x), will always be the same which is ��� Learn more about the Definition and Properties of Constant and Identity Function for IIT JEE exam at Vedantu.com. Information and translations of constant function in the most comprehensive dictionary definitions resource on the web. This is a function of the type f(x)=k, where k is any real number. credit by exam that is accepted by over 1,500 colleges and universities. | {{course.flashcardSetCount}} In mathematics, a constant function is a function whose (output) value is the same for every input value. You may wonder what a constant function would look like on a graph. Here x is the variable and the derivatives are with respect to a second variable t. The letters a, b, c and d are taken to be constants here. A function on a connected set is locally constant if and only if it is constant. Not sure what college you want to attend yet? For example, y = 7 or y = 1,094 are constant functions. The codomain of this function is just {2}. In this unit we learn how to solve constant coefficient second order linear differential equations, ... where x h is the general solution to the homogeneous equation, and x p is any particular solution to the inhomogeneous equation. What is the value of j(x) if x = 100? It may be a number on its own or a letter that stands for a fixed number in an equation. Virtual Nerd's patent-pending tutorial system provides in-context information, hints, and links to supporting tutorials, synchronized with videos, each 3 to 7 minutes long. If a( x), b( x), and c( x) are actually constants, a( x) ��� a ��� 0, b( x) ��� b, c( x) ��� c, then the equation becomes simply. This article was adapted from an original article by A.V. Let's look at a few more examples to really solidify our understanding of this concept. = When we graph all these points, we see that we get horizontal lines (shown in red on the graph). For instance, (-2, 7), (0, 7), (7, 7), (1000, 7), and (-1000, 7) are all on this graph. For example, let���s say you have the following equation for a power series: f(x) = 3 + 5x + 7x2 + 9x3 + 11x4+ ���. Solution for A1. These unique features make Virtual Nerd a viable alternative to private tutoring. Constant Function Constant Function is defined as the real valued function $f : R \rightarrow R$ , y = f(x) = c for each $x \in R$ and c is a constant So ,this function basically associate each real number to a constant value It is a linear function where $f(x_1) =f(x_2)$ for all $x_1,x_2 \in R$ Domain and Range of the Constant Function There are nontrivial di詮�erential equations which have some constant solutions. A. a = b B. a = c, a is not equal to b C. a = b = c D. a is not equal to c, Consider the function j(x) = 5. We can write this type of function as: f(x) = c. Where: c is a ��� A function is also neither increasing nor decreasing at extrema. This equation would be described as a second order, linear differential equation with constant coefficients. Note that we have to speak of local extrema, because any given local extremum as defined here is not necessarily the highest maximum or lowest minimum in the function���s entire domain. Using Linear Equations. If y is a constant, then dy/dt = 0. Constant Functions. imaginable degree, area of These three roots are the constant solutions to the differential equation. If y is increasing (as a function ��� 3) Why do some of the functions in this workshop specify "const" for the array parameters and not in others? On the other hand, the polynomial f(x) = 0 is the identically zero function. Furthermore, it contains the point (3,2) and (2,7), so we see that we get different outputs by varying our inputs. However, the function could be a constant function. All other trademarks and copyrights are the property of their respective owners. flashcard set{{course.flashcardSetCoun > 1 ? f ( x 1) = f ( x 2) for any x 1 and x 2 in the domain. The general power series can be defined as: f(x) = c0 + c1x + c2x + c3x + c4x + ��� As you might be able to tell, the only constant on its own (i.e. A differential equation is an equation which contains the derivatives of a variable, such as the equation. 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Into Math Grade 6 Module 8 Lesson 4 Answer Key Interpret and Evaluate Algebraic Expressions
We included HMH Into Math Grade 6 Answer Key PDF Module 8 Lesson 4 Interpret and Evaluate Algebraic Expressions to make students experts in learning maths.
HMH Into Math Grade 6 Module 8 Lesson 4 Answer Key Interpret and Evaluate Algebraic Expressions
I Can interpret and evaluate algebraic expressions using the order of operations.
Step It Out
You can evaluate an expression with a variable by substituting a known value for the variable.
Question 1.
Evaluate each expression for the given value of the variable.
A. The area of a rug can be represented by the expression 6w.
Find the area when w = 4 feet.
Substitute 4 for w. 6()
Multiply.
When w = 4 ft, 6w = square feet.
Given,
w = 4 feet
Now we have to substitute 4 for w.
Area of the rectangle = l × w
A = 6 × 4
6 × 4 = 24
When w = 4 ft, 6w = 24 square feet.
B. Evaluate 18y when y = $$\frac{2}{3}$$.
Substitute $$\frac{2}{3}$$ for y. 18
Multiply.
When y = $$\frac{2}{3}$$, 18y = .
Given,
18y when y = $$\frac{2}{3}$$
18 × $$\frac{2}{3}$$ = 6 × 2 = 12
When y = $$\frac{2}{3}$$, 18y = 12
C. Evaluate x – 12 when x = 18.6.
Substitute 18.6 for x. – 12
Subtract.
When x = 18.6, x – 12 = .
x – 12 when x = 18.6
We have to substitute 18.6 for x.
18.6 – 12 = 6.6
x – 12 = 6.6
Turn and Talk For the expression in Part A. how is the value of the expression related to the value of the variable? Explain.
The shape of the rug is in the form of a rectangle.
By seeing the figure we can say that the width is less than the length of the rug.
Area of the rectangle = l × w
A = 6 × 4
6 × 4 = 24
When w = 4 ft, 6w = 24 square feet.
Sometimes expressions have more than one variable.
Question 2.
The perimeter of a rectangular swimming pool can be written as 2l + 2w, where l is the length and w is the width. Find the perimeter when l = 7.5 meters and w = 4.5 meters.
Given,
l = 7.5 meters and w = 4.5 meters
The perimeter of a rectangular swimming pool can be written as 2l + 2w,
where l is the length and w is the width.
Substitute 7.5 for l and 4.5 for w.
P = 2(7.5 + 4.5)
P = 2 (12)
P = 24
Thus when l = 7.5 and w = 4.5, 2l + 2w = 24 meters
Question 3.
Evaluate each expression for the given values of the variables.
A. Evaluate 4x3 – 3y when x = 2 and y = 10.
4x3 – 3y when x = 2 and y = 10
Substitute x = 2 and y = 10
4(2)3 – 3(10)
4(8) – 30
32 – 30 = 2
So, when x = 2 and y = 10, 4x3 – 3y = 2.
B. Evaluate 6s + $$\frac{d}{2}$$ when s = $$\frac{5}{12}$$ and d = 15.
6s + $$\frac{d}{2}$$
when s = $$\frac{5}{12}$$ and
d = 15
6 ($$\frac{5}{12}$$) + $$\frac{15}{2}$$
$$\frac{5}{2}$$ + $$\frac{15}{2}$$ = $$\frac{20}{2}$$ = 10
So, when s = $$\frac{5}{12}$$ and d = 15
6s + $$\frac{d}{2}$$ = 10
Turn and Talk How is evaluating algebraic expressions similar to evaluating numerical expressions? How is it different? Explain.
C. Evaluate 4x²y when x = 3 and y = 0.5.
Substitute 3 for x and 0.5 for y. 4()2 ()
Evaluate the exponent. 4()()
Multiply.
When x = and y = 0.5, 4x2y = .
4x²y when x = 3 and y = 0.5
Substitute 3 for x and 0.5 for y
4(3)².0.5 = 4(9) × 0.5
= 36 × 0.5
= 18
So, When x = 2 and y = 0.5, 4x2y = 18.
Question 4.
The expression 1 .8C + 32 gives the temperature in degrees Fahrenheit (°F) for a given temperature C in degrees Celsius (°C). Find the temperature in degrees Fahrenheit that is equivalent to 30 °C.
Substitute 30 for C. 1.8( ) + 32
Multiply. + 32
30°C = °F
Substitute 30 for C.
1.8( 30) + 32
54 + 32 = 86
30°C = 86 °F
Check Understanding
Question 1.
Evaluate the expression $$\frac{6x}{5}$$ when x = 20.
$$\frac{6x}{5}$$ when x = 20
$$\frac{6(20)}{5}$$ = $$\frac{120}{5}$$ = 60
So, when x = 20, $$\frac{6x}{5}$$ = 60
Question 2.
The expression $$\frac{k}{8}$$ shows the cost per person of splitting a restaurant bill k among 8 people. Evaluate the expression when k = 72 dollars.
Given,
The expression $$\frac{k}{8}$$ shows the cost per person of splitting a restaurant bill k among 8 people.
Substitute 72 for k.
$$\frac{k}{8}$$ = $$\frac{72}{8}$$ = 9
So, when k = 72 dollars, $$\frac{k}{8}$$ = 9
For Problems 3-6, evaluate the expression when x = 4 and y = 3.
Question 3.
3x – 4y
Given,
3x – 4y, when x = 4 and y = 3
Substitute 4 for x and 3 for y
3(4) – 4(3) = 12 – 12 = 0
3x – 4y = 0
Question 4.
2x2
Given,
2x2
when x = 4
Substitute 4 for x in 2x2
2 (4²) = 2 × 16 = 32
2x2 = 32
Question 5.
$$\frac{30}{y}$$
Given,
$$\frac{30}{y}$$
Substitute 3 for y
$$\frac{30}{y}$$ = $$\frac{30}{3}$$ = 10
So, $$\frac{30}{y}$$ = 10
Question 6.
$$\frac{3x}{4y}$$
Given,
$$\frac{3x}{4y}$$
Substitute 4 for x and 3 for y
$$\frac{3x}{4y}$$ = $$\frac{3(4)}{4(3)}$$ = $$\frac{12}{12}$$ = 1
$$\frac{3x}{4y}$$ = 1
Question 7.
The volume of a right rectangular prism is found by multiplying length by width by height, or lwh. What is the volume of a right rectangular prism with length 6 inches, width 4 inches, and height $$\frac{3}{4}$$ inch?
Substitute 6 for l, 4 for w, and $$\frac{3}{4}$$ for h. (_______) (________) (_________)
Multiply. __________
Given,
The volume of a right rectangular prism is found by multiplying the length by width by height, or lwh.
Substitute 6 for l, 4 for w, and $$\frac{3}{4}$$ for h.
V = l . w . h
V = 6 × 4 × $$\frac{3}{4}$$
V = 6 × 3
V = 18 cubic in.
Question 8.
Jasper needs wood for a rectangular sandbox. The sandbox will be 8 feet by 4.5 feet. He knows the perimeter is found using the expression 2l + 2w, or 2(1 + w). How many feet of wood should Jasper use?
Given,
Jasper needs wood for a rectangular sandbox.
The sandbox will be 8 feet by 4.5 feet.
He knows the perimeter is found using the expression 2l + 2w, or 2(1 + w).
P = 2(8 + 4.5)
P = 2(12.5)
P = 25 feet.
Question 9.
A cube has an edge length, s, of 5 centimeters. The expression 6s2 gives the surface area of the cube and the expression s3 gives the volume. What are the volume and surface area of the cube?
surface area = cm2
volume = cm3
A cube has an edge length, s, of 5 centimeters.
The expression 6s2 gives the surface area of the cube
s = 5 cm
surface area = 6(5)² = 6 × 25 = 150 sq. cm
the expression s3 gives the volume.
Volume = 6(5)³ = 6 × 125 = 750 cubic centimeters
Question 10.
Use Structure A right triangle has a base, b, that is 6 inches. The area of the triangle, with h representing the height, is given by the expression $$\frac{bh}{2}$$. Complete the table to show how the area of the triangle changes with height:
How does the area change as the height increases? Why do you think this happens?
Given,
A right triangle has a base, b, that is 6 inches.
The area of the triangle, with h representing the height, is given by the expression $$\frac{bh}{2}$$.
b = 6 in.
h = 4 in
Area = $$\frac{bh}{2}$$
Area = $$\frac{6.4}{2}$$ = $$\frac{24}{2}$$ = 12 sq. in.
b = 6 in.
h = 5 in
Area = $$\frac{bh}{2}$$
Area = $$\frac{6.5}{2}$$ = $$\frac{30}{2}$$ = 15 sq. in.
b = 6 in.
h = 6 in
Area = $$\frac{bh}{2}$$
Area = $$\frac{6.6}{2}$$ = $$\frac{36}{2}$$ = 18 sq. in.
b = 6 in.
h = 7 in
Area = $$\frac{bh}{2}$$
Area = $$\frac{6.7}{2}$$ = $$\frac{42}{2}$$ = 21 sq. in.
b = 6 in.
h = 8 in
Area = $$\frac{bh}{2}$$
Area = $$\frac{6.8}{2}$$ = $$\frac{48}{2}$$ = 24 sq. in.
In Problems 11—14, evaluate the expression for n = 0.75.
Question 11.
4n
Given,
n = 0.75
4 × 0.75 = 3
Question 12.
6 – n
Given,
6 – n
n = 0.75
6 – 0.75 = 5.25
Question 13.
n + 12.5
Given,
n + 12.5
n = 0.75
0.75 + 12.5 = 13.25
Question 14.
0.5n
Given,
0.5n
n = 0.75
0.5 × 0.75 = 0.375
Question 15.
Critique Reasoning Bill and Tia are trying to evaluate the expression 5×2 when x = 3. They both agree that 3 should be substituted for x. Tia says they should multiply 3 by 5, and then square the result. Bill says they should square 3 and then multiply by 5. Who is correct and why? What is the value of the expression?
Given,
Bill and Tia are trying to evaluate the expression 5x² when x = 3. They both agree that 3 should be substituted for x.
Tia says they should multiply 3 by 5, and then square the result.
Bill says they should square 3 and then multiply by 5.
5x² when x = 3
5(3)² = 5(9) = 45
So, Bill is correct
Question 16.
STEM The expression (F – 32)$$\frac{5}{9}$$ gives the temperature in degrees Celsius (°C) for a given temperature F in degrees Fahrenheit (°F).
A. Find the temperature in degrees Celsius that is equivalent to 77 °F.
Given,
(F – 32)$$\frac{5}{9}$$
F = 77 °F
(77 – 32)$$\frac{5}{9}$$
45$$\frac{5}{9}$$ = 5 × 5 = 25°C
B. Water freezes at 32 °F. At what temperature does water freeze in degrees Celsius?
Given,
(F – 32)$$\frac{5}{9}$$
F = 32 °F
(32 – 32)$$\frac{5}{9}$$
0$$\frac{5}{9}$$ = 0°C
Question 17.
To find approximately how many pounds are equivalent to a given number of kilograms, use the expression 2.2k, where k represents kilograms. How many pounds are equivalent to 6.5 kilograms?
Given expression is 2.2k
where k = 6.5 kilograms
2.2k = 2.2(6.5) = 14.3 pounds
Therefore 14.3 pounds are equivalent to 6.5 kilograms.
Question 18.
To find the perimeter of a regular octagon, use the expression 8s, where s represents side length. If the side length of a regular octagon is 1.5 feet, what is its perimeter in feet?
Given,
length of a regular octagon is 1.5 feet
The perimeter of the octagon = 8s
P = 8 × 1.5
P = 12 feet
Thus the perimeter of the octagon is 12 feet.
Question 19.
A rectangle is twice as long as it is wide. If the width is w, the length is 2w. The area of the rectangle can be found using the expression 2w2. If the rectangle is 10 centimeters wide, what is its area in square centimeters?
Given,
A rectangle is twice as long as it is wide. If the width is w, the length is 2w.
The area of the rectangle can be found using the expression 2w2.
w = 10 cm
length = 2w = 2 × 10 = 20 cm
Area of the rectangle = 2w² = 2(10)² = 2 × 100 = 200 sq. cm
For Problems 20-25, evaluate the expression for the given value.
Question 20.
6w; w = 0.1
Given,
6w
w = 0.1
6 × 0.1 = 0.6
6w = 0.6
Question 21.
x + 5$$\frac{1}{4}$$; x = 3$$\frac{1}{2}$$
Given,
x + 5$$\frac{1}{4}$$
x = 3$$\frac{1}{2}$$
3$$\frac{1}{2}$$ + 5$$\frac{1}{4}$$
3 + 5 = 8
$$\frac{1}{2}$$ + $$\frac{1}{4}$$ = $$\frac{3}{4}$$
8 + $$\frac{3}{4}$$ = 8$$\frac{3}{4}$$
Question 22.
1.4y; y = 5
Given,
1.4y
y = 5
1.4 × 5 = 7
1.4y = 7
Question 23.
$$\frac{48}{k}$$, k = 3
Given,
$$\frac{48}{k}$$,
k = 3
$$\frac{48}{3}$$ = 16
$$\frac{48}{k}$$ = 16
Question 24.
z5; z = 2
Given,
z5; z = 2
25 = 2 × 2 × 2 × 2 × 2 = 32
So, z5 when z = 2
Question 25.
2.5g2; g = 4
Given,
2.5g2; g = 4
2.5(4)² = 2.5 × 16 = 40
So, 2.5g2 when g = 4 is 40.
Question 26.
Critique Reasoning To evaluate the expression (r + 6)2 for r = 7, Sayid says that r should be squared and 6 should be squared, and then the results should be added. Explain why Sayid is incorrect. Then find the value of the expression when r = 7.
Sayid says that r should be squared and 6 should be squared, and then the results should be added.
Sayid is incorrect because the first 7 and 6 should be added and then squared.
(r + 6)2 for r = 7
(7 + 6)² = (13)² = 169
Question 27.
Evaluate the expression 4a2 – $$\frac{b}{6}$$ when a = 6 and b = 36. Show your work.
4a2 – $$\frac{b}{6}$$
when a = 6 and b = 36
4(6)² – $$\frac{36}{6}$$
4(36) – 6
144 – 6 = 138
4a2 – $$\frac{b}{6}$$ = 138
Question 28.
Rosa sells pens. She pays $0.75 for each pen and sells them for$1.25 each. She uses the expression 1.25p – 0.75p, where p is the number of pens she sells, to calculate her profit. If Rosa sells 48 pens, what is her profit?
Given,
Rosa sells pens. She pays $0.75 for each pen and sells them for$1.25 each.
She uses the expression 1.25p – 0.75p, where p is the number of pens she sells, to calculate her profit.
p = 48
1.25(48) – 0.75(48) = 24
Thus if Rosa sells 48 pens then her profit is $24. Question 29. Steve is playing a carnival game. He wants to win a prize, but he thinks he has about a 20% chance of winning. He uses the expression 0.2t to calculate the number of games he can expect to win if he plays t times. If he plays the game 12 times, about how many times can he expect to win? Answer: Given, Steve is playing a carnival game. He wants to win a prize, but he thinks he has about a 20% chance of winning. He uses the expression 0.2t to calculate the number of games he can expect to win if he plays t times. 0.2(12) = 2.4 2.4 × 20% = 0.48 = 48% For Problems 30-35, evaluate the expression s3 for the given value. Question 30. s = 4 Answer: Given, s = 4 To find the expression s³ (4)³ = 4 × 4 × 4 = 64 Question 31. s = $$\frac{1}{2}$$ Answer: Given, s = $$\frac{1}{2}$$ To find the expression s³ ($$\frac{1}{2}$$)³ = $$\frac{1}{8}$$ Question 32. s = 0.3 Answer: Given, s = 0.3 To find the expression s³ (0.3)³ = 0.3 × 0.3 × 0.3 = 0.027 Question 33. s = 10 Answer: Given, s = 10 To find the expression s³ (10)³ = 10 × 10 × 10 = 1000 Question 34. s = 1.2 Answer: Given, s = 1.2 To find the expression s³ (1.2)³ = 1.2 × 1.2 × 1.2 = 1.728 Question 35. s = $$\frac{1}{6}$$ Answer: Given, s = $$\frac{1}{6}$$ To find the expression s³ ($$\frac{1}{6}$$)³ = $$\frac{1}{216}$$ Lesson 8.4 More Practice/Homework Question 1. Every day jin reads for 0.75 hour in the morning and 1.25 hours in the evening. He uses the expression 0.75d + 1 .25d to keep track of the number of hours he has read for any number of days, d. If jin reads for 20 days, how many hours has he read? Show your work. Answer: Given, Every day jin reads for 0.75 hour in the morning and 1.25 hours in the evening. He uses the expression 0.75d + 1 .25d to keep track of the number of hours he has read for any number of days, d. 0.75d + 1.25d = 2d d = 20 2 × 20 = 40 hours Question 2. There are 16 ounces in 1 pound, so the expression $$\frac{z}{16}$$, where z represents the number of ounces, can be used to find the number of pounds for any given number of ounces. How many pounds are in 40 ounces? Answer: Given, 1 pound = 16 ounces the expression $$\frac{z}{16}$$ z = 40 $$\frac{z}{16}$$ = $$\frac{40}{16}$$ = 2.5 pounds Thus there are 2.5 pounds in 40 ounces Question 3. Jeffrey is 5 years older than his brother. If j represents Jeffrey’s age, the expression j – 5 can be used to find his brother’s age. If Jeffrey is 23, how old is his brother? Answer: Given, Jeffrey is 5 years older than his brother. The expression j – 5 can be used to find his brother’s age. j = 23 23 – 5 = 18 Thus the age of his brother is 18 years. Math on the Spot For Problems 4-6, evaluate each expression. Question 4. 4x – 5 for x = 10 Answer: 4x – 5 for x = 10 4(10)-5 = 40 – 5 = 35 4x – 5 = 35 Question 5. w ÷ 5 + w for w = 20 Answer: w ÷ 5 + w for w = 20 20 ÷ 5 + 20 4 + 20 = 24 w ÷ 5 + w = 24 Question 6. 3z2 – 6z for z = 5 Answer: 3z2 – 6z for z = 5 z(3z – 6) 5(3(5) – 6) 5(15 – 6) 5(9) = 45 3z2 – 6z = 45 For Problems 7-10, evaluate each expression for b = 5. Question 7. 2.1b Answer: 2.1b b = 5 2.1(5) = 10.5 Question 8. 5b – 12.4 Answer: 5b – 12.4 b = 5 5(5) – 12.4 25 – 12.4 = 12.6 Question 9. 7.4b Answer: 7.4b b = 5 7.4(5) = 37 7.4b = 37 Question 10. 4b2 Answer: 4b2 = 4(5)² b = 5 4 × 25 = 100 Test Prep Question 11. What is the value of the expression 8w – 4j2 when w = 0.25 and j = 0.5? Answer: Given, The value of the expression 8w – 4j2 when w = 0.25 and j = 0.5 8w – 4j² = 8(0.25) – 4(0.5)² = 2 – 4(0.25) = 2 – 1 = 1 8w – 4j2 = 1 Question 12. What is the first step in evaluating an expression for given variable values? Answer: The first step in evaluating an expression for given variable values is to substitute the values 0.25 and 0.5 in the place of w and j. And then we have to multiply and simplify the expression. Question 13. Match the expression and its value to the value of x that gives the expression its value by filling in the box in the correct column. Answer: Explanation: 4x² – 2x for x = 2 4x² – 2x = 4(2)² – 2(2) 4(4) – 4 = 16 – 4 = 12 4x² – 2x for x = 3 4x² – 2x = 4(3)² – 2(3) 4(9) – 6 = 36 – 6 = 30 5x + 20 for x = 2 5(2) + 20 = 10 + 20 = 30 5x + 20 for x = 3 5(3) + 20 = 15 + 20 = 35 5x + 20 for x = 4 5(4) + 20 = 20 + 20 = 40 5x + 20 for x = 5 5(5) + 20 = 25 + 20 = 45 8x/2 – 3x for x = 2 8(2)/2 – 3(2) 8 – 6 = 2 8x/2 – 3x for x = 3 8(3)/2 – 3(3) 12 – 9 = 3 8x/2 – 3x for x = 4 8(4)/2 – 3(4) 16 – 12 = 4 Question 14. Evaluate the expression 4(n + 3) – 5r for n = $$\frac{1}{4}$$ and r = $$\frac{1}{5}$$. (A) 3 (B) 5 (C) 8 (D) 12 Answer: Given, 4(n + 3) – 5r where n = $$\frac{1}{4}$$ and r = $$\frac{1}{5}$$ 4(n + 3) – 5r = 4($$\frac{1}{4}$$ + 3) – 5$$\frac{1}{5}$$ 4(n + 3) – 5r = 4(3$$\frac{1}{4}$$) – 1 4(n + 3) – 5r = 4($$\frac{13}{4}$$) – 1 4(n + 3) – 5r = 13 – 1 4(n + 3) – 5r = 12 Option D is the correct answer. Question 15. The surface area for a rectangular prism with a square base is given by the expression 2s2 + 4sh, where s is the side length of the square base and h is the height of the prism. What is the surface area in square feet of a rectangular prism when s = 4 feet and h = 6 feet? Answer: Given, The surface area for a rectangular prism with a square base is given by the expression 2s2 + 4sh, Where s is the side length of the square base and h is the height of the prism. s = 4 feet and h = 6 feet A = 2s(s + 2h) A = 2(4)(4 + 2(6)) A = 8(4 + 12) A = 8(16) A = 128 sq. feet Spiral Review Question 16. Write a numerical expression to represent the phrase “the sum of 25 and the cube of 9.” Do not evaluate the expression. Answer: the sum of 25 and the cube of 9 is 25 + 9³ Question 17. A customer at a department store has$45.75. How many neckties can the customer buy if each tie costs $15.25? Answer: Given that, A customer at a department store has$45.75.
To find:
How many neckties can the customer buy if each tie costs \$15.25
We have to divide 45.75 by 15.25
45.75/15.25 = 3
Thus the customer can buy 3 neckties.
Question 18.
An expression is shown.
5[8(7 – 4) ÷ 6]
What is the value of the expression?
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# Percentage : Formula, Shortcut, Tips and Tricks to Solve Problems
Hello Aspirants, In this lesson " Percentage : Formula, Shortcut, Tips and Tricks to Solve Problems " we will cover basic concepts of percentage, formulas, tips, tricks and shortcuts used to solve questions based on percentage.
Now let's start the lesson !
Percent implies " for every hundred " and it is denoted by sign " % " pronounced as percentage.
Accordingly, X % = X / 100, Y % = Y / 100
Now let's see some examples :
2 % = 2 / 100 = 0.02 ( means 2 parts from 100 )
25 % = 25 / 100 = 1 / 4 = 0.25 ( means 25 parts out of 100 )
Basic Formula :
To find X % of Y
X % of Y = X / 100 of Y = XY / 100
Now let's see some examples :
25 % of 200 = ( 25 x 200 ) / 100 = 50
50 % of 30 = ( 50 x 30 ) / 100 = 15
Tips : X % of Y = Y % of X i.e. 50 % of 30 = 30 % of 50
To find percent of a number X with respect to Y
Percent of X / Y = ( X / Y ) 100
Now let's see some examples
Percent of ( 2 / 5 ) = ( 2 / 5 ) 100 = 40 %
Percent of ( 20 / 200 ) = ( 20 / 200 ) 100 = 10 %
Some Important Points To Remember :
• If new value of a quantity is n times the previous value, then percentage increase = ( n - 1 ) 100 %
• If the present value X of a quantity is increased by N % then the new value of the quantity = X ( 1 + N / 100 )
• If the present value X of a quantity is decreased by N % then the new value of the quantity = X ( 1 - N / 100 )
• If A is N% more than B, then B is { 100N/( 100 + N )} % less than A.
• If A is N% less than B, then B is { 100N/(100 - N )} % more than A.
• If the value of an item is increased by N%, the percentage reduction that needs to make to bring the original value of item is calculated using formula : { 100N/( 100 + N )} %
• If the value of an item is decreased by N%, the percentage increment that needs to make to bring the original value of item is calculated using formula : { 100N/(100 - N )} %
• If the price of goods increases by N %, then the reduction in consumption so as not to increase the expenditure can be calculated using the formula : { 100N/( 100 + N )} %
• If the price of goods decreases by N %, then the increase in consumption so as not to decrease the expenditure can be calculated using the formula : { 100N/(100 - N )} %
• Let the present population of a town be P. Suppose it increases at the rate of R% per annum. Then:
1. Population after n years = P 1 + R n 100
2. Population n years ago =P
1 + R n 100
• Let the present value of a machine be P. Suppose it depreciates at the rate of R% per annum. Then:
1. Value of the machine after n years = P 1 - R n 100
2. Value of the machine n years ago =P
1 - R n 100
Successive Percentage : It mainly refers to two or more percentage applied to same item.
Let the current value of an item is N, it goes a percentage change of X % and then Y %,
Then new value of item = N { ( 1 + X / 100 ) ( 1 + Y / 100 ) }
Let the current value of an item is N, it goes a percentage change of X % and then Y % and then Z %,
Then new value of item = N { ( 1 + X / 100 ) ( 1 + Y / 100 ) ( 1 + Z / 100 ) }
Percentage Change : The increase or decrease in percent due to change of successive percentage of an item is called Percentage Change.
At last, Thanks for choosing Loud Study.
If you have any doubts or confusion, comment in the comment section below, we will try to resolve it as soon as possible.
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Teacher Kimberlie Hymes Math 6th 12 Proportionality, Ratios, and Rates
Standard(s) Taught
MAFS.6.RP.1.3
Use ratio and rate reasoning to solve real-world and mathematical problems, e.g., by reasoning about tables of equivalent ratios, tape diagrams, double number line diagrams, or equations.
a. Make tables of equivalent ratios relating quantities with whole-number measurements, finding missing values in the tables, and plot the pairs of values on the coordinate plane. Use tables to compare ratios. (to be taught C and P)
b. Solve unit rate problems including those involving unit pricing and constant speed. (to be taught P and A)
c. Find a percent of a quantity as a rate per 100 (e.g. 30% of a quantity means 30/100 times the quantity); solve problems involving finding the whole, given a part and the percent. (to be taught P and A)
d. Use ratio reasoning to convert measurement units; manipulate and transform units appropriately when multiplying or dividing quantities. (to be taught C and P)
e. Understand the concept of Pi as a ratio of the circumference of a circle to its diameter. (to be taught C)
Learning Targets and Learning Criteria
• create and interpret a table of equivalent ratios.
• plot pairs of values from a table to a coordinate plane
• use a table to compare ratios and find missing values using ratios.
• explain the difference between a ratio and a unit rate.
• understand that rate problems compare two different units, such as revolutions per minute or miles per hour.
• solve real-world problems using ratios and rates.
• reason to determine the better buy.
• write a percent as a rate over 100, including percents greater than 100 and less than 1.
• find the percent of a number using rate methods.
• represent the relationship of part to whole to describe percents using models.
• convert units by multiplication or division. • understand and explain that Pi is a ratio of the circumference of a circle to its diameter. • determine whether two ratios are equivalent, based on the description of equivalent ratios, using tape diagrams, double number line diagrams, and/or equivalent ratio tables. • create and complete an equivalent ratio table, tape diagram, and/or double number line diagram to solve problems. • calculate the percent of a given number by using visual representations (e.g., strip diagrams, percent bars, one-hundred grids) • find the whole when given both the part and the percent using tape diagrams, double number line diagrams, and tables of equivalent ratios.
Classroom Activities
In class, we complete the MATH workshop each period. This is a blended, rotation model of teaching where your student is receiving the standards in multiple different settings while completing many different activities. This allows me to have time with your students in a very small group setting to help when necessary further develop the understanding of the standards..
Ratio Test: We will test on Ratios Next week (Period 2 – Tuesday November 6 and Period 5/7 – Thursday November 8)
DIA 2 (ratios) will be on 11.15-11.16 Thursday/Friday for all Periods
Assignments Due
The week students will be assigned the following to practice class concepts:
IXL: (Monday/Tuesday/Wednesday)
Period 2: S2
Period 5 and 7: R7, R9, S2
students are expected to reach at least an 80% on these assignments.
Mathspace: (Thursday/Friday) None this week
If additional help is needed, students are encouraged to view a lesson regarding a given topic on Khan Academy. This website is free and provides both video instruction as well as practice problems.
Students are able to complete additional practice on both Mathspace and IXL
www.mathspace.co
www.ixl.com
**** Accommodations provided as needed
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Code360 powered by Coding Ninjas X Naukri.com. Code360 powered by Coding Ninjas X Naukri.com
1.
Introduction
2.
Spanning Trees
3.
Minimum Spanning Tree
4.
Properties of Minimum Spanning Tree
4.1.
Possible Multiplicity
4.2.
Uniqueness
4.3.
Minimum Cost Subgraph
4.4.
Minimum Cost Edge
4.5.
Cycle Property
4.6.
Cut Property
5.
6.
Key Takeaways
Last Updated: Mar 27, 2024
Introduction and Properties of Minimum Spanning Tree
Introduction
We have probably already learned about graphs. There we saw what a tree is, but have you ever heard of a spanning tree?
Probably not, so let’s see what it is!
We already know that a graph consists of a set of vertices and edges like this:
Here the set of vertices is V = {1, 2, 3, 4, 5, 6, 7, 8}
and, the set of edges is E = {(1,2), (2,3), (3,4), (4,5), (5,6), (6,7), (7,8), (8,1)}
In this article, we will learn about the various properties of a minimum spanning tree. Let’s start by learning about spanning trees.
Spanning Trees
To obtain a spanning tree from the above graph, we’ll have to make a tree taking all the vertices of the graph and choosing only (N-1) edges from the tree. Here, N is the total number of vertices in the graph.
The only thing we need to make sure of is, all the vertices must be connected and it satisfies all the properties of a tree. A graph can have multiple spanning trees.
For example, a spanning tree from the above graph will be:
Thus, to formally define it,
A spanning tree is a connected and undirected graph (n - 1) number of edges, where n is the number of vertices.
Minimum Spanning Tree
Now, let’s consider a weighted graph.
We can find different spanning trees, each of which will have different weights for different paths. This sum of the weights is known as cost.
For example, the weight of the spanning tree shown below is 26.
Similarly, there are many other combinations of spanning trees possible.
Now, suppose that the paths in a spanning tree are the different routes from one place to another. Which road will we prefer?
Source: giphy
The shortest one, of course! So, we need to find the path with the least weight. This is called a minimum spanning tree
Let’s move on to the properties of a minimum spanning tree.
Properties of Minimum Spanning Tree
Some of the important properties of minimum spanning tree are as follows:
Possible Multiplicity
We already saw that by the definition of a spanning tree, it must have (n - 1) edges where n is the number of vertices in the graph. This property of a minimum spanning tree is formally called possible multiplicity.
Uniqueness
According to this property, if all the edges in a spanning tree have different weights, there will be only one unique minimum spanning tree.
Let us see an example of this property.
The graph The minimum spanning tree
Minimum Cost Subgraph
This property states that if the weights of all the edges in the spanning tree are positive, then the minimum spanning tree will also be a minimum cost subgraph that connects all the vertices.
To understand this point, let us consider our previous example again.
The graph and its minimum spanning tree.
If you notice, it is also the minimum cost sub-graph for the given graph.
Minimum Cost Edge
By this property, if the edge with the least weight is present only once, it must be included in the minimum spanning tree. We saw that in our example practically too, where the edge with the least weight (= 1) is unique and is present in the minimum spanning tree.
Graph MinimumSpanning Tree
Cycle Property
In the graph below, we can see a cycle between nodes 2, 6 and 5. In this cycle, the weights of the edges are 2, 2 and 4, respectively.
The edge between nodes 2 and 5 has the largest weight (= 4) out of these edges.
So, according to the cycle property, this edge will not be a part of the minimum spanning tree.
Thus, the cycle property states that in a cycle, the edge with the largest weight will never be a part of the minimum spanning tree
Cut Property
To understand the cut property, let us first cut our graph as shown below.
Here, the edges we are cutting through (also known as cut edges) are having weights 3, 4 and 2. Out of these edges, the edge with the least weight is 2.
So, by the cut property, this edge will be a part of the minimum spanning tree.
Thus, by the cut property, if we cut a graph, the cutting edge with the least weight will always be a part of the minimum spanning tree.
These were all the crucial properties of a minimum spanning tree, let’s move on to some of the frequently asked questions of this topic.
Q1) What is a minimum spanning tree?
A1) A spanning tree is a connected and undirected graph (n - 1) number of edges, where n is the number of vertices. If the edges have weights, the path with the least sum of weights is called the minimum spanning tree.
Q2)What are the properties of a minimum spanning tree?
A2)The properties of an MST are:
• Possible Multiplicity
• Uniqueness
• Minimum Cost Subgraph
• Minimum Cost Edge
• Cycle Property
• Cut Property
• Contraction
Q3)What does the possible multiplicity of an MST mean?
A3)Possible multiplicity means that an MST will have (n - 1) edges where n is the number of vertices in the graph.
Q4)State the cycle property and the cut property of an MST.
A4)The cycle property states that in a cycle, the edge with the largest weight will never be a part of an MST.
By the cut property, if we cut a graph, the cutting edge with the least weight will always be a part of the minimum spanning tree.
Q5)What are some applications of an MST?
A5)Some applications of an MST are:
• Taxonomy
• Designing networks
• Circuit design
• To describe financial markets
Key Takeaways
In this article, we learned what a minimum spanning tree is and what its different properties are. Now that we are well versed with the theoretical knowledge of minimum spanning trees, we should know how to find them from a given graph. To do that, there are two algorithms, Kruskal’s algorithm and Prim’s algorithm
Check out this problem - Duplicate Subtree In Binary Tree
Apart from this, there are many other algorithms that we need to learn and practice. Coding Ninjas Studio is a platform where you can find practice coding questions and the interview experience of people working in reputed companies to further your knowledge.
Happy Learning!
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# Money in the Bank: Two-Digit Addition
This little piggy said "wee! wee! wee!" all the way to the bank...because he was loaded! Ask your second grader to do some mental math to figure out how much moolah each pig is carrying.
### Money in the Bank: Two-Digit Addition #1
How much money is in each piggy bank? Give your second grader practice with double-digit addition as he gets more familiar with mental math.
### Money in the Bank: Two-Digit Addition #2
Use this colorful worksheet to boost your second grader's math skills. She'll add up all the problems to see how much money is in each piggy bank.
### Money in the Bank: Two-Digit Addition #3
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### Money in the Bank: Two-Digit Addition #5
How much money does each little piggy have? Give your second grader's double-digit addition skills a work out with this fun printable.
### Money in the Bank: Two-Digit Addition #6
Can your second grader find out how much money is in each piggy bank? He'll use his addition skills to count up the dollars!
The set is continued below.
### Money in the Bank: Two-Digit Addition #7
Which piggy bank has the most money? Your second grader will boost her double-digit addition and mental math ability as she works to find out.
### Money in the Bank: Two-Digit Addition #9
This charming piggy bank worksheet gives your second grader a chance to exercise his two-digit addition and mental math skills.
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Your second grader will get a mental math boost as he practices two-digit addition to figure out how much money is in each piggy bank.
### Money in the Bank: Two-Digit Addition #12
Watch your second grader boost his mental math skills as he works through these addition problems to find out how much money each piggy bank holds.
The set is continued below.
### Money in the Bank: Two-Digit Addition #13
In no time at all, you'll see your second grader boost her double-digit addition skills and practice mental math with this piggy bank worksheet.
### Money in the Bank: Two-Digit Addition #14
How many dollars are in each piggy bank? Watch your second grader use his addition skills to count up the dollars, which strengthens his mental math, too.
### Money in the Bank: Two-Digit Addition #15
Let your second grader work through a heap of double-digit addition problems to find the piggy bank that holds the most and least amount of money.
### Money in the Bank: Two-Digit Addition #16
Kids will solve math problems to find out how much money is in each piggy bank. Exercise mental math skills with this two-digit addition worksheet.
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### Money in the Bank: Two-Digit Addition #18
Use mental math to work through two-digit addition problems in this math worksheet. Kids will solve to find out how much money is in each piggy bank.
### Money in the Bank: Two-Digit Addition #19
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### Money in the Bank: Two-Digit Addition #20
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### Money in the Bank: Two-Digit Addition #21
Solve two-digit addition problems with piggy bank math. Find out how much money is in each piggy bank with this math worksheet.
### Money in the Bank: Two-Digit Addition #22
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### Money in the Bank: Two-Digit Addition #23
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### Money in the Bank: Two-Digit Addition #24
Kids will do two-digit addition problems to find out how much money is in each piggy bank in this worksheet. Practice math skills with this worksheet.
### Money in the Bank: Two-Digit Addition #25
Solve two-digit addition problems in this piggy-bank-themed worksheet. Use your math skills to figure out how much money is in each piggy bank.
### Money in the Bank: Two-Digit Addition #27
Kids will solve two-digit addition problems in this piggy-bank math worksheet. Find out how much money is in each piggy bank through addition.
### Money in the Bank: Two-Digit Addition #28
Use this piggy-bank math worksheet with your young mathematician. Find out how much money is in each piggy with double-digit addition.
### Money in the Bank: Two-Digit Addition #29
Try this piggy-bank-themed addition worksheet with your child. Your child will figure out how much money is in each piggy while practicing adding double digits.
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# Prime Factorization Of 70
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The method of calculating the factors of 70 is as follows. 1 70 2 35 5 14 and 7 10.
Christmas Prime Factorization Tree Mini Math Project Prime Factorization Math Math Projects
### The number 70 is a composite number because 70 can be divided by 1 by itself and at least by 2 5 and 7.
Prime factorization of 70. Prime Factorization of 2310 Prime Factorization of 945. 2 5 7 70 is not a prime number. 2 x 5 x 7 70.
The Prime Factors of 70. So it is possible to draw its prime tree. If you multiply all the prime factors of 70 you will get 70.
We can call this type of division as the perfect division. The prime factorization of 70 257. Prime factors of 70.
The GCF of 15 36 and 70 is 1The prime factorization of 15 is 35The prime factorization of 36 is 2233The prime factorization of 70 is 257Since they have no common factors they are coprime. The prime factorization of 70 is the list of prime factors of 70. Finding the prime factors of 70 To find the prime factors you start by dividing the number by the first prime number which is 2.
The prime factor of the 70 is 2 x 5 x 7. It is determined that the prime factors of number 70 are. 11 11 11 2 x 2 x 2 8.
We write number 70 above a 2-column table 2. Prime Factorization of 70. Prime factorization or prime factor decomposition is the process of finding which prime numbers can be multiplied together to make the original number.
So it means that 70 is divisible by 2 5 and 7. The prime factors of 70 are 2 5 and 7. Math mathantics factortree Factor Tree Of 12 httpsyoutubek8DmNR6sUeY Factor Tree Of 18 httpsyoutubeR_whbNNQMbM Factor Tree Of 20.
70 is divisible by the prime number 2 which results in 35. See its prime factors tree below. To find all the prime factors of 70 divide it by the lowest prime number possible.
The result 7 cannot be divided any further as it is a prime number. Factor tree or prime decomposition for 70. It is clear that prime numbers are numbers that can divide a number without a remainder.
The prime factorization of 70 257. The prime factorization of a positive integer is a list of the integers prime factors together with their multiplicities. 2 5 7 Prime Factorization Of 69 Prime Factorization Of 71 Is 70 A Prime Number.
First every number is divisible by itself and 1. Therefore the number 70 has 8 factors. Study the tree to see the step by step division proper divisors of 70 Prime divisor of 70 2 5 10 7.
The exponents in the prime factorisation are 1 1 and 1. Are all the factors of 70 composite. The prime factors of 70 are 2 5 and 7.
70 2 x 5 x 7 The solution above and all other available prime factorization solutions were provided by the Prime Factorization Application. We can check our work by multiplying all of the prime factors together. The prime factors of 70 are the prime numbers that can be divided into 70 exactly with no remainder.
The Factor Tree of 70 above shows the level of divisions carried out to get the factor numbers. The positive factors of 70 in pairs are. 2 5 7 In number theory the prime factors of a positive integer are the prime numbers that divide that integer exactly.
Hence the prime factors of 70 are 2 5 7. The process of determining these factors is called integer factorization. If you would like to know more or need additional explanation please review our prime factorization explanation.
A prime factor is a prime number. Continuing the number 35 is divisible by prime number 5 and the result after division will be 7. By prime factorization of 70 we follow 5 simple steps.
In other words a composite number is any integer greater than one that is not a prime number. Therefore the factors of 70 are 1 we can find all factors of a number by dividing it by 1 2 3 4. Prime factorization of a number is the determination of a set of prime integers which when multiplied together give the original number or integer.
What are the positive pair factors of 70. View this answer The prime factors of the number 70 are 2 5 and 7. Add the number 1 with the exponents and multiply it.
Here are some more numbers to try. Thus 2 x 5 x 7 70. This can also be written as 2 1 x 5 1 x 7 1 70.
Here We Are Providing The Table Of Factors Prime Factor 1 150 It Contains Following Thingsa Factors O Prime Numbers Composite Numbers Prime Factorization
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Videos and lessons to help Grade 4 students learn to understand a fraction a/b witha > 1 as a sum of fractions 1/b.
A. Understand addition and subtraction of fractions as joining and separating parts referring to the same whole.
B. Decompose a fraction into a sum of fractions with the same denominator in more than one way, recording each decomposition by an equation. Justify decompositions, e.g., by using a visual fraction model. Examples: 3/8 = 1/8 + 1/8 + 1/8 ; 3/8 = 1/8 + 2/8 ; 2 1/8 = 1 + 1 + 1/8 = 8/8 + 8/8 + 1/8.
C. Add and subtract mixed numbers with like denominators, e.g., by replacing each mixed number with an equivalent fraction, and/or by using properties of operations and the relationship between addition and subtraction.
D. Solve word problems involving addition and subtraction of fractions referring to the same whole and having like denominators, e.g., by using visual fraction models and equations to represent the problem.
Common Core: 4.NF.3
### Suggested Learning Targets
• I can use models to add and subtract fractions.
• I can use visual models to decompose a fraction. For example, 7/12 = 4/12 + 1/12 + 1/12 + 1/12.
• I can add or subtract mixed numbers.
• I can solve word problems with fractions.
Related Topics:
Understanding addition and subtraction of fractions --1 (4.NF.3a)
In this lesson you will learn how to add fractions by joining parts.
Add fractions with like denominators using shapes and sets - 4.NF.3
In this lesson you will learn how to add fractions with like denominators by finding equal parts and grouping them together.
Add fractions with like denominators in number sentences
In this lesson you will learn how to add fractions with like denominators in number sentences by adding adding the numerators and keeping the denominators the same.
Add fractions with like denominators from two different wholes
In this lesson you will learn how to add fractions with like denominators from two different wholes by determining the unit of the fraction that you are being asked to add and then counting those units.
Adding Fractions with Like Denominators on a Number Line
This video shows how add fractions with like denominators and also decompose them on a number line.
## Subtract Fractions
Understanding addition and subtraction of fractions - 2 (4.NF.3a)
In this lesson you will learn how to subtract fractions by separating parts to find the difference.
Subtract fractions with like denominators by labeling shapes - 4.NF.3
In this lesson you will learn how to subtract fractions of a shape with like denominators by labeling shapes broken into equal parts.
Subtract fractions with like denominators by labeling sets
In this lesson you will learn how to subtract fractions of a set with like denominators by labeling sets of objects.
Subtract fractions with like denominators starting from less than 1 whole - 4.NF.3
In this lesson you will learn how to subtract fractions with like denominators when you are not starting with one whole by using and labeling pictures.
## Decompose Fractions
Understand decomposing fractions (4.NF3b)
In this lesson you will learn how to decompose a fraction by breaking up the fraction into a sum of fractions.
You will learn that fractions with numerators greater than 1 are composed of a sum of fractions from the same whole with common denominators.
Decomposing fractions into unit fractions.
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# Graph Theory – Trees
Trees are graphs that do not contain even a single cycle. They represent hierarchical structure in a graphical form. Trees belong to the simplest class of graphs. Despite their simplicity, they have a rich structure.
Trees provide a range of useful applications as simple as a family tree to as complex as trees in data structures of computer science.
## Tree
connected acyclic graph is called a tree. In other words, a connected graph with no cycles is called a tree.
The edges of a tree are known as branches. Elements of trees are called their nodes. The nodes without child nodes are called leaf nodes.
A tree with ‘n’ vertices has ‘n-1’ edges. If it has one more edge extra than ‘n-1’, then the extra edge should obviously has to pair up with two vertices which leads to form a cycle. Then, it becomes a cyclic graph which is a violation for the tree graph.
### Example 1
The graph shown here is a tree because it has no cycles and it is connected. It has four vertices and three edges, i.e., for ‘n’ vertices ‘n-1’ edges as mentioned in the definition.
Note − Every tree has at least two vertices of degree one.
### Example 2
In the above example, the vertices ‘a’ and ‘d’ has degree one. And the other two vertices ‘b’ and ‘c’ has degree two. This is possible because for not forming a cycle, there should be at least two single edges anywhere in the graph. It is nothing but two edges with a degree of one.
## Forest
disconnected acyclic graph is called a forest. In other words, a disjoint collection of trees is called a forest.
### Example
The following graph looks like two sub-graphs; but it is a single disconnected graph. There are no cycles in this graph. Hence, clearly it is a forest.
## Spanning Trees
Let G be a connected graph, then the sub-graph H of G is called a spanning tree of G if −
• H is a tree
• H contains all vertices of G.
A spanning tree T of an undirected graph G is a subgraph that includes all of the vertices of G.
### Example
In the above example, G is a connected graph and H is a sub-graph of G.
Clearly, the graph H has no cycles, it is a tree with six edges which is one less than the total number of vertices. Hence H is the Spanning tree of G.
## Circuit Rank
Let ‘G’ be a connected graph with ‘n’ vertices and ‘m’ edges. A spanning tree ‘T’ of G contains (n-1) edges.
Therefore, the number of edges you need to delete from ‘G’ in order to get a spanning tree = m-(n-1), which is called the circuit rank of G.
This formula is true, because in a spanning tree you need to have ‘n-1’ edges. Out of ‘m’ edges, you need to keep ‘n–1’ edges in the graph.
Hence, deleting ‘n–1’ edges from ‘m’ gives the edges to be removed from the graph in order to get a spanning tree, which should not form a cycle.
### Example
Take a look at the following graph −
For the graph given in the above example, you have m=7 edges and n=5 vertices.
Then the circuit rank is
`G = m – (n – 1)`
` = 7 – (5 – 1)`
` = 3`
### Example
Let ‘G’ be a connected graph with six vertices and the degree of each vertex is three. Find the circuit rank of ‘G’.
By the sum of degree of vertices theorem,
n i=1 deg(Vi) = 2|E|
6 × 3 = 2|E|
|E| = 9
Circuit rank = |E| – (|V| – 1)
= 9 – (6 – 1) = 4
## Kirchoff’s Theorem
Kirchoff’s theorem is useful in finding the number of spanning trees that can be formed from a connected graph.
### Example
The matrix ‘A’ be filled as, if there is an edge between two vertices, then it should be given as ‘1’, else ‘0’.
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Can you find the pattern?
Can you find the pattern?. 3-2 Solving by Graphing. Objective: to find the root/solution/zero of a linear equation by graphing. Vocabulary. Root/ Zeros/ Solutions : the solutions of a equations; x-intercepts. Linear Function. Example 1 B.
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Can you find the pattern?
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Presentation Transcript
1. Can you find the pattern?
2. 3-2 Solving by Graphing Objective: to find the root/solution/zero of a linear equation by graphing.
3. Vocabulary • Root/ Zeros/ Solutions: the solutions of a equations; x-intercepts.
4. Linear Function
5. Example 1 B Method Solve by graphing. Using the x and y intercepts. B. Solve Where does the graph cross the x-axis?
6. Example 1 B The graph intersects the x-axis at –3. Answer: So, the solution is –3.
7. TOO Solve for the solution to the linear equation. A.x = –4 B.x = –9 C.x = 4 D.x = 9
8. A.x = 4; B.x = –4; C.x = –3; D.x = 3; Let’s try together Graph using the x and y intercepts.
9. What type of line would have no solution? So which line has no solution, y=3 or x=5? Explian your conclusion with your neighbor
10. Example 3 FUNDRAISING Kendra’s class is selling greeting cards to raise money for new soccer equipment. They paid \$115 for the cards, and they are selling each card for \$1.75. The function y = 1.75x – 115 represents their profit y for selling x greeting cards. Find the zero of this function. Describe what this value means in this context. The graph appears to intersect the x-axis at about 65. So…? That means what?
11. Example 3 Estimate by Graphing y = 1.75x – 115 Original equation 0 = 1.75x – 115 Replace y with 0. 115 = 1.75x Add 115 to each side. 65.71 ≈ x Divide each side by 1.75. Answer: The zero of this function is about 65.71. Since part of a greeting card cannot be sold, they must sell 66 greeting cards to make a profit.
12. Homework 3-2 Worksheet
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