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# FINDING THE SLOPE AND Y-INTERCEPT OF A LINEAR FUNCTION WORKSHEET
Finding the slope and y-intercept of a linear function worksheet :
Here we are going to see some practice questions on finding slope and y-intercept of a linear function.
Y-intercept :
The y-intercept of the line is the value of at the point where the line crosses the y axis.
How to find slope :
To find the slope from a graph first we have to mark any two points on the graph.
After marking two points we have to draw the right triangle which connects the points which have been marked. In the right triangle,
• the vertical direction at the line is called "rise"
• the horizontal direction at the line is called "run"
Alternate method :
We can use the formula (y-y)/(x-x) to find the slope of the line. Here (x₁, y₁) and (x₂, y₂) are any points lie on the line.
## Finding the slope and y-intercept of a linear function worksheet - Questions
Problem 1 :
Find the slope and y-intercept of a linear function from the graph given below.
Problem 2 :
Find the slope and y-intercept of a linear function from the graph given below.
Problem 3 :
Find the slope and y-intercept of a linear function from the graph given below.
## Finding the slope and y-intercept of a linear function worksheet - Answers
Problem 1 :
Find the slope and y-intercept of a linear function from the graph given below.
Solution :
Y-intercept :
The y-intercept of the line is the value of at the point where the line crosses the y axis.
The above line intersects the y-axis at the point 4. So y-intercept = 4 Finding the slope and y-intercept of a linear function worksheet
Slope :
Now we are going to mark two points (-2, 0) and (0, 4) on the line to find the slope.
Slope = Change of y/change of x
"We have to rise before we run"
We need to take downward direction from (0,4) for rise.Since we take downward direction we have to put negative sign.
Rise = -4
After reaching the common point we need to move towards to the left ward 2 units. So we have to take -2.
Run = -2
= (- 4) / (-2) = 2
Hence, slope of the given line is 2.
By using the alternative method, we will get the same answer.
m = (y-y)/(x-x)
(x₁,y₁) ==> (-2, 0) and (x₂, y₂) ==> (0, 4)
m = (4 - 0)/(0 + 2)
m = 4 /2 = 2
Problem 2 :
Find the slope and y-intercept of a linear function from the graph given below.
Solution :
Y-intercept :
The y-intercept of the line is the value of at the point where the line crosses the y axis.
The above line intersects the y-axis at the point -3. So y-intercept = -3
Slope :
Now we are going to mark two points (1, -1) and (2, 1) on the line to find the slope.
Slope = Change of y/change of x
"We have to rise before we run"
We need to take downward direction from (2,1) for rise.Since we take downward direction we have to put negative sign.
Rise = -2
After reaching the common point we need to move towards to the left ward 1 unit. So we have to take -1.
Run = -1
= (-2) / (-1) = 2
Hence, slope of the given line is 2.
By using the alternative method, we will get the same answer.
m = (y-y)/(x-x)
(x₁,y₁) ==> (1, -1) and (x₂, y₂) ==> (2, 1)
m = (1 + 1)/(2 - 1)
m = 2 / 1 = 2
Problem 3 :
Find the slope and y-intercept of a linear function from the graph given below.
Solution :
Y-intercept :
The y-intercept of the line is the value of at the point where the line crosses the y axis.
The above line intersects the y-axis at the point 3. So y-intercept = 3
Slope :
Now we are going to mark two points (0, 3) and (-5, 1) on the line to find the slope.
Slope = Change of y/change of x
"We have to rise before we run"
We need to take downward direction from (0,3) for rise.Since we take downward direction we have to put negative sign.
Rise = -2
After that we need to move towards to the left ward 5 unit. So we have to take -5.
Run = -5
= (-2) / (-5) = 2/5
Hence, slope of the given line is 2/5.
By using the alternative method, we will get the same answer.
m = (y-y)/(x-x)
(x₁,y₁) ==> (-5, 1) and (x₂, y₂) ==> (0, 3)
m = (3 - 1)/(0 - (-5))
m = 2 /(0 + 5) = 2/5
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Vertical and Horizontal Transformations
Shifts of parent functions produced by adding a constant term.
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Vertical and Horizontal Transformations
Horizontal and vertical transformations are two of the many ways to convert the basic parent functions in a function family into their more complex counterparts.
What vertical and/or horizontal shifts must be applied to the parent function of \begin{align*}y = x^{2}\end{align*} in order to graph \begin{align*}g(x) = (x - 3)^{2} + 4\end{align*}?
Embedded Video:
Guidance
Have you ever tried to draw a picture of a rabbit, or cat, or dog? Unless you are talented, even the most common animals can be a bit of a challenge to draw accurately (or even recognizably!). One trick that can help even the most "artistically challenged" to create a clearly recognizable basic sketch is demonstrated in nearly all "learn to draw" courses: start with basic shapes. By starting your sketch with simple circles, ellipses, rectangles, etc., the basic outline of the more complex figure is easily arrived at, then details can be added as necessary, but the figure is already recognizable for what it is.
The same trick works when graphing equations. By learning the basic shapes of different types of function graphs, and then adjusting the graphs with different types of transformations, even complex graphs can be sketched rather easily. This lesson will focus on two particular types of transformations: vertical shifts and horizontal shifts.
We can express the application of vertical shifts this way:
Formally: For any function f(x), the function g(x) = f(x) + c has a graph that is the same as f(x), shifted c units vertically. If c is positive, the graph is shifted up. If c is negative, the graph is shifted down.
Informally: Adding a positive number after the x, outside the parenthesis, shifts the graph up, whereas, adding a negative (or subtracting) shifts the graph down.
We can express the application of horizontal shifts this way:
Formally: given a function f(x), and a constant a > 0, the function g(x) = f(x - a) represents a horizontal shift a units to the right from f(x). The function h(x) = f(x + a) represents a horizontal shift a units to the left.
Informally: Adding a positive number after the x, inside the parenthesis, shifts the graph left, whereas, adding a negative (or subtracting) shifts the graph right.
Example 1
What must be done to the graph of y = x2 to convert it into the graphs of y = x2 - 3, and y = x2 + 4?
Solution:
At first glance, it may seem that the graphs have different widths. For example, it might look like y = x2 + 4, the uppermost of the three parabolas, is thinner than the other two parabolas. However, this is not the case. The parabolas are congruent, just on different locations on the graph.
If we shifted the graph of y = x2 up four units, we would have the exact same graph as y = x2 + 4. If we shifted y = x2 down three units, we would have the graph of y = x2 - 3.
Example 2
Identify the transformation(s) involved in converting the graph of f(x) = |x| into g(x) = |x - 3|.
Solution:
From the examples of vertical shifts above, you might think that the graph of g(x) is the graph of f(x), shifted 3 units to the left. However, this is not the case. The graph of g(x) is the graph of f(x), shifted 3 units to the right.
The direction of the shift makes sense if we look at specific function values.
x g(x) = |x - 3|
0 3
1 2
2 1
3 0
4 1
5 2
6 3
From the table we can see that the vertex of the graph is the point (3, 0). The function values on either side of x = 3 are symmetrical, and greater than 0.
Example 3
What transformations must be applied to \begin{align*}y = x^{2}\end{align*}, in order to graph \begin{align*}g(x) = (x + 2)^{2} - 2\end{align*}?
Solution
The graph of \begin{align*}g(x) = (x + 2)^{2} - 2\end{align*} is the graph of \begin{align*}y = x^{2}\end{align*} shifted 2 units to the left, and 2 units down.
Were you able to solve the question at the beginning of the lesson?
"What transformations must be applied to \begin{align*}y = x^{2}\end{align*}, in order to graph \begin{align*}g(x) = (x - 3)^{2} + 4\end{align*}?"
The graph of \begin{align*}g(x) = (x - 3)^{2} + 4\end{align*} is the graph of \begin{align*}y = x^{2}\end{align*} shifted 3 units to the right, and 4 units up.
If you were able to identify the translation before the review, congratulations! You are on your way to an excellent conceptual base for manipulating functions.
Vocabulary
A shift, also known as a translation or a slide, is a transformation applied to the graph of a function which does not change the shape of the graph, only the location.
Vertical shifts are a result of adding a constant term to the value of a function. A positive term results in an upward shift, and a negative term in a downward shift.
Horizontal shifts are produced by adding a constant term to the function inside the parenthesis. A positive term results in a shift to the left and a negative term results in a shift to the right (easily confused, pay attention!).
Example 4
Use the graph of y = x2 to graph the function y = x2 - 5.
Solution
The graph of y = x2 is a parabola with vertex at (0, 0).
The graph of y = x2- 5 is therefore a parabola with vertex (0, -5).
To quickly sketch y = x2 - 5, you can sketch several points on y = x2, and then shift them down 5 units.
Example 5
What is the relationship between f(x) = x2 and g(x) = (x - 2)2?
Solution
The graph of g(x) is the graph of f(x), shifted 2 units to the right.
Example 6
What is the relationship between f(x) = x2 - 6 and f(x) = x2?
Solution
Adding or subtracting a value outside the parenthesis results in a vertical shift.
Therefore, the graph of f(x) = x2 - 6 is the same as f(x) = x2 shifted 6 units down.
Example 7
Use the parent function f(x) = x2 to graph f(x) = x2 + 3.
Solution
The function f(x) = x2 is a parabola with the vertex at (0, 0).As we saw in example 3, adding outside the parenthesis shifts the graph vertically.Therefore, f(x) = x2 + 3 will be a parabola with the vertex 3 units up from (0, 0).
Example 8
Use the parent function f(x) = |x| to graph f(x) = |x - 2|.
Solution
The graph of the absolute value function family parent function f(x) = |x| is a large "V" with the vertex at the origin.
Adding or subtracting inside the parenthesis results in horizontal movement.
Recall that the horizontal shift is right for negative numbers, and left for positive numbers.
Therefore f(x) = |x - 2| is a large "V" with the vertex 4 units to the right of the origin.
Review Questions
1. a. Graph the function \begin{align*}f(x) = 2 |x - 1| - 3\end{align*} without a calculator.
b. What is the vertex of the graph and how do you know?
c. Does it open up or down and how do you know?
1. For the function: \begin{align*}f(x) = |x| + c\end{align*} if c is positive, the graph shifts in what direction?
2. For the function: \begin{align*}f(x) = |x| + c\end{align*} if c is negative, the graph shifts in what direction?
3. The function \begin{align*}g(x) = |x - a|\end{align*} represents a shift to the right or the left?
4. The function \begin{align*}h(x) = |x + a|\end{align*} represents a shift to the right or the left?
Describe the transformation that has taken place for the parent function \begin{align*} f(x) = |x|\end{align*}
1. \begin{align*}f(x) = |x| - 5\end{align*}
2. \begin{align*}f(x) = |x - 7| + 3\end{align*}
Write an equation that reflects the transformation that has taken place for the parent function \begin{align*} g(x) = \frac{1}{x}\end{align*}, for it to move in the following ways:
1. Move two spaces up
2. Move four spaces to the right
Write an Equation for each described transformation.
1. a V-shape shifted down 4 units.
2. a V-shape shifted left 6 units
3. a V-shape shifted right 2 units and up 1 unit.
The following graphs are transformations of the parent function \begin{align*}f(x) = |x|\end{align*} in the form of \begin{align*}f(x) = |x - h| = k\end{align*}. Graph or sketch each to observe the type of transformation.
1. \begin{align*}f(x) = |x| + 2\end{align*}. What happens to the graph when you add a number to the function? (i.e. f(x) + k).
2. \begin{align*}f(x) = |x| - 4\end{align*}. What happens to the graph when you subtract a number from the function? (i.e. f(x) - k).
3. \begin{align*}f(x) = |x - 4|\end{align*}. What happens to the graph when you subtract a number in the function? (i.e. f(x - h)).
4. \begin{align*}f(x) = |x + 2|\end{align*}. What happens to the graph when you add a number in the function? (i.e. f(x + h)).
Practice: Graph the following:
1. Let \begin{align*}f(x) = x^2\end{align*}. Let \begin{align*}g(x)\end{align*} be the function obtained by shifting the graph of \begin{align*}f(x)\end{align*} two units to the right and then up three units. Find the equation for \begin{align*}g(x)\end{align*} and then draw its graph
Suppose H(t) gives the height of high tide in Hawaii(H) on a Tuesday, (t) of the year. Use shifts of the function H(t) to find formulas of each of the following functions:
1. F(t), the height of high tide on Fiji on Tuesday (t), given that high tide in Fiji is always one foot higher than high tide in Hawaii.
2. S(d), the height of high tide in Saint Thomas on Tuesday (t), given that high tide in Saint Thomas is the same height as the previous day's height in Hawaii.
1. Braingenie: Graphing functions by translating f(x) = x
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This content is taken from the National STEM Learning Centre's online course, Maths Subject Knowledge: Fractions, Decimals, and Percentages. Join the course to learn more.
6.11
## National STEM Learning Centre
Skip to 0 minutes and 7 secondsPAULA KELLY: A more common way to charge interest is called compound interest. Most loans are charged using compound interest and most savings accounts pay interest using compound interest. If you have a savings account, the interest is added to the amount of money in your account. Future interest is then paid on the total in your account, not just your original investment. To explain how compound interest works, we'll need to make a little detour to have a look at repeated percentages.
Skip to 0 minutes and 41 secondsA rare postage stamp is a square of side length 5 centimetres. To make a larger picture of the stamp, the stamp is photocopied to make a photocopy in which the length of the stamp is increased by 10%. The new photocopy is itself increased by 10%. This process is repeated until the size of the poster required is obtained. What is the size of the stamp on the first five photocopies? So the original size is 5 centimetres. In photocopy one, we have 1.1 multiplied by 5, to give us a side length of 5.5 centimetres. Photocopy two, we have 1.1 times 5.5 to get 6.05 centimetres. Photocopy three is 1.1 times by 6.05 to give us 6.655 centimetres.
Skip to 1 minute and 40 secondsPhotocopy four is 1.1 multiplied by 6.655 to give us 7.3205 centimetres. And photocopy five is 1.1 multiplied by 7.3205. That gives us 8.05255 centimetres. This is quite long-winded, so let's look at what we have done to see if there's a more efficient method. We started with the length of 5 centimetres and multiplied this by 1.1 five times. So we had 5 multiplied by 1.1 by 1.1, 1.1, 1.1, and 1.1. That gave us 8.05255. A simpler way of writing this is 5 times 1.1 to the power of 5. That gives us 8.05255. This allows us to determine how large the 10th photocopy will be without calculating each intermediate stage.
Skip to 2 minutes and 44 secondsSo photocopy 10 will be 5 multiplied by 1.1 to the power of 10. This gives the 12.968712, so about 13 centimetres. So compound interest works in the same way.
Skip to 3 minutes and 0 secondsMICHAEL ANDERSON: So let's see how this applies to compound interest. Say we have 200 pounds, and we're going to invest it for 10 years with a compound interest rate of 3%.
Skip to 3 minutes and 11 secondsPAULA KELLY: OK. So this is where you more commonly would see compound interest. We have 200 pounds.
Skip to 3 minutes and 16 secondsPAULA KELLY: Was it 10 years?
Skip to 3 minutes and 22 secondsAnd then our percentage was 3%?
Skip to 3 minutes and 24 secondsPAULA KELLY: OK. So we could do our 200 multiplied by 0 and add it on, have our very long-winded way. But we've seen our quick way of doing it.
Skip to 3 minutes and 35 secondsPAULA KELLY: So we have our original amount of money, our 200 pounds.
Skip to 3 minutes and 39 secondsPAULA KELLY: We're going to multiply this by 1.03.
Skip to 3 minutes and 43 secondsPAULA KELLY: So we have our original investment--
Skip to 3 minutes and 48 secondsPAULA KELLY: At the moment, this would calculate how much you'd have in your account at the end of one year.
Skip to 3 minutes and 52 secondsPAULA KELLY: We're not looking for one year. You want 10 years.
Skip to 3 minutes and 55 secondsPAULA KELLY: So that repeated multiplier, we can write as a power of 10.
Skip to 3 minutes and 59 secondsMICHAEL ANDERSON: OK. Because we're doing it every year for 10 years?
Skip to 4 minutes and 2 secondsPAULA KELLY: Exactly. And that same profit is reinvested, the 3% of that amount.
Skip to 4 minutes and 8 secondsPAULA KELLY: That's why this is called compound interest. So on our calculator, again, this would give us 268.78 pounds and lots of other decimals. As we using money, we'd round to two decimal places.
Skip to 4 minutes and 21 secondsMICHAEL ANDERSON: OK. So after 10 years, my 200 pounds is now worth nearly 270 pounds?
Skip to 4 minutes and 26 secondsPAULA KELLY: Yes, a small increase. A small increase.
# Compound interest
The more common way to charge interest is called Compound Interest. Most loans are charged using compound interest and most savings accounts pay interest using compound interest. If you have a savings account the interest is added to the amount of money in your account, future interest is then paid on the total in your account, not just your original investment.
Paula and Michael explain compound interest by first looking at repeated percentages, with an example based upon repeated photocopying.
They then apply the same principles to the compound interest problem.
## Interest annually, monthly or daily?
Using the example in the video, we say in this case that the annual percentage rate, (APR) is 79.58%.
This type of interest is used when you pay off part of the loan each month and interest is calculated each month on the amount of money you still owe.
### Daily interest rates
You need to be careful to see when the interest rate is compounded. With some lending companies the interest is compounded daily.
I borrow £100 and pay back the full amount at the end of one year. The interest rate is 0.2% but this is compounded daily. To calculate the annual percentage rate:
$100 \times 1.002^{365} = 207.3568...$
This means I have to pay back £207.36, of which the interest paid is £107.36. So although an interest rate of 0.2% appears small, when compounded daily the annual percentage rate is over 107%. I will have to pay back over double the amount that I borrowed!
If you hunt around the small print of online loan companies you should be able to find the daily compound interest rate.
## Problem worksheet
Now complete the final questions on this week’s worksheet, questions 16 and 17. |
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2 Lesson The Expect students to say or do If students do not, then the Part 1 1. We are going to solve a circle puzzle. Circle puzzles have this format: What does product mean? What does sum mean? ab a b a+b Notice that the product of two numbers, a and b, is in the top section of the puzzle and the sum of the two numbers is in the lower section. If any two sections of the puzzle are filled in, it is possible to determine the remaining two sections.. Complete this circle puzzle. a = and b = 1 or a = 1 and b= What two numbers have a product of? Which of those number pairs have a sum of 3? a b 3 3. Repeat circle puzzles with a variety of number combinations. (See Circle Puzzles at the end of the lesson.)
3 The Expect students to say or do If students do not, then the Part 4. Write the area of the rectangle as a sum. x x x What is the area of each part of the rectangle? x x x 5. Can we combine any like terms? What do we get? 6. What are the dimensions of the large rectangle? 7. What is its area as a product? 8. Since the area of the rectangle is the same regardless of how we write it, we can say x 3x x 1 x. Multiply x 1 x to verify our results. 9. Writing an expression as a product is called factoring. 10. Repeat Steps 4 8 with a variety of rectangles. (See Rectangles at the end of the lesson.) x 3x Refer to Addition and Subtraction of Polynomials. x 1 and x Refer to Multiplying Polynomial Expressions. x 1 x How do we find the area of a rectangle? x 1 x x x x Refer to Multiplying x 3x Polynomial Expressions.
4 The 11. Factor this polynomial by writing the sum as a product: x 4x x 8 You may use the rectangle to help you. Expect students to say or do x x 4x 8 If students do not, then the Model for students. x 4 x 1. Since 4x and x are like terms, this polynomial can be written as x 6x 8. How do we know? 13. When we did the Circle Puzzles, we looked for numbers with a certain product and a certain sum. What numbers have a product of 8 and a sum of 6? 4 and 14. So we can say x 6x 8 x 4 x. 15. Repeat Steps with a variety of quadratic trinomials, moving from writing the trinomial with 4 terms to writing it with 3 terms. Part Now that we know how to factor a quadratic trinomial, let s see what the graph can tell us. 4x x 6x Refer to Addition and Subtraction of Polynomials. What numbers have a product of 8? (1 and 8 or and 4) Which of these have a sum of 6?
5 The 17. Using your graphing calculator, graph y x 6x 8 x 4 x. What do you notice about the graph? (See Teacher Notes.) 18. Repeat Step 17 with additional trinomials until students make the connection between the factors and the x intercepts. 19. Can someone summarize what we have discovered? 0. Use your graphing calculator to graph y x x. What do you notice? 1. What do you think this means about its factors? Expect students to say or do Answers will vary, but listen for the x intercepts are 4 and and the y intercept is 8. The x intercepts are the opposites of the numbers in the factors. The y intercept is the constant term in the trinomial. Answers may vary, but listen for, there are no x intercepts. There are not any factors. If students do not, then the Model for students. It may take several examples before students see the relationship. Where does the graph cross the x axis? Not all trinomials have real factors. Teacher Notes: 1. It is suggested that this lesson be taught over a number of days. Students seldom develop fluency in factoring quickly.. It is recommended that teachers use the circle puzzles in part one as warm up activities for several days prior to the actual factoring lesson. These puzzles will help students think about numbers in ways that will help them in the actual factoring lesson. 3. Use a variety of circle puzzles, including positive and negative numbers. 4. Help students remember the format of circle puzzles by placing the diagram in Step 1 where students can refer to it. 5. When solving circle puzzles, it may be helpful for students to list all of the factor pairs for a number. 6. Once students can solve circle puzzles, move to writing areas as sums and as products in Part. 7. Remind students that the word factoring means to write an expression as a product. 8. It may take several examples of graphing quadratic trinomials before students see the relationship between the factors and the x intercepts. Be patient and let the students discover this relationship.
6 Variations Algebra tiles may be used in place of, or in addition to, the rectangles. Formative Assessment Factor x 5x 36. Answer: x 9 x 4 References Russell Gersten, P. (n.d.). RTI and Mathematics IES Practice Guide Response to Intervention in Mathematics. Retrieved 5, 011, from rti4sucess.
7 Circle Puzzles Rectangles x 5x x 3x x 4x x 5 x 3 3x 1 x 5x x 4x x 3x x 10 5x 0 3x 9
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hexacordoK
2021-02-08
Find values of a and b such that the system of linear equations has (a) no solution, (b) exactly one solution, and (c) infinitely many solutions.
x + 2y = 3
ax + by = −9
irwchh
Given,
The system of linear equations, x + 2y = 3, ax + by = −9.
Part a)
To find values of a,b such that given system has no solution:
Linear equations with no solution are inconsistent equation or the graph of such equations do not intersect that is these lines are parallel.
Consider the first equation,
$x+2y=3$
$⇒2y=-x+3$
$⇒y=-\frac{x}{2}+3$
Above equation has slope $-\frac{1}{2}$.
Consider another equation,
$ax+by=-9$
$⇒by+=ax=9$
$⇒y=\frac{ax}{b}-\frac{9}{b}$
Slope of above equation is $-\frac{a}{b}$.
As given system of equations have no solution the given equations are parallel lines and they have same slopes.
That is,
$-\frac{a}{b}=-\frac{1}{2}$
$⇒a=1,b=2$
Thus, for a = 1 and b = 2 , the given system has no solution.
Part b)
Given system of equations has exactly one solution.
Linear equations with exactly one solution are lines which intersect at exactly one point.
Also, intersecting lines have negative reciprocal slopes.
From part a), first equation have slope $-\frac{1}{2}$ and its negative reciprocal is 2.
Also, slope of second equation is $-\frac{a}{b}$ .
Therefore,
$-\frac{a}{b}=\frac{2}{1}$
$⇒a=-2,b=1$
Algebra homework question answer, step 3, image 1
Thus, for a = -2 and b = 1 or a = 2 and b = -1 the given system has exactly one solution.
.
Part c)
Given system of equation has infinitely many solutions,
Linear equations with infinitely many solutions are lines that intersect at all points.
They are multiples of original equations.
To solve for a and b, multiple of the first equation's constant with the second equation's constant , Just multiply it to the coefficients of the first equations variable.
For multiple,
$-\frac{9}{3}=-3$
Then, multiply first equation by -3,
$-3\left(x+2y=3\right)$
$⇒-3x-6y=-9$
$⇒a=-3,b=-6$
Thus, for a = -3 and b = -6 the system of linear equations have infinitely many solutions.
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# Coplanarity Two Lines
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## Coplanarity of Two Lines In 3D Geometry
Coplanar lines in 3-dimensional geometry are a common mathematical theory. To recall, a plane is 2-D in nature stretching into infinity in the 3-D space, while we have employed vector equations to depict straight lines.
In this chapter, we will further look into what condition is mandatory to be fulfilled for two lines to be coplanar. We will learn to prove how two lines are coplanar using the condition in Cartesian form and vector form using important concepts and solved examples for your better understanding.
### How do we Identify Coplanar Lines?
Why do we want for example lines m →, n and MN→MN → to be coplanar? Let’s take into account the following two cases.
(1) If m ∥n m→∥n→, then the lines are parallel and thus coplanar. Remember that, in such a case, the 3 vectors are also coplanar irrespective of the 3rd vector.
(2) Otherwise, we would require differentiating between bisecting lines (coplanar) and skew lines (not coplanar). If the lines are bisecting, then all their points will lie in the same plane as m m→ and n n→, thus MN→MN→ should lie in that same plane.
### What is the Condition of Vectors Coplanarity?
• For 3-vectors: The 3 vectors are said to be coplanar if their scalar triple product equals 0. Also, if three vectors are linearly dependent, then they are coplanar.
• For n-vectors: Vectors are said to be coplanar if no more than two amongst those vectors are linearly independent.
### Coplanarity of Lines Using Condition in Vector Form
Let’s take into account the equations of two straight lines as below:
• r1 = p1 + λq1
• r2 = p2 + λq2
Wondering what the above equations suggest? It implies that the 1st line crosses through a point, L, whose position vector is provided by l1 and is parallel to m1. In the same manner, the 2nd line passes through another point whose position vector is provided by l2 and is parallel to m2.
The condition for coplanarity under the vector form is that the line connecting the 2 points should be perpendicular to the product of the two vectors namely, p1 and p2. To represent this, we know that the line connecting the two said points can be expressed in the vector form as (l2 – l1). So, we have:
(l2 – l1). (P1 x p2) = 0
### Coplanarity of Lines Using Condition in Vector Form
Coplanarity in Cartesian is a derivative of the vector form. Let’s take into account the two points L (a1, b1, c1) & P (a2, b2, c2) in the Cartesian plane. Let there be 2 vectors p1 and p2. Their direction ratios are provided as x1, y1, z1 and x2, y2, z2 respectively.
The vector equation of the line connecting L and P can be provided by:
LP = (a2 – a1)i + (b2 – b1)j + (c2– c1)k
p1 = x1i + y1j + z1k
p2 = x2i + y2j + z2k
We must now apply the above condition under the vector form in order to derive our condition in Cartesian form. By the condition stated above, the two lines are coplanar if LM. (p1 a p2) = 0. Hence, in the Cartesian form, the matrix representing this equation is provided as 0.
### Solved Examples
Question 1: Prove that the Lines [a + 3]/3 = [b – 1]/1 = [c – 5]/5 and [a + 1]/ -1 = [b – 2]/2 = [c – 5]/5 are Coplanar?
Answer: On comparing the equations, we get:
[a1, b1, c1] = {-3, 1, 5} and [a2, b2, c2] = {-1, 2, 5}.
Now, using the condition of Cartesian form, we shall solve the matrix:
= 2 [5 – 10] – 1 [-15 + 5] + 0 [-6 + 1]
= -10 + 10 = 0
Because the solution of the matrix provides a zero, we can say that the lines given are coplanar
## FAQs on Coplanarity Two Lines
Q1. What Do We Understand By Coplanar Lines?
Answer: Coplanar lines are simply the lines that lie on the same plane. Imagine a sheet of paper or cardboard. Whatever lines are constructed on that sheet will be coplanar since they are lying on the same plane, or on the same flat surface.
Q2. What Do We Understand By Non Coplanar Lines?
Answer: On contrary to the coplanar lines, these are the lines that do not lie on the same plane or a flat surface. Such a plane is said to be non-coplanar. Consider the image given below. The points E and D are non-coplanar because they lie on different planes or different surfaces while points A, B and C are coplanar given that they lie on the same surface. |
# Which Linear Function Represents A Slope Of
## Introduction
In mathematics, linear functions are essential to understanding the relationship between two variables. The slope of a linear function is a key parameter that helps us determine the rate of change between these variables. When analyzing linear functions, it is crucial to know which specific function represents a particular slope.
## Understanding Linear Functions
Linear functions are mathematical expressions that represent a straight line on a graph. They have a general form of y = mx + b, where y represents the dependent variable, x represents the independent variable, m is the slope of the line, and b is the y-intercept of the line.
## The Slope of a Linear Function
The slope of a linear function, denoted by m, is the measure of how steep the line is. It indicates the rate at which the dependent variable changes concerning the independent variable. The slope can be positive, negative, zero, or undefined, depending on the nature of the relationship between the variables.
## Identifying the Slope in Linear Functions
When given a linear function in the form of y = mx + b, the coefficient m represents the slope of the line. By examining the value of m, we can determine whether the slope is positive, negative, zero, or undefined.
## Examples of Linear Functions and Their Slopes
• y = 2x + 3
• In this linear function, the slope m is 2.
• Therefore, the slope of the line is positive.
• y = -3x + 5
• The slope m in this function is -3.
• Thus, the slope of the line is negative.
• y = 4
• For this function, the slope m is 0.
• As a result, the line is horizontal with a slope of 0.
• x = 3
• When the function is in this form, the slope is undefined.
• The line is vertical, which means the slope is undefined.
## Graphical Representation of Slopes
Graphically, the slope of a linear function can be visualized as the steepness of the line. A positive slope indicates an upward trend, a negative slope represents a downward trend, a zero slope results in a horizontal line, and an undefined slope corresponds to a vertical line.
## Calculating the Slope from Two Points
To determine the slope of a linear function, we can use the formula:
m = (y2 – y1) / (x2 – x1)
Where (x1, y1) and (x2, y2) are two points on the line. By substituting the coordinates of the points into the formula, we can calculate the slope accurately.
## Special Cases in Linear Functions
### Horizontal Line
When dealing with a horizontal line, the slope is always 0. This is because the line does not rise or fall; it remains parallel to the x-axis.
### Vertical Line
Vertical lines have an undefined slope since they are perpendicular to the x-axis and do not have a change in the x-coordinate values.
### Zero Slope
A line with a zero slope is horizontal, indicating that there is no change in the y-coordinate values as x changes. This means the line remains constant at a specific y-coordinate level.
### Unit Slope
A unit slope refers to a slope of 1 or -1, where the line increases or decreases by one unit for each unit change in the x-coordinate. The line rises or falls at a 45-degree angle.
## Conclusion
In summary, understanding the concept of slope in linear functions is crucial for analyzing the relationship between variables. By identifying the slope in a linear function, we can interpret the rate of change and determine the direction of the relationship. Whether the slope is positive, negative, zero, or undefined, it provides valuable information about the nature of the function. By examining the coefficient m in the linear function y = mx + b, we can easily determine which linear function represents a specific slope.
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# Problems from Laplace's rule
We throw two six-sided dice. Adding the two results, we want to know the probability of:
a) $$A =$$ "get a total higher than $$8$$"
b) $$B =$$ "a total less than or equal to $$3$$"
c) $$C =$$ "a total higher than $$8$$, or less than or equal to $$3$$"
See development and solution
### Development:
First, we have to describe our sample space. In each dice there may came out a number between $$1$$ and $$6$$, so that all possible outcomes are $$\Omega=\{1-1,1-2,1-3,1-4,1-5,1-6,2-1,2-2,\ldots\}$$$Note that $$36$$ items would be: we can see it thinking that in the first dice it can come out a number between $$1$$ and $$6$$, and in the second one the same, so in total there are $$6\cdot6=36$$ possible ways. All cases are equiprobable, and therefore the probability of each elementary event is $$\dfrac{1}{36}$$, in virtue of Laplace's rule. a) We want to see which ones are the results favorable to $$A =$$ "get more than $$8$$." The results that will meet are all those that turn out to be $$9, 10, 11,$$ and $$12$$. There cannot be results that are higher since the maximum when throwing both dices, if the both of them turn to be $$6$$, is $$6+6=12$$. So, we are saying that, $$A=A_1 \cup A_2 \cup A_3 \cup A_4$$, and we have to think about what events accomplish the following: $$A_1=$$"total addition $$=9$$"$$=\{3-6,4-5,5-4,6-3\}$$, that is, there are four favorable cases. $$A_2=$$"total addition $$=10$$"$$=\{4-6,5-5,6-4\}$$, three favorable cases. $$A_3=$$"total addition $$=11$$"$$=\{5-6,6-5\}$$, two favorable cases. $$A_4=$$"total addition $$=12$$"$$=\{6-6\}$$, one favorable case. As all the events are equiprobable, we can apply Laplace's rule, $$P(A_1)=\dfrac{4}{36}$$, $$P(A_2)=\dfrac{3}{36}$$, $$P(A_3)=\dfrac{2}{36}$$, $$P(A_4)=\dfrac{1}{36}$$. Therefore $$P(A)=P(A_1 \cup A_2 \cup A_3 \cup A_4)=$$$ $$=\dfrac{4}{36}+\dfrac{3}{36}+\dfrac{2}{36}+\dfrac{1}{36}=\dfrac{11}{36}$$$b) To calculate $$P(B)$$, we do as we have done before. Let $$B_1=$$"add up $$1$$", $$B_2=$$"add up $$2$$", $$B_3=$$"add up $$3$$". Then the possible cases that satisfy each of these events are: $$B_1=\emptyset$$, because it is an impossible event. The smallest result that we could obtain would be $$2$$, that is, one on both dices, $$1+1=2$$. $$B_2=\{1-1\}$$ $$B_3=\{1-2,2-1\}$$ So, $$P(B)=P(B_1 \cup B_2 \cup B_3)=$$$ $$=0+\dfrac{1}{36}+\dfrac{2}{36}=\dfrac{3}{36}=\dfrac{1}{12}.$$$As $$B_1$$ is an impossible event, it has no favorable outcome, and therefore, in accordance with Laplace's rule, its probability is $$P(B_1)=\dfrac{0}{36}=0$$. c) In this case, we can see all the events that satisfy $$C$$, but it's a long procedure. We can solve it easily if we note that $$C=A \cup B$$. We can calculate $$P(C)=P(A \cup B)=P(A)+P(B)=\dfrac{11}{36}+\dfrac{1}{12}=\dfrac{14}{36}=\dfrac{7}{18}.$$$
### Solution:
a) $$P(A)=\dfrac{11}{36}$$
b) $$P(B)=\dfrac{1}{12}$$
c) $$P(C)=\dfrac{7}{18}$$
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# wikiHow to Write Complex Functions in u+iv Form
Three Parts:Example 1Example 2Example 3Community Q&A
A complex function is a function that takes in and outputs complex numbers. Just as complex numbers can be written in the form ${\displaystyle z=x+iy,}$ where we are taking apart the complex number into its real and imaginary components, complex functions can also be written as ${\displaystyle w=u(x,y)+iv(x,y),}$ where ${\displaystyle u}$ and ${\displaystyle v}$ are real-valued functions. Writing the function in this way is as simple as substituting ${\displaystyle z=x+iy}$ and simplifying.
### Part 1 Example 1
1. 1
Write ${\displaystyle w=e^{z}}$ in terms of its real and imaginary components. The exponential function is one of the first functions introduced in complex analysis for many reasons, most notably being that it is its own derivative, and highlights the very important relationship between rotations and exponentials.
2. 2
Substitute ${\displaystyle z=x+iy}$ into the function. Use the exponent relation ${\displaystyle e^{a+b}=e^{a}e^{b}.}$
• ${\displaystyle w=e^{x+iy}=e^{x}e^{iy}}$
3. 3
Use Euler's formula to decompose the complex exponential.
• ${\displaystyle w=e^{x}(\cos y+i\sin y)}$
• The function is now in ${\displaystyle u+iv}$ form. Here, we have ${\displaystyle u(x,y)=e^{x}\cos y}$ and ${\displaystyle v(x,y)=e^{x}\sin y.}$
### Part 2 Example 2
1. 1
Write ${\displaystyle w=1/z}$ in terms of its real and imaginary components. Substitute ${\displaystyle z=x+iy}$ into the function.
• ${\displaystyle w={\frac {1}{x+iy}}}$
2. 2
Multiply the numerator and denominator by the complex conjugate and simplify.
• ${\displaystyle {\frac {1}{x+iy}}{\frac {x-iy}{x-iy}}={\frac {x-iy}{x^{2}+y^{2}}}}$
3. 3
Separate the real and imaginary components.
• ${\displaystyle w={\frac {x}{x^{2}+y^{2}}}-i{\frac {y}{x^{2}+y^{2}}}}$
### Part 3 Example 3
1. 1
Write ${\displaystyle w=e^{z}+e^{-z}}$ in terms of its real and imaginary components. Substitute ${\displaystyle z=x+iy}$ into the function.
• ${\displaystyle w=e^{x+iy}+e^{-x-iy}}$
2. 2
Use Euler's formula to decompose the complex exponentials. The expression ${\displaystyle e^{-iy}}$ is the conjugate of ${\displaystyle e^{iy}.}$
• ${\displaystyle w=e^{x}(\cos y+i\sin y)+e^{-x}(\cos y-i\sin y)}$
3. 3
Bring the real and imaginary components together.
• ${\displaystyle w=\cos y(e^{x}+e^{-x})+i\sin y(e^{x}-e^{-x})}$
4. 4
Simplify using the hyperbolic functions. Recall that the hyperbolic functions are defined as ${\displaystyle \cosh x={\frac {e^{x}+e^{-x}}{2}}}$ and ${\displaystyle \sinh x={\frac {e^{x}-e^{-x}}{2}}.}$
• ${\displaystyle w=2\cos y\cosh x+i2\sin y\sinh x}$ |
# Frequency Tables and Histograms
## Create and read data from histograms
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Frequency Tables and Histograms
The coach of the Markwell Cougars track team wants to compare the heights of his team to that of their rivals, the Sampson Hawks. He was wondering if there is a correlation between speed and height.
The coach wrote the following heights from smallest to largest.
Markwell Cougars:
170, 172, 175, 176, 176, 176, 178, 181, 182, 183, 183, 183, 185, 185, 187, 188, 188, 189, 190, 195
Sampson Hawks:
169, 175, 176, 176, 178, 179, 180, 183, 183, 186, 186, 186, 187, 187, 187, 187, 187, 188, 190, 191, 192
Create a visual display of this data.
In this concept, you will learn to use frequency tables and histograms.
### Frequency
A visual display is used to show data. Each type of visual tool has advantages and the best type of plot or graph depends on the situation. Indeed, sometimes it is a matter of preference as many different graphs could be used to illustrate the same data.
Let’s take a look at frequency tables and histograms.
Frequency is a measure of how often something occurs. A frequency table is used to measure and visually show how often a data value occurs.
Let’s look at an example.
A teacher is preparing for parent conferences. In order to provide parents with the most information possible about their children, he wants to organize the grades of the class so that they can compare the grades to the rest of the class.
The math percentages have been calculated and his students earned the following grades:
88, 86, 92, 65, 72, 75, 81, 84, 85, 93, 99, 50, 78, 80, 86, 76, 74, 95, 81, 87, 90, 72, 76, 61, 85, 84, 78, 83.
Grades are determined by percent where 0-59% is an F, 60-69% is a D, 70-79% is a C, 80-89% is a B, and 90-100% is an A. These values make the most logical intervals. Intervals are always chosen depending on the range of the data. He will make a frequency table to illustrate the information.
First, for each student who scored in the given range, he puts an X. Sometimes, frequency tables use X’s and other times, they can use lines for tally marks.
Interval Tally Frequency 90 - 100 XXXXX 5 80 - 89 XXXXXXXXXXXX 12 70 - 79 XXXXXXXX 8 60 - 69 XX 2 0 - 59 X 1
This tally is useful in the sense that it communicates to parents how many students in the class scored in the A range, B range, etc. It would not be as important for the parents to see the individual scores of each student as it would be to see the total number of students in each interval. That way, if their child earned a B, then they would know that the child falls in a category that most other students scored in. If a child earned a D, for example, it would indicate that they are below the general level of the other students and might need additional help.
Next, he creates a histogram. A histogram is similar to a bar graph in that it uses columns to illustrate data on - and -axes. In a histogram, you can use the same intervals as you did for the frequency table. The bars in the histogram will have no space between them.
The histogram shows the same information as the frequency table does. However, the histogram is a type of graph, meaning that it is visual representation. The bars on the histogram are interpreted more easily by size than numerical data.
### Examples
#### Example 1
Earlier, you were given a problem about the track teams and the heights of the runners.
The coach is comparing the heights of his team, the Markwell Cougars to the rival team, the Sampson Hawks. You need to create a double histogram for the data. The data collected is:
Markwell Cougars:
170, 172, 175, 176, 176, 176, 178, 181, 182, 183, 183, 183, 185, 185, 187, 188, 188, 189, 190, 195
Sampson Hawks:
169, 175, 176, 176, 178, 179, 180, 183, 183, 186, 186, 186, 187, 187, 187, 187, 187, 188, 190, 191, 192
First, create a frequency table for the data. Since you are comparing two sets of data, you have to put both data sets into the frequency table.
Markwell Cougars Sampson Hawks Interval Tally Frequency Tally Frequency 160 - 169 1 170 - 179 7 5 180 - 189 11 12 190 - 199 2 3
Next, use this data to create a histogram that compares the data.
You can see from the histogram that both teams have more players in the 180 - 189 interval. However, while the Cougars have more players in the 170 - 179 interval, the Hawks have slightly more in the taller interval. The Hawks have a slight height advantage.
#### Example 2
Create a histogram of the mass of geodes found at a volcanic site. Scientists measured 24 geodes in kilograms and got the following data:
0.8, 0.9, 1.1, 1.1, 1.2, 1.5, 1.5, 1.6, 1.7, 1.7, 1.7, 1.9, 2.0, 2.3, 5.3, 6.8, 7.5, 9.6, 10.5, 11.2, 12.0, 17.6, 23.9, and 26.8.
The minimum item is 0.8 kg and the maximum is 26.8. To get a good idea of the data, you could use intervals that encompass perhaps 4 kg intervals, 5 kg intervals, or 6 kg intervals. Let’s try intervals of 5 kg.
Begin with a frequency table.
Interval Tally Frequency 0 - 5 14 5.1 - 10 4 10.1 - 15 3 15.1 - 20 1 20.1 - 25 1 25.1 - 30 1
Next, create a histogram for this data.
#### Example 3
True or false: An interval is the frequency that an event happens.
Intervals are always chosen depending on the range of the data.
#### Example 4
True or false: To create a histogram, you first need a frequency table.
#### Example 5
True or false: If I wanted to create a histogram on the number of people who went to the town movie theater on the weekend, I would first need to figure out how many people went to the movies on each weekend day and night.
The data collected would help create a frequency table and then you could create your histogram.
### Review
Number of Hours Slept Tally Frequency 5 I 1 6 I I 2 7 I I I I 4 8 I I I 3 9 I I I 3 10 I I I 3 11 I I 2 12 I I 2
1. Create a histogram that illustrates this data.
2. Explain why you chose the intervals that you chose.
3. What can you interpret from your histogram?
Compare the stem-and-leaf plot to the histogram of Melanie’s Christmas gift expenses. She told her husband, “Most of the gifts were about \$60.”
4. Is she telling the truth?
5. Which tool is more useful in making a decision about her truthfulness?
6. Looking at this histogram, can you conclude that most people exercise between 6 - 11 hours per week?
7. What is the fewest number of hours?
8. What is the range of hours?
9. Why do you think they chose the interval that they did?
10 - 15
Conduct your own survey and collect data. Choose attendance rates in your class or vacation days per year for example. Then create a frequency table, histogram and analyze your data. Explain why you chose the interval that you did and which data set had the greatest and least results.
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
TermDefinition
frequency distribution table A frequency distribution table lists the data values, as well as the number of times each value appears in the data set.
Histogram A histogram is a display that indicates the frequency of specified ranges of continuous data values on a graph in the form of immediately adjacent bars. |
# 6.2 Graphs of exponential functions (Page 4/6)
Page 4 / 6
## Stretches and compressions of the parent function f ( x ) = b x
For any factor $\text{\hspace{0.17em}}a>0,$ the function $\text{\hspace{0.17em}}f\left(x\right)=a{\left(b\right)}^{x}$
• is stretched vertically by a factor of $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ if $\text{\hspace{0.17em}}|a|>1.$
• is compressed vertically by a factor of $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ if $\text{\hspace{0.17em}}|a|<1.$
• has a y -intercept of $\text{\hspace{0.17em}}\left(0,a\right).$
• has a horizontal asymptote at $\text{\hspace{0.17em}}y=0,$ a range of $\text{\hspace{0.17em}}\left(0,\infty \right),$ and a domain of $\text{\hspace{0.17em}}\left(-\infty ,\infty \right),$ which are unchanged from the parent function.
## Graphing the stretch of an exponential function
Sketch a graph of $\text{\hspace{0.17em}}f\left(x\right)=4{\left(\frac{1}{2}\right)}^{x}.\text{\hspace{0.17em}}$ State the domain, range, and asymptote.
Before graphing, identify the behavior and key points on the graph.
• Since $\text{\hspace{0.17em}}b=\frac{1}{2}\text{\hspace{0.17em}}$ is between zero and one, the left tail of the graph will increase without bound as $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ decreases, and the right tail will approach the x -axis as $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ increases.
• Since $\text{\hspace{0.17em}}a=4,$ the graph of $\text{\hspace{0.17em}}f\left(x\right)={\left(\frac{1}{2}\right)}^{x}\text{\hspace{0.17em}}$ will be stretched by a factor of $\text{\hspace{0.17em}}4.$
• Create a table of points as shown in [link] . $x$ $-3$ $-2$ $-1$ $0$ $1$ $2$ $3$ $f\left(x\right)=4\left(\frac{1}{2}{\right)}^{x}$ $32$ $16$ $8$ $4$ $2$ $1$ $0.5$
• Plot the y- intercept, $\text{\hspace{0.17em}}\left(0,4\right),$ along with two other points. We can use $\text{\hspace{0.17em}}\left(-1,8\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(1,2\right).$
Draw a smooth curve connecting the points, as shown in [link] .
The domain is $\text{\hspace{0.17em}}\left(-\infty ,\infty \right);\text{\hspace{0.17em}}$ the range is $\text{\hspace{0.17em}}\left(0,\infty \right);\text{\hspace{0.17em}}$ the horizontal asymptote is $\text{\hspace{0.17em}}y=0.$
Sketch the graph of $\text{\hspace{0.17em}}f\left(x\right)=\frac{1}{2}{\left(4\right)}^{x}.\text{\hspace{0.17em}}$ State the domain, range, and asymptote.
The domain is $\text{\hspace{0.17em}}\left(-\infty ,\infty \right);\text{\hspace{0.17em}}$ the range is $\text{\hspace{0.17em}}\left(0,\infty \right);\text{\hspace{0.17em}}$ the horizontal asymptote is $\text{\hspace{0.17em}}y=0.\text{\hspace{0.17em}}$
## Graphing reflections
In addition to shifting, compressing, and stretching a graph, we can also reflect it about the x -axis or the y -axis. When we multiply the parent function $\text{\hspace{0.17em}}f\left(x\right)={b}^{x}\text{\hspace{0.17em}}$ by $\text{\hspace{0.17em}}-1,$ we get a reflection about the x -axis. When we multiply the input by $\text{\hspace{0.17em}}-1,$ we get a reflection about the y -axis. For example, if we begin by graphing the parent function $\text{\hspace{0.17em}}f\left(x\right)={2}^{x},$ we can then graph the two reflections alongside it. The reflection about the x -axis, $\text{\hspace{0.17em}}g\left(x\right)={-2}^{x},$ is shown on the left side of [link] , and the reflection about the y -axis $\text{\hspace{0.17em}}h\left(x\right)={2}^{-x},$ is shown on the right side of [link] .
## Reflections of the parent function f ( x ) = b x
The function $\text{\hspace{0.17em}}f\left(x\right)=-{b}^{x}$
• reflects the parent function $\text{\hspace{0.17em}}f\left(x\right)={b}^{x}\text{\hspace{0.17em}}$ about the x -axis.
• has a y -intercept of $\text{\hspace{0.17em}}\left(0,-1\right).$
• has a range of $\text{\hspace{0.17em}}\left(-\infty ,0\right)$
• has a horizontal asymptote at $\text{\hspace{0.17em}}y=0\text{\hspace{0.17em}}$ and domain of $\text{\hspace{0.17em}}\left(-\infty ,\infty \right),$ which are unchanged from the parent function.
The function $\text{\hspace{0.17em}}f\left(x\right)={b}^{-x}$
• reflects the parent function $\text{\hspace{0.17em}}f\left(x\right)={b}^{x}\text{\hspace{0.17em}}$ about the y -axis.
• has a y -intercept of $\text{\hspace{0.17em}}\left(0,1\right),$ a horizontal asymptote at $\text{\hspace{0.17em}}y=0,$ a range of $\text{\hspace{0.17em}}\left(0,\infty \right),$ and a domain of $\text{\hspace{0.17em}}\left(-\infty ,\infty \right),$ which are unchanged from the parent function.
## Writing and graphing the reflection of an exponential function
Find and graph the equation for a function, $\text{\hspace{0.17em}}g\left(x\right),$ that reflects $\text{\hspace{0.17em}}f\left(x\right)={\left(\frac{1}{4}\right)}^{x}\text{\hspace{0.17em}}$ about the x -axis. State its domain, range, and asymptote.
Since we want to reflect the parent function $\text{\hspace{0.17em}}f\left(x\right)={\left(\frac{1}{4}\right)}^{x}\text{\hspace{0.17em}}$ about the x- axis, we multiply $\text{\hspace{0.17em}}f\left(x\right)\text{\hspace{0.17em}}$ by $\text{\hspace{0.17em}}-1\text{\hspace{0.17em}}$ to get, $\text{\hspace{0.17em}}g\left(x\right)=-{\left(\frac{1}{4}\right)}^{x}.\text{\hspace{0.17em}}$ Next we create a table of points as in [link] .
$x$ $-3$ $-2$ $-1$ $0$ $1$ $2$ $3$ $g\left(x\right)=-\left(\frac{1}{4}{\right)}^{x}$ $-64$ $-16$ $-4$ $-1$ $-0.25$ $-0.0625$ $-0.0156$
Plot the y- intercept, $\text{\hspace{0.17em}}\left(0,-1\right),$ along with two other points. We can use $\text{\hspace{0.17em}}\left(-1,-4\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(1,-0.25\right).$
Draw a smooth curve connecting the points:
The domain is $\text{\hspace{0.17em}}\left(-\infty ,\infty \right);\text{\hspace{0.17em}}$ the range is $\text{\hspace{0.17em}}\left(-\infty ,0\right);\text{\hspace{0.17em}}$ the horizontal asymptote is $\text{\hspace{0.17em}}y=0.$
how do I set up the problem?
what is a solution set?
Harshika
find the subring of gaussian integers?
Rofiqul
hello, I am happy to help!
Abdullahi
hi mam
Mark
find the value of 2x=32
divide by 2 on each side of the equal sign to solve for x
corri
X=16
Michael
Want to review on complex number 1.What are complex number 2.How to solve complex number problems.
Beyan
yes i wantt to review
Mark
use the y -intercept and slope to sketch the graph of the equation y=6x
how do we prove the quadratic formular
Darius
hello, if you have a question about Algebra 2. I may be able to help. I am an Algebra 2 Teacher
thank you help me with how to prove the quadratic equation
Seidu
may God blessed u for that. Please I want u to help me in sets.
Opoku
what is math number
4
Trista
x-2y+3z=-3 2x-y+z=7 -x+3y-z=6
can you teacch how to solve that🙏
Mark
Solve for the first variable in one of the equations, then substitute the result into the other equation. Point For: (6111,4111,−411)(6111,4111,-411) Equation Form: x=6111,y=4111,z=−411x=6111,y=4111,z=-411
Brenna
(61/11,41/11,−4/11)
Brenna
x=61/11 y=41/11 z=−4/11 x=61/11 y=41/11 z=-4/11
Brenna
Need help solving this problem (2/7)^-2
x+2y-z=7
Sidiki
what is the coefficient of -4×
-1
Shedrak
the operation * is x * y =x + y/ 1+(x × y) show if the operation is commutative if x × y is not equal to -1
An investment account was opened with an initial deposit of \$9,600 and earns 7.4% interest, compounded continuously. How much will the account be worth after 15 years?
lim x to infinity e^1-e^-1/log(1+x)
given eccentricity and a point find the equiation |
2.1.3.2.4 - Complements
2.1.3.2.4 - Complements
Complement
The probability that the event does not occur. The complement of $P(A)$ is $P(A^C)$. This may also be written as $P(A')$.
In the diagram below we can see that $A^{C}$ is everything in the sample space that is not A.
Complement of A
$P(A^{C})=1−P(A)$
Example: Coin Flip
When flipping a coin, one can flip heads or tails. Thus, $P(Tails^{C})=P(Heads)$ and $P(Heads^{C})=P(Tails)$
Example: Hearts
If you randomly select a card from a standard 52-card deck, you could pull a heart, diamond, spade, or club. The complement of pulling a heart is the probability of pulling a diamond, spade, or club. In other words: $P(Heart^{C})=P(Diamond,\; Spade,\;\;Club)$
The complement of any outcome is equal to one minus the outcome. In other words: $P(A^{C})=1-P(A)$
It is also true then that: $P(A)=1-P(A^{C})$
Example: Rain
According to the weather report, there is a 30% chance of rain today: $P(Rain) = .30$
Raining and not raining are complements.
$P(Not \:rain)=P(Rain^{C})=1-P(Rain)=1-.30=.70$
There is a 70% chance that it will not rain today.
Example: Winning
The probability that your team will win their next game is calculated to be .45, in other words:
$P(Winning)=.45$
Winning and losing are complements of one another. Therefore the probability that they will lose is:
$P(Losing)=P(Winning^{C})=1-.45=.55$
The sum of all of the probabilities for possible events is equal to 1.
Example: Cards
In a standard 52-card deck there are 26 black cards and 26 red cards. All cards are either black or red.
$P(red)+P(black)=\frac{26}{52}+\frac{26}{52}=1$
Example: Dominant Hand
Of individuals with two hands, it is possible to be right-handed, left-handed, or ambidextrous. Assuming that these are the only three possibilities and that there is no overlap between any of these possibilities:
$P(right\;handed)+P(left\;handed)+P(ambidextrous) = 1$
[1] Link ↥ Has Tooltip/Popover Toggleable Visibility |
Algebra 2 2-4 Independent Practice: More About Linear Equations
starstarstarstarstarstarstarstarstarstar
by Matthew Richardson
| 13 Questions
The outlined content above was added from outside of Formative.
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Graphing: What are the intercepts of 3x + y = 6? Graph the equation. Be sure to include relevant graph detail: label axes, indicate units and scale on both axes, and use arrows to represent end behavior, as appropriate.
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B
C
D
5
5
10
A
B
C
D
6
10
Vocabulary: Tell whether the equation is in slope-intercept, point-slope, or standard form.
y + 2 = -2(x - 1)
standard form
slope-intercept form
point-slope form
7
10
Vocabulary: Tell whether the equation is in slope-intercept, point-slope, or standard form.
y = -ΒΌx + 9
point-slope form
slope-intercept form
standard form
8
10
Vocabulary: Tell whether the equation is in slope-intercept, point-slope, or standard form.
-x - 2y = 1
point-slope form
standard form
slope-intercept form
9
10
Vocabulary: Tell whether the equation is in slope-intercept, point-slope, or standard form.
y - 3 = 4x
slope-intercept form
standard form
point-slope form
10
10
Understanding: Which form would you use to write the equation of a line if you knew its slope and x-intercept? Explain.
11
10
Understanding: If the intercepts of a line are (a, 0) and (0, b), what is the slope of the line? Assume that a and b are both greater than 0.
Enter only the slope, in simplified fraction form.
12
10
Error Analysis: Your friend says the line y = -2x + 3 is perpendicular to the line x + 2y = 8. Do you agree? Explain.
Yes; After converting the second equation to slope-intercept form, we can see that its slope is 1/2. Since -2 and 1/2 are opposite reciprocals, the lines are perpendicular.
No; After converting the second equation to slope-intercept form, we can see that its slope is -1/2. Since -2 and -1/2 are not opposite reciprocals, the lines are not perpendicular.
13
10
Reflection: Math Success |
## Numerical Systems
### Overview
In this article I’ll be covering some things that most individuals already know and some might not even be interested in. There are a few reasons I’m doing this, but the most important one if for me to have a central place where I can refer to some arbitrary things that we know, but that we all do in fact forget and need a refresher on every now and again. So, in this range of articles I’ll have some short summaries of various things including Mathematics, Networking, and Analysis & Design etc. If you’re not interested, move onto one of those articles that do interest you. J In this article I’ll move straight into the Number Systems where I’ll try and explain in a very brief way the difference of the different number systems with their respective Properties. I’ll cover it in the following manner:
1. Mind Map
2. 11 Laws of Numbers
• Law 1: Commutativity
• Law 2: Associativity
• Law 3: Distributivity
• Law 4: Multiplicative Identity
• Law 5: Linearity
• Law 6: Monotonicity
• Law 7: Transitivity
• Law 8: Additive Identity
• Law 9: Absence of zero divisors
• Law 10: Additive Inverse
• Law 11: Multiplicative Inverse
3. Number Systems
4. Overview Table
5. Summary
### Mind Map
I find Mind Maps as an easy way to view and remember certain things. I don’t use them as often as I should, but in this case I thought it wise to do so seeing that it gives us a very brief and high level overview of the numbers and the laws that pertain to them.
Number Systems - Mind Map
### 11 Laws of Numbers
##### Law 1: Commutativity
Definition: For all integers of m and n, m + n = n + m and m.n = n.m
Explanation: You can switch the numbers on both sides of the operand (+ or .) and the answer will be the same.
Example: 2 + 4 = 4 + 2 = 6
2 x 4 = 4 x 2 = 8
##### Law 2: Associativity
Definition: For all integers of m, n and k, m + (n + k) = (m + n) + k and m(nk) = (mn)k
Explanation: You can move the positions of the brackets and the answer will be the same.
Example: 2 + (4 + 3) = (2 + 4) + 3 = 9
2 x (4 x 3) = (2 x 4) x 3 = 24
##### Law 3: Distributivity
Definition: For all integers of m, n and k, m(n + k) = (mn) + (mk)
Explanation: If you have a multiplier on the outside of the brackets (m) and you have an addition (+) happening on the inside of the bracket then you will get the same answer if you multiply the items (n and k) on the inside with the multiplier (m) and then do the addition.
Example: 2(4 + 3) = (2 x 4) + (2 x 3) = 14
##### Law 4: Multiplicative Identity
Definition: For all integers of m, m x 1 = m
Explanation: If you multiply any number (m) with 1, the answer will be the original number (m).
Example: 4 x 1 = 4
##### Law 5: Linearity
Definition: For all integers of m and n, exactly one of the following statements is true
m < n
m = n
m > n
Explanation: A number can only be smaller than, equal to or bigger than any other number.
Example: 2 < 4
2 = 2
2 > 1
##### Law 6: Monotonicity
Definition: For all integers of m, n and k
if m = n, then m + k = n + k and m.k = k.m and
if m < n then m + k < n + k and
if k > 0, then m.k < n.k while
if k < 0, then m.k > n.k
Explanation: If you look at the logic above then you may think that this is the hardest law of them all. It’s not, so look again and look at it in the smallest pieces. If you look at the first section where m = n, you’ll see that this section is the same as Law 1 (Commutativity). Now if you look at the second section where m < n then you’ll see that it’s the same as stating that m < n if you take k out of the picture while doing addition. The third section states the same as the previous section, but with k not being able to be zero seeing that anything multiplied by 0 is in fact then 0.
Example: if m = 2 and n = m then
m + k = n + k
2 + k = 2 + k (substitution)
if m = 2 and n = 3 then
m < n and therefore
m + k < n + k
2 + k < 3 + k (Substitution)
if m = 2 and n = 3 and k = 4 then
m < n and therefore
m.k < n.k
2 x 4 < 3 x 4 (Substitution)
8 < 12
##### Law 7: Transitivity
Definition: For all integers of m, n and k,
if m = n and n = k then m = k and
if m < n and n < k then m < k
Explanation: If m is smaller than n and n is smaller than k then m is also smaller than k.
Example: if m = 1 and n = 2 and k = 3 then 1 < 2 < 3
##### Law 8: Additive Identity
Definition: For all integers of m, m + 0 = m
Explanation: If you add zero to any number the answer will be the original number. In this case the Additive Identity is the number zero.
Example: 1 + 0 = 1
##### Law 9: Absence of zero divisors
Definition: For all integers of m and n, m.n = 0 if and only if m = 0 or n = 0
Explanation: If you multiply zero to any number the answer will be the zero.
Example: 2 x 0 = 0
##### Law 10: Additive Inverse
Definition: For all positive integers m there exists an integer n such that, m + n = 0
Explanation: Every number has an opposite number that exists where if the two were added to each other the answer would be zero.
Example: if m = 2 and n = -2 then m + n = 2 + -2 = 0
##### Law 11: Multiplicative Inverse
Definition: For every non-zero rational number x there exists a rational number y such that, x.y = 1
Explanation: This is the same kind of logic as with Law 10 (Additive Inverse) but over multiplication. So, in fact it states that every number has a number where if the two were multiplied with each other the answer would be 1.
Example: if x = 3/7 and y = 7/3 then x.y = (3/7) x (7/3) = 1
### Number Systems
##### Positive Number: Z+
Definition: All positive integers i.e. numbers greater than zero i.e. n > 0.
Symbol: Z+
Example: {1, 2, 3, 4 …}
Applicable Laws: Law 1: Commutative
Law 2: Associative
Law 3: Distributive
Law 4: Multiplicative Identity
Law 5: Linearity
Law 6: Monotonicity
Law 7: Transitivity
Law 8: Additive Identity
##### Natural Number: N
Definition: All positive integers i.e. numbers greater including zero i.e. n >= 0.
Symbol: N
Example: {0, 1, 2, 3, 4 …}
Applicable Laws: Law 1: Commutative
Law 2: Associative
Law 3: Distributive
Law 4: Multiplicative Identity
Law 5: Linearity
Law 6: Monotonicity
Law 7: Transitivity
Law 8: Additive Identity
Law 9: Absence of zero divisors
##### Integer: Z
Definition: All integers including the numbers less than zero i.e. n >= 0 and n <= 0.
Symbol: Z
Example: {…, -2, -1, 0, 1, 2 …}
Applicable Laws: Law 1: Commutative
Law 2: Associative
Law 3: Distributive
Law 4: Multiplicative Identity
Law 5: Linearity
Law 6: Monotonicity
Law 7: Transitivity
Law 8: Additive Identity
Law 9: Absence of zero divisors
Law 10: Additive Inverses
##### Rational Number: Q
Definition: All numbers of the form p/q where p and q are integers and q is non-zero.
Symbol: Q
Example: 1/2, 1/3, 1/4 …
Applicable Laws: Law 1: Commutative
Law 2: Associative
Law 3: Distributive
Law 4: Multiplicative Identity
Law 5: Linearity
Law 6: Monotonicity
Law 7: Transitivity
Law 8: Additive Identity
Law 9: Absence of zero divisors
Law 10: Additive Inverses
Law 11: Multiplicative Inverses
##### Irrational Number: ‘Q
Note: Irrational Numbers are not seen as a Number system seeing that it not closed as the number systems are. These numbers are those numbers that have no end like pi or the answer of 22/7 symbolized by the sign – ∏.
Definition: Any Real Number which cannot be expressed as a fraction a/b, where a and b are integers, with b being non-zero, and is therefore not a Rational Number.
Symbol: ‘Q (I’m only adding a symbol to this number for personal notes. In fact Irrational Numbers have no symbol.)
Example: √2
Further Reading: Wikipedia
Applicable Laws: Law 1: Commutative
Law 2: Associative
Law 3: Distributive
Law 4: Multiplicative Identity
Law 5: Linearity
Law 6: Monotonicity
Law 7: Transitivity
Law 8: Additive Identity
Law 9: Absence of zero divisors
Law 10: Additive Inverses
Law 11: Multiplicative Inverses
##### Real Number: R
Definition: The expanded number system that consists of the combination of the rational and irrational numbers.
Symbol: R
Example: Same as Q and ‘Q
Applicable Laws: Law 1: Commutative
Law 2: Associative
Law 3: Distributive
Law 4: Multiplicative Identity
Law 5: Linearity
Law 6: Monotonicity
Law 7: Transitivity
Law 8: Additive Identity
Law 9: Absence of zero divisors
Law 10: Additive Inverses
Law 11: Multiplicative Inverses
### Overview Table
In this table you’ll see a short summary and yet another way to put what we already have above. For some people (like me) it helps to have more than one way to see things. I like short and concise, so here is a table showing the Number System and the properties that pertain to them.
Number Systems - Laws
### Summary
This is the first of a series of articles which will have a very wide range. I’m doing this seeing more for myself than anyone else, so if you’re not interested, please have a look at the other articles on this site. I’m sure you’ll find something interesting and more applicable to you. My next article in this range will be around Set Theory, which will include the basics of this field and which should also grow into something that I can use as reference point for the future. As mentioned earlier in this article, I love math and love reading up about things. Writing articles like this is one way for me to learn things and remember them and even though this was not much of a learning experience for me, I’m sure the articles that follow will be J Everything has a base, and if you don’t have that base or if your understanding is a bit rough and outdated then you will get to a point where you’ll have to go back. These kinds of articles are that base for me and will be used as reference point so as to always be a refresher in the most basic things.
I hope you enjoyed this one and if not, stay tuned as I start delving more and more into the interesting things 🙂
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# Lesson 5Negative Exponents with Powers of 10
Let’s see what happens when exponents are negative.
### Learning Targets:
• I can use the exponent rules with negative exponents.
• I know what it means if 10 is raised to a negative power.
## 5.1Number Talk: What's That Exponent?
Solve each equation mentally.
## 5.2Negative Exponent Table
Complete the table to explore what negative exponents mean.
1. As you move toward the left, each number is being multiplied by 10. What is the multiplier as you move right?
2. How does each of these multipliers affect the placement of the decimal?
3. Use the patterns you found in the table to write as a fraction.
4. Use the patterns you found in the table to write as a decimal.
5. Write using a single exponent.
6. Use the patterns in the table to write as a fraction.
1. Match the expressions that describe repeated multiplication in the same way:
\left(10^2\right)^3 \frac{1}{(10 \boldcdot 10)} \boldcdot \frac{1}{(10 \boldcdot 10)} \boldcdot \frac{1}{(10 \boldcdot 10)} \left(10^2\right)^{\text -3} \left(\frac{1}{10} \boldcdot \frac{1}{10}\right)\left(\frac{1}{10} \boldcdot \frac{1}{10}\right)\left(\frac{1}{10} \boldcdot \frac{1}{10}\right) \left(10^{\text -2}\right)^3 \frac{1}{ \frac{1}{10} \boldcdot \frac{1}{10} }\boldcdot \frac{1}{ \frac{1}{10} \boldcdot \frac{1}{10} } \boldcdot \frac{1}{ \frac{1}{10} \boldcdot \frac{1}{10} } \left(10^{\text -2}\right)^{\text-3} (10 \boldcdot 10)(10 \boldcdot 10)(10 \boldcdot 10)
2. Write as a power of 10 with a single exponent. Be prepared to explain your reasoning.
1. Match the expressions that describe repeated multiplication in the same way:
\frac{10^2}{10^5} \frac{ \frac{1}{10} \boldcdot \frac{1}{10} }{ \frac{1}{10} \boldcdot \frac{1}{10} \boldcdot \frac{1}{10}\boldcdot \frac{1}{10}\boldcdot \frac{1}{10} } \frac{10^2}{10^{\text -5}} \frac{10 \boldcdot 10}{10 \boldcdot 10 \boldcdot 10 \boldcdot 10 \boldcdot 10} \frac{10^{\text -2}}{10^5} \frac{ \frac{1}{10} \boldcdot \frac{1}{10} }{ 10 \boldcdot 10\boldcdot 10\boldcdot 10\boldcdot 10 } \frac{10^{\text -2}}{10^{\text -5}} \frac{ 10 \boldcdot 10 }{ \frac{1}{10} \boldcdot \frac{1}{10} \boldcdot \frac{1}{10}\boldcdot \frac{1}{10}\boldcdot \frac{1}{10}}
2. Write as a power of 10 with a single exponent. Be prepared to explain your reasoning.
1. Match the expressions that describe repeated multiplication in the same way:
10^4 \boldcdot 10^3 (10 \boldcdot 10 \boldcdot 10 \boldcdot 10) \boldcdot ( \frac{1}{10} \boldcdot \frac{1}{10}\boldcdot \frac{1}{10}) 10^4 \boldcdot 10^{\text -3} \left(\frac{1}{10} \boldcdot \frac{1}{10} \boldcdot \frac{1}{10} \boldcdot \frac{1}{10}\right) \boldcdot \left( \frac{1}{10} \boldcdot \frac{1}{10} \boldcdot \frac{1}{10}\right) 10^{\text -4} \boldcdot 10^3 \left(\frac{1}{10}\boldcdot \frac{1}{10} \boldcdot \frac{1}{10} \boldcdot \frac{1}{10}\right) \boldcdot \left(10 \boldcdot 10 \boldcdot 10\right) 10^{\text -4} \boldcdot 10^{\text -3} (10 \boldcdot 10 \boldcdot 10 \boldcdot 10) \boldcdot (10 \boldcdot 10 \boldcdot 10)
2. Write as a power of 10 with a single exponent. Be prepared to explain your reasoning.
### Are you ready for more?
Priya, Jada, Han, and Diego are playing a game. They stand in a circle in this order and take turns playing a game.
Priya says, SAFE. Jada, standing to Priya's left, says, OUT and leaves the circle. Han is next: he says, SAFE. Then Diego says, OUT and leaves the circle. At this point, only Priya and Han are left. They continue to alternate. Priya says, SAFE. Han says, OUT and leaves the circle. Priya is the only person left, so she is the winner.
Priya says, “I knew I’d be the only one left, since I went first.”
1. Record this game on paper a few times with different numbers of players. Does the person who starts always win?
2. Try to find as many numbers as you can where the person who starts always wins. What patterns do you notice?
## Lesson 5 Summary
When we multiply a positive power of 10 by , the exponent decreases by 1: This is true for any positive power of 10. We can reason in a similar way that multiplying by 2 factors that are decreases the exponent by 2:
That means we can extend the rules to use negative exponents if we make . Just as is two factors that are 10, we have that is two factors that are . More generally, the exponent rules we have developed are true for any integers and if we make
Here is an example of extending the rule to use negative exponents: To see why, notice that which is equal to .
Here is an example of extending the rule to use negative exponents: To see why, notice that . This means that
## Lesson 5 Practice Problems
1. Write with a single exponent: (ex: )
2. Write each expression as a single power of 10.
3. Select all of the following that are equivalent to :
4. Match each equation to the situation it describes. Explain what the constant of proportionality means in each equation.
Equations:
Situations:
1. A dump truck is hauling loads of dirt to a construction site. After 20 loads, there are 70 square feet of dirt.
2. I am making a water and salt mixture that has 2 cups of salt for every 6 cups of water.
3. A store has a “4 for \$10” sale on hats.
4. For every 48 cookies I bake, my students get 24.
1. Explain why triangle is similar to
2. Find the missing side lengths. |
# Is O is the smallest positive integer
## Understand math 3, textbook
Which two whole numbers are represented by markings on the number line? Compare the two numbers! Give reasons for the answer! Solution: The numbers ‒8 and ‒7 are shown on the number line. It can be seen that ‒8 <‒7, since the number ‒8 is on the number line to the left of the number ‒7. While with natural numbers every number has a successor, but not every number has a predecessor (the smallest natural number 0 has no predecessor among the natural numbers), this is different with whole numbers: Every whole number has both a predecessor and a successor . There is no smallest or largest integer. In Exercise 1.20 the number ‒8 is the predecessor of the number ‒7 or the number ‒7 is the successor of the number ‒8. Enter the predecessor and the successor of the number ‒15! Solution: The predecessor of the number ‒15 is ‒16, the successor of the number ‒15 is ‒14. Exercises Basics Which two whole numbers are indicated by markings on the number line? Compare the two numbers! Give reasons for the answer! a) 3 2 4 5 6 1 0 ‒1 ‒2 ‒6 ‒3 ‒4 ‒5 b) 1 0 c) 6 4 8 10 14 2 0 ‒2 12 ‒4 ‒14 ‒12 ‒6 ‒8 ‒10 d ) 50 10 0 ‒50 Enter the predecessor and the successor of the number! a) ‒5 c) ‒1 e) ‒99 b) ‒19 d) 0 f) ‒300 Write down all integers between a) ‒7 and 4, b) ‒11 and ‒2, c) ‒8 and 8 lie! How many numbers are there in each case? Tick all that apply! The predecessor of 0 is a negative number. The predecessor of 0 is a positive number. The successor of every negative number is negative. The successor of every positive number is positive. A 1.20 I 0 ‒1 1 2 3 ‒2 ‒3 ‒4 ‒5 ‒9 ‒6 ‒7 ‒8 1.21 D 1.22 IA 1.23 D Ó 1.24 D 1.25 I Ó exercise - gi73my 26 I1 Numbers and measurements For testing purposes only - property of the publisher öbv |
# Table of 16
multiplication Table
Table of 16 results in the number ‘16’ being multiplied by a set of whole numbers. Here is a table of 16 students up to 20 times that can help them with repetitive problems.
Repeating 16 times alone makes the 16th table. Let us see how;
16 + 16 = 32
16 + 16 + 16 = 48
16+16+16+16 = 64
And so on.
### Memorization Tips Table of 16 in Math
Students may find reading 16 tables a little challenging. So, here are some tips and tricks for students to read Table 16 so that they can memorize it easily.
A continuous study of tables during early education helps students solve long-standing problems in their academic as well as daily life using multiplication and division. The 16 Times Table helps you visualize repetitive patterns and apply them to real-world situations.
The pattern of table of 16 is that the digits in the unit area are followed by 6, 2, 8, 4, 0 and then repeated. All 16 multiples are even numbers
16 times any equal number will have the same digit in the unit area as an equal number (16 x 2 = 32, 16 x 4 = 64, etc.).
You can skip alternate reading in table of 8 to find numbers to get 16 multiplication tables, eg: 8 x 2 is 16,
8 x 4 is 32,
8 x 6 is 48,
8 x 8 is 64, 8 x 10 is 80 and so on.
As we can see, this gives the table of in sequence: 16, 32, 48, 64, 80
### Worksheet on Table 16:
Example 1: What is the number 16 times 8 plus 16 times 10?
Solution:
From the 16th table, we have;
16 times 8 that is 16 x 8 = 128
16 times 10 that is 16 x 10 = 160
Total = 128 + 160 = 288
Example 2: Solve the Equation: 16 x 11 – 100.
Solution:
From the table of 16, we have;
16 x 11 = 176
Therefore, the important difference is:
176 – 100 = 76
Example 3: Radha brought 5 packets of coffee. The cost of each packet is Rs 16. How much does the package cost?
Solution:
Given,
Number of packets of coffee 5
Pack of one toffee pack costs = Rs.16
Total cost of 5 packets amounts to Rs.16 x 5 = Rs.80
Example: 4 Miley’s mother made 4 pizzas. Miley divides each into 4 quadrants. You got 5 pieces of olives on each quadrant. Using 16 Times Table, find out how many pieces of olives are placed on top of 4 pizzas?
Solution:
4 pizza bars 4 quadrant bars 5 olives which means 16 quadrant bars 5 olive.
Let’s figure this out in terms of 16 × 5 = 80
So, there are 80 pieces of olive on top of the pizza. |
The LCM of 5 and 7 is 35. To find the LCM (least common multiple) of 5 and 7, we need to find the multiples of 5 and 7 (multiples of 5 = 5, 10, 15, 20 . . . .
Also, What is the LCM of 5 and 3?
What is the LCM of 3 and 5? The LCM of 3 and 5 is 15.
Hereof, What is the HCF of 5 10 and 15?
HCF of 5, 10 and 15 by Prime Factorization
As visible, 5, 10 and 15 have only one common prime factor i.e. 5. Hence, the HCF of 5, 10 and 15 is 5.
Also to know What is the LCM of 5 and 7 and 10? The LCM of 5, 7, and 10 is 70.
To find the LCM (least common multiple) of 5, 7, and 10, we need to find the multiples of 5, 7, and 10 (multiples of 5 = 5, 10, 15, 20 . . . .
What is the LCM of 3 and 7?
Answer: LCM of 3 and 7 is 21.
## What is the HCF of 27 and 30?
Therefore, the greatest common factor of 27 and 30 is 3.
## What is the LCM of 5 15 and 3?
Answer: LCM of 3, 5 and 15 is 15.
## What is the LCM for 15 and 30?
What is the LCM of 15 and 30? Answer: LCM of 15 and 30 is 30.
## What is the HCF of 5 15 and 30?
To find the HCF of 5 and 30, we will find the prime factorization of the given numbers, i.e. 5 = 5; 30 = 2 × 3 × 5. ⇒ Since 5 is the only common prime factor of 5 and 30. Hence, HCF (5, 30) = 5.
## What is the HCF of 5 and 10?
HCF of 5 and 10 by Prime Factorization
Prime factorization of 5 and 10 is (5) and (2 × 5) respectively. As visible, 5 and 10 have only one common prime factor i.e. 5. Hence, the HCF of 5 and 10 is 5.
## What is the HCF of 15?
Factors of 15 (Fifteen) = 1, 3, 5 and 15. Factors of 35 (Thirty five) = 1, 5, 7 and 35. Therefore, common factor of 15 (Fifteen) and 35 (Thirty five) = 1 and 5. Highest common factor (H.C.F) of 15 (Fifteen) and 35 (Thirty five) = 5.
## What is the LCM of 6 and 7 and 10?
Least Common Multiple of 10, 6, 7
Least common multiple (LCM) of 10, 6, 7 is 210.
## What’s the LCM of 10 and 7?
Answer: LCM of 7 and 10 is 70.
## What is the LCM for 15 and 10?
Answer: LCM of 10 and 15 is 30.
## What are 5 common multiples of 3 and 7?
The common multiples of 3 and 7 up to 70 are 21, 42, and 63. A multiple of a number a is a number that is the product of a and another number.
## What is the GCF of 3 and 7?
Answer: GCF of 3 and 7 is 1.
## What is the LCM of 7 and 21?
Answer: LCM of 7 and 21 is 21.
## What is the HCF of 30?
Factors of 30 (Thirty) = 1, 2, 3, 5, 6, 10, 15 and 30. Factors of 24 (Twenty four) = 1, 2, 3, 4, 6, 8, 12 and 24. Therefore, common factor of 30 (Thirty) and 24 (Twenty four = 1, 2, 3, and 6. Highest common factor (H.C.F) of 30 (Thirty) and 24 (Twenty four = 6.
## What is the HCF of 36 and 24?
Answer: HCF of 24 and 36 is 12.
## What is the HCF of 27 and 63?
Answer: HCF of 27 and 63 is 9.
## What is the HCF of 3 and 15?
⇒ Since 3 is the only common prime factor of 3 and 15. Hence, HCF (3, 15) = 3.
## What is the LCM of 10 15 and 20?
Answer: LCM of 10, 15, and 20 is 60.
## What are the LCM for 15?
What is the LCM of 15 and 15? The LCM of 15 and 15 is 15.
## What is the LCM of 15 and 20?
The smallest number to appear on both lists is 60, so 60 is the least common multiple of 15 and 20.
## What is the LCM of 15 and 25?
What is the LCM of 15 and 25? Answer: LCM of 15 and 25 is 75. |
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# Definitions and Introduction to Proof
## Formal arguments of mathematical statements written in paragraph, two-column, and flow diagram formats.
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Progress
Progress
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Theorems and Proofs
Most of the geometry concepts and theorems that are learned in high school today were first discovered and proved by mathematicians such as Euclid thousands of years ago. Given that these geometry concepts and theorems have been known to be true for thousands of years, why is it important that you learn how to prove them for yourself?
#### Watch This
http://www.youtube.com/watch?v=M6cbpQ_TUAQ James Sousa: Introduction to Proof Using Properties of Congruence
#### Guidance
In geometry, a postulate is a statement that is assumed to be true based on basic geometric principles. An example of a postulate is the statement “through any two points is exactly one line”. A long time ago, postulates were the ideas that were thought to be so obviously true they did not require a proof. A theorem is a mathematical statement that can and must be proven to be true. You've heard the word theorem before when you learned about the Pythagorean Theorem. Much of your future work in geometry will involve learning different theorems and proving they are true.
What does it mean to “prove” something? In the past you have often been asked to “justify your answer” or “explain your reasoning”. This is because it is important to be able to show your thinking to others so that ideally they can follow it and agree that you must be right. A proof is just a formal way of justifying your answer. In a proof your goal is to use given information and facts that everyone agrees are true to show that a new statement must also be true.
Suppose you are given the picture below and asked to prove that . This means that you need to give a convincing mathematical argument as to why the line segments MUST be congruent.
Here is an example of a paragraph-style proof. This is similar to a detailed explanation you might have given in the past.
because it is marked in the diagram. Also, and are both right angles because it is marked in the diagram. This means that and are right triangles because right triangles are triangles with right angles. Both triangles contain segment . because of the reflexive property that any segment is congruent to itself. by because they are right triangles with a pair of congruent legs and congruent hypotenuses. because they are corresponding segments and corresponding parts of congruent triangles must be congruent.
There are two key components of any proof -- statements and reasons.
• The statements are the claims that you are making throughout your proof that lead to what you are ultimately trying to prove is true. Statements are written in red throughout the previous proof.
• The reasons are the reasons you give for why the statements must be true. Reasons are written in blue throughout the previous proof. If you don't give reasons, your proof is not convincing and so is not complete.
When writing a proof, your job is to make everything as clear as possible, because you need other people to be able to understand and believe your proof. Skipping steps and using complicated words is not helpful!
There are many different styles for writing proofs. In American high schools, a style of proof called the two-column proof has traditionally been the most common (see Example A). In college and beyond, paragraph proofs are common. An example of a style of proof that is more visual is a flow diagram proof (see Example B). No matter what style is used, the key components of statements and reasons must be present. You should be familiar with different styles of proof, but ultimately can use whichever style you prefer.
Learning to write proofs can be difficult. One of the best ways to learn is to study examples to get a sense for what proofs look like.
Example A
Rewrite the proof from the guidance in a two-column format.
Using the picture below, prove that .
Solution: In a two-column proof, the statements and reasons are organized into two columns. All of the same logic that was used in the paragraph proof will be used here. Look at the proof below and compare it to the paragraph proof from the guidance.
Statements Reasons Given and are right angles Given and are right triangles definition of right triangles reflexive property CPCTC (corresponding parts of congurent triangles must be congurent)
There are a couple of points to note about two-column proofs.
1. For a two-column proof, instead of saying “it is marked in the diagram” as a reason, you just write “given”. You can use the reason “given” for anything that was stated up front or marked in a diagram. Typically, the first few rows of your proof will always be the “givens”.
2. In a two-column proof you will use less words than in a paragraph proof, because you are not writing in complete sentences.
• and the other criteria for triangle congruence are always acceptable reasons if you have shown in earlier rows that each part of the criteria has been met. You do not need to write a sentence explaining why you can use .
• Instead of stating right triangles are triangles with right angles as a reason, you can just say “definition of right triangles”. Definitions are always acceptable reasons.
• CPCTC is an abbreviation for the statement “corresponding parts of congruent triangles are congruent”. The abbreviation was developed because this reason is used often, and it can be cumbersome to write it over and over.
Example B
Rewrite the proof from the guidance in a flow diagram format.
Using the picture below, prove that .
Solution: In a flow diagram, the statements and reasons will be organized into boxes that are connected with arrows to show the flow of logic. Look at the proof below and compare it to the two-column and paragraph versions of the same proof. In the proof below, statements are written in red and reasons are written in blue.
There are a couple of points to note about flow diagram proofs.
1. Statements are written inside the boxes and the reasons the statements must be true are written below the boxes.
2. The arrows show the flow of logic. If two boxes are connected by arrows it means that the statement in the lower box can be made because the statement in the upper box is true. Notice that three boxes point towards the statement that . This is because all three of those statements were necessary for making the conclusion that the two triangles are congruent.
3. Just like in the two-column format, “given” is the reason used for anything that was stated up front or marked in the diagram. The “given” reasons will be towards the top of the flow diagram.
4. Just like in the two-column format, you use abbreviations where possible. , other triangle congruence criteria, CPCTC, and definitions are all acceptable reasons.
Example C
Each proof below has a mistake, can you figure out where the mistake is and why it is a mistake?
Using the picture below, prove that .
PROOF A:
Statements Reasons Given CPCTC (corresponding parts of congruent triangles must be congruent)
PROOF B:
looks to be the same size and shape as , so the two triangles are congruent because they are corresponding segments and corresponding parts of congruent triangles must be congruent.
Solution: PROOF A is incorrect because it is missing steps. You can't say that the two triangles are congruent by without having shown that all the parts of the criteria have been met (congruent leg pair, congruent hypotenuse pair, right triangles). Be careful when writing proofs that you don't skip over steps, even if the steps seem obvious.
PROOF B is incorrect because it did not convincingly explain why the two triangles have to be congruent. Looking congruent is not a good enough reason. For proving triangles are congruent, there are five triangle congruence criteria to use. If you don't have enough information to use one of those five criteria, you can't prove that the triangles are congruent.
Remember, your goal when writing a proof is to convince everyone else that what you are trying to show is true actually is true. If you skip steps or use reasons that aren't convincing, other people won't believe your proof.
##### Concept Problem Revisited
Most of the geometry concepts and theorems that are taught at the high school level today were first discovered and proved by mathematicians such as Euclid thousands of years ago. Given that these geometry concepts and theorems have been known to be true for thousands of years, why is it important that you learn how to prove them for yourself?
There are many reasons why it is valuable to learn to write proofs for yourself. Even though all of the theorems you will learn in geometry have already been proven, mathematicians today are working on trying to prove new ideas that will hopefully help to advance science/technology/medicine. Writing proofs in geometry class allows you to see what proofs are all about and practice writing them. That way, when you someday want to prove something new, you can feel confident in your proof writing abilities.
Writing proofs is all about logic. If you get good at writing proofs, this logical thinking can transfer to other subjects. Writing a persuasive essay about any topic is very similar to writing a paragraph proof. Knowing how to persuade others to believe your way of thinking can be very helpful in many careers and life in general.
#### Vocabulary
A postulate is a statement that is assumed to be true without proof.
A theorem is a true statement that must/can be proven.
A proof is a mathematical argument that shows step by step why a statement must be true. All proofs must contain statements and reasons.
A paragraph proof is a proof that is written out in words/sentences.
A two-column proof organizes statements and reasons into columns.
A flow diagram proof organizes statements in boxes with reasons underneath. Arrows show the flow of logic from the original assumptions and given statements to the conclusion.
The reflexive property states that anything is congruent to itself.
CPCTC is an abbreviation for “corresponding parts of congruent triangles are congruent”. It is used to show that two angles or line segments are congruent after it has been shown that two triangles are congruent.
#### Guided Practice
Given: is the midpoint of and of . .
Prove:
1. Write a paragraph proof that shows that .
2. Write a two-column proof that shows that .
3. Write a flow diagram proof that shows that .
No matter which style of proof you use, before starting to write you should brainstorm what you will say in your proof. Start by looking at the given information and thinking about what you know based on each given fact.
• The fact that is a midpoint means it is right in the middle of the two line segments. This means there are two pairs of segments that must be congruent. Mark these congruent segments on the diagram as you brainstorm. This will help you to keep track of what you know!
• You also are given that . This should be marked on the diagram as well.
Next think about what other conclusions you can make based on what you have now marked on the diagram. You have SAS criteria marked, so you can say that the two triangles are congruent. This will allow you to be able to say that , because they are corresponding parts of the triangles.
Once you have thought through the proof and your approach, start writing. In all proofs, the statements have been written in red and the reasons have been written in blue.
1. is the midpoint of and because it is given information. This means that and , because midpoints divide segments into two congruent segments. Also, because it is given information. by because they are triangles with two pairs of corresponding sides congruent and included angles congruent. because they are corresponding segments and corresponding parts of congruent triangles must be congruent.
2.
Statements Reasons is the midpoint of and Given and definition of midpoint Given CPCTC
3.
#### Practice
1. What’s the difference between a postulate and a theorem?
2. What are the two main components of any proof?
3. What does it mean when a reason in a proof is “given”?
4. What should the last line/sentence/box for any proof be?
5. What are three styles of proof?
For 6-8, consider the proof below.
Given triangles and as marked, prove that .
Statements Reasons Given Given ??? ??? CPCTC (corresponding parts of congruent triangles must be congruent)
6. Fill in the missing statements and reasons.
7. Rewrite this proof as a paragraph proof.
8. Rewrite this proof as a flow diagram proof.
For 9-11, consider the proof below.
Given: Circle with center . .
Prove:
_______________________ because it is given information. Point is the center of the circle because _______________________. , , are all radii of the circle, because they are segments that connect the center of the circle with the circle. and because all ______ are congruent. _______ because they are triangles with two pairs of corresponding sides congruent and included angles congruent.
9. Fill in the blanks.
10. Rewrite this proof as a two-column proof.
11. Rewrite this proof as a flow diagram proof.
For 12-14, consider the proof below.
Given: Square
Prove:
12. Fill in the missing boxes/reasons.
13. Rewrite this proof as a paragraph proof.
14. Rewrite this proof as a two-column proof.
15. Give an example of a real life situation where being able to persuade someone else that something is true would be helpful.
### Vocabulary Language: English
argument
argument
An argument in logical reasoning is a series of statements, progressing (usually in order) from the premises, which are the assumptions (true or untrue), to the conclusion.
postulate
postulate
A postulate is a statement that is accepted as true without proof.
proof
proof
A proof is a series of true statements leading to the acceptance of truth of a more complex statement.
Pythagorean Theorem
Pythagorean Theorem
The Pythagorean Theorem is a mathematical relationship between the sides of a right triangle, given by $a^2 + b^2 = c^2$, where $a$ and $b$ are legs of the triangle and $c$ is the hypotenuse of the triangle.
theorem
theorem
A theorem is a statement that can be proven true using postulates, definitions, and other theorems that have already been proven. |
$\newcommand{\N}{\mathbb{N}} \newcommand{\Z}{\mathbb{Z}} \newcommand{\R}{\mathbb{R}} \newcommand{\lt}{<} \newcommand{\gt}{>} \newcommand{\amp}{&}$
## SectionA.2Algebra Review
### SubsectionA.2.1Lines and Linear Functions
Lines are perhaps the most important elementary geometric object. A line captures the idea of following a given direction without turning. In ordinary language, we sometimes think of a line as a smooth curve that we could draw. Mathematically, a line would then be a straight line or a straight curve that does not bend. Algebraically, we can define a line using an equation involving two variables. This section reviews the basic principles of the algebraic properties of lines.
###### DefinitionA.2.1General Equation of a Line
Every line in the $(x,y)$ plane can be described as the set of points $(x,y)$ that satisfy an equation
$$Ax + By = C\tag{A.2.1}$$
where $A\text{,}$ $B$ and $C$ are constants.
There are some special cases that describe horizontal and vertical lines. The $(x,y)$ plane uses $x$ as the horizontal axis (independent variable) and $y$ as the vertical axis (dependent variable). So a horizontal line is a line where the dependent variable is constant while a vertical line is a line where the independent variable constant.
###### DefinitionA.2.2Horizontal Line
A horizontal line in the $(x,y)$ plane is the set of points that satisfy an equation
$$y=k\tag{A.2.2}$$
where $k$ is a constant.
###### DefinitionA.2.3Vertical Line
A vertical line in the $(x,y)$ plane is the set of points that satisfy an equation
$$x=h\tag{A.2.3}$$
where $h$ is a constant.
All other lines have an equation that involves both variables. We often wish to think of the line as describing the dependent variable as a function of the independent variable. These equations involve the calculation of the slope, which represents a rate (or ratio) of change.
###### DefinitionA.2.4Slope as Rate of Change
Given any two points $(x_1,y_1)$ and $(x_2,y_2)$ on a non-vertical line, the change in the dependent variable $\Delta y = y_2 - y_1$ is proportional to the change in the independent variable $\Delta x = x_2 - x_1\text{,}$ written $\Delta y = m \cdot \Delta x\text{.}$ The proportionality constant $m$ is called the slope of the line, calculated as the ratio of changes (rate of change)
$$m = \frac{\Delta y}{\Delta x} = \frac{y_2-y_1}{x_2-x_1}.\tag{A.2.4}$$
Knowing the slope and one point is enough to quickly find an equation of a line.
###### DefinitionA.2.5Point–Slope Equation of Line
Given that a line has slope $m$ and passes through a point $(x,y)=(h,k)\text{,}$ every point on the line satisfies the equation
$$y=m \, (x-h)+k.\tag{A.2.5}$$
We interpret $k$ as the starting value for $y$ and the expression $\Delta y = m\,(x-h)$ as the change in $y$ given the change in $x\text{,}$ $\Delta x = x-h\text{.}$
A special case of the point–slope equation of the line occurs when the point is on the $y$-axis or, in other words, is a $y$-intercept.
###### DefinitionA.2.6Slope–Intercept Equation of Line
Given that a line has slope $m$ and passes through a $y$-intercept $(x,y)=(0,b)\text{,}$ every point on the line satisfies the equation
$$y=mx+b.\tag{A.2.6}$$
###### RemarkA.2.7
Preparatory mathematics courses often emphasize the slope–intercept equation of a line as if it were the most important. However, the point–slope equation is the preferred equation to use in almost every circumstance.
Another special case of the point–slope equation of a line is when we know the slope and the $x$-intercept.
###### DefinitionA.2.8Slope and X-Intercept Equation of Line
Given that a line has slope $m$ and passes through an $x$-intercept $(x,y)=(a,0)\text{,}$ every point on the line satisfies the equation
$$y=m(x-a).\text{.}\tag{A.2.7}$$
###### DefinitionA.2.9
A quadratic polynomial in a variable $x$ is an algebraic function that is equal to a formula of the form
$$f(x) = ax^2+bx+c,\tag{A.2.8}$$
where $a\text{,}$ $b$ and $c$ are constants called coefficients.
The graph of a quadratic function, $y=ax^2+bx+c\text{,}$ is a parabola. Such a parabola has a mirror symmetry across a vertical line that passes through its vertex $x=-\frac{b}{2a}\text{.}$ Depending on whether the vertex is above, on or below the $x$-axis and whether the parabola opens up or down, the graph can cross the $x$-axis twice, once or never. The location of these points are called $x$-intercepts, roots or zeros of the function.
Zeros are closely related to factoring. If we know the zeros, then we can immediately rewrite the polynomial in a factored form. On the other hand, if we know the factors, then we can quickly solve for the zeros without using the quadratic formula. This is a consequence of the fundamental properties of numbers in Theorem Theorem A.1.6.
A quadratic polynomial that has complex roots is called irreducible because it can not be rewritten in a factored form involving only real roots.
There are some tricks to factoring that can be useful to know. Factoring is the reverse process of multiplying by distribution, so we start by noticing what happens when you multiply out two simple factors:
\begin{equation*} (x+a)(x+b) = x^2+(a+b)x+ab. \end{equation*}
Notice that the coefficient of $x$ is the sum $a+b$ and the constant term is the product $ab\text{.}$ When trying to factor a quadratic, look for numbers that multiply to give the product term and add to give the coefficient of $x\text{.}$ This is often a matter of trial and error.
Knowing one root $x=r$ of a quadratic $f(x)=ax^2+bx+c$ so that $f(r)=0\text{,}$ we know that $x-r$ is a factor. The other factor can be determined easily.
Synthetic division is a procedure that works for quadratics as well as higher order polynomials. This procedure uses a table that starts with the coefficients on the first row. For a more thorough discussion for higher-order polynomials, see Algorithm 20.
Every quadratic can be rewritten in a form $y=a(x-h)^2+k$ where $(h,k)$ is the vertex of the parabola and $a$ is the leading coefficient and scaling factor. The process of rewriting a quadratic $y=ax^2+bx+c$ in this vertex form is called completing the square. It is based on noticing what happens with expanding the square of a binomial, $(x+a)^2 = x^2+2ax+a^2\text{.}$ The strategy involves adding a term to form a perfect square and subtracting the same term to guarantee the expression does not change.
###### ExampleA.2.15
Complete the square for $3x^2-4x+1\text{.}$
Solution
1. Group the non-constant terms and factor out the leading coefficient.
\begin{align*} 3x^2-4x+1 &= (3x^2-4x) + 1\\ & = 3(x^2 - \frac{4}{3}x) + 1 \end{align*}
2. Recognize the coefficient $-\frac{4}{3}$ as double $-\frac{2}{3}$ and use this to complete the square.
\begin{align*} 3x^2-4x+1 & = 3\Big(x^2 + 2 \cdot \frac{-2}{3}x \Big) + 1 \\ & = 3\Big(x^2 + 2 \cdot \frac{-2}{3}x + (\frac{-2}{3})^2 - (\frac{-2}{3})^2 \Big) + 1 \\ & = 3 \Big(x^2 - \frac{4}{3} x + \frac{4}{9}\Big) - 3(\frac{4}{9}) + 1\\ & = 3 \Big(x - \frac{2}{3}\Big)^2 - \frac{4}{3} + 1 = 3\Big(x-\frac{2}{3}\Big)^2 - \frac{1}{3}. \end{align*}
3. Interpret the results as saying $h=\frac{2}{3}$ (because the completed square is always of the form $(x-h)^2$) and $k=-\frac{1}{3}\text{.}$ Thus the vertex of the parabola is at $(\frac{2}{3}, -\frac{1}{3})\text{.}$ The leading coefficient $a=3$ indicates that the parabola opens up and is three times steeper than the standard parabola $y=x^2\text{.}$
### SubsectionA.2.3Polynomials
Linear and quadratic formulas are special cases of polynomials. This section gives an overview of principles about polynomials that are likely to appear in calculus. First, we introduce some basic definitions.
###### DefinitionA.2.16Monomials
A monomial is an expression that is a constant multiple of a variable raised to a non-negative integer power, $a x^k\text{,}$ where $k=0,1,2,3,\ldots$ and $a \in \mathbb{R}\text{.}$
Examples include $4x^2$ (with $a=4$ and $k=2$), $\frac{1}{3}x^7$ (with $a=\frac{1}{3}$ and $k=7$) and $3$ (where $a=3$ and $k=0$). The following are not monomials: $3\sqrt{x} = 3x^{1/2}$ (since not an integer power) and $\frac{3}{x^2} = 3x^{-2}$ since the power is a negative integer.
###### DefinitionA.2.17Polynomials
An algebraic expression that can be rewritten as a sum of monomials is called a polynomial. The monomials are called the terms of the polynomial. The monomial with the highest power is called the leading term and its power is called the degree of the polynomial. The constant multiples in the monomials are called coefficients and the coefficient in the leading term is called the leading coefficient.
We usually write a polynomial with terms ordered by decreasing powers, called standard form. An abstract representation of a polynomial with degree $n$ is written
\begin{equation*} p(x) = a_n x^n + a_{n-1}x^{n-1} + \cdots + a_2 x^2 + a_1 x + a_0 \end{equation*}
where the symbols $a_n, a_{n-1}, \ldots, a_0$ represent the coefficients. A missing term is represented by a coefficient zero.
###### ExampleA.2.18
$x^4-2x^2+3x+1$ is a polynomial with degree $n=4\text{.}$ The coefficients are $a_4=1\text{,}$ $a_3=0$ (since no $x^3$ term), $a_2=-2\text{,}$ $a_1=3$ and $a_0=1\text{.}$
$2x^2(3x+1)(x-2)$ is a polynomial, but must be expanded (multiply out) to find the coefficients.
\begin{align*} 2x^2(3x+1)(x-2) &= 2x^2(3x^2-6x+x-2) \\ &= 2x^2(3x^2-5x-2) \\ & = 6x^4-10x^3-4x^2 \end{align*}
We see that the polynomial has degree $n=4$ and coefficients $a_4=6\text{,}$ $a_3=-10\text{,}$ $a_2=-4$ and $a_1=a_0=0\text{.}$
Every polynomial $p(x)$ is a function whose domain is all real numbers $(-\infty,\infty)\text{.}$ Values of $x$ for which $p(x)=0$ are called zeros or roots of the polynomial. These roots are related to factors.
Synthetic division is an algorithm that can both test if a value $x=r$ is a root and determine the coefficients of the factored polynomial $q(x)$ at the same time. Synthetic division for quadratic polynomials (degree $n=2$) is a special case of this process, described in Algorithm 13.
###### ExampleA.2.21
Use synthetic division with the polynomial $p(x)=x^3-6x+2$ with the test value $x=2$ and interpret the result.
Solution
Start by identifying the coefficients. Any missed terms have a coefficient of zero,
\begin{equation*} p(x) = x^3 + 0 x^2 + -6x + 2. \end{equation*}
We start the synthetic division table using the coefficients in the first row.
\begin{equation*} \begin{matrix} 1 & 0 & -6 & 2 \\ 0 & \underline{\phantom{WW}} & \underline{\phantom{WW}} & \underline{\phantom{WW}} \\ \underline{\phantom{WW}} & \underline{\phantom{WW}} & \underline{\phantom{WW}} & \underline{\phantom{WW}} \end{matrix} \end{equation*}
We then finish filling the table. To find values in the second row, we use the previous result in the third row and multiply by the test value 2. To find the values in the third row, we add the values in the column. The first value in the second row is always 0. The completed table is shown below.
\begin{equation*} \begin{matrix} 1 & 0 & -6 & 2 \\ 0 & 2 & 4 & -4 \\ 1 & 2 & -2 & -2 \end{matrix} \end{equation*}
Once the table is complete, we interpret the values in the third row as coefficients and a remainder. The last value is the remainder, $r=-2\text{,}$ and the other values are the coefficients of a polynomial whose degree is one smaller than the original, in this case $n-1=2\text{.}$ That is, the quotient polynomial is $q(x) = x^2 + 2x - 2\text{.}$ The original polynomial can be written
\begin{align*} p(x) &= (x-2)q(x) + r\\ x^3-6x+2 &= (x-2)(x^2+2x-2) + -2 \end{align*}
The non-zero remainder means that $x-2$ is not a factor and also tells us that $p(2) = -2\text{.}$
How do we know which numbers to try? If you have access to a graph of the polynomial, you should use the values for roots that you see. If you do not have access to a graph, then you might be able to use the results of the Integer Root Theorem or Rational Root Theorem so long as all of the coefficients of your polynomial are integers.
###### ExampleA.2.23
The polynomial $p(x) = x^3-6x+2$ has only integer coefficients and a constant coefficient $a_0=2\text{.}$ The only factors of $a_0$ are $\pm 1$ and $\pm 2\text{.}$ So the Integer Root Theorem guarantees that these four integers are the only four numbers that we need to check if they are roots. A quick test of each of those values (below) shows that $p(x)$ has no integer roots. (Without the theorem, we wouldn't know how many points we had to check.)
$x$ $p(x)$ 1 -3 -1 -5 2 -2 -2 6
The Integer Root Theorem is a special case of the Rational Root Theorem where $s=1$ (which is always a factor of $a_n\text{.}$
### SubsectionA.2.4Exponentials and Logarithms
Exponential and logarithmic functions are related to the algebraic ideas of exponents and powers. An exponential function is a function involving powers where the power is a variable while the base is a constant. A logarithmic function is the inverse of an exponential function.
###### DefinitionA.2.25Exponential Function
An exponential function is a function $f(x)$ that can be written in the form
\begin{equation*} f(x)=A\cdot b^x \end{equation*}
where $A$ is a real number (called the coefficient) and $b$ is a positive number (called the base). A base $b=1$ is usually not considered an exponential function because then $f(x)=A$ is a constant function.
Sometimes an exponential function is not written in its standard form. You need to know how to rewrite formulas involving exponents to see if they are exponential functions or not. This requires applying the properties of powers (or exponents) Theorem 28.
###### ExampleA.2.26
Show that $f(x) = 3^{2x-1}$ is an exponential function.
Solution
If we think of $2x-1$ as a sum $2x-1=2x+-1\text{,}$ then we can apply the property that adding exponents is equivalent to a product.
\begin{equation*} f(x) = 3^{2x-1} = 3^{2x+-1} = 3^{2x} \cdot 3^{-1} \end{equation*}
Because $3^{-1} = \frac{1}{3}$ and we can apply the property that multiplying exponents is equivalent to consecutive powers, we can rewrite this as
\begin{equation*} f(x) = \frac{1}{3} (3^2)^{x} = \frac{1}{3} \cdot 9^x. \end{equation*}
This is recognized as the standard form of an exponential function.
###### DefinitionA.2.27Elementary Exponential Functions
An elementary exponential function is an exponential function with coefficient $A=1\text{,}$ characterized only by the base $b \gt 0\text{.}$ We write
\begin{equation*} \exp_b(x) = b^x. \end{equation*}
Exponential functions inherit properties from the properties of powers or exponents.
Elementary exponential functions with $b \gt 1$ show exponential growth and are increasing functions. Elementary exponential functions with $b \lt 1$ show exponential decay and are decreasing functions. (When $b=1\text{,}$ the function is constant.) That exponential functions are either increasing or decreasing means that they are one-to-one. For $b \ne 1\text{,}$ an elementary exponential function has a range of $(0,\infty)$ so that for any value $y \gt 0\text{,}$ there is a unique value $x$ such that $b^x = y\text{.}$ This value of $x$ is how we define a logarithm.
###### DefinitionA.2.29Logarithm Functions
For any base $b \gt 0$ with $b \ne 1\text{,}$ we can define the logarithm with base $b$ as the function whose input is a positive number $u \gt 0$ and whose output is the number $x$ such that $b^x=u\text{.}$ We write
$$\log_b(u) = x \qquad \Leftrightarrow \qquad \exp_b(x) = b^x = u.\tag{A.2.18}$$
The equivalence of equations that defines the logarithm (Equation (A.2.18)) allows us to interpret logarithms in terms of powers. Logarithms are the inverse functions of elementary exponential functions, which allows us to solve some equations where the variable is in a power.
###### ExampleA.2.31
Find the exact value of $\log_4(32)\text{.}$
Solution
The value of $\log_4(32)$ is unknown, so we use a variable and write the exponential equivalent equation (see Equation (A.2.18)).
The next step is to recognize that 4 and 32 are both powers of 2, so we rewrite the equation in terms of powers of 2:
Now that we have an equation with the same base, because exponentials are one-to-one, the powers must be equal, $2x = 5\text{,}$ allowing us to solve for $x=\frac{5}{2}\text{,}$ which is the exact value for the logarithm.
###### ExampleA.2.32
Solve the equation
\begin{equation*} 2^{3x+1} = 5 \end{equation*}
for $x$ using a logarithm.
Solution
The equation we are solving can be rewritten in terms of exponential functions as
\begin{equation*} \exp_2(3x+1) = 5. \end{equation*}
This new equation emphasizes that the variable expression $3x+1$ is the input of an exponential. (It is not necessary to write this fact, just to recognize it.) The logarithm $\log_2$ is the inverse of $\exp_2\text{,}$ which in composition with one another yields the identity
\begin{equation*} \log_2(2^{3x+1}) = 3x+1. \end{equation*}
Applying this operation to both sides of the original equation, we have
The expression $\log_2(5)$ is just a value, so we can solve for $x\text{,}$
\begin{equation*} x = \frac{\log_2(5) - 1}{3}, \end{equation*}
giving us our desired solution.
The properties of powers translate to corresponding properties of logarithms.
It is most useful to remember the properties of logarithms conceptually. For example, Equation (A.2.21) states that the logarithm of a product is equivalent to the sum of the logarithms of the factors. Similarly, Equation (A.2.22) states that the logarithm of a quotient is equivalent to the difference of the logarithms (and the logarithm of the denominator is subtracted). Finally, Equation (A.2.23) states that the logarithm of a value raised to a power is equivalent to the product of the power times the logarithm of the value.
Although there are valid logarithm functions for every base $b \gt 0$ with $b \ne 1\text{,}$ calculators generally have only two specific bases. The common logarithm refers to the logarithm with base $b=10$ and is chosen to be conveniently related to scientific notation. The natural logarithm refers to the logarithm with base $b=e$ (a special transcendental number). These two logarithms have their own special notation. The common logarithm (base 10) is often written without any base,
\begin{equation*} \log(x) \equiv \log_{10}(x). \end{equation*}
The natural logarithm (base $e$) is often written
\begin{equation*} \ln(x) \equiv \log_{e}(x). \end{equation*}
The properties of logarithms allow us to express the logarithm of any base in terms of the logarithm of a base of your choice. This is accomplished through the change of base formula for logarithms.
The strategy to prove these formulas is the same as would be used to solve an equation with logarithms in terms of the natural or common logarithm.
###### ExampleA.2.35
Solve the equation $2^{3x} = 5$ in terms of the natural logarithm. (Note that this is equivalent to solving $8^x=5$ or $x=\log_8(5)\text{.}$)
Solution
Starting with the equation involving the exponential, we will apply the natural logarithm (as asked in the problem) to both sides of the equation.
The left side of the equation is the logarithm of a power, so we can use the corresponding logarithm property (Equation (A.2.23)) to say
\begin{equation*} \ln(2^{3x}) = 3x \cdot \ln(2) = \ln(5). \end{equation*}
Dividing both sides of the equation by the product $3 \ln(2)$ we find
\begin{equation*} x = \frac{\ln(5)}{3 \ln(2)} \end{equation*}
which matches what we would get using the change of base formula since $3 \ln(2) = \ln(2^3) = \ln(8)$ (using Equation (A.2.23).)
### SubsectionA.2.5Absolute Value
The absolute value operation takes a number and finds its magnitude (or distance from zero). Because magnitude is a non-negative value and positive and negative pairs are the same distance from zero, we often imagine that the role of absolute value is to remove a negative sign, $|-3| = 3\text{.}$ However, when a variable is involved, a negative sign means finding the inverse of a value for which we may not know if it is positive or negative. So it is incorrect to say that $|-x| = x$ (FALSE). The proper definition of absolute value uses a piecewise formula.
###### DefinitionA.2.36Absolute Value
\begin{equation*} |x| = \begin{cases} x, & x \ge 0 \\ -x, & x \lt 0 \end{cases} \end{equation*}
As a function, the graph of absolute value $y=|x|$ gives two lines: $y=x$ when $x \ge 0$ and $y=-x$ when $x \lt 0\text{.}$
It is sometimes useful to take advantage of an identity between the square root of a square and the absolute value. This is the source of the plus/minus when solving an equation with a square.
###### ExampleA.2.39
Solve the equation $x^2 = 16\text{.}$
Solution
Applying a square root to both sides of the equation, we then get to use the absolute value identity.
\begin{equation*} \sqrt{x^2} = \sqrt{16} \quad \Leftrightarrow \quad |x| = \sqrt{16} = 4 \end{equation*}
The source of plus/minus is that there are two numbers with magnitude 4,
\begin{equation*} x = \pm 4. \end{equation*}
The absolute value splits nicely with multiplication (and division). However, addition of two values with opposite signs shows that absolute values do not add: $|3+-4| = |-1| \ne |3| + |-4| = 7\text{.}$ Instead, we have an inequality called the triangle inequality.
The triangle inequality is used to show that the absolute value of a sum (or difference) is bounded by the sum of the magnitudes of the individual terms.
Occasionally we need to apply the triangle in reverse, showing that the absolute value of a sum (or difference) must be bigger than the difference in magnitudes of the parts.
Absolute value and subtraction is often used to describe the distance between two values. For example, the graph $y=|x-2|$ represents a shift of the graph $y=|x|$ two units to the right, so that instead of measuring the distance of $x$ from 0 it measure the distance of $x$ from 2.
Note that $|x+3| = |x--3|$ so that it represents the distance between $x$ and the value $-3\text{.}$ |
# My Favorite One-Liners: Part 15
In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them. Let me describe a one-liner that I’ll use when I want my class to figure out a pattern, thus developing a theorem by inductive logic rather than deductive logic.
Today’s one-liner is easily stated: “Gosh, I’ve seen that somewhere before.”
For example, In my statistics class, here’s the very first illustration that I show to demonstrate how to compute a standard deviation:
Find the standard deviation of the following data set: 1, 4, 6, 7, 8, 10.
The first step is finding the average:
$\overline{x} = \displaystyle \frac{1+4+6+7+8+10}{6} = 6$.
We then find the deviations from average by subtracting 6 from all of the original data values:
Deviations from average = -5, -2, 0, 1, 2, 4
With these numbers, we can compute the standard deviation:
$s = \displaystyle \sqrt {\frac{ (-5)^2 + (-2)^2 + 0^2 + 1^2 + 2^2 + 4^2}{5} } = \sqrt{10}$.
After asking if there are any questions of clarification about the nuts and bolts of this calculation, I’ll proceed to the next example:
Find the standard deviation of the following data set: 5, 8, 10, 11, 12, 14.
The first step is finding the average:
$\overline{x} = \displaystyle \frac{5+8+10+11+12+14}{6} = 10$.
We then find the deviations from average by subtracting 10 from all of the original data values and then constructing the square root as before:
$s = \displaystyle \sqrt {\frac{ (-5)^2 + (-2)^2 + 0^2 + 1^2 + 2^2 + 4^2}{5} } = \sqrt{10}$.
Then, in a loud obvious voice, I’ll declare, “Gosh, I’ve seen that somewhere before” and then wait a few seconds for the answer. Students can obviously see that the two answers are the same — which gets them thinking about why that happened.
Obviously, the two answers are the same. The real conceptual question that I want my students to figure out is why the two answers are same. Eventually, someone will come up with the correct answer — the second data set was made by adding 4 to all the values in the first data set, which may change the average but does not change how spread out the numbers are… so the standard deviation should be unchanged.
I love the “Gosh, I’ve seen that somewhere before” line after a couple of carefully chosen examples, as it cues my class that they really need to think a little harder than the dull and mechanical operations toward a deeper conceptual understanding of what’s really happening. |
Symbol
# Calculator search results
Formula
Number of solution
Relationship between roots and coefficients
Graph
$y = \left ( x + 4 \right ) ^ { 2 } - 25$
$y = 0$
$x$Intercept
$\left ( 1 , 0 \right )$, $\left ( - 9 , 0 \right )$
$y$Intercept
$\left ( 0 , - 9 \right )$
Minimum
$\left ( - 4 , - 25 \right )$
Standard form
$y = \left ( x + 4 \right ) ^ { 2 } - 25$
$\left( x+4 \right) ^{ 2 } -25 = 0$
$\begin{array} {l} x = 1 \\ x = - 9 \end{array}$
Find solution by method of factorization
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 4 } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 25 } = 0$
Expand the expression
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 8 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 9 } = 0$
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 8 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 9 } = 0$
$acx^{2} + \left(ad + bc\right)x +bd = \left(ax + b\right)\left(cx+d\right)$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 9 } \right ) = 0$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 9 } \right ) = \color{#FF6800}{ 0 }$
If the product of the factor is 0, at least one factor should be 0
$\begin{array} {l} \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } = \color{#FF6800}{ 0 } \\ \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 9 } = \color{#FF6800}{ 0 } \end{array}$
$\begin{array} {l} \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 1 } = \color{#FF6800}{ 0 } \\ \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 9 } = \color{#FF6800}{ 0 } \end{array}$
Solve the equation to find $x$
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ 1 } \\ \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 9 } \end{array}$
$\begin{array} {l} x = 1 \\ x = - 9 \end{array}$
Solve quadratic equations using the square root
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 4 } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 25 } = 0$
Organize the expression
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 8 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 9 } = 0$
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 8 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 9 } = \color{#FF6800}{ 0 }$
Convert the quadratic expression on the left side to a perfect square format
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 4 } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 9 } \color{#FF6800}{ - } \color{#FF6800}{ 4 } ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ 0 }$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 4 } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 9 } \color{#FF6800}{ - } \color{#FF6800}{ 4 } ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ 0 }$
Organize the expression
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 4 } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ 25 }$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 4 } \right ) ^ { \color{#FF6800}{ 2 } } = \color{#FF6800}{ 25 }$
Solve quadratic equations using the square root
$\color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 4 } = \pm \sqrt{ \color{#FF6800}{ 25 } }$
$\color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 4 } = \pm \sqrt{ \color{#FF6800}{ 25 } }$
Solve a solution to $x$
$\color{#FF6800}{ x } = \pm \color{#FF6800}{ 5 } \color{#FF6800}{ - } \color{#FF6800}{ 4 }$
$\color{#FF6800}{ x } = \pm \color{#FF6800}{ 5 } \color{#FF6800}{ - } \color{#FF6800}{ 4 }$
Separate the answer
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ + } \color{#FF6800}{ 5 } \\ \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ - } \color{#FF6800}{ 5 } \end{array}$
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ + } \color{#FF6800}{ 5 } \\ \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ - } \color{#FF6800}{ 5 } \end{array}$
Organize the expression
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ 1 } \\ \color{#FF6800}{ x } = \color{#FF6800}{ - } \color{#FF6800}{ 9 } \end{array}$
$\begin{array} {l} x = 1 \\ x = - 9 \end{array}$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 4 } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 25 } = 0$
Organize the expression
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 8 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 9 } = 0$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 8 \pm \sqrt{ 8 ^ { 2 } - 4 \times 1 \times \left ( - 9 \right ) } } { 2 \times 1 } }$
Organize the expression
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 8 \pm \sqrt{ 100 } } { 2 \times 1 } }$
$x = \dfrac { - 8 \pm \sqrt{ \color{#FF6800}{ 100 } } } { 2 \times 1 }$
Organize the part that can be taken out of the radical sign inside the square root symbol
$x = \dfrac { - 8 \pm \color{#FF6800}{ 10 } } { 2 \times 1 }$
$x = \dfrac { - 8 \pm 10 } { 2 \color{#FF6800}{ \times } \color{#FF6800}{ 1 } }$
Multiplying any number by 1 does not change the value
$x = \dfrac { - 8 \pm 10 } { \color{#FF6800}{ 2 } }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 8 \pm 10 } { 2 } }$
Separate the answer
$\begin{array} {l} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 8 + 10 } { 2 } } \\ \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { - 8 - 10 } { 2 } } \end{array}$
$\begin{array} {l} x = \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 8 } \color{#FF6800}{ + } \color{#FF6800}{ 10 } } { 2 } \\ x = \dfrac { - 8 - 10 } { 2 } \end{array}$
Add $- 8$ and $10$
$\begin{array} {l} x = \dfrac { \color{#FF6800}{ 2 } } { 2 } \\ x = \dfrac { - 8 - 10 } { 2 } \end{array}$
$\begin{array} {l} x = \color{#FF6800}{ \dfrac { 2 } { 2 } } \\ x = \dfrac { - 8 - 10 } { 2 } \end{array}$
Do the reduction of the fraction format
$\begin{array} {l} x = \color{#FF6800}{ \dfrac { 1 } { 1 } } \\ x = \dfrac { - 8 - 10 } { 2 } \end{array}$
$\begin{array} {l} x = \color{#FF6800}{ \dfrac { 1 } { 1 } } \\ x = \dfrac { - 8 - 10 } { 2 } \end{array}$
Reduce the fraction to the lowest term
$\begin{array} {l} x = \color{#FF6800}{ 1 } \\ x = \dfrac { - 8 - 10 } { 2 } \end{array}$
$\begin{array} {l} x = 1 \\ x = \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 8 } \color{#FF6800}{ - } \color{#FF6800}{ 10 } } { 2 } \end{array}$
Find the sum of the negative numbers
$\begin{array} {l} x = 1 \\ x = \dfrac { \color{#FF6800}{ - } \color{#FF6800}{ 18 } } { 2 } \end{array}$
$\begin{array} {l} x = 1 \\ x = \color{#FF6800}{ \dfrac { - 18 } { 2 } } \end{array}$
Do the reduction of the fraction format
$\begin{array} {l} x = 1 \\ x = \color{#FF6800}{ \dfrac { - 9 } { 1 } } \end{array}$
$\begin{array} {l} x = 1 \\ x = \dfrac { - 9 } { \color{#FF6800}{ 1 } } \end{array}$
If the denominator is 1, the denominator can be removed
$\begin{array} {l} x = 1 \\ x = \color{#FF6800}{ - } \color{#FF6800}{ 9 } \end{array}$
2 real roots
Find the number of solutions
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 4 } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 25 } = 0$
Organize the expression
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 8 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 9 } = 0$
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 8 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 9 } = \color{#FF6800}{ 0 }$
Determine the number of roots using discriminant, $D=b^{2}-4ac$ from quadratic equation, $ax^{2}+bx+c=0$
$\color{#FF6800}{ D } = \color{#FF6800}{ 8 } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 9 } \right )$
$D = \color{#FF6800}{ 8 } ^ { \color{#FF6800}{ 2 } } - 4 \times 1 \times \left ( - 9 \right )$
Calculate power
$D = \color{#FF6800}{ 64 } - 4 \times 1 \times \left ( - 9 \right )$
$D = 64 - 4 \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \times \left ( - 9 \right )$
Multiplying any number by 1 does not change the value
$D = 64 - 4 \times \left ( - 9 \right )$
$D = 64 \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 9 } \right )$
Multiply $- 4$ and $- 9$
$D = 64 + \color{#FF6800}{ 36 }$
$D = \color{#FF6800}{ 64 } \color{#FF6800}{ + } \color{#FF6800}{ 36 }$
Add $64$ and $36$
$D = \color{#FF6800}{ 100 }$
$\color{#FF6800}{ D } = \color{#FF6800}{ 100 }$
Since $D>0$ , the number of real root of the following quadratic equation is 2
2 real roots
$\alpha + \beta = - 8 , \alpha \beta = - 9$
Find the sum and product of the two roots of the quadratic equation
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 4 } \right ) ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 25 } = 0$
Organize the expression
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 8 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 9 } = 0$
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ + } \color{#FF6800}{ 8 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 9 } = \color{#FF6800}{ 0 }$
In the quadratic equation $ax^{2}+bx+c=0$ , if the two roots are $\alpha, \beta$ , then it is $\alpha + \beta =-\dfrac{b}{a}$ , $\alpha\times\beta=\dfrac{c}{a}$
$\color{#FF6800}{ \alpha } \color{#FF6800}{ + } \color{#FF6800}{ \beta } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { 8 } { 1 } } , \color{#FF6800}{ \alpha } \color{#FF6800}{ \beta } = \color{#FF6800}{ \dfrac { - 9 } { 1 } }$
$\alpha + \beta = - \dfrac { 8 } { \color{#FF6800}{ 1 } } , \alpha \beta = \dfrac { - 9 } { 1 }$
If the denominator is 1, the denominator can be removed
$\alpha + \beta = - \color{#FF6800}{ 8 } , \alpha \beta = \dfrac { - 9 } { 1 }$
$\alpha + \beta = - 8 , \alpha \beta = \dfrac { - 9 } { \color{#FF6800}{ 1 } }$
If the denominator is 1, the denominator can be removed
$\alpha + \beta = - 8 , \alpha \beta = \color{#FF6800}{ - } \color{#FF6800}{ 9 }$
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# Values that satisfy equations I
Lesson
"Values that satisfy an equation" is another way of saying "numbers that make a number sentence true." For example, if I wanted to find the value that satisfies the equation $x+1=3$x+1=3, we want to find a value for $x$x that makes that equation true. This statement is true when $x=2$x=2 because $2+1=3$2+1=3.
When we're working out whether or not a value satisfies the equation, we need to see whether the left hand side or the equation is the same as the right hand side. We can think of it as a see-saw.
For example, if I had the equation $x+12=20$x+12=20, we could think of it visually as:
If $12$12 was removed from the left hand side of the seesaw, it would look unbalanced like this:
So how do we balance the equation again? We need to remove $12$12 from the right hand side as well:
So $x=8$x=8 satisfies the equation $x+12=20$x+12=20 because $8+12=20$8+12=20
Remember!
Whatever you do to one side of the equation, you must do to the other. This keeps your equation balanced.
#### Examples
##### Question 1
Determine if $b=47$b=47 is a solution of $b+48=96$b+48=96.
a) Find the value of the left-hand side of the equation when $b=47$b=47.
Think: We need to substitute $47$47 into the equation for $b$b.
Do:
$\text{LHS }$LHS $=$= $b+48$b+48 $=$= $47+48$47+48 $=$= $95$95
b) Is $b=47$b=47 a solution of $b+48=96$b+48=96?
Think: Is the LHS equal to the RHS in this equation?
Do: No, $b=47$b=47 is not a solution of $b+48=96$b+48=96.
##### Question 2
John is paid $\$600$$600 a week as a carpenter. His friend Sean, who is paid \1200$$1200 a fortnight as a fireman, claims that he earns the same amount.
a) If weekly earnings is represented by $w$w and fortnightly earnings is represented by $f$f, which of the following equations represents the relationship between $w$w and $f$f?
A) $w=f-2$w=f2 B) $f=\frac{w}{2}$f=w2 C) $2f=w$2f=w D) $2w=f$2w=f
Think: How many weeks are in a fortnight? Which equation represents that?
Do: There are two weeks in a fortnight. This is represented by equation D) $2w=f$2w=f.
b) Find John's fortnightly earnings. That is, using the equation $2w=f$2w=f, find the value of the left-hand side of the equation when $w=600$w=600.
Think: We need to substitute this value into the equation.
Do:
$2w$2w $=$= $f$f $f$f $=$= $2w$2w $=$= $2\times600$2×600 $=$= $\$1200$$1200 ## Values that satisfy inequalities Again, just like when we're working with equations, values that satisfy inequalities are values that make the inequality true. In you need a refresher about inequalities, click here. #### Examples ##### Question 3 Neville is saving up to buy a plasma TV that is selling for \950$$950. He has $\$650$$650 in his bank account and expects a nice sum of money for his birthday next month. a) If the amount he is to receive for his birthday is represented by xx, which of the following inequalities models the situation where he is able to afford the plasma TV? A) x+650\le950x+650950 B) x+650\ge950x+650950 C) x-650\ge950x650950 D) x-650\le950x650950 Think: How would we write an equation for the total amount of money Neville will have after his birthday? How much money will he need in his account to buy the TV? Do: After his birthday, Neville will have his \650$$650 plus an additional $x$x dollars from his parents.
We could write this algebraically as $650+x$650+x or $x+650$x+650.
He needs at least $\$950$$950 in his account to buy the TV. Of the four options we have B) x+650\ge950x+650950 models the situation. b) How much money would he have in total if his parents were to give him \310$$310 for his birthday?
Think: We found the algebraic expression for Neville's total amount of money in part a.
Do:
$\text{Total amount of money }$Total amount of money $=$= $310+650$310+650 $=$= $\$960$$960 c) Would he have enough to buy the plasma TV if his parents were to give him \310$$310 for his birthday?
Think: What was the minimum amount of money Neville needs to buy the TV?
Do: He needs $\$950$$950 to buy the TV, so yes he would have enough money if his parents gave him \310$$310.
##### Question 4
There are two rectangular-shaped pools at the local aquatic centre. Each pool has a length that is triple its width. Pool 1 has a perimeter of $256$256 metres.
1. Let the width of the pools be represented by $w$w. Which of the following equations represents the perimeter of each pool in terms of $w$w?
$2w+3w=256$2w+3w=256
A
$3w+3w+3w=256$3w+3w+3w=256
B
$w+3w=256$w+3w=256
C
$w+w+3w+3w=256$w+w+3w+3w=256
D
$2w+3w=256$2w+3w=256
A
$3w+3w+3w=256$3w+3w+3w=256
B
$w+3w=256$w+3w=256
C
$w+w+3w+3w=256$w+w+3w+3w=256
D
2. The width of Pool 2 is $32$32 metres. Find its perimeter.
3. Is the width of Pool 1 also $32$32 metres?
no
A
yes
B
no
A
yes
B
##### Question 5
We want to determine if $b=8$b=8 is the solution of $8b=63$8b=63.
1. Find the value of the left-hand side of the equation when $b=8$b=8.
2. Is $b=8$b=8 the solution of $8b=63$8b=63?
yes
A
no
B
yes
A
no
B |
## Arithmetic Series
### Learning Outcomes
• Evaluate a sum given in summation notation.
• Find the partial sum of an arithmetic series.
• Solve an application problem using an arithmetic series.
## Using Summation Notation
To find the total amount of money in the college fund and the sum of the amounts deposited, we need to add the amounts deposited each month and the amounts earned monthly. The sum of the terms of a sequence is called a series. Consider, for example, the following series.
$3+7+11+15+19+\cdots$
The $n\text{th }$ partial sum of a series is the sum of a finite number of consecutive terms beginning with the first term. The notation
\begin{align}&{S}_{n}\text{ represents the partial sum.} \\ &{S}_{1}=3 \\ &{S}_{2}=3+7=10 \\ &{S}_{3}=3+7+11=21 \\ &{S}_{4}=3+7+11+15=36\end{align}
Summation notation is used to represent series. Summation notation is often known as sigma notation because it uses the Greek capital letter sigma, $\Sigma$, to represent the sum. Summation notation includes an explicit formula and specifies the first and last terms in the series. An explicit formula for each term of the series is given to the right of the sigma. A variable called the index of summation is written below the sigma. The index of summation is set equal to the lower limit of summation, which is the number used to generate the first term in the series. The number above the sigma, called the upper limit of summation, is the number used to generate the last term in a series.
If we interpret the given notation, we see that it asks us to find the sum of the terms in the series ${a}_{k}=2k$ for $k=1$ through $k=5$. We can begin by substituting the terms for $k$ and listing out the terms of this series.
\begin{align} &{a}_{1}=2\left(1\right)=2 \\ &{a}_{2}=2\left(2\right)=4 \\ &{a}_{3}=2\left(3\right)=6 \\ &{a}_{4}=2\left(4\right)=8 \\ &{a}_{5}=2\left(5\right)=10 \end{align}
We can find the sum of the series by adding the terms:
$\sum\limits _{k=1}^{5}2k=2+4+6+8+10=30$
### A General Note: Summation Notation
The sum of the first $n$ terms of a series can be expressed in summation notation as follows:
$\sum\limits _{k=1}^{n}{a}_{k}$
This notation tells us to find the sum of ${a}_{k}$ from
$k=1$ to $k=n$.
$k$ is called the index of summation, 1 is the lower limit of summation, and $n$ is the upper limit of summation.
### Q & A
#### Does the lower limit of summation have to be 1?
No. The lower limit of summation can be any number, but 1 is frequently used. We will look at examples with lower limits of summation other than 1.
### How To: Given summation notation for a series, evaluate the value.
1. Identify the lower limit of summation.
2. Identify the upper limit of summation.
3. Substitute each value of $k$ from the lower limit to the upper limit into the formula.
4. Add to find the sum.
### Example: Using Summation Notation
Evaluate $\sum\limits _{k=3}^{7}{k}^{2}$.
### Try It
Evaluate $\sum\limits _{k=2}^{5}\left(3k - 1\right)$.
### Arithmetic Series
Just as we studied special types of sequences, we will look at special types of series. Recall that an arithmetic sequence is a sequence in which the difference between any two consecutive terms is the common difference, $d$. The sum of the terms of an arithmetic sequence is called an arithmetic series. We can write the sum of the first $n$ terms of an arithmetic series as:
${S}_{n}={a}_{1}+\left({a}_{1}+d\right)+\left({a}_{1}+2d\right)+…+\left({a}_{n}-d\right)+{a}_{n}$.
We can also reverse the order of the terms and write the sum as
${S}_{n}={a}_{n}+\left({a}_{n}-d\right)+\left({a}_{n}-2d\right)+…+\left({a}_{1}+d\right)+{a}_{1}$.
If we add these two expressions for the sum of the first $n$ terms of an arithmetic series, we can derive a formula for the sum of the first $n$ terms of any arithmetic series.
\begin{align}{S}_{n}&={a}_{1}+\left({a}_{1}+d\right)+\left({a}_{1}+2d\right)+…+\left({a}_{n}-d\right)+{a}_{n} \\ +{S}_{n}&={a}_{n}+\left({a}_{n}-d\right)+\left({a}_{n}-2d\right)+…+\left({a}_{1}+d\right)+{a}_{1} \\ \hline 2{S}_{n}&=\left({a}_{1}+{a}_{n}\right)+\left({a}_{1}+{a}_{n}\right)+…+\left({a}_{1}+{a}_{n}\right) \end{align}
Because there are $n$ terms in the series, we can simplify this sum to
$2{S}_{n}=n\left({a}_{1}+{a}_{n}\right)$.
We divide by 2 to find the formula for the sum of the first $n$ terms of an arithmetic series.
${S}_{n}=\dfrac{n\left({a}_{1}+{a}_{n}\right)}{2}$
This is generally referred to as the Partial Sum of the series.
### A General Note: Formula for the Partial Sum of an Arithmetic Series
An arithmetic series is the sum of the terms of an arithmetic sequence. The formula for the partial sum of an arithmetic sequence is
${S}_{n}=\dfrac{n\left({a}_{1}+{a}_{n}\right)}{2}$
### How To: Given terms of an arithmetic series, find the partial sum
1. Identify ${a}_{1}$ and ${a}_{n}$.
2. Determine $n$.
3. Substitute values for ${a}_{1},{a}_{n}$, and $n$ into the formula ${S}_{n}=\dfrac{n\left({a}_{1}+{a}_{n}\right)}{2}$.
4. Simplify to find ${S}_{n}$.
### Example: Finding the partial sum of an Arithmetic Series
Find the partial sum of each arithmetic series.
1. $5 + 8 + 11 + 14 + 17 + 20 + 23 + 26 + 29 + 32$
2. $20 + 15 + 10 + \dots + -50$
3. $\sum\limits _{k=1}^{12}3k - 8$
### Try It
Use the formula to find the partial sum of each arithmetic series.
$1.4+1.6+1.8+2.0+2.2+2.4+2.6+2.8+3.0+3.2+3.4$
$12+21+29\dots + 69$
$\sum\limits _{k=1}^{10}5 - 6k$
### Example: Solving Application Problems with Arithmetic Series
On the Sunday after a minor surgery, a woman is able to walk a half-mile. Each Sunday, she walks an additional quarter-mile. After 8 weeks, what will be the total number of miles she has walked?
### Try It
A man earns $100 in the first week of June. Each week, he earns$12.50 more than the previous week. After 12 weeks, how much has he earned?
## Contribute!
Did you have an idea for improving this content? We’d love your input. |
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Teaching Fractions to Young Students
Teaching Fractions to Young Students
You own a bakery and the only thing you sell are loaves of bread. Not only that, but every loaf is identical. So when customers come into your shop they just say how many loaves they want..
{1,2,3,4,...}
One day a customer comes in and explains that they love your bread but your loaves are too big. They ask if you have smaller loaves. You don't. But then you have a creative idea. You take a knife and cut a loaf into 2 equal sized pieces. You sell one piece to the customer and they are happy.
But what did you just sell? It was not a loaf. It was something less. You chopped a loaf into 2 equal pieces and sold one of the pieces. You sold "one out of two", so you could represent that as 1/2.
This idea is popular with your customers. Pretty soon you are chopping your loaves into 5 equal pieces and selling customers 1, 2, 3 or 4 of the pieces. That's 1/5, 2/5, 3/5 or 4/5.
Of course, if you gave a customer 5 out of 5 then that's the same as the whole loaf, so 5/5=1.
If a really fussy customer comes in and asks for 7/13 you know exactly what to do. You take a loaf, chop it into exactly 13 equally sized pieces and then sell the customer 7 of the pieces.
These things are fractions. Mathematicians call them rational numbers.
We can think of these as numbers between the integers. It's pretty easy to see that between any two integers there are an infinite amount of rational numbers. For example, the infinite series of fractions {1/2, 1/3, 1/4, 1/5...} all lie between 0 and 1.
So our list of numbers is expanding.
First we had the positive integers {1,2,3..}
Then we added the negative integers {..-3,-2,-1,0,1,2,3..}
And now we've added an infinite number of rational numbers between any two integers.
Rational numbers, or fractions, are just pairs of integers that obey certain rules.
We can multiply two fractions. Here's the rule..
(a/b)*(c/d)=(a*c)/(b*d)
Here's an example (3/4)*(2/5)=(3*2)/(4*5)=6/20
We can add two fractions. Here's the rule..
(a/b)+(c/d)=(a*d+b*c)/(b*d)
Here's an example (3/4)+(2/5)=(3*5+4*2)/(4*5)=23/20
Rational numbers do a very important job, they extend the number system. We have an infinite number of integers, and between any two integers we have an infinite number of rational numbers. That's a lot of numbers, but it's still not enough! That's because rational numbers do not totally fill the gaps between the integers. Amazingly there's still room for other numbers.
That's right. Between every two integers there's an infinite amount of other numbers in addition to the rational numbers. These are yet another type of number. But the name that mathematicians picked for these number is not so creative, they called them irrational numbers!
Content written and posted by Ken Abbott abbottsystems@gmail.com |
# How to Construct a Triangle when Its Three Sides Are Given (SSS Criterion)
Author Info
Updated: January 13, 2020
This article teaches you how to construct a triangle when the lengths of all three of its sides are known.
## Steps
1. 1
Say you need to construct a triangle having side lengths 6 centimeter (2.4 in), 7 centimeter (2.8 in) and 4 centimeter (1.6 in).
2. 2
Draw a line segment measuring one of the given side lengths of the triangle. Let's draw the side having length 6 centimeter (2.4 in) and mark its end points as A and B. Use a ruler to carry out this step.[1]
3. 3
Set the width of your compass equal to another given side length. Let's set it equal to 7 centimeter (2.8 in). Retain this width for the next step.[2]
4. 4
Place the tip of the compass on one of the end points of the side AB (let's choose to place it on A) and draw an arc on either side of the line segment AB.
5. 5
Set the width of your compass equal to length of the third side, which is 4 centimeter (1.6 in) in this case. Retain this width for the next step.[3]
6. 6
Place the tip of the compass on B and draw another arc which cuts the previously drawn arc at some point (say C).[4]
7. 7
Join C to each of the points A and B to complete the triangle.[5]
## Community Q&A
Search
• Question
If the perimeter is 9.5 and the ratio among lengths of its sides is 2:3:5, what is the answer?
Donagan
2 + 3 + 5 = 10. Divide 9.5 by 10. That gives you 0.95. Multiply 0.95 separately by 2, 3, and 5. That gives you 1.9, 2.85, and 4.75, respectively. Those are the lengths of your triangle's sides.
• Question
If I have a triangle like ACB, should I first draw the line AC i.e : 7 cm?
Donagan
First you draw AB. Then you find point C. Then you construct ABC by drawing AC and BC.
• Question
What do you do if you don't have a pair of compasses, and need them badly?
Donagan
In a pinch, you could make a compass out of two pencils and some tape.
• Question
Why would I use a divider?
Donagan
It is to transfer a specific length from one place to another. It's similar to a compass, but it does not draw arcs.
• Question
What if I don't have a compass?
Donagan
You can use a ruler as a compass. It's a bit awkward, but if you work very carefully, you can do it.
• Question
How do I draw a triangle of sides 3, 4 and 6 cm?
Donagan
Draw a straight line 6 cm in length. Each end of the line is a vertex of the triangle. Using a compass, from one vertex draw an arc with a 3 cm radius. From the other vertex draw an arc with a 4 cm radius. The point where the two arcs intersect is the third vertex of the triangle. Connect the three vertices with straight lines.
• Question
How do I construct an RHS triangle?
Donagan
An RHS triangle is a right triangle with a known hypotenuse and one known leg. First draw a right angle. (See Construct a 90 Degrees Angle Using Compass and Ruler). Mark off the known length on one of the rays of the angle. From the far end of that ray, use a compass to draw an arc with a radius equal to the length of the hypotenuse. Assuming you've been given valid dimensions, the arc will intersect the other ray. From that point of intersection draw a straight line to the center point of the arc. That will complete the triangle.
200 characters left
## Things You'll Need
• Paper
• Pen or Pencil
• Ruler
• Compass |
Search by Topic
Resources tagged with Divisibility similar to Differences:
Filter by: Content type:
Stage:
Challenge level:
There are 54 results
Broad Topics > Numbers and the Number System > Divisibility
Differences
Stage: 3 Challenge Level:
Can you guarantee that, for any three numbers you choose, the product of their differences will always be an even number?
Going Round in Circles
Stage: 3 Challenge Level:
Mathematicians are always looking for efficient methods for solving problems. How efficient can you be?
What Numbers Can We Make?
Stage: 3 Challenge Level:
Imagine we have four bags containing a large number of 1s, 4s, 7s and 10s. What numbers can we make?
Counting Factors
Stage: 3 Challenge Level:
Is there an efficient way to work out how many factors a large number has?
What Numbers Can We Make Now?
Stage: 3 Challenge Level:
Imagine we have four bags containing numbers from a sequence. What numbers can we make now?
Stage: 3 Challenge Level:
Powers of numbers behave in surprising ways. Take a look at some of these and try to explain why they are true.
Repeaters
Stage: 3 Challenge Level:
Choose any 3 digits and make a 6 digit number by repeating the 3 digits in the same order (e.g. 594594). Explain why whatever digits you choose the number will always be divisible by 7, 11 and 13.
Take Three from Five
Stage: 4 Challenge Level:
Caroline and James pick sets of five numbers. Charlie chooses three of them that add together to make a multiple of three. Can they stop him?
Digital Roots
Stage: 2 and 3
In this article for teachers, Bernard Bagnall describes how to find digital roots and suggests that they can be worth exploring when confronted by a sequence of numbers.
Stage: 3 Challenge Level:
List any 3 numbers. It is always possible to find a subset of adjacent numbers that add up to a multiple of 3. Can you explain why and prove it?
Elevenses
Stage: 3 Challenge Level:
How many pairs of numbers can you find that add up to a multiple of 11? Do you notice anything interesting about your results?
Dozens
Stage: 3 Challenge Level:
Do you know a quick way to check if a number is a multiple of two? How about three, four or six?
Three Times Seven
Stage: 3 Challenge Level:
A three digit number abc is always divisible by 7 when 2a+3b+c is divisible by 7. Why?
Legs Eleven
Stage: 3 Challenge Level:
Take any four digit number. Move the first digit to the 'back of the queue' and move the rest along. Now add your two numbers. What properties do your answers always have?
Expenses
Stage: 4 Challenge Level:
What is the largest number which, when divided into 1905, 2587, 3951, 7020 and 8725 in turn, leaves the same remainder each time?
Remainders
Stage: 3 Challenge Level:
I'm thinking of a number. When my number is divided by 5 the remainder is 4. When my number is divided by 3 the remainder is 2. Can you find my number?
Oh! Hidden Inside?
Stage: 3 Challenge Level:
Find the number which has 8 divisors, such that the product of the divisors is 331776.
Powerful Factorial
Stage: 3 Challenge Level:
6! = 6 x 5 x 4 x 3 x 2 x 1. The highest power of 2 that divides exactly into 6! is 4 since (6!) / (2^4 ) = 45. What is the highest power of two that divides exactly into 100!?
Skeleton
Stage: 3 Challenge Level:
Amazing as it may seem the three fives remaining in the following `skeleton' are sufficient to reconstruct the entire long division sum.
Eminit
Stage: 3 Challenge Level:
The number 8888...88M9999...99 is divisible by 7 and it starts with the digit 8 repeated 50 times and ends with the digit 9 repeated 50 times. What is the value of the digit M?
Mod 3
Stage: 4 Challenge Level:
Prove that if a^2+b^2 is a multiple of 3 then both a and b are multiples of 3.
Flow Chart
Stage: 3 Challenge Level:
The flow chart requires two numbers, M and N. Select several values for M and try to establish what the flow chart does.
Digat
Stage: 3 Challenge Level:
What is the value of the digit A in the sum below: [3(230 + A)]^2 = 49280A
AB Search
Stage: 3 Challenge Level:
The five digit number A679B, in base ten, is divisible by 72. What are the values of A and B?
Square Routes
Stage: 3 Challenge Level:
How many four digit square numbers are composed of even numerals? What four digit square numbers can be reversed and become the square of another number?
Factoring Factorials
Stage: 3 Challenge Level:
Find the highest power of 11 that will divide into 1000! exactly.
Ewa's Eggs
Stage: 3 Challenge Level:
I put eggs into a basket in groups of 7 and noticed that I could easily have divided them into piles of 2, 3, 4, 5 or 6 and always have one left over. How many eggs were in the basket?
Just Repeat
Stage: 3 Challenge Level:
Think of any three-digit number. Repeat the digits. The 6-digit number that you end up with is divisible by 91. Is this a coincidence?
Obviously?
Stage: 4 and 5 Challenge Level:
Find the values of n for which 1^n + 8^n - 3^n - 6^n is divisible by 6.
Squaresearch
Stage: 4 Challenge Level:
Consider numbers of the form un = 1! + 2! + 3! +...+n!. How many such numbers are perfect squares?
Ben's Game
Stage: 3 Challenge Level:
Ben passed a third of his counters to Jack, Jack passed a quarter of his counters to Emma and Emma passed a fifth of her counters to Ben. After this they all had the same number of counters.
Knapsack
Stage: 4 Challenge Level:
You have worked out a secret code with a friend. Every letter in the alphabet can be represented by a binary value.
What an Odd Fact(or)
Stage: 3 Challenge Level:
Can you show that 1^99 + 2^99 + 3^99 + 4^99 + 5^99 is divisible by 5?
Remainder
Stage: 3 Challenge Level:
What is the remainder when 2^2002 is divided by 7? What happens with different powers of 2?
Transposition Fix
Stage: 4 Challenge Level:
Suppose an operator types a US Bank check code into a machine and transposes two adjacent digits will the machine pick up every error of this type? Does the same apply to ISBN numbers; will a machine. . . .
Divisively So
Stage: 3 Challenge Level:
How many numbers less than 1000 are NOT divisible by either: a) 2 or 5; or b) 2, 5 or 7?
The Remainders Game
Stage: 2 and 3 Challenge Level:
A game that tests your understanding of remainders.
American Billions
Stage: 3 Challenge Level:
Play the divisibility game to create numbers in which the first two digits make a number divisible by 2, the first three digits make a number divisible by 3...
Sixational
Stage: 4 and 5 Challenge Level:
The nth term of a sequence is given by the formula n^3 + 11n . Find the first four terms of the sequence given by this formula and the first term of the sequence which is bigger than one million. . . .
Gaxinta
Stage: 3 Challenge Level:
A number N is divisible by 10, 90, 98 and 882 but it is NOT divisible by 50 or 270 or 686 or 1764. It is also known that N is a factor of 9261000. What is N?
Check Codes
Stage: 4 Challenge Level:
Details are given of how check codes are constructed (using modulus arithmetic for passports, bank accounts, credit cards, ISBN book numbers, and so on. A list of codes is given and you have to check. . . .
Big Powers
Stage: 3 and 4 Challenge Level:
Three people chose this as a favourite problem. It is the sort of problem that needs thinking time - but once the connection is made it gives access to many similar ideas.
N000ughty Thoughts
Stage: 4 Challenge Level:
How many noughts are at the end of these giant numbers?
LCM Sudoku
Stage: 4 Challenge Level:
Here is a Sudoku with a difference! Use information about lowest common multiples to help you solve it.
The Chinese Remainder Theorem
Stage: 4 and 5
In this article we shall consider how to solve problems such as "Find all integers that leave a remainder of 1 when divided by 2, 3, and 5."
396
Stage: 4 Challenge Level:
The four digits 5, 6, 7 and 8 are put at random in the spaces of the number : 3 _ 1 _ 4 _ 0 _ 9 2 Calculate the probability that the answer will be a multiple of 396.
Multiplication Magic
Stage: 4 Challenge Level:
Given any 3 digit number you can use the given digits and name another number which is divisible by 37 (e.g. given 628 you say 628371 is divisible by 37 because you know that 6+3 = 2+7 = 8+1 = 9). . . .
Novemberish
Stage: 4 Challenge Level:
a) A four digit number (in base 10) aabb is a perfect square. Discuss ways of systematically finding this number. (b) Prove that 11^{10}-1 is divisible by 100.
Why 24?
Stage: 4 Challenge Level:
Take any prime number greater than 3 , square it and subtract one. Working on the building blocks will help you to explain what is special about your results. |
# Simplifying ExpressionsPage 1
#### WATCH ALL SLIDES
Slide 1
Simplifying Expressions
By: Karen Overman
Slide 2
Objective
This presentation is designed to give a brief review of simplifying algebraic expressions and evaluating algebraic expressions.
Slide 3
## Algebraic Expressions
An algebraic expression is a collection of real numbers, variables, grouping symbols and operation symbols.
Here are some examples of algebraic expressions.
Slide 4
Consider the example:
The terms of the expression are separated by addition. There are 3 terms in this example and they are .
The coefficient of a variable term is the real number factor. The first term has coefficient of 5. The second term has an unwritten coefficient of 1.
The last term , -7, is called a constant since there is no variable in the term.
Slide 5
Lets begin with a review of two important skills for simplifying expression, using the Distributive Property and combining like terms. Then we will use both skills in the same simplifying problem.
Slide 6
## Distributive Property
a ( b + c ) = ba + ca
To simplify some expressions we may need to use the Distributive Property
Do you remember it?
Distributive Property
Slide 7
Examples
Example 1: 6(x + 2)
Distribute the 6.
6 (x + 2) = x(6) + 2(6)
= 6x + 12
Example 2: -4(x 3)
Distribute the 4.
-4 (x 3) = x(-4) 3(-4)
= -4x + 12
Slide 8
## Practice Problem
Try the Distributive Property on -7 ( x 2 ) .
Be sure to multiply each term by a 7.
-7 ( x 2 ) = x(-7) 2(-7)
= -7x + 14
Notice when a negative is distributed all the signs of the terms in the ( )s change.
Slide 9
Examples with 1 and 1.
Example 3: (x 2)
= 1( x 2 )
= x(1) 2(1)
= x - 2
Notice multiplying by a 1 does nothing to the expression in the ( )s.
Example 4: -(4x 3)
= -1(4x 3)
= 4x(-1) 3(-1)
= -4x + 3
Notice that multiplying by a 1 changes the signs of each term in the ( )s.
Slide 10
## Like Terms
Like terms are terms with the same variables raised to the same power.
Go to page:
1 2 3 |
# Interesting Statsitcal Problems in Interviews
### Problem 1
If we have $n$ dice draws with a dice that has $k$ sides. Let $X_i$ to be the random variables that describes the number of times that the dice faces $i$. What is the correlation between $x_i$ and $x_j$ for $i\neq j$.
We first notice, $X_i$ are identitically distributed. Thus, the correlation between $X_i$ and $X_j$ does depend on $i$ and $j$ and they they have the same variance. Thus, we only need to calculate covariance first. We then notice
$X_1+X_2 + \dots + X_k = n$
Next we compute
\begin{aligned} &\text{Cov}[X_i, X_1+X_2+\dots + X_k] = \text{Cov}[X_i, n] =0 \\ &=\text{Cov}(X_i, X_i) + (k-1)\text{Cov}(x_i,x_j)\\ \end{aligned}
Thus, we find
$(k-1)\text{Cov}(x_i,x_j) = - \text{Var}(X_i)$
Thus,
$\text{Cov}(x_i,x_j) = - \frac{\text{Var}(x_i)}{k-1}$
Therefore, the correlation becomes
$\text{Corr}(X_i,X_j) = \frac{\text{Cov}(X_i,X_j)}{\text{Var}(X_i)} = - \frac{1}{k-1}$
### Problem 2
With a set of 52 cards, we draw without replacement from the card. What is the expected number of draws to see the first Ace?
Counting down from the top of the deck of cards. Let $X_1$ be the number of cards before the first Ace. Let $X_2$ be the number of cards between the first and the second Ace. Let $X_3$ be the number of cards between the second and third Ace. Let $X_4$ be the number of cards between the third Ace and the fourth Ace. Finally, let $X_5$ be the number of cards between the fourth Ace and the bottom of the deck. We find
$X_1+X_2+X_3+X_4+X_5 = 52-4 = 48$
We also find the distribution of $X_i$’s are the same and thus,
$5E[X_1] = 48\quad \Rightarrow \quad E[X_1] = \frac{48}{5}$
### Problem 3
A 2D random walk starts from $(1,1)$. If it will stop once it hits the $y$-axis. What is the probability that it will stop at the negative part of the $y$-axis?
Notice the symmetry of the problem. In 2D random walk, the problem is symmetric. At a given time, it will be uniformly distributed in the direction respect to the starting point. Given it hits $y$-axis, it will be uniformly distributed around within the angle ($0$,$\pi$) with respect to the starting point. However, the probability of going to the negative part is the last $\pi/4$ angle. Thus, the probability of landing in the negative part of $y$-axis is $1/4$.
### Problem 4
Give an example that two random variables are uncorrelated but dependent
Let $X = \sin(\tau)$ and $Y = \cos(\tau)$ and $\tau\sim U(0,2\pi)$. This is uncorrelated because, if we plot $(X,Y)$, it relys on the unit circle. If we run a linear regression on that, the line will have zero slope. Since correlation between $Y$ and $X$ is the $R^2$ score and it is zero in this case. They are dependent because, once we know $X$, $Y$ can only have at most two values.
### Problem 5
How to uniformly generate points on a disc with radiu $R$?
If we label points by $(r,\theta)$, the probability density function of $(r,\theta)$ is
$p(r,\theta)drd\theta = \frac{rd\theta dr }{\pi R^2} = \frac{d\theta}{2\pi} \cdot \frac{d r^2}{R^2} = p(\theta) p(r^2)$
Thus, we uniformly generate $r^2$ in the interval $[0,R^2]$ and uniform generate $\theta$ from $[0,2\pi)$.
### Problem 6
How do you use a standard normal distribution samples to estimate $\pi$ ?
This is different from using uniform to estimate $\pi$ where you try to create a monte carlo experiment in a unit square. But you could try to transform the normal to uniform by using the CDF. Here we could use the quantiles. The quantiles of normal samples are of uniform distribution and you could use the monte carlo to estimate the one quarter disk area in a unit square.
Here is another way: note that if $X\sim N(0,1)$, then $E[|X|] = \int_{-\infty}^\infty |x| \frac{1}{\sqrt{2\pi}} e^{-x^2/2}dx = \sqrt{\frac{2}{\pi}} \int_0^\infty xe^{-x^2/2} dx = \sqrt{\frac{2}{\pi}}$ Therefore, we have $\pi = \frac{2}{E[|X|]^2}$ Thus, if we are given a set of samples sampled from normal distribution, we could take 2/(np.mean(np.absolute(arr)))**2 as the estimation of $\pi$.
### Problem 7
If we have two bags containing $n$ balls each and we call the bags as bag A and bag B. We also have a fair coin. Then we start a process: at each step, we toss the coin. If the head faces up, we draw a ball from A and if tail faces up, we draw a ball from bag B. We continue the process until I find I cannot draw a ball from the bag indicated by the coin. Then I open the other bag and I find there are $X$ balls remaining. Then the question is what is $E[X]$ ? What is the dependence on $n$? why?
First, we could calculate that analytically. Since the two bags are symmetric we could write $E[X] = \frac{1}{2}E[X|\text{A is empty}] + \frac{1}{2} E[X|\text{B is empty}] = E[X|\text{A is empty}]$ Then if we draw $i$ balls from B before A is drain, then we totally draw $i +n$ balls with the last ball from $A$. Then there will be $n-i$ balls left in $B$. Thus, we could calculate this expectation as \begin{aligned} E[X|\text{A is empty}] &= \sum_{i=0}^{n-1} (n-i) {i+n-1 \choose i}\left(\frac{1}{2}\right)^{i+n-1} \\ &= \sum_{x = n}^{2n-1} (2n-x) { x-1 \choose n-1} \left(\frac{1}{2}\right)^{x-1} \\ &= 2n - 2n\sum_{x=n}^{2n-1} \frac{x!}{(x-n)!n!} \left(\frac{1}{2}\right)^x \\ &= 2n - 2n \left[\sum_{x=n}^{2n} \frac{x!}{(x-n)!n!}\left(\frac{1}{2}\right)^x - \frac{2n!}{n!n!} \left(\frac{1}{2}\right)^{2n} \right]\\ &= 2n {2n \choose n} \left(\frac{1}{2} \right)^{2n} \end{aligned} The power dependence on $n$ could be approximated using the exact formula but we can also argue from the other point of view.
Let $X_A - X_B$ be the number of left balls difference at each draw. Then we notice $X_A- X_B$ at each step can increase $1$ if we draw a ball from $B$ or decrease $1$ if we draw from $A$. Then $X_A-X_B$ forms a random walk. Also $E[X_A- X_B]$ will be zero at each step, since it is symmetric on A and B. Thus, at each step $i$ $E[(X_A-X_B)^2] = \text{Var}(X_A- X_B) \sim i$ Also we know $E[(X_A- X_B)^2] = E[X_A^2] + E[X_B^2] - 2E[X_AX_B]$ At final step, either $X_A$ or $X_B$ equal to $0$. if $X_B= 0$, then we have $E[X_A^2] \sim M$ where $M$ is the total number of steps used. But total number of steps is of order $n$ total number of balls. Thus, $E[X_A] \sim \sqrt{n}$.
Categories:
Updated: |
### Expected Value
```Section 7.4 (partially)
Section Summary
Expected Value
Linearity of Expectations
Independent Random Variables
Expected Value
Definition: The expected value (or expectation or mean) of
the random variable X(s) on the sample space S is equal to
Example-Expected Value of a Die: Let X be the number
that comes up when a fair die is rolled. What is the
expected value of X?
Solution: The random variable X takes the values 1, 2, 3, 4,
5, or 6. Each has probability 1/6. It follows that
Expected Value
Theorem 1: If X is a random variable and p(X = r) is
the probability that X = r, so that
then
see the text for the proof
Example: what is the expected sum of the numbers that appear
when two fair dice are rolled?
Hint: compute p(X=k) for each k from 2 to 12 and use Theorem 1
to get expected sum 7.
Expected Value
Theorem 2: The expected number of successes when
n mutually independent Bernoulli trials are performed
is np, where p is the probability of success on each
trial.
see the text for the proof
Example: what is the expected number of heads that come up when a
fair coin is flipped 5 times.
Solution: By Theorem 2 with p=1/2 and n=5, we see that the expected
number of heads is 2.5 .
Linearity of Expectations
The following theorem tells us that expected values are
linear. For example, the expected value of the sum of
random variables is the sum of their expected values.
Theorem 3: If Xi, i = 1, 2, …,n with n a positive integer, are
random variables on S, and if a and b are real numbers,
then
(i) E(X1 + X2 + …. + Xn) = E(X1 )+ E(X2) + …. + E(Xn)
(ii) E(aX + b) = aE(X) + b.
see the text for the proof
Independent Random Variables
Definition 3: The random variables X and Y on a
sample space S are independent if
p(X = r1 and Y = r2) = p(X = r1)∙ p(Y = r2).
Theorem 5: If X and Y are independent variables on a
sample space S, then E(XY) = E(X)E(Y).
see text for the proof
Independent Random Variables
Theorem 5: If X and Y are independent variables on
a sample space S, then E(XY) = E(X)E(Y).
Example: Let X be the number that comes up on the
first die when two fair dice are rolled and Y be the sum
of the numbers appearing on the two dice. Show that
E(X)E(Y) is not equal E(XY).
Answer: E(X)=7/2. E(Y)= 7. E(XY)= 329/12.
For example,
``` |
# Moments
## Key points
• When a force is applied to an object it can cause it to rotate.
• The rotational effect of a force is called a moment.
## What is a moment?
A moment is the turning effect of a force. Forces that create a moment act around a point called the pivot. The pivot is the point around which the object can rotate or turn.
On a seesaw the pivot is the point in the middle.
It makes calculations easier to try to measure the perpendicular distance between the of the force and the pivot.
For example, if you apply a force to a spanner it rotates. The pivot is at the bolt.
When you push open a door, you apply a force to the edge of the door furthest from the hinges.
This force has a turning effect on the door - a moment which causes the door to rotate around the hinges - the - and the door opens.
The size of a depends on two things:
• the size of the force that is applied
• the distance the force acts from the pivot
It is very important to remember that the distance from the pivot is measured at a right angle, or , to the line of action of the force.
## Moments and levers
Watch the video to understand more about moments and levers.
## Calculating a moment
Calculate the size of a moment using the following equation:
$$Moment~of~a~force = force \times perpendicular~distance~from~pivot$$
or
$$M = F \times d$$
where:
• moment (M) is measured in newton metres (Nm)
• force (F) is measured in newtons (N)
• perpendicular distance from pivot (d) is measured in metres (m)
For example:
To open a door, a person pushes on the edge of a door with a force of 20 N. The distance between their hand and the hinges is 0.7 metres. What is the moment used to open the door?
M = ?
F = 20 N
d = 0.7 m
Use the following equation to calculate the size of a moment:
$$M = F \times d$$
Substitute in the values you know:
$$M = 20 \times 0.7$$
$$M = 1.4$$
The moment used to open the door 1.4 Nm.
## Balancing moments
A seesaw is a good example of balancing . If a person sits on one end, the seesaw rotates around the .
However, if a second person sits on the other end, it is possible to balance the seesaw horizontally. This usually requires one person to adjust their distance from the pivot.
The states that for an object to be the total clockwise moment must be equal to the total anti-clockwise moment.
For example:
A seesaw needs to balance. One one side, 3 m from the pivot, is a box which has a weight of 5 N, and on the other is a box which has a weight of 3 N. Calculate the distance needed between the mass which has a weight of 3 N box and the pivot.
Total anti-clockwise moment = total clockwise moment
Step 1: Calculate the clockwise moment using the following equation:
$$M = F \times d$$
M = ?
F = 5 N
D = 3 m
Substitute in the values you know:
$$M = 5 \times 3$$
$$M = 15~Nm$$
The clockwise moment is 15 Nm
Step two - the seesaw needs to be balanced:
Remember - total anti-clockwise moment = total clockwise moment.
We have already calculated that the clockwise moment is 15 Nm.
Use the equation:
$$M = F \times d$$
Substitute in the values you already know:
M = 15 Nm
F = 3 N
d = ?
$$15 = 3 \times d$$
Now divide both sides by 3:
$$\frac{15}{3} = \frac{3 \times d}{3}$$
This cancels to give:
$$5 = d$$
So the distance between the box weighing 3 N and the pivot is 5 metres.
## Using moments
Spanners and levers both use moments.
### Spanners
Spanners are used to turn nuts and bolts. If you need to undo a nut that is very tight, you can:
• use a short spanner and apply a large force
or
• use a long spanner and apply a small force
Using the longer spanner increases the distance from the pivot. This reduces the amount of force needed to undo the nut from the bolt.
### Levers
Removing the lid from a can of paint requires a large lifting force on the lid. A screwdriver acts as a lever.
The pivot is the edge of the can and this is very close to where the strong push is needed to lift the lid to open the can.
A screwdriver with a long handle means that you can push down on the handle of the screwdriver with a small force and still open the can.
## Calculating an upwards force
Have a look at this calculation of an upwards force.
Calculate the upwards force on the lid if the distance from the pivot to the lid is 0.01 m and the horizontal distance from the pivot to the screwdriver handle is 0.15 m. The person is pushing down with a force of 10 N.
Calculate the anticlockwise moment:
F = 10 N
d = 0.15 m
M = ?
$$M = F \times d$$
$$M = 10 \times 0.15$$
Anticlockwise moment = 1.5 Nm
The clockwise moment must be the same:
F = ?
d = 0.01 m
M = 1.5 Nm
$$M = F \times d$$
$$1.5 = F \times 0.01$$
Divide both sides by 0.01:
$$\frac{1.5}{0.01} = {F \times 0.01}{0.01}$$
This cancels to give:
$$150 = F$$
The lifting force on the lid is 150 N. |
# What is meant by multiplicative identity property?
## What is meant by multiplicative identity property?
: an identity element (such as 1 in the group of rational numbers without 0) that in a given mathematical system leaves unchanged any element by which it is multiplied.
What is multiplicative identity example?
What do you Mean by Multiplicative Identity? Multiplicative identity states that if a number is multiplied to 1 the resultant will be the number itself. “1” is the multiplicative identity of a number and is represented as: p × 1 = p = 1 × p. For example, 27 × 1 = 27 = 1 × 27.
### What is the multiplicative of identity?
1
The multiplicative identity is 1 . When one multiplies any number, it does not change the value.
How do you explain multiplicative property?
The multiplicative property of -1 says that any time you multiply something by -1, you change it into its opposite. The opposite of a number is that same number on the opposite side of 0 on a number line. For example, if you multiply 5 by -1 you’ll get -5.
#### How do you solve multiplicative identity property?
Solution: According to the multiplicative identity property when we multiply any rational number by 1 the result will be the same rational number. a) 1/9 × 1 = 1/9, this equation satisfies the property because the result is the same number that is 1/9 and the multiplicative identity element is 1 in this case.
Is 0 called the multiplicative identity?
Zero (0) is the additive identity and one (0) is the multiplicative identity for all the numbers such as whole numbers, natural numbers, integers, etc.
## What is an example of an identity property?
The identity property of 1 says that any number multiplied by 1 keeps its identity. In other words, any number multiplied by 1 stays the same. The reason the number stays the same is because multiplying by 1 means we have 1 copy of the number. For example, 32×1=32.
What is the multiplicative identity of 3?
Since 3 = 3, we have proven the Multiplicative Identity Property using the number 3.
### What is an example of multiplicative?
A multiplicative inverse is a reciprocal. A reciprocal is one of a pair of numbers that when multiplied with another number equals the number 1. For example, if we have the number 7, the multiplicative inverse, or reciprocal, would be 1/7 because when you multiply 7 and 1/7 together, you get 1!
Why is 1 called the multiplicative identity?
The identity property of 1 says that any number multiplied by 1 keeps its identity. The reason the number stays the same is because multiplying by 1 means we have 1 copy of the number.
#### Does every ring have a multiplicative identity?
Every ring has a multiplicative identity. It is possible for a subset of some field to be a ring but not a subfield, under the induced operations.
What are some examples of identity property?
Identity Property. The identity property says that any number plus zero equals itself. For example, 3 + 0 = 3. The identity property also applies to subtraction since 3 – 0 = 3. Zero is known as the identity number because in addition and subtraction it does not affect other numbers.
## What does the word multiplicative identity mean in math?
Definition of multiplicative identity. : an identity element (such as 1 in the group of rational numbers without 0) that in a given mathematical system leaves unchanged any element by which it is multiplied.
What is the identify property of multiplication?
Identity property of multiplication. The identity property of multiplication, also called the multiplication property of one says that a number does not change when that number is multiplied by 1.
### What does multiplication identity property of one mean?
The identity property of multiplication, also called the multiplication property of one says that a number does not change when that number is multiplied by 1. You can see this property readily with a printable multiplication chart. |
Are ∠1 and ∠2 adjacent angles?
02:42
### Video Transcript
Are angles one and two adjacent angles?
In the diagram, we can see three lines. And the way that these lines have been drawn, we can also see three angles. To help us identify them, they’re labelled one, two, and three. Angle one is this angle here. Angle two is between these two lines here. And angle three is this larger angle that we can see here. Now, we don’t need to worry about angle three, because the question asks us simply about angles one and two. Are angles one and two adjacent angles?
Let’s remind ourselves what adjacent angles are. Adjacent angles have two things in common. Firstly, they share a common vertex. A vertex is simply the point from which all the rays that make up the angles come from. So they all come from one point. The second thing that adjacent angles have in common is that they share a common side. One of the rays or sides of the angle is shared by two angles if they’re adjacent. In this example, we can see that the blue line is a side that’s common between angle 𝑎, but also angle 𝑏.
Let’s use these two rules about adjacent angles to help us find out whether angles one and two in our diagram are adjacent. Firstly, do they share a common vertex? Well, if we begin by looking at angle one, we could mark the vertex by drawing an orange dot. The vertex for angle one is here. But if we look at angle two and mark the vertex here with a pink dot, we can see that the vertex for angle two is here. Both sides that make the angle come from this point. Angles one and two don’t share a common vertex. So they can’t be adjacent angles.
Let’s see now whether they share a common side. Well, you could say that they do. This part of the line from angle two is shared with angle one. But it’s not the whole of the line that’s used in angle one. And so we can’t really say that angles one and two share a common side. We’ve used the two rules for adjacent angles to prove that angles one and two are not adjacent angles. And so we can give a very simple answer to the question.
Are angles one and two adjacent angles? No. |
# 4TH GRADE FIRST QUARTER MATHEMATICS STANDARDS. Vocabulary. answers using mental computation and estimation strategies including rounding.
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1 4TH GRADE FIRST QUARTER MATHEMATICS STANDARDS Critical Area: Developing understanding and fluency with multi-digit multiplication, and developing understanding of dividing to find quotients involving multi-digit dividends. Place Value and Operations with Whole Numbers (Chapter 1) Generalize place value understanding for multi-digit whole MAFS.4.NBT1.1 Recognize that in a multi-digit whole number, a digit in one place represents ten times what it represents in the place to its right. For example, recognize that = 10 by applying concepts of place value and division. MAFS.4.NBT1.2 Read and write multi-digit whole numbers using base-ten numerals, number names, and expanded form. Compare two multi-digit numbers based on meanings of the digits in each place, using >, =, and < symbols to record the results of comparisons. MAFS.4.NBT1.3 Use place value understanding to round multidigit whole numbers to any place. MAFS.4.NBT.2.4 Fluently add and subtract multi-digit whole numbers using the standard algorithm. Multiply by 1-Digit Numbers (Chapter 2) MAFS.4.OA.1.1 Interpret a multiplication equation as a comparison, e.g., interpret 35 = 5 7 as a statement that 35 is 5 times as many as 7 and 7 times as many as 5. Represent verbal statements of multiplicative comparisons as multiplication equations. MAFS.4.OA.1.a Determine whether an equation is true or false by using comparative relational thinking. For example, without adding 60 and 24, determine whether the equation = is true or false. MAFS.4.OA.1.b Determine the unknown whole number in an equation relating four whole numbers using comparative relational thinking. For example, solve = n + 5 for n by arguing that nine is four more than five, so the unknown number must be four greater than 76. MAFS.4.OA.1.2 Multiply or divide to solve word problems involving multiplicative comparison, e.g., by using drawings and equations with a symbol for the unknown number to represent the problem, distinguishing multiplicative comparison from additive comparison. MAFS.4.NBT.2.5 Multiply a whole number of up to four digits by a one-digit whole number, and multiply two two-digit numbers, using strategies based on place value and the properties of operations. Illustrate and explain the calculation by using equations, rectangular Multiply 2-Digit Numbers (Chapter 3) MAFS.4.NBT.2.5 Multiply a whole number of up to four digits by a one-digit whole number, and multiply two two-digit numbers, using strategies based on place value and the properties of operations. Illustrate and explain the calculation by using equations, rectangular Divide by 1-Digit Numbers (Chapter 4) MAFS.4.NBT.2.6 Find whole-number quotients and remainders with up to four-digit dividends and one-digit divisors, using strategies based on place value, the properties of operations, and/or the relationship between multiplication and division. Illustrate and explain the calculation by using equations, rectangular arrays, and/or area models. Number and Operations in Base Ten (4.NBT): Digit Estimate Place value Round Expanded form Regroup Period Addend Standard form Addition Word form Difference Sum Compare Multiply Equal sign Number line Greater than sign Product Less than sign Distributive Property Number line Partial product Order Associative Property of Multiplication Compatible numbers Commutative Property of Multiplication Counting number Remainder Multiplication Divide Division Quotient Divisor Hundreds Ones Place Value Tens Thousands
2 4TH GRADE SECOND QUARTER MATHEMATICS STANDARDS Critical Area: Developing understanding and fluency with multi-digit multiplication, and developing understanding of dividing to find quotients involving multi-digit dividends. Critical Area: Developing an understanding of fraction equivalence, addition and subtraction of fractions with like denominators and multiplication of fractions by whole Divide by 1-Digit Numbers (Chapter 4) (continued) MAFS.4.NBT.2. 6 Find whole-number quotients and remainders with up to four-digit dividends and one-digit divisors, using strategies based on place value, the properties of operations, and/or the relationship between multiplication and division. Illustrate and explain the calculation by using equations, rectangular arrays, and/or area models. s, s, and Patterns (Chapter 5) Gain familiarity with factors and multiples. MAFS.4.OA.2.4 Investigate factors and multiples. a. Find all factor pairs for a whole number in the range b. Recognize that a whole number is a multiple of each of its factors. Determine whether a given whole number in the range is a multiple of a given one-digit number. c. Determine whether a given whole number in the range is prime or composite. Generate and analyze patterns MAFS.4.OA.3.5 Generate a number or shape pattern that follows a given rule. Identify apparent features of the pattern that were not explicit in the rule itself. For example, given the rule Add 3 and the starting number 1, generate terms in the resulting sequence and observe that the terms appear to alternate between odd and even Explain informally why the numbers will continue to alternate in this way. Fractions Equivalence and Comparison (Chapter 6) Extend understanding of fraction equivalence and ordering. MAFS.4.NF.1.1 Explain why a fraction a/b is equivalent to a fraction (n a)/(n b) by using visual fraction models, with attention to how the number and size of the parts differ even though the two fractions themselves are the same size. Use this principle to recognize and generate equivalent fractions. MAFS.4.NF.1.2 Compare two fractions with different numerators and different denominators, e.g., by creating common denominators or numerators, or by comparing to a benchmark fraction such as 1/2. Recognize that comparisons are valid only when the two fractions refer to the same whole. Record the results of comparisons with symbols >, =, or <, and justify the conclusions, e.g., by using a visual fraction model. Add and Subtract Fractions (Chapter 7) Build fractions from unit fractions by applying and extending previous understandings of operations on whole MAFS.4.NF.2.3 Understand a fraction a/b with a > 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 = /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. Multiply Fractions by Whole Numbers (Chapter 8) Build fractions from unit fractions by applying and extending previous understandings of operations on whole MAFS.4.NF.2.4 Apply and extend previous understandings of multiplication to multiply a fraction by a whole number. a. Understand a fraction a/b as a multiple of 1/b. For example, use a visual fraction model to represent 5/4 as the product 5 (1/4), recording the conclusion by the equation 5/4 = 5 (1/4). b. Understand a multiple of a/b as a multiple of 1/b, and use this understanding to multiply a fraction by a whole number. For example, use a visual fraction model to express 3 (2/5) as 6 (1/5), recognizing this product as 6/5. (In general, n (a/b) = (n a)/b.) c. Solve word problems involving multiplication of a fraction by a whole number, e.g., by using visual fraction models and equations to represent the problem. For example, if each person at a party will eat 3/8 of a pound of roast beef, and there will be 5 people at the party, how many pounds of roast beef will be needed? Between what two whole numbers does your answer lie? Number and Operations in Base Ten (4.NBT): Counting number Multiplication Divide Dividend Division divisor Quotient hundreds Ones Place value Tens Thousands Compatible numbers Distributive Property Partial quotient Operations and Algebraic Thinking (4.OA) Array Product Common factor Common multiple Composite number Prime number Pattern Term Number and Operations-Fractions (4.NF): Equivalent fractions Denominator Fraction Numerator Simplest form Common factor Common denominator Common multiple Benchmark Common numerator
5 MAFS.5.G.1.1 Use a pair of perpendicular number lines, called axes, to define a coordinate system, with the intersection of the lines (the origin) arranged to coincide with the 0 on each line and a given point in the plane located by using an ordered pair of numbers, called its coordinates. Understand that the first number indicates how far to travel from the origin in the direction of one axis, and the second number indicates how far to travel in the direction of the second axis, with the convention that the names of the two axes and the coordinates correspond (e.g., x-axis and x-coordinate, y-axis and y-coordinate). Geometric measurement: understand concepts of volume and relate volume to multiplication and to addition. MAFS.5.MD.3.5 Relate volume to the operations of multiplication and addition and solve real world and mathematical problems involving volume. a. Find the volume of a right rectangular prism with whole-number side lengths by packing it with unit cubes, and show that the volume is the same as would be found by multiplying the edge lengths, equivalently by multiplying the height by the area of the base. Represent threefold whole-number products as volumes, e.g., to represent the associative property of multiplication. b. Apply the formulas V = l w h and V = B h for rectangular prisms to find volumes of right rectangular prisms with whole-number edge lengths in the context of solving real world and mathematical problems. c. Recognize volume as additive. Find volumes of solid figures composed of two non-overlapping right rectangular prisms by adding the volumes of the non-overlapping parts, applying this technique to solve real world problems. 4TH GRADE FOURTH QUARTER MATHEMATICS STANDARDS
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Chapter 18
1. 8x2 + 8x – 6
Apply “FOIL” to multiply the terms in the binomials.
First: 2x(4x) = 8x2
Outer: 2x(–2) = –4x
Inner: 3(4x) = 12x
Last: 3(–2) = –6
Combine these products and simplify:
8x2 – 4x + 12x – 6 = 8x2 + 8x – 6
2. 4x2 + 3x – 4
First, multiply the two binomials together using “FOIL.”
(x + 4)(x – 1) = x2x + 4x – 4 = x2 + 3x – 4
Now add 3x2 to that product.
3x2 + x2 + 3x – 4 = 4x2 + 3x – 4
3. 8x2 – 2x – 11
First, perform the two multiplications.
(3x + 1)(x – 3) = 3x2 – 9x + x – 3 = 3x2 – 8x – 3
(x + 2)(5x – 4) = 5x2 – 4x + 10x – 8 = 5x2 + 6x – 8
3x2 – 8x – 3 + 5x2 + 6x – 8 = 8x2 – 2x – 11
4. 8x2 – 20x – 11
First, perform the two multiplications.
5(x – 3)(x + 2) = 5(x2x – 6) = 5x2 – 5x – 30
3(x – 3)(x – 2) = 3(x2 – 5x + 6) = 3x2 – 15x + 18
Then add the two products and the 1.
5x2 – 5x – 30 + 3x2 – 15x + 18 + 1 = 8x2 – 20x – 11
5. x3 + x2 – 7x + 20
Distribute the two terms in the binomial over the terms in the trinomial; then combine like terms.
x(x2 – 3x + 5) + 4(x2 – 3x + 5)
= x3 – 3x2 + 5x + 4x2 – 12x + 20
= x3 + x2 – 7x + 20
6. 3x3x2 – 3x + 1
Distribute the two terms in the binomial over the terms in the trinomial; then combine like terms.
x(3x2 + 2x – 1) – 1(3x2 + 2x – 1)
= 3x3 + 2x2x – 3x2 – 2x + 1
= 3x3x2 – 3x + 1
7. 2x3 + 3x2 – 23x – 12
First, multiply the second and third binomials together.
(2x + 1)(x – 3)(x + 4) =
(2x + 1)(x2 + x – 12)
Now distribute the two terms in the binomial ...
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# Combined events for AQA GCSE Maths
1. Combined events
2. Independent combined events
3. Dependent combined events
4. Tree diagrams
Systematic listing is when all possible outcomes of combined events are listed so that none are missed out. This can then be used to calculate the probability of any event happening. The product rule for counting states that multiplying the number of possible outcomes for each event together gives the total number of outcomes for combined events.
The probability of independent combined events can be found by multiplying the probabilities of the individual events.
If the first event that occurs affects the probability of the second event occuring, then the two events are dependent. Probabilities can be calculated by multiplying the individual probabilities together.
Tree diagrams show different combinations of events and the probabilities of each event occuring. To find the probability of a specific combination occuring, the probabilities of the individual events are multiplied together. If the probability of the first event occuring does not affect the probability of the second one occuring, the two events are independent.
# 1
A deck of 52 cards is made up of equal red and black cards. Two cards are drawn without replacement. State how the given tree diagram shows that the events are dependent.
The probability for the same outcome changes on the tree diagram after the first card is drawn, therefore the events are dependent.
# 2
An A-Level student can pick between 12 classes to take. The student can choose 3 classes. Calculate how many combinations of the classes the student can take.
Total number of combinations = 12 × 11 × 10 = 1320.
# 3
List all the possible ways in which a three-digit number can be made using the following digits, given that each digit is used once: 1, 2, 3.
The possible outcomes are: 123, 132, 213, 231, 312, 321.
# 4
Anna keeps 8 pairs of shorts and 12 T-shirts in the same drawer. She draws two random items from a drawer. Calculate the probability that she gets a T-shirt and a pair of shorts.
P(T-shirt and shorts) = 8/20 × 12/19 + 12/20 × 8/19 = 48/95.
# 5
Two cards are drawn, with replacement, out of an ordinary deck. Calculate the probability that the first card is an 8 and the second card is red.
P(first is 8 and second is red) = 4/52 × 26/52 = 1/26.
End of page |
RD Sharma Solutions: Operations on Whole Numbers (Exercise 4.3)
# Operations on Whole Numbers (Exercise 4.3) RD Sharma Solutions | Mathematics (Maths) Class 6 PDF Download
Page 1
Exercise 4.3 page: 4.14
1. Fill in the blanks to make each of the following a true statement:
(i) 785 × 0 = …..
(ii) 4567 × 1 = …..
(iii) 475 × 129 = 129 × …..
(iv) ….. × 8975 = 8975 × 1243
(v) 10 × 100 × …. = 10000
(vi) 27 × 18 = 27 × 9 + 27 × ….. + 27 × 5
(vii) 12 × 45 = 12 × 50 – 12 × …..
(viii) 78 × 89 = 78 × 100 – 78 × ….. + 78 × 5
(ix) 66 × 85 = 66 × 90 – 66 × ….. – 66
(x) 49 × 66 + 49 × 34 = 49 × (….. + …..)
Solution:
(i) 785 × 0 = 0
(ii) 4567 × 1 = 4567 based on multiplicative identity
(iii) 475 × 129 = 129 × 475 based on commutativity
(iv) 1243 × 8975 = 8975 × 1243 based on commutativity
(v) 10 × 100 × 10 = 10000
(vi) 27 × 18 = 27 × 9 + 27 × 4 + 27 × 5
(vii) 12 × 45 = 12 × 50 – 12 × 5
(viii) 78 × 89 = 78 × 100 – 78 × 16 + 78 × 5
(ix) 66 × 85 = 66 × 90 – 66 × 4 – 66
(x) 49 × 66 + 49 × 34 = 49 × (66 + 34)
2. Determine each of the following products by suitable rearrangements:
(i) 2 × 1497 × 50
(ii) 4 × 358 × 25
(iii) 495 × 625 × 16
(iv) 625 × 20 × 8 × 50
Solution:
(i) 2 × 1497 × 50
It can be written as
2 × 1497 × 50 = (2 × 50) × 1497
= 100 × 1497
= 149700
(ii) 4 × 358 × 25
It can be written as
Page 2
Exercise 4.3 page: 4.14
1. Fill in the blanks to make each of the following a true statement:
(i) 785 × 0 = …..
(ii) 4567 × 1 = …..
(iii) 475 × 129 = 129 × …..
(iv) ….. × 8975 = 8975 × 1243
(v) 10 × 100 × …. = 10000
(vi) 27 × 18 = 27 × 9 + 27 × ….. + 27 × 5
(vii) 12 × 45 = 12 × 50 – 12 × …..
(viii) 78 × 89 = 78 × 100 – 78 × ….. + 78 × 5
(ix) 66 × 85 = 66 × 90 – 66 × ….. – 66
(x) 49 × 66 + 49 × 34 = 49 × (….. + …..)
Solution:
(i) 785 × 0 = 0
(ii) 4567 × 1 = 4567 based on multiplicative identity
(iii) 475 × 129 = 129 × 475 based on commutativity
(iv) 1243 × 8975 = 8975 × 1243 based on commutativity
(v) 10 × 100 × 10 = 10000
(vi) 27 × 18 = 27 × 9 + 27 × 4 + 27 × 5
(vii) 12 × 45 = 12 × 50 – 12 × 5
(viii) 78 × 89 = 78 × 100 – 78 × 16 + 78 × 5
(ix) 66 × 85 = 66 × 90 – 66 × 4 – 66
(x) 49 × 66 + 49 × 34 = 49 × (66 + 34)
2. Determine each of the following products by suitable rearrangements:
(i) 2 × 1497 × 50
(ii) 4 × 358 × 25
(iii) 495 × 625 × 16
(iv) 625 × 20 × 8 × 50
Solution:
(i) 2 × 1497 × 50
It can be written as
2 × 1497 × 50 = (2 × 50) × 1497
= 100 × 1497
= 149700
(ii) 4 × 358 × 25
It can be written as
4 × 358 × 25 = (4 × 25) × 358
= 100 × 358
= 35800
(iii) 495 × 625 × 16
It can be written as
495 × 625 × 16 = (625 × 16) × 495
= 10000 × 495
= 4950000
(iv) 625 × 20 × 8 × 50
It can be written as
625 × 20 × 8 × 50 = (625 × 8) × (20 × 50)
= 5000 × 1000
= 5000000
3. Using distributivity of multiplication over addition of whole numbers, find each of the following
products:
(i) 736 × 103
(ii) 258 × 1008
(iii) 258 × 1008
Solution:
(i) 736 × 103
It can be written as
= 736 × (100 + 3)
By using distributivity of multiplication over addition of whole numbers
= (736 × 100) + (736 × 3)
On further calculation
= 73600 + 2208
We get
= 75808
(ii) 258 × 1008
It can be written as
= 258 × (1000 + 8)
By using distributivity of multiplication over addition of whole numbers
= (258 × 1000) + (258 × 8)
On further calculation
= 258000 + 2064
We get
= 260064
(iii) 258 × 1008
It can be written as
= 258 × (1000 + 8)
By using distributivity of multiplication over addition of whole numbers
= (258 × 1000) + (258 × 8)
On further calculation
= 258000 + 2064
Page 3
Exercise 4.3 page: 4.14
1. Fill in the blanks to make each of the following a true statement:
(i) 785 × 0 = …..
(ii) 4567 × 1 = …..
(iii) 475 × 129 = 129 × …..
(iv) ….. × 8975 = 8975 × 1243
(v) 10 × 100 × …. = 10000
(vi) 27 × 18 = 27 × 9 + 27 × ….. + 27 × 5
(vii) 12 × 45 = 12 × 50 – 12 × …..
(viii) 78 × 89 = 78 × 100 – 78 × ….. + 78 × 5
(ix) 66 × 85 = 66 × 90 – 66 × ….. – 66
(x) 49 × 66 + 49 × 34 = 49 × (….. + …..)
Solution:
(i) 785 × 0 = 0
(ii) 4567 × 1 = 4567 based on multiplicative identity
(iii) 475 × 129 = 129 × 475 based on commutativity
(iv) 1243 × 8975 = 8975 × 1243 based on commutativity
(v) 10 × 100 × 10 = 10000
(vi) 27 × 18 = 27 × 9 + 27 × 4 + 27 × 5
(vii) 12 × 45 = 12 × 50 – 12 × 5
(viii) 78 × 89 = 78 × 100 – 78 × 16 + 78 × 5
(ix) 66 × 85 = 66 × 90 – 66 × 4 – 66
(x) 49 × 66 + 49 × 34 = 49 × (66 + 34)
2. Determine each of the following products by suitable rearrangements:
(i) 2 × 1497 × 50
(ii) 4 × 358 × 25
(iii) 495 × 625 × 16
(iv) 625 × 20 × 8 × 50
Solution:
(i) 2 × 1497 × 50
It can be written as
2 × 1497 × 50 = (2 × 50) × 1497
= 100 × 1497
= 149700
(ii) 4 × 358 × 25
It can be written as
4 × 358 × 25 = (4 × 25) × 358
= 100 × 358
= 35800
(iii) 495 × 625 × 16
It can be written as
495 × 625 × 16 = (625 × 16) × 495
= 10000 × 495
= 4950000
(iv) 625 × 20 × 8 × 50
It can be written as
625 × 20 × 8 × 50 = (625 × 8) × (20 × 50)
= 5000 × 1000
= 5000000
3. Using distributivity of multiplication over addition of whole numbers, find each of the following
products:
(i) 736 × 103
(ii) 258 × 1008
(iii) 258 × 1008
Solution:
(i) 736 × 103
It can be written as
= 736 × (100 + 3)
By using distributivity of multiplication over addition of whole numbers
= (736 × 100) + (736 × 3)
On further calculation
= 73600 + 2208
We get
= 75808
(ii) 258 × 1008
It can be written as
= 258 × (1000 + 8)
By using distributivity of multiplication over addition of whole numbers
= (258 × 1000) + (258 × 8)
On further calculation
= 258000 + 2064
We get
= 260064
(iii) 258 × 1008
It can be written as
= 258 × (1000 + 8)
By using distributivity of multiplication over addition of whole numbers
= (258 × 1000) + (258 × 8)
On further calculation
= 258000 + 2064
We get
= 260064
4. Find each of the following products:
(i) 736 × 93
(ii) 816 × 745
(iii) 2032 × 613
Solution:
(i) 736 × 93
It can be written as
= 736 × (100 – 7)
By using distributivity of multiplication over subtraction of whole numbers
= (736 × 100) - (736 × 7)
On further calculation
= 73600 – 5152
We get
= 68448
(ii) 816 × 745
It can be written as
= 816 × (750 – 5)
By using distributivity of multiplication over subtraction of whole numbers
= (816 × 750) - (816 × 5)
On further calculation
= 612000 – 4080
We get
= 607920
(iii) 2032 × 613
It can be written as
= 2032 × (600 + 13)
By using distributivity of multiplication over addition of whole numbers
= (2032 × 600) + (2032 × 13)
On further calculation
= 1219200 + 26416
We get
= 1245616
5. Find the values of each of the following using properties:
(i) 493 × 8 + 493 × 2
(ii) 24579 × 93 + 7 × 24579
(iii) 1568 × 184 – 1568 × 84
(iv) 15625 × 15625 – 15625 × 5625
Solution:
(i) 493 × 8 + 493 × 2
It can be written as
= 493 × (8 + 2)
By using distributivity of multiplication over addition of whole numbers
Page 4
Exercise 4.3 page: 4.14
1. Fill in the blanks to make each of the following a true statement:
(i) 785 × 0 = …..
(ii) 4567 × 1 = …..
(iii) 475 × 129 = 129 × …..
(iv) ….. × 8975 = 8975 × 1243
(v) 10 × 100 × …. = 10000
(vi) 27 × 18 = 27 × 9 + 27 × ….. + 27 × 5
(vii) 12 × 45 = 12 × 50 – 12 × …..
(viii) 78 × 89 = 78 × 100 – 78 × ….. + 78 × 5
(ix) 66 × 85 = 66 × 90 – 66 × ….. – 66
(x) 49 × 66 + 49 × 34 = 49 × (….. + …..)
Solution:
(i) 785 × 0 = 0
(ii) 4567 × 1 = 4567 based on multiplicative identity
(iii) 475 × 129 = 129 × 475 based on commutativity
(iv) 1243 × 8975 = 8975 × 1243 based on commutativity
(v) 10 × 100 × 10 = 10000
(vi) 27 × 18 = 27 × 9 + 27 × 4 + 27 × 5
(vii) 12 × 45 = 12 × 50 – 12 × 5
(viii) 78 × 89 = 78 × 100 – 78 × 16 + 78 × 5
(ix) 66 × 85 = 66 × 90 – 66 × 4 – 66
(x) 49 × 66 + 49 × 34 = 49 × (66 + 34)
2. Determine each of the following products by suitable rearrangements:
(i) 2 × 1497 × 50
(ii) 4 × 358 × 25
(iii) 495 × 625 × 16
(iv) 625 × 20 × 8 × 50
Solution:
(i) 2 × 1497 × 50
It can be written as
2 × 1497 × 50 = (2 × 50) × 1497
= 100 × 1497
= 149700
(ii) 4 × 358 × 25
It can be written as
4 × 358 × 25 = (4 × 25) × 358
= 100 × 358
= 35800
(iii) 495 × 625 × 16
It can be written as
495 × 625 × 16 = (625 × 16) × 495
= 10000 × 495
= 4950000
(iv) 625 × 20 × 8 × 50
It can be written as
625 × 20 × 8 × 50 = (625 × 8) × (20 × 50)
= 5000 × 1000
= 5000000
3. Using distributivity of multiplication over addition of whole numbers, find each of the following
products:
(i) 736 × 103
(ii) 258 × 1008
(iii) 258 × 1008
Solution:
(i) 736 × 103
It can be written as
= 736 × (100 + 3)
By using distributivity of multiplication over addition of whole numbers
= (736 × 100) + (736 × 3)
On further calculation
= 73600 + 2208
We get
= 75808
(ii) 258 × 1008
It can be written as
= 258 × (1000 + 8)
By using distributivity of multiplication over addition of whole numbers
= (258 × 1000) + (258 × 8)
On further calculation
= 258000 + 2064
We get
= 260064
(iii) 258 × 1008
It can be written as
= 258 × (1000 + 8)
By using distributivity of multiplication over addition of whole numbers
= (258 × 1000) + (258 × 8)
On further calculation
= 258000 + 2064
We get
= 260064
4. Find each of the following products:
(i) 736 × 93
(ii) 816 × 745
(iii) 2032 × 613
Solution:
(i) 736 × 93
It can be written as
= 736 × (100 – 7)
By using distributivity of multiplication over subtraction of whole numbers
= (736 × 100) - (736 × 7)
On further calculation
= 73600 – 5152
We get
= 68448
(ii) 816 × 745
It can be written as
= 816 × (750 – 5)
By using distributivity of multiplication over subtraction of whole numbers
= (816 × 750) - (816 × 5)
On further calculation
= 612000 – 4080
We get
= 607920
(iii) 2032 × 613
It can be written as
= 2032 × (600 + 13)
By using distributivity of multiplication over addition of whole numbers
= (2032 × 600) + (2032 × 13)
On further calculation
= 1219200 + 26416
We get
= 1245616
5. Find the values of each of the following using properties:
(i) 493 × 8 + 493 × 2
(ii) 24579 × 93 + 7 × 24579
(iii) 1568 × 184 – 1568 × 84
(iv) 15625 × 15625 – 15625 × 5625
Solution:
(i) 493 × 8 + 493 × 2
It can be written as
= 493 × (8 + 2)
By using distributivity of multiplication over addition of whole numbers
= 493 × 10
On further calculation
= 4930
(ii) 24579 × 93 + 7 × 24579
It can be written as
= 24579 × (93 + 7)
By using distributivity of multiplication over addition of whole numbers =
24579 × 100
On further calculation
= 2457900
(iii) 1568 × 184 – 1568 × 84
It can be written as
= 1568 × (184 - 84)
By using distributivity of multiplication over subtraction of whole numbers
= 1568 × 100
On further calculation
= 156800
(iv)15625 × 15625 – 15625 × 5625
It can be written as
= 15625 × (15625 - 5625)
By using distributivity of multiplication over subtraction of whole numbers
= 15625 × 10000
On further calculation
= 156250000
6. Determine the product of:
(i) the greatest number of four digits and the smallest number of three digits.
(ii) the greatest number of five digits and the greatest number of three digits.
Solution:
(i) We know that
Largest four digit number = 9999
Smallest three digit number = 100
Product of both = 9999 × 100 = 999900
Hence, the product of the greatest number of four digits and the smallest number of three digits is 999900.
(ii) We know that
Largest five digit number = 99999
Largest three digit number = 999
Product of both = 99999 × 999
It can be written as
= 99999 × (1000 – 1)
By using distributivity of multiplication over subtraction of whole numbers
= (99999 × 1000) – (99999 × 1)
On further calculation
= 99999000 – 99999
Page 5
Exercise 4.3 page: 4.14
1. Fill in the blanks to make each of the following a true statement:
(i) 785 × 0 = …..
(ii) 4567 × 1 = …..
(iii) 475 × 129 = 129 × …..
(iv) ….. × 8975 = 8975 × 1243
(v) 10 × 100 × …. = 10000
(vi) 27 × 18 = 27 × 9 + 27 × ….. + 27 × 5
(vii) 12 × 45 = 12 × 50 – 12 × …..
(viii) 78 × 89 = 78 × 100 – 78 × ….. + 78 × 5
(ix) 66 × 85 = 66 × 90 – 66 × ….. – 66
(x) 49 × 66 + 49 × 34 = 49 × (….. + …..)
Solution:
(i) 785 × 0 = 0
(ii) 4567 × 1 = 4567 based on multiplicative identity
(iii) 475 × 129 = 129 × 475 based on commutativity
(iv) 1243 × 8975 = 8975 × 1243 based on commutativity
(v) 10 × 100 × 10 = 10000
(vi) 27 × 18 = 27 × 9 + 27 × 4 + 27 × 5
(vii) 12 × 45 = 12 × 50 – 12 × 5
(viii) 78 × 89 = 78 × 100 – 78 × 16 + 78 × 5
(ix) 66 × 85 = 66 × 90 – 66 × 4 – 66
(x) 49 × 66 + 49 × 34 = 49 × (66 + 34)
2. Determine each of the following products by suitable rearrangements:
(i) 2 × 1497 × 50
(ii) 4 × 358 × 25
(iii) 495 × 625 × 16
(iv) 625 × 20 × 8 × 50
Solution:
(i) 2 × 1497 × 50
It can be written as
2 × 1497 × 50 = (2 × 50) × 1497
= 100 × 1497
= 149700
(ii) 4 × 358 × 25
It can be written as
4 × 358 × 25 = (4 × 25) × 358
= 100 × 358
= 35800
(iii) 495 × 625 × 16
It can be written as
495 × 625 × 16 = (625 × 16) × 495
= 10000 × 495
= 4950000
(iv) 625 × 20 × 8 × 50
It can be written as
625 × 20 × 8 × 50 = (625 × 8) × (20 × 50)
= 5000 × 1000
= 5000000
3. Using distributivity of multiplication over addition of whole numbers, find each of the following
products:
(i) 736 × 103
(ii) 258 × 1008
(iii) 258 × 1008
Solution:
(i) 736 × 103
It can be written as
= 736 × (100 + 3)
By using distributivity of multiplication over addition of whole numbers
= (736 × 100) + (736 × 3)
On further calculation
= 73600 + 2208
We get
= 75808
(ii) 258 × 1008
It can be written as
= 258 × (1000 + 8)
By using distributivity of multiplication over addition of whole numbers
= (258 × 1000) + (258 × 8)
On further calculation
= 258000 + 2064
We get
= 260064
(iii) 258 × 1008
It can be written as
= 258 × (1000 + 8)
By using distributivity of multiplication over addition of whole numbers
= (258 × 1000) + (258 × 8)
On further calculation
= 258000 + 2064
We get
= 260064
4. Find each of the following products:
(i) 736 × 93
(ii) 816 × 745
(iii) 2032 × 613
Solution:
(i) 736 × 93
It can be written as
= 736 × (100 – 7)
By using distributivity of multiplication over subtraction of whole numbers
= (736 × 100) - (736 × 7)
On further calculation
= 73600 – 5152
We get
= 68448
(ii) 816 × 745
It can be written as
= 816 × (750 – 5)
By using distributivity of multiplication over subtraction of whole numbers
= (816 × 750) - (816 × 5)
On further calculation
= 612000 – 4080
We get
= 607920
(iii) 2032 × 613
It can be written as
= 2032 × (600 + 13)
By using distributivity of multiplication over addition of whole numbers
= (2032 × 600) + (2032 × 13)
On further calculation
= 1219200 + 26416
We get
= 1245616
5. Find the values of each of the following using properties:
(i) 493 × 8 + 493 × 2
(ii) 24579 × 93 + 7 × 24579
(iii) 1568 × 184 – 1568 × 84
(iv) 15625 × 15625 – 15625 × 5625
Solution:
(i) 493 × 8 + 493 × 2
It can be written as
= 493 × (8 + 2)
By using distributivity of multiplication over addition of whole numbers
= 493 × 10
On further calculation
= 4930
(ii) 24579 × 93 + 7 × 24579
It can be written as
= 24579 × (93 + 7)
By using distributivity of multiplication over addition of whole numbers =
24579 × 100
On further calculation
= 2457900
(iii) 1568 × 184 – 1568 × 84
It can be written as
= 1568 × (184 - 84)
By using distributivity of multiplication over subtraction of whole numbers
= 1568 × 100
On further calculation
= 156800
(iv)15625 × 15625 – 15625 × 5625
It can be written as
= 15625 × (15625 - 5625)
By using distributivity of multiplication over subtraction of whole numbers
= 15625 × 10000
On further calculation
= 156250000
6. Determine the product of:
(i) the greatest number of four digits and the smallest number of three digits.
(ii) the greatest number of five digits and the greatest number of three digits.
Solution:
(i) We know that
Largest four digit number = 9999
Smallest three digit number = 100
Product of both = 9999 × 100 = 999900
Hence, the product of the greatest number of four digits and the smallest number of three digits is 999900.
(ii) We know that
Largest five digit number = 99999
Largest three digit number = 999
Product of both = 99999 × 999
It can be written as
= 99999 × (1000 – 1)
By using distributivity of multiplication over subtraction of whole numbers
= (99999 × 1000) – (99999 × 1)
On further calculation
= 99999000 – 99999
We get
= 99899001
7. In each of the following, fill in the blanks, so that the statement is true:
(i) (500 + 7) (300 – 1) = 299 × …..
(ii) 888 + 777 + 555 = 111 × …..
(iii) 75 × 425 = (70 + 5) (….. + 85)
(iv) 89 × (100 – 2) = 98 × (100 - …..)
(v) (15 + 5) (15 – 5) = 225 - …..
(vi) 9 × (10000 + …..) = 98766
Solution:
(i) By considering LHS
(500 + 7) (300 – 1)
We get
= 507 × 299
By using commutativity
= 299 × 507
(ii) By considering LHS
888 + 777 + 555
We get
= 111 (8 + 7 + 5)
By using distributivity
= 111 × 20
(iii) By considering LHS
75 × 425
We get
= (70 + 5) × 425
It can be written as
= (70 + 5) (340 + 85)
(iv) By considering LHS
89 × (100 – 2)
We get
= 89 × 98
It can be written as
= 98 × 89
By using commutativity
= 98 × (100 – 11)
(v) By considering LHS
(15 + 5) (15 – 5)
We get
= 20 × 10
On further calculation
= 200
It can be written as
= 225 – 25
## Mathematics (Maths) Class 6
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## Mathematics (Maths) Class 6
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# Quick Subtraction Strategies Back Through 10
Quick Subtraction Strategies Back Through 10 is a fundamental math skill that helps students quickly subtract numbers from 10 or multiples of 10. These strategies are useful for mental math calculations and building a strong foundation in arithmetic. Here are some subtraction-back-through-10 for subtracting numbers back through 10:
1. Counting Backward:
• Start with the larger number and count backward to reach 10.
• For example, to subtract 7 from 10, count backward: 10, 9, 8, 7. The answer is 3.
2. Use Number Bonds:
• Visualize the number bonds of 10. For example, 10 can be split into 7 and 3 or 6 and 4.
• To subtract 7 from 10, think of it as taking away 3 from 10, resulting in 3.
3. Use Complements of 10:
• Recognize that numbers that add up to 10 are complements. For example, 7 and 3 are complements because 7 + 3 = 10.
• To subtract 7 from 10, use the complement concept and subtract 3 from 10, which equals 7.
4. Counting Up:
• If subtracting a number that’s larger than the starting number, you can count up from the smaller number to the larger number.
• For example, to subtract 8 from 10, start at 8 and count up: 8, 9, 10. The answer is 2.
5. Use a Number Line:
• Draw a number line with 10 as the endpoint and count backward or forward to perform the subtraction.
• For example, to subtract 6 from 10, start at 10 and move 6 units to the left on the number line.
6. Mental Math Techniques:
• Break down the subtraction into smaller steps. For example, to subtract 7 from 10, subtract 1 first (10 – 1 = 9), and then subtract 6 from 9 (9 – 6 = 3).
Practice these strategies regularly to become more proficient at Quick Subtraction Strategies Back Through 10. As you become more confident with these techniques, you’ll be able to perform mental math calculations quickly and accurately.
“Back through 10” Quick Subtraction Strategies are techniques that involve subtracting numbers by regrouping or borrowing to make the process easier. Here are 10 quick subtraction strategies that involve subtracting numbers by going back through 10:
## How do you make a 10 Quick Subtraction Strategies Back Through 10
1. Splitting Numbers: Break down one or both of the numbers into parts that are easier to subtract.
2. For example, to subtract 47 – 25, you can subtract 20 from 47 first, getting 27, and then subtract the remaining 5 to get 22.
3. Using Friendly Numbers: Round one or both numbers to a friendly number that is easier to work with. For example, to subtract 48 – 26, round 48 to 50 and subtract 26 from 50, which is 24.
4. Count Up: Start with the smaller number and count up to the larger number. For example, to subtract 9 – 5, start at 5 and count 6, 7, 8, 9. The answer is 4.
5. Number Line: Draw a number line and represent both numbers on it. Then move back through 10 to make the subtraction easier. For example, to subtract 43 – 17, start at 43 on the number line and move back to 33, then subtract 10 more to get 23.
6. Use Complements: Find the complement of one number to the nearest multiple of 10. For example, to subtract 68 – 37, find the complement of 37 to 40, which is 3, and subtract that from 68 to get 65.
7. Cross-Out Method: Write both numbers vertically and subtract the digits from right to left. If you can’t subtract the digits directly, borrow from the tens place. For example, to subtract 64 – 28, start with the ones place (4 – 8), which requires borrowing from the tens place to get 14 – 8 = 6. Then, subtract the tens place (6 – 2) to get 4.
8. Use Mental Math: Mentally subtract 10 from one number and add the difference to the other number. Repeat if necessary. For example, to subtract 79 – 38, mentally subtract 10 from 79 and add 10 to 38, making it 69 – 48, which is easier to calculate.
9. Regrouping with Zeros: If you’re subtracting a number with zeros in it, consider regrouping those zeros. For example, to subtract 600 – 398, think of it as 600 – 300 – 98, which simplifies the calculation.
10. Estimate First: Round both numbers to the nearest multiple of 10, subtract, and then adjust for the rounding. For example, to subtract 47 – 28, round 47 to 50 and 28 to 30, subtract to get 20, and then adjust for the rounding difference (3) to get 23.
Quick Subtraction Strategies Back Through 10 These strategies can help make subtraction through 10 quicker and more manageable, depending on the specific numbers you’re working with. Practice with different numbers to become proficient in using these techniques.
# Make Ten Addition Quick Subtraction Strategies?
### Learning Targets Quick Subtraction Strategies:
• I can add and subtract within 10 easily.
• I can add and subtract by making 10. (e.g. 7 + 3 = 10, 10 – 6 = 4)
The following diagrams show the Make Ten Subtraction Strategy. Scroll down the page for more examples and solutions.
Mental Math Quick Subtraction Strategies Back Through 10: Break Down To 10
How to use mental Quick Subtraction Strategies Back Through 10 to subtract within 20 using the “break down to 10” strategy?
This is one of the strategies suggested in the Common Core State Standards for Mathematics, 1.OA.6 and 2.OA.2.
With practice, this Quick Subtraction Strategies can help students gain fluency when subtracting within 20. Eventually students will memorize these facts, which is expected by the end of grade 2.
Example:
14 – 9 = 14 – 4 – 5 = 10 – 5 = 5
15 – 7 = 15 – 5 – 2 = 10 – 2 = 8
17 – 8 = 17 – 7 – 1 = 10 – 1 = 9
Make 10 Quick Subtraction Strategies Back Through 10
Example:
11 – 6 = 11 – 1 – 5 = 10 – 5 = 5
13 – 9 = 13 – 3 – 6 = 10 – 6 = 4
17 – 8 = 17 – 7 – 1 = 10 – 1 = 9
14 – 7 = 14 – 4 – 3 = 10 – 3 = 7 |
# Is a parallelogram always sometimes or never?
## Is a parallelogram always sometimes or never?
Explanation: For this question, all you need to know are the properties of each shape. Since the question is asking if a rectangle is a parallelogram, you would check to make sure all the properties of the parallelogram agree with those of a rectangle and since they all do, the answer is always.
Is a parallelogram a square always Sometimes Never?
Parallelograms are quadrilaterals with two sets of parallel sides. Since squares must be quadrilaterals with two sets of parallel sides, then all squares are parallelograms. This is always true. Squares are quadrilaterals with 4 congruent sides.
What is never a parallelogram?
An ordinary quadrilateral with no equal sides is not a parallelogram. A trapezium and and an isosceles trapezium have one pair of opposite sides parallel. A Concave quadrilateral or arrowhead does not have parallel sides.
### Is a parallelogram always sometimes or never a rhombus?
A rhombus is a quadrilateral with four congruent sides. Therefore, every rhombus is a parallelogram.
What is sometimes a parallelogram?
 A parallelogram is a four-sided figure with two sets of parallel sides. A square is a parallelogram. This is always true.  Squares are quadrilaterals with 4 congruent sides and 4 right angles, and they also have two sets of parallel sides. Parallelograms are quadrilaterals with two sets of parallel sides.
Are rectangles always parallelograms?
A rectangle is a parallelogram with four right angles, so all rectangles are also parallelograms and quadrilaterals. On the other hand, not all quadrilaterals and parallelograms are rectangles. A rectangle has all the properties of a parallelogram, plus the following: The diagonals are congruent.
## What shape is always a quadrilateral and never a parallelogram?
Trapezoids have only one pair of parallel sides; parallelograms have two pairs of parallel sides. A trapezoid can never be a parallelogram. The correct answer is that all trapezoids are quadrilaterals.
Are parallelograms always similar?
This is always true. Â Squares are quadrilaterals with 4 congruent sides and 4 right angles, and they also have two sets of parallel sides. Parallelograms are quadrilaterals with two sets of parallel sides.
Is a parallelogram never sometimes or always a trapezium?
A trapezoid is a parallelogram. The diagonals of a rectangle bisect eachother.
### Why every parallelogram is not a rhombus?
Only the opposite sides of a parallelogram are equal. However, a rhombus has all its sides equal. So, every parallelogram cannot be a rhombus, except those parallelograms that have all equal sides.
Which quadrilateral will never be a parallelogram?
trapezoid
Trapezoids have only one pair of parallel sides; parallelograms have two pairs of parallel sides. A trapezoid can never be a parallelogram. The correct answer is that all trapezoids are quadrilaterals.
Which of the following are not parallelograms?
Trapezium is not a parallelogram as it has only one pair of parallel sides.
## Is a square always a parallelogram?
A square is a parallelogram. This is always true. Squares are quadrilaterals with 4 congruent sides and 4 right angles, and they also have two sets of parallel sides. Parallelograms are quadrilaterals with two sets of parallel sides.
What are the opposite angles of a parallelogram?
In a parallelogram, opposite angles are supplementary. In a parallelogram, diagonals bisect each other. In a parallelogram, the diagonals create four isosceles triangles . In a rectangle, the diagonals are congruent . In a parallelogram, opposite angles are congruent . In a parallelogram, opposite angles are right angles .
How are the diagonals of a parallelogram and a rectangle similar?
In a parallelogram, the diagonals create four isosceles triangles . In a rectangle, the diagonals are congruent . In a parallelogram, opposite angles are congruent . In a parallelogram, opposite angles are right angles .
### What is the difference between a parallelogram and a rhombus?
In a parallelogram, you can find congruent alternate interior angles. In a rhombus, diagonals are perpendicular. In a square, the diagonals create four isosceles triangles. In a rhombus, the diagonals are congruent. A square is a parallelogram. A parallelogram is a rectangle. |
### b - Ms. Hatch -
```m Before b !
UNIT 5.06
I can solve any equation for y and put it into slopeintercept form. I can then identify the unit rate and
starting point and interpret the equation!
Vocabulary
Slope-Intercept Form of a Linear Equation:
A line on the coordinate plane can be described by the
equation y = mx + b, where m is the unit rate and b is the
starting point. This is the most common form of a Linear
Equation.
In order to identify the unit rate and/or the starting point
it is easiest if an equation is in this form. But, a lot of
times, they are not...
WE CAN FIX THAT!
y = mx + b
Example:
3y = –6 + 18x
3
3
3
y = –2 + 6x
All we need to do is
isolate y on the left
and make sure the
x-term is first on
the right followed
by the constant
term.
y = 6x - 2
• Find the Slope-Intercept form: _________________
6
• Find the Unit Rate, m: ____
-2
• Find the Starting Point, b: ____
Let’s Try It! y = mx + b
1)
–7 + y = 2 + 5x
+7
+7
y = 9 + 5x
y = 5x + 9
• Find the Slope-Intercept form: _________________
5
• Find the Unit Rate: ____
9
• Find the Starting Point: ____
2)
Let’s Try It! y = mx + b
3x + y = 8x – 6
-3x
-3x
y = 5x – 6
y = 5x - 6
• Find the Slope-Intercept form: _________________
5
• Find the Unit Rate: ____
-6
• Find the Starting Point: ____
Let’s Try It! y = mx + b
3)
20x + 5y = 10
-20x
-20x
5y = 10 – 20x
5
5
5
y = 2 – 4x
y = -4x + 2
• Find the Slope-Intercept form: _________________
-4
• Find the Unit Rate, m: ____
2
• Find the Starting Point, b: ____
Let’s Try It! y = mx + b
4)
6x – 4y = 8 + 12x
-6x
-6x
-4y = 8 + 6x
-4
-4
-4
y = -2 – 3/2x
y = -3/2x - 2
• Find the Slope-Intercept form: _________________
-3/2
• Find the Unit Rate, m: ______
-2
• Find the Starting Point, b: ____
Let’s Try It! y = mx + b
5)
9x + y – 7 = 2x + 14
-9x
-9x
y – 7 = -7x + 14
+7
y
+7
Cody found a
package of oreo
LOVES oreo
= -7x + 21
y = -7x + 21
• Find the Slope-Intercept form: _________________
-7
• Find the Unit Rate, m: ______
21
• Find the Starting Point, b: ____
• Interpret the Linear Equation: __________...
Homework Time!
5.07 M BEFORE B WS
I can solve any equation for y and put it into slopeintercept form. I can then identify the unit rate and
starting point and interpret the equation!
``` |
## RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.7
These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.7
Other Exercises
Question 1.
Divide :
Solution:
Question 2.
Find the value and express as a rational number in standard form :
Solution:
Question 3.
The product of two rational numbers is15. If one of the numbers is -10, find the other.
Solution:
Question 4.
The product of two rational numbers is$$\frac { -9 }{ 8 }$$ if one of the numbers is $$\frac { -4 }{ 15 }$$, find other.
Solution:
Question 5.
By what number should we multiply $$\frac { -1 }{ 6 }$$ so that the product may be $$\frac { -23 }{ 9 }$$ ?
Solution:
Question 6.
By what number should we multiply $$\frac { -15 }{ 28 }$$ so that the product may be $$\frac { -5 }{ 7 }$$ ?
Solution:
Question 7.
By what number should we multiply $$\frac { -8 }{ 13 }$$ so that the product may be 24 ?
Solution:
Question 8.
Bv what number should $$\frac { -3 }{ 4 }$$ multiplied in order to produce $$\frac { 2 }{ 3 }$$ ?
Solution:
Question 9.
Find (x +y) + (x – y), if
Solution:
Question 10.
The cost of 7 $$\frac { 2 }{ 3 }$$ metres of rope is Rs 12 $$\frac { 3 }{ 4 }$$.Find its cost per metre.
Solution:
Question 11.
The cost of 2 $$\frac { 1 }{ 3 }$$ metres of cloth is Rs. 75 $$\frac { 1 }{ 4 }$$Find the cost of cloth per metre.
Solution:
Question 12.
By what number should $$\frac { -33 }{ 16 }$$ be divided to get $$\frac { -11 }{ 4 }$$ ?
Solution:
Question 13.
Divide the sum of $$\frac { -13 }{ 5 }$$ and $$\frac { 12 }{ 7 }$$ by the product of $$\frac { -31 }{ 7 }$$ and $$\frac { -1 }{ 2 }$$.
Solution:
Question 14.
Divide the sum of $$\frac { 65 }{ 12 }$$ and $$\frac { 12 }{ 7 }$$ bv their difference.
Solution:
Question 15.
If 24 trousers of equal size can be prepared in 54 metres of cloth, what length of cloth is required for each trouser ?
Solution:
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## RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.8
These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.8
Other Exercises
Question 1.
Find a rational number between -3 and 1.
Solution:
Question 2.
Find any five rational number less than 1.
Solution:
Question 3.
Find four rational numbers between $$\frac { -2 }{ 9 }$$ and $$\frac { 5 }{ 9 }$$ .
Solution:
Question 4.
Find two rational numbers between $$\frac { 1 }{ 5 }$$ and $$\frac { 1 }{ 2 }$$ .
Solution:
Question 5.
Find ten rational numbers between $$\frac { 1 }{ 4 }$$ and $$\frac { 1 }{ 2 }$$ .
Solution:
Question 6.
Find ten rational numbers between $$\frac { -2 }{ 5 }$$ and $$\frac { 1 }{ 2 }$$ .
Solution:
Question 7.
Find ten rational numbers between $$\frac { 3 }{ 5 }$$ and $$\frac { 3 }{ 4 }$$ .
Solution:
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## RD Sharma Class 8 Solutions Chapter 2 Powers MCQS
These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 2 Powers MCQS
Other Exercises
Choose the correct alternative in each of the following :
Question 1.
Square of $$\left( \frac { -2 }{ 3 } \right)$$
Solution:
Question 2.
Cube of $$\frac { -1 }{ 2 }$$ is
Solution:
Question 3.
Which of the following is not equal to
Solution:
Question 4.
Which of the following in not reciprocal of
Solution:
Question 5.
Which of the following numbers is not equal to $$\frac { -8 }{ 27 }$$ ?
Solution:
Question 6.
Solution:
Question 7.
Solution:
Question 8.
Solution:
Question 9.
Solution:
Question 10.
Solution:
Question 11.
Solution:
Question 12.
Solution:
Question 13.
Solution:
Question 14.
Solution:
Question 15.
For any two non-zero rational numbers a and b,a4+b4 is equal to
(a) (a + b)1
(b) (a + b)0
(c) (a + b)4
(d) (a + b)8
Solution:
(c) {∵ a4 + b4 = (a + b)4}
Question 16.
For any two rational numbers a and b, a5 x b5 is equal to
(a) (a x b)0
(b) (a x b)10
(c) (a x b)5
(d) (a x b)25
Solution:
(c) {∵ a5 x b5 = (a x b)5}
Question 17.
For a non-zero rational number a, a7 + a12 is equal to
(a) a5
(b) a-19
(c) a-5
(d) a19
Solution:
(c) {a5 a12 = a7-12 =a-5}
Question 18.
For a non-zero rational number a, (a3)-2 is equal to
(a) a6
(b) a-6
(c) a-9
(d) a1
Solution:
(b) {(a3)-2 = a3 x (-2)= a6}
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## RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.5
These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.5
Other Exercises
Question 1.
Multiply:
Solution:
Question 2.
Multiply:
Solution:
Question 3.
Simplify each of the following and express the result as a rational number in standard form :
Solution:
Question 4.
Simplify :
Solution:
Question 5.
Simplify :
Solution:
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## RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.4
These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.4
Other Exercises
Question 1.
Simplify each of the following and write as a rational number of the form :
Solution:
Question 2.
Express each of the following as a rational number of the form $$\frac { p }{ q }$$:
Solution:
Question 3.
Simplify :
Solution:
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## RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.3
These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.3
Other Exercises
Question 1.
Subtract the first rational number from the second in each of the following :
Solution:
Question 2.
Evaluate each of the following :
Solution:
Question 3.
The sum of two numbers is $$\frac { 5 }{ 9 }$$. If one of the numbers is $$\frac { 1 }{ 3 }$$, find the other.
Solution:
Question 4.
The sum of two numbers is $$\frac { -1 }{ 3 }$$. If one of the numbers is $$\frac { -12 }{ 3 }$$, find the other.
Solution:
Question 5.
The sum of two numbers is $$\frac { -4 }{ 3 }$$. If one of the number is -5, find the
Solution:
Question 6.
The sum of two rational numbers is -8. If one of the numbers is $$\frac { -15 }{ 7 }$$ find the other.
Solution:
Question 7.
What should be added to so as to $$\frac { -7 }{ 8 }$$ get $$\frac { 5 }{ 9 }$$ ?
Solution:
Question 8.
What number should be added to $$\frac { -5 }{ 11 }$$ so as to get $$\frac { 26 }{ 3 }$$ ?
Solution:
Question 9.
What number should be added to $$\frac { -5 }{ 7 }$$ to get $$\frac { -2 }{ 3 }$$ ?
Solution:
Question 10.
What number should be subtracted from $$\frac { -5 }{ 3 }$$ to get $$\frac { 5 }{ 6 }$$ ?
Solution:
Question 11.
What number should be subtracted from $$\frac { 3 }{ 7 }$$ to get $$\frac { 5 }{ 4 }$$ ?
Solution:
Question 12.
What should be added to $$\left( \frac { 2 }{ 3 } +\frac { 3 }{ 5 } \right)$$ to get $$\frac { -12 }{ 15 }$$ ?
Solution:
Question 13.
What should be added to $$\left( \frac { 1 }{ 2 } +\frac { 1 }{ 3 } +\frac { 1 }{ 5 } \right)$$ to get 3 ?
Solution:
Question 14.
What should be subtracted from $$\left( \frac { 3 }{ 4 } -\frac { 2 }{ 3 } \right)$$ to get $$\frac { -1 }{ 6 }$$ ?
Solution:
Question 15.
Fill in the blanks :
Solution:
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## RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.2
These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.2
Other Exercises
Question 1.
Verify commutativity of addition of rational numbers for each of the following pairs of rational numbers :
Solution:
Question 2.
Verify associativity of addition of rational numbers i.e., (A: + y) + z = x + (y + z), when :
Solution:
Question 3.
Write the additive inverse of each of the following rational numbers :
Solution:
Question 4.
Write the negative (additive inverse) of each of the following :
Solution:
Question 5.
Using commutativity and associativity of addition of rational numbers, express ‘iach of the following as a rational number :
Solution:
Question 6.
Re-arrange suitably and find the sum in each of the following :
Solution:
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## RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.1
These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 1 Rational Numbers Ex 1.1
Other Exercises
Question 1.
Add the following rational numbers:
Solution:
Question 2.
Add the following rational numbers:
Solution:
Question 3.
Simplify:
Solution:
Question 4.
Add and Express the sum as mixed fraction:
Solution:
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## RD Sharma Class 8 Solutions Chapter 8 Division of Algebraic Expressions Ex 8.6
These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 8 Division of Algebraic Expressions Ex 8.6
Other Exercises
Divide :
Question 1.
x2 – 5x + 6 by x – 3
Solution:
Question 2.
ax2 – ay2 by ax + ay
Solution:
Question 3.
x– y4 by x– y2
Solution:
Question 4.
acx2 + (bc + ad)x + bd by (ax + b)
Solution:
Question 5.
(a2 + 2ab + b2)- (a2 + 2ac + c2) by 2a + b + c
Solution:
Question 6.
$$\frac { 1 }{ 4 }$$ x– $$\frac { 1 }{ 2 }$$ x- 12 by $$\frac { 1 }{ 2 }$$ x – 4
Solution:
Hope given RD Sharma Class 8 Solutions Chapter 8 Division of Algebraic Expressions Ex 8.6 are helpful to complete your math homework.
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## RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.5
These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.5
Other Exercises
Factorize each of the following expressions :
Question 1.
16x2-25y2
Solution:
16x2 – 25y2 = (4x)2 – (5y)2 {∵ a2 – b2 = (a + b) (a – b)}
= (4x + 5y) (4x – 5y)
Question 2.
27x2 12y2
Solution:
27x2 – 12y2 = 3 (9x2 – 4y2) {∵ a2 -b2 = (a + b) (a – b)}
= 3 [(3x)2 – (2y)2]
= 3 (3x + 2y) (3x – 2y)
Question 3.
144a– 289b2
Solution:
144a2 – 289b2 = (12a)2 – (17b)2 { ∵ a2b2 = (a + b) (a – b}
= (12a+ 17b) (12a- 17b)
Question 4.
12m2 – 27
Solution:
12m2 – 27 = 3 (4m2 – 9)
= 3 {(2m)2-(3)2} {∵ a2b2 = (a + b) (a – b)}
= 3 (2m + 3) (2m – 3)
Question 5.
125x2 – 45y2
Solution:
125x2 – 45y2 = 5 (25x2 – 9y2)
= 5 {(5x-)2 – (3y)2} {∵ a2 – b2 = (a + b) (ab}
= 5 (5x +
3y) (5x – 3y)
Question 6.
144a2 – 169b2
Solution:
144a2 – 169b2 = (12a)2 – (13b)2 {∵ a2 -b2 = (a + b) (a – b)}
= (12a + 13b) (12a-13b)
Question 7.
(2a – b)2 – 16c2
Solution:
(2a – b)2 – 16c2 = (2a – b)2 – (4c)2 {∵ a2 – b2 = (a + b) (a – b)}
= (2a – b + 4c) (2a – b – 4c)
Question 8.
(x + 2y)2 – 4 (2x -y)2
Solution:
(x + 2y)2 – 4 (2x – y)2
= (x + 2y)2 – {2 (2x –y)}2
=
(x + 2y)2 – (4x – 2y)2 {∵ a2– b2 = (a + b) (a – b)}
=
(a + 2y + 4x – 2y) (x + 2y – 4x + 2y)
= 5x (-3x + 4y)
Question 9.
3a5 – 48a3
Solution:
3a5 – 48a3 = 3a3 (a2– 16)
= 3a3 {(a)2 – (4)2} {∵ a2 – b2 = (a + b) (a – b)}
= 3a3 (a + 4) (a – 4)
Question 10.
a4 – 16b4
Solution:
a4 – 16b4 = (a2)2 – (4b2)2
= (a2 + 4b2) (a2 – 4b2)
= (a2 + 4b2) {(a)2 – (2b)2 } { ∵ a2 – b2 = (a + b) (a – b)}
= (a2 + 4b2) (a + 2b) (a – 2b)
Question 11.
x8 – 1
Solution:
x8 – 1 = (x4)2 – (1)2
= (x4 + 1) (x4 – 1)
= (x4+ 1) I (x2)2 – (1)2} { a2 – b2 = (a + b) (a – b)}
= (x4 + 1) (x2 + 1) (x2 – 1)
= (x4 + 1) (x2 + 1) {(x)2 – (1)2}
= (x4+ 1)(x2 + 1)(x+ 1)(x- 1)
= (x-1)(x+ 1) (x2 + 1) (x4 + 1)
Question 12.
64 – (a + 1)2
Solution:
64 – (a + 1)2 = (8)2 – (a + 1)2 {∵ a2 – b2 = (a + b) (a – b)}
= (8 + a + 1) (8 – a – 1)
= (9 + a) (7 – a)
Question 13.
36l2 – (m + n)2
Solution:
36l2 – (m + n)2 = (6l)2 – (m + n)2 {∵ a2 – b2 = (a + b) (a – b)}
= (6l + m + n) (6l – m – n)
Question 14.
25x4y4 – 1
Solution:
25x4y4 – 1 = (5x4y4)2 – (1)2 { ∵ a2 – b2 = (a + b) (a – b)}
= (5x4y4 + 1) (5x2y2 – 1)
Question 15.
Solution:
Question 16.
x3 – 144x
Solution:
x3 – 144x = x (x2 – 144)
= x {(x)2 – (12)2} { a2 – b2 = (a + b) (a – b)}
= x (x + 12) (x – 12)
Question 17.
(x – 4y)2 – 625
Solution:
(x – 4y)2 – 625
= (x – 4y)2 – (25)2 {∵ a2 – b2 = (a + b) (a – b)}
= (x – 4y + 25) (x -4y – 25)
Question 18.
9 (a – b)2 – 100 (x -y)2
Solution:
9(a-b)2– 100(x-y)2
= {3(a-b)}2-{10(x-y)}2 { a2 – b2 = (a + b) (a – b)}
= (3a – 3b)2 – (10x – 10y)2
= (3a – 3b + 10x – 10y) (3a – 3b – 10x + 10y)
Question 19.
(3 + 2a)2 – 25a2
Solution:
(3 + 2a)2 – 25a2
= (3 + 2a)2 – (5a)2 ( a2 – b2 = (a + b) (a – b)}
= (3 + 2a + 5a) (3 + 2a – 5a)
= (3 + 7a) (3 – 3a)
= (3 + 7a) 3 (1 – a)
= 3(1-a) (3 +7a)
Question 20.
(x + y)2 – (a – b)2
Solution:
Question 21.
Solution:
Question 22.
75a3b2 – 108ab4
Solution:
75a3b2 – 108ab4
= 3ab2 (25a2 – 36b2)
= 3ab2 {(5a)2 – (6b)2} { a2 – b2 = (a + b) (a – b)}
= 3ab2 (5a + 6b) (5a – 6b)
Question 23.
x5– 16x3
Solution:
x5 – 16x3 = x3 (x2 – 16)
= x3 {(x)2 – (4)2} { a2 – b2 = (a + b) (a – b)}
= x3 (x + 4) (x – 4)
Question 24.
Solution:
Question 25.
256x5 – 81x
Solution:
256x5– 81x = x(256x4– 81)
= x {(16x2)2 – (9)2} {∵ a2 – b2 = {a + b) (a – b)}
= x (16x2 + 9) (16x2 – 9)
= x (16x2 + 9) {(4x)2 – (3)2}
= x (16x2 + 9) (4x + 3) (4x-3)
Question 26.
a4 – (2b + c)4
Solution:
a4 – (2b + c)4
= (a2)2 – [(2b + c)2]2 { a2 – b2 = (a + b) (a – b)}
= {a2 + (2b + c)2} {a2 – (2b + c)2}
= {a2 + (2b + c)2} {(a)2 – (2b + c)2}
= {a2 + (2b + c)2} (a + 2b + c) (a -2b- c)
Question 27.
(3x + 4y)4 – x4
Solution:
(3x + 4y)4 – x4 – [(3x + 4y)2]2 – (x2)2
= [(3x + 4y)2 + x2] [(3x + 4y)2 – x2] {∵ a2 – b2 = (a + b) (a – b)
= [(3x + 4y)2 + x2] [(3x + 4y + x) (3x + 4y – x)]
= [(3x + 4y)2 + x2] (4x + 4y) (2x + 4y)
= [(3x + 4y)2 + x2] 4 (x + y) 2 (x + 2y)
= 8 (x + y) (x + 2y) [(3x + 4y)2 + x2]
Question 28.
p2q2 – p4q4
Solution:
p2q2– p4q4 =p2q2 (1 -p2q2)
=p2q2 [(1)2 – (pq)2] { a2 – b2 = (a + b) (a – b)
= p2q2 (1 +pq) (1 -pq)
Question 29.
3x3y – 243xy3
Solution:
3x3y – 243xy3
= 3xy (x2 – 81y2)
= 3xy [(x)2 – (9y)2]
= 3xy (x + 9y) (x – 9y)
Question 30.
a4b4 – 16c4
Solution:
a4b4 – 16c4 = (a2b2)2 – (4c2)2
= (a2b2 + 4c2) (a2b2 – 4c2)
= (a2b2 + 4c2) [(ab)2 – (2c)2] { a2 – b2 = (a + b) (a – b)
= (a2b2 + 4c2) (ab + 2c) (ab – 2c)
Question 31.
x4-625
Solution:
x4 – 625 = (x2)2 – (25)2 { a2 – b2 – (a + b) (a – b)
= (x2 + 25) (x2 – 25)
= (x2 + 25) [(x)2 – (5)2]
= (x2 + 25) (x + 5) (x – 5)
Question 32.
x4-1
Solution:
x4 – 1 = (x2)2 – (1)2 = (x2 + 1) (x2 – 1)
= (x2 + 1) [(x)2 – (1)2]
= (x2 + 1) (x + 1) (x – 1)
Question 33.
49 (a – b)2 -25 (a + b)2
Solution:
49 (a – by -25 (a + b)2
= [7 (a – b)]2 [5 (a + b)]2
= (7a – 7b)2 – (5a + 5b)2 { a2 – b2 = (a + b) (a – b)
= (7a -7b + 5a + 5b) (7a – 7b -5a- 5b)
=(12a – 2b)(2a – 12b)
= 2 (6a – b) 2 (a – 6b)
= 4 (6 a- b) (a – 6b)
Question 34.
x – y – x2 + y
Solution:
x-y-x2 + y2 = (x-y)-(x2-y2) {∵ a2 – b2 = (a + b) (a – b)
= {x-y)-(x + y)(x-y)
= (x-y)(1 – x – y)
Question 35.
16 (2x – 1)2 – 25y2
Solution:
16 (2x – 1)2 – 25y2
= [4 (2x – 1)]2 – (5y)2
= (8x – 4)2 – (5y)2
= (8x – 4 + 5y) (8x -4-5y)
= (8x + 5y – 4) (8x – 5y – 4)
Question 36.
4 (xy + 1)2 – 9 (x – 1)2
Solution:
4 (xy + 1)2 – 9 (x – 1)2
=
[2 (xy + 1)]2 – [3 (x – 1)]2
= (2xy + 2)2 – (3x – 3){∵ a2 – b2 = (a + b) (a – b)
= (2xy + 2 + 3x – 3) (2xy + 2 – 3x + 3)
= (2xy + 3x – 1) (2xy – 3x + 5)
Question 37.
(2x + 1)2 – 9x4
Solution:
(2x + 1)2 – 9x4 = (2x + 1)2 – (3x2)2 { a2 – b2 = (a + b) (a – b)
= (2x + 1 + 3x2) (2x + 1 – 3x2)
= (3x2 + 2x + 1) (-3x + 2x + 1)
Question 38.
x4 – (2y- 3z)2
Solution:
x4 – (2y – 3z)2 = (x2)2 – (2y – 3z)2
= (x2 + 2y- 3z) (x2 – 2y + 3z)
Question 39.
a2-b2 +a-b
Solution:
a2 – b2 + a – b
= (a + b) {a – b) + 1 (a – b)
= (a – b) (a + b + 1)
Question 40.
16a4 – b4
Solution:
16a4 – b4
= (4a2)2 – (b2)2 { a2 – b2 = (a + b) (a – b)
= (4a2 + b2) (4a2 – b2)
= (4a2 + b2) {(2a)2 – (b)2}
= (4a2 + b2) (2a + b) (2a – b)
Question 41.
a4 – 16 (b – c)4
Solution:
a4 – 16 (b- c)4 = (a2)2 – [4 (b – c)2]{ a2 – b2 = (a + b) (a – b)
= [a2 + 4 (b – c)2] [a2 – 4 (b – c)2]
= [a2 + 4 (b – c)2] [(a)2 – [2 (b – c)]2]
= [a2 + 4 (b – c)2] [(a)2 – (2b – 2c)2]
= [a2 + 4 (b – c)2] (a + 2b – 2c) (a – 2b + 2c)
Question 42.
2a5 – 32a
Solution:
2a5 – 32a = 2a (a4 – 16)
= 2a [(a2)2 – (4)2] {∵ a2 – b2 = (a + b) (a – b)
= 2a (a2 + 4) (a2 – 4)]
= 2a (a2 + 4) [(a)2 – (2)2]
= 2a (a2 + 4) (a + 2) (a – 2)
Question 43.
a4b4 – 81c4
Solution:
a4b4 – 81c4 = (a2b2)2 – (9c2)2
= (a2b2 + 9c2) (a2b2 – 9c2
{∵ a2 – b2 = (a + b) (a – b)
= (a2b2 + 9c2) {(ab)2 – (3c)2}
= (a2b2 + 9c2) (ab + 3c) (ab – 3c)
Question 44.
xy9-yx9
Solution:
xy9yx9 = xy (y8 – x8)
= xy [(y4)2 – (x4)2{∵ a2 – b2 = (a + b) (a – b)}
= xy(y4 + x4)(y4-x4)
= xy (y4 + x4) {(y2)2 – (x2)2}
= xy (y4 + x4) (y2 + x2) (y2 – x2)
= xy (y4 + x4) (y2 + x2) (y + x) (y – x)
Question 45.
x3 -x
Solution:
x3-x = x(x2 1)
= x [(x)2 – (1)2] = x (x + 1) (x – 1)
Question 46.
18a2x2 – 32
Solution:
18a2x2 – 32
= 2 [9a2x2 – 16]
= 2 [(3ax)2 – (4)2] { a2 – b2 = (a + b) (a – b)
= 2 (3ax + 4) (3ax – 4)
Hope given RD Sharma Class 8 Solutions Chapter 7 Factorizations Ex 7.5 are helpful to complete your math homework.
If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.
## RD Sharma Class 8 Solutions Chapter 8 Division of Algebraic Expressions Ex 8.5
These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 8 Division of Algebraic Expressions Ex 8.5
Other Exercises
Question 1.
Divide the first polynomial by the second polynomial in each of the following. Also, write the quotient and remainder.
(i) 3x2 + 4x + 5, x – 2
(ii) 10x2 – 7x + 8, 5x – 3
(iii) 5y3– 6y2 + 6y-1,5y-1
(iv)x4-x3 + 5x,x-1
(v) y4 +y2,y2-2
Solution:
(i) 3x2 + 4x + 5, x – 2
= 3x (x – 2) + 10x + 5
= 3x (x – 2) + 10 (x – 2) + 25
∴ Quotient = 3x + 10
Remainder = 25
(iii) 5y3 – 6y2 + 6y – 1, 5y – 1
= y(5y – 1) – 5y2 + 6y- 1
= y2 (5y – 1) -y (5y – 1) + 5y – 1
= y2 (5y- 1) -y (5y- 1) + 1 (5y- 1)
∴ Quotient = y2 – y + 1 and Remainder = 0
(iv) x4 – x3 + 5x, x – 1
= x3(x – 1) + 5x
= x3 (x – 1) + 5 (x – 1) + 5
∴ Quotient = x3 + 5, Remainder = 5
(v) y4+y2,y2– 2
= y2(y– 2) + 3y2
= y2 (y2 – 2) + 3 (y2 – 2) + 6
∴ Quotient =y2 + 3 and Remainder = 6
Question 2.
Find, whether or not the first polynomial is a factor of the second :
(i) x + 1, 2x2 + 5x + 4
(ii) y- 2, 3y3 + 5y2 + 5y + 2
(iii) 4x2 – 5, 4.x4 + 7x2 + 15
(iv) 4-z, 3z2 – 13z + 4
(v) 2a-3,10a2 – 9a – 5
(vi) 4y+1 ,8y2-2y + 1
Solution:
(i) x + 1, 2x2 + 5x + 4
2x2 + 5x + 4 = 2x (x + 1) + 3x + 4
= 2x (x + 1) + 3 (x + 1) + 1
∵ Remainder = 1
∴ x + 1 is not a factor of 2x2 + 5x + 4
(ii) y – 2, 3y3 + 5y2 + 5y + 2
3y3 + 5y2 + 5y + 2 = 3y2(y – 2)+11y2 + 5y + 2
= 3y2(y – 2)+11y (y – 2) + 27y + 2
= 3y2 (y – 2) + 11y (y – 2) + 27 (y – 2) + 56
∵ Remainder = 56
∴ y – 2 is not a factor of 3y3 + 5y2 + 5y + 2
(iii) 4x2 – 5, 4x4 + 7x2 + 15
4x4 + 7x2 + 15 = x2 (4x2 – 5) + 12x2 + 15
= x2 (4x2 – 5) + 3 (4x2 – 5) + 30
∵ Remainder = 30
∴ 4x2 – 5 is not a factor of 4x4 + 7x2 + 15
(iv) 4 – z, 3z2 – 13z + 4
3z2 – 13z + 4 = -3z (-z + 4) – z + 4
= -3z (-z + 4) + 1 (-z + 4)
∵ Remainder = 0
∴ 4 – z or – z + 4 is a factor of 3z2 – 13z + 4
(v) 2a – 3, 10a2 – 9a – 5
10a2 – 9a – 5 = 5a (2a – 3) + 6a – 5
= 5a (2a – 3) + 3 (2a – 3) + 4
∵ Remainder = 4
∴ 2a – 3 is not a factor of 10a2 – 9a – 5
(vi) 4y + 1, 8y2 – 2y + 1
8y2 – 2y + 1 = 2y (4y + 1) – 4y + 1
= 2y (4y + 1) – 1 (4y + 1) + 2
∵ Remainder = 2
∴ 4y + 1 is not a factor of 8y2 – 2y + 1
Hope given RD Sharma Class 8 Solutions Chapter 8 Division of Algebraic Expressions Ex 8.5 are helpful to complete your math homework.
If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you. |
# Unit 1: The Number System Review Material The unit 1 test will be
```Unit 1: The Number System
Review Material
The unit 1 test will be on the application of the following skills and concepts:
Rational Numbers
Rational numbers can be expressed as one integer divided by another. Rational numbers either terminate or
repeat.
Absolute Value
Absolute value is the distance from zero.
We need to use the absolute value when finding the distance between two points. This is because distance must
be positive, but differences can be positive or negative. Taking the absolute value makes the number positive.
Opposites
Opposites are numbers that are the same distance from zero. Their sum is always zero.
Comparing and Ordering Rational Numbers
Positive numbers are larger than negative numbers.
Positive numbers that are further from zero are larger.
Negative numbers that are further from zero are smaller.
Operations with Rational Numbers
Adding numbers with the same sign: add the absolute value of the numbers and keep the original sign.
Adding numbers with different signs: subtract the smaller absolute value from the larger absolute value. If the
positive number has the larger absolute value, the answer is positive. If the negative number has the larger
absolute value, the answer is negative.
Subtraction
Same-change-change
Keep the first number the same- change subtraction to addition- change the number to its opposite
Multiplication
Multiply the two numbers
If the numbers have the same sign, the answer is positive.
If the numbers have different signs, the answer is negative.
*Note: this applies to two numbers (-2)(-3)(-1) is negative.
(-2)(-3) is positive since a negative times a negative is positive. Now we have a positive times (-1) which is
negative since a positive times a negative is negative.
Divison
Divide the two numbers
If the numbers have the same sign, the answer is positive.
If the numbers have different signs, the answer is negative.
Dividing by zero:
You cannot divide by zero. Look at the example 13 ÷ 0.
We can rewrite this as multiplication: 0 x _____ = 13
But zero times anything is zero so there is no number that when multiplied by zero will equal 13.
Expressing fractions as decimals:
Use long division to express fractions as decimals
The decimals will terminate or repeat. We show repeating decimals by placing a bar above the repeating
number(s).
Fractions
A common denominator is needed to add and subtract fractions.
Once you have a common denominator, add or subtract the numerators. Put the answer over the denominator.
The denominator remains the same.
Multiplying
Multiply the numerators and multiply the denominators.
Put the product of the numerators over the product of the denominators
Dividing
Same-change-flip
Keep the first fraction the same- change division to multiplication- flip the second fraction
Then solve as multiplication
Mixed numbers
Turn mixed numbers into improper fractions to add, subtract, multiply or divide
Multiply the denominator by the whole number. Then add the numerator. Put the answer over the original
denominator.
Order of Operations
Parenthesis - Exponents - Multiplication and Division (left to right) - Addition and Subtraction (left to right)
Word Problems
Review your classwork and homework for examples
``` |
Tải bản đầy đủ - 0 (trang)
4: Integers: Multiplication and Division
# 4: Integers: Multiplication and Division
Tải bản đầy đủ - 0trang
22
Chapter 1 • Some Basic Concepts of Arithmetic and Algebra
Certainly, to continue this pattern, the product of Ϫ1 and Ϫ3 has to be 3. In general, this type of
reasoning helps us to realize that the product of any two negative integers is a positive integer.
Using the concept of absolute value, these three facts precisely describe the multiplication of integers:
1. The product of two positive integers or two negative integers is the product of their
absolute values.
2. The product of a positive and a negative integer (either order) is the opposite of the
product of their absolute values.
3. The product of zero and any integer is zero.
The following are examples of the multiplication of integers:
(Ϫ5)(Ϫ2) ϭ 0 Ϫ5 0 и 0 Ϫ2 0 ϭ 5 и 2 ϭ 10
(7)(Ϫ6) ϭ Ϫ( 0 7 0 и 0 Ϫ6 0 ) ϭ Ϫ(7 и 6) ϭ Ϫ42
(Ϫ8)(9) ϭ Ϫ(0 Ϫ8 0 и 0 9 0 ) ϭ Ϫ(8 и 9) ϭ Ϫ72
(Ϫ14)(0) ϭ 0
(0)(Ϫ28) ϭ 0
These examples show a step-by-step process for multiplying integers. In reality, however,
the key issue is to remember whether the product is positive or negative. In other words, we
need to remember that the product of two positive integers or two negative integers is a
positive integer; and the product of a positive integer and a negative integer (in either
order) is a negative integer. Then we can avoid the step-by-step analysis and simply write
the results as follows:
(7)(Ϫ9) ϭ Ϫ63
(8)(7) ϭ 56
(Ϫ5)(Ϫ6) ϭ 30
(Ϫ4)(12) ϭ Ϫ48
Division of Integers
By looking back at our knowledge of whole numbers, we can get some guidance for our work
8
with integers. We know, for example, that ϭ 4, because 2 и 4 ϭ 8. In other words, we can
2
find the quotient of two whole numbers by looking at a related multiplication problem. In
the following examples we use this same link between multiplication and division to determine the quotients.
8
ϭ Ϫ4 because (Ϫ2)(Ϫ4) ϭ 8
Ϫ2
Ϫ10
ϭ Ϫ2 because (5)(Ϫ2) ϭ Ϫ10
5
Ϫ12
ϭ 3 because (Ϫ4)(3) ϭ Ϫ12
Ϫ4
0
ϭ 0 because (Ϫ6)(0) ϭ 0
Ϫ6
Ϫ9
0
0
0
is undefined because no number times 0 produces Ϫ9
is indeterminate because any number times 0 equals 0
1.4 • Integers: Multiplication and Division
23
The following three statements precisely describe the division of integers:
1. The quotient of two positive or two negative integers is the quotient of their absolute
values.
2. The quotient of a positive integer and a negative integer (or a negative and a positive) is
the opposite of the quotient of their absolute values.
3. The quotient of zero and any nonzero number (zero divided by any nonzero number)
is zero.
The following are examples of the division of integers:
0 Ϫ8 0
Ϫ8
8
ϭ
ϭ ϭ2
Ϫ4
0 Ϫ4 0
4
0 Ϫ14 0
Ϫ14
14
ϭ Ϫa
b ϭ Ϫa b ϭ Ϫ7
2
02 0
2
0 15 0
15
15
ϭ Ϫa
b ϭ Ϫa b ϭ Ϫ5
Ϫ3
0 Ϫ3 0
3
0
ϭ0
Ϫ4
For practical purposes, when dividing integers the key is to remember whether the quotient is positive or negative. Remember that the quotient of two positive integers or two
negative integers is positive; and the quotient of a positive integer and a negative integer or a negative integer and a positive integer is negative. We can then simply write the
quotients as follows without showing all of the steps:
Ϫ18
ϭ3
Ϫ6
Ϫ24
ϭ Ϫ2
12
36
ϭ Ϫ4
Ϫ9
Remark: Occasionally, people use the phrase “two negatives make a positive.” We hope they
realize that the reference is to multiplication and division only; in addition the sum of two negative integers is still a negative integer. It is probably best to avoid such imprecise statements.
Simplifying Numerical Expressions
Now we can simplify numerical expressions involving any or all of the four basic operations
with integers. Keep in mind the order of operations given in Section 1.1.
Classroom Example
Simplify 4(Ϫ5) Ϫ 3(Ϫ6) Ϫ 7(4).
EXAMPLE 1
Simplify Ϫ4(Ϫ3) Ϫ 7(Ϫ8) ϩ 3(Ϫ9) .
Solution
Ϫ4(Ϫ3) Ϫ 7(Ϫ8) ϩ 3(Ϫ9) ϭ 12 Ϫ (Ϫ56) ϩ (Ϫ27)
ϭ 12 ϩ 56 ϩ (Ϫ27)
ϭ 41
Classroom Example
Simplify
Ϫ3 ϩ 6(Ϫ7)
.
Ϫ5
EXAMPLE 2
Simplify
Solution
Ϫ8 Ϫ 4152
Ϫ4
ϭ
Ϫ8 Ϫ 20
Ϫ4
ϭ
Ϫ28
Ϫ4
ϭ7
Ϫ8 Ϫ 4152
Ϫ4
.
24
Chapter 1 • Some Basic Concepts of Arithmetic and Algebra
Evaluating Algebraic Expressions
Evaluating algebraic expressions will often involve the use of two or more operations with
integers. We use the final examples of this section to represent such situations.
Classroom Example
Find the value of 5m ϩ 4n, when
m ϭ 3 and n ϭ Ϫ7.
EXAMPLE 3
Find the value of 3x ϩ 2y when x ϭ 5 and y ϭ Ϫ9.
Solution
3x ϩ 2y ϭ 3(5) ϩ 2(Ϫ9) when x ϭ 5 and y ϭ Ϫ9
ϭ 15 ϩ (Ϫ18)
ϭ Ϫ3
Classroom Example
Evaluate Ϫ3x ϩ 11y for x ϭ 5 and
y ϭ Ϫ2.
EXAMPLE 4
Evaluate Ϫ2a ϩ 9b for a ϭ 4 and b ϭ Ϫ3.
Solution
Ϫ2a ϩ 9b ϭ Ϫ2(4) ϩ 9(Ϫ3) when a ϭ 4 and b ϭ Ϫ3
ϭ Ϫ8 ϩ (Ϫ27)
ϭ Ϫ35
Classroom Example
3a Ϫ 7b
Find the value of
, when
5
a ϭ Ϫ2 and b ϭ Ϫ3.
EXAMPLE 5
Find the value of
x Ϫ 2y
when x ϭ Ϫ6 and y ϭ 5.
4
Solution
Ϫ6 Ϫ 2(5)
x Ϫ 2y
ϭ
when x ϭ Ϫ6 and y ϭ 5
4
4
Ϫ6 Ϫ 10
4
Ϫ16
ϭ
4
ϭ Ϫ4
ϭ
Concept Quiz 1.4
For Problems 1–10, answer true or false.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
The product of two negative integers is a positive integer.
The product of a positive integer and a negative integer is a positive integer.
When multiplying three negative integers the product is negative.
The rules for adding integers and the rules for multiplying integers are the same.
The quotient of two negative integers is negative.
The quotient of a positive integer and zero is a positive integer.
The quotient of a negative integer and zero is zero.
The product of zero and any integer is zero.
The value of Ϫ3x Ϫy when x ϭ Ϫ4 and y ϭ 6 is 6.
The value of 2x Ϫ5y Ϫxy when x ϭ Ϫ2 and y ϭ Ϫ7 is 17.
1.4 • Integers: Multiplication and Division
25
Problem Set 1.4
For Problems 1– 40, find the product or quotient (multiply or
divide) as indicated. (Objective 1)
1. 5(Ϫ6)
48. 7(Ϫ4) Ϫ 8(Ϫ7) ϩ 5(Ϫ8)
2. 7(Ϫ9)
49.
13 ϩ (Ϫ25)
Ϫ3
50.
15 ϩ (Ϫ36)
Ϫ7
3.
Ϫ27
3
4.
Ϫ35
5
51.
12 Ϫ 48
6
52.
16 Ϫ 40
8
5.
Ϫ42
Ϫ6
6.
Ϫ72
Ϫ8
53.
Ϫ7(10) ϩ 6(Ϫ9)
Ϫ4
54.
Ϫ6(8) ϩ 4(Ϫ14)
Ϫ8
55.
4(Ϫ7) Ϫ 8(Ϫ9)
11
56.
5(Ϫ9) Ϫ 6(Ϫ7)
3
7. (Ϫ7) (8)
9. (Ϫ5) (Ϫ12)
11.
96
Ϫ8
8. (Ϫ6)(9)
10. (Ϫ7)(Ϫ14)
12.
Ϫ91
7
13. 14(Ϫ9)
14. 17(Ϫ7)
15. (Ϫ11) (Ϫ14)
16. (Ϫ13)(Ϫ17)
135
17.
Ϫ15
Ϫ144
18.
12
19.
Ϫ121
Ϫ11
20.
Ϫ169
Ϫ13
57. Ϫ2(3) Ϫ 3(Ϫ4) ϩ 4(Ϫ5) Ϫ 6(Ϫ7)
58. 2(Ϫ4) ϩ 4(Ϫ5) Ϫ 7(Ϫ6) Ϫ 3(9)
59. Ϫ1(Ϫ6) Ϫ 4 ϩ 6(Ϫ2) Ϫ 7(Ϫ3) Ϫ 18
60. Ϫ9(Ϫ2) ϩ 16 Ϫ 4(Ϫ7) Ϫ 12 ϩ 3(Ϫ8)
For Problems 61–76, evaluate each algebraic expression for
the given values of the variables. (Objective 2)
61. 7x ϩ 5y for x ϭ Ϫ5 and y ϭ 9
21. (Ϫ15) (Ϫ15)
22. (Ϫ18)(Ϫ18)
62. 4a ϩ 6b for a ϭ Ϫ6 and b ϭ Ϫ8
112
23.
Ϫ8
112
24.
Ϫ7
64. 8a Ϫ 3b for a ϭ Ϫ7 and b ϭ 9
63. 9a Ϫ 2b for a ϭ Ϫ5 and b ϭ 7
65. Ϫ6x Ϫ 7y for x ϭ Ϫ4 and y ϭ Ϫ6
25.
0
Ϫ8
26.
Ϫ8
0
27.
Ϫ138
Ϫ6
28.
Ϫ105
Ϫ5
67.
5x Ϫ 3y
for x ϭ Ϫ6 and y ϭ 4
Ϫ6
29.
76
Ϫ4
30.
Ϫ114
6
68.
Ϫ7x ϩ 4y
for x ϭ 8 and y ϭ 6
Ϫ8
31. (Ϫ6) (Ϫ15)
32.
0
Ϫ14
33. (Ϫ56) Ϭ (Ϫ4)
34. (Ϫ78) Ϭ (Ϫ6)
71. Ϫ2x ϩ 6y Ϫ xy for x ϭ 7 and y ϭ Ϫ7
35. (Ϫ19) Ϭ 0
36. (Ϫ90) Ϭ 15
72. Ϫ3x ϩ 7y Ϫ 2xy for x ϭ Ϫ6 and y ϭ 4
37. (Ϫ72) Ϭ 18
38. (Ϫ70) Ϭ 5
73. Ϫ4ab Ϫ b for a ϭ 2 and b ϭ Ϫ14
39. (Ϫ36) (27)
40. (42)(Ϫ29)
74. Ϫ5ab ϩ b for a ϭ Ϫ1 and b ϭ Ϫ13
For Problems 41– 60, simplify each numerical expression.
(Objective 2)
41. 3(Ϫ4) ϩ 5(Ϫ7)
42. 6(Ϫ3) ϩ 5(Ϫ9)
43. 7(Ϫ2) Ϫ 4(Ϫ8)
44. 9(Ϫ3) Ϫ 8(Ϫ6)
66. Ϫ5x Ϫ 12y for x ϭ Ϫ5 and y ϭ Ϫ7
69. 3(2a Ϫ 5b) for a ϭ Ϫ1 and b ϭ Ϫ5
70. 4(3a Ϫ 7b) for a ϭ Ϫ2 and b ϭ Ϫ4
75. (ab ϩ c)(b Ϫ c) for a ϭ Ϫ2, b ϭ Ϫ3, and c ϭ 4
76. (ab Ϫ c)(a ϩ c) for a ϭ Ϫ3, b ϭ 2, and c ϭ 5
For Problems 77–82, find the value of
the given values for F. (Objective 2)
5(F Ϫ 32)
for each of
9
45. (Ϫ3) (Ϫ8) ϩ (Ϫ9)(Ϫ5)
77. F ϭ 59
78. F ϭ 68
46. (Ϫ7) (Ϫ6) ϩ (Ϫ4)(Ϫ3)
79. F ϭ 14
80. F ϭ Ϫ4
47. 5(Ϫ6) Ϫ 4(Ϫ7) ϩ 3(2)
81. F ϭ Ϫ13
82. F ϭ Ϫ22
26
Chapter 1 • Some Basic Concepts of Arithmetic and Algebra
9C
For Problems 83–88, find the value of
ϩ 32 for each of
5
the given values for C. (Objective 2)
83. C ϭ 25
84. C ϭ 35
90. In one week a small company showed a profit of \$475 for
one day and a loss of \$65 for each of the other four days.
Use multiplication and addition of integers to describe
this situation and to determine the company’s profit or
loss for the week.
91. At 6 P.M. the temperature was 5°F. For the next four
hours the temperature dropped 3° per hour. Use multiplication and addition of integers to describe this situation
and to find the temperature at 10 P.M.
85. C ϭ 40
86. C ϭ 0
87. C ϭ Ϫ10
88. C ϭ Ϫ30
For Problems 89–92, solve the problem by applying the
concepts of adding and multiplying integers. (Objective 3)
89. On Monday morning, Thad bought 800 shares of a stock
at \$19 per share. During that week, the stock went up \$2
per share on one day and dropped \$1 per share on each
of the other four days. Use multiplication and addition of
integers to describe this situation and to determine the
value of the 800 shares by closing time on Friday.
92. For each of the first three days of a golf tournament,
Jason shot 2 strokes under par. Then for each of the last
two days of the tournament he shot 4 strokes over par.
Use multiplication and addition of integers to describe
this situation and to determine how Jason shot relative to
par for the five-day tournament.
41–60.
Thoughts Into Words
94. Your friend keeps getting an answer of 27 when simplifying the expression Ϫ6 ϩ (Ϫ8) Ϭ 2. What mistake is
she making and how would you help her?
95. Make up a problem that could be solved using
6(24) ϭ 224.
1. True 2. False 3. True 4. False
1.5
5. False
96. Make up a problem that could be solved using
(Ϫ4)(Ϫ3) ϭ 12.
4
0
97. Explain why ϭ 0 but is undefined.
4
0
6. False
7. False
8. True
9. True
10. True
Use of Properties
OBJECTIVES
1
Recognize the properties of integers
2
Apply the properties of integers to simplify numerical expressions
3
Simplify algebraic expressions
We will begin this section by listing and briefly commenting on some of the basic properties
of integers. We will then show how these properties facilitate manipulation with integers and
also serve as a basis for some algebraic computation.
1.5 • Use of Properties
27
If a and b are integers, then
aϩbϭbϩa
Commutative Property of Multiplication
If a and b are integers, then
ab ϭ ba
Addition and multiplication are said to be commutative operations. This means that the order
in which you add or multiply two integers does not affect the result. For example,
3 ϩ 5 ϭ 5 ϩ 3 and 7(8) ϭ 8(7) . It is also important to realize that subtraction and division
are not commutative operations; order does make a difference. For example, 8 Ϫ 7 ϶ 7 Ϫ 8
and 16 Ϭ 4 ϶ 4 Ϭ 16.
If a, b, and c are integers, then
(a ϩ b) ϩ c ϭ a ϩ (b ϩ c)
Associative Property of Multiplication
If a, b, and c are integers, then
(ab)c ϭ a(bc)
Our arithmetic operations are binary operations. We only operate (add, subtract, multiply, or
divide) on two numbers at a time. Therefore, when we need to operate on three or more numbers, the numbers must be grouped. The associative properties can be thought of as grouping properties. For a sum of three numbers, changing the grouping of the numbers does not
affect the final result. For example, (Ϫ8 ϩ 3) ϩ 9 ϭ Ϫ8 ϩ (3 ϩ 9). This is also true for
multiplication as [ (Ϫ6)(5) ] (Ϫ4) ϭ (Ϫ6) [ (5)(Ϫ4) ] illustrates. Addition and multiplication are associative operations. Subtraction and division are not associative operations. For
example, (8 Ϫ 4) Ϫ 7 ϭ Ϫ3, whereas 8 Ϫ (4 Ϫ 7) ϭ 11 shows that subtraction is not an
associative operation. Also, (8 Ϭ 4) Ϭ 2 ϭ 1, whereas 8 Ϭ (4 Ϭ 2) ϭ 4 shows that division is not associative.
If a is an integer, then
aϩ0ϭ0ϩaϭa
We refer to zero as the identity element for addition. This simply means that the
sum of any integer and zero is exactly the same integer. For example, Ϫ197 ϩ 0 ϭ
0 ϩ (Ϫ197) ϭ Ϫ197.
28
Chapter 1 • Some Basic Concepts of Arithmetic and Algebra
Identity Property of Multiplication
If a is an integer, then
a(1) ϭ 1(a) ϭ a
We call one the identity element for multiplication. The product of any integer and one is
exactly the same integer. For example, (Ϫ573)(1) ϭ (1)(Ϫ573)ϭ Ϫ573.
For every integer a, there exists an integer Ϫa such that
a ϩ (Ϫa) ϭ (Ϫa) ϩ a ϭ 0
The integer Ϫa is called the additive inverse of a or the opposite of a. Thus 6 and Ϫ6 are
additive inverses, and their sum is 0. The additive inverse of 0 is 0.
Multiplication Property of Zero
If a is an integer, then
(a)(0) ϭ (0)(a) ϭ 0
The product of zero and any integer is zero. For example, 1Ϫ8732102 ϭ 1021Ϫ8732 ϭ 0.
Multiplication Property of Negative One
If a is an integer, then
(a)(Ϫ1) ϭ (Ϫ1)(a) ϭ Ϫa
The product of any integer and Ϫ1 is the opposite of the integer. For example,
(Ϫ1)(48) ϭ (48)(Ϫ1) ϭ Ϫ48.
Distributive Property
If a, b, and c are integers, then
a(b ϩ c) ϭ ab ϩ ac
The distributive property involves both addition and multiplication. We say that multiplication
distributes over addition. For example, 3(4 ϩ 7) ϭ 3(4) ϩ 3(7) . Since b Ϫ c ϭ b ϩ (Ϫc),
it follows that multiplication also distributes over subtraction. This could be stated as
a(b Ϫ c) ϭ ab Ϫ ac. For example, 7(8 Ϫ 2) ϭ 7(8) Ϫ 7(2).
Let’s now consider some examples that use these properties to help with various types of
manipulations.
1.5 • Use of Properties
Classroom Example
Find the sum (37 ϩ (Ϫ18)) ϩ 18.
EXAMPLE 1
29
Find the sum (43 ϩ (Ϫ24)) ϩ 24.
Solution
In this problem it is much more advantageous to group Ϫ24 and 24. Thus
(43 ϩ (Ϫ24)) ϩ 24 ϭ 43 ϩ ((Ϫ24) ϩ 24)
ϭ 43 ϩ 0
ϭ 43
Classroom Example
Find the product [(Ϫ26)(5)](20).
EXAMPLE 2
Find the product [(Ϫ17)(25)](4) .
Solution
In this problem it is easier to group 25 and 4. Thus
[(Ϫ17)(25)](4) ϭ (Ϫ17)[(25) (4)]
ϭ (Ϫ17)(100)
ϭ Ϫ1700
Classroom Example
Find the sum
(Ϫ32) ϩ 11 ϩ (Ϫ15) ϩ 16 ϩ 27 ϩ
(Ϫ23) ϩ 52.
EXAMPLE 3
Associative property for multiplication
Find the sum 17 ϩ (Ϫ24) ϩ (Ϫ31) ϩ 19 ϩ (Ϫ14) ϩ 29 ϩ 43.
Solution
Certainly we could add in the order that the numbers appear. However, since addition is commutative and associative, we could change the order and group any convenient way. For
example, we could add all of the positive integers and add all of the negative integers, and
then add these two results. In that case it is convenient to use the vertical format as follows.
17
19
29
43
108
Ϫ24
Ϫ31
Ϫ14
Ϫ69
108
Ϫ69
39
For a problem such as Example 3 it might be advisable to first add in the order that the
numbers appear, and then use the rearranging and regrouping idea as a check. Don’t forget the
link between addition and subtraction. A problem such as 18 Ϫ 43 ϩ 52 Ϫ 17 Ϫ 23 can be
changed to 18 ϩ (Ϫ43) ϩ 52 ϩ (Ϫ17) ϩ (Ϫ23) .
Classroom Example
Simplify (Ϫ25)(Ϫ3 ϩ 20).
EXAMPLE 4
Simplify 1Ϫ7521Ϫ4 ϩ 1002 .
Solution
For such a problem, it is convenient to apply the distributive property and then to simplify.
1Ϫ7521Ϫ4 ϩ 1002 ϭ 1Ϫ7521Ϫ42 ϩ 1Ϫ75211002
ϭ 300 ϩ 1Ϫ75002
ϭ Ϫ7200
30
Chapter 1 • Some Basic Concepts of Arithmetic and Algebra
Classroom Example
Simplify 24 (Ϫ16 ϩ 18).
EXAMPLE 5
Simplify 19(Ϫ26 ϩ 25) .
Solution
For this problem we are better off not applying the distributive property, but simply adding
the numbers inside the parentheses first and then finding the indicated product. Thus
19(Ϫ26 ϩ 25) ϭ 19(Ϫ1) ϭ Ϫ19
Classroom Example
Simplify 33(6) ϩ 33(Ϫ106).
EXAMPLE 6
Simplify 27(104) ϩ 27(Ϫ4) .
Solution
Keep in mind that the distributive property allows us to change from the form a(b ϩ c) to
ab ϩ ac or from ab ϩ ac to a(b ϩ c) . In this problem we want to use the latter conversion.
Thus
27(104) ϩ 27(Ϫ4) ϭ 27(104 ϩ (Ϫ4) )
ϭ 27(100) ϭ 2700
Examples 4, 5, and 6 demonstrate an important issue. Sometimes the form a(b ϩ c) is
the most convenient, but at other times the form ab ϩ ac is better. A suggestion in regard to
this issue—as well as to the use of the other properties—is to think first, and then decide
whether or not the properties can be used to make the manipulations easier.
Combining Similar Terms
Algebraic expressions such as
3x 5y 7xy Ϫ4abc and z
are called terms. A term is an indicated product, and it may have any number of factors. We
call the variables in a term literal factors, and we call the numerical factor the numerical
coefficient. Thus in 7xy, the x and y are literal factors, and 7 is the numerical coefficient. The
numerical coefficient of the term Ϫ4abc is Ϫ4. Since z ϭ 1(z), the numerical coefficient of the
term z is 1. Terms that have the same literal factors are called like terms or similar terms.
Some examples of similar terms are
3x and 9x
14abc and 29abc
7xy and Ϫ15xy
4z, 9z, and Ϫ14z
We can simplify algebraic expressions that contain similar terms by using a form of the distributive property. Consider the following examples.
3x ϩ 5x ϭ (3 ϩ 5)x
ϭ 8x
Ϫ9xy ϩ 7xy ϭ (Ϫ9 ϩ 7)xy
ϭ Ϫ2xy
18abc Ϫ 27abc ϭ (18 Ϫ 27)abc
ϭ (18 ϩ (Ϫ27) )abc
ϭ Ϫ9abc
4x ϩ x ϭ (4 ϩ 1)x
ϭ 5x
Don’t forget that x ϭ 1(x)
1.5 • Use of Properties
31
More complicated expressions might first require some rearranging of terms by using the
commutative property:
7x ϩ 3y ϩ 9x ϩ 5y ϭ 7x ϩ 9x ϩ 3y ϩ 5y
ϭ (7 ϩ 9)x ϩ (3 ϩ 5)y
ϭ 16x ϩ 8y
9a Ϫ 4 Ϫ 13a ϩ 6 ϭ 9a ϩ (Ϫ4) ϩ (Ϫ13a) ϩ 6
ϭ 9a ϩ (Ϫ13a) ϩ (Ϫ4) ϩ 6
ϭ (9 ϩ (Ϫ13)) a ϩ 2
ϭ Ϫ4a ϩ 2
As you become more adept at handling the various simplifying steps, you may want to do the
steps mentally and go directly from the given expression to the simplified form as follows.
19x Ϫ 14y ϩ 12x ϩ 16y ϭ 31x ϩ 2y
17ab ϩ 13c Ϫ 19ab Ϫ 30c ϭ Ϫ2ab Ϫ 17c
9x ϩ 5 Ϫ 11x ϩ 4 ϩ x Ϫ 6 ϭ Ϫx ϩ 3
Simplifying some algebraic expressions requires repeated applications of the distributive
property as the next examples demonstrate.
5(x Ϫ 2) ϩ 3(x ϩ 4) ϭ 5(x) Ϫ 5(2) ϩ 3(x) ϩ 3(4)
ϭ 5x Ϫ 10 ϩ 3x ϩ 12
ϭ 5x ϩ 3x Ϫ 10 ϩ 12
ϭ 8x ϩ 2
Ϫ7(y ϩ 1) Ϫ 4(y Ϫ 3) ϭ Ϫ7(y) Ϫ 7(1) Ϫ 4(y) Ϫ 4(Ϫ3)
ϭ Ϫ7y Ϫ 7 Ϫ 4y ϩ 12
Be careful with this sign
ϭ Ϫ7y Ϫ 4y Ϫ 7 ϩ 12
ϭ Ϫ11y ϩ 5
5(x ϩ 2) Ϫ (x ϩ 3) ϭ 5(x ϩ 2) Ϫ 1(x ϩ 3)
Remember Ϫa ϭ Ϫ1a
ϭ 5(x) ϩ 5(2) Ϫ 1(x) Ϫ 1(3)
ϭ 5x ϩ 10 Ϫ x Ϫ 3
ϭ 5x Ϫ x ϩ 10 Ϫ 3
ϭ 4x ϩ 7
After you are sure of each step, you can use a more simplified format.
5(a ϩ 4) Ϫ 7(a Ϫ 2) ϭ 5a ϩ 20 Ϫ 7a ϩ 14
ϭ Ϫ2a ϩ 34
9(z Ϫ 7) ϩ 11(z ϩ 6) ϭ 9z Ϫ 63 ϩ 11z ϩ 66
ϭ 20z ϩ 3
Ϫ(x Ϫ 2) ϩ (x ϩ 6) ϭ Ϫx ϩ 2 ϩ x ϩ 6
ϭ8
Back to Evaluating Algebraic Expressions
To simplify by combining similar terms aids in the process of evaluating some algebraic
expressions. The last examples of this section illustrate this idea.
32
Chapter 1 • Some Basic Concepts of Arithmetic and Algebra
Classroom Example
Evaluate 9s Ϫ 4t ϩ 3s ϩ 7t for s ϭ 2
and t ϭ Ϫ5.
EXAMPLE 7
Evaluate 8x Ϫ 2y ϩ 3x ϩ 5y for x ϭ 3 and y ϭ Ϫ4.
Solution
Let’s first simplify the given expression.
8x Ϫ 2y ϩ 3x ϩ 5y ϭ 11x ϩ 3y
Now we can evaluate for x ϭ 3 and y ϭ Ϫ4.
11x ϩ 3y ϭ 11(3) ϩ 3(Ϫ4)
ϭ 33 ϩ (Ϫ12) ϭ 21
Classroom Example
Evaluate 6x ϩ 3yz Ϫ 4x Ϫ 7yz for
x ϭ 4, y ϭ Ϫ6, and z ϭ 3.
EXAMPLE 8
Evaluate 2ab ϩ 5c Ϫ 6ab ϩ 12c for a ϭ 2, b ϭ Ϫ3, and c ϭ 7.
Solution
2ab ϩ 5c Ϫ 6ab ϩ 12c ϭ Ϫ4ab ϩ 17c
ϭ Ϫ4(2) (Ϫ3) ϩ 17(7) when a ϭ 2, b ϭ Ϫ3, and c ϭ 7
ϭ 24 ϩ 119 ϭ 143
Classroom Example
Evaluate 4(m ϩ 5) Ϫ 9(m Ϫ 3) for
m ϭ 7.
EXAMPLE 9
Evaluate 8(x Ϫ 4) ϩ 7(x ϩ 3) for x ϭ 6.
Solution
8(x Ϫ 4) ϩ 7(x ϩ 3) ϭ 8x Ϫ 32 ϩ 7x ϩ 21
ϭ 15x Ϫ 11
ϭ 15(6) Ϫ 11 when x ϭ 6
ϭ 79
Concept Quiz 1.5
For Problems 1–10, answer true or false.
1. Addition is a commutative operation.
2. Subtraction is a commutative operation.
3. [(2)(Ϫ3)](7) ϭ (2)[(Ϫ3)(7)] is an example of the associative property for
multiplication.
4. [(8)(5)](Ϫ2) ϭ (Ϫ2)[(8)(5)] is an example of the associative property for
multiplication.
5. Zero is the identity element for addition.
6. The integer Ϫa is the additive inverse of a.
7. The additive inverse of 0 is 0.
8. The numerical coefficient of the term Ϫ8xy is 8.
9. The numerical coefficient of the term ab is 1.
10. 6xy and Ϫ2xyz are similar terms.
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# WHAT TYPE OF QUADRILATERAL IS FORMED BY CONNECTING THE POINTS
What Type of Quadrilateral is Formed by Connecting the Points ?
In this section, we will see some examples to examine which type of quadrilateral formed by connecting the given points.
## Examining the Type of Quadrilateral is Formed by Connecting the Points
Example 1 :
Name the type of quadrilateral formed. If any, by the following points and give reasons for your answer:
(4,5) (7,6) (4,3) (1,2)
Solution :
Let the given points as A(4, 5) B(7, 6) C(4, 3) and D (1, 2)
Distance between two points = √(x2 - x1)2 + (y2 - y1)2
Length of the side AB :
Here, x1 = 4, y1 = 5, x2 = 7 and y2 = 6
= √(7-4)2 + (6-5)2
= √3² + 1²
= √9 + 1
= √10
Length of the side BC :
Here, x1 = 7, y1 = 6, x2 = 4 and y2 = 3
= √(4-7)² + (3-6)²
= √(-3)² + (-3)²
= √9 + 9
= √18
Length of the side CD :
Here, x1 = 4, y1 = 3, x2 = 1 and y2 = 2
= √(1-4)2 + (2-3)2
= √(-3)2 + (-1)2
= √9 + 1
= √10
Length of the side DA :
Here, x1 = 1, y1 = 2, x2 = 4 and y2 = 5
= √(4-1)2 + (5-2)2
= √32 + 32
= √9 + 9
= √18
AB = CD, BC = DA. length of opposite sides are equal.
Length of AC :
Here, x1 = 4, y1 = 5, x2 = 4 and y2 = 3
= √(4-4)² + (3-5)²
= √0² + (-2)²
= √4
= 2
Length of BD :
Here, x1 = 7, y1 = 6, x2 = 1 and y2 = 2
= √(1-7)2 + (2-6)2
= √(-6)2 + (-4)2
= √36 + 16
= √52
It can be observed that opposite sides of this quadrilateral are of the same length.
Since the diagonals are of different lengths, the given points are the vertices of the parallelogram.
Example 2 :
Find the points on the x-axis which is equidistant from (2, -5) and (-2, 9)
Solution :
We have to find the point on x-axis. So its y-coordinate will be 0.
Let the point on x axis be (x, 0)
Distance between (x, 0) and (2, -5) =
Distance between (x, 0) and (-2, 9)
Distance between two points
= √(x2 - x1)2 + (y2 - y1)2
Here, x1 = x, y = 0, x2 = 2 and y2 = -5
= √(2 - x)2 + (5 - 0)2
= √(2 - x)2 + 25 ---------------(1)
Here, x1 = x, y = 0, x2 = -2 and y2 = 9
= √(-2-x)2 + (9-0)2
= √(2+x)2 + 92
= (2+x)2 + 81
√(2 - x)2 + 25 (2 + x)2 + 81
Taking squares on both sides, we get
4 + x2 - 4 x + 25 = 4 + x2 + 4 x + 81
x² - x² - 4 x - 4 x + 4 - 4 = 81 - 25
-8 x = 56
x = -7
So, the required point is (-7, 0)
After having gone through the stuff given above, we hope that the students would have understood, what type of quadrilateral is formed by connecting the points.
Apart from the stuff given in this section, if you need any other stuff in math, please use our google custom search here.
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### Transcript
Example 1 The cost of 5 metres of a particular quality of cloth is Rs 210. Tabulate the cost of 2, 4, 10 and 13 metres of cloth of the same type. Given cost of 5 metres of cloth = ₹ 210 We need to find cost of 2, 4, 10, 13 meters of cloth Let’s draw a table Finding cost of 2 meters of cloth Let the Cost of cloth = Rs y As quantity of cloth increases, its cost also increases. ∴ Quantity and cost are in direct proportion. 5/210 = 2/𝑦 𝑦 × 5 = 2 × 210 𝑦 × 5 = 420 𝑦 = 420/5 = 84 ∴ Cost of 2 metres of cloth = Rs 84 Finding cost of 4 meters of cloth Let the Cost of cloth = Rs y Since Quantity and cost are in direct proportion. 5/210 = 4/𝑦 𝑦 × 5 = 4 × 210 𝑦 × 5 = 840 𝑦 = 840/5 𝑦 = 168 ∴ Cost of 4 metres of cloth = Rs 168 Finding cost of 10 meters of cloth Let the Cost of cloth = Rs y Since Quantity and cost are in direct proportion. 5/210 = 10/𝑦 𝑦 × 5 = 10 × 210 𝑦 × 5 = 2100 𝑦 = 2100/5 𝑦 = 420 ∴ Cost of 10 metres of cloth = Rs 420 Finding cost of 13 meters of cloth Let the Cost of cloth = Rs y Since Quantity and cost are in direct proportion. 5/210 = 13/𝑦 𝑦 × 5 = 13 × 210 𝑦 = (13 × 210)/5 𝑦 = 13 × 42 y = 546 ∴ Cost of 13 metres of cloth = Rs 546 Thus, our table looks like |
# Prove/Disprove: if $x^2 = a^2$, then $x = a$
From Prof. Charles Pinter's "A Book of Abstract Algebra"'s Chapter 4 exercises:
For each of the following rules, either prove that it is true in every group $$G$$, or give a counter-example.
$$\text{if } x^{2}=a^{2}, \text{then } x=a$$
I believe that this is true by:
$$xx = aa$$
by cancellation,
$$x=a$$
Is that right? Also, when the word "prove" is used, does that mean to use theorems to prove?
• You can't use cancellation like that. You need to cancel the same thing on either side. In any case, this is not true. (Hint: $(-1)^2 = 1$) – Prahlad Vaidyanathan May 13 '15 at 15:15
• Disproof can be by counterexample, for example: $x=-1 , a=1$ – Joffan May 13 '15 at 15:25
What is it that you are canceling there? It looks to me like you used that $x=a$ to show that $x=a$, which is not a good plan.
Think: is this true even in the real numbers?
In standard real numbers:
$$x^2 = a^2 \implies x^2-a^2=0$$ We can then factor this polynomial as $$(x-a)(x+a)=0$$ Thus $x=a$ or $x=-a$.
Thus in the group of real numbers $\mathbb{R}\setminus\{0\}$ with multiplication we see that this claim is not true.
In particular, $x^2=4=2^2$ has two solutions: $x=\pm 2$.
Look for a group with two distinct elements $a, b$ of order $2$.
Then $a^2 = b^2 = e$, but $a \neq b$.
For example, in the symmetric group $S_4$, we have $a = (1, 2), \;\;b = (1,3)(2,4)$. Each, when composed with itself yields $a^2 = b^2 = e$.
More immediately, in the Klein-4 group (order $4$) $\{e, a, b, c\}$, we have $a^2 = b^2 = e$, but $a\neq b$
Consider the group of the symmetries in the plane. There are infinitely many of these symmetries.
Take any of them $s$. We have that $s^2=i^2$ ($i$ is the identity). "Therefore", $s=i$. That is, every symmetry is the identity.
Noting all the other proofs that it is not a general result, your argument holds for odd cyclic groups (eg. $C_3, C_5$), where every square is distinct.
Just think $(R,*)$ as a counter-example. $(R,*)$ is a Group. then $2^2={(-2)}^2$, however $2 \neq -2$.
If you press twice the swith of the lamp in your bedroom, you get the same as if you do the same in the kitchen: nothing. (Ignore the cost of electricity).
But that does not mean that the lamp of your room and the lamp of your kitchen are the same lamp!
Cancellation is not possible, you must take the root of both sides. Square roots have the property $q = \pm\sqrt{q^2}$, therefore $a = \pm x$, which is not equivalent to $a = x$. The original statement has thus been disproved.
The free product $$\Bbb Z_2\ast\Bbb Z_2$$ of $$\Bbb Z_2$$ with itself has as a presentation $$\langle a, x\mid a^2=x^2=e\rangle,$$ from which it is clear that, if $$x=a$$, then the presentation becomes $$\langle a\mid a^2\rangle,$$ a presentation for $$\Bbb Z_2$$, implying $$\Bbb Z_2\ast\Bbb Z_2\cong \Bbb Z_2,$$ a contradiction.
You cannot cancel in the way you do, as you can only cancel the same element. However, what you can do is transform the equation to get more of an insight into what is needed for this to happen.
Note $$a^2 = x^2$$ is equivalent to $$a^2 x^{-2} = e_G$$ and further to $$(ax^{-1})^2= e_G$$.
Now $$a=x$$ is equivalent to $$ax^{-1}=e_G$$. So the question reduces to answering if there can be a group with an element $$b \neq e_G$$ such that $$b^2= e_G$$.
The answer to which is: yes. I assume you can find an explicit example. |
# How to Use Mental Math to Solve Equations
Instructor: Laura Pennington
Laura has taught collegiate mathematics and holds a master's degree in pure mathematics.
You may be used to solving arithmetic problems, like addition or subtraction, in your head. In this lesson, we are going to take that mental math a step further and look at how to use mental math to solve equations.
## Mental Math
Suppose I asked you what 2 plus 3 is. Most likely, it wouldn't take you long to tell me that it is 5 without using a calculator or anything else to figure it out. This is called mental math, and we use it to calculate math problems in our heads without having to write anything down or use any tools to do so.
There are many different ways to do mental math. For instance, in the scenario above, I asked you what 2 plus 3 is, and you answered 5. It could be that you have that problem memorized, or it could be that you have a mental trick to figure out what 2 plus 3 is like starting at 2 and then counting three more, or picturing 2 objects and 3 more objects, then counting them all up.
Whichever way you choose to do mental math, you are probably familiar with being able to add, subtract, multiply, or divide in your head without too much difficulty, especially with smaller numbers.
Now suppose I ask you what number, when multiplied by 4, gives you 20? Hmmm, that's a little trickier. Basically, I'm asking you for a number, x, that when multiplied by 4 gives 20, or what is x if 4x = 20? Ah-ha! You may recognize that as an equation, making the problem a bit more clear.
## Equations
Equations are statements in mathematics that are made by putting two mathematical expressions equal to each other. Consider our equation 4x = 20. We can use algebra to solve this equation by dividing both sides of the equation by 4.
We see that x = 5. So, the number, that when multiplied by 4, gives 20, is 5. We call x = 5 the solution to the equation. In general, a solution to an equation is a number that when plugged in for the variable, in our case x, makes the equation a true statement.
## Solving Simple Equations Using Mental Math
You may be familiar with solving equations using algebra as we just did, but I've got some exciting news! We can actually solve simple equations using mental math. Remember how we set up our equation from the statement 'a number that when multiplied by 4 gives 20'. These types of statements are the key to solving equations mentally. To solve an equation using mental math, we use the following steps:
1. Convert the equation into words.
2. Put those words in the form of a question, and answer the question using inverse operations.
So, what do you think this means? Well, consider our earlier example again. The equation 4x = 20 can be put into words by saying 'a number that when multiplied by 4 gives 20.' We put that into question form by asking 'what number times 4 equals 20?' That question is probably pretty easy for you to answer! It's 5!
If the answer to the question isn't obvious to you, you can make use of inverse operations to reword the question. Inverse operations are basically operations that are opposites of each other. In other words, the inverse operation of addition is subtraction and vice versa, and the inverse operation of multiplication is division and vice versa.
Let's consider some simple examples of equations involving addition, subtraction, multiplication, and division and see the question we want to ask when solving these types of equations mentally.
Equation x + 2 = 9 x - 7 = 3 3x = 15 x / 2 = 11
Words A number plus 2 equals 9 A number minus 7 equals 3 A number times 3 equals 15 A number divided by 2 equals 11
Question What number plus 2 equals 9? What number minus 7 equals 3? What number times 3 equals 15? What number divided by 2 equals 11?
Inverse Question What is 2 subtracted from 9? (9-2) What is 7 added to 3? (3+7) What is 3 divided into 15? (15/3) What is 2 multiplied by 11? (11*2)
Answer 9 - 2 = 7 3 + 7 = 10 15 / 3 = 5 11*2 = 22
We see that we can put an equation into question form and answer it, or if the answer isn't immediately obvious, we can use inverse operations to reword the question and answer it that way.
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# T Score Formula: Calculate in Easy Steps
## T Scores in Statistics
Watch the video or read on below:
## What is the T Score Formula?
A t score is one form of a standardized test statistic (the other you’ll come across in elementary statistics is the z-score). The t score formula enables you to take an individual score and transform it into a standardized form>one which helps you to compare scores.
You’ll want to use the t score formula when you don’t know the population standard deviation and you have a small sample (under 30).
The t score formula is:
Where
x̄ = sample mean
μ0 = population mean
s = sample standard deviation
n = sample size
If you have only one item in your sample, the square root in the denominator becomes √1. This means the formula becomes:
In simple terms, the larger the t score, the larger the difference is between the groups you are testing. It’s influenced by many factors including:
• How many items are in your sample.
• The means of your sample.
• The mean of the population from which your sample is drawn.
• The standard deviation of your sample.
## What is the T Score Formula used for?
You traditionally look up a t score in a t-table. The number of items in your sample, minus one, is your degrees of freedom. For example, if you have 20 items in your sample, then df = 19. You use the degrees of freedom along with the confidence level you are willing to accept, to decide whether to support or reject the null hypothesis.
The t score formula can also be used to solve probability questions. You won’t have an alpha level, but you can use the result from the formula, along with a calculator like the TI-83, to find probabilities.
The following example shows how to calculate a t-score formula for a single sample. Paired samples and independent samples use different formulas.
## Example of the T Score Formula
Sample question:
A law school claims it’s graduates earn an average of \$300 per hour. A sample of 15 graduates is selected and found to have a mean salary of \$280 with a sample standard deviation of \$50. Assuming the school’s claim is true, what is the probability that the mean salary of graduates will be no more than \$280?
Step 1: Plug the information into the formula and solve:
x̄ = sample mean = 280
μ0 = population mean = 300
s = sample standard deviation = 50
n = sample size = 15
t = (280 – 300)/ (50/√15) = -20 / 12.909945 = -1.549.
Step 2: Subtract 1 from the sample size to get the degrees of freedom:
15 – 1 = 14. The degrees of freedom lets you know which form of the t distribution to use (there are many, but you can solve these problems without knowing that fact!).
Step 3: Use a calculator to find the probability using your degrees of freedom (8). You have several options, including the TI-83 (see How to find a t distribution on a TI 83) and this online calculator. Here’s the result from that calculator. Note that I selected the radio button under the left tail, as we’re looking for a result that’s no more than \$280:
The probability is 0.0718, or 7.18%.
## T Scores in Psychometrics
A t score in psychometric (psychological) testing is a specialized term that is not the same thing as a t score that you get from a t-test.
T scores in t-tests can be positive or negative. T scores in psychometric testing are always positive, with a mean of 50.
A difference of 10 (positive or negative) from the mean is a difference of one standard deviation. For example, a score of 70 is two standard deviations above the mean, while a score of 0 is one standard deviations below the mean.
A t score is similar to a z score — it represents the number of standard deviations from the mean. While the z-score returns values from between -5 and 5 (most scores fall between -3 and 3) standard deviations from the mean, the t score has a greater value and returns results from between 0 to 100 (most scores will fall between 20 and 80). Many people prefer t scores because the lack of negative numbers means they are easier to work with and there is a larger range so decimals are almost eliminated. This table shows z-scores and their equivalent t scores.
## T Score Conversion in Psychometrics
Watch the video or read the article below:
Calculating a t score is really just a conversion from a z score to a t score, much like converting Celsius to Fahrenheit. The formula to convert a z score to a t score is:
T = (Z x 10) + 50.
Sample question: A candidate for a job takes a written test where the average score is 1026 and the standard deviation is 209. The candidate scores 1100. Calculate the t score for this candidate.
Note: If you are given the z-score for a question, skip to Step 2.
Step 1: Calculate the z score. (See: How to calculate a z-score).. The z-score for the data in this sample question is .354.
Step 2: Multiply the z score from Step 1 by 10:
10 * .354 = 3.54.
3.54 + 50 = 53.54.
That’s it!
Tips:
1. z-scores and t scores both represent standard deviations from the mean, but while “0” on a z-score is 0 standard deviations from the mean, a “50” on a t score represents the same thing. That’s because t scores use a mean of 50 and z-scores use a mean of 0.
2. A t score of over 50 is above average; below 50 is below average. In general, a t score of above 60 means that the score is in the top one-sixth of the distribution; above 63, the top one-tenth. A t score
below 40 indicates a lowest one-sixth position; below 37, the bottom one-tenth.
------------------------------------------------------------------------------
If you prefer an online interactive environment to learn R and statistics, this free R Tutorial by Datacamp is a great way to get started. If you're are somewhat comfortable with R and are interested in going deeper into Statistics, try this Statistics with R track. |
# Difference between revisions of "Mock AIME 3 Pre 2005 Problems/Problem 8"
## Problem
Let $N$ denote the number of $8$-tuples $(a_1, a_2, \dots, a_8)$ of real numbers such that $a_1 = 10$ and
$\left|a_1^{2} - a_2^{2}\right| = 10$
$\left|a_2^{2} - a_3^{2}\right| = 20$
$\cdots$
$\left|a_7^{2} - a_8^{2}\right| = 70$
$\left|a_8^{2} - a_1^{2}\right| = 80$
Determine the remainder obtained when $N$ is divided by $1000$.
## Solution
Here are some thoughts on the problem:
We can call $a_1^2$ through $a_8^2$ by $b_1$ through $b_8$ and the only restriction is that the $b_i$'s are positive. We can express $b_1=100$, $b_2=100 \pm 10$, ...$b_5=100 \pm 10 \pm 20 \pm 30 \pm 40$ and also $b_8=100 \pm 80$. Note that $b_7$ is either $90,110,250$. Note that regardless of how we choose these $\pm$'s all the $b_i$'s I've listed are positive so no restrictions are imposed here. There are restrictions imposed by $b_6$ being equal to $b_7 \pm 60$. We can now write $b_7/(10) \pm 6=b_6/(10)=10 \pm 1 \pm 2 \pm 3 \pm 4 \pm 5$ so the only restrictions are imposed by $10 \pm 1 \pm 2 \pm 3 \pm 4 \pm 5 \pm 6$ being equal to either $9,11,25$. If we find all the $\pm$ in this expression then $b_1$ through $b_8$ are all determined. We can reformulate now as find the number of choices of $\pm$ signs in the expression below:
$\pm 1 \pm 2 \pm 3 \pm 4 \pm 5 \pm 6$
which equals either $-1,1,15$.
If the expression equals $15$ then note that $\pm 1 \pm 2 \pm 3 \pm 4 \pm 5$ is at most 15 so we must have $\pm 1 \pm 2 \pm 3 \pm 4 \pm 5=9$, which forces $\pm 1 \pm 2 \pm 3 \pm 4 =4$ which forces $\pm 1 \pm 2 \pm 3 =0$ for which there are two possibilities of signs.
Now if the expression equals $1$ its symmetric to the case where it equals $-1$ so lets just consider
$\pm 1 \pm 2 \pm 3 \pm 4 \pm 5 \pm 6=1$
The signs in $\pm 5 \pm 6$ cannot both be positive. If they are both negative we get $\pm 1 \pm 2 \pm 3 \pm 4=10$ and there is obviously $1$ choice here only. Otherwise $\pm 5 \pm 6=\pm 1$ so
$\pm 1 \pm 2 \pm 3 \pm 4=0$
or
$\pm 1 \pm 2 \pm 3 \pm 4=2$
The latter case means $\pm 1 \pm 2 \pm 3 =6$ (obviously 1 choice) or $\pm 1 \pm 2 \pm 3 =2$ (1 choice). Thus $2$ choices total for the latter case. In the former case the number of choices is twice the number of choices for $\pm 1 \pm 2 \pm 3=4$ which forces $\pm 1 \pm 2=1$ for which there is 1 choice. Thus 2 choices total for the former case. Thus the number of choices when the expression equals $1$ is $1+2+2=5$. So the answer is $2*5+1=11$, so actually the conditions of the problem were quite restrictive.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page. |
## Combining square roots
Square roots, which use the radical symbol, are nonbinary operations — operations which involve just one number — that ask you, “What number times itself gives you this number under the radical?” Finding square roots is a relatively common operation in algebra, but working with and combining the roots isn’t always so clear.
Expressions with radicals can be multiplied or divided as long as the root power or value under the radical is the same. Expressions with radicals cannot be added or subtracted unless both the root power and the value under the radical are the same.
When you find square roots, the symbol for that operation is called a radical. The root power refers to the number outside and to the upper left of the radical. If there is no number, you assume that the root power is 2.
When it comes to combining radicals, there are just a couple of simple rules to remember:
• Addition and subtraction can be performed if the root power and the value under the radical are the same.
Examples:
These radicals cannot be combined because the operation is addition, and the value under the radical is not the same:
These radicals can be combined because the root power and the numbers under the radical are the same:
These radicals cannot be combined because the operation is subtraction, and the root power isn’t the same:
• Multiplication and division can be performed if the root powers are the same.
Examples:
These radicals can be combined because the operation is multiplication, and the root power is the same:
These radicals can be combined because the operation is division, and the root power is the same:
These radicals cannot be combined because the operation is division, and the root power isn’t the same:
## Converting square roots to exponents
Finding square roots and converting them to exponents is a relatively common operation in algebra. Square roots, which use the radical symbol, are nonbinary operations — operations which involve just one number — that ask you, “What number times itself gives you this number under the radical?” To convert the square root to an exponent, you use a fraction in the power to indicate that this stands for a root or a radical.
When you find square roots, the symbol for that operation is a radical, which looks like this:
When changing from radical form to fractional exponents, remember these basic forms:
• The nth root of a can be written as a fractional exponent with a raised to the reciprocal of that power.
• When the nth root of
is taken, it’s raised to the 1/n power. When a power is raised to another power, you multiply the powers together, and so the m (otherwise written as m/1) and the 1/n are multiplied together.
Use fractions in the powers to indicate that the expression stands for a root or a radical.
Here are some examples of changing radical forms to fractional exponents:
When raising a power to a power, you multiply the exponents, but the bases have to be the same.
Because raising a power to a power means that you multiply exponents (as long as the bases are the same), you can simplify the following expressions:
Leave the exponent as 9/4. Don’t write it as a mixed number.
The following example can’t be combined because the bases are not the same: |
### Greatest Common Factor (GCF)
```Greatest Common
Factor (GCF)
Created by: Mrs. J Couch
GCF
Finding the greatest of the common factors of two (or
more) numbers
Example
Find the GCF of 20 and 24 using prime factorization.
20 = 2 x 2 x 5
24 = 2 x 2 x 2 x 3
-Identify all common factors in the factorization of 20 and
24.
20 = 2 x 2 x 5
24 = 2 x 2 x 2 x 3
The greatest common factor of 20 and 24 is the product of
all common factors 2 x 2 = 4
Finding GCF using
Divide out all the common factors down the left side.
GCF= 2 * 2 * 3 = 12
What number
(smallest prime #)
goes into both
Does
#
24
andany
60?
besides 1
go into both
12 and 30?
6 and 15?
Anything else?
No?
You’re done.
Multiply left side
Only to get GCF
Practice
Find the greatest common factor of 24 and 18.
Practice
Find the greatest common factor of 96 and 72.
Practice
Find the greatest common factor of 32 and 52.
Practice
Xavier is visiting the hospital in order to give stuffed
animals and books to sick children. He has 16 stuffed
animals and 20 books to give. If he wants to give the
same combination of stuffed animals and books to
each child, with no gifts left over, what is the greatest
number of children Xavier can give gifts to?
sentence)
Practice
Find the GCF of 12 and 36
Practice
Two students are having a party. They want to make
treat bags for their guests. They want each bag to be
identical with nothing leftover. They have 36 Silly Bandz
and 72 pieces of bubble gum to put in the bags. What
is the greatest number of treat bags they can make and
how many of each item will be in each treat bag?
Practice
Mitzi is making trail mix out of 48 bags of nuts and 32
bags of dried cranberries. She wants each new portion
of trail mix to be identical containing the same
combination of nuts and cranberries with nothing left
over. What is the greatest number of portions of trail
mix Mitzi can make and how much of each ingredient
will be in each portion?
Connection to Distributive
Property
in the distributive property?
Ex.
Write a number sentence for 12 + 18
-think about the common factors of 12 and 18
Common factors: 1, 2, 3, 6
2 (6 + 9)
Use the common factors to write the
number sentence
3(
+
)
6(
+
)
Practice with
Distributive Property
Write a number sentence (using the GCF) for:
42 + 36
18 + 25
50 + 60
Exit Slip
Find the GCF of 27 and 36
Find the GCF of 12 and 22
Find the GCF of 42 and 80
Write a number sentence (using the GCF) for 20 + 25
``` |
# System of equations by elimination solver
System of equations by elimination solver is a software program that helps students solve math problems. We can help me with math work.
## The Best System of equations by elimination solver
Here, we will be discussing about System of equations by elimination solver. When it comes to solving math problems, there is no one-size-fits-all solution. The best approach depends on the nature of the problem, as well as the skills and knowledge of the person solving it. However, there are a few general tips that can help make solving math problems easier. First, it is important to take the time to understand the problem. What is being asked for? What information is given? Once you have a clear understanding of the problem, you can begin to consider different approaches. Sometimes, visual aids such as charts or diagrams can be helpful in solving math problems. Other times, it may be helpful to break the problem down into smaller steps. And sometimes, simply taking a step back and looking at the problem from a different perspective can make all the difference. There is no single right way to solve math problems. However, by taking the time to understand the problem and trying different approaches, it is usually possible to find a solution that works.
This involves making a change of variable in order to transform the integral equation into a differential equation, which is easier to solve. Another method is to use the Fourier transform, which converts the integral equation into an infinite series that can be solved using standard methods. In some cases, it may also be possible to use numerical methods to approximate the solution to an integral equation. Whichever method is used, solving an integral equation can be a challenging but rewarding experience.
We all know that exponents are a quick way to multiply numbers by themselves, but how do we solve for them? The answer lies in logs. Logs are basically just exponents in reverse, so solving for an exponent is the same as solving for a log. For example, if we want to find out what 2^5 is, we can take the log of both sides of the equation to get: 5 = log2(2^5). Then, we can just solve for 5 to get: 5 = log2(32). Therefore, 2^5 = 32. Logs may seem like a complicated concept, but they can be very useful in solving problems with exponents.
Luckily, there are a number of resources that can help. Online math tutors can work with students one-on-one to help them understand difficult concepts and work through different problems. In addition, there are a number of websites that provide step-by-step solutions to common math problems. With a little bit of effort, any student can get the help they need to succeed in math.
Solving matrix equations is a process of finding the values of unknown variables that satisfy a given set of constraints. In other words, it is a way of solving systems of linear equations. There are several different methods that can be used to solve matrix equations, and the choice of method will depend on the specific equation being solved. However, all methods involve manipulating the equation to achieve a more simplified form that can be solved using standard algebraic methods. Once the unknown variables have been determined, they can be substitued back into the original equation to verify that they are indeed solutions. Solving matrix equations is a powerful tool that can be used to solve a wide variety of problems in mathematics and science.
## Instant support with all types of math
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# Texas Go Math Kindergarten Lesson 7.6 Answer Key One More and One Less
Refer to our Texas Go Math Kindergarten Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Kindergarten Lesson 7.6 Answer Key One More and One Less.
## Texas Go Math Kindergarten Lesson 7.6 Answer Key One More and One Less
Unlock the Problem
DIRECTIONS: There are 14 children in the classroom. One more child walks into the classroom. How many children are in the classroom now? Draw to solve the problem. Write the number.
Explanation:
Number of children in the classroom = 14.
Number of children added in the classroom = 1.
Total number of children are in the classroom now = Number of children in the classroom + Number of children added in the classroom
= 14 + 1
= 15.
Try Another Problem
DIRECTIONS: 1. Cole has 13 marbles. Phillip has one fewer marble than Cole. How many marbles does Phillip have? Draw to solve the problem. Write the number.
Question 1.
Explanation:
Number of marbles Cole has = 13.
Phillip has one fewer marble than Cole.
=> Number of marbles Phillips has = Number of marbles Cole has – 1
= 13 – 1
= 12 or Twelve.
Share and Show
DIRECTIONS: 2. There are 12 children in the drink line. The snack line has one more child than the drink line. How many children are in the snack line? Draw to solve the problem. Write the number.
Question 2.
Explanation:
Number of children in the drink line = 12.
The snack line has one more child than the drink line.
=> Number of children in the snack line = Number of children in the drink line + 1
= 12 + 1
= 13 or Thirteen.
HOME ACTIVITY • Use drawing paper to hove your child draw pictures of one more or one less than any number 1 to 15.
Explanation:
Mt kid drawn different colors of stars.
Number of stars = 13 or Thirteen.
3. Choose the correct answer. The bakery has 13 muffins. Which set shows a number one less than 13? 4. Choose the correct answer. There are 13 red cubes. What number is one more?
Question 3.
Explanation:
Number of muffins less than 13 = 13 – 1 = 12 or Twelve.
Question 4.
Explanation:
Number of red cubes = 13.
Number of muffins more than 13 = 13 + 1 = 14 or Fourteen.
### Texas Go Math Kindergarten Lesson 7.6 Homework and Practice Answer Key
DIRECTIONS: 1. There are 15 dogs playing at the dog park. One dog goes home. How many dogs are playing at the dog park now? Draw to solve the problem. Write the number.
Question 1.
Explanation:
Number of dogs playing at the dog park = 15.
Number of dogs goes home = 1.
Number of dogs are playing at the dog park now = Number of dogs playing at the dog park – Number of dogs goes home
= 15 – 1
= 14 or Fourteen.
DIRECTIONS: Choose the correct answer. 2. Lisa has 14 bear counters. Which set shows a number one more than 14? 3. There are 12 red cubes. What number is one less?
Lesson Check
Question 2. |
# 2010 AIME I Problems/Problem 1
## Problem
Maya lists all the positive divisors of $2010^2$. She then randomly selects two distinct divisors from this list. Let $p$ be the probability that exactly one of the selected divisors is a perfect square. The probability $p$ can be expressed in the form $\frac {m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$.
## Solution 1
$2010^2 = 2^2\cdot3^2\cdot5^2\cdot67^2$. Thus there are $(2+1)^4$ divisors, $(1+1)^4$ of which are squares (the exponent of each prime factor must either be $0$ or $2$). Therefore the probability is $$\frac {2\cdot2^4\cdot(3^4 - 2^4)}{3^4(3^4 - 1)} = \frac {26}{81} \Longrightarrow 26+ 81 = \boxed{107}.$$
## Solution 2 (Using a Bit More Counting)
The prime factorization of $2010^2$ is $67^2\cdot3^2\cdot2^2\cdot5^2$. Therefore, the number of divisors of $2010^2$ is $3^4$ or $81$, $16$ of which are perfect squares. The number of ways we can choose $1$ perfect square from the two distinct divisors is $\binom{16}{1}\binom{81-16}{1}$. The total number of ways to pick two divisors is $\binom{81}{2}$
Thus, the probability is $$\frac {\binom{16}{1}\binom{81-16}{1}}{\binom{81}{2}} = \frac {16\cdot65}{81\cdot40} = \frac {26}{81} \Longrightarrow 26+ 81 = \boxed{107}.$$ |
# Probability of Simple Events
## Presentation on theme: "Probability of Simple Events"— Presentation transcript:
Probability of Simple Events
Probability of Simple Events
Objective: Students will be able to find the probability of a simple event. Students will be able to understand the distinction between simple events and compound events. Essential Question: (1) How do I find the probability of a simple event? (2) How can I distinguish between a simple and compound event?
Probability of Simple Events
Vocabulary: Outcome – one possible result of a probability. Sample Space – the list of possible outcomes for a probability event. Random – outcomes that occur at random if each outcome is equally likely to occur. Simple Event – a specific outcome or type of outcome. Complementary Events – the events of one outcome happening and that outcomes not happening are complimentary; the sum of the probabilities of complementary events is 1.
Probability of Simple Events
Real World Example: Best Buy is having an IPOD giveaway. They put all the IPOD Shuffles in a bag. Customers may choose an IPOD without looking at the color. Inside the bag are 4 orange, 5 blue, 6 green, and 5 pink IPODS. If Maria chooses one IPOD at random, what is the probability she will choose an orange IPOD?
Probability of Simple Events
Real World Example: Best Buy is having an IPOD giveaway. They put all the IPOD Shuffles in a bag. Customers may choose an IPOD without looking at the color. Inside the bag are 4 orange, 5 blue, 6 green, and 5 pink IPODS. If Maria chooses one IPOD at random, what is the probability she will choose an orange IPOD? P(orange) = 4/20 = 2/10 = 1/5 or 20%
Probability of Simple Events
What is a PROBABILITY? - Probability is the chance that some event will happen - It is the ratio of the number of ways a certain event can occur to the number of possible outcomes
Probability of Simple Events
What is a PROBABILITY? number of favorable outcomes number of possible outcomes Examples that use Probability: (1) Dice, (2) Spinners, (3) Coins, (4) Deck of Cards, (5) Evens/Odds, (6) Alphabet, etc. P(event) =
Probability of Simple Events
What is a PROBABILITY? 0% 25% 50% 75% % 0 ¼ or ½ 0r ¾ or Impossible Not Very Equally Likely Somewhat Certain Likely Likely
Probability of Simple Events
Example 1: Roll a dice. What is the probability of rolling a 4? # favorable outcomes # possible outcomes 1 6 The probability of rolling a 4 is 1 out of 6 P(event) = P(rolling a 4) =
Probability of Simple Events
Example 2: Roll a dice. What is the probability of rolling an even number? # favorable outcomes # possible outcomes 3 1 6 2 The probability of rolling an even number is 3 out of 6 or .5 or 50% P(event) = P(even #) = =
Probability of Simple Events
Example 3: Spinners. What is the probability of spinning green? # favorable outcomes # possible outcomes 1 1 4 4 The probability of spinning green is 1 out of 4 or .25 or 25% P(event) = P(green) = =
Probability of Simple Events
Example 4: Flip a coin. What is the probability of flipping a tail? # favorable outcomes # possible outcomes 1 1 2 2 The probability of spinning green is 1 out of 2 or .5 or 50% P(event) = P(tail) = =
Probability of Simple Events
Example 5: Deck of Cards. What is the probability of picking a heart? # favorable outcomes # possible outcomes The probability of picking a heart is 1 out of 4 or .25 or 25% What is the probability of picking a non heart? # favorable outcomes 3 out of 4 or .75 or 75% P(heart) = = = P(nonheart) = = =
Probability of Simple Events
Key Concepts: - Probability is the chance that some event will happen - It is the ratio of the number of ways a certain even can occur to the total number of possible outcomes
Probability of Simple Events
Guided Practice: Calculate the probability of each independent event. 1) P(black) = 2) P(1) = 3) P(odd) = 4) P(prime) =
Probability of Simple Events
Guided Practice: Answers. 1) P(black) = 4/8 2) P(1) = 1/8 3) P(odd) = 1/2 4) P(prime) = 1/2
Probability of Simple Events
Independent Practice: Calculate the probability of each independent event. 1) P(red) = 2) P(2) = 3) P(not red) = 4) P(even) =
Probability of Simple Events
Independent Practice: Answers. 1) P(red) = 1/2 2) P(2) = 1/4 3) P(not red) = 1/2 4) P(even) = 1/2
Probability of Simple Events
Real World Example: A computer company manufactures 2,500 computers each day. An average of 100 of these computers are returned with defects. What is the probability that the computer you purchased is not defective? P(not defective) = # not defective = 2,400 = 24 total # manufactured 2,
Probability of Simple Events
Homework: - Workbook Practice 11-1 |
Learning
is
simple
and
fun!
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### Theory:
Parallelogram:
A parallelogram is a quadrilateral (four-sided figure) in which the opposite sides are parallel.
In the figure, the side $$AB$$ is parallel to $$CD$$, and $$AD$$ is parallel to $$BC$$.
Important!
• The opposite sides are congruent.
• The two diagonals bisect each other.
• The opposite angles are congruent.
• The consecutive angles are supplementary.
Area of the parallelogram:
The area of any parallelogram is calculated using the formula $$A = b \times h$$ square units.
Where $$h$$ represents the height of the parallelogram and $$b$$ represents the base of the parallelogram.
Example:
If the height and the base of the parallelogram are $$12$$ $$cm$$ and $$6$$ $$cm$$, then find its area.
Solution:
Given $$h$$ $$=$$ $$12$$ $$cm$$ and $$b$$ $$=$$ $$6$$ $$cm$$.
Area of the parallelogram, $$A$$ $$=$$ $$b \times h$$
$$=$$ $$12 \times 6$$
$$=$$ $$72$$ $$cm^2$$
Ways to construct a parallelogram:
A parallelogram can be constructed if one of the following four measurements are given.
(i) Two adjacent sides and one angle.
(ii) Two adjacent sides and one diagonal.
(iii) Two diagonals and one included angle.
(iv) One side, one diagonal and one angle. |
# How To Multiply Fractions With Exponents
Multiply the two denominators (bottom numbers) to get the denominator of the answer. X 1/3 × x 1/3 × x 1/3 = x (1/3 + 1/3 + 1/3) = x 1 = x.
Exponent Rules Laws of Exponents Math interactive
### How to multiply fractions with exponents?
How to multiply fractions with exponents. If you want to multiply exponents with the same base, simply add the exponents together. The following formula is used to multiply variables with exponents. Keep in mind that performing these operations on fractional exponents is the same process as normal exponents, with the extra considerations we must have when operating with fractions.
Did you notice a relationship between all of the exponents in the example above? There are two ways to simplify a fraction exponent such $$\frac 2 3$$. When the bases and the exponents are different we have to calculate each exponent and then multiply:
To multiply two fractions, just do the following: When you multiply two proper fractions, the answer is always […] Notice that 3^2 multiplied by 3^3 equals 3^5.also notice that 2 + 3 = 5.
Simplifying exponents with fractions, variables, negative exponents, multiplication & division, math. Flip the exponents into their reciprocals, then multiply. The first is when the exponent itself is a fraction.
Multiply terms with exponents using the general rule: In the case of mixed fractions, simplify it. And then we multiply them.
When the bases and the exponents are different we have to calculate each exponent and then multiply: We already know a good bit about exponents. This lesson will cover how to find the power of a fraction as well as introduce how to w
Now that we have reviewed the rules for exponents, here are the steps required for simplifying exponential expressions notice that we apply the rules in the same order the rule were written above :. Then try m=2 and slide n up and down to see fractions like 2/3 etc Similarly, if the bases are different and the exponents are same, we first multiply the bases and use the exponent.
So you multiply by 2/3 once, twice, three times. Since x 1/3 implies “the cube root of x,” it shows that if x is multiplied 3 times, the product is x. Or you can also view it as starting with a 1, and then multiplying the 1 by 4, or multiplying that by 4, three times.
For example 7 to the third power × 7 to the fifth power = 7 to the eighth power because 3 + 5 = 8. Start activity javascript is required to view this activity. Video about how do you multiply fractions with exponents.
However, to solve exponents with different bases, you have to calculate the exponents and multiply them as regular numbers. There are 3 simple steps to multiply fractions: (a / b) n ⋅ (a / b) m = (a / b) n+m.
By using this website, you agree to our cookie policy. In this section we will go over how to add, subtract, multiply, and divide fractional exponents. X a + x b = x ( a + b ) and divide terms with exponents using the rule:
See how smoothly the curve changes when you play with the fractions in this animation, this shows you that this idea of fractional exponents fits together nicely: Remember me when you’re famous! If you need a reminder, check out our post on how to multiply fractions.
Start with m=1 and n=1, then slowly increase n so that you can see 1/2, 1/3 and 1/4; To multiply fractions, first simply the fraction to its lowest term. And the denominator’s going to be 3 times 3 times 3 times 3, which is equal to 27.
Simplify the fraction if needed. If there’s nothing in common, go directly to solving the equation. Multiplying terms having the same base and with fractional exponents is equal to adding together the exponents.
Why can’t everything in life be as easy as multiplying fractions? Multiply the top numbers (the numerators), 2. Multiply the bottom numbers (the denominators), 3.
For exponents with the same base, we can add the exponents: This relationship applies to multiply exponents with the same base whether the base is a number or a variable: When the numerator is not 1.
Below is a specific example illustrating the formula for fraction exponents when the numerator is not one. You can either apply the numerator. A n ⋅ b m.
$$\frac 1 n$$ is another way of asking: To multiply fractions, you need to multiply the numerators together and multiply the denominators together. If you like this site about solving math problems, please let google know by clicking the +1 button.
If a +1 button is dark blue, you have already +1’d it. 1 3 ⋅ 2 2 = 1 ⋅ 4 = 4. Where x is the variable being raised
After simplifying the fraction, multiply the numerator with the numerator and denominator with the denominator. If the bases are different but the exponents are the same, multiply the bases and leave the exponents the way they are. How to multiply fractional exponents with the same base.
What number can you multiply by itself n times to get x? Multiplying fractions with exponents with same fraction base: Now, the other way of viewing this is you start with a 1, and you multiply it by 2/3 three times.
There are two ways that fractions get involved in exponents. (if you are not logged into your google account (ex., gmail, docs), a login window opens when you click on +1. When multiplying fractions with the same base, we add the exponents.
The multiply fractions calculator will multiply fractions and reduce the fraction to its simplest form. Follow to get the latest 2020 recipes, articles and more! Then, the product of fractions is obtained in p/q form.
Find the how do you multiply fractions with exponents, including hundreds of ways to cook meals to eat. Multiply the two numerators (top numbers) to get the numerator of the answer; Enter the fraction you want to simplify.
For example, we know if we took the number 4 and raised it to the third power, this is equivalent to taking three fours and multiplying them. If you like this page, please click that +1 button, too. Thank you for your support!
The second is when the base is a fraction, and we’re raising that fractional base to an exponent.
Simplifying Expressions with Exponents (Number Bases) in
Exponents anchor chart Math instruction, Upper grades
Algebra in Practice Teaching math, Algebra, Homeschool math
Dividing Fractions Poster Dividing fractions, Anchor
Multiplying and Dividing by Powers of 10 Lesson {5th Grade
Exponent Rules Math pages, Multiplication facts practice
Properties of Exponents & Proof ♻keep on reduce reuse
Exponents with Decimal and Fraction Bases Boom Cards
Valentine’s Day Math Activity Simplifying Exponents 8th
Exponents with Decimal and Fraction Bases Boom Cards in
Simplify Exponents Negative Exponents Fractional Exponents
Order of Operations with Fractions Differentiated
Christmas Math Activity Worksheets Laws of Exponents
One of the newest (and often most difficult) elements of
Exponent Rules Radical Rules Rationalizing Fractions |
# Figure 7.21 shows a series LCR circuit connected to a variable frequency 230 V source.
Question:
Figure 7.21 shows a series LCR circuit connected to a variable frequency 230 V source. = 5.0 H, = 80μF, = 40 Ω
(a) Determine the source frequency which drives the circuit in resonance.
(b) Obtain the impedance of the circuit and the amplitude of current at the resonating frequency.
(c) Determine the rms potential drops across the three elements of the circuit. Show that the potential drop across the LC combination is zero at the resonating frequency.
Solution:
Inductance of the inductor, = 5.0 H
Capacitance of the capacitor, C = 80 μF = 80 × 10−6 F
Resistance of the resistor, R = 40 Ω
Potential of the variable voltage source, V = 230 V
(a) Resonance angular frequency is given as:
$\omega_{R}=\frac{1}{\sqrt{L C}}$
$=\frac{1}{\sqrt{5 \times 80 \times 10^{-6}}}=\frac{10^{3}}{20}=50 \mathrm{rad} / \mathrm{s}$
Hence, the circuit will come in resonance for a source frequency of 50 rad/s.
(b) Impedance of the circuit is given by the relation,
$Z=\sqrt{R^{2}+\left(\omega L-\frac{1}{\omega C}\right)^{2}}$
At resonance,
$\omega L=\frac{1}{\omega C}$
$\therefore Z=R=40 \Omega$
Amplitude of the current at the resonating frequency is given as: $I_{0}=\frac{V_{0}}{Z}$
Where,
V0 = Peak voltage
$=\sqrt{2} V$
$\therefore I_{0}=\frac{\sqrt{2} V}{Z}$
$=\frac{\sqrt{2} \times 230}{40}=8.13 \mathrm{~A}$
Hence, at resonance, the impedance of the circuit is 40 Ω and the amplitude of the current is 8.13 A.
(c) Rms potential drop across the inductor,
$\left(V_{L}\right)_{r m s}=I \times \omega_{R} L$
Where,
I = rms current
$=\frac{I_{0}}{\sqrt{2}}=\frac{\sqrt{2} V}{\sqrt{2} Z}=\frac{230}{40} \mathrm{~A}$
$\therefore\left(V_{L}\right)_{\mathrm{rms}}=\frac{230}{40} \times 50 \times 5=1437.5 \mathrm{~V}$
Potential drop across the capacitor,
$\left(V_{c}\right)_{\text {mes }}=I \times \frac{1}{\omega_{R} C}$
$=\frac{230}{40} \times \frac{1}{50 \times 80 \times 10^{-6}}=1437.5 \mathrm{~V}$
Potential drop across the resistor,
(VR)rms = IR
$=\frac{230}{40} \times 40=230 \mathrm{~V}$
Potential drop across the LC combination,
$V_{L C}=I\left(\omega_{R} L-\frac{1}{\omega_{R} C}\right)$
At resonance, $\omega_{R} L=\frac{1}{\omega_{R} C}$
$\therefore V_{L C}=0$
Hence, it is proved that the potential drop across the LC combination is zero at resonating frequency. |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# 5.4: Least Common Multiple
Difficulty Level: At Grade Created by: CK-12
## Introduction
The Decoration Committee
As the sixth grade has been planning for the social, each cluster formed a decoration committee. Each decoration committee was given the opportunity every few days to meet in the art room and make decorations for the social. Some students worked on banners, some worked on posters, some worked with streamers. All of the students had a terrific time.
The big conflict is that every few days both groups seem to be in the art room at the same time and there are never enough supplies for everyone. Mr. Caron the art teacher wants to figure out why this keeps happening.
Cluster 6A gets to work in the art room every two days.
Cluster 6B gets to work in the art room every three days.
If Mr. Caron could figure out when the groups are both in the art room on the same day, then he would have more art supplies ready. Or on those days, he could plan for the students to work on a bigger project.
If 6A works in the art room every two days and 6B works in the art room every three days, when is the first day that all of the students will be working in the art room together?
This problem may seem challenging to figure out, but if you use multiples and least common multiples, you will be able to help Mr. Caron figure out the schedule.
Pay attention and at the end of the lesson you will help solve the dilemma.
What You Will Learn
In this lesson, you will learn to:
• Find common multiples of different numbers.
• Find the least common multiple of given numbers using lists.
• Find the least common multiple of given numbers using prime factorization.
• Find two numbers given the greatest common factor and the least common multiple.
Teaching Time
I. Find the Common Multiples of Different Numbers
In mathematics, you have been working with multiples for a long time. One of the first things that you probably learned was how to count by twos or threes. Counting by twos and threes is counting by multiples. When you were small, you didn’t call it “counting by multiples,” but that is exactly what you were doing.
What is a multiple?
A multiple is the product of a quantity and a whole number.
What does that mean exactly?
It means that when you take a number like 3 that becomes the quantity. Then you multiply that quantity times different whole numbers.
3 ×\begin{align*}\times\end{align*} 2 =\begin{align*}=\end{align*} 6, 3 ×\begin{align*}\times\end{align*} 3 =\begin{align*}=\end{align*} 9, 3 ×\begin{align*}\times\end{align*} 4 =\begin{align*}=\end{align*} 12, 3 ×\begin{align*}\times\end{align*} 5 =\begin{align*}=\end{align*} 15, 3 ×\begin{align*}\times\end{align*} 6 =\begin{align*}=\end{align*} 18
Listing out these products is the same as listing out multiples.
3, 6, 9, 12, 15, 18.....
You can see that this is also the same as counting by threes.
The dots at the end mean that these multiples can go on and on and on. Each numbers has an infinite number of multiples.
Example
List five the multiples for 4.
To do this, we can think of taking the quantity 4 and multiplying it by 2, 3, 4, 5, 6.....
4 ×\begin{align*}\times\end{align*} 2 =\begin{align*}=\end{align*} 8, 4 ×\begin{align*}\times\end{align*} 3 =\begin{align*}=\end{align*} 12, 4 ×\begin{align*}\times\end{align*} 4 =\begin{align*}=\end{align*} 16, 4 ×\begin{align*}\times\end{align*} 5 =\begin{align*}=\end{align*} 20, 4 ×\begin{align*}\times\end{align*} 6 =\begin{align*}=\end{align*} 24
Our answer is 8, 12, 16, 20, 24....
Notice that we could keep on listing multiples of 4 forever.
What is a common multiple?
A common multiple is a multiple that two or more numbers have in common.
Example
What are the common multiples of 3 and 4?
To start to find the common multiples, we first need to write out the multiples for 3 and 4. To find the most common multiples that we can, we can list out multiples through multiplying by 12.
3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36
4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48
The common multiples of 3 and 4 are 12, 24, 36.
Now it is time for you to practice a few.
1. List out five multiples of 6.
2. List out five multiples of 8.
3. What are the common multiples of 6 and 8?
Take a few minutes to check your work with a peer.
II. Find the Least Common Multiple of Given Numbers Using Lists
We can also find the least common multiple of a pair of numbers.
What is the least common multiple?
The least common multiple (LCM) is just what it sounds like, the smallest multiple that two numbers have in common.
Let’s look back at the common multiples for 3 and 4.
3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36
4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48
Here we know that the common multiples are 12, 24 and 36.
The LCM of these two numbers is 12. It is the smallest number that they both have in common.
We used lists of multiples for 3 and 4 to find the common multiples and then the least common multiple.
Find the Least Common Multiple for each pair of numbers.
1. 5 and 3
2. 2 and 6
3. 4 and 6
III. Find the Least Common Multiple of Given Numbers Using Prime Factorization
Remember back to factoring numbers? We worked on using factor trees to factor numbers or to break down numbers into their primes. Take a look at this one.
12 / \ 4 3/ \ 2 222×3\begin{align*}& \ \ \ \quad 12\\ & \ \quad \ \big / \ \ \big\backslash\\ & \quad \ 4 \quad \ \ 3\\ & \quad \big / \ \big\backslash\\ & \ \ 2 \ \ \ 2\\ & 2^2 \times 3\end{align*}
We used a factor tree in this example to factor twelve down to the prime factors of 2 squared times 3.
We can also use prime factorization when looking for the least common multiple.
How can we use prime factorization to find the LCM?
If we wanted to find the LCM of two numbers without listing out all of the multiples, we could do it by using prime factorization.
Example
What is the LCM of 9 and 12?
First, we factor both numbers to their primes.
9/ \ 33 12/ \ 34 / \2 2\begin{align*}& \quad \ \ 9 && \quad \ 12\\ & \quad \big / \ \ \big\backslash && \quad \big / \ \ \big\backslash\\ & \ 3 \qquad 3 && \ 3 \qquad 4\\ &&& \ \qquad \big / \ \big\backslash\\ &&& \qquad 2 \quad \ 2\end{align*}
Next, we identify any shared primes. With 9 and 12, 3 is a shared prime number.
Then, we take the shared prime and multiply it with all of the other prime factors.
3 ×\begin{align*}\times\end{align*} 3 ×\begin{align*}\times\end{align*} 2 ×\begin{align*}\times\end{align*} 2
The first 3 is the shared prime factor.
The other numbers are the other prime factors.
Our answer is 36. The LCM of 9 and 12 is 36.
Now it is time for you to try.
1. Find the LCM of 4 and 10 using prime factorization.
Take a minute to check your work with a peer.
Before moving on take a few notes on multiples and finding the LCM.
IV. Find Two Numbers Given the GCF and the LCM
This section is a bit more advanced than some of the work that we have been doing. We are going to be playing detective. A detective is someone who uses clues to figure something out.
The task that you will have as a detective is to figure out two missing numbers if you have only been given the greatest common factor and the least common multiple.
If you were given the least common multiple of 10, you could think of possible numbers that would multiply to equal 10.
2 would be a possibility for one of the numbers since 2 ×\begin{align*}\times\end{align*} 5 =\begin{align*}=\end{align*} 10.
5 would be another possibility for one of the numbers since 5 ×\begin{align*}\times\end{align*} 2 =\begin{align*}=\end{align*} 10.
This one was easier to figure out because the numbers are small. We didn’t even need to know the greatest common factor.
What do we do when the numbers aren’t small?
When working with larger numbers, we can use a formula to figure out missing parts.
GCF(LCM)=ab\begin{align*}GCF(LCM) = ab\end{align*}
The GCF times the LCM is equal to number a\begin{align*}a\end{align*} times number b\begin{align*}b\end{align*}.
Remember that a\begin{align*}a\end{align*} and b\begin{align*}b\end{align*} are variables that represent unknown numbers.
Now let’s apply this formula with an example.
Example
GCF is 6. LCM = 36.
If one of the missing numbers is 12, can you find the other missing number?
First, we take our known quantities and put them into the formula.
6(36)=12b\begin{align*}6(36) = 12b\end{align*}
Next, we multiply the left side of the equation.
216=12b\begin{align*}216 = 12b\end{align*}
To solve for b\begin{align*}b\end{align*}, we can ask ourselves, “What number times 12 is 216?” Said another way, we can divide 216 by 12.
12)216 18\begin{align*}& \overset{\quad \ \overset{18} {\underline{\;\;\;\;\;\;\;\;\;\;}}}{12 \big ) 216}\end{align*}
Our answer is that b\begin{align*}b\end{align*} is 18.
## Real Life Example Completed
The Decoration Committee
Now that you have learned all about least common multiples, it is time to help Mr. Caron with the decoration committees.
Here is the problem once again.
As the sixth grade has been planning for the social, each cluster formed a decoration committee. Each decoration committee was given the opportunity every few days to meet in the art room and make decorations for the social. Some students worked on banners, some worked on posters, some worked with streamers. All of the students had a terrific time.
The big conflict is that every few days both groups seem to be in the art room at the same time and there are never enough supplies for everyone. Mr. Caron, the art teacher, wants to figure out why this keeps happening.
Cluster 6A gets to work in the art room every two days.
Cluster 6B gets to work in the art room every three days.
If Mr. Caron could figure out when the groups are both in the art room on the same day, then he would have more art supplies ready. Or on those days, he could plan for the students to work on a bigger project.
If 6A works in the art room every two days and 6B works in the art room every three days, when is the first day that all of the students will be working in the art room together?
First, let’s underline the important question that we are trying to solve.
Next, let’s think about how to solve this dilemma. We want to know the first common day that both 6A and 6B will meet in the art room. If you think about this, it is the same as a least common multiple.
Since 6A meets every two days, two will be the first quantity.
Since 6B meets every three days, three will be the second quantity.
Now let’s list the multiples of two and three. The common multiples will show the days that the students will both meet in the art room. The least common multiple will show the first day that the students will both meet in the art room.
2 4 6 8 10 12 14 16 18 20 22 24 26 28 30
3 6 9 12 15 18 21 24 27 30
The common multiples are 6, 12, 18, 24, 30.
The least common multiple is 6. The students will both be in the art room on these days.
If the students start the decoration committee on a Monday, what is the first day of the week that the students will both be in the art room?
We can make a list of days to figure this out.
Day 1 Monday
Day 2 Tuesday
Day 3 Wednesday
Day 4 Thursday
Day 5 Friday
Day 6 Monday – this the first day that both groups will be in the art room at the same time
Sometimes when you have a scheduling conflict like the one Mr. Caron had, using least common multiples is a great way to solve it!!
## Technology Integration
Other Videos:
1. http://www.mathplayground.com/howto_gcflcm.html – This video covers finding the greatest common factor and the least common multiple of two numbers.
2. http://www.teachertube.com/members/music.php?music_id=1351&title=Mr_Duey_GCF_LCM – This is a song only, but it is a great rap about greatest common factor and least common multiple. You'll need to register at the website to access this song.
3. http://www.teachertube.com/members/viewVideo.php?video_id=15601&title=LCM_and_GCF_Indian_Method – This is a different way of finding the greatest common factor and the least common multiple. You'll need to register at the website to access this video.
## Time to Practice
Directions: List the first five multiples for each of the following numbers.
1. 3
2. 5
3. 6
4. 7
5. 8
Directions: Find two common multiples of each pair of numbers.
6. 3 and 5
7. 2 and 3
8. 3 and 4
9. 2 and 6
10. 3 and 9
11. 5 and 7
12. 4 and 12
13. 5 and 6
14. 10 and 12
15. 5 and 8
Directions: Go back through the common multiples for numbers 6 – 15 and select the LCM for each pair of numbers.
16. LCM =
17. LCM =
18. LCM =
19. LCM =
20. LCM =
21. LCM =
22. LCM =
23. LCM =
24. LCM =
25. LCM =
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# How do you find the limit of (1/(x-1)+1/(x^2-3x+2)) as x->1?
Feb 9, 2017
$- 1$
#### Explanation:
${\lim}_{x \to 1} \left(\frac{1}{x - 1} + \frac{1}{{x}^{2} - 3 x + 2}\right)$
$= {\lim}_{x \to 1} \left(\frac{1}{x - 1} + \frac{1}{\left(x - 1\right) \left(x - 2\right)}\right)$
if we let $x - 1 = \delta$
$= {\lim}_{\delta \to 0} \left(\frac{1}{\delta} + \frac{1}{\delta \left(\delta - 1\right)}\right)$
$= {\lim}_{\delta \to 0} \left(\frac{\delta - 1}{\delta \left(\delta - 1\right)} + \frac{1}{\delta \left(\delta - 1\right)}\right)$
$= {\lim}_{\delta \to 0} \frac{\delta}{\delta \left(\delta - 1\right)}$
$= {\lim}_{\delta \to 0} \frac{1}{\delta - 1} = - 1$
Feb 9, 2017
${\lim}_{x \to 1} \frac{1}{x - 1} + \frac{1}{{x}^{2} - 3 x + 2} = - 1$
#### Explanation:
${\lim}_{x \to 1} \frac{1}{x - 1} + \frac{1}{{x}^{2} - 3 x + 2} = {\lim}_{x \to 1} \frac{1}{x - 1} + \frac{1}{\left(x - 1\right) \left(x - 2\right)}$
$\text{ } = {\lim}_{x \to 1} \frac{\left(x - 2\right) + 1}{\left(x - 1\right) \left(x - 2\right)}$
$\text{ } = {\lim}_{x \to 1} \frac{\left(x - 1\right)}{\left(x - 1\right) \left(x - 2\right)}$
$\text{ } = {\lim}_{x \to 1} \frac{1}{\left(x - 2\right)}$
$\text{ } = \frac{1}{\left(1 - 2\right)}$
$\text{ } = - 1$ |
Quick Answer: Do 2d Shapes Have Vertices Or Corners?
What shape has 3 sides and 3 corners?
triangletriangle A triangle has 3 straight sides and 3 corners..
Is a semi circle a 2d shape?
It has only one line of symmetry (reflection symmetry). In non-technical usage, the term “semicircle” is sometimes used to refer to a half-disk, which is a two-dimensional geometric shape that also includes the diameter segment from one end of the arc to the other as well as all the interior points.
What is 2d shapes with examples?
Any shape that can be laid flat on a piece of paper or any mathematical plane is a 2D shape. As a child, your first drawings probably used basic shapes, such as squares, triangles, and circles. Now you can find 2D shapes in the world all around you. Examples of 2D shapes include rectangles, octagons, and even hearts.
How many corners do 2d shapes have?
6 CornersA 2D shape. 6 straight sides • 6 Corners.
How many sides will a shape with 4 vertices have?
Its angles are 90 degrees. This is a parallelogram. A parallelogram has both pairs of opposite sides that are parallel. It has 4 sides and 4 vertices.
How do you find vertices?
We find the vertex of a quadratic equation with the following steps:Get the equation in the form y = ax2 + bx + c.Calculate -b / 2a. This is the x-coordinate of the vertex.To find the y-coordinate of the vertex, simply plug the value of -b / 2a into the equation for x and solve for y.
What are sides vertices?
The two parts of a flat shape are its sides and its vertices. Sides are lines. Vertices are the points where the sides meet. We say vertex when there’s just one.
Are there vertices on a 2d shape?
2D shapes have sides and vertices. A vertex is a point where two or more lines meet. The plural of vertex is vertices.
Do 3d shapes have corners or vertices?
3D shapes have faces (sides), edges and vertices (corners).
What shape has 4 sides and 3 vertices?
The geometric shape that is characterised by 4 sides and 3 corners is a pyramid or a triangular prism.
Do 2d shapes have the same number of sides and corners?
No, a 2d shape will always have the same number of vertices and edges.
What is a shape with 4 vertices?
In geometry, a quadrilateral can be defined as a closed, two-dimensional shape which has four straight sides. The polygon has four vertices or corners.
What 3d shapes do Year 1 need to know?
In Year 1, children need to be able to recognise and name: 2D shapes including rectangles, squares, circle and triangles. 3D shapes including cubes, cuboids, pyramids and spheres. sort, make and describe common 2D and 3D shapes.
Are corners the same as vertices?
Vertex typically means a corner or a point where lines meet. For example a square has four corners, each is called a vertex. The plural form of vertex is vertices.
What Is a corner on a 2d shape?
2D shapes A 2D shape is a flat shape. … When we talk about the properties of these shapes we look at the number of sides that each shape has and the number of corners. A corner is where 2 sides meet. E.g. a triangle has 3 straight sides and 3 corners, whereas a circle has 1 curved side but no corners. |
NCERT Solutions for Class 8 Maths Chapter 3- Understanding Quadrilaterals Exercise 3.2 - GMS - Learning Simply
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# NCERT Solutions for Class 8 Maths Chapter 3- Understanding Quadrilaterals Exercise 3.2
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# NCERT Solutions for Class 8 Maths Chapter 3- Understanding Quadrilaterals Exercise 3.2
The NCERT solutions for Class 8 maths Chapter 3- Understanding Quadrilaterals contains solutions for all exercise questions. NCERT Class 8 Exercise 3.2 is based on the measurement of the exterior angle of a polygon. Students can download the NCERT Solutions of Class 8 mathematics to enhance their problem solving skills.
### Access Answers of Maths NCERT Class 8 Chapter 3- Understanding Quadrilaterals Exercise 3.2 Page Number 44
1. Find x in the following figures.
Solution:
a)
125° + m = 180° ⇒ m = 180° – 125° = 55° (Linear pair)
125° + n = 180° ⇒ n = 180° – 125° = 55° (Linear pair)
x = m + n (exterior angle of a triangle is equal to the sum of 2 opposite interior 2 angles)
⇒ x = 55° + 55° = 110°
b)
Two interior angles are right angles = 90°
70° + m = 180° ⇒ m = 180° – 70° = 110° (Linear pair)
60° + n = 180° ⇒ n = 180° – 60° = 120° (Linear pair) The figure is having five sides and is a pentagon.
Thus, sum of the angles of pentagon = 540° 90° + 90° + 110° + 120° + y = 540°
⇒ 410° + y = 540° ⇒ y = 540° – 410° = 130°
x + y = 180° (Linear pair)
⇒ x + 130° = 180°
⇒ x = 180° – 130° = 50°
## 2. Find the measure of each exterior angle of a regular polygon of
(i) 9 sides (ii) 15 sides Solution:
Sum of angles a regular polygon having side n = (n-2)×180°
(i) Sum of angles a regular polygon having side 9 = (9-2)×180°= 7×180° = 1260°
Each interior angle=1260/9 = 140°
Each exterior angle = 180° – 140° = 40°
Or,
Each exterior angle = sum of exterior angles/Number of angles = 360/9 = 40°
(ii) Sum of angles a regular polygon having side 15 = (15-2)×180°
= 13×180° = 2340°
Each interior angle = 2340/15 = 156°
Each exterior angle = 180° – 156° = 24°
Or,
Each exterior angle = sum of exterior angles/Number of angles = 360/15 = 24°
## Each exterior angle = sum of exterior angles/Number of angles
24°= 360/ Number of sides
⇒ Number of sides = 360/24 = 15
Thus, the regular polygon has 15 sides.
## 4. How many sides does a regular polygon have if each of its interior angles is 165°? Solution:
Interior angle = 165°
Exterior angle = 180° – 165° = 15°
Number of sides = sum of exterior angles/ exterior angles
⇒ Number of sides = 360/15 = 24
Thus, the regular polygon has 24 sides.
5.
a) Is it possible to have a regular polygon with measure of each exterior angle as 22°?
b) Can it be an interior angle of a regular polygon? Why?
Solution:
a) Exterior angle = 22°
Number of sides = sum of exterior angles/ exterior angle
⇒ Number of sides = 360/22 = 16.36
No, we can’t have a regular polygon with each exterior angle as 22° as it is not divisor of 360.
b) Interior angle = 22°
Exterior angle = 180° – 22°= 158°
No, we can’t have a regular polygon with each exterior angle as 158° as it is not divisor of 360.
## 6.
a) What is the minimum interior angle possible for a regular polygon? Why?
b) What is the maximum exterior angle possible for a regular polygon?
Solution:
a) Equilateral triangle is a regular polygon with 3 sides has the least possible minimum interior angle because the regular with minimum sides can be constructed with 3 sides at least. Since, sum of interior angles of a triangle = 180°
Each interior angle = 180/3 = 60°
b) Equilateral triangle is a regular polygon with 3 sides has the maximum exterior angle because the regular polygon with least number of sides have the maximum exterior angle possible. Maximum exterior possible = 180 – 60° = 120°
At the helm of GMS Learning is Principal Balkishan Agrawal, a dedicated and experienced educationist. Under his able guidance, our school has flourished academically and has achieved remarkable milestones in various fields. Principal Agrawal’s visio… |
# Math photos
Math photos can be a helpful tool for these students. So let's get started!
## The Best Math photos
Keep reading to understand more about Math photos and how to use it. Once this has been accomplished, the resulting equation can be solved for the remaining variable. In some cases, it may not be possible to use elimination to solve a system of linear equations. However, by understanding how to use this method, it is usually possible to simplify a system of equations so that it can be solved using other methods.
College algebra word problems can be difficult to solve, but there are some tips that can help. First, read the problem carefully and make sure you understand what is being asked. Next, identify the key information and identify any variables that need to be solved for. Once you have all of the information, you can start solving the problem. College algebra word problems often require the use of equations, so it is important to be familiar with the various types of equations and how to solve them. With a little practice, solving college algebra word problems can become easier.
A differential equation is an equation that relates a function with one or more of its derivatives. In order to solve a differential equation, we must first find the general solution, which is a function that satisfies the equation for all values of the variable. The general solution will usually contain one or more arbitrary constants, which can be determined by using boundary conditions. A boundary condition is a condition that must be satisfied by the solution at a particular point. Once we have found the general solution and determined the values of the arbitrary constants, we can substitute these values back into the solution to get the particular solution. Differential equations are used in many different areas of science, such as physics, engineering, and economics. In each case, they can help us to model and understand complicated phenomena.
A rational function is a function that can be written in the form of a ratio of two polynomial functions. In other words, it is a fraction whose numerator and denominator are both polynomials. Solving a rational function means finding the points at which the function equals zero. This can be done by setting the numerator and denominator equal to zero and solving for x. However, this will only give you the x-intercepts of the function. To find the y-intercepts, you will need to plug in 0 for x and solve for y. The points at which the numerator and denominator are both equal to zero are called the zeros of the function. These points are important because they can help you to graph the function. To find the zeros of a rational function, set the numerator and denominator equal to zero and solve for x. This will give you the x-intercepts of the function. To find the y-intercepts, plug in 0 for x and solve for y. The points at which the numerator and denominator are both zero are called the zeros of the function. These points can help you to graph the function.
Simple solutions math is a method of teaching that focuses on breaking down complex problems into small, manageable steps. By breaking down problems into smaller pieces, students can better understand the concepts behind the problem and find more creative solutions. Simple solutions math also encourages students to think outside the box and approach problems from different angles. As a result, Simple solutions math can be an effective way to teach problem-solving skills. In addition, Simple solutions math can help to improve test scores and grades. Studies have shown that students who use Simple solutions math outperform those who do not use the method. As a result, Simple solutions math is an effective tool for helping students succeed in school.
## We cover all types of math problems
less problems that are yet to be solved, and much more helpful demonstrations. I know you say, hey, but there are infinite numbers of problems, and you would be correct. However, the app touches on the foundations and so much more!
Rebecca Jackson
This is an excellent program for showing the solution to equations step by step, letting you see where your mistakes are. Each step can be broken down to their simplest form and the rule applied in that step is well described. The UI is very simple yet complete & intuitive. The photo capture of equations is simple & accurate. Simply an excellent tool for anyone learning any level of math, grade school to university. The best math tutor app bar none.
Felicity Peterson |
# 1.15: Real Number Comparisons
Difficulty Level: At Grade Created by: CK-12
Can you order the following real numbers from least to greatest?
\begin{align*}\frac{22}{7},1.234 234 \ldots, - \sqrt{7}, -5, -\frac{21}{4}, 7,-1.666,0,8.32,\frac{\pi}{2},-\pi,-5.38\end{align*}
### Guidance
The simplest way to order numbers is to express them all in the same form – all fractions or all decimal numbers. With a calculator, it is easy to express every number as a decimal number. Watch your signs – don't drop any of the negative signs.
When plotting numbers on a number line, keep in mind that it is impossible to place decimal numbers in the exact location on the number line. Place them as close as you can to the appropriate spot on the line.
#### Example A
Draw a number line and place these numbers on the line.
\begin{align*}{\color{red}\sqrt{\frac{2}{5}}}, {\color{blue}0.6467},{\color{red}-\frac{3}{5}},{\color{red}\frac{1}{8}},{\color{green}0},{\color{red}\sqrt{0.5}},{\color{blue}-2.34},{\color{red}\pi},{\color{red}\frac{2 \pi}{3},{\color{green}-1}},{\color{green}2}\end{align*}
\begin{align*}\sqrt{\frac{2}{5}}=0.6324 \quad -\frac{3}{5}=-0.6 \quad \frac{1}{8}=0.125 \quad \sqrt{0.5}=0.7071 \quad -\pi=-3.1416 \quad \frac{2 \pi}{3}=2.0944\end{align*}
Start by placing the \begin{align*}{\color{green}\mathbf{integers}}\end{align*} on the line first. Next place the \begin{align*}{\color{blue}\mathbf{decimal \ numbers}}\end{align*} on the line.
Use your calculator to convert \begin{align*}{\color{red}\mathbf{the \ remaining \ numbers}}\end{align*} to decimal numbers. Place these on the line last.
#### Example B
For each given pair of real numbers, find another real number that is between each of the pairs.
i) \begin{align*}-2,1\end{align*}
ii) \begin{align*}3.5,3.6\end{align*}
iii) \begin{align*}\frac{1}{2},\frac{1}{3}\end{align*}
iv) \begin{align*}-\frac{1}{3}, -\frac{1}{4}\end{align*}
i) The number must be greater than -2 and less than 1. \begin{align*}-2, {\color{blue}0},1\end{align*}
ii) The number must be greater than 3.5 and less than 3.6. \begin{align*}3.5, {\color{blue}3.54},3.6\end{align*}
iii) The number must be greater \begin{align*}\frac{1}{3}\end{align*} than and less than \begin{align*}\frac{1}{2}\end{align*}. Write each fraction with a common denominator. \begin{align*}\frac{1}{2}=\frac{3}{6},\frac{1}{3}=\frac{2}{6}\end{align*}. If you look at \begin{align*}\frac{2}{6}\end{align*} and \begin{align*}\frac{3}{6}\end{align*}, there is no fraction, with a denominator of 6, between these values. Write the fractions with a larger common denominator. \begin{align*}\frac{1}{2}=\frac{6}{12}, \frac{1}{3}=\frac{4}{12}\end{align*}. If you look at \begin{align*}\frac{4}{12}\end{align*} and \begin{align*}\frac{6}{12}\end{align*}, the fraction \begin{align*}\frac{5}{12}\end{align*} is between them. \begin{align*}\frac{1}{3},{\color{blue}\frac{5}{12}},\frac{1}{2}\end{align*}
iv) The number must be greater than \begin{align*}-\frac{1}{3}\end{align*} and less than \begin{align*}-\frac{1}{4}\end{align*}. Write each fraction with a common denominator. \begin{align*}-\frac{1}{3}=-\frac{4}{12},-\frac{1}{4}=-\frac{3}{12}\end{align*}. If you look at \begin{align*}-\frac{3}{12}\end{align*} and \begin{align*}-\frac{4}{12}\end{align*}, there is no fraction, with a denominator of 12, between these values. Write the fractions with a larger common denominator. \begin{align*}-\frac{1}{3}=-\frac{8}{24},-\frac{1}{4}=-\frac{6}{24}\end{align*}. If you look at \begin{align*}-\frac{6}{24}\end{align*} and \begin{align*}-\frac{8}{24}\end{align*}, the fraction \begin{align*}-\frac{7}{24}\end{align*} is between them. \begin{align*}-\frac{8}{24}, {\color{blue}-\frac{7}{24}}, -\frac{6}{24}\end{align*}
#### Example C
Order the following fractions from least to greatest.
\begin{align*}\frac{2}{11},\frac{7}{9},\frac{8}{7},\frac{1}{11},\frac{5}{6}\end{align*}
The fractions do not have a common denominator. This makes it almost impossible to arrange the fractions from least to greatest. To determine the common denominator, may take some time. Let’s use the TI-83 to order these fractions.
The fractions were entered into the calculator as division problems. The decimal numbers were entered into List 1.
The calculator has sorted the data from least to greatest.
The data is sorted. The decimal numbers and the corresponding fractions can now be matched from the screen where they were first entered.
\begin{align*}\frac{1}{11},\frac{2}{11},\frac{7}{9},\frac{5}{6},\frac{8}{7}\end{align*}
#### Concept Problem Revisited
\begin{align*}\frac{22}{7},1.234 234 \ldots, - \sqrt{7}, -5, -\frac{21}{4}, 7,-1.666,0,8.32,\frac{\pi}{2},-\pi,-5.38\end{align*}
As you examine the above numbers, you can see that there are natural numbers, whole numbers, integers, rational numbers and irrational numbers. These numbers, as they are presented here, would be very difficult to order from least to greatest.
Now that all the numbers are in decimal form, make two lists of decimal numbers – negatives and positives. The most places after the decimal point in the given numbers is 6. The decimal numbers that you determined with your calculator need only have 6 numbers after the decimal point.
\begin{align*}-5.38, -\frac{21}{4}, -5, -\pi,\sqrt{7}, -1.666, 0, 1.234234, \frac{\pi}{2},\frac{22}{7}, 7, 8.32\end{align*}
### Vocabulary
Inequality
An inequality is a mathematical statement relating expressions by using one or more inequality symbols. The inequality symbols are \begin{align*}>,<,\ge,\le\end{align*}
Integer
All natural numbers, their opposites, and zero are integers. A number in the list \begin{align*}\ldots, -3, -2, -1, 0, 1, 2, 3 \ldots\end{align*}
Irrational Numbers
The irrational numbers are those that cannot be expressed as the ratio of two numbers. The irrational numbers include decimal numbers that are non-terminating decimals as well as non-periodic decimal numbers.
Natural Numbers
The natural numbers are the counting numbers and consist of all positive, whole numbers. The natural numbers are numbers in the list \begin{align*}1, 2, 3\ldots\end{align*} and are often referred to as positive integers.
Number Line
A number line is a line that matches a set of points and a set of numbers one to one. It is often used in mathematics to show mathematical computations.
Rational Numbers
The rational numbers are numbers that can be written as the ratio of two numbers \begin{align*}\frac{a}{b}\end{align*} and \begin{align*}b \ne 0\end{align*}. The rational numbers include all terminating decimals as well as periodic decimal numbers.
Real Numbers
The rational numbers and the irrational numbers make up the real numbers.
Set Notation
Set notation is a mathematical statement that shows an inequality and the set of numbers to which the variable belongs. \begin{align*}\{x|x \ge -3, x \ \varepsilon \ I\}\end{align*} is an example of set notation.
### Guided Practice
1. Arrange the following numbers in order from least to greatest and place them on a number line.
\begin{align*}-3.78, -\frac{11}{4},-4, \frac{\pi}{2}, -\sqrt{6},-1.888,0,0.2424,\pi,\frac{21}{15},2,1.75\end{align*}
2. For each given pair of real numbers, find another real number that is between each of the pairs.
i) \begin{align*}-3,-5\end{align*}
ii) \begin{align*}-3.4,-3.5\end{align*}
iii) \begin{align*}\frac{1}{5},\frac{3}{10}\end{align*}
iv) \begin{align*}-\frac{3}{4},-\frac{11}{6}\end{align*}
3. Use technology to sort the following numbers:
\begin{align*}\sqrt{\frac{3}{5}},\frac{15}{38},-\frac{7}{12},\frac{1}{4},0,\sqrt{8},-\frac{13}{21},-\pi,\frac{3 \pi}{5},-6,3\end{align*}
1. \begin{align*}-3.78, -\frac{11}{4},-4, \frac{\pi}{2}, -\sqrt{6},-1.888,0,0.2424,\pi,\frac{21}{15},2,1.75\end{align*}
2. i) The number must be greater than -5 and less than -3. \begin{align*}-5,{\color{blue}-4},-3\end{align*}
ii) The number must be greater than -3.5 and less than -3.4. \begin{align*}-3.5,{\color{blue}-3.45},-3.4\end{align*}
iii) The number must be greater than \begin{align*}\frac{1}{5}\end{align*} and less than \begin{align*}\frac{3}{10}\end{align*}. Write each fraction with a common denominator. \begin{align*}\frac{1}{5}=\frac{2}{10}\end{align*}. If you look at \begin{align*}\frac{2}{10}\end{align*} and \begin{align*}\frac{3}{10}\end{align*}, there is no fraction, with a denominator of 10, between these values. Write the fractions with a larger common denominator. \begin{align*}\frac{1}{5}=\frac{4}{20},\frac{3}{10}=\frac{6}{20}\end{align*}. If you look at \begin{align*}\frac{4}{20}\end{align*} and \begin{align*}\frac{6}{20}\end{align*}, the fraction \begin{align*}\frac{5}{20}=\frac{1}{4}\end{align*} is between them. \begin{align*}\frac{1}{5}, \frac{{\color{blue}1}}{{\color{blue}4}}, \frac{3}{10}\end{align*}
iv) The number must be greater than \begin{align*}-\frac{3}{4}\end{align*}and less than \begin{align*}-\frac{11}{16}\end{align*}. Write each fraction with a common denominator. \begin{align*}-\frac{3}{4}=-\frac{12}{16}\end{align*}. If you look at \begin{align*}-\frac{12}{16}\end{align*} and \begin{align*}-\frac{11}{16}\end{align*}, there is no fraction, with a denominator of 16 between these values. Write the fractions with a larger common denominator. \begin{align*}-\frac{3}{4}=-\frac{24}{32},-\frac{11}{16}=-\frac{22}{32}\end{align*}. If you look at \begin{align*}-\frac{24}{32}\end{align*} and \begin{align*}-\frac{22}{32}\end{align*}, the fraction \begin{align*}-\frac{23}{32}\end{align*} is between them. \begin{align*}-\frac{3}{4},-{\color{blue}\frac{23}{32}},-\frac{11}{16}\end{align*}
3. \begin{align*}\sqrt{\frac{3}{5}},\frac{15}{38},-\frac{7}{12},\frac{1}{4},0,\sqrt{8},-\frac{13}{21},-\pi,\frac{3 \pi}{5},-6,3\end{align*}
The numbers have been sorted. The numbers from least to greatest are:
\begin{align*}-6,-\pi,-\frac{13}{21},-\frac{7}{12},0,\frac{1}{4},\frac{15}{38},\sqrt{\frac{3}{5}},\frac{3 \pi}{5},\sqrt{8},3\end{align*}
### Practice
Arrange the following numbers in order from least to greatest and place them on a number line.
1. \begin{align*}\{0.5,0.45,0.65,0.33,0,2,0.75,0.28\}\end{align*}
2. \begin{align*}\{0.3,0.32,0.21,0.4,0.3,0,0.31\}\end{align*}
3. \begin{align*}\{-0.3,-0.32,-0.21,-0.4,-0.3,0,-0.31\}\end{align*}
4. \begin{align*}\{\frac{1}{2},-2,0,-\frac{1}{3},3,\frac{2}{3},-\frac{1}{2}\}\end{align*}
5. \begin{align*}\{0.3,-\sqrt{2},1,-0.25,0,1.8,-\frac{\pi}{3}\}\end{align*}
For each given pair of real numbers, find another real number that is between each of the pairs.
1. \begin{align*}8,10\end{align*}
2. \begin{align*}-12,-13\end{align*}
3. \begin{align*}-12.01,-12.02\end{align*}
4. \begin{align*}-7.6,-7.5\end{align*}
5. \begin{align*}\frac{1}{7},\frac{4}{21}\end{align*}
6. \begin{align*}\frac{2}{5},\frac{7}{9}\end{align*}
7. \begin{align*}-\frac{2}{9},-\frac{3}{18}\end{align*}
8. \begin{align*}-\frac{3}{5},-\frac{1}{2}\end{align*}
Use technology to sort the following numbers:
1. \begin{align*}\{-2,\frac{2}{3},0,\frac{3}{8},-\frac{7}{5},\frac{1}{2},4,-3.6\}\end{align*}
2. \begin{align*}\{\sqrt{10},-1,\frac{7}{12},3,-\frac{5}{4},-\sqrt{7},0,-\frac{2 \pi}{3},-\frac{3}{5}\}\end{align*}
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Elementary row/column operations on matrices
The elementary row/column operations on matrices are:
1) Multiplying the $i$-th row (column) by a number $\lambda \ne 0$.
$R_i := \lambda \cdot R_i$
2) Adding the $j$-th row/column multiplied by a number $\lambda$ to the $i$-th row/column.
$R_i := R_i + \lambda \cdot R_j$
3) Exchanging the rows/columns $i$ and $j$.
$R = R_i$
$R_i = R_j$
$R_j = R$
The interesting thing is that for square matrices each of these operations can be accomplished by matrix multiplication.
Let us define:
• $E$ - the identity matrix of order $n \times n$.
• $E_{ij}$ - the square matrix of order $n \times n$ which has an element $1$ at position $(i, j)$ and zeroes at all other positions.
• $A$ - any square matrix of order $n \times n$.
Then one can show that:
(1) Multiplying the i-th row/column of A by the number $\lambda \ne 0$ is accomplished by left/right multiplying A with the matrix $A_i(\lambda) = E + (\lambda-1)E_{ii}$
(2A) Adding the $j$-th row multiplied by a number $\lambda$ to the $i$-th row is accomplished by left multiplying A with the matrix $B_{ij}(\lambda) = E + \lambda E_{ij}$
(2B) Adding the $j$-th column multiplied by a number $\lambda$ to the $i$-th column is accomplished by right multiplying A with the matrix $B_{ji}(\lambda) = E + \lambda E_{ji}$
(3) Exchanging the rows/columns $i$ and $j$ is accomplished by left/right multiplying A with the matrix $C_{ij} = E - E_{ii} - E_{jj} + E_{ij} + E_{ji}$
The matrices $A_i(\lambda), B_{ij}(\lambda), C_{ij}$ are usually called matrices of the elementary transformations. |
# Placing and Identifying Fractions on The Number Line
## Presentation on theme: "Placing and Identifying Fractions on The Number Line"— Presentation transcript:
Placing and Identifying Fractions on The Number Line
In this Power Point students will learn to place and identify fractions on the number line. These strategies can be used to help students successfully use a number line to represent and order fractions. 4.NF.A.1, 4.NF.A.2, 4.NF.B.3
A number line helps us visualize the order of numbers.
Students have used a number line since early elementary years to count and understand values of numbers. As the number line continues from left to right they learn the value of the whole number increases. Students have used number lines to understand math concepts such as adding & subtracting, and greater than, less than in comparing whole numbers.
A number line helps us compare two or more numbers.
5 < 9
We can use a number line to place fractions
We can use a number line to place fractions. Let’s look at the section between 0 and 1 enlarged. A number line can be used to place fractions in order from least to greatest. It is important for the student to understand that a fraction is part of a whole number. Emphasize that we will be looking at an enlarged portion of the number line between 0 and 1 to place fractions. It is important to mention that fractions can also be larger than 1 but the whole is represented at the area between two whole numbers. 1
To place a fraction on a number line we have to think about the space between the whole numbers as the whole. 1
First, look at the denominator of the fraction and think…
Into how many equal parts will the whole be divided? The denominator tells us how many equal parts the whole is divided. 1 Divide the whole into four equal parts.
Next, look at the numerator of the fraction and think…
How many of the equal parts are we talking about? The numerator tells us how many equal parts we are talking about. 1 Shade one of the equal parts.
The point at which the shaded part ends on the number line is the placement for the fraction.
1 Place a point on the number line where the shaded part ends. Write the fraction under the point.
This is what the fraction on the number line would look like.
1
Let’s watch a video that shows this strategy for placing fractions on a number line.
This links to the LearnZillion lesson on the strategy.
Look at the denominator and divide the whole into equal pieces. Look at the numerator and shade in the number of pieces we are talking about. Place a point on the number line at the location where the shaded part ends. Label the fraction under the number line. On an interactive whiteboard, students can follow the steps to determine the location of the point for the given fraction on the number line. Be sure they follow each step to demonstrate the strategy. On the third click of the frame, the Answer Check will come up and should align with the student’s work if correct. 1 Answer Check
Look at the denominator and divide the whole into equal pieces. Look at the numerator and shade in the number of pieces we are talking about. Place a point on the number line at the location where the shaded part ends. Label the fraction under the number line. 1 Answer Check
Look at the denominator and divide the whole into equal pieces. Look at the numerator and shade in the number of pieces we are talking about. Place a point on the number line at the location where the shaded part ends. Label the fraction under the number line. 1 Answer Check
What if the numerator is greater than 1
What if the numerator is greater than 1? Watch the video explaining how to place a fraction with a numerator greater than 1 on the number line. (begin at 1 minute). Begin this video at 1 minute to bypass explanation of number lines that was covered in video 1.
Look at the denominator and divide the whole into equal pieces. Look at the numerator and shade in the number of pieces we are talking about. Place a point on the number line at the location where the shaded parts end. Label the fraction under the number line. On an interactive whiteboard, students can follow the steps to determine the location of the point for the given fraction on the number line. This example and the following two examples will show more than one part in the numerator. Be sure they follow each step to demonstrate the strategy. On the third click of the frame, the Answer Check will come up and should align with the student’s work if correct. 1 Answer Check
Look at the denominator and divide the whole into equal pieces. Look at the numerator and shade in the number of pieces we are talking about. Place a point on the number line at the location where the shaded parts end. Label the fraction under the number line. 1 Answer Check
Look at the denominator and divide the whole into equal pieces. Look at the numerator and shade in the number of pieces we are talking about. Place a point on the number line at the location where the shaded parts end. Label the fraction under the number line. 1 Answer Check
Now let’s identify fractions on the number line.
Follow these steps. Block the whole. Determine the denominator by counting the number of equal pieces the whole is divide into. Fill in the equal parts to the point on the number line. Determine the numerator by counting the number of equal pieces you have filled in. Write the fraction under the point on the number line. The last part of the Power Point is identifying fractions when the point is given on the number line. Students will use the same strategy in a different way in order to derive the name of the fraction represented. 1 3
You Try! Follow these steps. Block the whole.
Determine the denominator by counting the number of equal pieces the whole is divide into. Fill in the equal parts to the point on the number line. Determine the numerator by counting the number of equal pieces you have filled in. Write the fraction under the point on the number line. On an interactive white board, students can demonstrate use of the strategy by following the steps in order to name the fraction represented at the point on the number line. A fraction line is already positioned under the point for students to write in the answer. The Answer Check will come up on the third click of the frame. 1
You Try! Follow these steps. Block the whole.
Determine the denominator by counting the number of equal pieces the whole is divide into. Fill in the equal parts to the point on the number line. Determine the numerator by counting the number of equal pieces you have filled in. Write the fraction under the point on the number line. 1
You Try! Follow these steps. Block the whole.
Determine the denominator by counting the number of equal pieces the whole is divide into. Fill in the equal parts to the point on the number line. Determine the numerator by counting the number of equal pieces you have filled in. Write the fraction under the point on the number line. 1
Let's practice! Play “Battleship Number line”
This links to the BrainPop Battleship Number Line game. This activity can be done as a whole group activity on the interactive whiteboard. Select “play” then “fractions”.
More Practice! Play “Find Grampy”
This links to the interactive game Find Grampy. This activity can be done as a whole group activity on the interactive whiteboard. Students must estimate the location of Grampy in fraction form behind a set of hedges that represents a number line.
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# One of the most important applications of limits is the concept of the derivative of a function.
In calculus, the derivative of a function is used in a wide variety of problems, and understanding it is essential to applying it to such problems. The derivative of a function y = f( x) at a point ( x, f( x)) is defined as
if this limit exists. The derivative is denoted by f′ ( x), read “ f prime of x” or “ fprime at x,” and f is said to be differentiable at x if this limit exists (see Figure 1 ).
Figure 1 The derivative of a function as the limit of rise over run.
If a function is differentiable at x, then it must be continuous at x, but the converse is not necessarily true. That is, a function may be continuous at a point, but the derivative at that point may not exist. As an example, the function f( x) = x1/3 is continuous over its entire domain or real numbers, but its derivative does not exist at x = 0. Another example is the function f( x) = | x + 2|, which is also continuous over its entire domain of real numbers but is not differentiable at x = −2. The relationship between continuity and differentiability can be summarized as follows: Differentiability implies continuity, but continuity does not imply differentiability. Example 1: Find the derivative of f( x) = x2 − 5 at the point (2,−1).
hence, the derivative of f( x) = x2 − 5 at the point (2,−1) is 4.
Some others are y′. If this limit exists. f( x)) on the graph of y = f( x). hence. Example 2: Find the instantaneous velocity of at the time t = 3.One interpretation of the derivative of a function at a point is the slope of the tangent line at this point. and D x f( x). D x f. the instantaneous velocity of s( t) = 1/( t + 2) at time t = 3 is −1/25. Another interpretation of the derivative is the instantaneous velocity of a function representing the position of a particle along a line at time t. ( x. A number of different notations are used to represent the derivative of a function y= f( x) with f′ ( x) being most common. The negative velocity indicates that the particle is moving in the negative direction. f( x)) on the graph of y = f( x). where y = s( t). If this limit exists. it is defined to be the slope of the tangent line at the fixed point. df( x)/dx. The derivative may be thought of as a limit of the average velocities between a fixed time and other times that get closer and closer to the fixed time. y = s( t). The derivative may be thought of as the limit of the slopes of the secant lines passing through a fixed point on a curve and other points on the curve that get closer and closer to the fixed point. it is defined to be the instantaneous rate of change at the fixed point ( x. If this limit exists. . df/ dx. and you should be able to use any of these in selected problems. A third interpretation of the derivative is the instantaneous rate of change of a function at a point. it is defined to be the instantaneous velocity at time t for the function. The derivative may be thought of as the limit of the average rates of change between a fixed point and other points on the curve that get closer and closer to the fixed point. dy/dx.
AVERAGE RATE OF CHANGE ( SLOPE ) The average rate is the slope of the line. .Average and Instantaneous Rate of Change: September 7 Average and Instantaneous Rate of Change Hey guys so today in class we took a look at average rate of change and instantaneous rate of change and a little about limits. Now that is out of the way we can find the average velocity using the formula... First we have this very pretty graph that i made in paint! Next we need to make a best fit line or a secant line to make it easier for us to measure the average velocity.
. . but if we simplify it the equation becomes a very memorable one. INSTANTANEOUS VELOCITY ( Derivative of a function ) For Instantaneous velocity you need to use limits to find points that are as close tot he point you want to find the velocity of.. The rest is pretty simple you take your points from the chart you have and plug them in and solve for your slope.It may look a little weird. in the end the formula becomes . In the graph above this would be the information. Y2 = 3 Y1 = 0 X2 = 3 X1 = 1 That is how you find the average velocity ( change of rate)..
2012 Today's Notes: September 7 Here are the examples we looked at in class today. hope I helped a little .. limits. . MB. Canada Friday. September 7. derivative. kind of like a asymptote in a way because it can't touch the point but it gets as close as possible.In this unit a limit means as close you can get to a single point without being on the point. Am i supposed to pick a person after this? I think i am soooo the next scribe is ALEX Posted by Rayna at 4:57 PM No comments: Email ThisBlogThis!Share to TwitterShare to Facebook Labels: average velocity. instantaneous velocity. Rayna Location: Winnipeg. |
# How do you solve -7( x + 3) = - 63?
May 8, 2018
$\implies x = 6$
#### Explanation:
$- 7 \left(x + 3\right) = - 63$
We distribute the LHS
$- 7 x - 21 = - 63$
We add $21$ to both sides
$- 7 x = - 42$
We divide both sides by $- 7$
$\implies x = 6$
May 8, 2018
The answer to the problem is $\textcolor{red}{x = 6}$.
#### Explanation:
Start by distributing $- 7$ to $\left(x + 3\right)$.
$- 7 x - 21 = - 63$
Add $21$ to both sides.
$- 7 x = - 42$
And finally, divide both sides by $- 7$.
$\frac{- 7 x}{-} 7 = \frac{- 42}{-} 7$
$x = 6$
$- 7 \left(x + 3\right) = - 63$
$- 7 \left(6 + 3\right) = - 63$
$- 7 \left(9\right) = - 63$
$- 63 = - 63$
Which proves that the answer is $x = 6$. |
# How do you condense 1/3(log_8y+2log_8(y+4))-log_8(y-1)?
Jan 25, 2017
${\log}_{8} \left(\frac{\sqrt[3]{y {\left(y + 4\right)}^{2}}}{y - 1}\right)$
#### Explanation:
Since $k \log a = \log {a}^{k}$, the expression is equivalent to:
$\frac{1}{3} \left({\log}_{8} y + {\log}_{8} {\left(y + 4\right)}^{2}\right) - {\log}_{8} \left(y - 1\right)$
Since $\log a + \log b = \log \left(a b\right)$, you get:
$\frac{1}{3} {\log}_{8} y {\left(y + 4\right)}^{2} - {\log}_{8} \left(y - 1\right)$
$= {\log}_{8} {\left(y {\left(y + 4\right)}^{2}\right)}^{\frac{1}{3}} - {\log}_{8} \left(y - 1\right)$
Since $\log a - \log b = \log \left(\frac{a}{b}\right)$, you get
${\log}_{8} \left({\left(y {\left(y + 4\right)}^{2}\right)}^{\frac{1}{3}} / \left(y - 1\right)\right)$
that can be written as:
${\log}_{8} \left(\frac{\sqrt[3]{y {\left(y + 4\right)}^{2}}}{y - 1}\right)$ |
0
# Solving the system of equations by the addition method?
How do you solve?:
8x-3y=9
40x-15y=18
Thank you!
### 1 Answer by Expert Tutors
Tamara J. | Math Tutoring - Algebra and Calculus (all levels)Math Tutoring - Algebra and Calculus (al...
4.9 4.9 (51 lesson ratings) (51)
0
I think you mean the elimination method, which would ultimately require you to add the equations, but the main goal is to eliminate one of the variables first so that you can solve for the other variable then use the answer to solve for the other variable.
With this method, you want to manipulate one of the equations, or both is necessary, by multiplying it by a constant that will generate opposite coefficients on either the x or y variable so as to eliminate it when you add the equations.
8x - 3y = 9
40x - 15y = 18
You can choose to eliminate either variable, but we'll have to manipulate the first equation since the coefficients are smaller and we need to multiply them by another constant. Let's eliminate the y-variable. To do so, we will need its coefficient to be +15 since the y coefficient in the second equation is -15 and so we multiply the first equation by -5 to generate this result:
-5·(8x - 3y = 9)
(-5)·8x - (-5)·3y = (-5)·9
-40x + 15y = -45
Now, we add the two equations:
-40x + 15y = -45
+ 40x - 15y = 18
______________________
0x + 0y = -27
0 = -27 ==> not true because 0 ≠ -27
Therefore, since our solution is 0=-27 and that is obviously not true because 0 does not equal -27 then we can conclude that the system has no solution and is, thus, inconsistent.
Thus, |
# What is the answer to an equation?
## What is the answer to an equation?
To solve an equation is to determine the values of the variable that make the equation a true statement. Any number that makes the equation true is called a solution of the equation. It is the answer to the puzzle!
## What is P 4 math?
Explanation: 4P4 means the number of ways (permutations) of arranging 4 items from a collection of 4 items.
How do you find N in algebra?
In an equation, N represents a specific number, not any number. N + 9 = 12 means N is a number which, when added to 9, must give the answer 12. So N can only be the number 3 because only 3 + 9 is equal to 12. An algebraic expression tells us the relationship between numbers.
How do you solve this math problem?
Here are four steps to help solve any math problems easily:
1. Read carefully, understand, and identify the type of problem.
2. Draw and review your problem.
3. Develop the plan to solve it.
4. Solve the problem.
### How do you find the N in an arithmetic sequence?
What Is n in Arithmetic Sequence Formula? In the arithmetic sequence formula for finding the general term,an=a1+(n−1)d a n = a 1 + ( n − 1 ) d , n refers to the number of terms in the given arithmetic sequence.
### How to solve an equation using the equation solver?
To solve your equation using the Equation Solver, type in your equation like x+4=5. The solver will then show you the steps to help you learn how to solve it on your own. Need more problem types?
How to check correct solution for system of equations?
For system of equations x+y=8 and y=x+2, check (correct) solution x=3, y=5: x+y=8 and y=x+2 @ x=3, y=5 For 3xy=18, check (correct) solution x=2, y=3: 3xy=18 @ x=2, y=3
How does the Equation Calculator work in Excel?
The equation calculator allows you to take a simple or complex equation and solve by best method possible. Click the blue arrow to submit and see the result! |
# 10th CBSE math solution for exercise 7.1 part 5
This page 10th CBSE math solution for exercise 7.1 part 5 is going to provide you solution for every problems that you find in the exercise no 7.1
## 10th CBSE math solution for exercise 7.1 part 5
(8) Find the values of y for which the distance between the points P(2,-3) and Q(10,y) is 10 units.
Solution:
Distance between PQ = 10 units
Distance between two points = √(x₂ - x₁) ² + (y₂ - y₁) ²
Here x₁ = 2, y₁ = -3, x₂ = 10 and y₂ = y
= √(10-2)² + (y-(-3))²
√(8)² + (y + 3)² = 10
64 + y² + 6 y + 9 = 100
y² + 6 y + 9 + 64 - 100 = 0
y² + 6 y - 27 = 0
(y + 9) (y - 3) = 0
y + 9 = 0 y - 3 = 0
y = -9 y = 3
In the page 10th CBSE math solution for exercise 7.1 part 5 we are going to see the solution of next problem
(9) If Q(0,1) is equidistant from P(5,-3) and R(x,6),find the values of x.Also find the distances QR and PR.
Solution:
Distance between PQ = QR
Distance between two points = √(x₂ - x₁) ² + (y₂ - y₁) ²
Here x₁ = 5, y₁ = -3, x₂ = 0 and y₂ = 1
PQ = √(0-5)² + (1-(-3))²
= √(2 - x)² + 25
Here x₁ = x, y₁ = 0, x₂ = -2 and y₂ = 9
= √(-2-x)² + (9-0)²
= √(2+x)² + (9)²
= (2+x)² + 81
√(2 - x)² + 25 = (2+x)² + 81
4 + x² - 4 x + 25 = 4 + x² + 4 x + 81
x² - x² - 4 x - 4 x + 4 - 4 = 81 - 25
-8 x = 56
x = -7
Therefore the required point is (-7,0)
(10) Find a relation between x and y such that the points (x,y) is equidistant from the points (3,6) and (-3,4).
Solution:
Point (x,y) is equidistant from the points (3,6) and (-3,4)
√(x-3)² + (y-6)² = (x+3)² + (y-4)²
√(x²-6x+9+y²-12y+36) (x²+6x+9+y²-8y+16)
taking squares on both sides
x²-x²+y²-y²-6x-6x-12y+8y+45-25=0
-12x-4y+20 = 0
divide the whole equation by (-4)
3 x + y - 5 = 0 |
# What Is The Difference Between Single Variable And Multivariable Calculus?
What Is The Difference Between Single Variable And Multivariable Calculus? 4. Is The Difference Between These Different Types of Calculus? Regardless of either/or differences it is important to realize something on which the difference lies. To realize this you have to use some of the definitions of variables and other variations on the other, of the difference of Calculus, we will see some of the definitions of different variables in action. 4.1 Variables I am going to look at many different, different ways of meaning the variable of a value. Different ways of getting meaning from multiple words, the different ways of referencing the two values, we will go in this direction. You will understand this by considering one class of non-variables, the variable of “value”. The variable is just example if that variable is a type of continuous variable. 4.2 Variables For the sake of learning a different way of getting meaning from multiple words, the same way of referring to “values” will be the same way of referring to “valuation”. 4.3 Variables When writing, students do not have to worry about if that variable is “value” or a continuous variable. Instead students are taught that “value” and “valuation” are type of variable. The variable is chosen to evaluate each individual value of it. The method of judging that one variable is a “value” of the other variable, which is the one variable that gives the weight to “value” rather than to evaluation. 4.4 Variables used in Metric Calculus Four different “mechanisms” that are used in the Metric Calculus are: 4.1 The ratio of the objective my latest blog post number of points in a 1 × 2 matrix (2 × n) 4.2 Mean: Single variable. These are the values by 1 given: val1, val2,.
## Is Doing Someone’s Homework Illegal?
.., val8, and the 1 greatest common divisors are val3. 4.3 Variance The variable is the average of the two values val1 and val8. Volatility and variation are because a greater element of the variable is less chance to change another value, and thus it is more likely to change all values one way. 4.4 Double Variation, Ratio of Objective Variances 4.4.1.1: “Positives”: the objective variances are all one variable and the total amount of variables. Each two variable has a multiple of that variable — “multiplicity” is the constant. The differences, zero is an improvement but some difference is called “pushing” and “pushing increases” as more variables tend to push the concept of variable after each other. 4.4.2: “Integration”: the objective variances have weight of “value” as one variable, and they improve and decrease when more values are added to the concept. The magnitude of “multiplicity” is the one variable, the variable will increase with more numbers; they also achieve total increase in objective variances. 4.4.3: “Positives”: the objective variances are divided into variable and its proportional change in the sum of variables. |
# Properties of Subtraction of Rational Numbers
We will learn how to use the properties of subtraction of rational numbers to find the difference of two rational numbers.
In subtraction of rational numbers a/b and c/d, we define:
(a/b - c/d) = a/b + (-c/d) = a/b + (additive inverse of c/d)
How to use the properties to solve the subtraction of two rational numbers?
Solved examples using the properties of subtraction of rational numbers:
1. Find the additive inverse of:
(i) 2/3
(ii) -17/9
(iii) 6/-19
(iv) -5/-13
Solution:
(i) Additive inverse of 2/3 is -2/3
(ii) Additive inverse of -17/9 is 17/9.
(iii) In standard form, we write 6/-19 as 6/19.
Hence, its additive inverse is 6/19.
(iv) We may write, -5/-13 = (-5) × (-1)/(-13) × (-1) = 5/13
Hence, its additive inverse is -5/13
2. Subtract 5/7 from 4/5
Solution:
Subtract 5/7 from 4/5
= (4/5 – 5/7)
= 4/5 + (additive inverse of 5/7)
= (4/5 + -5/7)
= {28 + (-25)}/35
= 3/35
3. Subtract -3/5 from -3/4
Solution:
Subtract -3/5 from -3/4
= {-3/4 - (-3/5)}
= -3/4 + (additive inverse of -3/5)
= {-3/4 + 3/5)}, [since, additive inverse of -3/5 is 3/5]
= (-15 + 12)/20
= -3/20
4. The sum of two rational numbers is -7. If one of them is -11/3, find the other.
Solution:
Let the other number be x. Then,
x + -11/3 = -7
⇒ x = -7 + (additive inverse of -11/3)
⇒ x = (-7 + 11/3), [since, additive inverse of -11/3 is 11/3]
⇒ x = (-7/1 + 11/3)
⇒ x = (-21 + 11)/3
⇒ x = -10/3
Hence, the required number is -10/3.
5. What number should be added to -5/6 to get 13/15?
Solution:
Let the required number to be added be x. Then,
-5/6 + x = 13/15
⇒ x = 13/15 + (additive inverse of -5/6)
⇒ x = (13/15 + 5/6), [since, additive inverse of -5/6 is 5/6]
⇒ x = (26 + 25)/30
⇒ x = 51/30
⇒ x = 17/10
Hence, the required number is 17/10.
Rational Numbers
What is Rational Numbers?
Is Every Rational Number a Natural Number?
Is Zero a Rational Number?
Is Every Rational Number an Integer?
Is Every Rational Number a Fraction?
Positive Rational Number
Negative Rational Number
Equivalent Rational Numbers
Equivalent form of Rational Numbers
Rational Number in Different Forms
Properties of Rational Numbers
Lowest form of a Rational Number
Standard form of a Rational Number
Equality of Rational Numbers using Standard Form
Equality of Rational Numbers with Common Denominator
Equality of Rational Numbers using Cross Multiplication
Comparison of Rational Numbers
Rational Numbers in Ascending Order
Representation of Rational Numbers on the Number Line
Addition of Rational Number with Same Denominator
Addition of Rational Number with Different Denominator
Properties of Addition of Rational Numbers
Subtraction of Rational Number with Different Denominator
Subtraction of Rational Numbers
Properties of Subtraction of Rational Numbers
Simplify Rational Expressions Involving the Sum or Difference
Multiplication of Rational Numbers
Properties of Multiplication of Rational Numbers
Rational Expressions Involving Addition, Subtraction and Multiplication
Division of Rational Numbers
Rational Expressions Involving Division
Properties of Division of Rational Numbers
To Find Rational Numbers
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How will Adam Brody do on 12/04/2019 and the days ahead? Let’s use astrology to conduct a simple analysis. Note this is not at all guaranteed – take it with a grain of salt. I will first find the destiny number for Adam Brody, and then something similar to the life path number, which we will calculate for today (12/04/2019). 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 experts.
PATH NUMBER FOR 12/04/2019: We will consider the month (12), the day (04) and the year (2019), turn each of these 3 numbers into 1 number, and add them together. We’ll show you how it works now. First, for the month, we take the current month of 12 and add the digits together: 1 + 2 = 3 (super simple). Then do the day: from 04 we do 0 + 4 = 4. Now finally, the year of 2019: 2 + 0 + 1 + 9 = 12. Now we have our three numbers, which we can add together: 3 + 4 + 12 = 19. This still isn’t a single-digit number, so we will add its digits together again: 1 + 9 = 10. This still isn’t a single-digit number, so we will add its digits together again: 1 + 0 = 1. Now we have a single-digit number: 1 is the path number for 12/04/2019.
DESTINY NUMBER FOR Adam Brody: The destiny number will take the sum of all the letters in a name. Each letter is assigned a number per the below chart:
So for Adam Brody we have the letters A (1), d (4), a (1), m (4), B (2), r (9), o (6), d (4) and y (7). Adding all of that up (yes, this can get tedious) gives 38. This still isn’t a single-digit number, so we will add its digits together again: 3 + 8 = 11. This still isn’t a single-digit number, so we will add its digits together again: 1 + 1 = 2. Now we have a single-digit number: 2 is the destiny number for Adam Brody.
CONCLUSION: The difference between the path number for today (1) and destiny number for Adam Brody (2) is 1. That is less than the average difference between path numbers and destiny numbers (2.667), indicating that THIS IS A GOOD RESULT. But don’t go jumping for joy yet! As mentioned earlier, this is not at all guaranteed. If you want a forecast that we do recommend taking seriously, 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. |
# What is 30 percent of 828?
WRITTEN BY: supportmymoto.com STAFF
#### Answer for What’s 30 % of 828:
30 % *828 =
( 30:100)*828 =
( 30*828):100 =
24840:100 = 248.4
Now we’ve got: 30 % of 828 = 248.4
Query: What’s 30 % of 828?
Step 1: Our output worth is 828.
Step 2: We characterize the unknown worth with {x}.
Step 3: From step 1 above,{828}={100%}.
Step 4: Equally, {x}={ 30%}.
Step 5: This leads to a pair of straightforward equations:
{828}={100%}(1).
{x}={ 30%}(2).
Step 6: By dividing equation 1 by equation 2 and noting that each the RHS (proper hand facet) of each
equations have the identical unit (%); we’ve got
frac{828}{x}=frac{100%}{ 30%}
Step 7: Once more, the reciprocal of each side offers
frac{x}{828}=frac{ 30}{100}
Rightarrow{x} = {248.4}
Due to this fact, { 30%} of {828} is {248.4}
#### Answer for What’s 828 % of 30:
828 % * 30 =
(828:100)* 30 =
(828* 30):100 =
24840:100 = 248.4
Now we’ve got: 828 % of 30 = 248.4
Query: What’s 828 % of 30?
Step 1: Our output worth is 30.
Step 2: We characterize the unknown worth with {x}.
Step 3: From step 1 above,{ 30}={100%}.
Step 4: Equally, {x}={828%}.
Step 5: This leads to a pair of straightforward equations:
{ 30}={100%}(1).
{x}={828%}(2).
Step 6: By dividing equation 1 by equation 2 and noting that each the RHS (proper hand facet) of each
equations have the identical unit (%); we’ve got
frac{ 30}{x}=frac{100%}{828%}
Step 7: Once more, the reciprocal of each side offers
frac{x}{ 30}=frac{828}{100}
Rightarrow{x} = {248.4}
Due to this fact, {828%} of { 30} is {248.4}
NOTE : Please do not copy - https://supportmymoto.com |
# 1964 AHSME Problems/Problem 19
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Problem 19
If $2x-3y-z=0$ and $x+3y-14z=0, z \neq 0$, the numerical value of $\frac{x^2+3xy}{y^2+z^2}$ is:
$\textbf{(A)}\ 7\qquad \textbf{(B)}\ 2\qquad \textbf{(C)}\ 0\qquad \textbf{(D)}\ -20/17\qquad \textbf{(E)}\ -2$
## Solution 1
If the value of $\frac{x^2+3xy}{y^2+z^2}$ is constant, as the answers imply, we can pick a value of $z$, and then solve the two linear equations for the corresponding $(x, y)$. We can then plug in $(x,y, z)$ into the expression to get the answer.
If $z=1$, then $2x - 3y = 1$ and $x + 3y = 14$. We can solve each equation for $3y$ and set them equal, which leads to $2x - 1 = 14 - x$. This leads to $x = 5$. Plugging in $x=5$ into $x + 3y = 14$ gives $y = 3$. Thus, $(5, 3, 1)$ is one solution to the intersection of the two planes given.
Plugging $(5, 3, -1)$ into the expression gives $\frac{x^2+3xy}{y^2+z^2}$ gives $\frac{25 + 45}{10}$, or $7$, which is answer $\boxed{\textbf{(A)}}$
## Solution 2
If we think of $z$ as a parameter, we get $2x - 3y = z$ and $x + 3y = 14z$. Adding the equations leads to $3x = 15z$, or $x = 5z$. Plugging that into $x + 3y = 14z$ gives $5z + 3y = 14z$, or $y = 3z$. Thus, the intersection of the two planes is given by the parametric line $(5z, 3z, z)$, where $z$ varies along all real numbers.
We plug this in to $\frac{x^2+3xy}{y^2+z^2}$ to get $\frac{25z^2 + 45z^2}{10z^2}$, or $7$, which is answer $\boxed{\textbf{(A)}}$. |
# Accounting Final
Topics: Combination, Permutation, Binomial coefficient Pages: 15 (3525 words) Published: September 26, 2010
Section 5
Permutations and Combinations
In preceding sections we have solved a variety of counting problems using Venn diagrams and the generalized multiplication principle. Let us now turn our attention to two types of counting problems that occur very frequently and that can be solved using formulas derived from the generalized multiplication principle. These problems involve what are called permutations and combinations, which are particular types of arrangements of elements of a set. The sorts of arrangements we have in mind are illustrated in two problems:
Problem A How many words (by which we mean strings of letters) of two distinct letters can be formed from the letters {a, b, c}?
Problem B A construction crew has three members. A team of two must be chosen for a particular job. In how many ways can the team be chosen?
Each of the two problems can be solved by enumerating all possibilities.
Solution of Problem A There are six possible words, namely
abacbabccacb.
Solution of Problem B Designate the three crew members by a, b, and c. Then there are three possible two- person teams, namely
abacbc.
(Note that ba, the team consisting of b and a, is the same as the team ab.)
We deliberately set up both problems using the same letters in order to facilitate comparison. Both problems are concerned with counting the numbers of arrangements of the elements of the set {a, b, c}, taken two at a time, without allowing repetition (for example, aa was not allowed). However, in Problem A the order of the arrangement mattered, whereas in Problem B it did not. Arrangements of the sort considered in Problem A are called permutations, whereas those in Problem B are called combinations.
More precisely, suppose that we are given a set of n objects1.
Then a permutation of n objects taken r at a time is an arrangement of r of the n objects in a specific order. So, for example, Problem A was concerned with permutations of the three objects, a, b, c (n = 3), taken two at a time (r = 2).
A combination of n objects taken r at a time is a selection of r objects from among the n, with order disregarded. Thus, for example, in Problem B we considered combinations of the three objects a, b, c (n = 3), taken two at a time (r = 2).
1All are assumed to be different.
It is convenient to introduce the following notation for counting permutations and combinations. Let
P(n, r) = the number of permutations of n objects taken r at a time
C(n, r) = the number of combinations of n objects taken r at a time.
Thus, for example, from our solutions to Problems A and B, we have
P(3, 2) = 6C(3, 2) = 3.
Very simple formulas for P(n, r) and C(n, r) allow us to calculate these quantities for any n and r. Let us begin by stating the formula for P( n, r). For r = 1, 2, 3 we have, respectively,
|[pic] | | |[pic] |(two factors) | |[pic] |(three factors) |
and, in general,
| | |[pic] |
This formula is verified at the end of this section.
EXAMPLE 1 Applying the permutation formula Compute the following numbers.
(a) P(100, 2)(b) P(6, 4)(c) P(5, 5)
Solution (a) Here n = 100, r = 2. So we take the product of two factors, beginning with 100:
P(100, 2) = [pic] = 9900.
(b) P(6, 4) = 6 5 4 3 = 360
(c) P(5, 5) = 5 4 3 2 1 = 120
Now Try Exercise 1
In order to state the formula for C(n, r), we must introduce some further notation. Suppose that r is any positive... |
# Question Video: Converting Trigonometric Parametric Equations into Rectangular Form Mathematics • Higher Education
Convert the parametric equations π₯ = cos (π‘) and π¦ = sin (π‘) to rectangular form.
01:54
### Video Transcript
Convert the parametric equations π₯ is equal to the cos of π‘ and π¦ is equal to the sin of π‘ to rectangular form.
Weβre given a pair of parametric equations, and weβre asked to convert this into the rectangular form. Remember, this means we need to rewrite this as an equation in terms of π₯ and π¦. We want to eliminate our variable π‘. The first thing weβll do is look at our parametric equations. We can see we have π₯ is equal to the cos of π‘ and π¦ is equal to the sin of π‘. To eliminate the variable π‘, weβre going to want to find some kind of relationship between the cos of π‘ and the sin of π‘.
And luckily, we do know a relationship between these two functions. The Pythagorean theorem tells us the sin squared of π plus the cos squared of π will be equivalent to one for any value of π. And we used the equivalent sign here because this is true for any value π. However, you might see this written with an equal sign. Of course, it doesnβt matter what we label our variable in the Pythagorean theorem. However, because in the question weβre working with the cos of π‘ and the sin of π‘, weβll relabel our variable π‘.
So now we have for any value π‘, the sin squared of π‘ plus the cos squared of π‘ is equal to one. Of course, we can then rewrite the sin squared of π‘ as the sin of π‘ all squared and the cos squared of π‘ as the cos of π‘ all squared. And the cos of π‘ is equal to π₯, and the sin of π‘ is equal to π¦. So what we need to do is replace the sin of π‘ with π¦ and the cos of π‘ with π₯. This gives us π¦ squared plus π₯ squared is equal to one. And itβs worth pointing out here this is no longer an identity. Itβs now an equation. This is because itβs only true for certain pairs π₯ and π¦ whereas, before, our Pythagorean identity was true for any real value of π‘.
We could leave our answer like this. However, weβll also reorder our π₯- and π¦-terms, giving us π₯ squared plus π¦ squared is equal to one. Therefore, we were able to convert the parametric equations π₯ is equal to the cos of π‘ and π¦ is equal to sin of π‘ into rectangular form. We got π₯ squared plus π¦ squared will be equal to one. |
## What is the value of sin 45 in trigonometry?
= 1 / 2
Sine 0° 0
Sine 30° or Sine π/6 1/2
Sine 45° or Sine π/4 1 / 2
Sine 60°or Sine π/3 3 / 2
Sine 90° or Sine π/2 1
What is Sohcatoa?
SOHCAHTOA is a mnemonic device used to remember the ratios of sine, cosine, and tangent in trigonometry.
### What does tan 45 mean?
Explanation: For tan 45 degrees, the angle 45° lies between 0° and 90° (First Quadrant). Since tangent function is positive in the first quadrant, thus tan 45° value = 1. Since the tangent function is a periodic function, we can represent tan 45° as, tan 45 degrees = tan(45° + n × 180°), n ∈ Z.
What is the exact value of tan 45?
The exact value of tan(45°) tan ( 45 ° ) is 1 .
#### How do you solve cos 45 degrees?
How can you evaluate the value of cos 45? We can use Pythagoras theorem to find the value of cos 45. Since, for a right angle, if the adjacent angles are 45 degrees, then the adjacent side and opposite side will be equal. Thus, we can find the value of cos 45 equal to 1/√2.
What is the exact value of sin 45 enter your answer as a simplified fraction?
The exact value of sin(45) is √22 .
## What is cosec in math?
Cosecant Formula. Cosecant is one of the six trigonometric ratios which is also denoted as cosec or csc. The cosecant formula is given by the length of the hypotenuse divided by the length of the opposite side in a right triangle.
What is csc on calculator?
The cosecant formula is: csc(α) = hypotenuse c opposite a. Thus, the cosecant of angle α in a right triangle is equal to the length of the hypotenuse c divided by the opposite side a. To solve csc, simply enter the length of the hypotenuse and opposite side, then solve.
### What is the value of cos0?
The value of cos 0 is 1. Here, we will discuss the value for cos 0 degrees and how the values are derived using the quadrants of a unit circle. The trigonometric functions are also known as an angle function that relates the angles of a triangle to the length of the triangle sides.
How do you find the value of sin 45?
How to Find the Value of Sin 45 Degrees? The value of sin 45 degrees can be calculated by constructing an angle of 45° with the x-axis, and then finding the coordinates of the corresponding point (0.7071, 0.7071) on the unit circle. The value of sin 45° is equal to the y-coordinate (0.7071). ∴ sin 45° = 0.7071.
#### How do you solve tan 45?
To find the value of tan 45 degrees using the unit circle:
1. Rotate ‘r’ anticlockwise to form 45° angle with the positive x-axis.
2. The tan of 45 degrees equals the y-coordinate(0.7071) divided by x-coordinate(0.7071) of the point of intersection (0.7071, 0.7071) of unit circle and r.
How do you write tan 45 degrees?
The value of tan 45 degrees is 1. Tan 45 degrees in radians is written as tan (45° × π/180°), i.e., tan (π/4) or tan (0.785398. . .). In this article, we will discuss the methods to find the value of tan 45 degrees with examples. Tan 45° in radians: tan (π/4) or tan (0.7853981 . . .) |
# Class 11 RD Sharma Solutions – Chapter 29 Limits – Exercise 29.8 | Set 1
• Last Updated : 08 May, 2021
### Question 1. limx→π/2[π/2 – x].tanx
Solution:
We have,
limx→π/2[π/2 – x].tanx
Let us considered, y = [π/2 – x]
Here, x→π/2, y→0
= limy→0[y.tan(π/2 – y)]
= limy→0[y.{sin(π/2 – y)/cos(π/2 – y)]
= limy→0[y.{cosy/siny}]
= limy→0[y/siny].cosy
= limy→0[cosy] [Since, limy→0[siny/y] = 1]
= 1
### Question 2. limx→π/2[sin2x/cosx]
Solution:
We have,
limx→π/2[sin2x/cosx]
= limx→π/2[2sinx.cosx/cosx]
= 2Limx→π/2[sinx]
= 2
### Question 3. limx→π/2[cos2x/(1 – sinx)]
Solution:
We have,
limx→π/2[cos2x/(1 – sinx)]
= limx→π/2[(1 – sin2x)/(1 – sinx)]
= limx→π/2[(1 – sinx)(1 + sinx)/(1 – sinx)]
= limx→π/2[(1 + sinx)]
= 1 + 1
= 2
### Question 4. limx→π/2[(1 – sinx)/cos2x]
Solution:
We have,
limx→π/2[(1 – sinx)/cos2x]
= limx→π/2[(1 – sinx)/(1 – sin2x)]
= limx→π/2[(1 – sinx)/(1 – sinx)(1 + sinx)]
= limx→π/2[1/(1 + sinx)]
= 1/(1 + 1)
= 1/2
### Question 5. limx→a[(cosx – cosa)/(x – a)]
Solution:
We have,
limx→a[(cosx – cosa)/(x – a)]
=
= -2sin[(a + a)/2] × 1 × (1/2)
= -sina
### Question 6. limx→π/4[(1 – tanx)/(x – π/4)]
Solution:
We have,
limx→π/4[(1 – tanx)/(x – π/4)]
Let us considered, y = [x – π/4]
Here, x→π/4, y→0
= -2 × 1 × [1/(1 – 0)]
= -2
### Question 7. limx→π/2[(1 – sinx)/(π/4 – x)2]
Solution:
We have,
limx→π/2[(1 – sinx)/(π/4 – x)2]
Let us considered, y = [π/2 – x]
Here, x→π/2, y→0
= limy→0[(1 – cosy)/y2]
= 2 × 1 × (1/4)
= (1/2)
### Question 8. limx→π/3[(√3 – tanx)/(π – 3x)]
Solution:
We have,
limx→π/3[(√3 – tanx)/(π – 3x)]
Let us considered, y = [π/3 – x]
When, x→π/3, y→0
= (4/3) × 1 × [1/(1 + 0)]
= (4/3)
### Question 9. limx→a[(asinx – xsina)/(ax2 – xa2)]
Solution:
We have,
limx→a[(asinx – xsina)/(ax2 – xa2)]
= limx→a[(asinx – xsina)/{ax(x – a)}]
Let us considered, y = [x – a]
When, x→a, y→0
= limy→0[{asin(y + a) – (y + a)sina)}/{a(y + a)y}]
= limy→0[(a.siny.cosa + asina.cosy – ysina – asina)/{a(y + a)y}]
= limy→0[{a.siny.cosa + a.sina.(cosy – 1) – y.sina}/{a(y + a)y}]
= limy→0[{a.siny.cosa + a.sina.2sin2(y/2) – t.sina}/{a(y + a)y}]
= limy→0[a.siny.cosa/a(y + a)y] – limy→0[2.a.sina.sin2(y/2)/a(y + a)y] + limy→0[y.sina/a(a + y)y]
= [(a.cosa)/a2] – [(sina)/a2] + 0
= [(a.cosa – sina)/a2]
### Question 10. limx→π/2[{√2 – √(1 + sinx)}/cos2x]
Solution:
We have,
limx→π/2[{√2 – √(1 + sinx)}/cos2x]
On rationalizing the numerator, we get
= limx→π/2[{2 – (1 + sinx)}/cos2x{√2 + √(1 + sinx)}]
= limx→π/2[(1 – sinx)/(1 – sin2x){√2 + √(1 + sinx)}]
= limx→π/2[(1 – sinx)/(1 – sinx)(1 + sinx){√2 + √(1 + sinx)}]
= limx→π/2[1/(1 + sinx){√2 + √(1 + sinx)}]
= 1/{(1 + 1)(√2 + √2)}
= 1/4√2
### Question 11. limx→π/2[{√(2 – sinx) – 1}/(π/2 – x)2]
Solution:
We have,
limx→π/2[{√(2 – sinx) – 1}/(π/2 – x)2]
Let us considered, y = [π/2 – x]
Here, x→π/2, y→0
=
= limy→0[{√(2 – cosy) – 1}/y2]
On rationalizing the numerator, we get
= limy→0[{(2 – cosy) – 1}/y2{√(2 – cosy) – 1}]
= limy→0[{1 – cosy}/y2{√(2 – cosy) – 1}]
= 2/4(1 + 1)
= 1/4
### Question 12. limx→π/4[(√2 – cosx – sinx)/(π/4 – x)2]
Solution:
We have,
limx→π/4[(√2 – cosx – sinx)/(π/4 – x)2]
= 2√2/4
= (1/√2)
### Question 13. limx→π/8[(cot4x – cos4x)/(π – 8x)3]
Solution:
We have,
limx→π/8[(cot4x – cos4x)/(π – 8x)3]
= limx→π/8[(cot4x – cos4x)/83(π/8 – x)3]
Let us considered, (π/8 – x) = y
When x→π/8, y→0
=
= limx→0[(tan4x-sin4x)/83(π/8-x)3]
= limx→0[(sin4x/cos4x-sin4x)/83(π/8-x)3]
=
=
=
=
=
= (2 × 4 × 1 × 4 × 1)/(83)
= 1/16
### Question 14. limx→a[(cosx – cosa)/(√x – √a)]
Solution:
We have,
limx→a[(cosx – cosa)/(√x – √a)]
On rationalizing the denominator, we get
= -2 × sina × 1 × (1/2) × 2√a
= -2√a.sina
### Question 15. limx→π[{√(5 + cosx) – 2}/(π – x)2]
Solution:
We have,
limx→π[{√(5 + cosx) – 2}/(π – x)2]
Let us considered, y = [π – x]
When, x→π, y→0
= limy→0[{√(5 – cosy) – 2}/y2]
On rationalizing the numerator, we get
= limy→0[{1 – cosy}/y2{√(5 – cosy)-2}]
= 2 × (1/4) × {1/(2 + 2)}
= (1/8)
### Question 16. limx→a[(cos√x – cos√a)/(x – a)]
Solution:
We have,
limx→a[(cos√x – cos√a)/(x – a)]
= -2sin√a × 1 × (1/2√a) × (1/2)
= -(sin√a/2√a)
### Question 17. limx→a[(sin√x – sin√a)/(x – a)]
Solution:
We have,
limx→a[(sin√x – sin√a)/(x – a)]
= 2cos√a × 1 × (1/2√a) × (1/2)
= (cos√a/2√a)
### Question 18. limx→1[(1 – x2)/sin2πx]
Solution:
We have,
limx→1[(1 – x2)/sin2πx]
When, x→1, h→0
= limh→0[{1-(1-h)2}/sin2π(1-h)]
= limh→0[(2h-h2)/-sin2πh]
= limh→0[{h(2-h)}/sin2πh]
=
= -2/2π
= -1/π
### Question 19. limx→π/4[{f(x) – f(π/4)}/{x – π/4}]
Solution:
We have,
limx→π/4[{f(x) – f(π/4)}/{x – π/4}]
When, x→π/4, h→0
= limh→0[{f(π/4 + h) – f(π/4)}/{π/4 + h – π/4}]
It is given that f(x) = sin2x
= limh→0[{sin(π/2 + 2h) – sin(π/2)}/h]
= limh→0[(cos2h – 1)/h]
= limh→0[{-2sin2h}/h]
= -2Limh→0[(sinh/h)2] × h
= -2 × 1 × 0
= 0
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# Class 9 RD Sharma Solutions – Chapter 15 Areas of Parallelograms and Triangles- Exercise 15.3 | Set 2
### Question11. If P is any point in the interior of a parallelogram ABCD, then prove that area of the triangle APB is less than half the area of parallelogram.
Solution:
According to the question
ABCD is a parallelogram and P is any point in the interior of parallelogram
Prove: ar(ΔAPB) < 1/2 ar(|| gm ABCD)
Proof:
Construction: Draw DN ⊥ AB and PM ⊥ AB
Now, we find the area of ∥gm ABCD
= AB × DN,
Now, the area of ΔAPB
= (1/2) (AB × PM)
Now, PM < DN
AB × PM < AB × DN
(1/2)(AB × PM) < (1/2)(AB × DN)
ar(ΔAPB) < 1/2 ar(∥ gm ABCD)
Hence proved
### Question 12. If AD is a median of a triangle ABC, then prove that triangles ADB and ADC are equal in area. If G is the mid-point of the median AD, prove that ar(ΔBGC) = 2ar(ΔAGC).
Solution:
According to the question
AD is a median of a triangle ABC
Now, construct AM ⊥ BC
It is given that AD is the median of ΔABC
So, BD = DC
BD = AM = DC × AM
(1/2)(BD × AM) = (1/2)(DC × AM)
ar(ΔABD) = ar(ΔACD) ….(i)
In ΔBGC,
GD is the median
So, ar(ΔBGD) = ar(ΔCGD) …..(ii)
In ΔACD,
CG is the median
So, ar(ΔAGC) = ar(ΔCGD) ……..(iii)
From equation (ii) and (iii), we get
ar(ΔBGD) = ar(ΔAGC)
But, ar(ΔBGC) = 2ar(ΔBGD)
Hence, ar(ΔBGC) = 2ar(ΔAGC)
Hence proved
### Question 13. A point D is taken on the side BC of a ΔABC, such that BD = 2DC. Prove that ar(ΔABD) = 2ar(ΔADC).
Solution:
According to the question
BD = 2DC(In ΔABC)
Proof:
Now draw a point E on BD such that BE = ED
Since, BE = ED and BD = 2 DC
So, BE = ED = DC
As we know that, in triangles, median divides the triangle into two equal triangles.
In ΔABD,
AE is the median.
So, ar(ΔABD) = 2ar(ΔAED) …..(i)
In ΔAEC,
From eq(i) and (ii), we get
Hence proved
### (ii) ar(ΔABP) = 2ar(ΔCBP).
Solution:
According to the question
ABCD is the parallelogram, whose diagonals intersect at O
So, AO = OC and BO = OD
(i) Prove that ar(ΔADO) = ar(ΔCDO)
Proof:
In ΔDAC,
DO is a median.
Hence proved
(ii) Prove that ar(ΔABP) = 2ar(ΔCBP)
Proof:
In ΔBAC,
BO is a median.
So, ar(ΔBAO) = ar(ΔBCO) …..(i)
In ΔPAC,
PO is a median.
So, ar(ΔPAO) = ar(ΔPCO) …..(ii)
Now, subtract eq(ii) from (i), we get
ar(ΔBAO) − ar(ΔPAO) = ar(ΔBCO) − ar(ΔPCO)
ar(ΔABP) = 2ar(ΔCBP)
Hence proved
### (ii) If the area of ΔDFB = 3 cm2, find the area of ∥ gm ABCD.
Solution:
According to the question
ABCD is a parallelogram in which BC is produced to E
such that CE = BC and AE intersects CD at F
(i) Prove that ar(ΔADF) = ar(ΔECF)
Proof:
∠ADF = ∠ECF (Alternate interior angles)
∠DFA = ∠CFA (Vertically opposite angle)
So, by AAS congruence
By c.p.c.t
DF = CF
Hence proved
(ii) From the question area of ΔDFB = 3 cm2
Find the area of ||gm ABCD
Now, DF = CF (Proved above)
BF is a median in Δ BCD.
ar(ΔBCD) = 2ar(ΔDFB)
ar(ΔBCD) = 2 × 3 cm2 = 6 cm2
Now we find the area of a parallelogram = 2ar(ΔBCD)
= 2 × 6 cm2 = 12 cm2
Hence the area of parallelogram is 12 cm2
### Question 16. ABCD is a parallelogram whose diagonals AC and BD intersect at O. A line through O intersects AB at P and DC at Q. Prove that ar(ΔPOA) = ar(ΔQOC).
Solution:
Prove: ar(ΔPOA) = ar(ΔQOC)
In ΔPOA and QOC,
∠AOP = ∠COQ (Vertically opposite angle)
AO = OC
∠PAC = ∠QCA (Alternate angle)
So, by ASA congruence criterion, we have
ΔPOA ≅ ΔQOC
Hence ar(ΔPOA) = ar(ΔQOC)
Hence proved
### Question 17. ABCD is a parallelogram. E is a point on BA such that BE = 2EA and F is point on DC such that DF = 2FC. Prove that AECF is a parallelogram whose area is one third of the area of parallelogram ABCD.
Solution:
Prove: ar(∥gm AECF) = 1/3 ar(||gm ABCD)
Proof:
First construct FG ⊥ AB
Now, according to the question
BE = 2 EA and DF = 2FC
AB – AE = 2 AE and DC – FC = 2 FC
AB = 3 AE and DC = 3 FC
AE = (1/3) AB and FC = (1/3)DC …….(i)
But AB = DC
So, AE = FC (Opposite sides of the parallelogram)
AE = FC and AE ∥ FC
So, AECF is a parallelogram
Now, we find the area of parallelogram AECF
= AE × FG
Put the value of AE from eq(i), we get
ar(||gm AECF) = 1/3 AB × FG
3ar (||gm AECF) = AB × FG ….(ii)
and ar(||gm ABCD) = AB × FG ….(iii)
From eq(ii) and (iii), we get
3ar(||gm AECF) = ar(||gm ABCD)
Hence ar(||gm AECF) = 1/3 ar(||gm ABCD)
Hence proved
### (iii) ar(ΔRQC) = 3/8 ar(ΔABC).
Solution:
According to the question
ABC is a triangle, in which P, Q, and R are the mid-points of AB, BC and AP
(i) Prove that ar(ΔPBQ) = ar(ΔARC)
Proof:
CR is the median of ΔCAP
So, ar(ΔCRA) = (1/2) ar(ΔCAP) ….(i)
CP is the median of a ΔCAB
So, ar(ΔCAP) = ar(ΔCPB) ….(ii)
So, from eq(i) and (ii), we conclude that
ar(ΔARC) = (1/2) ar(ΔCPB) ….(iii)
Now, PQ is the median of a ΔPBC
So, ar(ΔCPB) = 2ar(ΔPBQ) ….(iv)
From eq(iii) and (iv), we conclude that
ar(ΔARC) = ar(ΔPBQ) ….(v)
Hence proved
(ii) Now QP and QR are the medians of triangles QAB and QAP
So, ar(ΔQAP) = ar(ΔQBP) ….(vi)
And ar(ΔQAP) = 2ar(ΔQRP) ….(vii)
From eq(vi) and (vii), we conclude that
ar(ΔPRQ) = (1/2) ar(ΔPBQ) ….(viii)
And from eq (v) and (viii), we conclude that
ar(ΔPRQ) = (1/2) ar(ΔARC)
Hence proved
(iii) Now, LR is a median of ΔCAP
= 1/2 × (1/2) ar(ΔABC)
= (1/4) ar(ΔABC)
and RQ is the median of ΔRBC.
So, ar(ΔRQC) = (1/2) ar(ΔRBC)
= (1/2) {ar(ΔABC) − ar(ΔARC)}
= (1/2) {ar(ΔABC) – (1/4) ar(ΔABC)}
= (3/8) ar(ΔABC)
Hence proved
### Question 19. ABCD is a parallelogram. G is a point on AB such that AG = 2GB and E is point on DC such that CE = 2DE and F is the point of BC such that BF = 2FC. Prove that:
(ii) ar(ΔEGB) = (1/6) ar(ABCD)
(iii) ar(ΔEFC) = (1/2) ar(ΔEBF)
(iv) ar(ΔEGB) = 3/2 × ar(ΔEFC)
(v) Find what portion of the area of parallelogram is the area of ΔEFG.
Solution:
According to the question
ABCD is a parallelogram
So, AG = 2 GB, CE = 2 DE and BF = 2 FC
Now, draw a parallel line to AB that pass through point F and a perpendicular line to AB from C.
(i) Prove that ar(ADEG) = ar(GBCE)
Since it is given that ABCD is a parallelogram
So, AB = CD and AD = BC
Now, let us consider the two trapezium ADEG and GBCE
Since AB = DC, EC = 2DE, AG = 2GB
So, ED = (1/3) CD = (1/3) AB and EC = (2/3) CD = (2/3) AB
AG = (2/3) AB and BG = (1/3) AB
So, DE + AG = (1/3) AB + (2/3) AB = AB and EC + BG = (2/3) AB + (1/3) AB = AB
Since these two trapeziums have same heights, also their sum of two parallel sides are equal,
So, the area of trapezium = sum of parallel side/2 x height
Hence proved
(ii) Prove that ar(ΔEGB) = (1/6) ar(ABCD)
From the above we know that
BG = (1/2) AB
So, ar(ΔEGB) = (1/2) × GB × height
ar(ΔEGB) = (1/2) × (1/3) × AB × height
ar(ΔEGB) = (1/6) × AB × height
Then, ar(ΔEGB) = (1/6) ar(ABCD)
Hence proved
(iii) Prove that ar(ΔEFC) = (1/2) ar(ΔEBF)
As we know that the height of triangle EFC and EBF are equal,
So, ar(ΔEFC) = (1/2) × FC × height
ar(ΔEFC) = (1/2) × (1/2) × FB × height
ar(ΔEFC) = (1/2) ar(EBF)
Then, ar(ΔEFC) = (1/2) ar(ΔEBF)
Hence proved
(iv) Prove that ar(ΔEGB) = 3/2 × ar(ΔEFC)
Let us consider the trapezium EGBC
So, ar(EGBC) = ar(ΔEGB) + ar(ΔEBF) + ar(ΔEFC)
Now put all these values, we get
(1/2) ar(ABCD) = (1/6) ar(ABCD) + 2ar(ΔEFC) + ar(ΔEFC)
(1/3) ar(ABCD) = 3 ar(ΔEFC)
ar(ΔEFC) = (1/9) ar(ABCD)
Now from option (ii), we get,
ar(ΔEGB) = (1/6) ar(ΔEFC)
ar(ΔEGB) = (3/2) × (1/9) ar(ABCD)
ar(ΔEGB) = (3/2) ar(ΔEFC)
Hence, ar(ΔEGB) = (3/2) ar(ΔEFC)
Hence proved
(v) From the figure, we have FB = 2CF,
So, Let us considered CF = x and FB = 2x.
Now, we take triangle CFI and CBH which are similar triangle.
So, by the property of similar triangle, we get
CI = k and IH = 2k
Now, we take ΔEGF in which,
ar(ΔEFG) = ar(ΔESF) + ar(ΔSGF)
ar(ΔEFG) = (1/2) SF × k + (1/2) SF × 2k
ar(ΔEFG) = (3/2) SF × k …….(i)
Now,
ar(ΔEGBC) = ar(SGBF) + ar(ESFC)
ar(ΔEGBC) = (1/2)(SF + GB) × 2k + (1/2)(SF + EC) × k
ar(ΔEGBC) = (3/2) k × SF + (GB + (1/2)EC) × k
ar(ΔEGBC) = (3/2) k × SF + (1/3 AB + (1/2) × (2/3) AB) × k
(1/2) ar(ΔABCD) = (3/2) k × SF + (2/3) AB × k
ar(ΔABCD) = 3k × SF + (4/3) AB × k
ar(ΔABCD) = 3k × SF + 4/9 ar(ABCD)
k × SF = (5/27)ar(ABCD) …….(ii)
From eq(i) and (ii), we conclude that
ar(ΔEFG) = (3/2) × (5/27) ar(ABCD)
ar(ΔEFG) = (5/18) ar(ABCD)
Hence proved
### (iii) Prove that ar(BCZY) = ar(ΔEDZ)
Solution :
According to the question
CD || AE and CY || BA
(i) Here, triangle BCA and triangle BYA are on the same base BA and
between same parallel lines BA and CY.
So, ar(ΔBCA) = ar(ΔBYA) ….(i)
Now as we know that ar(ΔBCA) = ar(ΔCBX) + ar(ΔBXA)
ar(ΔBYA) = ar(ΔBXA) + ar(ΔAXY)
So, put all these values in eq(i), we get
ar(ΔCBX) + ar(ΔBXA) = ar(ΔBXA) + ar(ΔAXY)
So, ar(ΔCBX) = ar(ΔAXY)
(ii) Here, triangles ACE and ADE are on the same base AE and
between same parallels CD and AE
Now as we know that ar(ΔACE) = ar(ΔCZA) + ar(ΔAZE)
and ar(ΔADE) = ar(ΔAZE) + ar(ΔDZE)
So, put all these values in eq(ii), we get
ar(ΔCZA) + ar(ΔAZE) = ar(ΔAZE) + ar(ΔDZE)
ar(ΔCZA) = ar(ΔDZE) ……(iii)
Hence proved
(iii) As we know that ar(ΔCBX) = ar(ΔAXY)
Now, add ar(ΔCYG) on both sides, we get
ar(ΔCBX) + ar(ΔCYZ) = ar(ΔCAY) + ar(ΔCYZ)
ar(BCZY) = ar(ΔCZA) ….(iv)
From eq(iii) and (iv), we conclude that
ar(BCZY) = ar(ΔDZE)
Hence proved
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# Grade 3 Common Core Math: Number and Operations/Fractions 3.NF.A.1-3
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Common Core State Standards - Grade 3 Mathematics
These 13 worksheets contain math problems aligned with the 3rd grade Common Core State Standards (CCSS) listed below.
[3.NF Number and Operations-Fractions]
CCSS.Math.Content.3.NF.A.1 Understand a fraction 1/b as the quantity formed by 1 part when a whole is partitioned into b equal parts; understand a fraction a/b as the quantity formed by a parts of size 1/b.
CCSS.Math.Content.3.NF.A.2 Understand a fraction as a number on the number line; represent fractions on a number line diagram.
CCSS.Math.Content.3.NF.A.2a Represent a fraction 1/b on a number line diagram by defining the interval from 0 to 1 as the whole and partitioning it into b equal parts. Recognize that each part has size 1/b and that the endpoint of the part based at 0 locates the number 1/b on the number line.
CCSS.Math.Content.3.NF.A.2b Represent a fraction a/b on a number line diagram by marking off a lengths 1/b from 0. Recognize that the resulting interval has size a/b and that its endpoint locates the number a/b on the number line.
CCSS.Math.Content.3.NF.A.3 Explain equivalence of fractions in special cases, and compare fractions by reasoning about their size.
CCSS.Math.Content.3.NF.A.3a Understand two fractions as equivalent (equal) if they are the same size, or the same point on a number line.
CCSS.Math.Content.3.NF.A.3b Recognize and generate simple equivalent fractions, e.g., 1/2 = 2/4, 4/6 = 2/3. Explain why the fractions are equivalent, e.g., by using a visual fraction model.
CCSS.Math.Content.3.NF.A.3d Compare two fractions with the same numerator or the same denominator by reasoning about their size. Recognize that comparisons are valid only when the two fractions refer to the same whole. Record the results of comparisons with the symbols >, =, or <, and justify the conclusions, e.g., by using a visual fraction model.
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# LessonAmortization Formula
#### MarkFL
##### La Villa Strangiato
Staff member
Moderator
Math Helper
Let $$P$$ = monthly payment, $$A$$ = amount borrowed, $$i$$ = monthly interest rate, and $$n$$ = the number of payments.
Also, let $$\displaystyle D_n$$ be the debt amount after payment $$n$$.
Consider the recursion:
(1) $$\displaystyle D_{n}=(1+i)D_{n-1}-P$$
(2) $$\displaystyle D_{n+1}=(1+i)D_{n}-P$$
Subtracting (1) from (2) yields the homogeneous recursion:
$$\displaystyle D_{n+1}=(2+i)D_{n}-(1+i)D_{n-1}$$
whose associated auxiliary equation is:
$$\displaystyle r^2-(2+i)r+(1+i)=0$$
$$\displaystyle \left(r-(1+i)\right)\left(r-1\right)=0$$
Thus, the closed-form for our recursion is:
$$\displaystyle D_n=k_1(1+i)^n+k_2$$
Using initial values, we may determine the coefficients $$\displaystyle k_i$$:
$$\displaystyle D_0=k_1+k_2=A$$
$$\displaystyle D_1=k_1(1+i)+k_2=(1+i)A-P$$
Solving this system, we find:
$$\displaystyle k_1=\frac{Ai-P}{i},\,k_2=\frac{P}{i}$$ and so we have:
$$\displaystyle D_n=\left(\frac{Ai-P}{i}\right)(1+i)^n+\left(\frac{P}{i}\right)=\frac{(Ai-P)(1+i)^n+P}{i}$$
Now, equating this to zero, we can solve for P:
$$\displaystyle \frac{(Ai-P)(1+i)^n+P}{i}=0$$
$$\displaystyle (Ai-P)(1+i)^n+P=0$$
$$\displaystyle (P-Ai)(1+i)^n=P$$
$$\displaystyle P\left((1+i)^n-1\right)=Ai(1+i)^n$$
$$\displaystyle P=\frac{Ai(1+i)^n}{(1+i)^n-1}$$
$$\displaystyle P=\frac{Ai}{1-(1+i)^{-n}}$$
harpazo
#### harpazo
##### Pure Mathematics
Let $$P$$ = monthly payment, $$A$$ = amount borrowed, $$i$$ = monthly interest rate, and $$n$$ = the number of payments.
Also, let $$\displaystyle D_n$$ be the debt amount after payment $$n$$.
Consider the recursion:
(1) $$\displaystyle D_{n}=(1+i)D_{n-1}-P$$
(2) $$\displaystyle D_{n+1}=(1+i)D_{n}-P$$
Subtracting (1) from (2) yields the homogeneous recursion:
$$\displaystyle D_{n+1}=(2+i)D_{n}-(1+i)D_{n-1}$$
whose associated auxiliary equation is:
$$\displaystyle r^2-(2+i)r+(1+i)=0$$
$$\displaystyle \left(r-(1+i)\right)\left(r-1\right)=0$$
Thus, the closed-form for our recursion is:
$$\displaystyle D_n=k_1(1+i)^n+k_2$$
Using initial values, we may determine the coefficients $$\displaystyle k_i$$:
$$\displaystyle D_0=k_1+k_2=A$$
$$\displaystyle D_1=k_1(1+i)+k_2=(1+i)A-P$$
Solving this system, we find:
$$\displaystyle k_1=\frac{Ai-P}{i},\,k_2=\frac{P}{i}$$ and so we have:
$$\displaystyle D_n=\left(\frac{Ai-P}{i}\right)(1+i)^n+\left(\frac{P}{i}\right)=\frac{(Ai-P)(1+i)^n+P}{i}$$
Now, equating this to zero, we can solve for P:
$$\displaystyle \frac{(Ai-P)(1+i)^n+P}{i}=0$$
$$\displaystyle (Ai-P)(1+i)^n+P=0$$
$$\displaystyle (P-Ai)(1+i)^n=P$$
$$\displaystyle P\left((1+i)^n-1\right)=Ai(1+i)^n$$
$$\displaystyle P=\frac{Ai(1+i)^n}{(1+i)^n-1}$$
$$\displaystyle P=\frac{Ai}{1-(1+i)^{-n}}$$
Very impressive work.
MarkFL |
# Elastic Collision of Two Masses – It Can Be Shown Exercise
An elastic collision is a collision where total momentum and total kinetic energy is conserved.
This illustration shows two objects A and B traveling towards each other. The mass of A is mA and the moving with velocity VAi. The second object has a mass of mB and velocity VBi. The two objects collide elastically. Mass A moves away at a velocity VAf and mass B has a final velocity of VBf.
Given these conditions, textbooks give the following formulas for VAf and VBf.
and
where
mA is the mass of the first object
VAi is the initial velocity of the first object
VAf is the final velocity of the first object
mB is the mass of the second object
VBi is the initial velocity of the second object and
VBf is the final velocity of the second object.
These two equations are often just presented in this form in the textbook with little or no explanations. Very early in your science education, you will encounter the phrase “It can be shown …” between two steps of mathematics or “left as an exercise for the student”. This almost always translates into “homework problem”. This “It can be shown” example shows how to find the final velocities of two masses after an elastic collision.
This is a step by step derivation of these two equations.
First, we know total momentum is conserved in the collision.
total momentum before collision = total momentum after collision
mAVAi + mBVBi = mAVAf + mBVBf
Rearrange this equation so the same masses are on the same side as each other
mAVAi – mAVAf = mBVBf – mBVBi
Factor out the masses
mA(VAi – VAf) = mB(VBf – VBi)
Let’s call this Equation 1 and come back to it in a minute.
Since we were told the collision was elastic, the total kinetic energy is conserved.
kinetic energy before collision = kinetic energy after collection
½mAVAi2 + ½mBVBi2 = ½mAVAf2 + ½mBVBf2
Multiply the entire equation by 2 to get rid of the ½ factors.
mAVAi2 + mBVBi2 = mAVAf2 + mBVBf2
Rearrange the equation so the like masses are together.
mAVAi2 – mAVAf2 = mBVBf2 – mBVBi2
Factor out the common masses
mA(VAi2 – VAf2) = mB(VBf2 – VBi2)
Use the “difference between two squares” relationship (a2 – b2) = (a + b)(a – b) to factor out the squared velocities on each side.
mA(VAi + VAf)(VAi – VAf) = mB(VBf + VBi)(VBf – VBi)
Now we have two equations and two unknowns, VAf and VBf.
Divide this equation by equation 1 from before (the total momentum equation from above) to get
Now we can cancel out most of this
This leaves
VAi + VAf = VBf + VBi
Solve for VAf
VAf = VBf + VBi – VAi
Now we have one of our unknowns in terms of the other unknown variable. Plug this into the original total momentum equation
mAVAi + mBVBi = mAVAf + mBVBf
mAVAi + mBVBi = mA(VBf + VBi – VAi) + mBVBf
Now, solve this for the final unknown variable, VBf
mAVAi + mBVBi = mAVBf + mAVBi – mAVAi + mBVBf
subtract mAVBi from both sides and add mAVAi to both sides
mAVAi + mBVBi – mAVBi + mAVAi = mAVBf + mBVBf
2mAVAi + mBVBi – mAVBi = mAVBf + mBVBf
factor out the masses
2 mAVAi + (mB – mA)VBi = (mA + mB)VBf
Divide both sides by (mA + mB)
Now we know the value of one of the unknowns, VBf. Use this to find the other unknown variable, VAf. Earlier, we found
VAf = VBf + VBi – VAi
Plug in our VBf equation and solve for VAf
Group the terms with the same velocities
The common denominator for both sides is (mA + mB)
Be careful of your signs in the first half of the expressions in this step
Now we’ve solved for both unknowns VAf and VBf in terms of known values.
Note these match the equations we were supposed to find.
This was not a difficult problem, but there were a couple spots to trip you up.
First, all the subscripts can get tangled up if you aren’t careful or neat in your handwriting.
Second, sign errors. Subtracting a pair of variables inside parentheses will change the sign on BOTH variables. It is all too easy to carelessly turn – (a + b) into -a + b instead of -a – b.
Last, learn the difference between two squares factor. a2 – b2 = (a + b)(a – b) is an extremely useful factoring trick when trying to cancel something out of an equation. |
# Chapter 10: Surface Area and Volume
## Presentation on theme: "Chapter 10: Surface Area and Volume"— Presentation transcript:
Chapter 10: Surface Area and Volume
Objectives: Students will be able to find the surface area and volume of three dimensional figures.
Prism: 2 congruent, parallel faces (called the bases).
The other faces of the prism are called lateral faces Named by the shape of its base Altitude: perpendicular segment joining the bases Height: the length of the altitude
Lateral Surface Area: For a Right Prism: LA = ph p=perimeter of base
The sum of the areas of the lateral faces Area of the sides (not including the bases) For a Right Prism: LA = ph p=perimeter of base h= height
Cylinder (bases are circles)
LA = 2∏rh
Surface Area: For a Right Prism: SA= LA + 2B B= area of the base
Sum of lateral surface area and the area of the 2 bases Find lateral surface area and then add the area of the bases For a Right Prism: SA= LA + 2B B= area of the base
Surface Area of Cylinders:
SA= 2∏rh + 2∏r2
Pyramids and Cones Pyramid: The base is any polygon and the other faces (lateral faces) are triangles that meet at a common vertex Cone: Pyramid whose base is a circle.
Slant Height (l): The length of the altitude of the lateral face of the pyramid
Lateral Surface Area of Pyramids and Cones:
Lateral Surface Area of Pyramid: LA = ½ pl p= perimeter of base l= slant height Lateral Surface Area of Cone: LA = ∏rl
Total Surface Area of Pyramids and Cones:
Pyramid: SA = LA + B B= area of the base Cone: SA = LA + ∏r2
Surface Area of a Sphere:
Cross Section The shape you get when you “slice” a 3 dimensional figure
VOLUME The space that a figure occupies Measured in cubic units
When finding volume of pyramids and cones, need to use the height, not slant height. The height is perpendicular to the base.
VOLUME FORMULAS: Right Rectangular Prism: Any Pyramid: V= lwh Any Right Prism: B = area of the base V= Bh B= area of the base Cone: Square Pyramid: Sphere: |
# √(cosx)=2cosx-1solve the equation
embizze | Certified Educator
Solve for x: `sqrt(cosx)=2cosx-1` Square both sides:
`cosx=4cos^2x-4cosx+1`
`4cos^2x-5cosx+1=0`
`(4cosx-1)(cosx-1)=0`
`cosx=1/4` or cosx=1
However, since we squared both sides we must check for extraneous solutions:
If `cosx=1/4` then `sqrt(cosx)=sqrt(1/4)=1/2` and 2cosx-1=`2(1/4)-1=-1/2` so this is an extraneous solution.
If cosx=1 then `sqrt(cosx)=1` and 2cosx-1=2(1)-1=1 which works.
cosx=1==>` ` `x=0+-2kpi`
----------------------------------------------------------------
The solutions to `sqrt(cosx)=2cosx-1` are `x=0+-2kpi,k in ZZ`
----------------------------------------------------------------
The graph of `sqrt(cosx)` in red; the graph of 2cosx-1 in black:
** The straight line portions of the red graph are artifacts of the grapher: you cannot take the square root of a negative number in the reals, so anywhere the cosine is negative is not in the domain.
sciencesolve | Certified Educator
You need to raise to square both sides, such that:
`(sqrt(cos x)) = (2cos x - 1)^2`
Expanding the square yields:
`cos x = 4cos^2 x - 4 cos x + 1`
Moving all the terms to one side yields:
`4cos^2 x - 4 cos x - cos x + 1 = 0`
`4cos^2 x - 5 cos x + 1 = 0`
You need to substitute t for `cos x` such that:
`4t^2 - 5t + 1 = 0`
You need to split the term `-5t` in two terms such that :
`4t^2 - 4t - t + 1 = 0`
Grouping the terms yields:
`(4t^2 - 4t) - (t - 1) = 0`
Factoring out `4t` in the first group yields:
`4t(t - 1) - (t - 1) = 0`
Factoring out `(t-1)` yields:
`(t-1)(4t-1) = 0 => {(t-1=0),(4t-1=0):} => {(t=1),(t=1/4):}`
Substituting back `cos x` for t yields:
`cos x = 1 > 0 => x = 2npi`
`cos x = 1/4 > 0 => x = +-cos^(-1)(1/4) + 2npi`
Hence, evaluating the solutions to the given equation yields `x = 2npi` and `x = +-cos^(-1)(1/4) + 2npi.` |
# Ordinary Differential Equations/Frobenius Solution to the Hypergeometric Equation
In the following we solve the second-order differential equation called the hypergeometric differential equation using Frobenius method, named after Ferdinand Georg Frobenius. This is a method that uses the series solution for a differential equation, where we assume the solution takes the form of a series. This is usually the method we use for complicated ordinary differential equations.
The solution of the hypergeometric differential equation is very important. For instance, Legendre's differential equation can be shown to be a special case of the hypergeometric differential equation. Hence, by solving the hypergeometric differential equation, one may directly compare its solutions to get the solutions of Legendre's differential equation, after making the necessary substitutions. For more details, please check the hypergeometric differential equation.
We shall prove that this equation has three singularities, namely at x = 0, x = 1 and around x = infinity. However, as these will turn out to be regular singular points, we will be able to assume a solution on the form of a series. Since this is a second-order differential equation, we must have two linearly independent solutions.
The problem however will be that our assumed solutions may or not be independent, or worse, may not even be defined (depending on the value of the parameters of the equation). This is why we shall study the different cases for the parameters and modify our assumed solution accordingly.
## The equationEdit
Solve the hypergeometric equation around all singularities:
${\displaystyle x(1-x)y''+\left\{\gamma -(1+\alpha +\beta )x\right\}y'-\alpha \beta y=0}$
## Solution around x = 0Edit
Let
{\displaystyle {\begin{aligned}P_{0}(x)&=-\alpha \beta ,\\P_{1}(x)&=\gamma -(1+\alpha +\beta )x,\\P_{2}(x)&=x(1-x)\end{aligned}}}
Then
${\displaystyle P_{2}(0)=P_{2}(1)=0.}$
Hence, x = 0 and x = 1 are singular points. Let's start with x = 0. To see if it is regular, we study the following limits:
{\displaystyle {\begin{aligned}\lim _{x\to a}{\frac {(x-a)P_{1}(x)}{P_{2}(x)}}&=\lim _{x\to 0}{\frac {(x-0)(\gamma -(1+\alpha +\beta )x)}{x(1-x)}}=\lim _{x\to 0}{\frac {x(\gamma -(1+\alpha +\beta )x)}{x(1-x)}}=\gamma \\\lim _{x\to a}{\frac {(x-a)^{2}P_{0}(x)}{P_{2}(x)}}&=\lim _{x\to 0}{\frac {(x-0)^{2}(-\alpha \beta )}{x(1-x)}}=\lim _{x\to 0}{\frac {x^{2}(-\alpha \beta )}{x(1-x)}}=0\end{aligned}}}
Hence, both limits exist and x = 0 is a regular singular point. Therefore, we assume the solution takes the form
${\displaystyle y=\sum _{r=0}^{\infty }a_{r}x^{r+c}}$
with a0 ≠ 0. Hence,
{\displaystyle {\begin{aligned}y'&=\sum _{r=0}^{\infty }a_{r}(r+c)x^{r+c-1}\\y''&=\sum _{r=0}^{\infty }a_{r}(r+c)(r+c-1)x^{r+c-2}.\end{aligned}}}
Substituting these into the hypergeometric equation, we get
${\displaystyle x\sum _{r=0}^{\infty }a_{r}(r+c)(r+c-1)x^{r+c-2}-x^{2}\sum _{r=0}^{\infty }a_{r}(r+c)(r+c-1)x^{r+c-2}+\gamma \sum _{r=0}^{\infty }a_{r}(r+c)x^{r+c-1}-(1+\alpha +\beta )x\sum _{r=0}^{\infty }a_{r}(r+c)x^{r+c-1}-\alpha \beta \sum _{r=0}^{\infty }a_{r}x^{r+c}=0}$
That is,
${\displaystyle \sum _{r=0}^{\infty }a_{r}(r+c)(r+c-1)x^{r+c-1}-\sum _{r=0}^{\infty }a_{r}(r+c)(r+c-1)x^{r+c}+\gamma \sum _{r=0}^{\infty }a_{r}(r+c)x^{r+c-1}-(1+\alpha +\beta )\sum _{r=0}^{\infty }a_{r}(r+c)x^{r+c}-\alpha \beta \sum _{r=0}^{\infty }a_{r}x^{r+c}=0}$
In order to simplify this equation, we need all powers to be the same, equal to r + c − 1, the smallest power. Hence, we switch the indices as follows:
{\displaystyle {\begin{aligned}&\sum _{r=0}^{\infty }a_{r}(r+c)(r+c-1)x^{r+c-1}-\sum _{r=1}^{\infty }a_{r-1}(r+c-1)(r+c-2)x^{r+c-1}+\gamma \sum _{r=0}^{\infty }a_{r}(r+c)x^{r+c-1}\\&\qquad -(1+\alpha +\beta )\sum _{r=1}^{\infty }a_{r-1}(r+c-1)x^{r+c-1}-\alpha \beta \sum _{r=1}^{\infty }a_{r-1}x^{r+c-1}=0\end{aligned}}}
Thus, isolating the first term of the sums starting from 0 we get
{\displaystyle {\begin{aligned}&a_{0}(c(c-1)+\gamma c)x^{c-1}+\sum _{r=1}^{\infty }a_{r}(r+c)(r+c-1)x^{r+c-1}-\sum _{r=1}^{\infty }a_{r-1}(r+c-1)(r+c-2)x^{r+c-1}\\&\qquad +\gamma \sum _{r=1}^{\infty }a_{r}(r+c)x^{r+c-1}-(1+\alpha +\beta )\sum _{r=1}^{\infty }a_{r-1}(r+c-1)x^{r+c-1}-\alpha \beta \sum _{r=1}^{\infty }a_{r-1}x^{r+c-1}=0\end{aligned}}}
Now, from the linear independence of all powers of x, that is, of the functions 1, x, x2, etc., the coefficients of xk vanish for all k. Hence, from the first term, we have
${\displaystyle a_{0}(c(c-1)+\gamma c)=0}$
which is the indicial equation. Since a0 ≠ 0, we have
${\displaystyle c(c-1+\gamma )=0.}$
Hence,
${\displaystyle c_{1}=0,c_{2}=1-\gamma }$
Also, from the rest of the terms, we have
${\displaystyle ((r+c)(r+c-1)+\gamma (r+c))a_{r}+(-(r+c-1)(r+c-2)-(1+\alpha +\beta )(r+c-1)-\alpha \beta )a_{r-1}=0}$
Hence,
{\displaystyle {\begin{aligned}a_{r}&={\frac {(r+c-1)(r+c-2)+(1+\alpha +\beta )(r+c-1)+\alpha \beta }{(r+c)(r+c-1)+\gamma (r+c)}}a_{r-1}\\&={\frac {(r+c-1)(r+c+\alpha +\beta -1)+\alpha \beta }{(r+c)(r+c+\gamma -1)}}a_{r-1}\end{aligned}}}
But
{\displaystyle {\begin{aligned}(r+c-1)(r+c+\alpha +\beta -1)+\alpha \beta &=(r+c-1)(r+c+\alpha -1)+(r+c-1)\beta +\alpha \beta \\&=(r+c-1)(r+c+\alpha -1)+\beta (r+c+\alpha -1)\end{aligned}}}
Hence, we get the recurrence relation
${\displaystyle a_{r}={\frac {(r+c+\alpha -1)(r+c+\beta -1)}{(r+c)(r+c+\gamma -1)}}a_{r-1},{\text{ for }}r\geq 1.}$
Let's now simplify this relation by giving ar in terms of a0 instead of ar−1. From the recurrence relation (note: below, expressions of the form (u)r refer to the Pochhammer symbol).
{\displaystyle {\begin{aligned}a_{1}&={\frac {(c+\alpha )(c+\beta )}{(c+1)(c+\gamma )}}a_{0}\\a_{2}&={\frac {(c+\alpha +1)(c+\beta +1)}{(c+2)(c+\gamma +1)}}a_{1}={\frac {(c+\alpha +1)(c+\alpha )(c+\beta )(c+\beta +1)}{(c+2)(c+1)(c+\gamma )(c+\gamma +1)}}a_{0}={\frac {(c+\alpha )_{2}(c+\beta )_{2}}{(c+1)_{2}(c+\gamma )_{2}}}a_{0}\\a_{3}&={\frac {(c+\alpha +2)(c+\beta +2)}{(c+3)(c+\gamma +2)}}a_{2}={\frac {(c+\alpha )_{2}(c+\alpha +2)(c+\beta )_{2}(c+\beta +2)}{(c+1)_{2}(c+3)(c+\gamma )_{2}(c+\gamma +2)}}a_{0}={\frac {(c+\alpha )_{3}(c+\beta )_{3}}{(c+1)_{3}(c+\gamma )_{3}}}a_{0}\end{aligned}}}
As we can see,
${\displaystyle a_{r}={\frac {(c+\alpha )_{r}(c+\beta )_{r}}{(c+1)_{r}(c+\gamma )_{r}}}a_{0},{\text{ for }}r\geq 0}$
Hence, our assumed solution takes the form
${\displaystyle y=a_{0}\sum _{r=0}^{\infty }{\frac {(c+\alpha )_{r}(c+\beta )_{r}}{(c+1)_{r}(c+\gamma )_{r}}}x^{r+c}.}$
We are now ready to study the solutions corresponding to the different cases for c1 − c2 = γ − 1 (this reduces to studying the nature of the parameter γ: whether it is an integer or not).
## Analysis of the solution in terms of the difference γ − 1 of the two rootsEdit
### γ not an integerEdit
Then y1 = y|c = 0 and y2 = y|c = 1 − γ. Since
${\displaystyle y=a_{0}\sum _{r=0}^{\infty }{\frac {(c+\alpha )_{r}(c+\beta )_{r}}{(c+1)_{r}(c+\gamma )_{r}}}x^{r+c},}$
we have
{\displaystyle {\begin{aligned}y_{1}&=a_{0}\sum _{r=0}^{\infty }{\frac {(\alpha )_{r}(\beta )_{r}}{(1)_{r}(\gamma )_{r}}}x^{r}=a_{0}\cdot {{}_{2}F_{1}}(\alpha ,\beta ;\gamma ;x)\\y_{2}&=a_{0}\sum _{r=0}^{\infty }{\frac {(\alpha +1-\gamma )_{r}(\beta +1-\gamma )_{r}}{(1-\gamma +1)_{r}(1-\gamma +\gamma )_{r}}}x^{r+1-\gamma }\\&=a_{0}x^{1-\gamma }\sum _{r=0}^{\infty }{\frac {(\alpha +1-\gamma )_{r}(\beta +1-\gamma )_{r}}{(1)_{r}(2-\gamma )_{r}}}x^{r}\\&=a_{0}x^{1-\gamma }{{}_{2}F_{1}}(\alpha -\gamma +1,\beta -\gamma +1;2-\gamma ;x)\end{aligned}}}
Hence, ${\displaystyle y=A'y_{1}+B'y_{2}.}$ Let A′ a0 = a and Ba0 = B. Then
${\displaystyle y=A{{}_{2}F_{1}}(\alpha ,\beta ;\gamma ;x)+Bx^{1-\gamma }{{}_{2}F_{1}}(\alpha -\gamma +1,\beta -\gamma +1;2-\gamma ;x)\,}$
### γ = 1Edit
Then y1 = y|c = 0. Since γ = 1, we have
${\displaystyle y=a_{0}\sum _{r=0}^{\infty }{\frac {(c+\alpha )_{r}(c+\beta )_{r}}{(c+1)_{r}^{2}}}x^{r+c}.}$
Hence,
{\displaystyle {\begin{aligned}y_{1}&=a_{0}\sum _{r=0}^{\infty }{\frac {(\alpha )_{r}(\beta )_{r}}{(1)_{r}(1)_{r}}}x^{r}=a_{0}{{}_{2}F_{1}}(\alpha ,\beta ;1;x)\\y_{2}&=\left.{\frac {\partial y}{\partial c}}\right|_{c=0}.\end{aligned}}}
To calculate this derivative, let
${\displaystyle M_{r}={\frac {(c+\alpha )_{r}(c+\beta )_{r}}{(c+1)_{r}^{2}}}.}$
Then
${\displaystyle \ln(M_{r})=\ln \left({\frac {(c+\alpha )_{r}(c+\beta )_{r}}{(c+1)_{r}^{2}}}\right)=\ln(c+\alpha )_{r}+\ln(c+\beta )_{r}-2\ln(c+1)_{r}}$
But
${\displaystyle \ln(c+\alpha )_{r}=\ln \left((c+\alpha )(c+\alpha +1)\cdots (c+\alpha +r-1)\right)=\sum _{k=0}^{r-1}\ln(c+\alpha +k).}$
Hence,
{\displaystyle {\begin{aligned}\ln(M_{r})&=\sum _{k=0}^{r-1}\ln(c+\alpha +k)+\sum _{k=0}^{r-1}\ln(c+\beta +k)-2\sum _{k=0}^{r-1}\ln(c+1+k)\\&=\sum _{k=0}^{r-1}\left(\ln(c+\alpha +k)+\ln(c+\beta +k)-2\ln(c+1+k)\right)\end{aligned}}}
Differentiating both sides of the equation with respect to c, we get:
${\displaystyle {\frac {1}{M_{r}}}{\frac {\partial M_{r}}{\partial c}}=\sum _{k=0}^{r-1}\left({\frac {1}{c+\alpha +k}}+{\frac {1}{c+\beta +k}}-{\frac {2}{c+1+k}}\right).}$
Hence,
${\displaystyle {\frac {\partial M_{r}}{\partial c}}={\frac {(c+\alpha )_{r}(c+\beta )_{r}}{(c+1)_{r}^{2}}}\sum _{k=0}^{r-1}\left({\frac {1}{c+\alpha +k}}+{\frac {1}{c+\beta +k}}-{\frac {2}{c+1+k}}\right).}$
Now,
${\displaystyle y=a_{0}x^{c}\sum _{r=0}^{\infty }{\frac {(c+\alpha )_{r}(c+\beta )_{r}}{(c+1)_{r}^{2}}}x^{r}=a_{0}x^{c}\sum _{r=0}^{\infty }M_{r}x^{r}.}$
Hence,
{\displaystyle {\begin{aligned}{\frac {\partial y}{\partial c}}&=a_{0}x^{c}\ln(x)\sum _{r=0}^{\infty }{\frac {(c+\alpha )_{r}(c+\beta )_{r}}{(c+1)_{r}^{2}}}x^{r}+a_{0}x^{c}\sum _{r=0}^{\infty }\left({\frac {(c+\alpha )_{r}(c+\beta )_{r}}{(c+1)_{r}^{2}}}\left\{\sum _{k=0}^{r-1}\left({\frac {1}{c+\alpha +k}}+{\frac {1}{c+\beta +k}}-{\frac {2}{c+1+k}}\right)\right\}\right)x^{r}\\&=a_{0}x^{c}\sum _{r=0}^{\infty }{\frac {(c+\alpha )_{r}(c+\beta )_{r}}{(c+1)_{r})^{2}}}\left(\ln x+\sum _{k=0}^{r-1}\left({\frac {1}{c+\alpha +k}}+{\frac {1}{c+\beta +k}}-{\frac {2}{c+1+k}}\right)\right)x^{r}.\end{aligned}}}
For c = 0, we get
${\displaystyle y_{2}=a_{0}\sum _{r=0}^{\infty }{\frac {(\alpha )_{r}(\beta )_{r}}{(1)_{r}^{2}}}\left(\ln x+\sum _{k=0}^{r-1}\left({\frac {1}{\alpha +k}}+{\frac {1}{\beta +k}}-{\frac {2}{1+k}}\right)\right)x^{r}.}$
Hence, y = Cy1 + Dy2. Let Ca0 = C and Da0 = D. Then
${\displaystyle y=C{{}_{2}F_{1}}(\alpha ,\beta ;1;x)+D\sum _{r=0}^{\infty }{\frac {(\alpha )_{r}(\beta )_{r}}{(1)_{r}^{2}}}\left(\ln(x)+\sum _{k=0}^{r-1}\left({\frac {1}{\alpha +k}}+{\frac {1}{\beta +k}}-{\frac {2}{1+k}}\right)\right)x^{r}}$
### γ an integer and γ ≠ 1Edit
#### γ ≤ 0Edit
The value of ${\displaystyle \gamma }$ is ${\displaystyle \gamma =0,-1,-2,\cdots }$ . To begin with, we shall simplify matters by concentrating a particular value of ${\displaystyle \gamma }$ and generalise the result at a later stage. We shall use the value ${\displaystyle \gamma =-2}$ . The indicial equation has a root at ${\displaystyle c=0}$ , and we see from the recurrence relation
${\displaystyle a_{r}={\frac {(r+c+\alpha -1)(r+c+\beta -1)}{(r+c)(r+c-3)}}a_{r-1},}$
that when ${\displaystyle r=3}$ that that denominator has a factor ${\displaystyle c}$ which vanishes when ${\displaystyle c=0}$ . In this case, a solution can be obtained by putting ${\displaystyle a_{0}=b_{0}c}$ where ${\displaystyle b_{0}}$ is a constant.
With this substitution, the coefficients of ${\displaystyle x^{r}}$ vanish when ${\displaystyle c=0}$ and ${\displaystyle r<3}$ . The factor of ${\displaystyle c}$ in the denominator of the recurrence relation cancels with that of the numerator when ${\displaystyle r\geq 3}$ . Hence, our solution takes the form
${\displaystyle y_{1}={\frac {b_{0}}{(-2)\times (-1)}}\left({\frac {(\alpha )_{3}(\beta )_{3}}{(3!0!}}x^{3}+{\frac {(\alpha )_{4}(\beta )_{4}}{4!1!}}x^{4}+{\frac {(\alpha )_{5}(\beta )_{5}}{5!2!}}x^{5}+\cdots \right)}$
${\displaystyle ={\frac {b_{0}}{(-2)_{2}}}\sum _{r=3}^{\infty }{\frac {(\alpha )_{r}(\beta )_{r}}{r!(r-3)!}}x^{r}={\frac {b_{0}}{(-2)_{2}}}{\frac {(\alpha )_{3}(\beta )_{3}}{3!}}\sum _{r=3}^{\infty }{\frac {(\alpha +3)_{r-3}(\beta +3)_{r-3}}{(1+3)_{r-3}(r-3)!}}x^{r}.}$
If we start the summation at ${\displaystyle r=0}$ rather than ${\displaystyle r=3}$ we see that
${\displaystyle y_{1}=b_{0}{\frac {(\alpha )_{3}(\beta )_{3}}{(-2)_{2}\times 3!}}x^{3}{_{2}F_{1}}(\alpha +3,\beta +3;(1+3);x).}$
The result (as we have written it) generalises easily. For ${\displaystyle \gamma =1+m}$ , with ${\displaystyle m=1,2,3,\cdots }$ then
${\displaystyle y_{1}=b_{0}{\frac {(\alpha )_{m}(\beta )_{m}}{(1-m)_{m-1}\times m!}}x^{m}{_{2}F_{1}}(\alpha +m,\beta +m;(1+m);x).}$
Obviously, if ${\displaystyle \gamma =-2}$ , then ${\displaystyle m=3}$ . The expression for ${\displaystyle y_{1}(x)}$ we have just given looks a little inelegant since we have a multiplicative constant apart from the usual arbitrary multiplicative constant ${\displaystyle b_{0}}$ . Later, we shall see that we can recast things in such a way that this extra constant never appears
The other root to the indicial equation is ${\displaystyle c=1-\gamma =3}$ , but this gives us (apart from a multiplicative constant) the same result as found using ${\displaystyle c=0}$ . This means we must take the partial derivative (w.r.t. ${\displaystyle c}$ ) of the usual trial solution in order to find a second independent solution. If we define the linear operator ${\displaystyle L}$ as
${\displaystyle L=x(1-x){\frac {d^{2}}{dx^{2}}}-(\alpha +\beta +1)x{\frac {d}{dx}}+\gamma {\frac {d}{dx}}-\alpha \beta ,}$
then since ${\displaystyle \gamma =-2}$ in our case,
${\displaystyle Lc\sum _{r=0}^{\infty }b_{r}(c)x^{r}=b_{0}c^{2}(c-3).}$
(We insist that ${\displaystyle b_{0}\neq 0}$ .) Taking the partial derivative w.r.t ${\displaystyle c}$ ,
${\displaystyle L{\frac {\partial }{\partial c}}c\sum _{r=0}^{\infty }b_{r}(c)x^{r+c}=b_{0}(3c^{2}-6c).}$
Note that we must evaluate the partial derivative at ${\displaystyle c=0}$ (and not at the other root ${\displaystyle c=3}$ ). Otherwise the right hand side is non-zero in the above, and we do not have a solution of ${\displaystyle Ly(x)=0}$ . The factor ${\displaystyle c}$ is not cancelled for ${\displaystyle r=0,1}$ and ${\displaystyle r=2}$ . This part of the second independent solution is
${\displaystyle {\bigg [}{\frac {\partial }{\partial c}}b_{0}{\bigg (}c+c{\frac {(c+\alpha )(c+\beta )}{(c+1)(c-2)}}x+c{\frac {(c+\alpha )(c+\alpha +1)(c+\beta )(c+\beta +1)}{(c+1)(c+2)(c-2)(c-1)}}x^{2}{\bigg )}{\bigg ]}{\bigg \vert }_{c=0}.}$ ${\displaystyle =b_{0}\left(1+{\frac {\alpha \beta }{1!\times (-2)}}x+{\frac {\alpha (\alpha +1)\beta (\beta +1)}{2!\times (-2)\times (-1)}}x^{2}\right)=b_{0}\sum _{r=0}^{3-1}{\frac {(\alpha )_{r}(\beta )_{r}}{r!(1-3)_{r}}}x^{r}.}$
Now we can turn our attention to the terms where the factor ${\displaystyle c}$ cancels. First
${\displaystyle cb_{3}={\frac {b_{0}}{(c-1)(c-2)}}{\cancel {c}}{\frac {(c+\alpha )(c+\alpha +1)(c+\alpha +2)(c+\beta )(c+\beta +1)(c+\beta +2)}{{\cancel {c}}(c+1)(c+2)(c+3)}}.}$
After this, the recurrence relations give us
${\displaystyle cb_{4}=cb_{3}(c){\frac {(c+\alpha +3)(c+\beta +3)}{(c+1)(c+4))}}.}$
${\displaystyle cb_{5}=cb_{3}(c){\frac {(c+\alpha +3)(c+\alpha +4)(c+\beta +3)(c+\beta +4))}{(c+2)(c+1)(c+5)(c+4)}}.}$
So, if ${\displaystyle r\geq 3}$ we have
${\displaystyle cb_{r}={\frac {b_{0}}{(c-1)(c-2)}}{\frac {(c+\alpha )_{r}(c+\beta )_{r}}{(c+1)_{r-3}(c+1)_{r}}}.}$
We need the partial derivatives
${\displaystyle {\frac {\partial cb_{3}(c)}{\partial c}}{\bigg \vert }_{c=0}={\frac {b_{0}}{(1-3)_{3-1}}}{\frac {(\alpha )_{3}(\beta )_{3}}{0!3!}}{\bigg [}{\frac {1}{1}}+{\frac {1}{2}}+{\frac {1}{\alpha }}+{\frac {1}{\alpha +1}}+{\frac {1}{\alpha +2}}}$ ${\displaystyle +{\frac {1}{\beta }}+{\frac {1}{\beta +1}}+{\frac {1}{\beta +2}}-{\frac {1}{1}}-{\frac {1}{2}}-{\frac {1}{3}}{\bigg ]}.}$
Similarly, we can write
${\displaystyle {\frac {\partial cb_{4}(c)}{\partial c}}{\bigg \vert }_{c=0}={\frac {b_{0}}{(1-3)_{3-1}}}{\frac {(\alpha )_{4}(\beta )_{4}}{1!4!}}{\bigg [}{\frac {1}{1}}+{\frac {1}{2}}}$ ${\displaystyle +\sum _{k=0}^{k=3}{\frac {1}{\alpha +k}}+\sum _{k=0}^{k=3}{\frac {1}{\beta +k}}-{\frac {1}{1}}-{\frac {1}{2}}-{\frac {1}{3}}-{\frac {1}{4}}-{\frac {1}{1}}{\bigg ]},}$
and
${\displaystyle {\frac {\partial cb_{5}(c)}{\partial c}}{\bigg \vert }_{c=0}={\frac {b_{0}}{(1-3)_{3-1}}}{\frac {(\alpha )_{5}(\beta )_{5}}{2!5!}}{\bigg [}{\frac {1}{1}}+{\frac {1}{2}}}$ ${\displaystyle +\sum _{k=0}^{k=4}{\frac {1}{\alpha +k}}+\sum _{k=0}^{k=4}{\frac {1}{\beta +k}}-{\frac {1}{1}}-{\frac {1}{2}}-{\frac {1}{3}}-{\frac {1}{4}}-{\frac {1}{5}}-{\frac {1}{1}}-{\frac {1}{2}}{\bigg ]}.}$
It becomes clear that for ${\displaystyle r\geq 3}$
${\displaystyle {\frac {\partial cb_{r}(c)}{\partial c}}{\bigg \vert }_{c=0}={\frac {b_{0}}{(1-3)_{3-1}}}{\frac {(\alpha )_{r}(\beta )_{r}}{(r-3)!r!}}{\bigg [}H_{2}+\sum _{k=0}^{k=r-1}{\frac {1}{\alpha +k}}+\sum _{k=0}^{k=r-1}{\frac {1}{\beta +k}}-H_{r}-H_{r-3}{\bigg ]}.}$
Here, ${\displaystyle H_{k}}$ is the ${\displaystyle k}$ th partial sum of the harmonic series, and by definition ${\displaystyle H_{0}=0}$ and ${\displaystyle H_{1}=1}$ .
Putting these together, for the case ${\displaystyle \gamma =-2}$ we have a second solution
${\displaystyle y_{2}(x)=\log x\times {\frac {b_{0}}{(-2)_{2}}}\sum _{r=3}^{\infty }{\frac {(\alpha )_{r}(\beta )_{r}}{r!(r-3)!}}x^{r}+b_{0}\sum _{r=0}^{3-1}{\frac {(\alpha )_{r}(\beta )_{r}}{r!(1-3)_{r}}}x^{r}}$
${\displaystyle +{\frac {b_{0}}{(-2)_{2}}}\sum _{r=3}^{\infty }{\frac {(\alpha )_{r}(\beta )_{r}}{(r-3)!r!}}{\bigg [}H_{2}+\sum _{k=0}^{k=r-1}{\frac {1}{\alpha +k}}+\sum _{k=0}^{k=r-1}{\frac {1}{\beta +k}}-H_{r}-H_{r-3}{{\bigg ]}x^{r}}.}$
The two independent solutions for ${\displaystyle \gamma =1-m}$ (where ${\displaystyle m}$ is a positive integer) are then
${\displaystyle y_{1}(x)={\frac {1}{(1-m)_{m-1}}}\sum _{r=m}^{\infty }{\frac {(\alpha )_{r}(\beta )_{r}}{r!(r-m)!}}x^{r}}$
and
${\displaystyle y_{2}(x)=\log x\times y_{1}(x)+\sum _{r=0}^{m-1}{\frac {(\alpha )_{r}(\beta )_{r}}{r!(1-m)_{r}}}x^{r}}$
${\displaystyle +{\frac {1}{(1-m)_{m-1}}}\sum _{r=m}^{\infty }{\frac {(\alpha )_{r}(\beta )_{r}}{(r-m)!r!}}{\bigg [}H_{m-1}+\sum _{k=0}^{k=r-1}{\frac {1}{\alpha +k}}+\sum _{k=0}^{k=r-1}{\frac {1}{\beta +k}}-H_{r}-H_{r-m}{\bigg ]}x^{r}.}$
The general solution is as usual ${\displaystyle y(x)=Ay_{1}(x)+By_{2}(x)}$ where ${\displaystyle A}$ and ${\displaystyle B}$ are arbitrary constants. Now, if the reader consults a standard solution" for this case, such as given by Abramowitz and Stegun [1] in §15.5.21 (which we shall write down at the end of the next section) it shall be found that the ${\displaystyle y_{2}}$ solution we have found looks somewhat different from the standard solution. In our solution for ${\displaystyle y_{2}}$ , the first term in the infinite series part of ${\displaystyle y_{2}}$ is a term in ${\displaystyle x^{m}}$ . The first term in the corresponding infinite series in the standard solution is a term in ${\displaystyle x^{m+1}}$ . The ${\displaystyle x^{m}}$ term is missing from the standard solution. Nonetheless, the two solutions are entirely equivalent.
#### the standard form of the solution where γ ≤ 0Edit
The reason for the apparent discrepancy between the solution given above and the standard solution in Abramowitz and Stegun [1] §15.5.21 is that there are an infinite number of ways in which to represent the two independent solutions of the hypergeometric ODE. In the last section, for instance, we replaced ${\displaystyle a_{0}}$ with ${\displaystyle b_{0}c}$ . Suppose though, we are given some function ${\displaystyle h(c)}$ which is continuous and finite everywhere in an arbitrarily small interval about ${\displaystyle c=0}$ . Suppose we are also given
${\displaystyle h(c)\vert _{c=0}\neq 0,}$ and ${\displaystyle {\frac {dh}{dc}}{\bigg \vert }_{c=0}\neq 0.}$
Then, if instead of replacing ${\displaystyle a_{0}}$ with ${\displaystyle b_{0}c}$ we replace ${\displaystyle a_{0}}$ with ${\displaystyle b_{0}h(c)c}$ , we still find we have a valid solution of the hypergeometric equation. Clearly, we have an infinity of possibilities for ${\displaystyle h(c)}$ . There is however a natural choice" for ${\displaystyle h(c)}$ . Suppose that ${\displaystyle cb_{N}(c)=b_{0}f(c)}$ is the first non zero term in the first ${\displaystyle y_{1}(x)}$ solution with ${\displaystyle c=0}$ . If we make ${\displaystyle h(c)}$ the reciprocal of ${\displaystyle f(c)}$ , then we won't have a multiplicative constant involved in ${\displaystyle y_{1}(x)}$ as we did in the previous section. From another point of veiw, we get the same result if we insist" that ${\displaystyle a_{N}}$ is independent of ${\displaystyle c}$ , and find ${\displaystyle a_{0}(c)}$ by using the recurrence relations backwards.
For the first ${\displaystyle (c=0)}$ solution, the function ${\displaystyle h(c)}$ gives us (apart from multiplicative constant) the same ${\displaystyle y_{1}(x)}$ as we would have obtained using ${\displaystyle h(c)=1}$ . Suppose that using ${\displaystyle h(c)=1}$ gives rise to two independent solutions ${\displaystyle y_{1}(x)}$ and ${\displaystyle y_{2}(x)}$ . In the following we shall denote the solutions arrived at given some ${\displaystyle h(c)\neq 1}$ as ${\displaystyle {\tilde {y}}_{1}(x)}$ and ${\displaystyle {\tilde {y}}_{2}(x)}$ .
The second solution requires us to take the partial derivative w.r.t ${\displaystyle c}$ , and substituting the usual trial solution gives us
${\displaystyle L{\frac {\partial }{\partial c}}\sum _{r=0}^{\infty }ch(c)b_{r}x^{r+c}=b_{0}\left({\frac {dh}{dc}}c^{2}(c-1)+2ch(c)(c-1)+h(c)c^{2}\right).}$
The operator ${\displaystyle L}$ is the same linear operator discussed in the previous section. That is to say, the hypergeometric ODE is represented as ${\displaystyle Ly(x)=0}$ .
Evaluating the left hand side at ${\displaystyle c=0}$ will give us a second independent solution. Note that this second solution ${\displaystyle {{\tilde {y}}_{2}}}$ is in fact a linear combination of ${\displaystyle y_{1}(x)}$ and ${\displaystyle y_{2}(x)}$ . Any two independent linear combinations (${\displaystyle {\tilde {y}}_{1}}$ and ${\displaystyle {\tilde {y}}_{2}}$ ) of ${\displaystyle y_{1}}$ and ${\displaystyle y_{2}}$ are independent solutions of ${\displaystyle Ly=0}$ . The general solution can be written as a linear combination of ${\displaystyle {\tilde {y}}_{1}}$ and ${\displaystyle {\tilde {y}}_{2}}$ just as well as linear combinations of ${\displaystyle y_{1}}$ and ${\displaystyle y_{2}}$ . We shall review the special case where ${\displaystyle \gamma =1-3=-2}$ that was considered in the last section. If we insist" ${\displaystyle a_{3}(c)=const.}$ , then the recurrence relations yield
${\displaystyle a_{2}=a_{3}{\frac {c(3+c)}{(2+\alpha +c)(2+\beta +c)}},}$ ${\displaystyle a_{1}=a_{3}{\frac {c(2+c)(3+c)(c-1)}{(1+\alpha +c)(2+\alpha +c)(1+\beta +c)(2+\beta +c)}},}$
and
${\displaystyle a_{0}=a_{3}{\frac {c(1+c)(2+c)(3+c)(c-1)(c-2)}{(\alpha +c)_{3}(\beta +c)_{3}}}=b_{0}ch(c).}$
These three coefficients are all zero at ${\displaystyle c=0}$ as expected. We have three terms involved in ${\displaystyle y_{2}(x)}$ by taking the partiial derivative w.r.t ${\displaystyle c}$ , we denote the sum of the three terms involving these coefficients as ${\displaystyle S_{3}}$ where
${\displaystyle S_{3}=\left[{\frac {\partial }{\partial c}}\left(a_{0}(c)x^{c}+a_{1}(c)x^{c+1}+a_{2}(c)x^{c+2}\right)\right]_{c=0},}$ ${\displaystyle =a_{3}\left[{\frac {3\times 2\times 1(-2)\times (-1)}{(\alpha )_{3}(\beta )_{3}}}x^{3-3}+{\frac {3\times 2\times (-1)}{(\alpha +1)(\alpha +2)(\beta +1)(\beta +2)}}x^{3-2}+{\frac {3}{(\alpha +2)(\beta +2)_{1}}}x^{3-1}\right].}$
The reader may confirm that we can tidy this up and make it easy to generalise by putting
${\displaystyle S_{3}=-a_{3}\sum _{r=1}^{3}{\frac {(-3)_{r}(r-1)!}{(1-\alpha -3)_{r}(1-\beta -3)_{r}}}x^{3-r}.}$
Next we can turn to the other coefficients, the recurrence relations yield
${\displaystyle a_{4}=a_{3}{\frac {(3+c+\alpha )(3+c+\beta )}{(4+c)(1+c)}}}$ ${\displaystyle a_{5}=a_{3}{\frac {(4+c+\alpha )(3+c+\alpha )(4+c+\beta )(3+c+\alpha }{(5+c)(4+c)(1+c)(2+c)}}}$
Setting ${\displaystyle c=0}$ gives us
${\displaystyle {\tilde {y}}_{1}(x)=a_{3}x^{3}\sum _{r=0}^{\infty }{\frac {(\alpha +3)_{r}(\beta +3)_{r}}{(3+1)_{r}r!}}x^{r}=a_{3}x^{3}{_{2}F_{1}}(\alpha +3,\beta +3;(1+3);z).}$
This is (apart from the multiplicative constant${\displaystyle (a)_{3}(b)_{3}/2}$ ) the same as ${\displaystyle y_{1}(x)}$ . Now, to find ${\displaystyle {\tilde {y}}_{2}}$ we need partial derivatives
${\displaystyle {\frac {\partial a_{4}}{\partial c}}{\bigg \vert }_{c=0}=a_{3}{\bigg [}{\frac {(3+c+\alpha )(3+c+\beta )}{(4+c)(1+c)}}{\bigg (}{\frac {1}{\alpha +3+c}}+{\frac {1}{\beta +3+c}}-{\frac {1}{4+c}}-{\frac {1}{1+c}}{\bigg )}{\bigg ]}_{c=0}}$
${\displaystyle =a_{3}{\frac {(3+\alpha )_{1}(3+\beta )_{1}}{(1+3)_{1}\times 1}}{\bigg (}{\frac {1}{\alpha +3}}+{\frac {1}{\beta +3}}-{\frac {1}{4}}-{\frac {1}{1}}{\bigg )}.}$
Then
${\displaystyle {\frac {\partial a_{5}}{\partial c}}{\bigg \vert }_{c=0}=a_{3}{\frac {(3+\alpha )_{2}(3+\beta )_{2}}{(1+3)_{2}\times 1\times 2}}{\bigg (}{\frac {1}{\alpha +3}}+{\frac {1}{\alpha +4}}+{\frac {1}{\beta +3}}+{\frac {1}{\beta +4}}-{\frac {1}{4}}-{\frac {1}{5}}-{\frac {1}{1}}-{\frac {1}{2}}{\bigg )}.}$
we can re-write this as
${\displaystyle {\frac {\partial a_{5}}{\partial c}}{\bigg \vert }_{c=0}=a_{3}{\frac {(3+\alpha )_{2}(3+\beta )_{2}}{(1+3)_{2}\times 2!}}{\bigg [}\sum _{k=0}^{1}\left({\frac {1}{\alpha +3+k}}+{\frac {1}{\beta +3+k}}\right)+\sum _{k=1}^{3}{\frac {1}{k}}-\sum _{k=1}^{5}{\frac {1}{k}}-{\frac {1}{1}}-{\frac {1}{2}}{\bigg ]}.}$
The pattern soon becomes clear, and for ${\displaystyle r=1,2,3,\cdots }$
${\displaystyle {\frac {\partial a_{r+3}}{\partial c}}{\bigg \vert }_{c=0}=a_{3}{\frac {(3+\alpha )_{r}(3+\beta )_{r}}{(1+3)_{r}\times r!}}{\bigg [}\sum _{k=0}^{r-1}\left({\frac {1}{\alpha +3+k}}+{\frac {1}{\beta +3+k}}\right)+\sum _{k=1}^{3}{\frac {1}{k}}-\sum _{k=1}^{r+3}{\frac {1}{k}}-\sum _{k=1}^{r}{\frac {1}{k}}{\bigg ]}.}$
Clearly, for ${\displaystyle r=0}$ ,
${\displaystyle {\frac {\partial a_{3}}{\partial c}}{\bigg \vert }_{c=0}=0.}$
The infinite series part of ${\displaystyle {\tilde {y}}_{2}}$ is ${\displaystyle S_{\infty }}$ , where
${\displaystyle S_{\infty }=x^{3}\sum _{r=1}^{\infty }{\frac {\partial a_{r+3}}{\partial c}}{\bigg \vert }_{c=0}x^{r}.}$
Now we can write (disregarding the arbitrary constant) for${\displaystyle \gamma =1-m}$
${\displaystyle {\tilde {y}}_{1}(x)=x^{3}{_{2}F_{1}}(\alpha +m,\beta +m;1+m;z)}$
${\displaystyle {\tilde {y}}_{2}(x)={\tilde {y}}_{1}(x)\log x-\sum _{r=1}^{m}{\frac {(-m)_{r}(r-1)!}{(1-\alpha -m)_{r}(1-\beta -m)_{r}}}x^{m-r}.}$ ${\displaystyle +x^{3}\sum _{r=0}^{\infty }{\frac {(\alpha +m)_{r}(\beta +m)_{r}}{(1+m)_{r}\times r!}}{\bigg [}\sum _{k=0}^{r-1}\left({\frac {1}{\alpha +m+k}}+{\frac {1}{\beta +m+k}}\right)+\sum _{k=1}^{3}{\frac {1}{k}}-\sum _{k=1}^{r+3}{\frac {1}{k}}-\sum _{k=1}^{r}{\frac {1}{k}}{{\bigg ]}x^{r}}.}$
Some authors prefer to express the finite sums in this last result using the digamma function ${\displaystyle \psi (x)}$ . In particular, the following results are used
${\displaystyle H_{n}=\psi (n+1)+\gamma _{em}.}$
Here, ${\displaystyle \gamma _{em}=0.5772156649=\psi (1)}$ is the Euler-Mascheroni constant. Also
${\displaystyle \sum _{k=0}^{n-1}{\frac {1}{z+k}}=\psi (z+n)-\psi (z).}$
With these results we obtain the form given in Abramamowitz and Stegun §15.5.21, namely
${\displaystyle {\tilde {y}}_{2}(x)={\tilde {y}}_{1}(x)\log x-\sum _{r=1}^{m}{\frac {(-m)_{r}(r-1)!}{(1-\alpha -m)_{r}(1-\beta -m)_{r}}}x^{m-r}.}$ ${\displaystyle +x^{3}\sum _{r=0}^{\infty }{\frac {(\alpha +m)_{r}(\beta +m)_{r}}{(1+m)_{r}\times r!}}{\bigg [}\psi (\alpha +r+m)-\psi (\alpha +m)+\psi (\beta +r+m)-\psi (\beta +m)}$ ${\displaystyle -\psi (r+1+m)-\psi (r+1)+\psi (1+m)+\psi (1){{\bigg ]}x^{r}}.}$
#### The Standard" Form of the Solution γ > 1Edit
In this section, we shall concentrate on the standard solution", and we shall not replace ${\displaystyle a_{0}}$ with ${\displaystyle b_{0}(c-1+\gamma )}$ . We shall put ${\displaystyle \gamma =1+m}$ where ${\displaystyle m=1,2,3,\cdots }$ . For the root ${\displaystyle c=1-\gamma }$ of the indicial eqauation we had
${\displaystyle A_{r}=\left[A_{r-1}{\frac {(r+\alpha -1+c)(r+\beta -1+c)}{(r+c)(r+c+\gamma -1)}}\right]_{c=1-\gamma }=A_{r-1}{\frac {(r+\alpha -\gamma )(r+\beta -\gamma )}{(r+1-\gamma )(r)}},}$
where ${\displaystyle r\geq 1}$ in which case we are in trouble if ${\displaystyle r=\gamma -1=m}$ . For instance, if ${\displaystyle \gamma =4}$ , the denominator in the recurrence relations vanishes for ${\displaystyle r=3}$ . We can use exactly the same methods that we have just used for the standard solution in the last section. We shall not (in the instance where ${\displaystyle \gamma =4}$ ) replace ${\displaystyle a_{0}}$ with ${\displaystyle b_{0}(c+3)}$ as this will not give us the standard form of solution that we are after. Rather, we shall insist" that ${\displaystyle A_{3}=const.}$ as we did in the standard solution for ${\displaystyle \gamma =-2}$ in the last section. (Recall that this defined the function ${\displaystyle h(c)}$ and that ${\displaystyle a_{0}}$ will now be replaced with ${\displaystyle b_{0}(c+3)h(c)}$ .) Then we may work out the coefficients of ${\displaystyle x^{0}}$ to ${\displaystyle x^{2}}$ as functions of ${\displaystyle c}$ using the recurrence relations backwards. There is nothing new to add here, and the reader may use the same methods as used in the last section to find the results of [1]§15.5.18 and §15.5.19, these are
${\displaystyle y_{1}={_{2}F_{1}}(\alpha ,\beta ;1+m;x),}$
and
${\displaystyle y_{2}={_{2}F_{1}}(\alpha ,\beta ;1+m;x)\log x+z^{m}\sum _{r=1}^{\infty }{\frac {(\alpha )_{r}(\beta )_{r}}{r!(1+m)_{r}}}[\psi (\alpha +r)-\psi (\alpha )+\psi (\beta +k)-\psi (\beta )}$ ${\displaystyle -\psi (m+1+r)+\psi (m+1)-\psi (r+1)+\psi (1)]z^{r}-\sum _{k=1}^{m}{\frac {(k-1)!(-m)_{k}}{(1-\alpha )_{k}(1-\beta )_{k}}}z^{-r}.}$
Note that the powers of ${\displaystyle z}$ in the finite sum part of ${\displaystyle y_{2}(x)}$ are now negative so that this sum diverges as ${\displaystyle z\rightarrow 0\}$
## Solution around x = 1Edit
Let us now study the singular point x = 1. To see if it is regular,
{\displaystyle {\begin{aligned}\lim _{x\to a}{\frac {(x-a)P_{1}(x)}{P_{2}(x)}}&=\lim _{x\to 1}{\frac {(x-1)(\gamma -(1+\alpha +\beta )x)}{x(1-x)}}=\lim _{x\to 1}{\frac {-(\gamma -(1+\alpha +\beta )x)}{x}}=1+\alpha +\beta -\gamma \\\lim _{x\to a}{\frac {(x-a)^{2}P_{0}(x)}{P_{2}(x)}}&=\lim _{x\to 1}{\frac {(x-1)^{2}(-\alpha \beta )}{x(1-x)}}=\lim _{x\to 1}{\frac {(x-1)\alpha \beta }{x}}=0\end{aligned}}}
Hence, both limits exist and x = 1 is a regular singular point. Now, instead of assuming a solution on the form
${\displaystyle y=\sum _{r=0}^{\infty }a_{r}(x-1)^{r+c},}$
we will try to express the solutions of this case in terms of the solutions for the point x = 0. We proceed as follows: we had the hypergeometric equation
${\displaystyle x(1-x)y''+(\gamma -(1+\alpha +\beta )x)y'-\alpha \beta y=0.}$
Let z = 1 − x. Then
{\displaystyle {\begin{aligned}{\frac {dy}{dx}}&={\frac {dy}{dz}}\times {\frac {dz}{dx}}=-{\frac {dy}{dz}}=-y'\\{\frac {d^{2}y}{dx^{2}}}&={\frac {d}{dx}}\left({\frac {dy}{dx}}\right)={\frac {d}{dx}}\left(-{\frac {dy}{dz}}\right)={\frac {d}{dz}}\left(-{\frac {dy}{dz}}\right)\times {\frac {dz}{dx}}={\frac {d^{2}y}{dz^{2}}}=y''\end{aligned}}}
Hence, the equation takes the form
${\displaystyle z(1-z)y''+(\alpha +\beta -\gamma +1-(1+\alpha +\beta )z)y'-\alpha \beta y=0.}$
Since z = 1 − x, the solution of the hypergeometric equation at x = 1 is the same as the solution for this equation at z = 0. But the solution at z = 0 is identical to the solution we obtained for the point x = 0, if we replace each γ by α + β − γ + 1. Hence, to get the solutions, we just make this substitution in the previous results. Note also that for x = 0, c1 = 0 and c2 = 1 − γ. Hence, in our case, c1 = 0 while c2 = γ − α − β. Let us now write the solutions. In the following we replaced each z by 1 - x.
## Analysis of the solution in terms of the difference γ − α − β of the two rootsEdit
To simplify notation from now on denote γ − α − β by Δ, therefore γ = Δ + α + β.
### Δ not an integerEdit
${\displaystyle y=A\left\{{{}_{2}F_{1}}(\alpha ,\beta ;-\Delta +1;1-x)\right\}+B\left\{(1-x)^{\Delta }{{}_{2}F_{1}}(\Delta +\beta ,\Delta +\alpha ;\Delta +1;1-x)\right\}}$
### Δ = 0Edit
${\displaystyle y=C\left\{{{}_{2}F_{1}}(\alpha ,\beta ;1;1-x)\right\}+D\left\{\sum _{r=0}^{\infty }{\frac {(\alpha )_{r}(\beta )_{r}}{(1)_{r}^{2}}}\left(\ln(1-x)+\sum _{k=0}^{r-1}\left({\frac {1}{\alpha +k}}+{\frac {1}{\beta +k}}-{\frac {2}{1+k}}\right)\right)(1-x)^{r}\right\}}$
### Δ is a non-zero integerEdit
#### Δ > 0Edit
{\displaystyle {\begin{aligned}y&=E\left\{{\frac {1}{(-\Delta +1)_{\Delta -1}}}\ \sum _{r=1-\Delta -\alpha -\beta }^{\infty }{\frac {(\alpha )_{r}(\beta )_{r}}{(1)_{r}(1)_{r-\Delta }}}(1-x)^{r}\right\}+\\&\quad +F\left\{(1-x)^{\Delta }\ \sum _{r=0}^{\infty }{\frac {(\Delta )(\Delta +\alpha )_{r}(\Delta +\beta )_{r}}{(\Delta +1)_{r}(1)_{r}}}\left(\ln(1-x)+{\frac {1}{\Delta }}+\sum _{k=0}^{r-1}\left({\frac {1}{\Delta +\alpha +k}}+{\frac {1}{\Delta +\beta +k}}-{\frac {1}{\Delta +1+k}}-{\frac {1}{1+k}}\right)\right)(1-x)^{r}\right\}\end{aligned}}}
#### Δ < 0Edit
{\displaystyle {\begin{aligned}y&=G\left\{{\frac {(1-x)^{\Delta }}{(\Delta +1)_{-\Delta -1}}}\ \sum _{r=-\Delta }^{\infty }{\frac {(\Delta +\alpha )_{r}(\Delta +\beta )_{r}}{(1)_{r}(1)_{r+\Delta }}}(1-x)^{r}\right\}+\\&\quad +H\left\{\sum _{r=0}^{\infty }{\frac {(\Delta )(\Delta +\alpha )_{r}(\Delta +\beta )_{r}}{(\Delta +1)_{r}(1)_{r}}}\left(\ln(1-x)-{\frac {1}{\Delta }}+\sum _{k=0}^{r-1}\left({\frac {1}{\alpha +k}}+{\frac {1}{\beta +k}}-{\frac {1}{-\Delta +1+k}}-{\frac {1}{1+k}}\right)\right)(1-x)^{r}\right\}\end{aligned}}}
## Solution around infinityEdit
Finally, we study the singularity as x → ∞. Since we can't study this directly, we let x = s−1. Then the solution of the equation as x → ∞ is identical to the solution of the modified equation when s = 0. We had
{\displaystyle {\begin{aligned}&x(1-x)y''+\left(\gamma -(1+\alpha +\beta )x\right)y'-\alpha \beta y=0\\&{\frac {dy}{dx}}={\frac {dy}{ds}}\times {\frac {ds}{dx}}=-s^{2}\times {\frac {dy}{ds}}=-s^{2}y'\\&{\frac {d^{2}y}{dx^{2}}}={\frac {d}{dx}}\left({\frac {dy}{dx}}\right)={\frac {d}{dx}}\left(-s^{2}\times {\frac {dy}{ds}}\right)={\frac {d}{ds}}\left(-s^{2}\times {\frac {dy}{ds}}\right)\times {\frac {ds}{dx}}=\left((-2s)\times {\frac {dy}{ds}}+(-s^{2}){\frac {d^{2}y}{ds^{2}}}\right)\times (-s^{2})=2s^{3}y'+s^{4}y''\end{aligned}}}
Hence, the equation takes the new form
${\displaystyle {\frac {1}{s}}\left(1-{\frac {1}{s}}\right)\left(2s^{3}y'+s^{4}y''\right)+\left(\gamma -(1+\alpha +\beta ){\frac {1}{s}}\right)(-s^{2}y')-\alpha \beta y=0}$
which reduces to
${\displaystyle \left(s^{3}-s^{2}\right)y''+\left((2-\gamma )s^{2}+(\alpha +\beta -1)s\right)y'-\alpha \beta y=0.}$
Let
{\displaystyle {\begin{aligned}P_{0}(s)&=-\alpha \beta ,\\P_{1}(s)&=(2-\gamma )s^{2}+(\alpha +\beta -1)s,\\P_{2}(s)&=s^{3}-s^{2}.\end{aligned}}}
As we said, we shall only study the solution when s = 0. As we can see, this is a singular point since P2(0) = 0. To see if it is regular,
{\displaystyle {\begin{aligned}\lim _{s\to a}{\frac {(s-a)P_{1}(s)}{P_{2}(s)}}&=\lim _{s\to 0}{\frac {(s-0)((2-\gamma )s^{2}+(\alpha +\beta -1)s)}{s^{3}-s^{2}}}\\&=\lim _{s\to 0}{\frac {(2-\gamma )s^{2}+(\alpha +\beta -1)s}{s^{2}-s}}\\&=\lim _{s\to 0}{\frac {(2-\gamma )s+(\alpha +\beta -1)}{s-1}}=1-\alpha -\beta .\\\lim _{s\to a}{\frac {(s-a)^{2}P_{0}(s)}{P_{2}(s)}}&=\lim _{s\to 0}{\frac {(s-0)^{2}(-\alpha \beta )}{s^{3}-s^{2}}}=\lim _{s\to 0}{\frac {(-\alpha \beta )}{s-1}}=\alpha \beta .\end{aligned}}}
Hence, both limits exist and s = 0 is a regular singular point. Therefore, we assume the solution takes the form
${\displaystyle y=\sum _{r=0}^{\infty }{a_{r}s^{r+c}}}$
with a0 ≠ 0. Hence,
{\displaystyle {\begin{aligned}y'&=\sum \limits _{r=0}^{\infty }{a_{r}(r+c)s^{r+c-1}}\\y''&=\sum \limits _{r=0}^{\infty }{a_{r}(r+c)(r+c-1)s^{r+c-2}}\end{aligned}}}
Substituting in the modified hypergeometric equation we get
${\displaystyle \left(s^{3}-s^{2}\right)y''+\left((2-\gamma )s^{2}+(\alpha +\beta -1)s\right)y'-(\alpha \beta )y=0}$
And therefore:
${\displaystyle \left(s^{3}-s^{2}\right)\sum _{r=0}^{\infty }{a_{r}(r+c)(r+c-1)s^{r+c-2}}+\left((2-\gamma )s^{2}+(\alpha +\beta -1)s\right)\sum _{r=0}^{\infty }{a_{r}(r+c)s^{r+c-1}}-(\alpha \beta )\sum _{r=0}^{\infty }{a_{r}s^{r+c}}=0}$
i.e.,
${\displaystyle \sum _{r=0}^{\infty }{a_{r}(r+c)(r+c-1)s^{r+c+1}}-\sum _{r=0}^{\infty }{a_{r}(r+c)(r+c-1)x^{r+c}}+(2-\gamma )\sum _{r=0}^{\infty }{a_{r}(r+c)s^{r+c+1}}+(\alpha +\beta -1)\sum _{r=0}^{\infty }{a_{r}(r+c)s^{r+c}}-\alpha \beta \sum _{r=0}^{\infty }{a_{r}s^{r+c}}=0.}$
In order to simplify this equation, we need all powers to be the same, equal to r + c, the smallest power. Hence, we switch the indices as follows
{\displaystyle {\begin{aligned}&\sum _{r=1}^{\infty }{a_{r-1}(r+c-1)(r+c-2)s^{r+c}}-\sum _{r=0}^{\infty }{a_{r}(r+c)(r+c-1)s^{r+c}}+(2-\gamma )\sum _{r=1}^{\infty }{a_{r-1}(r+c-1)s^{r+c}}+\\&\qquad \qquad +(\alpha +\beta -1)\sum _{r=0}^{\infty }{a_{r}(r+c)s^{r+c}}-\alpha \beta \sum _{r=0}^{\infty }{a_{r}s^{r+c}}=0\end{aligned}}}
Thus, isolating the first term of the sums starting from 0 we get
{\displaystyle {\begin{aligned}&a_{0}\left(-(c)(c-1)+(\alpha +\beta -1)(c)-\alpha \beta \right)s^{c}+\sum _{r=1}^{\infty }{a_{r-1}(r+c-1)(r+c-2)s^{r+c}}-\sum _{r=1}^{\infty }{a_{r}(r+c)(r+c-1)x^{r+c}}+\\&\qquad \qquad +(2-\gamma )\sum _{r=1}^{\infty }{a_{r-1}(r+c-1)s^{r+c}}+(\alpha +\beta -1)\sum _{r=1}^{\infty }{a_{r}(r+c)s^{r+c}}-\alpha \beta \sum _{r=1}^{\infty }{a_{r}s^{r+c}}=0\end{aligned}}}
Now, from the linear independence of all powers of s (i.e., of the functions 1, s, s2, ...), the coefficients of sk vanish for all k. Hence, from the first term we have
${\displaystyle a_{0}\left(-(c)(c-1)+(\alpha +\beta -1)(c)-\alpha \beta \right)=0}$
which is the indicial equation. Since a0 ≠ 0, we have
${\displaystyle (c)(-c+1+\alpha +\beta -1)-\alpha \beta )=0.}$
Hence, c1 = α and c2 = β.
Also, from the rest of the terms we have
${\displaystyle \left((r+c-1)(r+c-2)+(2-\gamma )(r+c-1)\right)a_{r-1}+\left(-(r+c)(r+c-1)+(\alpha +\beta -1)(r+c)-\alpha \beta \right)a_{r}=0}$
Hence,
${\displaystyle a_{r}=-{\frac {\left((r+c-1)(r+c-2)+(2-\gamma )(r+c-1)\right)}{\left(-(r+c)(r+c-1)+(\alpha +\beta -1)(r+c)-\alpha \beta \right)}}a_{r-1}={\frac {\left((r+c-1)(r+c-\gamma )\right)}{\left((r+c)(r+c-\alpha -\beta )+\alpha \beta \right)}}a_{r-1}}$
But
{\displaystyle {\begin{aligned}(r+c)(r+c-\alpha -\beta )+\alpha \beta &=(r+c-\alpha )(r+c)-\beta (r+c)+\alpha \beta \\&=(r+c-\alpha )(r+c)-\beta (r+c-\alpha ).\end{aligned}}}
Hence, we get the recurrence relation
${\displaystyle a_{r}={\frac {(r+c-1)(r+c-\gamma )}{(r+c-\alpha )(r+c-\beta )}}a_{r-1},\quad \forall r\geq 1}$
Let's now simplify this relation by giving ar in terms of a0 instead of ar−1. From the recurrence relation,
{\displaystyle {\begin{aligned}a_{1}&={\frac {(c)(c+1-\gamma )}{(c+1-\alpha )(c+1-\beta )}}a_{0}\\a_{2}&={\frac {(c+1)(c+2-\gamma )}{(c+2-\alpha )(c+2-\beta )}}a_{1}={\frac {(c+1)(c)(c+2-\gamma )(c+1-\gamma )}{(c+2-\alpha )(c+1-\alpha )(c+2-\beta )(c+1-\beta )}}a_{0}={\frac {(c)_{2}(c+1-\gamma )_{2}}{(c+1-\alpha )_{2}(c+1-\beta )_{2}}}a_{0}\end{aligned}}}
As we can see,
${\displaystyle a_{r}={\frac {(c)_{r}(c+1-\gamma )_{r}}{(c+1-\alpha )_{r}(c+1-\beta )_{r}}}a_{0}\quad \forall r\geq 0}$
Hence, our assumed solution takes the form
${\displaystyle y=a_{0}\sum _{r=0}^{\infty }{\frac {(c)_{r}(c+1-\gamma )_{r}}{(c+1-\alpha )_{r}(c+1-\beta )_{r}}}s^{r+c}}$
We are now ready to study the solutions corresponding to the different cases for c1 − c2 = α − β.
## Analysis of the solution in terms of the difference α − β of the two rootsEdit
### α − β not an integerEdit
Then y1 = y|c = α and y2 = y|c = β. Since
${\displaystyle y=a_{0}\sum _{r=0}^{\infty }{\frac {(c)_{r}(c+1-\gamma )_{r}}{(c+1-\alpha )_{r}(c+1-\beta )_{r}}}s^{r+c},}$
we have
{\displaystyle {\begin{aligned}y_{1}&=a_{0}\sum _{r=0}^{\infty }{\frac {(\alpha )_{r}(\alpha +1-\gamma )_{r}}{(1)_{r}(\alpha +1-\beta )_{r}}}s^{r+\alpha }=a_{0}s^{\alpha }\ {}_{2}F_{1}(\alpha ,\alpha +1-\gamma ;\alpha +1-\beta ;s)\\y_{2}&=a_{0}\sum _{r=0}^{\infty }{\frac {(\beta )_{r}(\beta +1-\gamma )_{r}}{(\beta +1-\alpha )_{r}(1)_{r}}}s^{r+\beta }=a_{0}s^{\beta }\ {}_{2}F_{1}(\beta ,\beta +1-\gamma ;\beta +1-\alpha ;s)\end{aligned}}}
Hence, y = Ay1 + By2. Let Aa0 = A and Ba0 = B. Then, noting that s = x−1,
${\displaystyle y=A\left\{x^{-\alpha }\ {}_{2}F_{1}\left(\alpha ,\alpha +1-\gamma ;\alpha +1-\beta ;x^{-1}\right)\right\}+B\left\{x^{-\beta }\ {}_{2}F_{1}\left(\beta ,\beta +1-\gamma ;\beta +1-\alpha ;x^{-1}\right)\right\}}$
### α − β = 0Edit
Then y1 = y|c = α. Since α = β, we have
${\displaystyle y=a_{0}\sum _{r=0}^{\infty }{{\frac {(c)_{r}(c+1-\gamma )_{r}}{\left((c+1-\alpha )_{r}\right)^{2}}}s^{r+c}}}$
Hence,
{\displaystyle {\begin{aligned}y_{1}&=a_{0}\sum _{r=0}^{\infty }{{\frac {(\alpha )_{r}(\alpha +1-\gamma )_{r}}{(1)_{r}(1)_{r}}}s^{r+\alpha }}=a_{0}s^{\alpha }\ {}_{2}F_{1}(\alpha ,\alpha +1-\gamma ;1;s)\\y_{2}&=\left.{\frac {\partial y}{\partial c}}\right|_{c=\alpha }\end{aligned}}}
To calculate this derivative, let
${\displaystyle M_{r}={\frac {(c)_{r}(c+1-\gamma )_{r}}{\left((c+1-\alpha )_{r}\right)^{2}}}}$
Then using the method in the case γ = 1 above, we get
${\displaystyle {\frac {\partial M_{r}}{\partial c}}={\frac {(c)_{r}(c+1-\gamma )_{r}}{\left((c+1-\alpha )_{r}\right)^{2}}}\sum _{k=0}^{r-1}\left({\frac {1}{c+k}}+{\frac {1}{c+1-\gamma +k}}-{\frac {2}{c+1-\alpha +k}}\right)}$
Now,
{\displaystyle {\begin{aligned}y&=a_{0}s^{c}\sum _{r=0}^{\infty }{\frac {(c)_{r}(c+1-\gamma )_{r}}{\left((c+1-\alpha )_{r}\right)^{2}}}s^{r}\\&=a_{0}s^{c}\sum _{r=0}^{\infty }{M_{r}s^{r}}\\&=a_{0}s^{c}\left(\ln(s)\sum _{r=0}^{\infty }{\frac {(c)_{r}(c+1-\gamma )_{r}}{\left((c+1-\alpha )_{r}\right)^{2}}}s^{r}+\sum _{r=0}^{\infty }{\frac {(c)_{r}(c+1-\gamma )_{r}}{\left((c+1-\alpha )_{r}\right)^{2}}}\left\{\sum _{k=0}^{r-1}{\left({\frac {1}{c+k}}+{\frac {1}{c+1-\gamma +k}}-{\frac {2}{c+1-\alpha +k}}\right)}\right\}s^{r}\right)\end{aligned}}}
Hence,
${\displaystyle {\frac {\partial y}{\partial c}}=a_{0}s^{c}\sum _{r=0}^{\infty }{\frac {(c)_{r}(c+1-\gamma )_{r}}{\left((c+1-\alpha )_{r}\right)^{2}}}\left(\ln(s)+\sum _{k=0}^{r-1}{\left({\frac {1}{c+k}}+{\frac {1}{c+1-\gamma +k}}-{\frac {2}{c+1-\alpha +k}}\right)}\right)s^{r}}$
Therefore:
${\displaystyle y_{2}=\left.{\frac {\partial y}{\partial c}}\right|_{c=\alpha }=a_{0}s^{\alpha }\sum _{r=0}^{\infty }{\frac {(\alpha )_{r}(\alpha +1-\gamma )_{r}}{(1)_{r}(1)_{r}}}\left(\ln(s)+\sum _{k=0}^{r-1}\left({\frac {1}{\alpha +k}}+{\frac {1}{\alpha +1-\gamma +k}}-{\frac {2}{1+k}}\right)\right)s^{r}}$
Hence, y = C′y1 + D′y2. Let C′a0 = C and D′a0 = D. Noting that s = x−1,
${\displaystyle y=C\left\{x^{-\alpha }{}_{2}F_{1}\left(\alpha ,\alpha +1-\gamma ;1;x^{-1}\right)\right\}+D\left\{x^{-\alpha }\sum _{r=0}^{\infty }{\frac {(\alpha )_{r}(\alpha +1-\gamma )_{r}}{(1)_{r}(1)_{r}}}\left(\ln \left(x^{-1}\right)+\sum _{k=0}^{r-1}\left({\frac {1}{\alpha +k}}+{\frac {1}{\alpha +1-\gamma +k}}-{\frac {2}{1+k}}\right)\right)x^{-r}\right\}}$
### α − β an integer and α − β ≠ 0Edit
#### α − β > 0Edit
From the recurrence relation
${\displaystyle a_{r}={\frac {(r+c-1)(r+c-\gamma )}{(r+c-\alpha )(r+c-\beta )}}a_{r-1}}$
we see that when c = β (the smaller root), aα−β → ∞. Hence, we must make the substitution a0 = b0(cci), where ci is the root for which our solution is infinite. Hence, we take a0 = b0(c − β) and our assumed solution takes the new form
${\displaystyle y_{b}=b_{0}\sum _{r=0}^{\infty }{\frac {(c-\beta )(c)_{r}(c+1-\gamma )_{r}}{(c+1-\alpha )_{r}(c+1-\beta )_{r}}}s^{r+c}}$
Then y1 = yb|c = β. As we can see, all terms before
${\displaystyle {\frac {(c-\beta )(c)_{\alpha -\beta }(c+1-\gamma )_{\alpha -\beta }}{(c+1-\alpha )_{\alpha -\beta }(c+1-\beta )_{\alpha -\beta }}}s^{\alpha -\beta }}$
vanish because of the c − β in the numerator.
But starting from this term, the c − β in the numerator vanishes. To see this, note that
${\displaystyle (c+1-\alpha )_{\alpha -\beta }=(c+1-\alpha )(c+2-\alpha )\cdots (c-\beta ).}$
Hence, our solution takes the form
{\displaystyle {\begin{aligned}y_{1}&=b_{0}\left({\frac {(\beta )_{\alpha -\beta }(\beta +1-\gamma )_{\alpha -\beta }}{(\beta +1-\alpha )_{\alpha -\beta -1}(1)_{\alpha -\beta }}}s^{\alpha -\beta }+{\frac {(\beta )_{\alpha -\beta +1}(\beta +1-\gamma )_{\alpha -\beta +1}}{(\beta +1-\alpha )_{\alpha -\beta -1}(1)(1)_{\alpha -\beta +1}}}s^{\alpha -\beta +1}+\cdots \right)\\&={\frac {b_{0}}{(\beta +1-\alpha )_{\alpha -\beta -1}}}\sum _{r=\alpha -\beta }^{\infty }{\frac {(\beta )_{r}(\beta +1-\gamma )_{r}}{(1)_{r}(1)_{r+\beta -\alpha }}}s^{r}\end{aligned}}}
Now,
${\displaystyle y_{2}=\left.{\frac {\partial y_{b}}{\partial c}}\right|_{c=\alpha }.}$
To calculate this derivative, let
${\displaystyle M_{r}={\frac {(c-\beta )(c)_{r}(c+1-\gamma )_{r}}{(c+1-\alpha )_{r}(c+1-\beta )_{r}}}.}$
Then using the method in the case γ = 1 above we get
${\displaystyle {\frac {\partial M_{r}}{\partial c}}={\frac {(c-\beta )(c)_{r}(c+1-\gamma )_{r}}{(c+1-\alpha )_{r}(c+1-\beta )_{r}}}\left({\frac {1}{c-\beta }}+\sum _{k=0}^{r-1}\left({\frac {1}{c+k}}+{\frac {1}{c+1-\gamma +k}}-{\frac {1}{c+1-\alpha +k}}-{\frac {1}{c+1-\beta +k}}\right)\right)}$
Now,
${\displaystyle y_{b}=b_{0}\sum _{r=0}^{\infty }{\left({\frac {(c-\beta )(c)_{r}(c+1-\gamma )_{r}}{(c+1-\alpha )_{r}(c+1-\beta )_{r}}}s^{r+c}\right)}=b_{0}s^{c}\sum _{r=0}^{\infty }{M_{r}s^{r}}}$
Hence,
{\displaystyle {\begin{aligned}{\frac {\partial y}{\partial c}}&=b_{0}s^{c}\ln(s)\sum _{r=0}^{\infty }{\frac {(c-\beta )(c)_{r}(c+1-\gamma )_{r}}{(c+1-\alpha )_{r}(c+1-\beta )_{r}}}s^{r}\\&\quad +b_{0}s^{c}\sum _{r=0}^{\infty }{\frac {(c-\beta )(c)_{r}(c+1-\gamma )_{r}}{(c+1-\alpha )_{r}(c+1-\beta )_{r}}}\left({\frac {1}{c-\beta }}+\sum _{k=0}^{r-1}\left({\frac {1}{c+k}}+{\frac {1}{c+1-\gamma +k}}-{\frac {1}{c+1-\alpha +k}}-{\frac {1}{c+1-\beta +k}}\right)\right)s^{r}\end{aligned}}}
Hence,
${\displaystyle {\frac {\partial y}{\partial c}}=b_{0}s^{c}\sum _{r=0}^{\infty }{\frac {(c-\beta )(c)_{r}(c+1-\gamma )_{r}}{(c+1-\alpha )_{r}(c+1-\beta )_{r}}}\left(\ln(s)+{\frac {1}{c-\beta }}+\sum _{k=0}^{r-1}\left({\frac {1}{c+k}}+{\frac {1}{c+1-\gamma +k}}-{\frac {1}{c+1-\alpha +k}}-{\frac {1}{c+1-\beta +k}}\right)\right)s^{r}}$
At c = α we get y2. Hence, y = Ey1 + Fy2. Let Eb0 = E and Fb0 = F. Noting that s = x−1 we get
{\displaystyle {\begin{aligned}y&=E\left\{{\frac {1}{(\beta +1-\alpha )_{\alpha -\beta -1}}}\sum _{r=\alpha -\beta }^{\infty }{\frac {(\beta )_{r}(\beta +1-\gamma )_{r}}{(1)_{r}(1)_{r+\beta -\alpha }}}x^{-r}\right\}+\\&\quad +F\left\{x^{-\alpha }\sum _{r=0}^{\infty }{\frac {(\alpha -\beta )(\alpha )_{r}(\alpha +1-\gamma )_{r}}{(1)_{r}(\alpha +1-\beta )_{r}}}\left(\ln \left(x^{-1}\right)+{\frac {1}{\alpha -\beta }}+\sum _{k=0}^{r-1}\left({\frac {1}{\alpha +k}}+{\frac {1}{\alpha +1+k-\gamma }}-{\frac {1}{1+k}}-{\frac {1}{\alpha +1+k-\beta }}\right)\right)x^{-r}\right\}\end{aligned}}}
#### α − β < 0Edit
From the symmetry of the situation here, we see that
{\displaystyle {\begin{aligned}y&=G\left\{{\frac {1}{(\alpha +1-\beta )_{\beta -\alpha -1}}}\sum _{r=\beta -\alpha }^{\infty }{\frac {(\alpha )_{r}(\alpha +1-\gamma )_{r}}{(1)_{r}(1)_{r+\alpha -\beta }}}x^{-r}\right\}+\\&\quad +H\left\{x^{-\beta }\sum _{r=0}^{\infty }{\frac {(\beta -\alpha )(\beta )_{r}(\beta +1-\gamma )_{r}}{(1)_{r}(\beta +1-\alpha )_{r}}}\left(\ln \left(x^{-1}\right)+{\frac {1}{\beta -\alpha }}+\sum _{k=0}^{r-1}\left({\frac {1}{\beta +k}}+{\frac {1}{\beta +1+k-\gamma }}-{\frac {1}{1+k}}-{\frac {1}{\beta +1+k-\alpha }}\right)\right)x^{-r}\right\}\end{aligned}}}
## ReferencesEdit
• Ian Sneddon (1966). Special functions of mathematical physics and chemistry. OLIVER B. ISBN 978-0-05-001334-2.
Abramowitz, Milton; Stegun, Irene A. (1964). Handbook of Mathematical Functions. New York: Dover. ISBN 978-0486612720.
1. a b c Abramowitz and Stegun Invalid <ref> tag; name "AbSteg" defined multiple times with different content Invalid <ref> tag; name "AbSteg" defined multiple times with different content |
### Writing linear equations
There are two ways of using information to write the equation of a line. Both work equally as well, and unless are you asked to use a specific method, you can use whichever one makes more sense to you.
Method One:Slope-Intercept Form ($$y = mx + b$$) and solving for “b”.
Example 1:
Write the equation of the line with a slope of $$\Large \frac{9}{4}$$ and goes through the point (4,5).
We want to get our equation in the form $$y = mx + b$$. “m” stands for slope, so we have that part already! Since they also gave us a point, we can use these numbers and plug them in for x and y.
$$m = \Large \frac{9}{4}$$ $$x = 4$$ $$y = 5$$
$$y = mx + b$$
$${\text{5 }} = \Large \frac{9}{4}\left( {\text{4}} \right){\text{ }} + b$$
Simplify and solve for b.
$${\text{5 }} = {\text{ 9 }} + b$$
$$- {\text{4 }} = b$$
Now, plug “m” and “b” into your slope-intercept form
Answer: $$y = \Large \frac{9}{4}x--{\text{ 4}}$$
Method Two:Point-Slope Form
This way might be a little less confusing, but you have to remember a formula in order to use this method.
Point-Slope Form ►$$(y - {y_1}) = m(x - {x_1})$$
Example 2:
Write the equation of a line with a slope of $$\Large \frac{5}{3}$$ and goes through point (3,1)
We don’t have to worry about “b” in this method, so we already have all the parts that we need!
m = $$\Large \frac{5}{3}$$ $${x_1} = 3$$ $${y_1} = 1$$
$$(y - {y_1}) = m(x - {x_1})$$ Plug in values.
$$(y - 1) = {\Large \frac{5}{3}}(x - 3)$$ Distributive.
$$y - 1 = {\Large \frac{5}{3}x} - 5$$ Add 1 to both sides.
Answer: $$y = {\Large \frac{5}{3}x} - 4$$
You can also figure out the equation of a line if you are only given 2 points. There is one extra step and then you can choose one of the above methods to complete it. Here’s an example:
Example 3:
Write the equation of a line that passes through points (4, -2) & (-6, 0)
It doesn’t matter which method we would like to use, for both of them we first need a slope! You can use both points to find the slope.
Slope = $${\Large \frac{{{y_2} - {y_1}}}{{{x_2} - x}}_1}$$
Let’s label our points:
$$(4, - 2)( - 6,0)$$
$$({x_1},{y_1})({x_2},{y_2})$$
Now, plug them in.
$${\Large \frac{{{y_2} - {y_1}}}{{{x_2} - x}}_1} = \Large \frac{{0 - ( - 2)}}{{ - 6 - 4}} = \Large \frac{2}{{ - 10}} = - \Large \frac{1}{5}$$
So, the slope is
$$- \Large \frac{1}{5}$$
We will finish this problem using both methods, so you can decide which way you like best.
Slope-Intercept Method
Choose one of the points that we started with. (-6,0)
$$m = - \Large \frac{1}{5}$$ $$x = - 6$$ $$y = 0$$
$$y = mx + b$$
$$0 = - {\Large \frac{1}{5}}( - 6) + b$$
$$0 = {\Large \frac{6}{5}} + b$$
$$- \Large \frac{6}{5} = b$$
Answer: $$y = - \Large \frac{1}{5}x - \Large \frac{6}{5}$$
Point-Slope Method
$$m = - \Large \frac{1}{5}$$ $${x_1} = - 6$$ $${y_1} = 0$$
$$(y - {y_1}) = m(x - {x_1})$$
$$(y - 0) = - \Large \frac{1}{5}(x + 6)$$
$$y = - \Large \frac{1}{5}x - \Large \frac{6}{5}$$
Below you can download some free math worksheets and practice.
2062 x
Write the slope-intercept form of the equation of each line given the slope and y-intercept.
This free worksheet contains 10 assignments each with 24 questions with answers.
Example of one question:
Watch below how to solve this example:
1915 x
Write the slope-intercept form of the equation of the line through the given point with the given slope.
This free worksheet contains 10 assignments each with 24 questions with answers.
Example of one question:
Watch below how to solve this example:
1919 x
Write the standard form of the equation of the line described.
This free worksheet contains 10 assignments each with 24 questions with answers.
Example of one question:
Watch below how to solve this example:
### Geometry
Circles
Congruent Triangles
Constructions
Parallel Lines and the Coordinate Plane
Properties of Triangles
### Algebra and Pre-Algebra
Beginning Algebra
Beginning Trigonometry
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# Proof of commutativity and associativity for binary error pattern
• Oct 31st 2012, 12:35 PM
x3bnm
Proof of commutativity and associativity for binary error pattern
There's a math problem on page-34 of Pinter's "A Book of Abstract Algebra" book which is(Problem F):
If a word is $\displaystyle \mathbf{a} = a_1 a_2 \cdots a_n$ is sent, but a word $\displaystyle \mathbf{b} = b_1 b_2 \cdots b_n$ is received(where the $\displaystyle a_i$
and $\displaystyle b_j$ are $\displaystyle 0s$ or $\displaystyle 1s$), then the error pattern is the word $\displaystyle e = e_1 e_2 ... e_n$ where:
$\displaystyle e_i = \begin{cases} 0 & \text{ if } a_i = b_i \\ 1 & \text{ if } a_i \neq b_i \end{cases}$
With this motivation, we define an operation of adding words, as follows: If $\displaystyle \mathbf{a}$ and $\displaystyle \mathbf{b}$ are both of length
$\displaystyle 1$, we add them according to the rules
$\displaystyle 0 + 0 = 0$, $\displaystyle 1 + 1 = 0$, $\displaystyle 0 + 1 = 1$ and $\displaystyle 1 + 0 = 1$
If $\displaystyle \mathbf{a}$ and $\displaystyle \mathbf{b}$ are both of length $\displaystyle n$, we add them by adding corresponding digits. That is(let us
introduce commas for convenience),
$\displaystyle (a_1, a_2, \cdots, a_n) + (b_1, b_2, \cdots, b_n) = (a_1 + b_1, a_2 + b_2, \cdots, a_n + b_n)$
Thus the sum of $\displaystyle \mathbf{a}$ and $\displaystyle \mathbf{b}$ is the error pattern $\displaystyle \mathbf{e}$.
For example,
$\displaystyle 0010110 + 0011010 = 0001100$ and $\displaystyle 10100111 + 11110111 = 01010000$
The symbol $\displaystyle \mathbb{B}^n$ will designate the set of all the binary words of length $\displaystyle n$.
We will prove that the operation of word addition has the following properties on $\displaystyle \mathbb{B}^n$:
1. It is commutative.
2. It is associative.
3. There is an identity element for word addition.
4. Every word has an inverse under word addition.
.....
1) Show that
$\displaystyle (a_1, a_2, \cdots, a_n) + (b_1, b_2, \cdots b_n) = (b_1, b_2, \cdots, b_n) + (a_1, a_2, \cdots, a_n)$
3) Show that
\displaystyle \begin{align*}(a_1, a_2, \cdots, a_n) + [(b_1, b_2, \cdots, b_n) + (c_1, c_2, \cdots, c_n)] =& [ (a_1, a_2, \cdots, a_n) + (b_1, b_2, \cdots, b_n) ] \\ + & (c_1, c_2, \cdots, c_n) \end{align*}.
How can I prove that the commutativity and associativity holds in this case?
• Oct 31st 2012, 01:58 PM
emakarov
Re: Proof of commutativity and associativity for binary error pattern
For each digit, this is addition modulo 2. It;s a well-known fact that addition modulo n is commutative and associative.
• Oct 31st 2012, 02:36 PM
x3bnm
Re: Proof of commutativity and associativity for binary error pattern
Quote:
Originally Posted by emakarov
For each digit, this is addition modulo 2. It;s a well-known fact that addition modulo n is commutative and associative.
Thanks emakarov for help. |
# How To Graph Tangent and Cotangent F07
```Trig F07 O’Brien
How to Graph y a tan (bx c) d and y a cot (bx - c) d
1.
Identify a, b, c, and d.
2.
5.
Determine if there has been an x-axis reflection: if a is negative, there has been an x-axis reflection.
π
Find the period: period = .
b
1
Find the x-increment: x-increment = period .
4
Find the phase shift by setting the argument, bx – c, equal to zero and solving for x:
c
c
c
c
P.S. =
right (if we see bx – c and
is positive) or P.S. =
left (if we see bx + c and
is negative).
b
b
b
b
Find the vertical shift: |d| up if d is positive or |d| down if d is negative.
6.
Identify the left asymptote, the three key points, and the right asymptote.
3a.
b.
4.
cotangent: To find the left asymptote, let x = phase shift, x
LA: x
c
b
(x of LA + x-increment, d + a)
c
.
b
(x1 + x-inc, d)
(x2 + x-inc, d – a)
RA: x = x3 + x-inc
Note: The left asymptote of a tangent function is NOT x = phase shift.
π
tangent: To find the left asymptote, set the argument, bx – c, equal to and solve for x.
2
c π
LA: x
(x of LA + x-increment, d – a) (x1 + x-inc, d) (x2 + x-inc, d + a) RA: x = x3 + x-inc
b 2b
7.
Plot the asymptotes and the three key points. Connect the points with a smooth, continuous curve. Extend the graph
to two full periods.
Example 1: y = –3 cot (2x + π) +1
1.
a = –3
2.
There is an x-axis reflection.
3.
period =
4.
phase shift =
5.
vertical shift = 1 up
6.
LA: x
b=2
π
2
c = –π
d=1
x-increment =
π
left
2
π
4π
2
8
1 π π
4 2 8
2x π 0 2x π x
3π
, 2
8
2π
, 1
8
π
, 4
8
π
4π
2
8
RA: x = 0
7.
1
Trig F07 O’Brien
1
Example 2: y = 2 tan x π – 3
4
b=
1
4
c=π
d = –3
1.
a=2
2.
There is no x-axis reflection.
3.
period =
π
1
4π
x-increment =
4
4.
phase shift = 4π right
5.
vertical shift = 3 down
6.
LA:
1
4π π
4
1
xπ 0
4
1
π
xπ
x 4π 2π x 2 π
4
2
x 4π 0
(3π, –5)
x 4π
(4π, –3)
(5π, –1)
RA: x = 6π
7.
2
``` |
This is the first of a series of posts in which a problem is presented to the reader, who in invited to solve it by following a proposed path. The background knowledge required is mostly that of a second year high school student. The focus, in fact, is rather on finding simple and efficient solutions.
Given a square $$ABCD$$, let $$E$$, $$F$$, and $$G$$ be the midpoints of segments $$AD$$, $$EC$$, and $$BF$$, respectively. Find the area of $$\triangle BDG$$.
An accurate drawing is often a good starting point. While drawing, in fact, geometrical constraints might arise that can be of help once they are “translated” into algebraic expressions. For exampe, you may notice, in our case, the symmetrical position of point $$F$$ with respect to sides $$AD$$ and $$BC$$ of the square. Too many lines, though, can be confusing.
1. Concentrate first on triangle $$\triangle BDF$$ and note that $$DG$$ is the median relative to side $$BF$$.
2. What is the relationship, then, between the area we want to calculate, i.e. that of $$\triangle BDG$$, and the area of $$\triangle BDF$$?
3. Thus, we can try to calculate the area of $$\triangle BDF$$ and then divide by two to get our final result…
4. Let us now go back to the symmetry of $$F$$ and try to express it in mathematical terms. Draw from $$F$$ the lines parallel to the square sides.
5. Notice that the lines you just drew bisect $$EC$$ and use Thales Theorem to determine the distance of $$F$$ from $$BC$$ and $$CD$$.
6. Using these results, you can now calculate the area of $$\triangle BDF$$ by subracting, from the area of the square, that of three triangles. Which triangles? What are their areas?
7. Finally divide by two to get $$\mathcal A_{BDG} = \frac{1}{16}$$. |
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# How to Convert cm to mm
Both centimeters and millimeters are derived from the "meter," a measurement of distance used in the metric system. The prefix centi- means "one-hundredth," so there are 100 centimeters in every meter. The prefix milli- means "one-thousandth," so there are 1000 millimeters in every meter. Millimeters and centimeters are only separated by one tens place, which means that there are ultimately 10 millimeters for every centimeter.
### Part 1 Mathematical Formula
#### Centimeters to Millimeters
1. 1
Examine the problem. It must describe a measurement of length in centimeters (cm) and instruct you to find the equivalent value in millimeters (mm).
• Example: The width of a particular table measures 58.75 centimeters. What is the width of the same table when taken in millimeters?
2. 2
Multiply the centimeter value by 10. There are 10 millimeters in every 1 centimeter. This means that you will need to find the number of millimeters per centimeter by multiplying the centimeter measurement by 10.[1]
• The "millimeter" is a smaller unit than the "centimeter," even though both are derived from the basic "meter." When you convert any larger metric unit to a smaller one, you must multiply the original value.
• Example: 58.75 cm * 10 = 587.5 mm
• The width of the table in the problem is 587.5 millimeters.
#### Millimeters to Centimeters
1. 1
Examine the problem. Read through the problem and make sure that it contains a measurement of length provided in millimeters (mm). It must also instruct you to convert that measurement to its equivalent in centimeters (cm).
• Example: The height of a certain door is 1780.9 millimeters. Find the height of the same door when described in centimeters.
2. 2
Divide the millimeter value by 10. There are 10 millimeters for every 1 centimeter, so you'll need to divide the number of millimeters by 10 to find its equivalent centimeter value.[2]
• The "centimeter" is larger than the "millimeter," and whenever you need to convert a smaller metric unit to a larger one, you must divide the original value.
• Example: 1780.9 mm / 10 = 178.09 cm
• The height of the door in this problem is 178.09 centimeters.
### Part 2 Moving the Decimal Point
#### Centimeters to Millimeters
1. 1
Look at the problem. Verify that the problem provides a measurement of length in centimeters (cm). It should also indicate that the measurement must be converted into its equivalent number of millimeters (mm).
• Example: The length of a certain television screen is 32.4 centimeters. Find the length of the same screen in millimeters.
2. 2
Move the decimal one place to the right. There are 10 millimeters in every 1 centimeter, so the centimeter value will be smaller by one decimal place. You should be able to convert centimeters into millimeters by shifting the decimal point one place to the right.
• Moving the decimal point of a number to the right increases its value, and every place holder shift equals a value of 10. As such, shifting the decimal to the right once will increase the resulting value by a factor of 10.
• Example: Moving the decimal point in "32.4" to the right once would result in a value of "324.0," so the length of the screen in this problem equals 324.0 millimeters.
#### Millimeters to Centimeters
1. 1
Look at the problem. Look over the problem and verify that it provides a measurement of length described in millimeters (mm). It should also instruct you to convert that value into its equivalent in centimeters (cm).
• Example: The height of a particular chair measures 958.3 millimeters. What is the height of the same chair in centimeters?
2. 2
Move the decimal one place to the left. There are 10 millimeters for every 1 centimeter, so the millimeter value will be larger by one decimal place. As such, you can convert millimeters into centimeters by shifting the decimal point one place to the left.
• Moving the decimal point to the left makes the resulting value smaller, and each decimal place holder represents a factor of 10. This means that shifting the decimal one place to the left will decrease the resulting value by a factor of 10.
• Example: Moving the decimal point in "958.3" to the left once would result in a value of "95.83," so the height of the chair in this problem measures 95.83 centimeters.
1. 1
Convert 184 centimeters to millimeters. In this problem, you need to change the centimeter value into its equivalent in millimeters. Do so by either multiplying the centimeter value by 10 or by shifting the decimal one place to the right.
• Mathematical Conversion:
• 184 cm * 10 = 1840 mm
• Decimal Shift:
• 184.0 cm => shift the decimal to the right once => 1840 mm
2. 2
Convert 90.5 millimeters to centimeters. This problem asks you to find the equivalent centimeter amount for a value provided in millimeters. You can do so by dividing the original millimeter value by 10. Alternatively, you can shift the decimal point in the original millimeter value one place to the left.
• Mathematical Conversion:
• 90.5 mm / 10 = 9.05 cm
• Decimal Shift:
• 90.5 mm => shift the decimal to the left once => 9.05 cm
3. 3
Convert 72.6 centimeters to millimeters. For this problem, you'll need to find the equivalent millimeter value of a number currently described in centimeters. Accomplish this by either multiplying the centimeter value by 10 or by shifting its decimal point one place to the right.
• Mathematical Conversion:
• 72.6 cm * 10 = 726 mm
• Decimal Shift:
• 72.6 cm => shift the decimal to the right once => 726 mm
4. 4
Convert 315 millimeters to centimeters. This problem instructs you to change the value of a number provided in millimeters to its equivalent in centimeters. Divide the original millimeter value by 10 to accomplish this task, or simply shift the decimal point one place to the left.
• Mathematical Conversion:
• 315 mm / 10 = 31.5 cm
• Decimal Shift:
• 315.0 mm => shift the decimal to the left once => 31.5 cm |
## Step 1: Determine the least common denominator
If the denominators are not the same, determine the least common denominator (LCD) shared by the fractions. For example, let’s look for the denominator of the following:
## Step 2: Express the fractions to share the lowest common denominator
Express the fractions so that they share the LCD as their denominator. Hence, we express 21/25 and 17/50 so that they share 50 as the denominator.
## Step 3: Subtract the numerators
Now that the denominators are the same, we can subtract the numerators directly.
## Step 4: Cancel the common factors
Simplify the difference of the fractions by cancelling out common factors.
## Examples of how to subtract fractions
Q1) What is the result if we subtract 11/40 from 19/40?
Since they share the same denominators, we can subtract the numerators right away.
Simplify the difference by cancelling out the common factors.
Q2) Blake has ordered 4/5 tons of bananas for his banana chip business. His first batch of banana chips required 11/30 tons of bananas. How much of the bananas are left?
To find the remaining amount of bananas, we subtract 11/30 from 4/5.
Since the denominators are not the same, we’ll need to find the LCD shared by the fractions.
Now, rewrite the fractions so that they both have 30 as their denominator.
Once the fractions share a common denominator, we can now subtract the numerators right away.
This means that Blake would still have 13/30 tons of bananas left after the first batch. |
# Precalculus : Divide Polynomials by Binomials Using Synthetic Division
## Example Questions
### Example Question #1 : Divide Polynomials By Binomials Using Synthetic Division
Divide the polynomial by .
Explanation:
Our first step is to list the coefficients of the polynomials in descending order and carry down the first coefficient.
We multiply what's below the line by and place the product on top of the line. We find the sum of this number with the next coefficient and place the sum below the line. We keep repeating these steps until we've reached the last coefficients.
To write the answer, we use the numbers below the line as our new coefficients. The last number is our remainder.
with remainder
This can be rewritten as
Keep in mind: the highest degree of our new polynomial will always be one less than the degree of the original polynomial.
### Example Question #1 : Synthetic Division And The Remainder And Factor Theorems
Divide the polynomial by .
Explanation:
Our first step is to list the coefficients of the polynomials in descending order and carry down the first coefficient.
We multiply what's below the line by and place the product on top of the line. We find the sum of this number with the next coefficient and place the sum below the line. We keep repeating these steps until we've reached the last coefficients.
To write the answer, we use the numbers below the line as our new coefficients. The last number is our remainder.
with remainder
This can be rewritten as:
Keep in mind: the highest degree of our new polynomial will always be one less than the degree of the original polynomial.
### Example Question #1 : Synthetic Division And The Remainder And Factor Theorems
Divide the polynomial by .
Explanation:
Our first step is to list the coefficients of the polynomials in descending order and carry down the first coefficient.
We mulitply what's below the line by 1 and place the product on top of the line. We find the sum of this number with the next coefficient and place the sum below the line. We keep repeating these steps until we've reached the last coefficients.
To write the answer, we use the numbers below the line as our new coefficients. The last number is our remainder.
with remainder
Keep in mind: the highest degree of our new polynomial will always be one less than the degree of the original polynomial.
### Example Question #1 : Divide Polynomials By Binomials Using Synthetic Division
Divide the polynomial by .
Explanation:
Our first step is to list the coefficients of the polynomials in descending order and carry down the first coefficient.
Remember to place a when there isn't a coefficient given.
We multiply what's below the line by and place the product on top of the line. We find the sum of this number with the next coefficient and place the sum below the line. We keep repeating these steps until we've reached the last coefficients.
To write the answer, we use the numbers below the line as our new coefficients. The last number is our remainder.
with remainder
This can be rewritten as:
Keep in mind: the highest degree of our new polynomial will always be one less than the degree of the original polynomial.
### Example Question #1 : Synthetic Division And The Remainder And Factor Theorems
Use synthetic division to divide by .
Remainder
Remainder
Remainder
Explanation:
To divide synthetically, we begin by drawing a box. On the inside separated by spaces, we write the coefficients of the terms of our polynomial being divided. On the outside, we write the root that would satisfy our binomial , namely . Leaving a space for another row of numbers, we then draw a line below our row of coefficients.
We then begin dividing by simply carrying our first coefficient (1) down below the line.
We then multiply this 1 by our divisor (3) and write the resulting product (3) below our next coefficient.
We then add the two numbers in that column and write the sum (5) below the line.
We then simply continue the process by multiplying this 5 by our divisor 3 and writing that product in the next column, adding it to the next coefficient, and continuing until we finish the columns.
We then need to translate our bottom row of numbers into the coefficients of our new quotient. Since the first column originally corresponded to our cubic term, it will now correspond to the quadratic term meaning that our 1 can be translated as . Similarly, our second column transitions from quadratic to linear, making our 5 become . Finally, our third column becomes the constant term, meaning 8 simply remains the constant 8. Finally, our former constant column becomes the column for our remainder. However, since we have a 0, we have no remainder and can disregard it.
Putting all of this together gives us a final answer of
### Example Question #6 : Divide Polynomials By Binomials Using Synthetic Division
Divide using synthetic division:
Explanation:
First, set up the synthetic division problem by lining up the coefficients. There are a couple of different strategies - for this one, we will put a -7 in the top corner and add the columns.
_________________________
The first step is to bring down the first 1. Then multiply what is below the line by the -7 in the box, write it below the next coefficient, and then add the columns:
_________________________
We can interpret this answer as meaning
### Example Question #7 : Divide Polynomials By Binomials Using Synthetic Division
What is the result when is divided by ? |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
10.4: Circumference and Arc Length
Difficulty Level: At Grade Created by: CK-12
Learning Objectives
• Find the circumference of a circle.
• Define the length of an arc and find arc length.
Review Queue
1. Find a central angle in that intercepts \begin{align*}\widehat{CE}\end{align*}
2. Find an inscribed angle that intercepts \begin{align*}\widehat{CE}\end{align*}.
3. How many degrees are in a circle? Find \begin{align*}m \widehat{ECD}\end{align*}.
4. If \begin{align*}m \widehat{CE} =26^\circ\end{align*}, find \begin{align*}m \widehat{CD}\end{align*} and \begin{align*}m \angle CBE\end{align*}.
Know What? A typical large pizza has a diameter of 14 inches and is cut into 8 pieces. Think of the crust as the circumference of the pizza. Find the length of the crust for the entire pizza. Then, find the length of the crust for one piece of pizza if the entire pizza is cut into 8 pieces.
Circumference of a Circle
Circumference: The distance around a circle.
The circumference can also be called the perimeter of a circle. However, we use the term circumference for circles because they are round. In order to find the circumference of a circle, we need to explore \begin{align*}\pi\end{align*} (pi).
Investigation 10-1: Finding \begin{align*}\pi\end{align*} (pi)
Tools Needed: paper, pencil, compass, ruler, string, and scissors
1. Draw three circles with radii of 2 in, 3 in, and 4 in. Label the centers of each \begin{align*}A, B\end{align*}, and \begin{align*}C\end{align*}.
2. Draw in the diameters and determine their lengths.
3. Take the string and outline each circle with it. Cut the string so that it perfectly outlines the circle. Then, lay it out straight and measure it in inches. Round your answer to the nearest \begin{align*}\frac{1}{8}\end{align*}-inch. Repeat this for the other two circles.
4. Find \begin{align*}\frac{circumference}{diameter}\end{align*} for each circle. Record your answers to the nearest thousandth.
You should see that \begin{align*}\frac{circumference}{diameter}\end{align*} approaches 3.14159... We call this number \begin{align*}\pi\end{align*}, the Greek letter “pi.” When finding the circumference and area of circles, we must use \begin{align*}\pi\end{align*}.
\begin{align*}\pi\end{align*}, or “pi”: The ratio of the circumference of a circle to its diameter. It is approximately equal to 3.14159265358979323846...
To see more digits of \begin{align*}\pi\end{align*}, go to http://www.eveandersson.com/pi/digits/.
From Investigation 10-1, we found that \begin{align*}\frac{circumference}{diameter}=\pi\end{align*}. Let’s solve for the circumference, \begin{align*}C\end{align*}.
\begin{align*}\frac{C}{d} &= \pi\\ C &= \pi d\end{align*}
We can also say \begin{align*}C=2 \pi r\end{align*} because \begin{align*}d=2r\end{align*}.
Circumference Formula: \begin{align*}C=\pi d\end{align*} or \begin{align*}C=2 \pi r\end{align*}
\begin{align*}d=2r\end{align*}
Example 1: Find the circumference of a circle with a radius of 7 cm.
Solution: Plug the radius into the formula.
\begin{align*}C=2 \pi (7)=14 \pi \approx 44 \ cm\end{align*}
Example 2: The circumference of a circle is \begin{align*}64 \pi\end{align*}. Find the diameter.
Solution: Again, you can plug in what you know into the circumference formula and solve for \begin{align*}d\end{align*}.
\begin{align*}64 \pi &= \pi d\\ 64 &= d\end{align*}
Example 3: A circle is inscribed in a square with 10 in. sides. What is the circumference of the circle? Leave your answer in terms of \begin{align*}\pi\end{align*}.
Solution: From the picture, we can see that the diameter of the circle is equal to the length of a side. \begin{align*}C=10 \pi \ in\end{align*}.
Example 4: Find the perimeter of the square. Is it more or less than the circumference of the circle? Why?
Solution: The perimeter is \begin{align*}P=4(10)=40 \ in\end{align*}. In order to compare the perimeter with the circumference we should change the circumference into a decimal.
\begin{align*}C=10 \pi \approx 31.42 \ in\end{align*}. This is less than the perimeter of the square, which makes sense because the circle is inside the square.
Arc Length
In Chapter 9, we measured arcs in degrees. This was called the “arc measure” or “degree measure.” Arcs can also be measured in length, as a portion of the circumference.
Arc Length: The length of an arc or a portion of a circle’s circumference.
The arc length is directly related to the degree arc measure.
Example 5: Find the length of \begin{align*}\widehat{PQ}\end{align*}. Leave your answer in terms of \begin{align*}\pi\end{align*}.
Solution: In the picture, the central angle that corresponds with \begin{align*}\widehat{PQ}\end{align*} is \begin{align*}60^\circ\end{align*}. This means that \begin{align*}m \widehat{PQ}=60^\circ\end{align*}. Think of the arc length as a portion of the circumference. There are \begin{align*}360^\circ\end{align*} in a circle, so \begin{align*}60^\circ\end{align*} would be \begin{align*}\frac{1}{6}\end{align*} of that \begin{align*}\left(\frac{60^\circ}{360^\circ}=\frac{1}{6}\right)\end{align*}. Therefore, the length of \begin{align*}\widehat{PQ}\end{align*} is \begin{align*}\frac{1}{6}\end{align*} of the circumference. length of \begin{align*}\widehat{PQ}=\frac{1}{6} \cdot 2 \pi (9)=3 \pi\end{align*}
Arc Length Formula: The length of \begin{align*}\widehat{AB}=\frac{m \widehat{AB}}{360^\circ} \cdot \pi d\end{align*} or \begin{align*}\frac{m \widehat{AB}}{360^\circ} \cdot 2 \pi r\end{align*}.
Another way to write this could be \begin{align*}\frac{x^\circ}{360^\circ} \cdot 2 \pi r\end{align*}, where \begin{align*}x\end{align*} is the central angle.
Example 6: The arc length of \begin{align*}\widehat{AB}=6 \pi\end{align*} and is \begin{align*}\frac{1}{4}\end{align*} the circumference. Find the radius of the circle.
Solution: If \begin{align*}6 \pi\end{align*} is \begin{align*}\frac{1}{4}\end{align*} the circumference, then the total circumference is \begin{align*}4(6 \pi )=24 \pi\end{align*}. To find the radius, plug this into the circumference formula and solve for \begin{align*}r\end{align*}.
\begin{align*}24 \pi &= 2 \pi r\\ 12 &= r\end{align*}
Example 7: Find the measure of the central angle or \begin{align*}\widehat{PQ}\end{align*}.
Solution: Let’s plug in what we know to the Arc Length Formula.
\begin{align*}15 \pi &= \frac{m \widehat{PQ}}{360^\circ} \cdot 2 \pi (18)\\ 15 &= \frac{m \widehat{PQ}}{10^\circ}\\ 150^\circ &= m \widehat{PQ}\end{align*}
Example 8: The tires on a compact car are 18 inches in diameter. How far does the car travel after the tires turn once? How far does the car travel after 2500 rotations of the tires?
Solution: One turn of the tire is the circumference. This would be \begin{align*}C=18 \pi \approx 56.55 \ in\end{align*}. 2500 rotations would be \begin{align*}2500 \cdot 56.55 \ in=141371.67 \ in\end{align*}, 11781 ft, or 2.23 miles.
Know What? Revisited The entire length of the crust, or the circumference of the pizza is \begin{align*}14 \pi \approx 44 \ in\end{align*}. In \begin{align*}\frac{1}{8}\end{align*} of the pizza, one piece would have \begin{align*}\frac{44}{8} \approx 5.5\end{align*} inches of crust.
Review Questions
• Questions 1-10 are similar to Examples 1 and 2.
• Questions 11-14 are similar to Examples 3 and 4.
• Questions 15-20 are similar to Example 5.
• Questions 21-23 are similar to Example 6.
• Questions 24-26 are similar to Example 7.
• Questions 27-30 are similar to Example 8.
Fill in the following table. Leave all answers in terms of \begin{align*}\pi\end{align*}.
1. 15
2. 4
3. 6
4. \begin{align*}84 \pi\end{align*}
5. 9
6. \begin{align*}25\pi\end{align*}
7. \begin{align*}2\pi\end{align*}
8. 36
1. Find the radius of circle with circumference 88 in.
2. Find the circumference of a circle with \begin{align*}d=\frac{20}{\pi} \ cm\end{align*}.
Square \begin{align*}PQSR\end{align*} is inscribed in \begin{align*}\bigodot T\end{align*}. \begin{align*}RS=8 \sqrt{2}\end{align*}.
1. Find the length of the diameter of \begin{align*}\bigodot T\end{align*}.
2. How does the diameter relate to \begin{align*}PQSR\end{align*}?
3. Find the perimeter of \begin{align*}PQSR\end{align*}.
4. Find the circumference of \begin{align*}\bigodot T\end{align*}.
Find the arc length of \begin{align*}\widehat{PQ}\end{align*} in \begin{align*}\bigodot A\end{align*}. Leave your answers in terms of \begin{align*}\pi\end{align*}.
Find \begin{align*}PA\end{align*} (the radius) in \begin{align*}\bigodot A\end{align*}. Leave your answer in terms of \begin{align*}\pi\end{align*}.
Find the central angle or \begin{align*}m \widehat{PQ}\end{align*} in \begin{align*}\bigodot A\end{align*}. Round any decimal answers to the nearest tenth.
For questions 27-30, a truck has tires with a 26 in diameter.
1. How far does the truck travel every time a tire turns exactly once? What is this the same as?
2. How many times will the tire turn after the truck travels 1 mile? (1 mile = 5280 feet)
3. The truck has travelled 4072 tire rotations. How many miles is this?
4. The average recommendation for the life of a tire is 30,000 miles. How many rotations is this?
1. \begin{align*}\angle CAE\end{align*}
2. \begin{align*}\angle CBE\end{align*}
3. \begin{align*}360^\circ, 180^\circ\end{align*}
4. \begin{align*}m\widehat{CD} = 180^\circ - 26^\circ = 154^\circ, m \angle CBE = 13^\circ\end{align*}
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Refer to our Texas Go Math Grade 6 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 6 Lesson 5.2 Answer Key Adding Integers with Different Signs.
Reflect
Question 1.
Make a Prediction Predict the sum of – 2 + 2. Explain your prediction and check it using the number line.
The sum of 2 + 2 is 0 because you start at – 2 and move |2| = 2 units right on the number line.
First graph number line with units.
Start at – 2 and move 2 units to the right.
You are 0 units left from 0 on the number line.
The sum is 0.
Number line:
The sum of – 2 + 2 is 0 because you start at – 2 and move 2 units right, or in the positive direction.
Make a Prediction Kyle models a sum of two integers. He uses more negative (red) counters than positive (yellow) counters. What do you predict about the sign of the sum? Explain.
The sign of the sum that Kyle models is – (minus).
One positive (yellow) counter and one negative (red) counter form a zero pair, so when you eliminate alt zero pairs, there are some negative (red) counters left.
The sign of the sum that Kyle models is minus, or – because he uses more negative counters.
Model and find each sum using counters.
Question 3.
5 + (- 1) ____________
The sum you want to find is:
5 + (- 1)
Begin with 5 yellow counters to represent 5 (see the picture bellow).
Eliminate zero pairs.
When you remove the zero pairs, there are 4 yellow counters left.
The final result is:
5 + (- 1) = 4
There are 5 positive counters and 1 negative and thus 5 + (-1) = 4.
Question 4.
4 + (- 6) ____________
The sum you want to find is:
4 + (- 6)
After 6 red counter to represent adding – 6.
Eliminate zero pairs.
When you remove the zero pairs, there are 2 red counters left.
The final result is:
4 + (- 6) = – 2
There are 4 positive and 6 negative counters and thus 4 + (-6) = – 2.
Question 5.
1 + (- 7) ____________
The sum you want to find is:
1 + (- 7)
After 7 red counter to represent adding – 7.
Eliminate zero pairs.
When you remove the zero pairs, there are 6 red counters left.
The final result is:
1 + (- 7) = – 6
There are 1 positive and 5 negative counters and thus 1 + (-7) = – 6.
3 + (- 4) ____________
The sum you want to find is:
3 + (- 4)
After 4 red counter represents adding – 4.
Eliminate zero pairs.
When you remove the zero pairs, there are 1 red counters left.
The final result is:
3 + (-4) = – 1
There are 3 positive and 4 negative counters and thus 3 + (-4) = – 1.
Find each sum.
Question 7.
– 51 + 23 = ____________
Find absolute values of both integers and subtract lesser absolute value from greater:
|- 51| – |23| = 28
The final result would be 28 or – 28, depending on signs of given integers.
To find the final result, use the sign of number with greater absolute value:
– 51 + 23 = – 28
The sum is – 28.
Find absolute values, subtract them and add minus if the integer with the greater absolute value was negative.
Question 8.
10 + (- 18) = ____________
Find absolute values of both integers and subtract lesser absolute value from greater:
|- 18| – |10| = 8
The final result would be 8 or – 8, depending on signs of given integers.
To find the final result, use the sign of number with greater absolute value:
10 + (- 18) = – 8
The sum is – 8.
Find absolute values, subtract them and add minus if the integer with the greater absolute value was negative.
Question 9.
13 + (- 13) = ____________
The opposite of an integer is called its additive inverse.
Sum of integer and its additive inverse is 0
so final result is:
13 + (- 13) = 0
The sum is 0 because the sum of an integer and its additive inverse is 0.
Question 10.
25 + (- 26) = ____________
Find absolute values of both integers and subtract lesser absolute value from greater:
|- 26| – |25| = 1
The final result would be 1 or – 1, depending on signs of given integers.
To find the final result, use the sign of number with greater absolute value:
25 + (- 26) = – 1
The sum is – 1.
Find absolute values, subtract them and add minus if the integer with the greater absolute value was negative.
Use a number line to find each sum.
Question 1.
9 + (- 3) = __________
First graph number line with units.
Start at 9 and move |- 3| = 3 units in the negative direction (to the left) on the number line.
To find the final result read sum from number line.
Number line:
The sum is 6 because it is 6 units right from 0 on the number line.
Question 2.
– 2 + 7 = ____________
First graph number line with units.
Start at – 2 and move |7| = 7 units in the negative direction (to the left) on the number line.
To find the final result, read sum from number line.
Number line:
The sum is 5 because it is 5 units right from 0 on the number line.
– 15 + 4 = _____________
First graph number line with units.
Start at – 15 and move |4| = 4 units in the negative direction (to the left) on the number line.
To find the final result, read sum from number line.
Number line:
The sum is – 11 because it is 11 units right from 0 on the number line.
Question 4.
1 + (- 4) = _____________
First graph number line with units.
Start at 1 and move |-4| = 4 units in the negative direction (to the left) on the number line.
To find the final result, read sum from number line.
Number line:
The sum is – 3 because it is 3 units right from 0 on the number line.
Circle the zero pairs in each model. Find the sum.
Question 5.
– 4 + 5 = _____________
The sum you want to find is:
– 4 + 5
Start with 4 red counters to represent – 4 and add 5 yellow counters to represent 5.
Find zero pairs and circle them.
When you remove the zero pairs, there is 1 yellow counter left so the sum is:
– 4 + 5 = 1
There are 4 negative counters and 5 positive counters and thus – 4 + 5 = 1.
Question 6.
– 6 + 6 = _____________
The sum you want to find is:
– 6 + 6
Start with 6 red counters to represent – 6 and add 6 yellow counters to represent 6.
Find zero pairs and circle them.
When you remove the zero pairs, there is 0 yellow counter left so the sum is:
– 6 + 6 = 0
There are 6 negative counters and 6 positive counters and thus – 6 + 6 = 0.
2 + (- 5) = _____________
The sum you want to find is:
2 + (- 5)
Find zero pairs and circle them.
When you remove the zero pairs, there are 3 red counters left so the sum is:
2 + (- 5) = – 3
There are 2 positive counters and 5 negative and thus 2 + (- 5) = – 3.
Question 8.
– 3 + 7 = ______________
The sum you want to find is:
– 3 + 7
Find zero pairs and circle them.
When you remove the zero pairs, there are 4 yellow counters left so the sum is:
– 3 + 7 = 4
There are 3 negative and 7 positive counters and thus – 3 + 7 = 4
Find each sum.
Question 9.
– 8 + 14 = ______________
Find absolute values and subtract lesser absolute value from greater:
|14| – |- 8| = 6
The final result would be 6 or – 6, depending on signs of given integers.
To find the final result, use the sign of number with greater absolute value:
– 8 + 14 = 6
The sum is 6.
Find absolute values, subtract them and add minus if the integer with the greater absolute value was negative.
Question 10.
7 + (- 5) = ______________
Find absolute values and subtract lesser absolute value from greater:
|7| – |- 5| = 2
The final result would be 2 or – 2, depending on signs of given integers.
To find the final result, use the sign of number with greater absolute value:
7 + (- 5) = 2
The sum is 2.
Find absolute values, subtract them and add minus if the integer with the greater absolute value was negative.
5 + (- 21) = ______________
Find absolute values and subtract lesser absolute value from greater:
|- 21| – |5| = 16
The final result would be 16 or – 16, depending on signs of given integers.
To find the final result, use the sign of a number with greater absolute value:
5 + (- 21) = – 16
The sum is 16.
Find absolute values, subtract them and add minus if the integer with the greater absolute value was negative.
Question 12.
14 + (- 14) = ______________
The opposite of an integer is called its additive inverse.
Sum of integer and its additive inverse is 0
So final result is:
14 + (- 14) = 0
The sum is 0 because the sum of an integer and its additive inverse is 0.
Question 13.
0 + (- 5) = ______________
Find absolute values and subtract lesser absolute value from greater:
|- 5| – |0| = 5
The final result would be 5 or – 5, depending on signs of given integers.
To find the final result, use the sign of number with greater absolute value:
0 + (- 5) = – 5
The sum is – 5.
Find absolute values, subtract them and add minus if the integer with the greater absolute value was negative.
Question 14.
32 + (- 8) = ______________
Find absolute values and subtract lesser absolute value from greater:
|32| – |- 8| = 24
The final result would be 24 or – 24, depending on signs of given integers.
To find the final result, use the sign of number with greater absolute value:
32 + (- 8) = 24
The sum is 24.
Find absolute values, subtract them and add minus if the integer with the greater absolute value was negative.
Essential Question Check-In
Question 15.
Describe how to find the sums – 4 + 2 and – 4 + (- 2) on a number line.
First sum is:
– 4 + 2
Graph number line with units.
Start at – 4 and move |2| = 2 units to the right to represent adding 2.
Read the sum from number line
Second sum is:
– 4 + (-2)
Start at – 4 and move |- 2| = 2 units to the left to represent adding – 2.
Again, read the sum from number line
Number line:
First sum is – 2 because it is 2 units left from 0 on the number line.
Second sum is – 6 because it is 6 units left from 0 on the number line.
To find these sums, start from – 4 on the number line and move 2 units to the right to represent the first sum, and 2 unit to the left to represent the second sum.
The first sum is – 2 and the second sum is – 6.
Find each sum.
Question 16.
– 15 + 71 = ______________
Find absolute values and subtract lesser absolute value from greater:
|71| – |- 15| = 56
The final result would be 56 or – 56, depending on signs of given integers.
To find the final result use the sign of a number with greater absolute value:
– 15 + 71 = 56
The sum is 56.
Find absolute values, subtract them, and add minus if the integer with the greater absolute value was negative.
– 53 + 45 = ______________
Find absolute values and subtract lesser absolute value from greater:
|- 53| – |- 45| = 8
The final result would be 8 or – 8, depending on signs of given integers.
To find the final result use the sign of number with greater absolute value:
– 53 + 45 = – 8
The sum is – 8.
Find absolute values, subtract them and add minus if the integer with the greater absolute value was negative.
Question 18.
– 79 + 79 = ______________
The opposite of an integer is called its additive inverse.
The sum of an integer and its additive inverse is 0
so the final result is
– 79 + 79 = 0
The sum is 0 because the sum of an integer and its additive inverse is 0.
Question 19.
– 25 + 50 = ______________
Find absolute values and subtract lesser absolute value from greater:
|50| – |- 25| = 25
The final result would be 25 or – 25, depending on signs of given integers.
To find the final result use the sign of number with greater absolute value:
– 25 + 50 = 25
The sum is 25.
Find absolute values, subtract them and add minus if the integer with the greater absolute value was negative.
Question 20.
18 + (- 32) = ______________
Find absolute values and subtract lesser absolute value from greater:
|- 32| – |18| = 14
The final result would be 14 or – 14, depending on signs of given integers.
To find the final result use the sign of number with greater absolute value:
18 + (- 32) = – 14
The sum is – 14.
Find absolute values, subtract them and add minus if the integer with the greater absolute value was negative.
Question 21.
5 + (- 100) = ______________
Find absolute values and subtract lesser absolute value from greater:
|- 100| – | 5 | = 95
The final result would be 95 or – 95, depending on signs of given integers.
To find the final result use the sign of number with greater absolute value:
5 + (- 100) = – 95
The sum is – 95.
Find absolute values, subtract them and add minus if the integer with the greater absolute value was negative.
Question 22.
– 12 + 8 + 7 = ______________
First find the sum:
– 12 + 8
Find absolute values and subtract lesser absolute value from greater:
|- 12| – 8 = 4
The result would be 4 or – 4, depending on signs of given integers.
To find the result, use the sign of number with greater absolute value:
– 12 + 8 = – 4
Now, find the sum:
– 4 + 7
Find absolute values and subtract lesser absolute value from greater:
|7| – |- 4| = 3
The final result would be 3 or – 3, depending on signs of given integers.
To find the result, use the sign of number with greater absolute value:
– 4 + 7 = 3
The final result is 3.
The sum is 3.
First, find the sum – 12 + 8 (find absolute values, subtract them and minus if the integer with the greater absolute value was negative) and then add 7 to that sum.
Question 23.
– 8 + (- 2) + 3 = ______________
First find the sum:
– 8 + (- 2)
Find absolute values of given integers:
|- 8| = 8 and |- 2| = 2
Then, sum that absolute values
8 + 2 = 10
The result would be 10 or – 10, depending on sign of given integers (both positive then +, both negative then -)
Given integers are negative so the result is – 10.
Now, find the sum:
– 10 + 3
Subtract the lesser absolute value from the greater:
|- 10| – |3| = 7
The final result would be 7 or – 7, depending on signs of given integers.
To find the result, use the sign of the number with the greater absolute value:
– 10 + 3 = – 7
The final result is – 7.
The sun is – 7
First, find the sum – 8 + (- 2) (find absolute values, sum them and add minus if all integers were negative) and then add 3 to that sum (find absolute values, subtract them and add minus if the integer with the greater absolute value was negative).
Go Math Lesson 5.2 6th Grade Adding Integers with Different Signs Question 24.
15 + (- 15) + 200 = ______________
First, find the sum:
15 + (- 15)
The sum is 0 because
the sum of a number and its additive inverse is 0
Now, find the sum:
0 + 200
The final result is 200.
The sum is 200
First, find the sum 15 + (-15) (the sum of 15 and its additive inverse is 0) and then add 200 to that sum.
Question 25.
– 500 + (- 600) + 1200 = ______________
First, find the sum:
– 500 + (- 600)
Find absolute values of given integers:
|- 500| = 500 and |- 600| = 600
Then, sum that absolute values
500 + 600 = 1100
The result would be 1100 or – 1100, depending on sign of given integers (both positive then +, both negative then -).
Given integers are negative the result is – 1100
Now, find the sum:
– 1100 + 1200
Subtract the lesser absolute value from the greater:
|1200| – |- 1100| = 100
The final result would be 100 or – 100, depending on signs of given integers
To find the result, use the sign of the number with the greater absolute value:
– 1100 + 1200 = 100
The final result is 100.
The sun is 100
First, find the sum – 500 + (- 600) (find absolute values, sum them and add minus if all integers were negative) and then add 1200 to that sum (find absolute values, subtract them and add minus if the integer with the greater absolute value was negative).
Question 26.
A football team gained 9 yards on one play and then lost 22 yards on the next. Write a sum of integers to find the overall change in field position. Explain your answer.
You can represent the change in field position by writing the sum:
9 + (- 22)
9 shows gain on one play and adding – 22 shows loss on the next
Now, to find the overall change, subtract lesser absolute value from greater:
|- 22| – |9| = 13
Final result would be 13 or – 13, depending on signs of integers.
Since the number with the greater absolute value is – 22, final result is:
9 + (- 22) = – 13
Gain on one play is represented with positive integer 9 and loss on the next game is represented with negative integer – 22 so the sum is 9 + (- 22).
The overall change in field position after two plays is – 13.
Question 27.
A soccer team is having a car wash. The team spent $55 on supplies. They earned$275, including tips. The team’s profit is the amount the team made after paying for supplies. Write a sum of integers that represents the team’s profit.
A soccer team spout $55 on supplies and earned$275 so the sum that represents the team’s profit is:
– 55 + 275
Find absolute values and subtract lesser absolute value from greater:
|275| – |- 55|= 220
Final result would be 220 or – 220. depending on sighs of given integers.
Since 275 ix integer with greater absolute value, the sum is:
– 55 + 275 = 220
The sum that represents soccer team’s profit is – 55 + 275 and the profit is $220. Find absolute values, subtract them and add minus if integer with greater absolute value was negative. Question 28. As shown in the illustration, Alexa had a negative balance in her checking account before depositing a$47.00 check. What is the new balance of Alexa’s checking account? What property did you use to find the sum?
The new balance in Alexa’s account is the sum of:
– 47 + 47
Since 47 is additive inverse of – 47 and the sum of integer and its additive inverse is 0, new balance in her account is:
– 47 + 47 = 0
New balance of Alexa’s account is 0.
A property that is used to find the sum is the Inverse Property of Addition.
Question 29.
The sum of two integers with different signs is 8. Give two possible integers that fit this description.
Two possible integers with different signs whose sum is 8 are
You can check it:
Find absolute values and subtract lesser absolute value from greater:
|20| – |- 12| = 8
Final result would be 8 or – 8, depending on signs of given integers.
To find the sum, use the sign of number with greater absolute value:
20 + (- 12) = 8
Two possible integers with different signs whose sum is 8 are 20 and – 12.
Find absolute values, subtract them and add minus it the integer with the greater absolute value was negative.
Question 30.
Multistep Bart and Sam played a game in which each player earns or loses points in each turn. A player’s total score after two turns is the sum of his points earned or lost. The player with the greater score after two turns wins. Bart earned 123 points and lost 180 points. Sam earned 185 points and lost 255 points. Use a problem-solving model to find which person won the game. Explain.
Bait’s total score is the sum:
123 + (- 180)
To find his score, subtract lesser absolute value from greater:
|- 180| – |123| = 57
Final result would be 57 or – 57, depending on signs of given integers.
Use the sign of number with greater absolute value to find the result:
123 + (- 180) = – 57
Bait’s total score after two turns is – 57 points
Sam’s total score is the sum:
185 + (- 255)
To find his score, subtract lesser absolute value from greater:
|- 255| – |185| = 70
Final result would be 70 or – 70, depending on signs of given integers.
Use the sign of number with greater absolute value to find the result:
185 + (- 255) = – 70
Sam’s total score after two turns is – 70 points.
Since both Bart and Sam have negative score after two turns, winner is the player who lost less points.
Bart lost 5 points and Sam lost 70 points so Bart won the game.
H.O.T. FOCUS ON HIGHER ORDER THINKING
Question 31.
Critical Thinking Explain how you could use a number line to show that – 4 + 3 and 3 + (- 4) have the same value. Which property of addition states that these sums are equivalent?
Use number line to represent sums:
– 4 + 3 and 3 + (- 4)
First graph number line with units.
To find first sum, start at – 4 and move 3 units to the right to represent adding 3 (yellow arrow).
To find second sum, start at 3 and move 4 units to the left to represent adding – 4 (red arrow).
Read both sums from number line
The Commutative Property states that these sums are equivalent
Number line:
First sum is -1 because it is 1 unit left from 0 on the number line.
Second sum is -1 because it is 1 unit left from 0 on the number line.
Question 32.
Represent Real-World Problems Jim is standing beside a pool. He drops a weight from 4 feet above the surface of the water in the pool. The weight travels a total distance of 12 feet down before landing on the bottom of the pool. Explain how you can write a sum of integers to find the depth of the water.
Jim drops a weight from 4 feet above the surface and it travels 12 feet down until landing on the bottom so the sum that represents the depth of the water is:
4 + (- 12)
To calculate the sum, subtract lesser absolute value from greater:
|- 12| – |4| = 8
Result would be 8 or – 8, depending on signs of given integers
Use the sign of number with greater absolute value to find result:
4 + (- 12) = – 8
Since you want to find the depth of the water, use the absolute value of the result because the depth of the water is positive number. The depth of the water is 8 feet.
Sum that represents the depth is 4 + (-12).
The absolute value of this sum is the depth of the water and that would be 8 feet
Question 33.
Communicate Mathematical Ideas Use counters to model two integers with different signs whose sum is positive. Explain how you know the sum is positive.
Yellow counters represent positive integer and red counters represent negative integer.
To model positive sum of two integers with different sign, take some yellow and some red counters and let there be more yellow counters.
Form and eliminate zero pairs.
When you remove zero pairs, there are some yellow counters left so the sum is positive.
For example:
There are more yellow than red counters When you remove zero pairs, there would be some yellow counters left so the sum is positive.
Question 34.
Analyze Relationships You know that the sum of – 5 and another integer is a positive integer. What can you conclude about the sign of the other integer? What can you conclude about the value of the other integer? Explain. |
# Introducing Power Series 1
The next few posts will discuss a way to introduce Taylor and Maclaurin series to students. We will kind of sneak up on the idea without mentioning where we are going or using any special terms. In this post we will find a way of approximation a function with a polynomial of any degree we choose. In the next post we will look at the graph of these polynomials and finally suggest some questions for further thought.
Making Better Approximations
Students already know and have been working with the tangent line approximation of a function at a point (a, f(a)):
$f(x)\approx f\left( a \right)+{f}'\left( a \right)\left( x-a \right)$
ln(x):
For the function $f\left( x \right)=\ln \left( x \right)$ at the point (1, 0) ask your students to write the tangent line approximation: $y=0+(1)(x-1)$ .Point out that this line has the same value as ln(xand its derivative as at (1, 0).
Then suggest that maybe having a polynomial that has the same value, first derivative and second derivative might be a better approximation. Suggest they start with $y=a+b\left( x-1 \right)+c{{\left( x-1 \right)}^{2}}$ and see if they can find values of a, b and c that will make this happen.
Since $f\left( 1 \right)=0,{f}'\left( 1 \right)=1\text{ and }{{f}'}'\left( x \right)=-1$ we can write
$y=a+b\left( x-1 \right)+c{{\left( x-1 \right)}^{2}};\quad y\left( 1 \right)=a+0+0=0;\quad a=0$
${y}'=b+2c\left( x-1 \right);\quad {y}'\left( 1 \right)=b+0=\tfrac{1}{1};\quad b=1$
${{y}'}'=2c;\quad {{y}'}'\left( 1 \right)=c=-\tfrac{1}{{{1}^{2}}}=-1;\quad c=-\tfrac{1}{2}$
$y=0+\left( x-1 \right)-\tfrac{1}{2}{{\left( x-1 \right)}^{2}}$
Then suggest they try a third degree polynomial starting with $y=a+b\left( x-1 \right)+c{{\left( x-1 \right)}^{2}}+d{{\left( x-1 \right)}^{3}}$. Proceeding as above, all the numbers come out the same and we find that
$\ln \left( x \right)\approx 0+\left( x-1 \right)+\left( -\tfrac{1}{2} \right){{\left( x-1 \right)}^{2}}+\left( \tfrac{1}{3} \right){{\left( x-1 \right)}^{3}}$
Then go for a fourth and fifth degree polynomial until they discover the patterns. (The signs alternate and the denominators are the factorial of the exponent.)
See if the class can write a general the polynomial of degree N :
$\displaystyle \ln \left( x \right)\approx \sum\limits_{k=1}^{N}{\frac{{{\left( -1 \right)}^{k+1}}}{k}{{\left( x-1 \right)}^{k}}}$
sin(x):
Then have the class repeat all this for a new function such as $f\left( x \right)=\sin \left( x \right)$ at the point (0, 0). This could be assigned as homework or group work. Ask them to do enough terms until they see the pattern. There will be patterns similar to ln(x ) and every other term (the even powers) will have a coefficient of zero.
$\sin \left( x \right)\approx x-\tfrac{1}{3!}{{x}^{3}}+\tfrac{1}{5!}{{x}^{5}}-\tfrac{1}{7!}{{x}^{7}}+\tfrac{1}{9!}{{x}^{9}}$
or in general the polynomial of degree N is
$\displaystyle \sin \left( x \right)\approx \sum\limits_{k=1}^{N}{\frac{{{\left( -1 \right)}^{k+1}}}{\left( 2k-1 \right)!}{{x}^{2k-1}}}$
How good is this approximation? Using only the first three terms of the polynomial above you will tell you that. Pretty close: correct to 5 decimal places. Using four terms gives correct to 7 decimal places when rounded.
Finally, see if they can generalize this idea to any function f at any point on the function $\left( {{x}_{0}},f\left( {{x}_{0}} \right) \right)$. This time you will not have the various derivatives as numbers, rather they will be expressions like . Work through the powers one at a time to go from $y=a+b\left( x-{{x}_{0}} \right)+c{{\left( x-{{x}_{0}} \right)}^{2}}+d{{\left( x-{{x}_{0}} \right)}^{3}}+e{{\left( x-{{x}_{0}} \right)}^{4}}$
and so on, until you get to
$f\left( x \right)\approx f\left( {{x}_{0}} \right)+\frac{{f}'\left( {{x}_{0}} \right)}{1!}\left( x-{{x}_{0}} \right)+\frac{{{f}'}'\left( {{x}_{0}} \right)}{2!}{{\left( x-{{x}_{0}} \right)}^{2}}$
$\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad +\cdots +\frac{{{f}^{\left( N \right)}}\left( {{x}_{0}} \right)}{N!}{{\left( x-{{x}_{0}} \right)}^{N}}$
For example the third derivative computation would look like this:
${{{y}'}'}'=3\cdot 2\cdot 1d+4\cdot 3\cdot 2e\left( x-{{x}_{0}} \right)$
${{{y}'}'}'\left( {{x}_{0}} \right)=3\cdot 2\cdot 1d+4\cdot 3\cdot 2e(0)={{{f}'}'}'\left( {{x}_{0}} \right)$
$d=\frac{{{{f}'}'}'\left( {{x}_{0}} \right)}{3!}$
The computations here are perhaps a little different than what students have seen, so take your time doing this. Two or even three class days may be necessary.
Notice these things:
• The first two terms are the tangent line approximation.
• The various derivatives are numbers that must be calculated.
• All the terms of any degree are the same as the terms of the previous degree with one additional term.
Next post in this series: Looking at all this graphically.
(Typos in an earlier version of this post have been corrected – LMc)
## 2 thoughts on “Introducing Power Series 1”
1. SL says:
Lin –
Taught AB for years, but this is my first go with BC. Anxious for series as I remember this being where the wheels fell off for me as a student. I’m been hammering tangent lines this year, and we recently did Euler’s Method where we adapted the typical point-slope format for y=y_1 + m(x-x_1). Think with that launching point, this structure will be well worth my time to use in class. Appreciate the ideas.
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2. Lin
Thanks for this post. Just about ready to dive into this with my BC gang this week and this will help me unpack this all.
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# Basics of Exponents
An exponent is the power a number or variable is raised to in an expression. This power can be positive, negative, zero, and fractional. Indeed, the exponent can be any real number. If the exponent is 0, then the value reduces to 1; that is, \begin{align}x^0 = 1\end{align}, for every non-zero real number x. Other examples:
• \begin{align}2^3 = 2 \times 2 \times 2 = 8\end{align}
• \begin{align}2^{-3} = \frac{1}{2^3} = \frac{1}{8}\end{align}
• \begin{align}2^\frac{1}{2} = \sqrt{2}\end{align}
## Rules of Exponents
Exponent problems can usually be solved by manipulating the component parts according to the rules of exponents, which we summarize as:
1. xa ∙ xb = xa+b Example: 23 ∙ 24 = 27, or 8 ∙ 16 = 128 = 27
2. (xa)b = xab Example: (32)3 = 323 = 36, or 93 = 729, and 36 = 729
3. xa / xb = xa-b Example: 24/23 = 24-3 = 21, or 16/8 = 2
4. x-a = 1/xa Example: 2-3 = 1/23 = 1/8
5. 1/x-a = xa Example: 1/3-2 = 32 = 9
6. \begin{align}x^\frac{m}{n} = \sqrt[n]{x^m}\end{align} Example: \begin{align}2^\frac{2}{3} = \sqrt[3]{2^2} = \sqrt[3]{4}\end{align}
7. xa ∙ ya = (xy)a Example: 3² · 4² = (3 · 4)² = 12² = 144 = 9 · 16 = 3² · 4²
Sample Exponent Problems |
# ONE STEP EQUATIONS Addition and Subtraction Equations Multiplication and Division Equations.
## Presentation on theme: "ONE STEP EQUATIONS Addition and Subtraction Equations Multiplication and Division Equations."— Presentation transcript:
ONE STEP EQUATIONS Addition and Subtraction Equations Multiplication and Division Equations
ONE STEP EQUATIONS What you do to one side of the equation must also be done to the other side to keep it balanced. An equation is like a balance scale because it shows that two quantities are equal.
Definitions Term : a number, variable or the product or quotient of a number and a variable. 12 X 2w c 3
Terms are separated by addition (+) or subtraction (-) signs. 3a – 2b + 7x – 4z + 52 5 How many Terms do you see? 5
Definitions Constant : a term that is a number. Coefficient : the number value in front of a variable in a term.
3x – 6y + 18 = 0 What are your coefficients? What is your constant? 3, -6 18
ONE STEP EQUATIONS What is the variable? To solve one step equations, you need to ask three questions about the equation: What operation is performed on the variable? What is the inverse operation? (The one that will undo what is being done to the variable)
ONE STEP EQUATIONS Example 1 Solve x + 4 = 12 What is the variable? Using the subtraction property of equality, subtract 4 from both sides of the equation. - 4 x x + 4 = 12 The variable is x. Addition. Subtraction. What is the inverse operation (the one that will undo what is being done to the variable)? What operation is being performed on the variable? - 4 8= The subtraction property of equality tells us to subtract the same thing on both sides to keep the equation equal.
ONE STEP EQUATIONS Example 1 Solve y - 7 = -13 What is the variable? Using the addition property of equality, add 7 to both sides of the equation. y - 7 = -13 + 7 y The variable is y. Subtraction. Addition. What operation is being performed on the variable? What is the inverse operation (the one that will undo what is being done to the variable)? + 7 =-6 The addition property of equality tells us to add the same thing on both sides to keep the equation equal.
ONE STEP EQUATIONS Example 1 Solve –6a = 12 What is the variable? Using the division property of equality, divide both sides of the equation by –6. –6a = 12 -6 a The variable is a. What operation is being performed on the variable?Multiplication. What is the inverse operation (the one that will undo what is being done to the variable)? Division -6 =-2 The division property of equality tells us to divide the same thing on both sides to keep the equation equal.
ONE STEP EQUATIONS Example 1 Solve What is the variable? Using the multiplication property of equality, multiply both sides of the equation by 2. = -10 b The variable is b. What operation is being performed on the variable?Division. What is the inverse operation (the one that will undo what is being done to the variable)? Multiplication = -102 2 =-20 =-10 The multiplication property of equality tells us to multiply the same thing on both sides to keep the equation equal.
1) 26 = 8 + v What is the variable? What operation is being performed on the variable? What is the inverse operation (the one that will undo what is being done to the variable)? Solve The variable is v. Addition. Subtraction. 26 = 8 + v -8 18 = v or v = 18
2) x - 7 = 13 What is the variable? What operation is being performed on the variable? What is the inverse operation (the one that will undo what is being done to the variable)? Solve The variable is x. Subtraction. Addition. x – 7 = 13 +7 x = 20
3) v / 8 = 2 What is the variable? What operation is being performed on the variable? What is the inverse operation (the one that will undo what is being done to the variable)? Solve The variable is v. Division. Multiplication. v = 2 8 8 8 v = 16
4) -126 = 14k What is the variable? What operation is being performed on the variable? What is the inverse operation (the one that will undo what is being done to the variable)? Solve The variable is k. Multiplication. Division. -126 = 14k ____ ___ 14 -9 = k or k = -9
Works Cited Charles, Randall I., Bonnie McNemar, and Alma Ramirez. Pre-algebra. Boston, Massachusetts: Pearson Prentice Hall, 2007. Print. Free Algebra 1 Worksheets. (n.d.). Create Custom Pre-Algebra, Algebra 1, Algebra 2, and Geometry Worksheets. Retrieved October 17, 2011, from http://www.kutasoftware.com/free.html |
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