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### Course: 6th grade (Illustrative Mathematics)>Unit 1
Lesson 6: Lesson 11: Polygons
# Area of composite shapes
We can sometimes calculate the area of a complex shape by dividing it into smaller, more manageable parts. In this example, we can determine the area of two triangles, a rectangle, and a trapezoid, and then add up the areas of the four shapes to get the total area. Created by Sal Khan.
## Want to join the conversation?
• Can't you just think of the bottom part as a trapezoid? You can add 6.5 and 3.5 and divide by 2 and multiply that by 9 to get the bottom part. That would save you the hassle of finding the triangle's and the rectangle's area.
• There can be many ways to do composite figures, and your way is just as valid since you divide it into known shapes especially since the area of trapezoids is the first in this string of videos. For fun, I might start with a large 10 by 9 rectangle and take away the three triangles that are cut off from the corner 90 - 3.5 - 13.5 - 12.25 = 60.75.
• What if the shape has a half circle in it?
• Then use the circle formula and divide it by 2.
• "Say when I grow up, what is this useful for?"
• geometry is useful for architecture, design,renovation, building, art, and sometimes just daily activities. This information will also be helpful if you end up needing or wanting to help someone else that has curriculum revolving around this subject. There many uses for geometry in life.
• I don't understand how to do this can someone explain?!
-\'_'/-
• So what he's saying is that you need to break up the shape to make it easy to divide
• can some one explain i'm not giving up but i want to can any one help. Thanks.
• So to find the area of an oddly shaped figure that you don't have a formula for, you split it into lots of smaller figures that you already know how to find the area of. Then you add them all together to find the total area of the original larger figure. Does that make sense? Let me know if there's anything you still don't understand.
• im brainig so hard my think hurts.
• I am sooooo confused about why he did 1 half times 7
• He is doing the formula for a triangle, 1/2(base)(height). This can be done in different orders. You can find (base)(height) and then divide by 2, or you can find 1/2(base) and multiply by (height) (or vice versa), which is what he is doing. I've always done it the first way, and this may be the way you do it, which is why you don't recognize his method.
(1 vote)
• am i slow, or did this video make no sense at all
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# Lesson 11
Zeros of Functions and Intercepts of Graphs
• Let’s see what happens when a function’s input or output is 0.
### 11.1: Which Output is 0?
Which of these functions have an output of 0 when the input is -4?
• $$v(x)=4x$$
• $$w(x)=\text-4x$$
• $$y(x)=8+2x$$
• $$z(x)=2x-8$$
### 11.2: Intercept Detective
Here are the definitions of some functions, followed by some possible inputs for the functions.
$$a(x)=x - 5$$
$$b(x)=x + 5$$
$$c(x)=x-3$$
$$d(x)=x+1$$
$$f(x)=3x - 6$$
$$g(x)=3x + 6$$
$$h(x)=(x+5)(x+3)$$
$$m(x)=(x+1)(x-3)$$
$$n(x)=(3x-6)(x-5)$$
Possible inputs: -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, and 5.
1. For each function, decide which input or inputs would give an output of 0.
2. Here are graphs of $$b$$, $$f$$, and $$m$$. Label each intercept with its coordinates, and be prepared to explain how you know.
### 11.3: Making More Connections
1. For each function, identify the input that would give an output of 0.
• $$p(x) = x + 10$$
• $$q(x) = x - 10$$
• $$r(x) = 8 - x$$
• $$s(x) = \text-8 - x$$
• $$t(x) = 2x - 8$$
• $$u(x) = 2x + 8$$
2. Match each graph to a function in the previous question. Be prepared to explain your matches.
3. Label the intercepts on each graph with their coordinates.
4. For each function, identify the inputs that would give an output of 0.
• $$v(x) = (x + 10)(2x - 8)$$
• $$w(x) = (2x + 8)(10 - x)$$
5. Create three different functions whose output is 0 when the input is 7. At least one of your functions must be quadratic.
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# Inverse of Functions through Algebraic Manipulation
## "Undo" a function by switching x and y values and solving for y.
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Inverse of Functions through Algebraic Manipulation
If you were given a function, such as f(x)=2xx+7\begin{align*}f(x) = \frac{2x}{x+7}\end{align*}, can you tell if the function has an inverse? Is there a way that you could find its inverse through algebraic manipulation?
### Finding the Inverse of a Function
An "inverse" is something that undoes a function, giving back the original argument. For example, a function such as y=13x\begin{align*}y = \frac{1}{3}x\end{align*} has an inverse function of y=3x\begin{align*}y = 3x\end{align*}, since any value placed into the first function will be returned as what it originally was if it is input into the second function. In this case, it is easy to see that to "undo" multiplication by 13\begin{align*}\frac{1}{3}\end{align*}, you should multiply by 3. However, in many cases it may not be easy to infer by examination what the inverse of a function is.
To start, let's examine what is required for a function to have an inverse. It is important to remember that each function has an inverse relation and that this inverse relation is a function only if the original function is one-to-one. A function is one-to-one when its graph passes both the vertical and the horizontal line test. This means that every vertical and horizontal line will intersect the graph in exactly one place.
This is the graph of f(x)=xx+1\begin{align*}f(x) = \frac{x}{x+1}\end{align*}. The graph suggests that f\begin{align*}f\end{align*} is one-to-one because it passes both the vertical and the horizontal line tests. To find the inverse of f\begin{align*}f\end{align*}, switch the x\begin{align*}x\end{align*} and y\begin{align*}y\end{align*} and solve for y\begin{align*}y\end{align*}.
First, switch x\begin{align*}x\end{align*} and y\begin{align*}y\end{align*}.
x=yy+1\begin{align*}x = \frac{y}{y+1}\end{align*}
Next, multiply both sides by (y+1)\begin{align*}(y + 1)\end{align*}.
(y+1)xx(y+1)=yy+1(y+1)=y\begin{align*}(y+1)x & = \frac{y}{y+1} (y+1)\\ x(y+1) & = y\end{align*}
Then, apply the distributive property and put all the \begin{align*}y\end{align*} terms on one side so you can pull out the \begin{align*}y\end{align*}.
\begin{align*}xy+x & = y\\ xy-y & = -x\\ y(x-1) & = -x\end{align*}
Divide by \begin{align*}(x - 1)\end{align*} to get \begin{align*}y\end{align*} by itself.
\begin{align*}y = \frac{-x}{x-1}\end{align*}
Finally, multiply the right side by \begin{align*}\frac{-1}{-1}\end{align*}.
\begin{align*}y = \frac{x}{1-x}\end{align*}
Therefore the inverse of \begin{align*}f\end{align*} is \begin{align*}f^{-1}(x)=\frac{x}{1-x}\end{align*}.
The symbol \begin{align*}f^{-1}\end{align*} is read “\begin{align*}f\end{align*} inverse” and is not the reciprocal of \begin{align*}f\end{align*}.
#### Finding the Inverse of a Function
1. Find the inverse of \begin{align*}f(x) = \frac{1}{x-5}\end{align*} algebraically.
To find the inverse algebraically, switch \begin{align*}f(x)\end{align*} to \begin{align*}y\end{align*} and then switch \begin{align*}x\end{align*} and \begin{align*}y\end{align*}.
\begin{align*}y & =\frac{1}{x-5}\\ x & = \frac{1}{y-5}\\ x(y-5) & = 1\\ xy-5x & = 1\\ xy & = 5x+1\\ y & = \frac{5x+1}{x}\end{align*}
2. Find the inverse of \begin{align*}f(x) = 5 \sin^{-1} \left ( \frac{2}{x-3} \right )\end{align*}
\begin{align*}f(x) & = 5 \sin^{-1} \left ( \frac{2}{x-3} \right )\\ x & = 5 \sin^{-1} \left ( \frac{2}{y-3} \right )\\ \frac{x}{5} & = \sin^{-1} \left ( \frac{2}{y-3} \right )\\ \sin \frac{x}{5} & = \left ( \frac{2}{y-3} \right )\\ (y-3)\sin \frac{x}{5} & = 2\\ (y-3) & = \frac{2}{\sin \frac{x}{5} }\\ y & = \frac{2}{\sin \frac{x}{5} } + 3\end{align*}
3. Find the inverse of the trigonometric function \begin{align*}f(x) = 4 \tan^{-1}(3x + 4)\end{align*}
\begin{align*}x & = 4\tan^{-1}(3y+4)\\ \frac{x}{4} & = \tan^{-1}(3y+4)\\ \tan \frac{x}{4} & = 3y+4\\ \tan \frac{x}{4} - 4 & = 3y\\ \frac{\tan \frac{x}{4} -4}{3} & = y\\ f^{-1} (x) & = \frac{\tan \frac{x}{4} -4}{3}\end{align*}
### Examples
#### Example 1
Earlier, you were asked to find the inverse of a function.
Since the original function is:
\begin{align*}f(x) = y = \frac{2x}{x+7}\end{align*}
You can first switch all of the "x" and "y" values:
\begin{align*}x = \frac{2y}{y+7}\end{align*}
You can then rearrange the equation and isolate "y":
\begin{align*} x(y+7) = 2y\\ xy + 7x = 2y\\ xy - 2y = -7x\\ y(x - 2) = -7x\\ y = \frac{-7x}{x - 2}\\ \end{align*}
The inverse function is written as \begin{align*}f^{-1}(x) = \frac{-7x}{x - 2}\end{align*}
#### Example 2
Find the inverse of \begin{align*}f(x) = 2x^3-5\end{align*}
\begin{align*}f(x) & = 2x^3-5\\ y & = 2x^3-5\\ x & = 2y^3-5\\ x + 5 & = 2y^3\\ \frac{x+5}{2} & = y^3\\ \sqrt[3]{\frac{x+5}{2}} & = y\end{align*}
#### Example 3
Find the inverse of \begin{align*}y = \frac{1}{3}\tan^{-1} \left ( \frac{3}{4}x-5 \right )\end{align*}
\begin{align*}y & = \frac{1}{3} \tan^{-1}\left ( \frac{3}{4}x-5 \right )\\ x & = \frac{1}{3}\tan^{-1}\left ( \frac{3}{4}y-5 \right )\\ 3x & = \tan^{-1}\left ( \frac{3}{4}y-5 \right )\\ \tan (3x) & = \frac{3}{4}y-5\\ \tan (3x) + 5 & = \frac{3}{4}y\\ \frac{4(\tan (3x)+5)}{3} & = y\end{align*}
#### Example 4
Find the inverse of \begin{align*}g(x)=2\sin (x-1)+4\end{align*}
\begin{align*}g(x) & = 2 \sin (x-1)+4\\ y & = 2 \sin (x-1)+4\\ x & = 2 \sin (y-1)+4\\ x-4 & = 2 \sin (y-1)\\ \frac{x-4}{2} & = \sin (y-1)\\ \sin^{-1} \left ( \frac{x-4}{2} \right ) & = y-1\\ 1 + \sin^{-1} \left ( \frac{x-4}{2} \right ) & = y\end{align*}
### Review
Find the inverse of each function.
1. \begin{align*}f(x)=3x+5\end{align*}
2. \begin{align*}g(x)=0.2x-7\end{align*}
3. \begin{align*}h(x)=0.1x^2\end{align*}
4. \begin{align*}k(x)=5x+6\end{align*}
5. \begin{align*}f(x)=\sqrt{x-4}\end{align*}
6. \begin{align*}g(x)=(x)^{\frac{1}{3}}+1\end{align*}
7. \begin{align*}h(x)=(x+1)^3\end{align*}
8. \begin{align*}k(x)=\frac{x^2}{3}\end{align*}
9. \begin{align*}f(x)=-2+4 \sin^{-1}(x + 7)\end{align*}
10. \begin{align*}g(x)=1+3 \tan^{-1}(2x + 1)\end{align*}
11. \begin{align*}h(x)=4 \cos^{-1}(3x)\end{align*}
12. \begin{align*}k(x)=-1 \tan^{-1}(6x)\end{align*}
13. \begin{align*}j(x)=5+2 \sin^{-1}(x + 5)\end{align*}
14. \begin{align*}m(x)=-2 \tan(3x + 1)\end{align*}
15. \begin{align*}p(x)=5-6 \sin(\frac{x}{2})\end{align*}
To see the Review answers, open this PDF file and look for section 4.3.
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
Horizontal Line Test
The horizontal line test says that if a horizontal line drawn anywhere through the graph of a function intersects the function in more than one location, then the function is not one-to-one and not invertible.
One to One Function
A function is one to one if its inverse is also a function. A one-to-one function passes both the horizontal and vertical line tests.
Vertical Line Test
The vertical line test says that if a vertical line drawn anywhere through the graph of a relation intersects the relation in more than one location, then the relation is not a function.
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# In what ratio is the line segment joining the points (2, 3) and (3, 7) divided by Y-axis?
Mathematics is a subject that is associated with numbers and calculations. And, according to the type of calculation mathematics is divided into different branches like algebra, geometry, arithmetic, etc. Geometry is the branch of mathematics that deals with shape and their properties. Geometry that deals with a point, lines, and planes in which coordinates are involved is called coordinate geometry.
### Coordinates
The location of any point on a plane can be expressed as (x, y) and these pairs are known as the coordinates, x is the horizontal value of a point on the plane. This value can also be called the x-coordinate or Abscissa, y is the vertical value of a point on the plane. This value can be called the y-coordinate or ordinate. In coordinate geometry, the point is represented on the cartesian plane.
Cartesian plane
Divided by y-axis means the point which lies on the y-axis.
The point which lies on the y-axis will always have its x-coordinate to be zero.
### Section formula
If a line AB has coordinates (x1, y1) and (x2, y2) of point A and B respectively and there is a point C on line AB which divides the line in ratio m:n, then the coordinate of C can be obtained from below formula, which is known as section formula:
1. x = (mx2 + nx1) /( m + n)
2. y = (my2 + ny1) / (m + n)
### In what ratio is the line segment joining the points (2, 3) and (3, 7) divided by Y-axis?
Solution:
In the above problem statement, there is one line segment say AB having coordinates (2,3) and (3,7) of A and B respectively. Let’s consider a point C having coordinate (x, y) which lies on the y-axis, then the x coordinate of point C is 0, the coordinates of C can be written as (0, y).
Find the ratio in which this C point will divide the line segment AB. To find the ratio, use the section formula of coordinate geometry. Let’s assume that point C divides AB into a k:1 ratio.
As known, the x-coordinate of point C, we can use it to obtain the value of k and get the ratio. Here,
m = k, n =1
x1 = 2, y1 = 3
x2 = 3, y2 = 7
x- coordinate using section formula:
x = (mx2 + nx1) / (m + n)
After substituting the values,
x = ((k)(3) + (1)(2))/ (k +1)
0 = (3k + 2)/ (k + 1)
0 = 3k + 2
– 2 = 3k
-2/3 = k
C divides the point in ratio k:1 = -2/3:1, therefore ratio is 2:3
A negative sign represents that point C divides the line segment externally in ratio 2:3.
### Sample Problems
Question 1: In what ratio is the line segment joining the points A and B have coordinates (0, -2) and (3, 1) respectively divided by X-axis?
Solution:
Let C be the point on X-axis that divides AB in ratio k: 1. As C lies on the x-axis, the y-coordinate of the C should be 0. Therefore, the coordinate of C : (x, 0). As known as the y-coordinate of point C, we can use it to obtain the value of k and get the ratio. Here,
m = k, n = 1
x1 = 0, y1 = -2
x2 = 3, y2 = 1
y- coordinate using section formula:
y = (my2 + ny1) / (m + n)
After substituting the values,
y = ((k)(1) + (1)(-2))/ (k +1)
0 = ( k – 2)/ (k + 1)
0 = k – 2
2 = k
2 = k
C divides the point in ratio k:1 = 2:1, therefore ratio is 2:1.
Question 2: In what ratio does a point (2,2) divides the line segment AB of which (0, 0) and (5, 5) are the coordinates of A and B respectively.
Solution:
Let (2, 2) divide AB in ratio k: 1. As known the a and y-coordinate of point, use it to obtain the value of k and get the ratio. Here,
m = k, n = 1
x1 = 0, y1 = 0
x2 = 5, y2 = 5
y- coordinate using section formula:
y = (my2 + ny1) / (m + n)
After substituting the values,
2 = ((k)(5) + (1)(0)) / (k + 1)
2 = ( 5k ) / (k+1)
2(k + 1) = 5k
2k + 2 = 5k
2 = 3k
2/3 = k
Hence, (0, 0 ) divides the point in ratio k:1 = 2/3:1, therefore ratio is 2:3.
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# How to Do Graphs in Math Homework: Beginner Tips
Over the years, one major aspect of mathematics that students often struggle with is graphs, especially when given an assignment. Whether you are studying algebra, geometry, calculus, or any other branch of mathematics, you are most likely to encounter graphs in your homework or the course of your study.
In this essay, we will take a cursory look at the different beginner tips on how to do graphs for math homework. Also, different types of graphs in math, including line graphs, bar graphs, and pie charts will be discussed. With these beginner tips, you will understand that graphs are not as difficult as most people presume them to be.
## What exactly does the graph mean?
Graphs are special pictorial tools that are mostly used in mathematics to represent data and illustrate mathematical concepts. Graphs are used to analyze and understand difficult mathematical relationships by displaying them clearly and simply for easy understanding.
## What are the different graphs in math?
There are many types of graphs but some of the most common math graphs are:
• Pie chart
It is also known as a circle graph and it is partitioned into slices to show numerical proportions. Each slice represents a part of the whole chart and the total of all slices is usually 100%. The angle of each slice is corresponding to the size of the category it represents.
• Line graph
A line graph uses points and lines to represent change over time. It is a combination of several points that explain the relation between the points. The graph represents measurable data between two changing variables with a line or curve that joins a series of successive data points.
• Bar graph
A bar graph uses rectangular bars of equal or varying width and height to represent numerical grouped data. A bar chart can either be horizontal or vertical, and the gap between one bar and another is uniform throughout. The height or length relates directly to its value and it is used to compare various sets of data among different groups.
## Beginner tips on how to do graphs in math homework
Are you thinking of where to start your graph math homework but can’t seem to know where? Do not overthink it; instead, engage the following tips:
1. Choose the right type of graph
Choosing the right types of graphs math will help improve communication as different types of graphs are used to represent different types of data. By doing so, you can communicate your findings more effectively to your audience. The second advantage of choosing the right type of graph for math homework is that it increases accuracy, and better analysis, reduces the risk of error and it helps save time.
Labeling your axis (X and Y axis) in the graph is very important as it helps your audience to better understand the data being presented. Labeling your graph provides clarity, accuracy, context, and accessibility. To label your math types of graphs accurately, you should identify which variables the X-axis and Y-axis each represent. Remember to include a unit of measure called scale to help your audience to understand your data.
1. Use appropriate scale
It is important to choose an appropriate scale for axes so that the data given can be vividly represented. The Scale of the graph determines the range of values that will be shown on the X-axis and Y-axis. If the scale is too large or too small, it can affect the data and even mislead the viewer.
1. Use colors and symbols
Using different colors and symbols for your graph homework makes it easy to differentiate between data sets. It is also used to highlight differences, add visual interest, enhance readability, convey meaning, and support accessibility. However, be careful not to use too many colors or symbols as this can make your homework difficult to read.
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# Derivative of exponential functions
In Chapter 1, we talked about Limits and we discussed about how important it is to understand the concepts behind this topic. In this chapter, we will talk about the derivative, which is a subject that you will deal with, all throughout your course in Calculus.
In derivation, you need to remember your lessons about the slope of tangent line or the instantaneous rate of change, as this is the main concept applied in this chapter. As we have learned in the previous courses, slope is defined as the ratio between the change in Y and the change in X. In other words, slope is equal to $\frac{\Delta rise}{\Delta run}$.
It is theorized that we can find the average slope between two points. So, if we want to know the slope between a given unknown point, we need to find the $\Delta$run or the $\Delta$ x that approaches zero. “The derivative of” is written mathematically as, $\frac{d}{dx}$.
In this chapter, we will have an 11 part discussion about derivative. For the first three chapters, we will look at the definition of derivative, the power rule and the slope and equation of a tangent line. We need to review about the power rule to help us derive long and complicated polynomial functions with large exponents.
For the next parts of the chapter, we will look at the chain rule. This is used in solving for the derivative of a composite function. We will discuss about the derivative of trigonometric function, exponential function, inverse trigonometric, and logarithmic functions. In order for us to understand these two topics, we will also study the differential rules for trigonometric function, logarithmic function, exponential function, and inverse trig-function (arc function).
We will also discuss about the product rule and quotient rule which are used in simplifying functions that we need to derive. Apart from the topics mentioned above, we will also talk about implicit differentiation. To let us understand the implicit functions, we will also look at the explicit function through the set of exercises we will have.
### Derivative of exponential functions
An exponential function is a function containing a numerical base with at least one variable in its exponent. In this section, we will learn how to differentiate exponential functions, including natural exponential functions and other composite functions that require the application of the Chain Rule.
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# Rational Exponents
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Education
Published on July 3, 2009
Author: Josephil
Source: slideshare.net
Rational Exponents
Definition of Rational Exponents For any nonzero number b and any integers m and n with n > 1, ( ) m m n m n b = b = n b except when b < 0 and n is even
NOTE: There are 3 different ways to write a rational exponent ( ) 4 4 3 4 3 27 = 27 = 3 27
Examples: 3 ( ) 3 36 = (6) = 216 3 2= 1. 36 4 2. 27 = ( 27) = (3) = 81 4 4 3 3 3 = ( 81 = (3) = 27 ) 3 3 4 3. 814
Evaluating and Simplifying Expressions with Rational Exponents
Evaluate vs. Simplify Evaluate – finding the numerical value. Simplify – writing the expression in simplest form.
Simplifying Expressions No negative exponents No fractional exponents in the denominator No complex fractions (fraction within a fraction) The index of any remaining radical is the least possible number
In this case, we use the laws of exponents to simplify expressions with rational exponents.
Properties of Rational Exponents n Definition 1 −n 1 1 − 1 1 3 1 1 a = = n of negative 8 3 3 = = = a a exponent 8 8 2 1 1 1 =a n Definition = 36 = 36 = 6 2 of negative 1 − −n a exponent 36 2
Properties of Rational Exponents (a ) 1 Power-to- m n 1 =a mn 2 1 1 1 a-power 2 3 = ( 2) 3⋅ 2 = 2 6 Law ( ab ) m Product-t0 1 1 =a b m m -a-power ( xy ) 1 2 =x 2 y 2 Law
Properties of Rational Exponents Quotient- 1 m m to-a- 1 a a 4 2 4 2 4 2 = m power = 1 = 25 = 25 5 b b Law 25 2 −m m Quotient- 1 1 a b − to-a- 4 25 2 25 5 2 = negative- = = = b a 25 4 4 2 power Law
Exampl e: Write as a Radical and Evaluate 1 49 = 49 = 2 2 49 = 49 = 7
Example: Simplify each expression 1 1 5 2 3 5 4 ⋅a ⋅b = 4 ⋅a ⋅b 3 2 6 6 6 6 6 2 6 3 6 5 Rewrite = 4 ⋅ a ⋅ b as a Get a 6 3 5 radical common = 16a b denominator - this is going
Example: Simplify each expression 1 3 1 1 3 1 + + x ⋅x ⋅x = x 2 4 5 2 4 5 10 15 4 29Remember + + =x we add 20 20 20 =x 20 exponents 20 9 20 9 = x ⋅x20 20 =x x
Example: Simplify each expression 4 4 1 −4 1 1 1 w 5 5 5 5 = = 4 = 4 ⋅ 1 w w 5 w5 w 5 w 1 1 1 5 w 5 w 5 w w 5 rationalize To = 4 1 = = = the + 5 w w denominator w5 5 w 5 we want an
Example: Simplify each expression 1 7 −1 1 8 1 x y 8 8 = x = x⋅ 1 = 1 ⋅ 7 xy y y 8 y y 8 8 7 7 xy 8x y 8 To rationalize = = the y y denominator we want an
Example: Simplify each expression 1 1 (2 ) = 2 5 10 5 10 32 32 10 10 = 1 = 1 2 (2 ) 2 8 4 2 8 1 4 8 8 1 1 2 1 1 2 2 − − 4 = 1 =2 2 4 =2 4 4 =2 = 2 4 2 4
Example: Simplify each expression −1 1 1 − − 5 2 5 1 5 2 2 = 1 = ⋅ 1 2 5 2 2 2⋅ 5 2 5 1 1 1 − − 1 −1 1 1 1 = ⋅5 2 2 = ⋅5 = ⋅ = 2 2 2 5 10
Example: Simplify each expression 1 Multiply by 1 m + 1 conjugate and 2 1 ⋅ 1 use FOIL 2 m − 1 m + 1 2 1 m +1 2 m +1 = = m− 1 m− 1
Example: Simplify each expression −2 −2 −2 3 −2 2x 3 −2 2x = 2x 2 2 −3 = 2 x2 x −2 −1 −1 ⋅−2 1 x = 2x = 2 x = 2 ⋅ x = 2 −2 2 2 4
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### Rational exponents and the properties of exponents
Evaluate numerical exponential expressions and simplify variable exponential expressions by using the properties of exponents.
### Rational Exponents - Concept - Algebra 2 Video by Brightstorm
Time-saving video that shows how to write radicals as rational exponents. Example problems write roots as exponents and convert fractional exponents into ...
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## Featured Post
Welcome Back to School …you little Mon...
## Monday, October 29, 2012
### GoMath Chapter 4
Our class will be learning how to model division, and use the division algorithm to divide up to 3-digit dividends by 1-digit divisors. The class will learn different methods to divide, including using models, repeated subtraction, and the standard division algorithm. We will also learn to divide with remainders.
Our homework will provide practice modeling division and using the division algorithm. There are twelve sections in this chapter.
Important Chapter Vocabulary:
1. Compatible Mumbers; Numbers that are easy to compute mentally.
2. Multiple; A number that is the product of a given number and a counting number.
3. Partial Quotient; A method of dividing in which multiples of the divisor are subtracted from the dividend and then the quotients are added together.
4. Remainder; The amount left over when a number cannot be divided equally.
5. Dividend; The number that is to be divided in a division problem.
6. Divisor; The number that divides the dividend.
### GoMath Chapter 3
Our class, during the next few weeks, will be learning to multiply by 2-digit whole numbers. We will also learn how to describe the reasonableness of an estimate.
The homework will provide practice with estimation and multiplication of numbers with more than 1 digit. There are seven sections in this chapter.
Important Chapter Vocabulary:
1. Compatible Numbers; Numbers that are easy to compute with mentally.
2. Associative Property of Multiplication; The property that states that when the grouping of factors is changed, the product remains the same.
3. Commutative Property of Multiplication; The property that states when the order of two factors is changed, the product remains the same.
4. Estimate; To find an answer that is close to the exact amount.
5. Partial Product; A method of multiplying in which the ones, tens, hundreds, and so on are multiplied separately and then the products are added together.
6. Product; The answer in a multiplication problem.
7. Regroup; To exchange amounts of equal value to rename a number.
### GoMath Chapter 2
During the next few weeks, our class will be learning about multiplying by 1-digit whole numbers. We will investigate strategies for multiplying 2-, 3-, and 4-digit numbers by the numbers 2-9.
You can expect to see homework that provides practice with multiplication by 1-digit numbers. There are twelve sections in this chapter.
Important Chapter Vocabulary:
1. Distributive Property; The property that states that multiplying a sum by a number is the same as multiplying each addend by the number and then adding the products.
2. Partial Product; A method of multiplying in which the ones, tens, hundreds, and so on are multiplied separately and then the products are added together.
3. Estimate; To find an answer that is close to the exact amount.
4. Expanded Form; A way to write numbers by showing the value of each digit.
5. Factor; A number that is multiplied by another number to find a product.
6. Round; To replace a number with another number that tells about how many or how much.
## Wednesday, October 3, 2012
### New Math Program, GoMath
The Cache County School District has adopted a new math program this year for grades 1-4. This is because of the new Utah Common Core that is being implemented. Our students are going to be tested by the standards of this new core, and they need to be prepared for what is expected. GoMath will help them be ready.
This is different from any other math program I have taught, but I love it! It is challenging, high-level thinking and exactly what our children need to be practicing. At the end of math time, both the students and I are mentally tired, but isn't that what we want? We want them digging and searching to figure things out. For so many years we have been simplifying mathematics by just learning steps or formulas without learning the "how" and "why". GoMath teaches these skills. We work daily, in math, for 60-90 minutes together, with partners, and alone. Then they come home with homework to practice again what we have been learning.
It is critical as parents, that you take the the time each night to go over their homework with them and see what your child is doing in class. The methods and strategies can be very different from what you and I learned growing up, but remember it is new, it can be difficult, but they are doing it! As the program works its way up through the grades it will make this type of learning easier for them each year. Please be patient with them and me. We are ALL learning math this year!
As a result of these new methods, sometimes the homework is difficult for the parents as well as the children. I will be posting a simple video tutorial each time a new lesson is taught. You and your child can watch the video together to assist in helping with homework. Hopefully this will alleviate a little stress during math homework time. :-)
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# SIXTH GRADE MATH PRACTICE ONLINE
Sixth Grade Math Practice Online :
In this section, we will see some practice questions for 6th grade students.
Question 1 :
The sum of numbers divisible by 7 between 50 and 100 ?
(A) 539 (B) 600 (C) 425
Solution :
Numbers which are divisible by 7 between 50 and 100 are
56, 63, 70, 77, 84, 91, 98
Now we have to find the sum of these numbers
For that we have to add these numbers
56 + 63 + 70 + 77 + 84 + 91 + 98 = 539
So the answer is 539.
Question 2 :
A bag contains red, green and blue balls. The total number of balls is 80. If 60% of the total balls are blue, what is the probability of taking a blue ball?
(A) 1/6 (B) 5/8 (C) 3/5
Solution :
Total number of balls n (S) = 80
60% of the total balls are blue
Number of blue balls = 60% of 80
= (60/100) 80
= (4800/100)
= 48
Let A be the event of getting blue balls
n(A) = 48
Probability of getting blue balls = n(A)/n(S)
= 48/80
= 3/5
So the answer is 3/5.
Question 3 :
2 is
(A) Rational (B) Irrational (C) Real
Solution :
Let √2 be rational
Since √2 is rational we shall write it as fraction
√2 = a/b
Taking squares on both sides
2 = a2/b2
2b2 = a2
2 divides a2
2 divides a
let a = 2 c
2b2 = (2c)2
2b2 = 4c2
c2 = 2b2/4
c2 = b2/2
2 divides b
from this we come to know that a and b are having common factor 2 other than 1. So we shall decide that a/b is not rational number.
It is irrational.So, √2 is irrational
Question 4 :
The distance between A to B is 5 cm, B to C is 6 cm, C to D is 9 cm, D to E is 12 cm, E to F is 6.5 cm and F to G is 4.8 cm. Then what is the distance between B to F?
(A) 30.4 cm (B) 33.5 cm (C) 12.5 cm
From this we need to find the distance between the points B and F
BF = BC + CD + DE + EF
BC = 6 cm, CD = 9 cm, DE = 12 cm, EF = 6.5 cm
BF = 6 + 9 + 12 + 6.5
BF = 33.5 cm
So the answer is 33.5 cm
Question 5 :
The greatest common factor of x² y² z², x² y z² and x y z² is
(A) xy2z2 (B) xy2z (C) xyz2
Solution :
x2 y2 z2, x2 y z2, x y z2
To find GCF, we have to select the common terms only
GCF = x y z2
So the correct answer is x y z2.
Question 6 :
The sum of a number and its square is 90. Find the required numbers.
(A) 9 and 81 (B) 45 and 45 (C) 20 and 70
Solution :
Let x be the required number and x2 be its square
x + x2 = 90
x + x2 - 90 = 0
(x + 10) (x - 9) = 0
x + 10 = 0 x - 9 = 0
x = -10 x = 9
So, the answer is 9 and 81.
Question 7 :
In how many ways can five people sit in a round table?
(A) 60 (B) 150 (C) 120
Solution :
The number of arrangements in a round table is (n - 1)!
= (5 - 1)!
= 4!
= 24
Hence the number of ways is 24.
Question 8 :
If two dice are rolled, what is the probability of getting same number on both the dies?
(A) 1/2 (B) 1/3 (C) 1/6
Solution :
Sample space
S = { {1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6) (2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6) (3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6) (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6) (5, 1) (5, 2) (5, 3) (5, 4)(5, 5) (5, 6) (6, 1) (6, 2) (6, 3) (6, 4)(6, 5) (6, 6)}
n(S) = 36
Let A be the event of getting the same number on both the dies
A = { (1, 1) (2, 2) (3, 3) (4, 4) (5, 5) (6, 6) }
n(A) = 6
p(A) = n(A)/n(S)
= 6/36
= 1/6
So, the required probability is 1/6.
Question 9 :
Compare 4/5 and 5/6
(A)(5/6) < (4/5) (B) (5/6) > (4/5) (C) (5/6) = (4/5)
Solution :
Since the denominators are not same, first we have to make the denominators same.
L.C.M of 5 and 6 = 30
= (4/5) (6/6)
= (24/30)
= (5/6) (5/5)
= (25/30)
Therefore 4/5 is less than 5/6
So the answer is (5/6) > (4/5).
Question 10 :
Evaluate :
25! / 24!
(A) 25 (B) 24 (C) 1/24
Solution :
25! / 24! = 25(24!)/24!
= 25
Hence the value is 25.
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# RATIO AND PROPORTION
RATIO: RATIO MEANS COMPARISON. THE NUMBER OF TIMES ONE QUANTITY CONTAINS ANOTHER QUANTITY OF THE SAME KIND.
Thus the ratio between 5 litres of oil and 15 litres of oil can be possible, but not between 10 apples and 25 kg of rice.
* The ratio between one quantity to another is measured by a : b or a/b
Ex: 8 : 9 or 5 : 7 etc.
* The two quantities in the ratio are called its terms. The first is called the antecedent and the second term is called consequent.
* The terms of the ratio can be multiplied or divided by the same number.
Types of Ratios:
1. Duplicate ratio: The ratio of the squares of the two numbers.
Ex: 9 : 16 is the duplicate ratio of 3 : 4.
2. Triplicate Ratio: The ratio of the cubes of the two numbers.
Ex: 27 : 64 is the triplicate ratio of 3 : 4.
3. Sub-duplicate Ratio: The ratio between the square roots of the two numbers.
Ex: 4 : 5 is the sub-duplicate ratio of 16 : 25.
4. Sub-triplicate Ratio: The ratio between the cube roots of the two numbers.
Ex: 4 : 5 is the sub-triplicate ratio of 64 : 125.
5.Inverse ratio: If the two terms in the ratio interchange their places, then the new ratio is inverse ratio of the first.
Ex: 9 :5 is the inverse ratio of 5 : 9.
6. Compound ratio: The ratio of the product of the first terms to that of the second terms of two or more ratios.
Ex: The compound ratio of
PROPORTION: If two ratios are equal, then they make a proportion.
Thus
Each term of the ratios is called proportional.
The middle terms 5 and 8 are called means and the end terms 4 and 10 are called extrems.
Product of Means = Product of Extremes
Continued Proportion: In the proportion 8, 12, 18 are in the continued proportion.
Fourth proportion: If a : b = c : x, then x is called fourth proportion of a,b and c.
There fore fourth proportion of a, b, c =
Third proportion: If a : b = b : x, then x is called third proportion of a and b.
Therefore third proportion of a, b =
Second or mean proportion: If a : x = x : b , then x is called second or mean proportion of a and b.
Therefore mean proportion of a and b =
EXAMPLES
1. a : b = 3: 4; b : c = 6 : 7. Find a : b : c.
Sol: a b c
3 4
6 7
a : b : c = 3 × 6 : 6 × 4 : 4 × 7 = 9 : 12 : 14
2. A sum of Rs.4960 has been divided among A, B and C in the ratio of 5:4:7. Find the share of B.
Sol: B's share =
= Rs.1240
3. 36% of first number is 28% of the second number. What is the respective ratio of the first number to the second number?
Sol: Let the numbers be x and y.
36% of x = 28% of y
∴ x : y = 7 : 9
4. Two numbers are in 4:7 ratio. The difference between them is 27. What is the bigger number?
Sol: Let the numbers be 4x and 7x.
7x − 4x = 27
⇒ 3x = 27 ⇒ x = 9
∴ Bigger number is 7x = 7 × 9 = 63
Short cut: The difference of the terms of the ratio = 7 − 4 = 3.
But the actual difference between the numbers is 27
∴ 3 parts is equal to 27
7 parts (Bigger number) = × 27 = 63
5. The ratio of the ages of a man and his son is 7 : 3. The average of their ages is 30 years. What will be the ratio of their ages after 4 years?
Sol: Average age = 30 years
Total age = 2 × 30 = 60 years.
Let their present ages be 7x and 3x years
∴ 7x + 3x = 60 ⇒ x = = 6
∴ Their present ages are
7 × 6 and 3 × 6 = 42 and 18.
∴ Their ages after 4 years
= 42 + 4 and 18 + 4 = 46 and 22 years
∴ ratio = 46 : 22 = 23 : 11
6. Two numbers are in the ratio of 3:4. If 4 is subtracted from each, the remainders are in the ratio of 5:7. What are the numbers?
Sol: Let the numbers be 3x and 4x.
If 4 is subtracted from each, the numbers will be (3x −4) and (4x −4).
∴ (3x −4) : (4x −4) = 5: 7
Product of means = Product of extremes
(3x –4) 7 = (4x – 4) 5
⇒ 21x – 28 = 20x – 20
⇒ x = 8
∴ The numbers are 3 × 8 and 4 × 8
= 24 and 32
7. In a bowl there is 30 litre mixture of milk and water. The ratio of milk and water is 7:3. How much water must be added to it so that the ratio of milk to the water be 3:7?
Sol : Milk quantity in the mixture
× 30 = 21 litres
∴ Water = 30 - 21 = 9 litres
New ratio = 3 : 7
∴ 3 parts of milk is 21 litres (There is no difference in the milk quantity of new mixture)
∴ Water quantity in the mixture
× 21 = 49 litres
∴ 49 - 9 = 40 litres water is to be added in the new mixture
8. A bag contains of one rupee, 50 paise and 25 paise coins. if these coins are in the ratio of 5 : 6 : 8, and the total amount of coins is Rs. 210, find the number of 50 paise coins in the bag.
Sol : Let the number of one rupee, 50 paise, 25 paise coins be 5, 6 and 8 respectively
The value of one rupee coins
= Rs. 1 × 5 = Rs. 5
The value of fifty paise coins
= Rs. 0.50 × 6 = Rs. 3
The value of twenty five paise coins
= Rs. 0.25 × 8 = Rs. 2
Total value = 5 + 3 + 2 = Rs. 10
If the total value is Rs. 10
there are 6 coins of fifty paise
if the total value is Rs. 210, then the number of 50 coins = × 6 = 126
9. If a sum of Rs.3150 were distributed among Ravi, Vijay and Suresh in the ratio of 12:9:14 respectively, then find the share of Vijay.
Ans: Rs.810
Sol: Vijay's Share = × 3150 = Rs.810
10. The total number of students in a school is 2850. If the number of boys in the school is 1650, then what is the respective ratio of the total number of boys to the total number of girls in the school?
Ans: 11:8
Sol: Total number of students = 2850
Number of boys = 1650
Number of girls = 2850-1650 = 1200
Ratio between boys and girls
=1650 : 1200 = 11 : 8
11. A sum of money is divided among A, B, C and D in the ratio of 5 : 6 : 12 : 15 respectively. If the share of C is Rs. 4092, then what is the total amount of money?
Ans: Rs. 12958
Sol: Let the share of A, B, C and D be Rs. 5x, 6x, 12x and 15x respectively. C's share is Rs.4092
⇒ 12x = 4092 ⇒ x= = 341
∴ Total money = 38x = 38 × 341= Rs.12958
12. Asum of Rs. 2820 has been distributed among A, B and C in the ratio respectively. What is the share of B?
Ans: Rs. 900
Sol: LCM of 3, 4 and 5 is 60
∴ ratio = : 20 : 15 : 12
B's share = × 2820
= × 2820 = Rs. 900
13. A, B and C divide an amount of Rs. 6300 amongst themselves in the ratio of 7:6:8 respectively. If an amount of Rs.300 is added to each of their shares, what will be the new respective ratio of their shares of amount?
Ans: 8 : 7 : 9
Sol: Total shares = 7 + 6 + 8 = 21
∴ 21 parts = 6300
each part = = 300
∴ Their shares are
7 × 300, 6 × 300 and 8 × 300
⇒ 2100, 1800 and 2400
If 300 is added to each of them then their shares are 2400, 2100 and 2700
Their ratio = 2400 : 2100 : 2700
= 8 : 7 : 9
14. Find out the two quantities whose difference is 30 and the ratio between them is 5/11.
Sol: The difference of quantities, which are in the ratio 5:11, is 6. To make the difference 30, we should Multiply them by 5.
Therefore
15. A factory employs skilled workers, unskilled workers and clerks in the ratio 8:5:1 and the wages of a skilled worker, an unskilled worker and a clerk are in the ratio 5:2:3 when 20 unskilled workers are employed the total daily wages fall amount to Rs. 318. Find out the daily wages paid to each category of employees.
Sol: Number of skilled worker: unskilled worker: clerks = 8:5:1 and the ratio of their respective Wages = 5:2:3
Hence the amount will be paid in the ratio 8 × 5 : 5 × 2 : 3 × 1 = 40 : 10 : 3
Hence total amount distributed among unskilled workers
But the number of unskilled workers is 20, so the daily wages of unskilled worker
The wages of a skilled worker, an unskilled worker and a clerk are in the ratio = 5:2:3
Multiplying the ratio by we get = 7.50 : 3 : 4.50
So, if an unskilled worker gets Rs.3 a day then a skilled worker gets Rs. 7.50 per day a clerks Rs. 4.50 a day
16. Two numbers are in the ratio of 11:13. If 12 be subtracted from each, the remainders are in the ratio of 7:9 Find out the numbers.
Sol: Since the numbers are in the ratio of 11:13. Let the numbers be 11x and 13x. Now if 2 is subtracted from each, the numbers become (11x -12) and (13x-12). As they are in the ratio of 7:9 (11x-12): (13x-12):: 7: 9
⇒ (11x – 12) 9 = (13x – 12) 7
⇒ 99x – 108 = 91x – 84
⇒ 9x = 24 or x = 3
17. In what ratio the two kinds of tea must be mixed together one at Rs. 48 per kg. and another at Rs. 32 per kg. So that the mixture may cost Rs. 36 per kg. ?
Sol: Mohan and Sohan = 5:6 or
Sohan and Rakesh = 8:5 or
Mohan and Rakesh =
= 4 : 3
18. What should be subtracted from 15, 28, 20 and 38 so that the remaining numbers may be proportional?
19. If Rs. 279 were distributed among Ram, Mohan and Sohan in the ratio of 15:10:6 respectively, then how many rupees did Mohan obtain?
Sol: Ratio in which Ram, Mohan and Sohan got = 15 :10 : 6
∴ Sum of ratios = 15 + 10 + 6 = 31
∴ Share of Mohan
= Rs. 90
20. A bag contains of one rupee, 50 paise and 25 paise coins. If these coins are in the ratio of 2:3:10, and the total amount of coins is Rs288, find the number of 25 paise coins in the bag.
Sol: Ratio of one rupee, 50 paise and 25 paise coins
= 2:3:10
∴ Ratio of their values = 8:6:10 = 4:3:5
And sum of the ratios of their values = 4 + 3 + 5 = 12
∴ Value of 25 paise coins = Rs. 120
∴ No. of 25 paise coins = 120 × 4 = 480
Posted Date : 07-02-2021
గమనిక : ప్రతిభ.ఈనాడు.నెట్లో కనిపించే వ్యాపార ప్రకటనలు వివిధ దేశాల్లోని వ్యాపారులు, సంస్థల నుంచి వస్తాయి. మరి కొన్ని ప్రకటనలు పాఠకుల అభిరుచి మేరకు కృత్రిమ మేధస్సు సాంకేతికత సాయంతో ప్రదర్శితమవుతుంటాయి. ఆ ప్రకటనల్లోని ఉత్పత్తులను లేదా సేవలను పాఠకులు స్వయంగా విచారించుకొని, జాగ్రత్తగా పరిశీలించి కొనుక్కోవాలి లేదా వినియోగించుకోవాలి. వాటి నాణ్యత లేదా లోపాలతో ఈనాడు యాజమాన్యానికి ఎలాంటి సంబంధం లేదు. ఈ విషయంలో ఉత్తర ప్రత్యుత్తరాలకు, ఈ-మెయిల్స్ కి, ఇంకా ఇతర రూపాల్లో సమాచార మార్పిడికి తావు లేదు. ఫిర్యాదులు స్వీకరించడం కుదరదు. పాఠకులు గమనించి, సహకరించాలని మనవి.
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# DAV Class 5 Maths Chapter 10 Brain Teasers Solutions
The DAV Books Solutions Class 5 Maths and DAV Class 5 Maths Chapter 10 Brain Teasers Solutions of Averages offer comprehensive answers to textbook questions.
## DAV Class 5 Maths Ch 10 Brain Teasers Solutions
Question 1.
Tick (✓) the correct answer.
(a) The average of five numbers is 6. The sum of five numbers is _________
(i) 20
(ii) 10
(iii) 30
(iv) 35
Solution:
Average = $$\frac{Sum of numbers}{Total number of addends}$$
Given average = 6, no. of addends = 5
6 = $$\frac{Sum of numbers}{5}$$
Sum of numbers = 30
So option (iii).
(b) The average of the first five odd numbers is _________
(i) 5
(ii) 3
(iii) 7
(iv) 9
Solution:
First five odd numbers = 1,3, 5, 7, 9
Average = $$\frac{1+3+5+7+9}{5}$$
= $$\frac{25}{5}$$
So option (i)
(c) The average of $$\frac{1}{4}$$ and $$\frac{3}{4}$$ is _________
(i) $$\frac{1}{2}$$
(ii) $$\frac{1}{4}$$
(iii) 1
(iv) $$\frac{3}{2}$$
Solution:
Average = $$\frac{\frac{1}{4}+\frac{3}{4}}{2}$$
= $$\frac{4}{4 \times 2}$$
= $$\frac{1}{2}$$
So option (i).
(d) The average of 4.8, 3.9, 8.6, 2.3 is _________
(i) 3.8
(ii) 4.9
(iii) 6.2
(iv) 5.9
Solution:
Average = $$\frac{4.8+3.9+8.6+2.3}{4}$$
= $$\frac{19.6}{4}$$
= 4.9
So option (ii).
(e) The average of three numbers is 6. If two of the numbers are 6 and 7, the third number is _________
(i) 5
(ii) 6
(iii) 4
(iv) 3
Solution:
Let the third number be x.
Average = 6, no. of addends = 3
Average = $$\frac{6+7+x}{3}$$
6 × 3 = 6 + 7 + x
18 = 13 + x
x = 18 – 13 = 5
So option (i).
Question 2.
Find the average of the following sets of numbers:
(a) 60, 24, 126, 200, 185, 92
Solution:
(b) $$\frac{2}{10}, \frac{5}{10}, \frac{6}{10}, \frac{7}{10}, \frac{8}{10}$$
Solution:
(c) 26.5, 15.4, 16.8, 20.9, 10.9
Solution:
Question 3.
Find the average of the first ten counting numbers.
Solution:
Average = $$\frac{1+2+3+4+5+6+7+8+9+10}{10}$$
= $$\frac{55}{10}$$
= 5.5
Question 4.
Find the average of the first five multiples of 5.
Solution:
First five multiples = 5, 10, 15, 20, 25
Average = $$\frac{Sum of no.}{No. of addends}$$
= $$\frac{5+10+15+20+25}{5}$$
= $$\frac{75}{5}$$
= 15
Question 5.
The average age of three brothers is 15 years. If the age of the two brothers is 15 years and 12 years, find the age of the third brother.
Solution:
Let the age of the third brother = x yrs
Average age of three brothers = $$\frac{15+12+x}{3}$$
15 = $$\frac{27+x}{3}$$
15 × 3 = 27 + x
45 – 27 = x
x = 18 yrs
The age of the third brother is 18 years.
Question 6.
The heights of four members of a family are 1 m 60 cm, 1 m 50 cm, 1 m 10 cm, and 1 m 16 cm respectively. Find the average height of one member of the family.
Solution:
Average of one member of family = $$\frac{Sum of heights}{No. of members}$$
= $$\frac{1.60+1.50+1.10+1.16}{4}$$
= $$\frac{5.36}{4}$$
= 1.34 m
1.34 m is the average height of one member of the family.
Question 7.
In a school, the number of girls and boys in three classes are given below:
(a) Find the average number of girls in a class.
Solution:
Average number of girls in a class = $$\frac{Sum of no. of girl in a class}{Number of classes}$$
= $$\frac{24+26+28}{3}$$
= $$\frac{78}{3}$$
= 26
26 is the average number of girls in a class.
(b) Find the average number of boys in a class.
Solution:
Average number of boys in a class = $$\frac{Sum of no. of boys in a class}{Number of classes}$$
= $$\frac{34+38+39}{3}$$
= $$\frac{111}{3}$$
= 37
37 is the average no. of boys in a class.
(c) Find the total number of students in all three classes.
Solution:
Total number of students = No. of girls + No. of boys
= 78 + 111
= 189
189 is the total number of students in all three classes.
(d) What is the average number of students in a class?
Solution:
Average number of students in a class = $$\frac{189}{3}$$ = 63
63 students is the average number of students in a class.
Question 1.
Find the average of the following set of numbers:
(a) 3, 13, 20, 25, 8
Solution:
(b) $$\frac{6}{20}, \frac{7}{20}, \frac{9}{20}, \frac{5}{20}, \frac{6}{20}$$
Solution:
(c) 20, 24, 25, 26, 27, 28
Solution:
Question 2.
Complete the table by finding the average of these sets.
Solution:
Question 3.
Find the average of the following:
(i) 3.2, 2.2, 2.6, 3.8, 2.8
Solution:
Average = $$\frac{3.2 + 2.2 + 2.6 + 3.8 + 2.8}{5}$$
= $$\frac{14.6}{5}$$
= 2.92
(ii) 0.5, 5.5, 2.5, 3.5, 4.5
Solution:
Average = $$\frac{0.5+5.5+2.5+3.5+4.5}{5}$$
= $$\frac{16.5}{5}$$
= 3.3
(iii) 1 m 50 cm, 2 m 30 cm, 4 m 20 cm, 3 m 25 cm, 5 m 20 cm
Solution:
Average = $$\frac{1.50+2.30+4.20+3.25+5.20}{5}$$
= $$\frac{16.45}{5}$$
= 3.29
Question 4.
Fill in the blanks:
(i) The average of 5, 15, and 10 is _________
Solution:
10
(ii) We divide _________ by _________ to find the average of 2, 3, 4.
Solution:
9, 3
(iii) We find the average by adding all the numbers and dividing the number of _________
Solution:
Average = $$\frac{Sum of numbers}{Number of addends}$$
8 = $$\frac{7+9+x}{3}$$
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# Application of infinite geometric progression
In this section we will discuss about the application of infinite geometric progression. 1) One side of an equilateral triangle is 18 cm. The mid points of its sides are joined to form another triangle whose mid points, in turn , are joined to form still another triangle. The process is continued indefinitely. Find the sum of the perimeters of all the triangles.
Solution : Let $A_{1}A_{2}A_{3}$ be the first equilateral triangle with each side 18 cm. Let $B_{1},B_{2},B_{3}$ are the mid points on its sides and $C_{1},C_{2},C_{3}$ are further midpoints of triangle.
All sides of triangle = 18 cm ⇒ $A_{1}A_{2} = A_{1}A_{3} =A_{2}A_{3}$ = 18 cm
∴ perimeter of $\Delta A_{1}A_{2}A_{3} = 3 \times$ 18 ----- (1)
∴ $B_{1}B_{2} = \frac{A_{2}A_{3}}{2}$;
$B_{1}B_{2} = \frac{18}{2}$
As $\Delta B_{1}B_{2}B_{3}$ is an equilateral triangle so each side of this triangle is $\frac{18}{2}$
∴ perimeter of $\Delta B_{1}B_{2}B_{3} = 3(\frac{18}{2}$) ----- (2)
Now we will consider inner triangle $C_{1}C_{2}C_{3}$
$C_{1}C_{2} = \frac{B_{2}B_{3}}{2}$
OR
$C_{1}C_{2} = \frac{A_{1}A_{2}}{4}$
∴ perimeter of $\Delta C_{1}C_{2}C_{3} = 3(\frac{18}{4}$) ----- (3)
∴ Sum of all the perimeters of the triangle will be
= $\left \{ 3 \times 18 + 3(\frac{18}{2})+ 3(\frac{18}{4}) + ...+ \infty \right \}$
= 3$\left \{ 18 + \frac{18}{2}+ \frac{18}{4} + ...+ \infty \right \}$
2) Find the rational number whose decimal expansion is $0.3\overline{56}$.
Solution : We have,
$0.3\overline{56}$ = 0.3 + 0.056 + 0.0056 + 0.00056 + 0.000056 + ...+ $\infty$
⇒ $0.3\overline{56}$ = 0.3 +$\left [ \frac{56}{10^{3}} + \frac{56}{10^{5}} + \frac{56}{10^{7}} + ...\infty \right ]$
First term = a = $\frac{56}{10^{3}}$ and common ratio = r = $\frac{1}{10^{2}}$
⇒ $0.3\overline{56} = \frac{\frac{56}{10^{3}}}{1-\frac{1}{10^{2}}}$
⇒ $0.3\overline{56} = \frac{3}{10} + \frac{56}{990}$
= $\frac{353}{990}$
## Practice questions on application of infinite geometric progression
1) A person has 2 parents, 4 grandparents, 8 great grandparents, and so on. Find the number of his ancestors during the ten generations preceding his own. (Ans- 2046)
2) A man deposited Rs 10000 in a bank at the rate of 5% simple interest annually. Find the amount in 15th year since he deposited the amount and also calculate the total amount after 20 years.
(Ans- Rs 17000; 295000)
3) Find the sum of the following series up to n terms:
(i) 5 + 55 +555 + … (ii) .6 +. 66 +. 666+…
4) A manufacturer reckons that the value of a machine, which costs him Rs. 15625, will depreciate each year by 20%. Find the estimated value at the end of 5 years. (Ans-. Rs 5120)
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# A train travelling 25 km an hour leaves Aligarh at 9a.m. and another train travelling 35 km and hour starts at 2 p.m. in the same direction. How many km from Aligarh will they be together?$\left( a \right)750km \\ \left( b \right)437\dfrac{1}{2}km \\ \left( c \right)417\dfrac{1}{2}km \\ \left( d \right)407\dfrac{1}{2}km \\$
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Hint: In this question, we use the concept of relative speed. Relative speed is defined as the speed of a moving object with respect to another. When two objects are moving in the same direction, relative speed is calculated as their difference.
Speed of the first train is 25km/h and speed of the second train is 35km/h.
Now, the first train leaves the Aligarh station at 9a.m. and travels by 2p.m. so the first train takes 5 hours.
Distance cover by first train in 5 hours,
${\text{Distance = }}\left( {{\text{speed of first train}}} \right) \times \left( {{\text{time taken by first train}}} \right) \\ \Rightarrow {\text{Distance = }}25 \times 5 \\ \Rightarrow {\text{Distance = }}125km \\$
Distance covered by the first train in 5 hours is 125km.
Now, relative speed of two trains $= \left( {35 - 25} \right)km/h = 10km/h$
Now we find time taken by second train will meet the first train,
${\text{Time taken}} = \dfrac{{{\text{Distance covered by first train }}}}{{{\text{Relative speed}}}} \\ \Rightarrow {\text{Time taken}} = \dfrac{{125}}{{10}} \\ \Rightarrow {\text{Time taken}} = 12.5{\text{hours}} \\$
Now the distance from Aligarh when both the trains are together,
$= \left( {{\text{Time taken}}} \right) \times \left( {{\text{Speed of second train}}} \right) \\ = 12.5 \times 35 = 437.5km = 437\dfrac{1}{2}km \\$
So, the correct option is (b).
Note: Whenever we face such types of problems we use some important points. First we find the distance covered by the first train with respect to the second train then find relative speed by using the concept of relative motion and also find time. So, we will get the required answer.
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# Are the following pair of linear
Question:
(i) $-3 x-4 y=12$ and $4 y+3 x=12$
(ii) $\frac{3}{5} x-y=\frac{1}{2}$ and $\frac{1}{5} x-3 y=\frac{1}{6}$
(iii) $2 a x+b y=a$ and $4 a x+2 b y-2 a=0 ; a, b \neq 0$
(iv) $x+3 y=11$ and $2(2 x+6 y)=22$
Solution:
Conditions for pair of linear equations are consistent
$\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$ [unique solution]
and $\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$ [infinitely many solutions]
(I) No, the given pair of linear equations
$-3 x-4 y \neq 12$ and $3 x+4 y=12$
Here, $a_{1}=-3, b_{1}=-4, c_{1}=-12 ;$
$a_{2}=3, b_{2}=4, c_{2}=-12$
Now, . $\frac{a_{1}}{a_{2}}=-\frac{3}{3}=-1, \frac{b_{1}}{b_{2}}=-\frac{4}{4}=-1, \frac{c_{1}}{c_{2}}=\frac{-12}{-12}=1$
$\because$ $\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$
Hence, the pair of linear equations has no solution, i.e., inconsistent.
(ii) Yes, the given pair of linear equations
$\frac{3}{5} x-y=\frac{1}{2}$ and $\frac{1}{5} x-3 y=\frac{1}{6}$
Here, $a_{1}=\frac{3}{5}, b_{1}=-1_{1} c_{1}=-\frac{1}{2}$
and $a_{2}=\frac{1}{5}, b_{2}=-3, c_{2}=-\frac{1}{6}$
Now, $\frac{a_{1}}{a_{2}}=\frac{3}{1}, \frac{b_{1}}{b_{2}}=\frac{-1}{-3}=\frac{1}{3}, \frac{c_{1}}{c_{2}}=\frac{3}{1}$ $\left[\because \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}\right]$
Hence, the given pair of linear equations has unique solution, i.e., consistent.
(iii) Yes, the given pair of linear equations
$2 a x+b y-a=0$
and $\quad 4 a x+2 b y-2 a=0, a, b \neq 0$
Here, $\quad a_{1}=2 a, b_{1}=b, c_{1}=-a_{i}$
$a_{2}=4 a, b_{2}=2 b, c_{2}=-2 a$
Now, $\frac{a_{1}}{a_{2}}=\frac{2 a}{4 a}=\frac{1}{2}, \frac{b_{1}}{b_{2}}=\frac{b}{2 b}=\frac{1}{2}, \frac{c_{1}}{c_{2}}=\frac{-a}{-2 a}=\frac{1}{2}$
$\because$ $\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}^{}}=\frac{c_{1}}{c_{2}}=\frac{1}{2}$
Hence, the given pair of linear equations has infinitely many solutions; i.e., consistent or dependent.
(iv) No, the given pair of linear equations
$x+3 y=11$ and $2 x+6 y=11$
Here, $a_{1}=1, b_{1}=3, c_{1}=-11$ ...(i)
$a_{2}=2, b_{2}=6, c_{2}=-11$
Now, $\frac{a_{1}}{a_{2}}=\frac{1}{2}, \frac{b_{1}}{b_{2}}=\frac{3}{6}=\frac{1}{2}, \frac{c_{1}}{c_{2}}=\frac{-11}{-11}=1$
$\therefore$ $\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$
Hence, the pair of linear equation have no solution $i . e .$, inconsistent.
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The Z Score Formula Example: Understanding Z Scores and How to Calculate Them
When it comes to statistical analysis, understanding the concept of z scores is crucial. Whether you’re a student of statistics, a researcher, or simply someone interested in data analysis, knowing how to calculate and interpret z scores can be incredibly valuable. In this article, we’ll delve into the z score formula with a comprehensive example to illustrate its application. By the end of this article, you’ll have a clear understanding of what z scores are, how they are calculated, and how they can be used in practical scenarios.
What is a Z Score?
Before we dive into the formula and example, let’s briefly cover the basics. A z score, also known as a standard score, is a statistical measurement that describes a value’s relationship to the mean of a group of values. In simpler terms, it tells us how many standard deviations a particular value is from the mean of the dataset.
Formula for Calculating Z Score
The z score formula is relatively straightforward. It is calculated using the following formula:
Z = (X - μ) / σ
Where:
• Z = z score
• X = individual data point
• μ = mean of the dataset
• σ = standard deviation of the dataset
Now, let’s see the z score formula in action with an example.
Z Score Formula Example
Suppose we have a dataset of exam scores from a class of 50 students. The mean (μ) score is 75, and the standard deviation (σ) is 10. We want to calculate the z score for a student who scored 82 on the exam.
Using the z score formula:
Z = (X - μ) / σ
We can plug in the values:
Z = (82 - 75) / 10
Z = 0.7
So, the z score for the student who scored 82 is 0.7. But what does this z score signify?
Interpreting Z Scores
A z score of 0.7 indicates that the student’s score is 0.7 standard deviations above the mean score of the class. In other words, the student performed better than the average, but not exceptionally so.
It’s essential to understand that z scores can be both positive and negative. A positive z score indicates a value above the mean, while a negative z score indicates a value below the mean.
Using Z Scores for Analysis
Z scores are valuable for various analytical purposes. They allow us to compare values from different datasets, identify outliers, and understand the relative position of a data point within a distribution. Whether it’s in finance, psychology, sports, or any other field that deals with data, z scores are a universal tool for standardizing and comparing values.
What is the significance of a z score?
Z scores help us understand where a particular value stands in relation to the mean of the dataset. They indicate how many standard deviations a data point is from the mean.
Can a z score be greater than 3?
Yes, a z score can be greater than 3, especially in large datasets. This simply means that the data point is several standard deviations away from the mean.
How are z scores used in hypothesis testing?
Z scores are used to determine the likelihood of obtaining a particular value in a normal distribution. This is crucial in hypothesis testing and assessing the statistical significance of results.
Conclusion
Understanding the z score formula and its application through an example is fundamental to mastering statistical analysis. Whether you’re dealing with test scores, economic indicators, or any other form of quantitative data, z scores provide a standardized method for comparing values. By grasping the concept of z scores and their interpretation, you gain a powerful tool for deriving insights from data and making informed decisions.
If you want to know other articles similar to The Z Score Formula Example: Understanding Z Scores and How to Calculate Them you can visit the category Sciences.
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Perpendicular lines (Coordinate Geometry)
When two lines are perpendicular, the slope of one is the negative reciprocal of the other.
If the slope of one line is m, the slope of the other is
Try this Drag points C or D. Note the slopes when the lines are at right angles to each other.
When two lines are perpendicular to each other (at right angles or 90°), their slopes have a particular relationship to each other. If the slope of one line is m then the slope of the other line is the negative reciprocal of m, or
So for example in the figure above, the line AB has a slope of 0.5, meaning it goes up by a half for every one across. The line CD if it is perpendicular to AB has a slope of -1/0.5 or -2. Adjust points C or D to make CD perpendicular to AB and verify this result.
Fig 1. Lines are still perpendicular
Remember that the equation works both ways, so it doesn't matter which line you start with. In the figure above the slope of CD is -2. So the slope of AB when perpendicular is
Note too that the lines to do not have to intersect to be perpendicular. In Fig 1, the two lines are perpendicular to each other even though they do not touch. The slope relationship still holds.
## Example 1. Are two lines perpendicular?
Fig 1. Are these lines perpendicular?
In Fig 1, the line AB and a line segment CD appear to be at right angles to each other. Determine if this is true.
To do this, we find the slope of each line and then check to see if one slope is the negative reciprocal of the other.
If the lines are perpendicular, each will be the negative reciprocal of the other. It doesn't matter which line we start with, so we will pick AB:
So, the slope of CD is -2.22, and the negative reciprocal of the slope of AB is -2.79. These are not the same, so the lines are not perpendicular, even though they look it. If you look carefully at the diagram, you can see that the point C is a little too far to the left for the lines to be perpendicular.
## Example 2. Define a line through a point perpendicular to a line
In Fig 1, find a line through the point E that is perpendicular to CD.
The point E is on the y-axis and so is the y-intercept of the desired line. Once we know the slope of the line, we can express it using its equation in slope-intercept form y=mx+b, where m is the slope and b is the y-intercept.
First find the slope of the line CD:
The line we seek will have a slope which is the negative reciprocal of this:
The intercept is 10, the point where the line will cross the y-axis. Substituting these values into the equation, the line we need is described by the equation
y = 0.45x + 10
This is one of the ways a line can be defined and so we have solved the problem. If we wanted to plot the line, we would find another point on the line using the equation and then draw the line through that point and the intercept. For more on this see Equation of a Line (slope - intercept form)
## Things to try
1. In the diagram at the top of the page, press 'reset'.
2. Note that because the slope of one line is the negative reciprocal of the other, the lines are perpendicular.
3. Adjust one of the points C,D. The lines are no longer perpendicular.
4. Click on "hide details". Determine the slope of both lines and prove they are not perpendicular. Click "show coordinates" if you wish to know them accurately. Click "show details" to verify.
## Limitations
In the interest of clarity in the applet above, the coordinates are rounded off to integers and the lengths rounded to one decimal place. This can cause calculatioons to be slightly off.
For more see Teaching Notes
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# Arithmetic Progressions-Exercise 5.3-Class 10
1. Find the sum of the following APs.
(i) 2, 7, 12 ,…., to 10 terms.
(ii) − 37, − 33, − 29 ,…, to 12 terms
(iii)0.6, 1.7, 2.8 ,…….., to 100 terms
(iv) 1/15 , 1/12 , 1/10 ,………, to 11 terms
Solution:
(i) 2, 7, 12 ,…, to 10 terms
For this A.P., a = 2, d = a2 – a1 = 7 − 2 = 5 and n = 10 We know that,
Sn = (n/2) [2a +(n-1)d]
S10 = (10/2)[2(2)+(10-1)5]
=5[4+(9)x5]
=5×49
245
(ii) −37, −33, −29 ,…, to 12 terms
For this A.P., a = −37, d = a2 – a1 = (−33) − (−37) = − 33 + 37 = 4
n = 12
W know that
Sn = (n/2) [2a +(n-1)d]
S12 = (12/2) [2(-37) +(12-1)4]
=6[-74+11×4]
=6[-74+44]
=6(-30)
=-180
(iii) 0.6, 1.7, 2.8 ,…, to 100 terms For this A.P., a = 0.6, d = a2 – a1 = 1.7 − 0.6 = 1.1 and n = 100 We know that,
Sn = (n/2) [2a +(n-1)d]
S100 = (100/2) [2(0.6) +(100-1)1.1]
=50[1.2 +(99)x(1.1)]
=50[1.2 + 108.9]
=50[110.1]
= 5505
(iv) 1 15 , 1 12 , 1 10 ,………, to 11 terms For this A.P.,
a = 1/15
n =11
d = a2 – a1 = 1/12 – 1/15 = (5-4)/60 = 1/60
We know that
Sn = (n/2) [2a +(n-1)d]
2. Find the sums given below
(i) 7 + 10 1 2 + 14 + ………… + 84
(ii) 34 + 32 + 30 + ……….. + 10
(iii) − 5 + (− 8) + (− 11) + ………… + (− 230)
Solution:
(i) 7 + 10 1/2 + 14 + …………+ 84
For this A.P., a = 7 and l = 84
d = a2 – a1 = 10 1/2 – 7 = 7/2
Let 84 be the n th term of this A.P.
l = a + (n − 1)d
84 = 7 + (n – 1)(7/2)
77 = (n – 1)(7/2)
22 = n – 1
n = 23
We know that
Sn = (n/2)[a + l]
S23 = (23/2)[7 + 84]
= (23×91)/2 = 2093/2
= 1046 ½
(ii)34 + 32 + 30 + ……….. + 10
For this A.P., a = 34, d = a2 – a1 = 32 − 34 = −2 and l = 10
Let 10 be the n th term of this A.P.
l = a + (n − 1) d
10 = 34 + (n − 1) (−2)
−24 = (n − 1)
(−2) 12 = n − 1
n = 13
Sn = (n/2)[a + l]
Sn = 13/2 [34+10]
= (13×44)/2 = 13×22
= 286
(iii)(−5) + (−8) + (−11) + ………… + (−230)
For this A.P., a = −5, l = −230 and d = a2 – a1 = (−8) − (−5) = − 8 + 5 = −3
Let −230 be the n th term of this A.P.
l = a + (n − 1)d
−230 = − 5 + (n − 1) (−3)
−225 = (n − 1) (−3)
(n − 1) = 75
n = 76
And,
Sn = (n/2)[a + l]
Sn = (76/2)[(-5) + (-230)]
=38(-235)
= -8930
3. In an AP
(i) Given a = 5, d = 3, an = 50, find n and Sn.
(ii) Given a = 7, a13 = 35, find d and S13.
(iii) Given a12 = 37, d = 3, find a and S12.
(iv) Given a3 = 15, S10 = 125, find d and a10.
(v) Given d = 5, S9 = 75, find a and a9.
(vi) Given a = 2, d = 8, Sn = 90, find n and an.
(vii) Given a = 8, an = 62, Sn = 210, find n and d.
(viii) Given an = 4, d 2, Sn = − 14, find n and a.
(ix) Given a = 3, n = 8, S = 192, find d.
(x)Given l = 28, S = 144 and there are total 9 terms. Find a.
Solution:
(i) Given that, a = 5, d = 3, an = 50
As an = a + (n − 1)d,
∴ 50 = 5 + (n − 1)3
45 = (n − 1)3
15 = n − 1
n = 16
Sn = (n/2)[a + an]
S16 = (16/2)[5 + 50]
= 8 x 55
= 440
(ii) Given that, a = 7, a13 = 35
As an = a + (n − 1) d,
∴ a13 = a + (13 − 1) d
35 = 7 + 12 d
35 − 7 = 12d
28 = 12d
d = 7/3
Sn = (n/2)[a + an]
S13 = (13/2)[7 + 35]
= (13×42)/2 = 13×21
= 273
(iii)Given that, a12 = 37, d = 3
As an = a + (n − 1)d,
a12 = a + (12 − 1)3
37 = a + 33
a = 4
an = a + (n – 1)d
a12 = a + (12 – 1)3
37 = a + 33
a = 4
Sn = (n/2)[a + an]
Sn = (12/2)[4 + 37]
= 6(41)
= 246
(iv) Given that, a3 = 15, S10 = 125
As an = a + (n − 1)d,
a3 = a + (3 − 1)d
15 = a + 2d ……………………….(i)
Sn = (n/2)[2a+(n – 1)d]
S10 = (10/2)[2a+(10 – 1)d]
125 = 5(2a + 9d)
25 = 2a + 9d ————-(ii)
On multiplying equation (i) by 2, we obtain
30 = 2a + 4d …………………..(iii)
On subtracting equation (iii) from (ii), we obtain
−5 = 5d d = −1
From equation (i),
15 = a + 2(−1)
15 = a − 2
a = 17
a10 = a + (10 − 1)d
a10 = 17 + (9) (−1)
a10 = 17 − 9 = 8
(v)Given that, d = 5, S9 = 75
As,
Sn = (n/2)[2a+(n – 1)d]
S9 = (9/2)[2a+(9 – 1)5]
75 = 9/2[2a+40]
25 = 3(a+20)
25 = 3a + 60
3a = 25 – 60
a = -35/3
an = a + (n – 1)d
a9 = a + (9 – 1)5
= (-35/3) +8(5)
= (-35/3) + 40
= (-35+120)/3
= 85/3
(vi) Given that, a = 2, d = 8, Sn = 90
Sn = (n/2)[2a+(n – 1)d]
90 = (n/2)[4+(n – 1)8]
90 = n[2+(n-1)4]
90 = n[ 2+4n-4]
90 = n (4n − 2) = 4n2 − 2n
4n2 − 2n + 18n − 90 = 0
4n2 − 20n + 18n − 90 = 0
n (n − 5) + 18 (n − 5) = 0
(n − 5) (4n + 18) = 0
Either n − 5 = 0 or 4n + 18 = 0
n = 5 or n = -18/4 = -9/2
However, n can neither b negative noe fractional.
Therefore, n = 5
an = a + (n – 1)d
a5 = 2 + (5 – 1)8
= 2 + 4×8
= 2 + 32
= 34
(vii) Given that, a = 8, an = 62, Sn = 210
Sn = (n/2)[a + an]
210 = (n/2)[8 + 62]
210 = n/2 (70)
n = 6
an = a + (n − 1)d
62 = 8 + (6 − 1)d
62 − 8 = 5d
54 = 5d
d = 54/5
(viii) Given that, an = 4, d = 2, Sn = −14
an = a + (n − 1)d
4 = a + (n − 1)2
4 = a + 2n − 2
a + 2n = 6
a = 6 − 2n …………………………….(i)
Sn = (n/2)[a + an]
-14 = n/2[a+4]
−28 = n (a + 4)
−28 = n (6 − 2n + 4) {From equation (i)}
−28 = n (− 2n + 10)
−28 = − 2n2 + 10n
2n2 − 10n − 28 = 0
n2 − 5n −14 = 0
n2 − 7n + 2n − 14 = 0
n (n − 7) + 2(n − 7) = 0
(n − 7) (n + 2) = 0
Either n − 7 = 0 or n + 2 = 0
n = 7 or n = −2
However, n can neither be negative nor fractional.
Therefore, n = 7
From equation (i), we obtain
a = 6 − 2n
a = 6 − 2(7)
= 6 – 14
= −8
(ix)Given that, a = 3, n = 8, S = 192
Sn = (n/2)[2a+(n – 1)d]
192 = (8/2)[2×3+(8 – 1)d]
192 = 4 [6 + 7d]
48 = 6 + 7d
42 = 7d
d = 6
(x)Given that, l = 28, S = 144 and there are total of 9 terms.
Sn = (n/2)[a+l]
144 = (9/2)[a+28]
16×2 = a + 28
32 = a + 28
a = 4
4. How many terms of the AP. 9, 17, 25 … must be taken to give a sum of 636?
Solution:
Let there be n terms of this A.P.
For this A.P., a = 9 and d = a2 – a1 = 17 − 9 = 8
Sn = (n/2)[2a+(n – 1)d]
636 = (n/2)[2×9+(n – 1)8]
636 = (n/2)[18+(n-1)8]
636 = n [9 + 4n − 4]
636 = n (4n + 5)
4n2 + 5n − 636 = 0
4n2 + 53n − 48n − 636 = 0
n(4n + 53) − 12 (4n + 53) = 0
(4n + 53) (n − 12) = 0
Either 4n + 53 = 0 or n − 12 = 0
𝑛 = −53/4 or n = 12
n cannot be −53/4 . As the number of terms can neither be negative nor fractional, therefore, n = 12 only.
5. The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.
Solution;
Given that, a = 5, l = 45 and Sn = 400
Sn = n/2 [a+l]
400 = n/2 (5+45)
400= n/2[50]
n = 16l = a+(n-1)d
45 = 5+(16-1)d
40 = 15d
d= 40/15 = 8/3
6. The first and the last term of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?
Solution;
Given that, a = 17, l = 350 and d = 9
Let there be n terms in the A.P.
l = a + (n − 1) d
350 = 17 + (n − 1)9
333 = (n − 1)9
(n − 1) = 37
n = 38
Sn = (n/2)[a+l]
Sn = (38/2)[17+350] = 19(367) = 6973
Thus, this A.P. contains 38 terms and the sum of the terms of this A.P. is 6973
7. Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.
Solution:
d = 7 and a22 = 149. S22 =?
an = a + (n − 1)d
a22 = a + (22 − 1)d
149 = a + 21 × 7
149 = a + 147
a = 2
Sn = (n/2)[a + an]
= (22/2)[2 + 149]
=11(151)
=1661
8. Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
Solution:
Given that, a2 = 14 and a3 = 18
d = a3 – a2 = 18 − 14 = 4
a2 = a + d
14 = a + 4
a = 10
Sn = (n/2)[2a+(n – 1)d]
Sn = (51/2)[2×10+(51 – 1)4]
= (51/2)[20+(50)4]
=(51×220)/2
= 51(110)
=5610
9. If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms
Solution:
Given that, S7 = 49
S17 = 289
Sn = (n/2)[2a+(n – 1)d]
S7 = (7/2)[2a+(7 – 1)d]
49 = (7/2)[2a+6d]
7 = (a+3d)
a+3d =7———–(1)
Similarly, S17 = (17/2)[2a+(17 – 1)d]
289 = (17/2)[2a+16d]
17 = (a+8d)
a+8d = 17 ———–(2)
Subtracting equation (i) from equation (ii),
5d = 10
d = 2
From equation (i),
a + 3(2) = 7
a + 6 = 7
a = 1
Sn = (n/2)[2a+(n – 1)d]
= (n/2)[2(1)+(n – 1)(2)]
= (n/2)(2+2n-2)
=(n/2)(2n)
=n2
10. Show that a1, a2 … , an , … form an AP where an is defined as below (i) an = 3 + 4n (ii)an = 9 − 5n Also find the sum of the first 15 terms in each case.
Solution:
(i) an = 3 + 4n
a1 = 3 + 4(1) = 7
a2 = 3 + 4(2) = 3 + 8 = 11
a3 = 3 + 4(3) = 3 + 12 = 15
a4 = 3 + 4(4) = 3 + 16 = 19
It can be observed that
a2 − a1 = 11 − 7 = 4
a3 − a2 = 15 − 11 = 4
a4 − a3 = 19 − 15 = 4
i.e., ak + 1 − ak is same every time.
Therefore, this is an AP with common difference as 4 and first term as 7.
Sn = (n/2)[2a+(n – 1)d]
S15 = (15/2)[2(7)+(15 – 1)4]
=(15/2)[(14)+56]
=15/2(70)
=15×35
=525
(ii) an = 9 − 5n
a1 = 9 − 5 × 1 = 9 − 5 = 4
a2 = 9 − 5 × 2 = 9 − 10 = −1
a3 = 9 − 5 × 3 = 9 − 15 = −6
a4 = 9 − 5 × 4 = 9 − 20 = −11
It can be observed that
a2 − a1 = − 1 − 4 = −5
a3 − a2 = − 6 − (−1) = −5
a4 − a3 = − 11 − (−6) = −5
i.e., ak + 1 − ak is same every time. Therefore, this is an A.P. with common difference as −5 and first term as 4.
Sn = (n/2)[2a+(n – 1)d]
S15 = (15/2)[2(4)+(15 – 1)(-5)]
=(15/2)[8+14(-5)]
= (15/2)[8-70]
= (15/2)[-62]
=15(-31)
=-465
11. If the sum of the first n terms of an AP is 4n – n2 , what is the first term (that is S1)? What is the sum of first two terms? What is the second term? Similarly find the 3rd, the 10th and the nth terms
Solution:
Given that, Sn = 4n – n2
First term, a = S1 = 4(1) − (1)2 = 4 − 1 = 3
Sum of first two terms = S2 = 4(2) − (2)2 = 8 − 4 = 4
Second term, a2 = S2 – S1 = 4 − 3 = 1
d = a2 − a = 1 − 3 = −2
an = a + (n − 1)d
= 3 + (n − 1) (−2)
= 3 − 2n + 2
= 5 − 2n
Therefore,
a3 = 5 − 2(3) = 5 − 6 = −1
a10 = 5 − 2(10) = 5 − 20 = −15
Hence, the sum of first two terms is 4. The second term is 1. 3rd, 10th , and n th terms are −1, −15, and 5 − 2n respectively.
12. Find the sum of first 40 positive integers divisible by 6.
Solution:
The positive integers that are divisible by 6 are 6, 12, 18, 24 …
It can be observed that these are making an A.P. whose first term is 6 and common difference is 6.
a = 6 and d = 6
S40 =?
Sn = (n/2)[2a+(n-1)d]
S40 = (40/2)[2(6)+(40-1)6]
= 20[12+(39)(6)]
= 20 x246
= 4920
13. Find the sum of first 15 multiples of 8.
Solution:
The multiples of 8 are 8, 16, 24, 32…
These are in an A.P., having first term as 8 and common difference as 8. Therefore, a = 8 and d = 8
S15 =?
Sn = (n/2)[2a+(n-1)d]
Sn = (15/2)[2(8)+(15-1)8]
= (15/2)[16+14(8)]
=(15/2)[16+112]
=(15×128)/2 = 15 x 64
= 960
14. Find the sum of the odd numbers between 0 and 50.
Solution:
The odd numbers between 0 and 50 are 1, 3, 5, 7, 9 … 49
Therefore, it can be observed that these odd numbers are in an A.P. a = 1, d = 2 and l = 49
l = a + (n − 1) d
49 = 1 + (n − 1)2
48 = 2(n − 1)
n − 1 = 24
n = 25
Sn = (n/2)[a+l]
S25 = (25/2)[1+49]
=(25×50)/2 = 25 x 25
= 625
15. A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: Rs. 200 for the first day, Rs. 250 for the second day, Rs. 300 for the third day, etc., the penalty for each succeeding day being Rs. 50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by 30 days.
Solution:
It can be observed that these penalties are in an A.P. having first term as 200 and common difference as 50.
a = 200 and d = 50 Penalty that has to be paid if he has delayed the work by 30 days = S30
= (30/2)[2(200)+(30-1)50]
= 15[400+1450]
= 15(1850)
= 27750
Therefore, the contractor has to pay Rs 27750 as penalty.
16. A sum of Rs 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is Rs 20 less than its preceding prize, find the value of each of the prizes.
Solution:
Let the cost of 1st prize be P. Cost of 2nd prize = P − 20
And cost of 3rd prize = P − 40 It can be observed that the cost of these prizes are in an A.P. having common difference as −20 and first term as P.
a = P and d = −20 Given that, S7 = 700
(7/2) [2a+(7-1)d] = 700
[2a+(6)(-20)]/2 = 100
a + 3(−20) = 100
a − 60 = 100
a = 160
Therefore, the value of each of the prizes was Rs 160, Rs 140, Rs 120, Rs 100, Rs 80, Rs 60, and Rs 40.
17. In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g., a section of class I will plant 1 tree, a section of class II will plant 2 trees and so on till class XII. There are three sections of each class. How many trees will be planted by the students?
Solution:
It can be observed that the number of trees planted by the students is in an AP.
1, 2, 3, 4, 5………………..12
First term, a = 1
Common difference, d = 2 − 1 = 1
Sn = (n/2)[2a+(n-1)d]
S12 = (12/2)[2(1)+(12-1)(1)]
= 6 (2 + 11)
= 6 (13)
= 78
Therefore, number of trees planted by 1 section of the classes = 78 Number of trees planted by 3 sections of the classes = 3 × 78 = 234 Therefore, 234 trees will be planted by the students.
18. A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A of radii 0.5, 1.0 cm, 1.5 cm, 2.0 cm, ……… as shown in figure. What is the total length of such a spiral made up of thirteen consecutive semi-circles?
Solution:
Semi-perimeter of circle = 𝜋𝑟
I1 = 𝜋 (0.5) = 𝜋/2 𝑐𝑚
I2 = 𝜋 (1) = 𝜋 𝑐𝑚
I3 = 𝜋 (1.5) = 3𝜋/2 𝑐𝑚
Therefore, I1, I2, I3 ,i.e. the lengths of the semi-circles are in an A.P.,
𝜋 /2, 𝜋, 3 𝜋/2, 2 𝜋….
a = 𝜋/2
d = 𝜋 – (𝜋/2) = 𝜋/2
s13 = ?
We know that the sum of n terms of an a A.P. is given by
= 143
Therefore, the length of such spiral of thirteen consecutive semi-circles will be 143 cm.
19. 200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on. In how many rows are the 200 logs placed and how many logs are in the top row?
Solution:
It can be observed that the numbers of logs in rows are in an A.P. 20, 19, 18…
For this A.P., a = 20 and d = a2 – a1 = 19 − 20 = −1
Let a total of 200 logs be placed in n rows. Sn = 200
Sn = (n/2)[2a+(n-1)d
200 = (n/2)[2(20)+(n-1)(-1)]
400 = n (40 − n + 1)
400 = n (41 − n)
400 = 41n – n2
n2 − 41n + 400 = 0
n2 − 16n − 25n + 400 = 0
n (n − 16) −25 (n − 16) = 0
(n − 16) (n − 25) = 0
Either (n − 16) = 0 or n − 25 = 0
n = 16 or n = 25
an = a + (n − 1)d
a16 = 20 + (16 − 1) (−1)
a16 = 20 – 15
a16 = 5
Similarly, a25 = 20 + (25 − 1) (−1)
a25 = 20 − 24 = −4
Clearly, the number of logs in 16th row is 5. However, the number of logs in 25th row is negative, which is not possible. Therefore, 200 logs can be placed in 16 rows and the number of logs in the 16th row is 5.
20. In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato and other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line.
A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run? [Hint: to pick up the first potato and the second potato, the total distance (in metres) run by a competitor is 2 × 5 + 2 ×(5 + 3)]
Solution:
The distances of potatoes are as follows. 5, 8, 11, 14…
It can be observed that these distances are in A.P. a = 5; d = 8 − 5 = 3
Sn = (n/2)[2a+(n-1)d]
S10 = (10/2)[2(5)+(10-1)3]
= 5[10+9×3]
= 5(10+27)
=5(37)
=185
As every time she has to run back to the bucket, therefore, the total distance that the competitor has to run will be two times of it. Therefore, total distance that the competitor will run = 2 × 185 = 370 m
## 3 thoughts on “Arithmetic Progressions-Exercise 5.3-Class 10”
1. Bandu Hulgunde says:
55/2=[2(6)+(55-1)-6]
Liked by 1 person
1. What happened? We didn’t get you.. can you tell us clearly?
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# When Is Triangle Isosceles: Miguel Ochoa Sanchez's Criterion
### Problem
In $\Delta ABC,\;$ point $P\;$ is on the angle bisector $AD.\;$ $BP\;$ meets $AC\;$ in $E;\;$ $CP\;$ meets $AB\;$ in $F.$
Prove that if
$\displaystyle \frac{1}{AB^3}+\frac{1}{AE^3}=\frac{1}{AC^3}+\frac{1}{AF^3},$
the triangle is isosceles.
### Lemma 1
Under the conditions of the above statement, if $AC\gt AB,\;$ then $[\Delta BCE]\gt [\Delta BCF],\;$ where $[X]\;$ denotes the area of shape $X.$
The lemma is equivalent to the assertion that point $E\;$ is "above" point $F.$
The lemma has an important consequence.
If $AC\gt AB,\;$ then
$AB\cdot AE\lt AC\cdot AF,$
Indeed, the lemma implies that
$\displaystyle\frac{1}{2}\sin A\cdot AB\cdot AE=[ABE]\lt [ACF]=\frac{1}{2}\sin A\cdot AC\cdot AF.$ Thus we have $AB\cdot AE\lt AC\cdot AF.$
### Lemma 2
Under the conditions of the above statement,
$\displaystyle\frac{1}{AB}+\frac{1}{AE}=\frac{1}{AC}+\frac{1}{AF}.$
Indeed, by Ceva's theorem, $\displaystyle\frac{AF}{FB}\cdot\frac{BD}{DC}\cdot\frac{CE}{EA}=1.\;$ In addition, using the fact that $AD\;$ is an angle bisector, $\displaystyle\frac{AB}{AC}=\frac{BD}{DC}.\;$ Combining the two we get,
$\displaystyle\frac{AB-AF}{AB\cdot AF}=\frac{AC-AE}{AC\cdot AE}.$
This reduces to the required identity.
The two lemmas combined give the following:
If $AC\gt AB,\;$ then
$AB+AE\lt AC+AF,$
Indeed, by Lemma 2, $\displaystyle\frac{AB+AE}{AB\cdot AE}=\frac{AC+AF}{AC\cdot AF}.\;$ So, since $AB\cdot AE\lt AC\cdot AF,\;$ so too $AB+AE\lt AC+AF.$
### A Simplified Version
First we prove the following
In $\Delta ABC,\;$ point $P\;$ is on the angle bisector $AD.\;$ $BP\;$ meets $AC\;$ in $E;\;$ $CP\;$ meets $AB\;$ in $F.$
Prove that if
$\displaystyle \frac{1}{AB^2}+\frac{1}{AE^2}=\frac{1}{AC^2}+\frac{1}{AF^2},$
the triangle is isosceles.
The proof is by contradiction. WLOG, assume $AC\gt AB.\;$ Then, by Lemma 2
$\displaystyle\left(\frac{1}{AB}+\frac{1}{AE}\right)^2=\left(\frac{1}{AC}+\frac{1}{AF}\right)^2,$
implying, under our assumption, $\displaystyle\frac{2}{AB\cdot AE}=\frac{2}{AC\cdot AF}.$ But this contradicts Lemma 1.
### Proof 1
We shall assume, as before, $AC\gt AB\;$ and now also $\displaystyle \frac{1}{AB^3}+\frac{1}{AE^3}=\frac{1}{AC^3}+\frac{1}{AF^3}.\;$ By Lemma 2,
$\displaystyle\left(\frac{1}{AB}+\frac{1}{AE}\right)^3=\left(\frac{1}{AC}+\frac{1}{AF}\right)^3,$
and subsequently,
$\displaystyle \frac{3}{AB\cdot AE\cdot (AB+AE)}=\frac{3}{AC\cdot AF\cdot (AC+AF)}.$
This is false by the corollaries to the two lemmas.
### Proof 2
This is the original proof by Miguel Ochoa Sanchez.
### Proof 3
Denote $\displaystyle\frac{1}{AB}+\frac{1}{AE}=s,\;$ $\displaystyle\frac{1}{AB}\cdot \frac{1}{AE}=p,\;$ $\displaystyle\frac{1}{AC}+\frac{1}{AF}=q,\;$ $\displaystyle\frac{1}{AC}\cdot\frac{1}{AF}=r.\;$
In $\Delta ABE,\;$ $\displaystyle AP=\frac{2\cos\phi}{s}\;$ and in $\Delta ACF,\;$ $\displaystyle AP=\frac{2\cos\phi}{q},\;$ implying $q=s.\;$ The given condittion is equivalent to $s^3-3sp=s^3-3sr\;$ such that $r=p.\;$ From here
$\displaystyle\left\{\frac{1}{AB},\frac{1}{AE}\right\}=\left\{\frac{1}{AC},\frac{1}{AF}\right\}.$
There are two possibilities: $AB=AF,\;AC=AE\;$ or $AB=AC,\;AE=AF.\;$ The former may only hold when $P=D,\;$ in which case the condtion of the problem becomes an inconsequential tautology and needs to be excluded. The latter possibility confirms the conclusion of the problem.
### Extra Challenge
In $\Delta ABC,\;$ point $P\;$ is on the angle bisector $AD.\;$ $BP\;$ meets $AC\;$ in $E;\;$ $CP\;$ meets $AB\;$ in $F.$
Prove that if
$\displaystyle \frac{1}{AB^4}+\frac{1}{AE^4}=\frac{1}{AC^4}+\frac{1}{AF^4},$
the triangle is isosceles.
Think of other exponents.
### Answer to the Extra Challenge
As in Proof 3, $q=s.\;$ From the given condition, $s^4-4s^2p+2p^2=s^4-4s^2r+2r^2,\;$ implying $4s^2p-2p^2=4s^2r-2r^2.\;$ But $\displaystyle 0\lt p,r\lt\frac{s^2}{4}\;$ and the quadratic $f(x)=x^2-2s^2x\;$ is injectivebon $[0,s^2]\;$ such that $r=p.\;$ The ending as in Proof 3.
### Acknowledgment
The problem above has been posted by Miguel Ochoa Sanchez at the Peru Geometrico facebook page. Proof 2 is by Miguel; Proof 3 and the resoponse to the Extra Challenge is by Leo Giugiuc.
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Pedals and Arcs
by
Patty Wagner
My goal today is to convince you that that the pedal triangle of a pedal triangle of a pedal triangle of a point is similar to the original triangle. That is, ΔLMK is similar to ΔABC.
We begin by constructing the pedal triangle of point P with ΔABC. Recall that the pedal triangle is formed by constructing the perpendicular to each side of the triangle passing through point P. The points of intersection of the perpendicular with the sides of the triangle form the vertices of the pedal triangle. ΔDEF is our first pedal triangle.
We can construct the circumcircle of ΔDAF and see that ∠PAF inscribes arc PF. Notice that ∠FDP inscribes the same arc. Since inscribed angles in the same circle subtending the same arc are equal, we know that ∠PAF ≅ ∠FDP.
We mark these angles and construct our next pedal triangle:
By again constructing the circumcircle, we can see that ∠GHP ≅ ∠GDP for the same reasons as we saw above.
Continuing the pattern with the pedal triangle, ΔKLM, we get the following:
Since we are only concerned with our inner pedal triangle and how it relates to ΔABC, we can rid ourselves of the extraneous information and note that ∠CAP ≅ ∠KLP. We can call that angle, θ. Now we shift our focus and see that we can make the same argument for the other angles:
Angle BAP ≅ to angle DFP. Continuing in the same manner for every angle in our pedal triangles, we end up with:
If we remove all the extraneous information again and look simply at ΔABC and ΔKLM, we can see that we have two sets of corresponding angles that are congruent:
• α + θ = ∠BAC ≅ ∠MLK = α + θ
• γ + β = ∠ACB ≅ ∠LKM = γ + β
The Angle-Angle Postulate says that when two angles of one triangle are equal to two corresponding angles of the other triangle, the two triangles must be similar. Therefore, ΔABC ∼ ΔLMK.
Q.E.D.
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# What is the point-slope form of the three lines that pass through (0,2), (4,5), and (0,0)?
Oct 4, 2016
The equations of three lines are $y = \frac{3}{4} x + 2$, $y = \frac{5}{4} x$ and $x = 0$.
#### Explanation:
The equation of line joining x_1,y_1) and x_2,y_2) is given by
$\frac{y - {y}_{1}}{{y}_{2} - {y}_{1}} = \frac{x - {x}_{1}}{{x}_{2} - {x}_{1}}$
while equation in pint slope form is of the type $y = m x + c$
Hence equation of line joining $\left(0 , 2\right)$ and $\left(4 , 5\right)$ is
$\frac{y - 2}{5 - 2} = \frac{x - 0}{4 - 0}$
or $\frac{y - 2}{3} = \frac{x}{4}$ or $4 y - 8 = 3 x$ or $4 y = 3 x + 8$ and
in point slope form it is $y = \frac{3}{4} x + 2$
and equation of line joining $\left(0 , 0\right)$ and $\left(4 , 5\right)$ is
$\frac{y - 0}{5 - 0} = \frac{x - 0}{4 - 0}$
or $\frac{y}{5} = \frac{x}{4}$ or $4 y = 5 x$ and
in point slope form it is $y = \frac{5}{4} x$
For equation of line joining $\left(0 , 0\right)$ and $\left(0 , 2\right)$, as ${x}_{2} - {x}_{1} = 0$ i.e. ${x}_{2} = {x}_{1}$, the denominator becomes zero and it is not possible to get equation. Similar would be the case if ${y}_{2} - {y}_{1} = 0$. In such cases as ordinates or abscissa are equal, we will have equations as $y = a$ or $x = b$.
Here, we have to find the equation of line joining $\left(0 , 0\right)$ and $\left(0 , 2\right)$. As we have common abscissa, the equation is
$x = 0$
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# Divisibility rule of 12 explained with examples
Learn and know the divisibility rule of 12 which is very useful for the everyone, especially for those who are preparing/attempting the bank exams. We know that in the bank exams there won’t be any sufficient time for checking the number whether it is divisible by 12 or not by regular division process. So learning this divisibility rule will help in solving problem in less time.
Every one of us knows the divisibility rules of the numbers 2 to 11 except 7. Only a few of us know the divisibility rule of 7. Generally, in the school curriculum, divisibility rules will be up to 11 only. Some of the students may be having doubt that there are any divisibility rules above 11 or not? Then the answer will be definitely “yes “only. So, it is clear that there are some divisibility rules for the numbers, which are above 11”. Now we will learn the divisibility rule of 12 in detail with examples.
## Divisibility rule of 12 as follows:
To check whether the given number is divisible by 12 or not, just check the given number is divisible by 4 and also by 3. If the given number is divisible by both 4 and 3 then it will be also divisible by the number 12. To understand this, let us consider the some examples.
Ex 1:
Find out 96 is divisible by 12 or not?
Sol:
According to the divisibility rule of 12, check 96 is divisible by the numbers 4 and 3. We know that, the divisibility rule of 4 i.e. last 2 digits should be divisible by the number 4. On observation it is clear that 96 is divisible by 4. Now check 96 is divisible by 3 or not. Divisibility rule of 3 is sum of the digits should be divisible by the 3. In 96, sum of the two digits i.e. 9 + 6 = 15. As the result 15 is divisible by 3 so 96 is divisible by 3.
So finally it is clear that 96 is divisible by 4 and 3. So, we say the number 96 is divisible by the number12.
Ex 2:
Find out 189 is divisible by 12 or not
Sol:
In 189, last 2 digits i.e. 89 is not divisible by 4. So the number 189 is not divisible by the 12.
NOTE:
In the above problem, I didn’t check 189 is divisible by 3 or not because according divisibility rule of 12 the given number 189 should be divisible by the both numbers 3 and 4. If one number not satisfying then no need to check other.
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# Unlocking the Power of Proportions: A Comprehensive Guide to Using a Proportion Calculator
## Introduction
In the world of mathematics, proportions play a crucial role in solving a wide range of real-world problems. Whether you’re a student struggling with your math homework, a scientist conducting experiments, or a chef adjusting a recipe, understanding proportions is essential. Fortunately, modern technology has made it easier than ever to work with proportions, thanks to the advent of proportion calculators. In this article, we will delve into the world of proportions, explore their significance, and guide you on how to effectively utilize a proportion calculator to simplify your calculations.
Understanding Proportions
Before we dive into the world of proportion calculators, let’s establish a solid understanding of what proportions are. A proportion is an equation that states two ratios are equal. It is often expressed in the form a:b = c:d, where ‘a’ and ‘c’ are the first ratios, and ‘b’ and ‘d’ are the second ratios. The key to solving proportion problems is to find the missing value within the equation, which is often denoted by a variable.
Proportions are widely used in various fields, such as:
1. Mathematics
Proportions are fundamental in solving many mathematical problems. They are commonly used in geometry, algebra, and statistics to compare quantities and find missing values. For instance, proportions can be used to determine the length of an unknown side in a right triangle or to calculate probabilities in statistics.
2. Science
In scientific research, proportions are vital for making accurate measurements and comparisons. Scientists often use proportions to express the relationship between different variables, enabling them to draw meaningful conclusions from their experiments.
3. Cooking
In the culinary world, proportions are essential for creating delicious dishes. Adjusting ingredient quantities based on the number of servings or personal preferences relies heavily on proportion calculations. A slight variation in proportions can make the difference between a mouthwatering meal and a culinary disaster.
The Role of Proportion Calculators
Proportion calculators have revolutionized the way we work with proportions. These user-friendly tools simplify complex calculations, making them accessible to individuals with varying levels of mathematical proficiency. Whether you are dealing with simple ratios or intricate proportions, a proportion calculator can save you time and reduce the risk of errors in your calculations.
Using a Proportion Calculator
Here’s a step-by-step guide on how to effectively use a proportion calculator:
Step 1: Input the Known Values
Begin by entering the known values into the calculator. These are the numbers you are sure of in your proportion equation. For example, if you have the proportion a:b = 2:5 and you know that ‘a’ is 10, you would input 10 as the first known value.
Step 2: Identify the Unknown Value
Determine which value in the proportion equation you need to find. Let’s say you want to find ‘b’ in the proportion a:b = 2:5; ‘b’ is your unknown value.
Step 3: Calculate the Unknown Value
Click the calculate button, and the proportion calculator will instantly provide you with the value of ‘b’ in this example. It will do the necessary calculations to ensure that the ratios remain equal.
Always double-check the results obtained from the proportion calculator to ensure accuracy. This step is especially important when working on critical tasks like scientific research or recipe adjustments.
Benefits of Using Proportion Calculators
Now that we’ve discussed how to use a proportion calculator, let’s explore the benefits of incorporating these tools into your problem-solving arsenal:
1. Speed and Efficiency
Proportion calculators eliminate the need for manual calculations, which can be time-consuming and error-prone. With just a few clicks, you can obtain accurate results in seconds.
2. Reduced Human Error
Human error is a common factor in mathematical calculations. Proportion calculators significantly reduce the risk of mistakes, ensuring the reliability of your results.
3. Versatility
Proportion calculators can handle a wide range of proportions, from simple to complex. Whether you’re dealing with basic arithmetic or intricate scientific equations, these tools have you covered.
4. Educational Aid
For students, proportion calculators can serve as valuable learning tools. They help students grasp the concept of proportions and provide step-by-step solutions to practice problems.
Real-World Applications
Proportion calculators have a multitude of real-world applications. Let’s explore a few scenarios where these tools prove invaluable:
1. Recipe Scaling
When you want to adjust a recipe to cater to a larger or smaller group, a proportion calculator can help you determine the correct ingredient quantities, ensuring your dish retains its flavor and consistency.
2. Medication Dosage
In healthcare, calculating medication dosages based on a patient’s weight or age is critical. Proportion calculators ensure that healthcare professionals administer the correct amount of medication.
3. Engineering and Construction
Engineers and builders rely on proportions to design structures and ensure their stability. Proportion calculators aid in precise measurements and material calculations.
Conclusion
Proportions are a fundamental concept in mathematics and science, and they find application in various aspects of our daily lives. Proportion calculators have emerged as powerful tools to simplify these calculations, offering speed, accuracy, and versatility. Whether you’re a student, scientist, chef, or professional in any field, understanding how to use a proportion calculator can enhance your problem-solving abilities and streamline your work. Embrace the convenience of these tools and unlock the power of proportions in your endeavors.
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Review question
# For which $x$-values does a function equal its inverse? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource
Ref: R6735
## Solution
A function $f$ is defined by $f:x\rightarrow \dfrac{1}{x+1}$. Write down in similar form expressions for $f^{-1}$
To find the inverse of a function we can write $y=f(x)$ and rearrange to make $x$ the subject. This gives
\begin{align*} y &= \dfrac{1}{x+1}\\ x+1 &= \dfrac{1}{y}\\ x &= \dfrac{1}{y} - 1\\ &= \dfrac{1-y}{y} \end{align*}
Replacing $y$ with $x$ in the right hand side, we obtain the inverse function $f^{-1}:x\rightarrow\frac{1-x}{x}.$
… and $ff$.
The composite function of $f$ with itself can be found by ‘replacing the $x$ in $f(x)$ with $f(x)$’. This gives $f\bigl( f(x) \bigr) = \dfrac{1}{ \left( \dfrac{1}{x+1} \right)+1 }.$
We can now simplify by multiplying top and bottom by $x+1$, yielding
\begin{align*} ff(x) &= \dfrac{1}{ \left( \dfrac{1}{x+1} \right)+1 } \times \dfrac{x+1}{x+1} \\ &= \dfrac{x+1}{1 + (x + 1)} \\ &= \dfrac{x+1}{x+2}. \end{align*}
Our solution then reads $ff:x\rightarrow \frac{x+1}{x+2}.$
It is required to find the values of $x$ for which $(i)$ $f=f^{-1}$, $(ii)$ $f=ff$.
Show that, in each case, the values of $x$ are given by the equation $x^2+x-1=0.$
Using the result from above, we set $f(x) = f^{-1}(x)$, giving
\begin{align*} \dfrac{1}{x+1}&=\dfrac{1-x}{x}\\ 1&=\dfrac{(1-x)(1+x)}{x}\\ x&=1-x^2\\ x^2+x-1&=0 \end{align*}
as required.
Similarly, setting $f(x) = ff(x)$ we have
\begin{align*} \dfrac{1}{x+1} &= \dfrac{x+1}{x+2}\\ 1&=\dfrac{(x+1)^2}{x+2}\\ x+2 &= (x+1)^2\\ x+2 &=x^2+2x+1\\ x^2+x-1&=0. \end{align*}
The question does not require us to solve this equation for $x$, but we could do so using the quadratic formula. We have
\begin{align*} x^2+x-1&=0\\ x &= \dfrac{-1 \pm \sqrt{1^2 - 4 \times 1 \times -1}}{2 \times 1} \\ &= \dfrac{-1 \pm \sqrt{5}}{2} . \end{align*}
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# Question #881a0
Feb 24, 2017
$h = - \frac{4}{7}$
#### Explanation:
To eliminate the fractions in the equation, multiply ALL terms on both sides by the $\textcolor{b l u e}{\text{lowest common multiple}}$ (LCM) of the denominators 4, 2 and 8
the LCM of 4, 2 and 8 is 8
$\Rightarrow \left({\cancel{8}}^{2} \times \frac{5 h}{\cancel{4}} ^ 1\right) + \left({\cancel{8}}^{4} \times \frac{1}{\cancel{2}} ^ 1\right) = \left({\cancel{8}}^{1} \times \frac{3 h}{\cancel{8}} ^ 1\right)$
$\Rightarrow 10 h + 4 = 3 h \leftarrow \textcolor{red}{\text{no fractions}}$
subtract 3h from both sides.
$10 h - 3 h + 4 = \cancel{3 h} \cancel{- 3 h}$
$\Rightarrow 7 h + 4 = 0$
subtract 4 from both sides.
$7 h \cancel{+ 4} \cancel{- 4} = 0 - 4$
$\Rightarrow 7 h = - 4$
divide both sides by 7
$\frac{\cancel{7} h}{\cancel{7}} = \frac{- 4}{7}$
$\Rightarrow h = - \frac{4}{7} \text{ is the solution}$
|
# 7 Is 20 Percent of What Number
7 is 20 Percent of What Number?
When faced with a percentage problem such as “7 is 20 percent of what number,” it can be challenging to determine the unknown value. However, with a basic understanding of percentages and some simple calculations, you can easily find the solution. In this article, we will explore the concept of percentages, provide step-by-step instructions on how to solve this problem, and answer some frequently asked questions to clarify any doubts you may have.
Understanding Percentages:
Before diving into the problem at hand, it’s crucial to understand what percentages represent. A percentage is a way of expressing a portion or a fraction of a whole value. It is denoted by the symbol “%.” For instance, if you have 50% of a pizza, it means you have half of the pizza, which is equivalent to 0.5 in decimal form.
Solving the Problem:
To find the number when given a percentage, you can set up a proportion. In this case, we are given that 7 is 20 percent of an unknown number. Let’s call this number “x.” The proportion can be set up as follows:
7 / x = 20 / 100
To solve this proportion, we need to cross multiply:
7 * 100 = 20 * x
Simplifying further:
700 = 20x
Now, isolate the variable by dividing both sides of the equation by 20:
700 / 20 = x
x = 35
Therefore, 7 is 20 percent of 35.
FAQs:
Q1: How did you set up the proportion?
A1: When given a percentage, you can set up a proportion by equating the given percentage to the corresponding fraction. In this case, 20 percent is equivalent to 20/100 or 0.2 in decimal form. By setting up the proportion, we can solve for the unknown number.
Q2: Can I use other methods to solve this problem?
A2: Yes, there are alternative methods to find the unknown number. One such method is multiplying the known number by the reciprocal of the percentage. In this case, you would multiply 7 by 100/20, which also yields the answer of 35.
Q3: What if the percentage is a fraction rather than a whole number?
A3: Percentages can be expressed as fractions as well. For example, if the given percentage is 1/4, you can convert it to a decimal by dividing the numerator (1) by the denominator (4) to get 0.25. Then, follow the same steps outlined above to solve the problem.
Q4: Can I solve this problem using a calculator?
A4: While a calculator can certainly help with the calculations, it’s essential to understand the underlying concepts and steps involved in finding the unknown number. Relying solely on a calculator may hinder your ability to grasp the fundamentals of percentages.
Q5: How can I check if my answer is correct?
A5: To verify your answer, you can calculate what percentage the known number is of the unknown number. In this case, divide 7 by 35 and multiply by 100. The result should be 20%, confirming that your answer is correct.
In conclusion, finding the unknown number when given a percentage can be accomplished by setting up a proportion and solving for the variable. By understanding the fundamentals of percentages and following the step-by-step instructions provided, you can easily determine that 7 is 20 percent of 35. Remember to practice these concepts regularly to strengthen your skills in percentage calculations.
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# Equations and Formulas
## What is an Equation?
An equation says that two things are equal. It will have an equals sign "=" like this:
x + 2 = 6
That equations says: what is on the left (x + 2) is equal to what is on the right (6)
So an equation is like a statement "this equals that"
## What is a Formula?
A formula is a special type of equation that shows the relationship between different variables.
(A variable is a symbol like x or V that stands in for a number we don't know yet).
### Example: The formula for finding the volume of a box is:
V = lwh
V stands for volume, l for length, w for width, and h for height.
When l=10, w=5, and h=4, then V = 10 × 5 × 4 = 200
A formula will have more than one variable.
These are all equations, but only some are formulas:
x = 2y - 7 Formula (relating x and y) a2 + b2 = c2 Formula (relating a, b and c) x/2 + 7 = 0 Not a Formula (just an equation)
## Without the Equals
Sometimes a formula is written without the "=":
Example: The formula for the volume of a box is:
lwh
But in a way the "=" is still there, because you could write V = lwh if you wanted to.
## Subject of a Formula
The "subject" of a formula is the single variable (usually on the left of the "=") that everything else is equal to.
Example: in the formula
s = ut + ½ at2
"s" is the subject of the formula
## Changing the Subject
One of the very powerful things that Algebra can do is to "rearrange" a formula so that another variable is the subject.
Rearrange the volume of a box formula (V = lwh) so that the width is the subject:
Start with: V = lwh divide both sides by h: V/h = lw divide both sides by l: V/(hl) = w swap sides: w = V/(hl)
So now if you have a box with a length of 2m, a height of 2m and a volume of 12m3, you can calculate its width:
w = V/(hl)
w = 12m3 / (2m × 2m) = 12/4 = 3m
|
Home > standard error > regression prediction standard error
# Regression Prediction Standard Error
the estimate from a scatter plot Compute the standard error of the estimate based on errors of prediction Compute the standard error using Pearson's correlation Estimate the standard error of the estimate based on a sample Figure standard error of prediction formula 1 shows two regression examples. You can see that in Graph A, the
## Standard Error Of Prediction Linear Regression
points are closer to the line than they are in Graph B. Therefore, the predictions in Graph A are more accurate standard error of prediction in r than in Graph B. Figure 1. Regressions differing in accuracy of prediction. The standard error of the estimate is a measure of the accuracy of predictions. Recall that the regression line is the line
## Standard Error Of Prediction Excel
that minimizes the sum of squared deviations of prediction (also called the sum of squares error). The standard error of the estimate is closely related to this quantity and is defined below: where σest is the standard error of the estimate, Y is an actual score, Y' is a predicted score, and N is the number of pairs of scores. The numerator is the sum of squared differences between standard error of prediction definition the actual scores and the predicted scores. Note the similarity of the formula for σest to the formula for σ.  It turns out that σest is the standard deviation of the errors of prediction (each Y - Y' is an error of prediction). Assume the data in Table 1 are the data from a population of five X, Y pairs. Table 1. Example data. X Y Y' Y-Y' (Y-Y')2 1.00 1.00 1.210 -0.210 0.044 2.00 2.00 1.635 0.365 0.133 3.00 1.30 2.060 -0.760 0.578 4.00 3.75 2.485 1.265 1.600 5.00 2.25 2.910 -0.660 0.436 Sum 15.00 10.30 10.30 0.000 2.791 The last column shows that the sum of the squared errors of prediction is 2.791. Therefore, the standard error of the estimate is There is a version of the formula for the standard error in terms of Pearson's correlation: where ρ is the population value of Pearson's correlation and SSY is For the data in Table 1, μy = 2.06, SSY = 4.597 and ρ= 0.6268. Therefore, which is the same value computed previously. Similar formulas are used when the standard error of the estimate is computed from a sample rather than a population. The only difference
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## Standard Error Of Prediction Interval
about hiring developers or posting ads with us Cross Validated Questions Tags Users Badges Unanswered Ask Question _ Cross prediction error formula statistics Validated is a question and answer site for people interested in statistics, machine learning, data analysis, data mining, and data visualization. Join them; it only takes a minute: Sign up Here's how http://onlinestatbook.com/lms/regression/accuracy.html it works: Anybody can ask a question Anybody can answer The best answers are voted up and rise to the top How are the standard errors computed for the fitted values from a logistic regression? up vote 17 down vote favorite 16 When you predict a fitted value from a logistic regression model, how are standard errors computed? I mean for the fitted values, not http://stats.stackexchange.com/questions/66946/how-are-the-standard-errors-computed-for-the-fitted-values-from-a-logistic-regre for the coefficients (which involves Fishers information matrix). I only found out how to get the numbers with R (e.g., here on r-help, or here on Stack Overflow), but I cannot find the formula. pred <- predict(y.glm, newdata= something, se.fit=TRUE) If you could provide online source (preferably on a university website), that would be fantastic. r regression logistic mathematical-statistics references share|improve this question edited Aug 9 '13 at 15:14 gung 74.5k19162311 asked Aug 9 '13 at 14:41 user2457873 8814 add a comment| 1 Answer 1 active oldest votes up vote 19 down vote accepted The prediction is just a linear combination of the estimated coefficients. The coefficients are asymptotically normal so a linear combination of those coefficients will be asymptotically normal as well. So if we can obtain the covariance matrix for the parameter estimates we can obtain the standard error for a linear combination of those estimates easily. If I denote the covariance matrix as $\Sigma$ and and write the coefficients for my linear combination in a vector as $C$ then the standard error is just $\sqrt{C' \Sigma C}$ # Making fake data and fitting the model and getting a prediction set.seed(
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Regression Standard Error Interpretation p Ana-Maria imundi Editor-in-ChiefDepartment of Medical Laboratory DiagnosticsUniversity Hospital Sveti Duh Sveti Duh Zagreb CroatiaPhone - e-mail address editorial office at biochemia-medica dot com Useful standard error of estimate interpretation links Events Follow us on Facebook Home Standard error meaning and p Standard Error Of Regression Formula p interpretation Lessons in biostatistics Mary L McHugh Standard error meaning and interpretation Biochemia Medica - http dx doi org BM standard error of regression coefficient School of Nursing University of Indianapolis Indianapolis Indiana USA Corresponding author Mary dot McHugh at uchsc dot edu Abstract Standard error statistics are
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Regression Coefficients Standard Error p The standard error of the coefficient is always positive Use the standard error of the coefficient to measure the precision of the estimate of the coefficient The smaller standard error of coefficient in linear regression the standard error the more precise the estimate Dividing the coefficient by standard error of coefficient multiple regression its standard error calculates a t-value If the p-value associated with this t-statistic is less than your standard error of regression coefficient excel alpha level you conclude that the coefficient is significantly different from zero For example a materials engineer at a
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Regression Analysis Error p model Generalized linear model Discrete choice Logistic regression Multinomial logit Mixed logit Probit Multinomial probit Ordered logit Ordered probit Poisson Multilevel model Fixed effects Random effects Mixed model Nonlinear regression standard error of regression Nonparametric Semiparametric Robust Quantile Isotonic Principal components Least angle Local Segmented Errors-in-variables standard error of regression coefficient Estimation Least squares Ordinary least squares Linear math Partial Total Generalized Weighted Non-linear Non-negative Iteratively reweighted Ridge regression Least p Standard Error Of Estimate p absolute deviations Bayesian Bayesian multivariate Background Regression model validation Mean and predicted response Errors and residuals Goodness of fit Studentized
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Relation Between Standard Error And Variance p average squared deviation from the mean Variance in a population is x is a value from the population is standard error formula the mean of all x n is the number of x in p Standard Error Excel p the population is the summation Variance is usually estimated from a sample drawn from a population p Standard Error Regression p The unbiased estimate of population variance calculated from a sample is xi is the ith observation from a sample of the population x-bar is the sample mean n sample size - p Standard
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Regression Standard Error Deviation p it comes to determining how well a linear model fits the data However I've stated previously that R-squared is overrated Is there a p Standard Error Of Regression Formula p different goodness-of-fit statistic that can be more helpful You bet Today standard error of regression coefficient I rsquo ll highlight a sorely underappreciated regression statistic S or the standard error of the regression S provides important p Standard Error Of Estimate Interpretation p information that R-squared does not What is the Standard Error of the Regression S S becomes smaller when the data points are
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Regression Standard Error Of Estimate Formula p the estimate from a scatter plot Compute the standard error of the estimate based on errors of prediction Compute the standard error using Pearson's correlation Estimate the standard error of the estimate based on a sample Figure shows two how to calculate standard error of regression coefficient regression examples You can see that in Graph A the points are closer to standard error of estimate interpretation the line than they are in Graph B Therefore the predictions in Graph A are more accurate than in Graph B Figure p Standard Error Of The
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Regression Error Standard Deviation p the estimate from a scatter plot Compute the standard error of the estimate based on errors of prediction Compute the standard error using Pearson's correlation Estimate the standard error of the estimate based on a sample Figure shows two regression examples standard error of regression formula You can see that in Graph A the points are closer to the line p Standard Error Of Regression Coefficient p than they are in Graph B Therefore the predictions in Graph A are more accurate than in Graph B Figure Regressions standard error of estimate interpretation differing in
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Regression Average Error p deviation MSD of an estimator of a procedure for estimating an unobserved quantity measures the average of the squares of the errors or deviations that is the difference between the estimator and what is estimated MSE is a risk root mean square error interpretation function corresponding to the expected value of the squared error loss or quadratic loss The standard error of regression formula difference occurs because of randomness or because the estimator doesn't account for information that could produce a more accurate estimate The MSE p Standard Error Of The Regression p is a measure
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Relationship Between Standard Error And Variance p average squared deviation from the mean Variance in a population is x is a value from the population is the standard error formula mean of all x n is the number of x in the population p Standard Error Excel p is the summation Variance is usually estimated from a sample drawn from a population The unbiased estimate standard error regression of population variance calculated from a sample is xi is the ith observation from a sample of the population x-bar is the sample mean n sample size - is degrees of p
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Regression Standard Error p the estimate from a scatter plot Compute the standard error of the estimate based on errors of prediction Compute the standard error using Pearson's correlation Estimate the standard error of the estimate based on a sample Figure shows two p Standard Error Of Coefficient p regression examples You can see that in Graph A the points are closer to the standard error of regression formula line than they are in Graph B Therefore the predictions in Graph A are more accurate than in Graph B Figure standard error of regression interpretation Regressions differing in accuracy of
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Regression Standard Error Values p The standard error of the coefficient is always positive Use the standard error of the coefficient to measure the precision of the estimate of the standard error of coefficient coefficient The smaller the standard error the more precise the estimate how to calculate standard error of regression Dividing the coefficient by its standard error calculates a t-value If the p-value associated with this t-statistic standard error of the regression is less than your alpha level you conclude that the coefficient is significantly different from zero For example a materials engineer at a furniture manufacturing site
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Relationship Between Standard Error And Confidence Interval p DisclaimerPublic Health TextbookResearch Methods a - Epidemiology b - Statistical Methods c - Health Care Evaluation and Health Needs Assessment d - Qualitative MethodsDisease p Standard Error And Confidence Limits Worked Example p Causation and Diagnostic a - Epidemiological Paradigms b - Epidemiology of Diseases of calculate confidence interval from standard error in r Public Health Significance c - Diagnosis and Screening d - Genetics e - Health and Social Behaviour f - Environment g - p Error Interval Maths p Communicable Disease h - Principles and Practice of Health Promotion i
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Regression Standard Error Estimation p the estimate from a scatter plot Compute the standard error of the estimate based on errors of prediction Compute the standard error using Pearson's correlation Estimate the standard error of the estimate based on a sample Figure p Standard Error Of Coefficient p shows two regression examples You can see that in Graph A the standard error of estimate interpretation points are closer to the line than they are in Graph B Therefore the predictions in Graph A are more accurate p Standard Error Of The Regression p than in Graph B Figure Regressions differing
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Relationship Between Standard Error Of Estimate And R Squared p it comes to determining how well a linear model fits the data However I've stated previously that R-squared is overrated Is there a p Standard Error Of Estimate Formula p different goodness-of-fit statistic that can be more helpful You bet Today standard error of the regression I rsquo ll highlight a sorely underappreciated regression statistic S or the standard error of the regression S provides important p Standard Error Of Estimate Interpretation p information that R-squared does not What is the Standard Error of the Regression S S becomes smaller
relationship between r squared and standard error
Relationship Between R Squared And Standard Error p using regression analysis ANOVA or design of experiments DOE you need to determine how well the model fits the data To help you out Minitab p Standard Error Of Regression Formula p statistical software presents a variety of goodness-of-fit statistics In this post we rsquo ll standard error of the regression explore the R-squared R statistic some of its limitations and uncover some surprises along the way For standard error of regression coefficient instance low R-squared values are not always bad and high R-squared values are not always good What Is Goodness-of-Fit
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Regression Standard Error Of The Estimate p the estimate from a scatter plot Compute the standard error of the estimate based on errors of prediction Compute the standard error using Pearson's correlation Estimate the standard error of the estimate based on a sample Figure shows two regression examples You standard error of estimate interpretation can see that in Graph A the points are closer to the line than they standard error of estimate calculator are in Graph B Therefore the predictions in Graph A are more accurate than in Graph B Figure Regressions differing in accuracy standard error of estimate
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Regression Beta Standard Error Formula p For linear regression on a single variable see simple linear regression For the computation of least squares curve fits see numerical methods for linear least squares standard error of beta coefficient Part of a series on Statistics Regression analysis Models Linear regression Simple p Standard Error Of Coefficient In Linear Regression p regression Ordinary least squares Polynomial regression General linear model Generalized linear model Discrete choice Logistic regression Multinomial p Standard Error Of Beta Linear Regression p logit Mixed logit Probit Multinomial probit Ordered logit Ordered probit Poisson Multilevel model Fixed effects Random effects
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relation of standard deviation to standard error
Relation Of Standard Deviation To Standard Error p transformation standard errors must be of means calculated from within an intervention group and not standard errors of the difference standard error to standard deviation calculator in means computed between intervention groups Confidence intervals for means calculate standard error from standard deviation in excel can also be used to calculate standard deviations Again the following applies to confidence intervals for mean values p Convert Standard Deviation To Standard Error In Excel p calculated within an intervention group and not for estimates of differences between interventions for these see Section Most confidence intervals
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Regression What Is Standard Error p it comes to determining how well a linear model fits the data However I've stated previously that R-squared is overrated Is there a different goodness-of-fit statistic standard error of regression formula that can be more helpful You bet Today I rsquo ll highlight a sorely p Standard Error Of Regression Coefficient p underappreciated regression statistic S or the standard error of the regression S provides important information that R-squared does not p Standard Error Of Estimate Interpretation p What is the Standard Error of the Regression S S becomes smaller when the data points
relationship between reliability and standard error of measurement
Relationship Between Reliability And Standard Error Of Measurement p than the score the student should actually have received true score The difference between the observed score and the true score is called the error score S standard error of measurement calculator true S observed S error In the examples to the right Student standard error of measurement and confidence interval A has an observed score of His true score is so the error score would be Student p Standard Error Of Measurement Interpretation p B has an observed score of His true score is so the error score would be
relationship between variance and standard error
Relationship Between Variance And Standard Error p average squared deviation from the mean Variance in a population is x is a value from the population is the p Standard Error Formula p mean of all x n is the number of x in the population standard error regression is the summation Variance is usually estimated from a sample drawn from a population The unbiased estimate standard error symbol of population variance calculated from a sample is xi is the ith observation from a sample of the population x-bar is the sample mean n sample size - is degrees of p
relationship between standard error estimate correlation
Relationship Between Standard Error Estimate Correlation p the estimate from a scatter plot Compute the standard error of the estimate based on errors of prediction Compute the standard error using Pearson's correlation Estimate the standard error of the estimate based on standard error of correlation coefficient formula a sample Figure shows two regression examples You can see that in types of correlation pdf Graph A the points are closer to the line than they are in Graph B Therefore the predictions in Graph weak correlation A are more accurate than in Graph B Figure Regressions differing in accuracy of prediction
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Regression Standard Error Of Mean p proportion of samples that would fall between and standard deviations above and below the actual value The standard error SE is the standard deviation of the sampling distribution of a statistic most commonly of the mean The term may also be used p Standard Error Of Regression Formula p to refer to an estimate of that standard deviation derived from a particular sample used to standard error of regression coefficient compute the estimate For example the sample mean is the usual estimator of a population mean However different samples drawn from that same population
relationship between standard error of the mean and sample size
Relationship Between Standard Error Of The Mean And Sample Size p describe the parameters of a population based on the description statistics of a set of elements drawn from the population In terms of the climate and attitudes survey that many of you were recently asked to complete we can describe the population sampling frame study population and sample What is the primary strength of using a the relationship between sample size and sampling error is quizlet probability sampling method First if many independent random samples are selected from a population the sample statistics purpose of sampling in research provided
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Relationship Between Sample Variance And Standard Error p average squared deviation from the mean Variance in a population is x is a value from the population is the p Standard Error Formula p mean of all x n is the number of x in the population standard error excel is the summation Variance is usually estimated from a sample drawn from a population The unbiased estimate p Standard Error Regression p of population variance calculated from a sample is xi is the ith observation from a sample of the population x-bar is the sample mean n sample size - is
relationship between standard deviation and standard error of mean
Relationship Between Standard Deviation And Standard Error Of Mean p Health Search databasePMCAll DatabasesAssemblyBioProjectBioSampleBioSystemsBooksClinVarCloneConserved DomainsdbGaPdbVarESTGeneGenomeGEO DataSetsGEO ProfilesGSSGTRHomoloGeneMedGenMeSHNCBI Web SiteNLM when to use standard deviation vs standard error CatalogNucleotideOMIMPMCPopSetProbeProteinProtein ClustersPubChem BioAssayPubChem CompoundPubChem SubstancePubMedPubMed HealthSNPSparcleSRAStructureTaxonomyToolKitToolKitAllToolKitBookToolKitBookghUniGeneSearch termSearch Advanced difference between standard error and standard deviation pdf Journal list Help Journal ListClin Orthop Relat Resv SepPMC Clin Orthop Relat difference between standard deviation and standard error formula Res Sep Published online May doi s - - - PMCID PMC In Brief Standard Deviation and Standard ErrorDavid J standard error of the mean excel Biau MD PhDDepartement de Biostatistique et Informatique Medicale H x f
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Relationship Between Standard Deviation Standard Error p transformation standard errors must be of means calculated from within an intervention group and not standard errors of the difference in means computed between intervention groups Confidence intervals calculate standard error from standard deviation in excel for means can also be used to calculate standard deviations Again the following p Convert Standard Deviation To Standard Error In Excel p applies to confidence intervals for mean values calculated within an intervention group and not for estimates of differences between interventions when to use standard deviation vs standard error for these see Section Most confidence
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Relative Standard Error Of The Mean p estimates for Labour Force data News Media Media Centre ABS in the Media ABS Statistics relative standard error excel Publications Release Dates Resources for Reporting Statistics Conferences Seminars and Events relative standard error proportion Contact Us Quick Links Consumer Price Index Unemployment Rate Retail Trade Figures Building Approvals Figures GDP standard error formula Trend Figures What is a Standard Error and Relative Standard Error Reliability of estimates for Labour Force data On this page Why do we have Standard Errors standard error vs standard deviation Importance of Standard Errors Standard Error versus Relative
relationship between standard deviation and standard error of the mean
Relationship Between Standard Deviation And Standard Error Of The Mean p Health Search databasePMCAll DatabasesAssemblyBioProjectBioSampleBioSystemsBooksClinVarCloneConserved DomainsdbGaPdbVarESTGeneGenomeGEO DataSetsGEO ProfilesGSSGTRHomoloGeneMedGenMeSHNCBI Web SiteNLM CatalogNucleotideOMIMPMCPopSetProbeProteinProtein ClustersPubChem BioAssayPubChem when to use standard deviation vs standard error CompoundPubChem SubstancePubMedPubMed HealthSNPSparcleSRAStructureTaxonomyToolKitToolKitAllToolKitBookToolKitBookghUniGeneSearch termSearch Advanced Journal list Help Journal standard error vs standard deviation example ListBMJv Oct PMC BMJ Oct doi bmj PMCID PMC Statistics standard error in excel NotesStandard deviations and standard errorsDouglas G Altman professor of statistics in medicine and J Martin Bland professor of health statistics Cancer Research p Standard Error In R p UK NHS Centre for Statistics in Medicine Wolfson College Oxford OX UD
relative standard deviation error bars
Relative Standard Deviation Error Bars p proportion of samples that would fall between and standard deviations above and below the actual value The standard error SE is the standard deviation of the sampling distribution of a p Standard Error Formula p statistic most commonly of the mean The term may also be used to refer standard error vs standard deviation to an estimate of that standard deviation derived from a particular sample used to compute the estimate For example the sample p Standard Error Calculator p mean is the usual estimator of a population mean However different samples drawn from
relationship between standard error and deviation
Relationship Between Standard Error And Deviation p Health Search databasePMCAll DatabasesAssemblyBioProjectBioSampleBioSystemsBooksClinVarCloneConserved DomainsdbGaPdbVarESTGeneGenomeGEO DataSetsGEO ProfilesGSSGTRHomoloGeneMedGenMeSHNCBI Web SiteNLM CatalogNucleotideOMIMPMCPopSetProbeProteinProtein ClustersPubChem BioAssayPubChem CompoundPubChem SubstancePubMedPubMed HealthSNPSparcleSRAStructureTaxonomyToolKitToolKitAllToolKitBookToolKitBookghUniGeneSearch termSearch Advanced Journal list convert standard error to standard deviation Help Journal ListClin Orthop Relat Resv SepPMC Clin Orthop p When To Use Standard Deviation Vs Standard Error p Relat Res Sep Published online May doi s - - - PMCID PMC In Brief calculate standard error from standard deviation in excel Standard Deviation and Standard ErrorDavid J Biau MD PhDDepartement de Biostatistique et Informatique Medicale H x f pital Saint-Louis avenue Claude Vellefaux Paris Cedex France p
relative standard error equation
Relative Standard Error Equation p estimates for Labour Force data News Media Media Centre ABS in the Media ABS Statistics Publications Release Dates Resources for Reporting Statistics Conferences Seminars and Events Contact Us Quick Links Consumer Price Index Unemployment p Relative Standard Error Excel p Rate Retail Trade Figures Building Approvals Figures GDP Trend Figures What is a Standard Error standard error formula and Relative Standard Error Reliability of estimates for Labour Force data On this page Why do we have Standard Errors Importance of Standard standard error vs standard deviation Errors Standard Error versus Relative Standard Error Example Further
relationship between confidence intervals and standard error
Relationship Between Confidence Intervals And Standard Error p DisclaimerPublic Health TextbookResearch Methods a - Epidemiology b - Statistical standard error and confidence limits worked example Methods c - Health Care Evaluation and Health Needs calculate confidence interval from standard error in r Assessment d - Qualitative MethodsDisease Causation and Diagnostic a - Epidemiological Paradigms b - Epidemiology of error intervals bitesize Diseases of Public Health Significance c - Diagnosis and Screening d - Genetics e - Health and Social Behaviour f - Environment g - Communicable Disease h p Error Intervals Maths p - Principles and Practice of Health Promotion
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Relative Standard Error Proportion p apply ABS Universities Australia agreement Published research About products and systems TableBuilder CURFs Census products Remote Access Data Laboratory RADL DataLab Log into your accounts TableBuilder - Population Census TableBuilder - Other data RADL Registration Centre MiCRO User guides and support relative standard error formula Census TableBuilder user guide tutorials TableBuilder user guide RADL user guide Responsible use of relative standard error excel microdata guide Microdata FAQs Terms and conditions Microdata contacts Sources of Variability There are two sources of uncertainty or variability associated with p Relative Standard Error Vs Relative Standard Deviation p survey
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## Get ready for Algebra 2
### Course: Get ready for Algebra 2>Unit 1
Lesson 8: Factoring quadratics with difference of squares
# Factoring quadratics: Difference of squares
Learn how to factor quadratics that have the "difference of squares" form. For example, write x²-16 as (x+4)(x-4).
Factoring a polynomial involves writing it as a product of two or more polynomials. It reverses the process of polynomial multiplication.
In this article, we'll learn how to use the difference of squares pattern to factor certain polynomials. If you don't know the difference of squares pattern, please check out our video before proceeding.
## Intro: Difference of squares pattern
Every polynomial that is a difference of squares can be factored by applying the following formula:
start color #11accd, a, end color #11accd, squared, minus, start color #1fab54, b, end color #1fab54, squared, equals, left parenthesis, start color #11accd, a, end color #11accd, plus, start color #1fab54, b, end color #1fab54, right parenthesis, left parenthesis, start color #11accd, a, end color #11accd, minus, start color #1fab54, b, end color #1fab54, right parenthesis
Note that a and b in the pattern can be any algebraic expression. For example, for a, equals, x and b, equals, 2, we get the following:
\begin{aligned}\blueD{x}^2-\greenD{2}^2=(\blueD x+\greenD 2)(\blueD x-\greenD 2)\end{aligned}
The polynomial x, squared, minus, 4 is now expressed in factored form, left parenthesis, x, plus, 2, right parenthesis, left parenthesis, x, minus, 2, right parenthesis. We can expand the right-hand side of this equation to justify the factorization:
\begin{aligned}(x+2)(x-2)&=x(x-2)+2(x-2)\\\\&=x^2-2x+2x-4\\ \\ &=x^2-4\end{aligned}
Now that we understand the pattern, let's use it to factor a few more polynomials.
## Example 1: Factoring $x^2-16$x, squared, minus, 16
Both x, squared and 16 are perfect squares, since x, squared, equals, left parenthesis, start color #11accd, x, end color #11accd, right parenthesis, squared and 16, equals, left parenthesis, start color #1fab54, 4, end color #1fab54, right parenthesis, squared. In other words:
x, squared, minus, 16, equals, left parenthesis, start color #11accd, x, end color #11accd, right parenthesis, squared, minus, left parenthesis, start color #1fab54, 4, end color #1fab54, right parenthesis, squared
Since the two squares are being subtracted, we can see that this polynomial represents a difference of squares. We can use the difference of squares pattern to factor this expression:
start color #11accd, a, end color #11accd, squared, minus, start color #1fab54, b, end color #1fab54, squared, equals, left parenthesis, start color #11accd, a, end color #11accd, plus, start color #1fab54, b, end color #1fab54, right parenthesis, left parenthesis, start color #11accd, a, end color #11accd, minus, start color #1fab54, b, end color #1fab54, right parenthesis
In our case, start color #11accd, a, end color #11accd, equals, start color #11accd, x, end color #11accd and start color #1fab54, b, end color #1fab54, equals, start color #1fab54, 4, end color #1fab54. Therefore, our polynomial factors as follows:
left parenthesis, start color #11accd, x, end color #11accd, right parenthesis, squared, minus, left parenthesis, start color #1fab54, 4, end color #1fab54, right parenthesis, squared, equals, left parenthesis, start color #11accd, x, end color #11accd, plus, start color #1fab54, 4, end color #1fab54, right parenthesis, left parenthesis, start color #11accd, x, end color #11accd, minus, start color #1fab54, 4, end color #1fab54, right parenthesis
We can check our work by ensuring the product of these two factors is x, squared, minus, 16.
1) Factor x, squared, minus, 25.
2) Factor x, squared, minus, 100.
### Reflection question
3) Can we use the difference of squares pattern to factor x, squared, plus, 25?
## Example 2: Factoring $4x^2-9$4, x, squared, minus, 9
The leading coefficient does not have to equal to 1 in order to use the difference of squares pattern. In fact, the difference of squares pattern can be used here!
This is because 4, x, squared and 9 are perfect squares, since 4, x, squared, equals, left parenthesis, start color #11accd, 2, x, end color #11accd, right parenthesis, squared and 9, equals, left parenthesis, start color #1fab54, 3, end color #1fab54, right parenthesis, squared. We can use this information to factor the polynomial using the difference of squares pattern:
\begin{aligned}4x^2-9 &=(\blueD {2x})^2-(\greenD{3})^2\\ \\ &=(\blueD {2x}+\greenD 3)(\blueD {2x}-\greenD 3) \end{aligned}
A quick multiplication check verifies our answer.
4) Factor 25, x, squared, minus, 4.
5) Factor 64, x, squared, minus, 81.
6) Factor 36, x, squared, minus, 1.
## Challenge problems
7*) Factor x, start superscript, 4, end superscript, minus, 9.
8*) Factor 4, x, squared, minus, 49, y, squared.
## Want to join the conversation?
• If the question is x^2 + 25 its not factorable? Why?
• The difference of squares: (a+b)(a-b). x^2 + 25 is not factorable since you're adding 25, not subtracting. A positive multiplied by a negative is always a negative. If you were to factor it, you would have to use imaginary numbers such as i5. The factors of 25 are 5 and 5 besides 1 and itself. Since the formula: (a-b)(a+b), it uses a positive and negative sign, making the last term always a negative.
• i really don't understand factoring AT ALL even after watching the videos so can someone help me because i really suck at math! thank you in advance!
• Are you talking about just this video (difference of squares) or about any factoring at all? If it is factoring in general, you have to go through a sequence of factoring to support your learning. So just reaching out says that you are not as bad as you think. If you say all factoring, then we will start with a=1 and go one step at a time.
• Does the order of the signs matter? I have gotten it right when I do (x+y)(x-y). If I did it (x-y)(x+y) would it be wrong?
• While it is the same answer, sometimes questions ask for a specific order.
• in the video how did he know what number to divide by ?
• I think he looked for their square roots. Like when he had 25 5*5 = 25 so that's how he figured it out. Unless that's not what you're talking about then I feel embarrassed
• So when you have 25x-4 you have to make the 4 into a 2
• Yes, because the square root of 4 is 2. You should also notice the 25x^2. The square root of 25x^2 is 5x.
• how do you explain difference of squares to someone in words
• factoris 27-3x to the power of 2
(1 vote)
• Same as your other question. Factor out the common factor as your first step. Then you will have a difference of 2 squares. Give it a try.
• Why can’t you factorise the difference of squares in the expression: (4a^2-9c^2)/ 12abc?
(1 vote)
• How would I solve for a question like xp^2-4x and give a number for x to make the binomial a difference of perfect squares?
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Jaden Smith Under The Microscope (02/14/2020)
How will Jaden Smith perform on 02/14/2020 and the days ahead? Let’s use astrology to perform a simple analysis. Note this is not scientifically verified – don’t get too worked up about the result. I will first find the destiny number for Jaden Smith, and then something similar to the life path number, which we will calculate for today (02/14/2020). By comparing the difference of these two numbers, we may have an indication of how well their day will go, at least according to some astrology people.
PATH NUMBER FOR 02/14/2020: We will take the month (02), the day (14) and the year (2020), 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 02 and add the digits together: 0 + 2 = 2 (super simple). Then do the day: from 14 we do 1 + 4 = 5. Now finally, the year of 2020: 2 + 0 + 2 + 0 = 4. Now we have our three numbers, which we can add together: 2 + 5 + 4 = 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 path number for 02/14/2020.
DESTINY NUMBER FOR Jaden Smith: The destiny number will calculate the sum of all the letters in a name. Each letter is assigned a number per the below chart:
So for Jaden Smith we have the letters J (1), a (1), d (4), e (5), n (5), S (1), m (4), i (9), t (2) and h (8). Adding all of that up (yes, this can get tiring) gives 40. This still isn’t a single-digit number, so we will add its digits together again: 4 + 0 = 4. Now we have a single-digit number: 4 is the destiny number for Jaden Smith.
CONCLUSION: The difference between the path number for today (2) and destiny number for Jaden Smith (4) is 2. That is lower than the average difference between path numbers and destiny numbers (2.667), indicating that THIS IS A GOOD RESULT. But don’t get too happy yet! As mentioned earlier, this is for entertainment purposes only. If you want really means something, check out your cosmic energy profile here. Go ahead and see what it says for you – you’ll be glad you did.
Abigale Lormen
Abigale is a Masters in Business Administration by education. After completing her post-graduation, Abigale jumped the journalism bandwagon as a freelance journalist. Soon after that she landed a job of reporter and has been climbing the news industry ladder ever since to reach the post of editor at Tallahasseescene.
Latest posts by Abigale Lormen (see all)
Abigale Lormen
Abigale is a Masters in Business Administration by education. After completing her post-graduation, Abigale jumped the journalism bandwagon as a freelance journalist. Soon after that she landed a job of reporter and has been climbing the news industry ladder ever since to reach the post of editor at Tallahasseescene.
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### Course: 6th grade (Illustrative Mathematics)>Unit 8
Lesson 13: Lesson 15: Quartiles and interquartile range
# Interquartile range review
## Interquartile Range (IQR)
Interquartile range is the amount of spread in the middle 50, percent of a dataset.
In other words, it is the distance between the first quartile left parenthesis, start text, Q, end text, start subscript, 1, end subscript, right parenthesis and the third quartile left parenthesis, start text, Q, end text, start subscript, 3, end subscript, right parenthesis.
start text, I, Q, R, end text, equals, start text, Q, end text, start subscript, 3, end subscript, minus, start text, Q, end text, start subscript, 1, end subscript
Here's how to find the IQR:
Step 1: Put the data in order from least to greatest.
Step 2: Find the median. If the number of data points is odd, the median is the middle data point. If the number of data points is even, the median is the average of the middle two data points.
Step 3: Find the first quartile left parenthesis, start text, Q, end text, start subscript, 1, end subscript, right parenthesis. The first quartile is the median of the data points to the left of the median in the ordered list.
Step 4: Find the third quartile left parenthesis, start text, Q, end text, start subscript, 3, end subscript, right parenthesis. The third quartile is the median of the data points to the right of the median in the ordered list.
Step 5: Calculate IQR by subtracting start text, Q, end text, start subscript, 3, end subscript, minus, start text, Q, end text, start subscript, 1, end subscript.
### Example
Essays in Ms. Fenchel's class are scored on a 6 point scale.
Find the IQR of these scores:
1, 3, 3, 3, 4, 4, 4, 6, 6
Step 1: The data is already in order.
Step 2: Find the median. There are 9 scores, so the median is the middle score.
1, 3, 3, 3, 4, 4, 4, 6, 6
The median is 4.
Step 3: Find start text, Q, end text, start subscript, 1, end subscript, which is the median of the data to the left of the median.
There is an even number of data points to the left of the median, so we need the average of the middle two data points.
1, 3, 3, 3
start text, Q, end text, start subscript, 1, end subscript, equals, start fraction, 3, plus, 3, divided by, 2, end fraction, equals, 3
The first quartile is 3.
Step 4: Find start text, Q, end text, start subscript, 3, end subscript, which is the median of the data to the right of the median.
There is an even number of data points to the right of the median, so we need the average of the middle two data points.
4, 4, 6, 6
start text, Q, end text, start subscript, 3, end subscript, equals, start fraction, 4, plus, 6, divided by, 2, end fraction, equals, 5
The third quartile is 5.
Step 5: Calculate the IQR.
\begin{aligned} \text{IQR} &= \text{Q}_3-\text{Q}_1 \\ \\ &= 5-3 \\ \\ &= 2 \end{aligned}
The IQR is 2 points.
### Practice problem
The following data points represent the number of classes that each teacher at Broxin High School teaches.
Sort the data from least to greatest.
Find the interquartile range (IQR) of the data set.
classes
Want to practice more problems like these? Check out this exercise on interquartile range (IQR).
## Want to join the conversation?
• When will I even need to use this in the real world?...
• To calculate the spread of the middle 50 percent of the ACT scores at a college that you are considering.
• what if there are two numbers in the middle?
• it's just the first number with a '.5' so its like half. Like if the two numbers was 13,14 its 13.5
Or what ever number in in the middle of them like 13,15 then its 14. :)
• wait what? when will we ever use this irl?
• never, so idk why we need to learn this :')
• how can you find the median
• The median is simply the middle number in the data set (if you have your data set ordered from least to greatest). This is easy if you have an odd number of data.
If your number set has an even amount of data, then there's no central number. You would then take the average (or mean) of the two middle numbers to obtain the median for the data set.
Someone else gave an example of 1,2,2,3,5. Since there are an odd number of data, the median would simply be the (third) middle number of '2'.
Had the data set looked like this (with an even number of data)-
1,2,2,3,5,9
...then you would take the middle two number, and find the average (mean) of them. In this case, 2 & 3, the median would be 2+3, divided by 2, which would be 2.5.
• Hey people. You may not be interested but I just took down a bunch of vapers in my school!
• You are my hero <3
(1 vote)
• this is stupid
• when and how would i use this in the real world
• What if there are a lot of numbers?
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Reader Alex Freuman passed this along — a simple method of establishing any row in Pascal’s triangle, attributed to Edric Cane. To establish, for example, the seventh row (after the initial solitary 1), create a row of fractions in which the numerators are 7, 6, 5, 4, 3, 2, 1 and the denominators are 1, 2, 3, 4, 5, 6, 7:
$\displaystyle \frac{7}{1} \times \frac{6}{2} \times \frac{5}{3} \times \frac{4}{4} \times \frac{3}{5} \times \frac{2}{6} \times \frac{1}{7}$
Now multiply these in sequence, cumulatively, to get the numbers for the seventh row of the triangle:
$\displaystyle 1 \quad 7 \quad 21 \quad 35 \quad 35 \quad 21 \quad 7 \quad 1$
These are the coefficients for
$\displaystyle \left ( a+b \right )^{7}=a^{7} + 7a^{6}b + 21a^{5}b^{2} + 35a^{4}b^{3} + 35a^{3}b^{4} + 21a^{2}b^{5} + 7ab^{6} + b^{7}$.
Cane writes, “It couldn’t be easier to remember or to implement.” Another example — row 10:
$\displaystyle \frac{10}{1} \times \frac{9}{2} \times \frac{8}{3} \times \frac{7}{4} \times \frac{6}{5} \times \frac{5}{6} \times \frac{4}{7} \times \frac{3}{8} \times \frac{2}{9} \times \frac{1}{10}$
$\displaystyle 1 \quad 10 \quad 45 \quad 120 \quad 210 \quad 252 \quad 210 \quad 120 \quad 45 \quad 10 \quad 1$
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# Rewriting Expressions using Math Laws
Previously, you learned about converting words to equations and expressions. You now know the difference between expression and equation.
In this lesson, you will learn about rewriting expressions, using mathematical fundamental laws.
If two expressions are written differently, are they equal ?
That is the kind of question we are trying to answer here. I assume that you know basic addition, and multiplication.
## Mathematical Laws
To ensure that expression written in different forms are equal, we will test them against mathematical laws. These laws applies to expressions involving addition and multiplication. These are:
• Commutative law
• Associative law
• Distributive law
## Commutative Law
Let me discuss the Commutative laws. Let a, b and c be three whole numbers, and is the expression.
The commutative laws states that the order of addition does not affect the result. It means that is equivalent to .
You add two number in any order, the result is going to be same. This is resulting from commutative law.
### Example
Our first example is a simple equation which is same as . Both results in .
4 + 3 = 3 + 4 = 7
Commutative law for Multiplication
If we multiply two number in any order , the result will be same which means is equivalent to . So, commutative laws applies to multiplication as well.
Let’s consider another example,
Suppose 5 and 3 are two whole numbers, if we multiply 5 with 3, the answer is 15. Let us rearrange the expression into 3 multiplied by 5 which is also 15. therefore, commutative law is true for multiplication as well.
5 \times 3 = 3 \times 5 = 15
## Associative Laws
The associative laws works on addition and multiplication. If three or more numbers are added or multiplied, different groupings of numbers does not affect the results.
If a, b and c are three numbers. Then is same as .
### Example:
let , and , then
\begin{aligned}
&(a + b ) + c = (3 + 5) + 7 \\\\
& Evaluate \hspace{3px} first \hspace{3px} 3 + 5 = 8\\\\
&8 + 7 = 15
\end{aligned}
Now we must evaluate
\begin{aligned}
&a + (b + c) = 3 + (5 + 7) \\\\
&Evaluate \hspace{3px} first \hspace{3px}5 + 7 = 12\\\\
&3 + 12 = 15
\end{aligned}
Our expression becomes which is equal to . so, by associative law addition of three numbers with different grouping is equal and does not affect the results.
Associative Law for Multiplication
The different grouping does not affect multiplication as well. You can multiply variable by grouping them in any order and it still gives same results.
Let a, b and c be three numbers multiplied together. (a multiplied by b) multiplied by c is equal to d. similarly, a multiplied by (b multiplied by c) is also equal to d, where d is a whole number.
(a \times b ) \times c = (b \times a ) \times c = d
Let us see an example of associativity of multiplication.
### Example:
(2 \times 6) \times 5 = 2 \times (6 \times 5)
In the left hand expression, we must evaluate first which is and then multiply with which gives us .
\begin{aligned}
&(2 \times 6) \times 5\\\\
&=12 \times 5 \\\\
&=60
\end{aligned}
Similarly, in the right hand side expression, we must evaluate first which is and then multiplied by will give us .
\begin{aligned}
&2 \times (6\times 5)\\\\
&=2 \times 30 \\\\
&=60
\end{aligned}
Therefore, the equation is true and associative law is true. You can rewrite any expression using the commutative and associative laws.
## Identity Law
Next I will talk about another law called the identity law. This mathematical law applies to addition and multiplication only. The identity law requires identity element in the expression. The identity element for addition is 0 and the identity element for multiplication is 1.
If a is whole number, then according to identity law, where 0 is the identity element gives the same number a as result.
By commutative law, is also equal to a.
a + 0 = 0 + a = a
I illustrated this with an example below.
### Example:
Let us see an example, suppose is a whole number , then . Also, .
Identity Law for Multiplication
For multiplication, let a be any whole number multiplied by its identity element 1, then
a \times 1 = a
we get number a itself. By commutative law, is also, the number a..
### Summary
Let me summarize what you learned.
• Commutative law says, if a and b are two whole numbers then
\begin{aligned}
&a + b = b + a\\\\
&a \times b = b \times a
\end{aligned}
• The associative law says , if a, b, and c are 3 numbers,then
\begin{aligned}
&(a + b) + c = a + (b + c)\\\\
&( a \times b ) \times c = a \times ( b \times c )
\end{aligned}
• The identity element involve the identity element. The identity element for addition is 0 and 1 is identity element for multiplication. Then,
\begin{aligned}
&a + 0 = 0 + a = a\\\\
&a \times 1 = 1 \times a = a
\end{aligned}
Note: examples in this article uses only two or three variables, however, if you try any of these examples with more than three variables, the result would be true only. The commutative, associative and identity laws applied to expressions with more variables.But, for simplicity I have chosen fewer than two or three variables.
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0
795
The quadratic formula is the best possible way of ensuring that kids can evaluate the solution of quadratic equations very easily and this is the best possible replacement of the factorisation method. The quadratic equation will always be in the form of AX square + BX + C =0 where A, B and C are the real numbers and also known as the numerical coefficients.
The quadratic formula has been perfectly utilised in terms of finding out the roots of the quadratic equation and has been explained as follows:
• X = – B plus + under the root B square – 4AC/2A
The 2nd-degree polynomial will always have at most two zeros which is the main reason that quadratic equations will always have at most two roots. Following are some of the basic points to be noted by the people in this particular area:
The roots of the quadratic equation are the same as the zeros of the polynomial.
By splitting the middle term people can very easily factorise the quadratic polynomials without any kind of problem.
## What is the process to find out the roots of the quadratic equation?
Roots of the quadratic equation can be very easily found by factoring the polynomial and equating it with zero. This is considered to be the best possible way of finding out the roots of the quadratic equation and further being clear about every step is very much vital for the kids so that there is no error in the whole process. Practice is the key to success over here which is the main reason that kids need to be very much clear about practising different kinds of questions and working on different kinds of worksheets so that they can deal with things very easily and can have a good command over the chapter of quadratic equations without any kind of problem. Further being clear about the derivation of the quadratic formula is another very important aspect to be undertaken so that there is no issue in the whole process and everything has been carried out very easily. It can also be categorised depending upon the value of the determinant and has been explained as:
• If the determinant value is zero then two roots will be real and equal
• If the determinant value is greater than zero then the roots will be real and unequal
• If the determinant value is less than zero then roots are not real or they can be imaginary.
Being clear about every basic formula element is very much important in the whole process so that people can have a good command over the entire chapter and further be clear about different kinds of tips and tricks in the whole process is very much important so that people can solve the questions very efficiently.
It is very much vital on behalf of people to be clear about the algebraic method of solving the quadratic equations so that there is no problem on the behalf of people and everything has been carried out very easily. Apart from this, being clear about the graphical solution of the quadratic equation is very much important so that people are able to interpret things very easily and are further very much capable of dealing with the determination of the solutions without any kind of hassle.
Apart from all the above-mentioned points, it is vital on the behalf of people to visit the Cuemath website because they will be having proper access to the experts over here who will be teaching the students with a higher level of professionalism. The worksheets provided by the experts over here will always help in making sure that kids will be able to indulge in the right kind of practice and will be having a good command over the chapter of quadratic equations very easily.
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# How do you express tan theta - cot theta +sintheta in terms of cos theta ?
Jan 9, 2016
$= \frac{- {\cos}^{3} \left(\theta\right) - 2 {\cos}^{2} \left(\theta\right) + \cos \left(\theta\right) + 1}{\cos \left(\theta\right) \sqrt{1 - {\cos}^{2} \theta}}$
#### Explanation:
$\tan \left(\theta\right) - \cot \left(\theta\right) + \sin \left(\theta\right)$
We have to write in terms of $\cos \left(\theta\right)$
$\textcolor{b l u e}{\text{Let us start by using the identity}}$
$\tan \left(\theta\right) = \sin \frac{\theta}{\cos} \left(\theta\right)$ and $\cot \left(\theta\right) = \cos \frac{\theta}{\sin} \left(\theta\right)$
We get
$\tan \left(\theta\right) - \cot \left(\theta\right) + \sin \left(\theta\right)$
$= \sin \frac{\theta}{\cos} \left(\theta\right) - \cos \frac{\theta}{\sin} \left(\theta\right) + \sin \left(\theta\right)$
$\textcolor{b l u e}{\text{In order to simplify we need to use Least Common Denominator for all the fractions}}$
$= \frac{\sin \left(\theta\right) \sin \left(\theta\right)}{\cos \left(\theta\right) \sin \left(\theta\right)} - \frac{\cos \left(\theta\right) \cos \left(\theta\right)}{\cos \left(\theta\right) \sin \left(\theta\right)} + \frac{\sin \left(\theta\right) \cos \left(\theta\right) \sin \left(\theta\right)}{\cos \left(\theta\right) \sin \left(\theta\right)}$
$= \frac{{\sin}^{2} \left(\theta\right) - {\cos}^{2} \left(\theta\right) + {\sin}^{2} \left(\theta\right) \cos \left(\theta\right)}{\cos \left(\theta\right) \sin \left(\theta\right)}$
$= \frac{1 - {\cos}^{2} \left(\theta\right) - {\cos}^{2} \left(\theta\right) + \left(1 - {\cos}^{2} \left(\theta\right)\right) \cos \left(\theta\right)}{\cos \left(\theta\right) \sin \left(\theta\right)}$
$= \frac{\left(1 - 2 {\cos}^{2} \left(\theta\right)\right) + \cos \left(\theta\right) - {\cos}^{3} \left(\theta\right)}{\cos \left(\theta\right) \sin \left(\theta\right)}$
$= \frac{- {\cos}^{3} \left(\theta\right) - 2 {\cos}^{2} \left(\theta\right) + \cos \left(\theta\right) + 1}{\cos \left(\theta\right) \sin \left(\theta\right)}$
$= \frac{- {\cos}^{3} \left(\theta\right) - 2 {\cos}^{2} \left(\theta\right) + \cos \left(\theta\right) + 1}{\cos \left(\theta\right) \sqrt{1 - {\cos}^{2} \theta}}$
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#### Concept of Conditional Probability
Conditional Probability:
Remember that the probability of an event occurring given that the other event has already occurred is termed as a conditional probability.
The probability which event B takes place, given that event A has already occurred is:
P(B|A) = P(A and B)/P(A)
The above formula is derived from the general multiplication principle and a little bit of algebra.
As we are given that the event A has occurred, we have a decreased sample space. Rather than the whole sample space S, we now contain a sample space of A as we know A has occurred. Therefore the old rule about being the number in the event divided by the number in sample space still applies. This is the number in A and B (should be in A as A has occurred) divided by the number in A. When divided numerator and denominator of the right hand side by the number in sample space S, then we have the probability of A and B divided by the probability of A.
Example: The question, ‘Do you smoke?’ was asked of 100 people. The outcomes are shown in the table below.
Yes No Total
Male 19 41 60
Female 12 28 40
Total 31 69 100
1) Find out the probability of a randomly chosen individual being a male who smokes?
This is simply a joint probability. The number of Male and Smoke divided by the total = 19/100 = 0.19
2) Find out the probability of a randomly chosen individual being a male?
This is a total for male divided by the total = 60/100 = 0.60. As no mention is made up of smoking or not smoking, it comprises all the cases.
3) Find the probability of a randomly chosen individual smoking?
Again, as no mention is made of gender, this is the marginal probability, the total who smoke divided by the total = 31/100 = 0.31.
4) Find out the probability of a randomly chosen male smoking?
This time, you are told that you encompass a male - think of stratified sampling. Determine the probability that the male smokes? Well, 19 males smoke out of 60 males, therefore 19/60 = 0.31666...
5) Find out the probability that a randomly chosen smoker is male?
This time, you are told that you encompass a smoker and asked to determine the probability that the smoker is as well male. There are 19 male smokers out of 31 total smokers, therefore 19/31 = 0.6129
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2Rings
IB Groups, Rings and Modules
2.4 Factorization in integral domains
We now move on to tackle the problem of factorization in rings. For sanity,
we suppose throughout the section that
R
is an integral domain. We start by
Definition
(Unit)
.
An element
a R
is a unit if there is an
b R
such that
ab = 1
R
. Equivalently, if the ideal (a) = R.
Definition
(Division)
.
For elements
a, b R
, we say
a
divides
b
, written
a | b
,
if there is a c R such that b = ac. Equivalently, if (b) (a).
Definition
(Associates)
.
We say
a, b R
are associates if
a
=
bc
for some unit
c. Equivalently, if (a) = (b). Equivalently, if a | b and b | a.
In integers, this can only happen if
a
and
b
differ by a sign, but in more
interesting rings, more interesting things can happen.
When considering division in rings, we often consider two associates to be
“the same”. For example, in Z, we can factorize 6 as
6 = 2 · 3 = (2) · (3),
but this does not violate unique factorization, since 2 and
2 are associates (and
so are 3 and 3), and we consider these two factorizations to be “the same”.
Definition
(Irreducible)
.
We say
a R
is irreducible if
a 6
= 0,
a
is not a unit,
and if a = xy, then x or y is a unit.
For integers, being irreducible is the same as being a prime number. However,
“prime” means something different in general rings.
Definition
(Prime)
.
We say
a R
is prime if
a
is non-zero, not a unit, and
whenever a | xy, either a | x or a | y.
It is important to note all these properties depend on the ring, not the
element itself.
Example. 2 Z is a prime, but 2 Q is not (since it is a unit).
Similarly, the polynomial 2
X Q
[
X
] is irreducible (since 2 is a unit), but
2X Z[X] not irreducible.
We have two things called prime, so they had better be related.
Lemma.
A principal ideal (
r
) is a prime ideal in
R
if and only if
r
= 0 or
r
is
prime.
Proof.
(
) Let (
r
) be a prime ideal. If
r
= 0, then done. Otherwise, as prime
ideals are proper, i.e. not the whole ring,
r
is not a unit. Now suppose
r | a · b
.
Then
a ·b
(
r
). But (
r
) is prime. So
a
(
r
) or
b
(
r
). So
r | a
or
r | b
. So
r
is
prime.
(
) If
r
= 0, then (0) =
{
0
} C R
, which is prime since
R
is an integral
domain. Otherwise, let
r 6
= 0 be prime. Suppose
a · b
(
r
). This means
r | a · b
.
So r | a or r | b. So a (r) and b (r). So (r) is prime.
Note that in
Z
, prime numbers exactly the irreducibles, but prime numbers
are also prime (surprise!). In general, it is not true that irreducibles are the
same as primes. However, one direction is always true.
Lemma. Let r R be prime. Then it is irreducible.
Proof.
Let
r R
be prime, and suppose
r
=
ab
. Since
r | r
=
ab
, and
r
is
prime, we must have
r | a
or
r | b
. wlog,
r | a
. So
a
=
rc
for some
c R
. So
r
=
ab
=
rcb
. Since we are in an integral domain, we must have 1 =
cb
. So
b
is
a unit.
We now do a long interesting example.
Example. Let
R = Z[
5] = {a + b
5 : a, b Z} C.
By definition, it is a subring of a field. So it is an integral domain. What are
the units of the ring? There is a nice trick we can use, when things are lying
inside C. Consider the function
N : R Z
0
given by
N(a + b
5) 7→ a
2
+ 5b
2
.
It is convenient to think of this as
z 7→ z¯z
=
|z|
2
. This satisfies
N
(
z · w
) =
N
(
z
)
N
(
w
). This is a desirable thing to have for a ring, since it immediately
implies all units have norm 1 if
r ·s
= 1, then 1 =
N
(1) =
N
(
rs
) =
N
(
r
)
N
(
s
).
So N (r) = N(s) = 1.
So to find the units, we need to solve
a
2
+ 5
b
2
= 1, for
a
and
b
units. The
only solutions are
±
1. So only
±
1
R
can be units, and these obviously are
units. So these are all the units.
Next, we claim 2
R
is irreducible. We again use the norm. Suppose 2 =
ab
.
Then 4 =
N
(2) =
N
(
a
)
N
(
b
). Now note that nothing has norm 2.
a
2
+ 5
b
2
can
never be 2 for integers
a, b Z
. So we must have, wlog,
N
(
a
) = 4
, N
(
b
) = 1.
So
b
must be a unit. Similarly, we see that 3
,
1 +
5,
1
5
are irreducible
(since there is also no element of norm 3).
We have four irreducible elements in this ring. Are they prime? No! Note
that
(1 +
5)(1
5) = 6 = 2 · 3.
We now claim 2 does not divide 1 +
5 or 1
5. So 2 is not prime.
To show this, suppose 2
|
1 +
5
. Then
N
(2)
| N
(1 +
5
). But
N
(2) = 4
and
N
(1 +
5
) = 6, and 4
-
6. Similarly,
N
(1
5
) = 6 as well. So
2 - 1 ±
5.
There are several life lessons here. First is that primes and irreducibles are
not the same thing in general. We’ve always thought they were the same because
we’ve been living in the fantasy land of the integers. But we need to grow up.
The second one is that factorization into irreducibles is not necessarily unique,
since 2 · 3 = (1 +
5)(1
5) are two factorizations into irreducibles.
However, there is one situation when unique factorizations holds. This is
when we have a Euclidean algorithm available.
Definition
(Euclidean domain)
.
An integral domain
R
is a Euclidean domain
(ED) if there is a Euclidean function φ : R \ {0} Z
0
such that
(i) φ(a · b) φ(b) for all a, b 6= 0
(ii) If a, b R, with b 6= 0, then there are q, r R such that
a = b · q + r,
and either r = 0 or φ(r) < φ(b).
What are examples? Every time in this course where we said “Euclidean
algorithm”, we have an example.
Example. Z is a Euclidean domain with φ(n) = |n|.
For any field F, F[X] is a Euclidean domain with
φ(f) = deg(f).
The Gaussian integers
R
=
Z
[
i
]
C
is a Euclidean domain with
φ
(
z
) =
N
(
z
) =
|z|
2
. We now check this:
(i) We have φ(zw) = φ(z)φ(w) φ(z), since φ(w) is a positive integer.
(ii) Given a, b Z[i], b 6= 0. We consider the complex number
a
b
C.
Consider the following complex plane, where the red dots are points in
Z[i].
Re
Im
a
b
By looking at the picture, we know that there is some
q Z
[
i
] such that
a
b
q
< 1. So we can write
a
b
= q + c
with |c| < 1. Then we have
a = b · q + b · c
|{z}
r
.
We know
r
=
a bq Z
[
i
], and
φ
(
r
) =
N
(
bc
) =
N
(
b
)
N
(
c
)
< N
(
b
) =
φ
(
b
).
So done.
This is not just true for the Gaussian integers. All we really needed was that
R C
, and for any
x C
, there is some point in
R
that is not more than 1 away
from x. If we draw some more pictures, we will see this is not true for Z[
5].
Before we move on to prove unique factorization, we first derive something
we’ve previously mentioned. Recall we showed that every ideal in
Z
is principal,
and we proved this by the Euclidean algorithm. So we might expect this to be
true in an arbitrary Euclidean domain.
Definition
(Principal ideal domain)
.
A ring
R
is a principal ideal domain (PID)
if it is an integral domain, and every ideal is a principal ideal, i.e for all
I C R
,
there is some a such that I = (a).
Example. Z is a principal ideal domain.
Proposition.
Let
R
be a Euclidean domain. Then
R
is a principal ideal domain.
We have already proved this, just that we did it for a particular Euclidean
domain Z. Nonetheless, we shall do it again.
Proof.
Let
R
have a Euclidean function
φ
:
R \ {
0
} Z
0
. We let
I C R
be a
non-zero ideal, and let
b I \ {
0
}
be an element with
φ
(
b
) minimal. Then for
any a I, we write
a = bq + r,
with
r
= 0 or
φ
(
r
)
< φ
(
b
). However, any such
r
must be in
I
since
r
=
a bq I
.
So we cannot have
φ
(
r
)
< φ
(
b
). So we must have
r
= 0. So
a
=
bq
. So
a
(
b
).
Since this is true for all
a I
, we must have
I
(
b
). On the other hand, since
b I, we must have (b) I. So we must have I = (b).
This is exactly, word by word, the same proof as we gave for the integers,
except we replaced the absolute value with φ.
Example. Z
is a Euclidean domain, and hence a principal ideal domain. Also,
for any field F, F[X] is a Euclidean domain, hence principal ideal domain.
Also, Z[i] is a Euclidean domain, and hence a principal ideal domain.
What is a non-example of principal ideal domains? In
Z
[
X
], the ideal
(2
, X
)
C Z
[
X
] is not a principal ideal. Suppose it were. Then (2
, X
) = (
f
). Since
2
(2
, X
) = (
f
), we know 2
(
f
) , i.e. 2 =
f · g
for some
g
. So
f
has degree
zero, and hence constant. So f = ±1 or ±2.
If
f
=
±
1, since
±
1 are units, then (
f
) =
Z
[
X
]. But (2
, X
)
6
=
Z
[
X
], since,
say, 1
6∈
(2
, X
). If
f
=
±
2, then since
X
(2
, X
) = (
f
), we must have
±
2
| X
,
but this is clearly false. So (2, X) cannot be a principal ideal.
Example.
Let
A M
n×n
(
F
) be an
n × n
matrix over a field
F
. We consider
the following set
I = {f F[X] : f(A) = 0}.
This is an ideal if
f, g I
, then (
f
+
g
)(
A
) =
f
(
A
) +
g
(
A
) = 0. Similarly, if
f I and h F[X], then (fg)(A) = f(A)g(A) = 0.
But we know
F
[
X
] is a principal ideal domain. So there must be some
m F[X] such that I = (m) for some m.
Suppose
f F
[
X
] such that
f
(
A
) = 0, i.e.
f I
. Then
m | f
. So
m
is
a polynomial that divides all polynomials that kill
A
, i.e.
m
is the minimal
polynomial of A.
We have just proved that all matrices have minimal polynomials, and that
the minimal polynomial divides all other polynomials that kill
A
. Also, the
minimal polynomial is unique up to multiplication of units.
Let’s get further into number theory-like things. For a general ring, we
cannot factorize things into irreducibles uniquely. However, in some rings, this
is possible.
Definition
(Unique factorization domain)
.
An integral domain
R
is a unique
factorization domain (UFD) if
(i) Every non-unit may be written as a product of irreducibles;
(ii)
If
p
1
p
2
···p
n
=
q
1
···q
m
with
p
i
, q
j
irreducibles, then
n
=
m
, and they can
be reordered such that p
i
is an associate of q
i
.
This is a really nice property, and here we can do things we are familiar with
in number theory. So how do we know if something is a unique factorization
domain?
Our goal is to show that all principal ideal domains are unique factorization
domains. To do so, we are going to prove several lemmas that give us some
really nice properties of principal ideal domains.
Recall we saw that every prime is an irreducible, but in
Z
[
5
], there are
some irreducibles that are not prime. However, this cannot happen in principal
ideal domains.
Lemma.
Let
R
be a principal ideal domain. If
p R
is irreducible, then it is
prime.
Note that this is also true for general unique factorization domains, which
we can prove directly by unique factorization.
Proof.
Let
p R
be irreducible, and suppose
p | a · b
. Also, suppose
p - a
. We
need to show p | b.
Consider the ideal (
p, a
)
C R
. Since
R
is a principal ideal domain, there is
some d R such that (p, a) = (d). So d | p and d | a.
Since
d | p
, there is some
q
1
such that
p
=
q
1
d
. As
p
is irreducible, either
q
1
or d is a unit.
If
q
1
is a unit, then
d
=
q
1
1
p
, and this divides
a
. So
a
=
q
1
1
px
for some
x
.
This is a contradiction, since p - a.
Therefore
d
is a unit. So (
p, a
) = (
d
) =
R
. In particular, 1
R
(
p, a
). So
suppose 1
R
=
rp
+
sa
, for some
r, s R
. We now take the whole thing and
multiply by b. Then we get
b = rpb + sab.
We observe that
ab
is divisible by
p
, and so is
p
. So
b
is divisible by
p
. So
done.
This is similar to the argument for integers. For integers, we would say if
p - a
,
then
p
and
a
are coprime. Therefore there are some
r, s
such that 1 =
rp
+
sa
.
Then we continue the proof as above. Hence what we did in the middle is to do
something similar to showing p and a are “coprime”.
Another nice property of principal ideal domains is the following:
Lemma.
Let
R
be a principal ideal domain. Let
I
1
I
2
I
3
···
be a chain
of ideals. Then there is some N N such that I
n
= I
n+1
for some n N .
So in a principal ideal domain, we cannot have an infinite chain of bigger
and bigger ideals.
Definition
(Ascending chain condition)
.
A ring satisfies the ascending chain
condition (ACC) if there is no infinite strictly increasing chain of ideals.
Definition
(Noetherian ring)
.
A ring that satisfies the ascending chain condition
is known as a Noetherian ring.
So we are proving that every principal ideal domain is Noetherian.
Proof.
The obvious thing to do when we have an infinite chain of ideals is to
take the union of them. We let
I =
[
n1
I
n
,
which is again an ideal. Since
R
is a principal ideal domain,
I
= (
a
) for some
a R. We know a I =
S
n=0
I
n
. So a I
N
for some N. Then we have
(a) I
N
I = (a)
So we must have I
N
= I. So I
n
= I
N
= I for all n N.
Notice it is not important that
I
is generated by one element. If, for some
reason, we know
I
is generated by finitely many elements, then the same argument
work. So if every ideal is finitely generated, then the ring must be Noetherian.
It turns out this is an if-and-only-if if you are Noetherian, then every ideal is
finitely generated. We will prove this later on in the course.
Finally, we have done the setup, and we can prove the proposition promised.
Proposition.
Let
R
be a principal ideal domain. Then
R
is a unique factoriza-
tion domain.
Proof. We first need to show any (non-unit) r R is a product of irreducibles.
Suppose
r R
cannot be factored as a product of irreducibles. Then it is
certainly not irreducible. So we can write
r
=
r
1
s
1
, with
r
1
, s
1
both non-units.
Since
r
cannot be factored as a product of irreducibles, wlog
r
1
cannot be
factored as a product of irreducibles (if both can, then
r
would be a product of
irreducibles). So we can write
r
1
=
r
2
s
2
, with
r
2
, s
2
not units. Again, wlog
r
2
cannot be factored as a product of irreducibles. We continue this way.
By assumption, the process does not end, and then we have the following
chain of ideals:
(r) (r
1
) (r
2
) ··· (r
n
) ···
But then we have an ascending chain of ideals. By the ascending chain condition,
these are all eventually equal, i.e. there is some
n
such that (
r
n
) = (
r
n+1
) =
(
r
n+2
) =
···
. In particular, since (
r
n
) = (
r
n+1
), and
r
n
=
r
n+1
s
n+1
, then
s
n+1
is a unit. But this is a contradiction, since
s
n+1
is not a unit. So
r
must be a
product of irreducibles.
To show uniqueness, we let
p
1
p
2
···p
n
=
q
1
q
2
···q
m
, with
p
i
, q
i
irreducible.
So in particular
p
1
| q
1
···q
m
. Since
p
1
is irreducible, it is prime. So
p
1
divides
some
q
i
. We reorder and suppose
p
1
| q
1
. So
q
1
=
p
1
·a
for some
a
. But since
q
1
is irreducible,
a
must be a unit. So
p
1
, q
1
are associates. Since
R
is a principal
ideal domain, hence integral domain, we can cancel p
1
to obtain
p
2
p
3
···p
n
= (aq
2
)q
3
···q
m
.
We now rename aq
2
as q
2
, so that we in fact have
p
2
p
3
···p
n
= q
2
q
3
···q
m
.
We can then continue to show that
p
i
and
q
i
are associates for all
i
. This also
shows that
n
=
m
, or else if
n
=
m
+
k
, saw, then
p
k+1
···p
n
= 1, which is a
We can now use this to define other familiar notions from number theory.
Definition
(Greatest common divisor)
. d
is a greatest common divisor (gcd) of
a
1
, a
2
, ··· , a
n
if
d | a
i
for all
i
, and if any other
d
0
satisfies
d
0
| a
i
for all
i
, then
d
0
| d.
Note that the gcd of a set of numbers, if exists, is not unique. It is only
well-defined up to a unit.
This is a definition that says what it means to be a greatest common divisor.
However, it does not always have to exist.
Lemma.
Let
R
be a unique factorization domain. Then greatest common
divisors exists, and is unique up to associates.
Proof.
We construct the greatest common divisor using the good-old way of
prime factorization.
We let
p
1
, p
2
, ··· , p
m
be a list of all irreducible factors of
a
i
, such that no
two of these are associates of each other. We now write
a
i
= u
i
m
Y
j=1
p
n
ij
j
,
where n
ij
N and u
i
are units. We let
m
j
= min
i
{n
ij
},
and choose
d =
m
Y
j=1
p
m
j
j
.
As, by definition, m
j
n
ij
for all i, we know d | a
i
for all i.
Finally, if d
0
| a
i
for all i, then we let
d
0
= v
m
Y
j=1
p
t
j
j
.
Then we must have
t
j
n
ij
for all
i, j
. So we must have
t
j
m
j
for all
j
. So
d
0
| d.
Uniqueness is immediate since any two greatest common divisors have to
divide each other.
|
Uniform Circular Motion Problems with Answers
In the following, some problems on uniform circular motion are provided with detailed answers. These circular motion questions are helpful for high school students and colleges.
Circular Motion Problems: Kinematic
Problem (1): An 5-kg object moves around a circular track with a radius of 18 cm with a constant speed of 6 m/s. Find
(a) The magnitude and direction of the acceleration of the object.
(b) The net force acting upon the object causing this acceleration.
Solution: When an object moves around a circular path at a constant speed, the only acceleration that experiences is centripetal acceleration or radial acceleration.
This kind of acceleration is always toward the center of the circle and its magnitude is found by the following formula $a_c=\frac{v^2}{r}$ where $v$ is the constant speed with which the object revolves the circle, and $r$ is the radius of the circle.
(a) The track is circular and the speed of the object is constant, so a centripetal acceleration directed toward the center is applied to the object whose magnitude is as follows $a_c=\frac{6^2}{0.18\,{\rm m}}=50\,{\rm \frac{m}{s^2}}$ In the figure below, a top view of the motion is sketched.
(b) By applying Newton's second law along the direction of the centripetal acceleration, we can find the magnitude of the net force causing the acceleration as follows $F_{net}=m\frac{v^2}{r}$ Therefore, $F_{net}=5\times 50=250\quad {\rm N}$
Note: At each point along the circular path, the instantaneous velocity of the revolving object is tangent to the path. The direction of this velocity changes, but its magnitude remains constant.
Problem (2): In a merry-go-round moving with a speed of 3 m/s a 25-kg child is sitting 3 m from its center. Calculate
(a) The centripetal acceleration of the child
(b) The net horizontal force acted upon the child
(c) Compare the above force with the child's weight
Solution: (a) The child has a circular motion with a centripetal acceleration as $a_c=v^2/r$ where $v$ is the constant speed of the revolving object. Therefore, $a_c=\frac{3^2}{3}=3\quad {\rm \frac{m}{s^2}}$
(b) The net force is found using Newton's second law as $F_{net}=ma_c$ which yields $F_{net}=25\times 3=75\quad {\rm N}$ This force is in the same direction as the acceleration, toward the center of the circle.
(c) Weight is mass times the gravitational acceleration at that place ($g$) or $w=mg$. The ratio between these two forces is $\frac{F_{net}}{w}=\frac{75}{25\times 10}=0.3$
Download Now: 550 solved high school physics midterm and final exams ($4). Problem (3): A cyclist moves around a circular path with a radius of 50 m at a speed of 10 m/s. (a) Find the acceleration of the cyclist? (b) Considering the combined mass of the cyclist and cycle to be 120 kg, what is the net force applied to them? Solution: (a) A movement with constant speed around a circular path yields a radially inward acceleration called centripetal acceleration whose magnitude is found as below $a_c=\frac{v^2}{r}=\frac{10^2}{50}=2\quad {\rm \frac{m}{s^2}}$ (b) The net force that acts on the object to keep it moving along a circular track is called centripetal force whose magnitude is found using Newton's second law as below $F_{net}=ma_c=120\times 2=240\quad N$ Problem (4): A jet plane is maneuvering in a circular path with a radius of 5.5 km at a constant speed of 2160 km/h. What is the acceleration of the plane in g's? Solution: This is a circular motion problem with constant speed so the only acceleration is due to centripetal acceleration. Because we are going to compare this acceleration with gravitational acceleration$g=9.8\,{\rm m/s^2}$, it is better first to convert all units into SI units. This conversion for constant speed is done as follows $\rm{1\,\frac{km}{h}}=\rm{\frac{1000\,m}{3600\,s}}=\frac{10}{36} {\rm \frac{m}{s}}$ So, 2160 km/h in SI units is converted as $2160\times \frac{10}{36}=600\quad {\rm \frac{m}{s}}$ Therefore, the magnitude of the centripetal acceleration is $a_c=\frac{v^2}{r}=\frac{600^2}{5000\,m}=72\quad {\rm \frac{m}{s^2}}$ To compare it with gravity, we should divide it by$g$$\frac{a_c}{g}=\frac{72}{9.8}=7.35$ Thus, this acceleration is about seven times greater than the gravitational acceleration,$a_c=7.35\,g$. Problem (5): A race car traveling around a circular path of a radius of 400 m with a speed of 50 m/s. Find the centripetal acceleration of the car. Solution: In a circular motion with constant speed the centripetal acceleration is$a_c=v^2/r. Therefore, substituting the numerical values into this equation we have $a_c=\frac{v^2}{r}=\frac{50^2}{400}=6.25\,{\rm \frac{m}{s^2}}$ This acceleration is toward the center of the circle. Problem (6): A crankshaft of radius 8 cm rotates at 2400 rpm (revolutions per minute). What is the speed of a point at its surface? Solution: The aim of this circular motion problem is the conversion of rpm into SI units of speed (m/s) which is done as below \begin{align*} 1 \frac{rev}{minute}&=\frac{\text{circle's circumference}}{60\,\text{seconds}}\\\\&=\frac{2\pi r}{60}\\\\&=\frac{\pi r}{30}\end{align*} Hence, to convert rpm tom/s$, we can use the following conversion ratio $\rm{1\,rpm}=\frac{\pi r}{30}\,{\rm \frac ms}$ In this problem, the circular motion has a radius of$r=0.08\,{\rm m}$and angular speed of 2400 rpm, so we get $2400 rpm=2400\times\frac{\pi(0.08)}{30}=20.1\,{\rm \frac ms}$ Problem (7): A 20-kg child sitting in a cart to which a 2-m rope is attached. The rope is tied to a motor that rotates the cart. At the instant that the tension in the rope is 100 N, how many revolutions per minute does the cart make? Solution: This circular motion question aims to find the method of conversion of$m/s$into$\rm rpm$. The child revolves in a circular path so the net centripetal force it experiences is determined by$F_c=\frac{mv^2}{r}$. First, use the above equation and solve for the constant speed$v$as below $v=\sqrt{\frac{rF_c}{m}}=\sqrt{\frac{2\times 100}{20}}=\sqrt{10}\,{\rm \frac ms}$ Next, convert the above linear speed in ($\rm \frac ms$) into angular speed in revolutions per minute ($\rm rpm$). The simple relation between the two speeds is as$v=r\omega$, where$\omega$is the angular speed measured in radians per second. $\omega=\frac{v}{r}=\frac{\sqrt{10}}{2} \,{\rm \frac {rad}{s}}$ Next, convert$\rm \frac{rad}{s}$into$\rm rpm$as below $\frac{1\,rad}{s}=\frac{\frac{1}{2\pi}\,rev}{\frac {1}{60}\,min}=\frac {30}{\pi}\, \rm rpm$ In the above, we used this fact that each revolution around a circle is equal to$2\pi$radians. Therefore, we have $\sqrt{10}\,{\rm \frac ms}=\sqrt{10}\left(\frac{30}{\pi}\right)\,{\rm rpm}=30.2\,{\rm rpm}$ Hence, the cart makes about 30 revolutions per minute. Problem (8): A radially inward constant force of 300 N is exerted on a 2-kg ball as it revolves around a circle of radius of 85 cm. What is the speed of the ball? Solution: The ball revolves around a circle, so the only radially inward force acting upon the object is the centripetal force whose magnitude is$F_c=mv^2/r$. Therefore, putting the numerical values into it and solving for unknown$v$, we have $v=\sqrt{\frac{rF_c}{m}}=\sqrt{\frac{0.85\times 300}{2}}=11.3\,{\rm \frac{m}{s}}$ Problem (9): A 35-kg child makes a turn which is a portion of a circle with a radius of 12 meters. She covers one-quarter of the circular path in 1.6 seconds. Determine the speed, acceleration, and net force applied to the child. Solution: The child turns one-quarter of a circle in$1.6$seconds. Using the definition of speed, as distance traveled divided by time elapsed, we can find its speed. But here, the distance traveled is the circumference of one-quarter of a circle of radius$12$meters so,$L=\frac 14 (2\pi r)=\frac{\pi r}{2}. \begin{align*} speed&=\frac{distance}{\text{time elapsed}}\\\\&=\frac{\pi r/2}{time}\\\\&=\frac{\pi\times (12/2)}{1.6}\\\\&=11.8\quad {\rm m/s}\end{align*} Because the child is moving around a portion of a circular path, a centripetal acceleration is applied upon the child whose magnitude isa_c=v^2/r$. $a_c=\frac{v^2}{r}=\frac{(11.8)^2}{12}=11.6\quad{\rm \frac{m}{s^2}}$ The net force is also found using Newton's second law of motion as$F_{net}=ma_c$. $F_{net}=ma_c=35\times 11.6=406\quad {\rm N}$ Problem (10): A ball of mass 2 kg is attached to a rope having a breaking strength of 1500 N and is whirled around a horizontal circle with a radius of 85 cm. Calculate the maximum speed that the ball can have. Solution: There is a circular motion so the only radially inward force that acted upon the ball is a centripetal force whose magnitude is$F_c=mv^2/r$where$r$is the radius of the circular path. Putting everything into this equation and solving for the unknown velocity, we have $v=\sqrt{\frac{rF_c}{m}}=\sqrt{\frac{0.85\times 1500}{2}}=25.25\,{\rm \frac{m}{s}}$ The following circular motion questions are helpful for the AP physics exam. Problem (11): The speed of a 515-kg roller-coaster at the bottom of a loop of radius 10 m is 20 m/s. Find the net vertical force pushing up on the object at this point of the circular path. Solution: Note that because of gravity the speed of the object changes at each point of the track, so the object is not undergoing uniform circular motion. Nonetheless, we can use the centripetal acceleration formula to find the velocity at each point along the circle. At the bottom, two forces are applied to the object: the downward force of gravity,$mg, and the upward tension force. The resultant of these two forces makes a centripetal acceleration toward the center of the circle. Applying Newton's second law in this direction (by choosing upward as positive direction since acceleration at this point is upward), we have \begin{align*} F_c-mg&=\frac{mv^2}{r}\\\\ \Rightarrow F_c &=mg+\frac{mv^2}{r}\\\\&=(515)(10)+\frac{515\times (20)^2}{10}\\\\&=25750\quad {\rm N}\end{align*} Problem (12): A 2000-kg car moving a curve of a radius of 200 m with a speed of 25 m/s. Calculate (a) The centripetal acceleration of the car. (b) The force causing this kind of acceleration. (c) The minimum coefficient of static friction between the tires and the surface of the road guarantees a safe turning. Solution: The free-body diagram below (side view) shows all forces acting on the car. (a) Centripetal acceleration is found by the following formula $a_c=\frac{v^2}{r}=\frac{25^2}{200}=3.125\quad {\rm \frac{m}{s^2}}$ (b) The force along this direction is found using Newton's second law of motion as below $F_c=ma_c=2000\times 3.125=6250\quad {\rm N}$ (c) When a car turns a curve, it experiences a centripetal acceleration toward the center of the curve. The force causing this acceleration for a car around a curve is static friction force. If the static friction force is not great enough, the car skid out of the curve and follows a nearly straight line. In this situation, the static friction must be balanced with the force needed to produce the given centripetal acceleration. On the other hand, recall that the maximum value of static friction force is given byf_{s, max}=\mu_s N$, where$N$is the normal force on the car which is equal to the weight$mg(only for flat circular tracks). Thus, \begin{align*} \mu_s (mg)&=\frac{mv^2}{r}\\\\ \Rightarrow \mu_s&=\frac{v^2}{rg}\\\\&=\frac{25^2}{200\times 9.8}\\\\&=0.32\end{align*} Therefore, the car can make a safe round if the coefficient of static friction is greater than this value. Practice these questions to understand friction force: Problems on the coefficient of friction Problem (13): A 1500-kg car moves around a flat circular track with a radius of 30 m. The coefficient of friction between the car's tires and the road is 0.3. Find the maximum speed at which the car turns the track. Solution: The static friction between the car's tires and the road provides the force required for turning the car around the circular track. The maximum value of the static friction force formula isf_{s, max}=\mu_s N$, where$Nis the force exerted on the car from the surface of the road which is called the normal force. By setting Newton's second law in the vertical direction zero (because the car does not fly!), we can find the normal force as below $N-mg=0 \Rightarrow N=mg$ Next, apply Newton's second law of motion again to the radial direction. Because the car moves around a curve, it experiences centripetal acceleration which provides by static friction. Thus, \begin{align*} f_{s,max}&=\frac{mv^2}r \\ \\\mu_s (mg)&=\frac{mv^2}r\\\\ \Rightarrow\quad v&=\sqrt{r\mu_s g}\\\\&=\sqrt{30\times (0.3)\times 9.8}\\\\&=9.4\quad {\rm \frac ms}\end{align*} Problem (14): A car is to turn a curve track of a radius of 120 m at a speed of 85 km/h. How large must the coefficient of static friction be between the car's tire and the road to maintain safe traveling? Solution: The forces on the car are downward gravity force,mg$, the normal force$N$exerted upward by the road, and the static friction force due to the road. All these forces are shown in the free-body diagram below. In the vertical direction, there is no acceleration. By applying Newton's second law, we find that the gravity force is equal to the normal force. $\Sigma F_y=0 \rightarrow N-mg=0 \Rightarrow N=mg$ In the radial direction, the net force required to keep the car in a circular path is found using the centripetal force formula,$F_c=\frac{mv^2}{r}$. This force must be balanced with the static friction force to guarantee a safe turning. Thus, the required condition for not skidding on the curve is as follows $F_c=f_{s,max}$ where$f_{s,max}=\mu_s Nis the maximum value of the friction force. Therefore, \begin{align*} \frac{mv^2}{r}&=\mu_s (mg) \\\\\Rightarrow \quad \mu_s&=\frac{v^2}{r\,g}\\\\&=\frac{(23.6)^2}{10\times 120}\\\\&=0.4\end{align*} In the above, we converted km/h'' to m/s'' as below $85\,{\rm \frac{km}{h}}=85\,\rm{\left(\frac{1000\,m}{3600\,s}\right)}=23.6\,{\rm \frac ms}$ Problem (15): A 2-kg bucket at the end of a rope is circulated in a vertical plane of a radius of 1.5 m. The tension of the rope at the lowest point of the path is 30 N. Calculate (a) The speed of the bucket. (b) How fast must the bucket at the top of the path move so that the rope does not go slack? Solution: The gravity force,mg, and the tension in the rope are exerted on the bucket moving in a circular motion. A free-body diagram shows these forces below. (a) Apply Newton's second law at the lowest point of the path to find the equation governs at that point. We choose the upward direction as positive because is in the direction of centripetal acceleration (radially inward). \begin{align*} T-mg&=\frac{mv^2}{r}\\\\ 30-2(10)&=\frac{2\times v^2}{1.5}\\\\ 10\times 1.5&=2v^2\\\\\Rightarrow v&=2.8\,{\rm \frac ms}\end{align*} (b) Sum of the forces at the highest point of the circle, using Newton's second law, is written as below (down is considered as positive)\begin{align*} T+mg&=\frac{mv^2}r\\\\ \Rightarrow v&= \sqrt{\frac{r(T+mg)}{m}}\end{align*} If we setT=0$in the above, the rope is on the verge of going slack. Therefore, $v= \sqrt{\frac{r(T+mg)}{m}}= \sqrt{\frac{(1.5)(0+2\times 10)}{2}}=3.8\,{\rm \frac ms}$ Problem (16): A jet plane moves around a vertical circular loop. Its speed at the lowest point of the circle is 950 km/h. Find the minimum radius of the circular path so that the (centripetal) acceleration exerted on it does not exceed 5 g's. Solution: The jet moves around a circular path, so it must experience a centripetal acceleration. Set this acceleration to$5g$and solve for$r. Thus, we have \begin{align*} a_c&=5g \\\\\frac{v^2}{r}&=5g \\\\\Rightarrow r&=\frac{v^2}{5g}\\\\&=\frac{(950\,\frac {km}{h})\left(\frac{1000\,m}{3600\,s}\right)}{5(9.8)}\\\\&=1421\,{\rm m}\end{align*} (b) The weight of the pilot is 75 kg. What is its apparent weight at the bottom and top of the circle? The apparent weight is equal to the net force on the pilot in the plane. Two forces act on the pilot. One is the downward gravity force,mg, and the other is the upward normal force at the lowest point of the circular path. These two forces acting on the pilot are shown in the free-body diagram below. The vector sum of these two forces provides the centripetal acceleration around the curved path. $\Sigma F_r=N-mg=m\frac{v^2}{r}$ where we have chosen up as the direction of the acceleration (radially inward) at this point, as the positive direction. Therefore, \begin{align*} N&=mg+mv^{2}/r\\ &=mg+m(5g)\\&=6mg\\&=6(75)(9.8)\\&=4410\,{\rm N}\end{align*} In all such vertical circular problems, the normal force is the apparent weight. Problem (17): The radius of a highway curve banked at an angle of13^\circ$is 75 m. At what speed can a car make a turn this curve without the help of friction? Solution Problem (18): An electron moving in a circular track of radius 2 mm at a speed of$2\times 10^{6}\,{\rm m/s}$. Find (a) The period of the revolutions (b) The centripetal acceleration experienced by the electron Solution: (a) This electron, every 1 s covers a distance of$2\times 10^6 \,{\rm m/s}$which is the circumference of the circle. Recall that the circumference of a circle of radius$r$is$2\pi r$, thus we must first find how many numbers the electron revolves around this circular track. $L=N(2\pi r) \Rightarrow N=\frac{L}{2\pi}$ On the other hand, the period is the time the electron takes to make one revolution in 1 second. According to this definition, the period of an object with$N$revolutions is obtained as $T=\frac{1}{N}=\frac{2\pi r}{L}$ Using this formula, we can find the period of this electron $T=\frac{2\pi\times (0.002)}{2\times 10^6}=6.3\times 10^{-9}\,{\rm s}$ Hence, the period of the electron is about$6\,{\rm ns}$. (b) The electron moves along a circular motion whose centripetal acceleration is computed as below $a_c=\frac{v^2}{r}=\frac{(2\times 10^6)^2}{0.002}=10^9\quad {\rm \frac {m}{s^2}}$ Problem (19): A 0.45-kg ball is fastened to a rope and is swung in a horizontal plane (without friction) around a circle of radius of 1.3 m. When the tension in the rope exceeds 75 N, the rope is broken. What is the maximum speed the ball can have at this point? Solution: In this circular motion problem, the only force causing the centripetal acceleration is tension in the rope. By applying centripetal force formula,$F_c=mv^2/r$, and solving for unknown speed$v$, we get $v=\sqrt{\frac{rF_c}{m}}=\sqrt{\frac{1.3\times 75}{0.45}}=14.7\,{\rm \frac ms}$ Problem (20): How fast must a centrifuge revolve if a point particle located 5 cm from its axis of rotation is to experience a centripetal acceleration of 100,000 g's? Solution: The particle in a centrifuge has a circular motion with a constant speed. Solving for$v$in the centripetal acceleration,$a_c=v^2/r$, we have $v=\sqrt{ra_c}=\sqrt{0.05\times 100,000}=32\,{\rm \frac ms}$ Problem (21): What is the acceleration of a stone attached to the end of a 2-m-long rope revolving at 45 revolutions per minute (rpm)? Solution: The stone has a circular motion with the given constant speed. In this problem, the speed of a revolving object is given in rpm or angular speed. To use the centripetal acceleration formula,$a_c=v^2/r$, we need speed in$m/s$. To convert angular speed (rpm) into linear speed (m/s), we proceed as below formula $v(m/s)=r\times \frac{rpm \times 2\pi}{60}$ Therefore, an angular speed of "45 rpm" is converted into "m/s" as below $v=2\times \frac{2\pi \times 45}{60}=9.5 \,{\rm m/s}$ Now, we can plug this value into the centripetal acceleration formula $a_c=\frac{v^2}{r}=\frac{(9.5)^2}{2}=45.1\quad {\rm \frac {m}{s^2}}$ Problem (22): A car moves along a curved path as in the following figure. When the car reaches the bottom of the track, find (a) The normal force exerted on the car (b) The effective weight (apparent weight) of the car. Solution: At the bottom of the track, two forces act on the car. The downward gravity force,$mg$, and the upward force due to the contact between the car and the surface which is known as normal force,$N$. The car round a curve so it experiences a centripetal force$F_c$radially inward in the same direction of centripetal acceleration. At the bottom of the track, this force is to the up. It is better to choose this direction as positive and apply Newton's second law along this direction as below \begin{gather*} N-mg=\frac{mv^2}{r} \\\\ \Rightarrow N=mg+\frac{mv^2}{r}\end{gather*} Problem (23): A centripetal force$F$is exerted on an object moving around a circular track with a constant speed$v$. If the speed of the object is tripled and the radius of the track is quadrupled, what happened to the centripetal force? Solution: The centripetal force in a circular motion is given by equation$F_c=\frac{mv^2}{r}. As you can see, this kind of force is proportional directly to the speed of the object and inversely to the radius of the curved path. Therefore, \begin{align*} \frac{F_2}{F_1}&=\left(\frac{v_2}{v_1}\right)^2 \frac{r_1}{r_2}\\\\&=\left(\frac{3v_1}{v_1}\right)^2 \frac{r_1}{4r_1}\\\\&=\frac{9}{4}\end{align*} Hence, the force becomes 2.25 times the original force. $F_2=\frac 94 F_1$ Problem (24): A stone is attached to the end of a rope of lengthL$and rotates in a vertical circle. At the top of the path, the tension in the rope is twice the weight of the stone. (a) Determine the net force on the stone when it reaches the highest point of the path. (b) What is the speed of the stone at this point? After a few circles, when the stone is at the top of the circle, the rope is broken. (c) Determine the time it takes the stone to hit the ground. (d) Find the horizontal distance traveled by the stone before striking the ground. Solution: This is left up to you to solve. Summary: In this article, we learned how to solve circular motion problems. The most important notes we practiced are as follows: 1. An object moving in a circular path of radius$r$at constant (uniform) speed$v$experiences an acceleration whose magnitude is$a_c=\frac{v^2}{r}$and whose direction is toward the center of the circle. 2. Uniform circular motion is an example that shows that acceleration and velocity are not always in the same direction. 3. In such motions,$\vec{a}$and$\vec{v}\$ are perpendicular.
Author: Dr. Ali Nemati
Date Published: 7-31-2021
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# Multiplication of Polynomials – Full chapter – Class IX
After studying chapter Multiplication of Polynomials you learn to find the product of a monomial and a binomial; find the product of two polynomials; find the product of binomials using some special identities; prove some of the important identities; expand the binomial terms using the identities; solve problems under identities; prove problems under identities; prove some of the important conditional identities.
## 3.1.1 Introduction to Multiplication of Polynomials
Definition: An algebraic expression of the form,
f(x) = a0 + a1x + a2x2 + a3x3 + … + anxn
where, a0, a1, a2, … an are real numbers and n a non-negative integer, is called a polynomial with real coefficients or (real polynomial)
Here a0, a1, a2, … an are called the coefficients of the polynomial f(x) and a0 ,a1x , a2x2 ,a3x3, … ,anxn are called the terms of f(x). If an ≠ 0, we say that n is the degree of the polynomial f(x).
## 3.1.2 Some Products – Multiplication of Polynomials
1. ### Product of a monomial by a binomial:
Consider a monomial ab and a binomial 2a + 1. We multiply them as follows:
ab(2a + 1) = ab(2a) + ab = 2a2b + ab
Example: Multiply caa and b2+2bc
Solution: Using distributive property
c2a(b2 + 2bc) = c2ab2 + c2a(2bc) = c2ab2 + 2abc3
1. ### Product of a binomial by a binomial:
Example: Find the product (x + 2) and (x + 3)
Solution:
We use our rule: multiply term by term. Thus,
(x + 2)(x + 3) = x(x + 2) + 3(x + 2) = x2 + 2x + 3x + 6 = x2 + 5x + 6
1. ### Some special products of binomials:
Some products occur so frequently In algebra that it is advantageous to recognize them by sight and write their product by memory. This will be particularly useful when we talk about factoring.
(a) Difference of two squares:
(a+b)(a – b) = a(a – b) + b(a – b)= a2 – ab + ab – b2 = a2 – b2
(b) Square of a binomial:
(a + b)2 = (a+b)(a+b) = a(a+b)+b(a+b) = a2 + ab + ab + b2 =a2 + 2ab + b2
(a – b)2 = (a-b)(a-b) = a(a-b)-b(a-b) = a2 – ab – ab + b2 =a2 – 2ab + b2
Example : Find the product (3 + √2)(3 + √2)
Solution:
(3 + √2)(3 + √2) = 3(3 + √2)+ √2(3 + √2) = 9 +3√2 + 3√2 + 2 = 9 + 6√2 + 2 = 11 + 6√2
## Multiplication of Polynomial – Exercise 3.1.2
1. Evaluate the following products:
(i) ax2 ( bx + c)
= ax2 (bx) + ax2 (c)
= abx3 + acx2
(ii) ab (a+b)
= ab (a) + ab (b)
= a2b + ab2
(iii) a2b2 (ab2+a2b)
= a2b2 (ab2) + a2b2 (a2b)
= a3b4 + a4b3
(iv) b4(b6 + b8)
= b4 (b6) + b4 (b8)
= b4 (b6) + b4 (b8)
= b10 + b12
1. Evaluate the following products:
(i) (x+3) (x+2)
= (x+3) x + (x+2) x
= x2 + 3x + 2x + 6
= x2 + 5x + 6
(ii) (x+5) (x–2)
= (x+5) x + (x+5) (–2)
= x2 +5x – 2x –10
= x2 + 3x –10
(iii) ( y – 4 ) ( y + 6 )
= (y – 4) y +(y – 4)6
= y2 – 4y + 6y – 24
= y2 + 2y – 24
(iv) (a–5) (a–6)
= (a–5) a + (a–6) (–6)
= a2 – 5a – 6a + 30
= a2 – 11a +30
(v) (2x+1) (2x–3)
= (2x+1)2x + (2x+1) (–3)
= 4x2 + 2x – 6x – 3x
= 4x2 – 4x –3
(vi) ( a + b ) ( c + d )
= (a + b) c + (a+b) d
= ac + bc + ad + bd
(vii) ( 2x – 3y ) ( x – y )
= (2x – 3y) x + (2x – 3y) (–y)
= 2x2 – 3xy – 2xy +3y2
= 2x2 – 5xy + 3y2
(viii) ( √𝟕𝐱 + √𝟓 ) (√𝟓𝐱 + √𝟕 )
= (√𝟕𝐱 + √𝟓 ) (√𝟓𝐱) + (√𝟕𝐱 + √𝟓 ) (√𝟕)
= √𝟑𝟓 x2 + 5x + 7x + √𝟑𝟓
= √𝟑𝟓x2 + 12x + √𝟑𝟓
(xi) (2a+3b) (2a–3b)
= (2a+3b) 2a + (2a+3b) (–3b)
= 4a2+ 6ab – 6ab – 9b2
= 4a2 – 9b2
(xii) (6xy–5) (6xy+5)
= (6xy – 5) (6xy) + (6xy – 5) 5
= 36x2y2 – 30xy + 30xy – 25
= 36x2y2 – 25
(xiii)(2/x+3) (2/x –7)
= (2/x+3)2/x +(2/x+3)(-7)
= 4/x2 + 6/x14/x – 21
= 4/x28/x –21
1. Expand the following using appropriate identity:
(i) (a +5)2
Using (a + b)2 = a2 +2ab +b2 we get
a = a b = 5
(a +5)2 = a2 +2.a.5 +b2
= a2 +10a +25
(ii) (2a +3)2
Using (a + b)2 = a2 +2ab +b2 we get
a = 2a; b = 3
(2a +3)2 = (2a)2 +2.2a.3 +b2
= 4a2 + 12a + 9
(iii) ( x + 𝟏/𝐱 )2
Using (a + b)2 = a2 +2ab +b2 we get
a = 2a; b = 1/x
(X + 𝟏/ )2 = x2 + 2.x. 1x + (𝟏/𝐱 )2
= x2 + 2 + (𝟏/x )2
(iv) ( √(12a) + √(6b) )2
Using (a + b)2 = a2 +2ab +b2 we get
a = √(12a) and b = √(6b)
(√(12a) + √ (6b))2= (√12a)2 + 2.√12 a + √(6b) + (√(6b))2
= 12a2 +2√72 ab +6b2
= 12a2 +2 √ (36 ×2) ab +6b2
= 12a2 + 12 √𝟐ab +6b2
(v) (𝛑 + 𝟐𝟐/7 )2
Using (a + b)2 = a2 +2ab +b2 we get
a = 𝛑 and b = 22/7
(𝛑 + 22/7 )2 = 𝛑2 + 2. 𝛑 22/7 + (22/7)2
= 𝛑2 + 44π/7 + ( 22/7)2
= 𝛑2 + 𝟒𝟒𝛑/7 + 𝟒𝟖𝟒/49
(vi) (y – 3)2
Using (a – b)2 = a2 – 2ab + b2
a = y and b = –3
(y – 3)2 = y2 – 2. y.3 + 32
= y2 – 6y + 9
(vii) (3a – 2b)2
Using (a – b)2 = a2 – 2ab + b2
a = 3a and b = –2b
(3a – 2b)2 = (3a)2 – 2.3a.2b + (2b)2
= 9a2 – 12ab + 4b2
(viii) ( y – 𝟏/y )2
Using (a – b)2 = a2 – 2ab + b2
a = y and b = 𝟏/y
(y – 𝟏/y)2 = y2 – 2.y. 𝟏/y + (𝟏/y )2
= y2 – 2 + 𝟏/
(ix) ( √(10x) – √(5y))2
Using (a – b)2 = a2 – 2ab + b2
a = √(10x) and b = √(5y)
(√(10x) – √(5y))2
= (√(10x))2 – 2. √(10x).√(5y)+ √(5y))2
= 10x2 – 2√50 xy + 5y2
= 10x2 – 2.5√2 xy + 5y2
= 10x2 – 10√2 xy + 5y2
(x) (𝛑 – 𝟐𝟐/7 )2
Using (a + b)2 = a2 +2ab +b2 we get
a = 𝛑 and b = 𝟐𝟐/𝟕
(𝛑 – 𝟐𝟐/ )2 = 𝛑2 – 2. 𝛑 𝟐𝟐/𝟕 + (𝟐𝟐/𝟕)2
= 𝛑244/𝟕 π + (𝟐𝟐/𝟕)2
= 𝛑2𝟒𝟒𝛑/7 + 𝟒𝟖𝟒/𝟒𝟗
(xi) (2x+3) (2x+5)
Using (x + a) (x + b) = x2 + x (a + b) ab we get
x = 2x, a = 3 and b = 5
(2x+3) (2x+5) = (2x)2 +2x (3 + 5) + 3.5
= 4x2 +16x +15
(xii) (3x – 3) (3x + 4)
Using (x + a) (x + b) = x2 + x (a + b) ab we get
x = 3x, a = –3 and b = 4
(3x – 3) (3x + 4) = (3x)2 + 3x [(–3)+(4)] + (-3)4
= 9x2 + 3x –12
= 9x2 + 3x –12
1. Expand :
(i) (x + 3 ) (x – 3)
Using (a + b) (a – b) = a2 – b2 we get
a = x, b = 3
(x + 3) (x – 3) = x2 – 32
= x2 – 9
(ii) (3x – 5y) (3x + 5y)
Using (a + b) (a – b) = a2 – b2 we get
a = 3x, b = 5y
(3x – 5y) (3x + 5y) = (3x)2 – (5y)2
= 9x2 – 25y2
(iii) (x/3+y/2)( x/3y/2)
Using (a + b) (a – b) = a2 – b2 we get
a = x/3, b = y/2
(x/3+ y/2)( x/3y/2)= (x/3 )2 – (x/3)2
= 𝐱𝟐/9-𝐲2/𝟒
(iv) (x2 + y2) (x2 – y2)
Using (a + b) (a – b) = a2 – b2 we get
a = x2, b = y2
(x2 + y2) (x2 – y2) = (x2)2 – (y2)2
= x4 – y4
(v) (a2 + 4b2) (a + 2b) (a – 2b)
Using (a + b) (a – b) = a2 – b2 for 2nd and 3rd term we get
(a2 + 4b2) (a + 2b) (a – 2b) = (a2 + 4b2) [a2 – (2b)2]
= (a2 + 4b2) (a2 – 4b2)
Using the above identity once again we get
= (a2)2 – (4b2)2
= a4 – 16b4
(vi) (x – 4) (x + 4) (x – 3) (x + 4)
Using (a + b) (a – b) = a2 – b2 we get
(x – 4) (x + 4) (x – 3) (x + 4) = (x2 – 42) (x2 – 32)
= (x2 – 16) (x2 – 9)
Using (a + b) (a – b) = x2 – x (a + b) + ab
= (x2)2 – x (16+9) +16×9
= x4 – 25x2 +144
(vii) (x – a) (x + a)(𝟏/𝐱𝟏/𝐚)( 𝟏/𝐱 +𝟏/𝐚 )
(x2 – a2) )(( /𝐱 )2– (1/a )2)
x2 x 𝟏/𝐱 2 – x2 x 𝟏/𝐚 2 – a2 x 𝟏/𝐱 2 + a2 x 𝟏/𝐚 2
1 – x2/a2 − a2/ x2 + 1
2 – 𝐱𝟐/a𝟐 − 𝐚𝟐 /x𝟐
1. Simplify the following:
(i) (2x – 3y)2 + 12xy
= (2x)2 + (3y)2 – 2.2x.3y + 12xy
= 4x2 + 9y2 -12xy +12xy
= 4x2 + 9y2
(ii) (3m + 5n)2 – (2n)2
= (3m)2 + (5n)2 + 2.3m.5n – 4n2
= 9m2 + 25n2 + 30mn – 4n2
= 9m2 + 30 mn +21n2
(iii) (4a – 7b)2 – (3a)2
= (4a)2 – 2.4a.7b + (7b)2 – (3a)2
= 16a2 – 56ab + 49b2 – 9a2
= 7a2 -56ab + 49b2
(iv) (x + 𝟏/x)2 (m + 𝟏/𝐦 )2
= (x2 + 2. x. 𝟏/x + 𝟏/x2)2 – (m2 + 2. m. 𝟏/𝐦 + 𝟏/𝐦 2)2
= x2 + 2 + 𝟏/x 2 – (m2 + 2 + 𝟏/𝐦 2)
= x2 + 2 + 𝟏/x 2 – m2 + 2𝟏/𝐦 2
= x2 – m2+ 𝟏/x 𝟐𝟏/𝐦 𝟐 + 4
(v) (m2 + 2n2)2 – 4m2n2
= m4 +2m4.2n2 +4n4 – 4m2 n2
= m4+ 4m2n2 + 4n2 – 4m2n2
= m4 + 4n2
(vi) (3a – 2)2 – (2a -3)2
= (9a2 – 2.3a.2 + 22) – (4a2 – 2.3a.2 + 92)
= 9a2 –12a + 4 – 4a2 + 12a – 9
= 5a2 – 5
=5(a2 – 1)
## 3.1.3 Identities – Multiplication of Polynomials
Consider 5x + 10 = 15 and (2x + 3)2 = 4x2 + 12x + 9. Only x = 1 satisfies the first relation, whereas any value of x satisfies the second relation. We say 5x + 10 = 15 is an equation and (2x + 3)2 = 4x2 + 12x + 9 is an identity. Thus an identity is an equation which is true for all values of the variables in it.
### Proof of identities:
Identity: (x+a)(x+b)(x+c) = x3 + x2(a+b+c) + x(ab+bc+ca) + abc
Let us expand LHS, (x + a)(x + b)(x + c) = [x2 + xb + ax + ab](x + c)
= x3 + x2b + ax2 + abx + x2c + xbc + axc + abc
= x3 + x2(a + b + c) + x(ab + bc + ac) + abc
This proves the identity.
Example 7: Find the product of the binomials (x+2), (x + 3) and (x + 4)
Solution:
(x+a)(x+b)(x+c) = x3 + x2(a+b+c) + x(ab+bc+ca) + abc
Here a = 2; b = 3; c = 4 then,
(x + 2)(x + 3)(x + 4) = x3 + x2(2+3+4)+x(2×3 + 3×4 + 4×2) + (2x3x4)
= x3 + x2(9) + x(6 + 12 + 8) + (24)
= x3 + 9x2 + 26x + 24
Identity: (a+b)3 = a3 + 3a2b + 3ab2 + b3
If a = b = c then, the product (x+a)(x+b)(x+c) simply reduces to (x+a)3. In this case a+ b+ c = a+a+a = 3a, ab+bc+ca = a2 + a2 + a2 = 3a2 and abc = a3 thus we obtain,
(x + a)3 = x3 + 3x2a + 3xa2 + a3
Taking x = b in this, we obtain,
(a+b)3 = a3 + 3a2b + 3ab2 + b3
This can also be written as,
(a + b)3 = a3 + 3ab(a+ b) + b3
Example: Expand (2x + 3y)3
Solution: We use the above identity: (a + b)3 = a3 + 3a2b + 3ab2 + b3 . Take a = 2x and b = 3y and we get,
(2x + 3y)3 = (2x)3 + 3(2x)2(3y) + 3(2x)(3y)2 + (3y)3
= 8x3 + 36x2y + 54xy2 + 27y3
## Multiplication of Polynomials – Exercise 3.1.3
1. Find the following products:
(i)(x + 4) (x + 5) (x + 2)
Using (x + a) (x + b) (x + c) = x3 + x2 (a + b + c) + x (ab + bc + ca) + abc
we get
a = 4, b = 5 and c =2
(x + 4) (x + 5) (x + 2) = x3 + x2(4 + 5 + 2) + x (4.5 + 5.2 + 2.4) + 4.5.2
= x3 + 11x2 + x (20 + 10 + 8) +40
= x3 + 11x2 + 38x +40
(ii) (y + 3) (y + 2) ( y – 1)
Using (x + a) (x + b) (x + c) = x3 + x2 (a + b + c) + x (ab + bc + ca) + abc we get
x = y, a = 3, b = 2 and c = -1
(y + 3) (y + 2) (y – 1) = y3 + y2 (3 + 2 – 1) + y (3.2 + 2(-1) + (-1)3 + 3.2(-1)
= y3 + 4y2 + y (6 – 2 – 3) – 6
= y3 + 4y2 + y – 6
(iii) (a + 2) (a – 3) (a + 4)
Using (x + a) (x + b) (x + c) = x3 + x2 (a + b + c) + x (ab + bc + ca) + abc
we get
x = a, a = 2, b = –3 and c = 4
(a + 2) (a – 3) (a – 4) = a3 + a2 (2 – 3 + 4) + a [2(–3) + (–3)4 + 4.2) + 2 (–3) 4
= a3 + 3a2 + a (–6 – 12 + 8) – 24
= a3 + 3a2 – 10a – 24
(iv) (m – 1) (m – 2) (m – 3)
Using (x + a) (x + b) (x + c) = x3 + x2 (a + b + c) + x (ab + bc + ca) + abc
we get
x = m, a = –1, b = –2 and c = –3
(m – 1) (m – 2) (m – 3) = m3 + m2 (–1 – 2 – 3) + m [(–1) (–2) + (–2) (–3) +
(– 3)(–1)] + (–1) (–2) (–3)
= m3 + m2 (–6) + m [2 + 6 + 3] – 6
= m3 – 6m2 + 11m – 6
(v) ( √𝟐 + √𝟑) (√𝟐+ √𝟓) (√𝟐+ √𝟕 )
Using (x + a) (x + b) (x + c) = x3 + x2 (a + b + c) + x (ab + bc + ca) + abc
we get
x = 2, a = 3, b = 5 and c = 7
(√2 + √3) (√2+ √5) (√2+ √7 ) = (√2)3 + (√2)2 [√3 + √5+ √7]+ √2 [√3. √5 + √5. √7 + √7. √3] + √3. √5. √7
= 2√2 + 2(√3 + √5 + √7) + √2 (15+ √35 + √21) + √105
(vi) 105 x 101 x 102
We can write this as
(100 + 5) (100 + 1) (100 + 2)
Using (x + a) (x + b) (x + c) = x3 + x2 (a + b + c) + x (ab + bc + ca) + abc
we get
x = 100, a = 5, b = 1 and c = 2
(100 + 5) (100 + 1) (100 + 2) = 1003 + 1002 (5 + 1 + 2) + 100 (5.1 + 1.2
+ 2.5) + 5.1.2
= 1000000 + 10000 (8) + 100(5 + 2 +10) + 10
= 1000000 + 80000 + 1700 +10
= 1081710
(vii) 95 x 98 x 103
We can write this as
(100 – 5) (100 – 2) (100 + 3)
Using (x + a) (x + b) (x + c) = x3 + x2 (a + b + c) + x (ab + bc + ca) + abc
we get
x = 100, a = -5, b = -2 and c = 3
(100 – 5) (100 – 2) (100 + 3) = 1003 + 1002 (–5 – 2 +3) + 100(–5) (–2) + (–2) 3
+ 3 (-5) + (-5) (-2) 3
= 1000000 + 10000(–4) +100 (10 – 6 –16) + 30
= 1000000 – 40000 – 1100 +30
= 958930
(viii) 1.01 x 1.02 x 1.03
We can write this as
(1 + 0.01) (1 + 0.02) (1 + 0.03)
Using (x + a) (x + b) (x + c) = x3 + x2 (a + b + c) + x (ab + bc + ca) + abc
we get
x = 1, a = 0.01, b = 0.02 and c = 0.03
(1 + 0.01) (1 + 0.02) (1 + 0.03) = 13 + 12 (0.01 + 0.02 + 0.03) + 1 [(0.01) (0.02) +
(0.02) (0.03) + (0.03) (0.01)] + (0.01) (0.02) (0.03)
= 1 + 0.06 + (0.0002 + 0.0006 + 0.0003) + 0.000006
= 1.061106
1. Find the coefficients of x2 and x in the following:
(i) (x + 4) (x + 1) (x + 2)
= x3 + x2 (4 + 1 + 2) + x (4.1 + 1.2 + 2.4) + 4.1.2
= x3 + 7x2 + 14x + 8
Coefficient of x2 is 7
Coefficient of x is 14
(ii) (x – 5) (x – 6) (x – 1)
= x3 + x2 (–5 – 6 –1) + x [(–5) (–6) + (–6) (–1) – 1(–5)] + (–5) (–6) (–1)
= x3– 12x2 + x (30 + 6 + 5) – 30
= x3– 12x2 + 41x – 30
Coefficient of x2 is –12 and x is 41
(iii) (2x + 1) (2x – 2) (2x – 5)
= (2x)3 + (2x)2 [1–2–5] + 2x [(1)(–2) + (–2)(–5) + (–5)(1)] + 1 (–2) (–5)
= 8x3 + 4x2 (–6) + 2x [–2 +10–5] + 10
= 8x3 – 24x2 + 6x + 10
Coefficient of x2 is –24 and x is 6
(iv) ( 𝐱/𝟐 + 1) ( 𝐱/𝟐 + 2) ( 𝐱/𝟐 + 3)
= (𝐱/ )3 + (𝐱/𝟐)2 [1 + 2 + 3] + 𝐱/𝟐 [1.2 + 2.3 + 3.1] + 1.2.3
= x3/8 + (x2/4 )(6) + 𝐱/𝟐 (2 + 6 + 3) + 6
= x3/8 + 3/2 x2 + 11/2 x + 6
Coefficient of x2 is 𝟑𝟐 and x is 𝟏𝟏𝟐
1. The length, breadth and height of a cuboids are (x +3), (x – 2) and (x -1) respectively. Find its volume.
Solution:
Volume of a cuboids = length x breadth x height
V = (x +3) (x – 2) (x – 1)
Using the identity (x + a) (x + b) (x + c) = x3 + x2 (a + b + c) + x (ab + bc + ca) + abc we get
V = x3 + x2 (3 – 2 – 1) + x [3(–2) + (–2) (–1) + (–1)3] + 3(–2) (–1)
= x3 – 0x2 + X (–6 + 2 – 3) + 6
= x3 – 7x2 + 6
1. The length, breadth and height of a metal box are cuboid are (x +5), (x – 2) and (x – 1) respectively. What is its volume?
Solution:
Volume of the metal box = length x breadth x height
V = (x +5) (x – 2) (x – 1)
Using the identity (x + a) (x + b) (x + c) = x3 + x2 (a + b + c) + x (ab + bc + ca) + abc we get
V = x3 + x2 + (–5 – 2 –1) + x [5(–2) + (–2) (–1) + (–1)5] + 5 (–2) (–1)
= x3 + 2x2 + x [–10 + 2 – 5] + 10
= x3 + 2x2 – 13x + 10
1. Prove that
(a + b) (b + c) (c + a) = (a + b + c) (ab + bc + ca) – abc
[Hint: write a + b = a + b + c – c, b + c = a + b + c – a, c + a = a + b + c – d]
Solution:
x3 + x2 (a + b + c) + x (ab + bc + ca) + abc we get
1. H.S. = (a + b + c)3 + (a + b + c)2 (–c – a – b) + (a + b + c) [(–c) (–a) +
(–a) (–b) + (–b) (–c)] – (–c) (–a) (–b)
= (a + b + c)3-(a + b + c)2[(a + b + c)] +(a + b + c) (ac + ab + bc) – abc
= (a + b + c)3– (a + b + c)3 + (a + b + c) (ac + ab + bc) – abc
= (a + b + c) (ac + ab + bc) – abc
= R. H. S.
1. Find the cubes of the following:
(i) (2x +y)3
Solution:
Using (a + b)3 = a3 + 3a2b + 3ab2 + b3 we get
a = 2x, b = y
(2x +y)3 = (2x)3 + 3(2x)2 y + 3 (2x) y2 +y3
= 8x3 + 12 x2y + 6 xy2 +y3
(ii) (2x + 3y)3
Solution:
Using (a + b)3 = a3 + 3a2b + 3ab2 + b3 we get
a = 2x, b = 3y
(2x + 3y)3 = (2x)3 + 3(2x)2 (3y) + 3 (2x) (3y)2 + (3y)3
= 8x3 + 36 x2y + 54 xy2 + 27y3
(viii) 1013
Solution:
We write 101 as (100 + 1)3
Using identity (a + b)3 = a3 + 3a2b + 3ab2 + b3 we get
a = 100, b = 1
(100 + 1)3 = 1003 + 3. 1002 + 3. 100.12 + 13
= 1000000 + 30000 + 300 + 1
= 1030301
(viii) 2.13
Solution:
We write 2.1 (2 + 0.1)3
Using identity (a + b)3 = a3 + 3a2b + 3ab2 + b3 we get
a = 2, b = 0.1
(2 + 0.1)3 = 23 + 3 x 22(0.1) + 3 x 2 x (0.1)2 + (0.1)3
= 8 + 1.2 + 0.06 + 0.001
= 9.261
1. Find the cubes of the following:
(i) (2a – 3b)3
Solution:
Using (a – b)3 = a3 – 3a2b + 3ab2 – b3 we get
a = 2a, b = 3b
(2a – 3b)3 = (2a)3 – 3 (2a)2(3b) + 3 (2a)(3b)2 – (3b)3
= 8a3 – 36a2b + 54ab2 -27b³
(ii) ( x – 𝟏/𝐱 )3
Solution:
Using (a – b)3 = a3 – 3a2b + 3ab2 – b3 we get
a = x, b = 𝟏/𝐱
(x – 𝟏/ )3 = x3 – 3x2 𝟏/𝐱 + 3x(𝟏/𝐱)3 – (𝟏/𝐱)3 = x3 – 3x + 3x/ x21/x3
= x3 – 3x + 3/x1/x3
(iii) (√3 x – 2)2
Solution:
Using (a – b)3 = a3 – 3a2b + 3ab2 – b3 we get
a = √3 x, b = 2
(√3 x – 2)2 = (√3x)3 – 3 (√3x)2 .2 + 3. √3x x 22 – 23
= 3√3 x3 – 6. 3 x2 + 12√3 x – 8
=3√3 x3 – 18x2 + 12√3 x – 8
(iv) (2x – √5)3
Solution:
Using (a – b)3 = a3 – 3a2b + 3ab2 – b3 we get
a = 2x, b = √5
(2x – √5)3 = (2x)3 – 3(2x)2 √5 + 3. 2x. √5)2 – (√5)3
= 8x3 – 12√5x2 + 30x – 5√5
(v) 493
Solution:
We can write 49 = 50 – 1
Using (a – b)3 = a3 – 3a2b + 3ab2 – b3 we get
a = 50, b = 1
(50 – 1)3 = 503 – 3.502.1 + 3.50.12 – 13
= 125000 – 3 x 2500 + 150 – 1
= 125000 – 7500 +149
= 117649
(vi) 183
Solution:
Let us write 18 = 20 – 2
Using (a – b)3 = a3 – 3a2b + 3ab2 – b3 we get
a = 20, b = 2
(20 – 2)3 = 203 – 3.202.2 + 3.20.22 – 23
= 8000 – 6×400 + 60×4 – 8
= 8000 – 2400 +240 – 8
= 5832
(vii) 953
Solution:
We write 95 = 100 – 5
Using (a – b)3 = a3 – 3a2b + 3ab2 – b3 we get
a = 100, b = 5
(100 – 5)3 = 1003 – 3.1002.5 + 3.100.52 – 53
= 1000000 – 150000 + 7500 – 125
= 857375
(viii) 1083
Solution:
We write 1083 = (110 – 2)
Using (a – b)3 = a3 – 3a2b + 3ab2 – b3 we get
a = 110, b = -2
(110 – 2)3 = 1103 – 3. (110)2.2 + 3.110×22 – 23
= 1331000 – 72600 + 1320 -8
= 1259712
1. If x + 𝟏/𝐱 = 3, prove that x3 + 𝟏/𝐱𝟑 = 18.
Solution:
Given x + 1/x = 3
Cubing both sides we get
(x + 1/x )3 = 33
Using (a + b)3 = a3 + 3a2b + 3ab2 + b3 we get
a = x b = 1/x
(x + 1/x )3 = (x)3 + ( 1/x )3 + 3x. 1/x (x + 1/x )
27 = x3 + 1/x3 + 3 (3)
x3 + 1/x3 = 27 – 9
= x3 + 1/x3 = 18
1. If p + q = 5 and pq = 6, find p3 + q3
Solution:
(p + q)3 = p3 + 3pq (p + q) + q3
53 = p3 + 3.6 (5) + q3
125 = p3 + 90 + q3
p3 + q3 = 125 – 90
p3 + q3 = 35
1. If a – b = 3 and ab = 10, find a3 – b3
Solution:
Given a – b = 3 and ab = 10
(a – b)3 = a3 – b3 – 3ab (a – b)
33 = a3 – b3 – 3.10 (3)
27 = a3 – b3 – 90
a3 – b3 = 27 + 90
a3 – b3 = 117
1. If a2 + 𝟏/𝐚𝟐 = 20 and a3 + 𝟏/𝐚𝟑 = 30, find a + 𝟏/𝐚
Solution:
a3 + 1/a3 = (a + 1/a) (a2 + 1/a2 – a x 1/a )
30 = (a + 1/a ) = (20 – 1)
30 = (a +1/a ) x 19
30/19 = a + 1/a
a + 1/a = 𝟑𝟎/𝟏𝟗
## 3.1.4 Square of a trinomial – Multiplication of Polynomials
(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
Example: Expand (x + 2y + 3z)2
Solution:
We take a = x; b = 2y ; c = 3z
(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
(x + 2y + 3z)2 = x2 + (2y)2 + (3z)2 + 2(x)(2y) + 2(2y)(3z) + 2(3z)(x)
= x2 + 4y2 + 9z2 + 4xy + 12yz + 6xz
### Multiplication of Polynomials – Exercise 3.1.4
1. Expand the following:
(i) (a + b + 2c)2
Using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc +2ca we get
a = a, b = b and c = 2c
(a + b + 2c)2 = a2 + b2 + (2c)2 + 2ab + 2b (2c) + 2(2c)a
= a2 + b2 + 4c2 + 2ab + 4bc + 4ca
(ii) (x + y + 3z)2
Using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc +2ca
a = x, b = y and c = 3z
(x + y + 3z)2 = x2 + y2 + (3z)2 + 2.x.y + 2y(3z) + 2.(3z)x
= x2 + y2 + 9z2 + 2xy + 6yz + 6zx
(iii) (p + q – 2r)2
Using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc +2ca
a = p, b = q and c = -2r
(p + q – 2r)2 = p2 + q2 + (-2r)2 + 2.p.q + 2q(-2r) +2(-2r)p
= p2 + q2 + 4r2 + 2pq – 4pr – 4pr
(iv) ( a/2+b/2+c/2 )2
Using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc +2ca
a = a/2, b = b/2 and c = c/2
(a/2+ b/2+ c/2 )2 =( a/2 )2 + (b/2 )2 + (c/2 )2 + 2 (a/2 ) (b/2) + 2 (b/2 ) (c/2) +
2 (c/2 ) (a/2 )
= a2/4+b2/4+c2/4 + ab/2+bc/2+ca/2
(v) (x2 + y2 + z)2
Using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc +2ca
a = x2, b = y2 and c = z
(x2 + y2 + z)2 = (x2)2 + (y2)2 + (z)2 + 2x2y2 + 2y2z +2zx2
= x4 + y4 + z2 + 2x2y2 + 2y2z +2zx2
(vi) (m – 3 – 1/m )2
Using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc +2ca
a = m, b = -3 and c = – 1/m
(m – 3 – 1/m )2 = m2 + (-3)2 + (1/m )2 + 2.m(-3) + 2(-3)(- 1/m ) + 2 (-1/m) m
= m2 + 9 + 1/m 2 – 6m + 6m – 2
= m2 + 1/m2 + 6/m – 6m + 7
(vii) (-a + b – c)2
Using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc +2ca
a = –a b = b c = –c
(-a + b – c)2 = (–a)2 + b2 + (–c)2 + 2(–a)b + 2b(–c) +2(–c)a
= a2 + b2 + c2 – 2ab – 2bc +2ca
(viii) (x + 5 + 1/2x )2
Using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc +2ca
a = x; b = 5; c = 1/2x
(x + 5 + 1/2x )2 = x2 + 52 + (1/2x)2 + 2.x.5 + 2.5. 1/2x + 2(1/2x)x
= x2 + 25 + 1/4x2 +10x + 5/x + 1
= x2 + 1/4x2 +10x + 5/x + 26
1. Simplify the following:
(i) (a – b + c)2 – (a – b – c)2
Using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc +2ca (a – b + c)2 – (a – b – c)2
= [ a2 + (-b)2 + c2 + 2a(-b) + 2(-b)c + 2ca ] – [a + (-b)2 + (–c)2 + 2a(–b)
+ 2(–b)(–c) + 2(–c)a]
= a2 + b2 + c2 – 2ab – 2bc +2ca – [a2 + b2 + c2 – 2ab + 2bc –2ca]
= a2 + b2 + c2 – 2ab – 2bc +2ca – a2 – b2 – c2 + 2ab – 2bc +2ca
= 4ac – 4bc
= 4c (a – b)
(ii) (3x + 4y + 5)2 – (x + 5y – 4)2
Using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc +2ca
(3x + 4y + 5)2 – (x + 5y – 4)2
= [(3x)2 + (4y)2 + 52 + 2.3x.4y + 2.4y.5 + 2.5(3x)] – [x2 + (5y)2 + (-4)2 +
1. X.5y + 2.5y (-4) + 2 (-4). x]
= 9x2 + 16y2 + 25 + 24xy + 40y + 30x – [x2 + 25y2 + 16 + 10xy – 40y-8x
= 9x2 + 16y2 + 25 + 24xy + 40y + 30x – x2 – 25y2 – 16 – 10xy + 40y + 8x
= 8x2 – 9y2 + 14xy + 80y + 38x + 9
(iii) (2m – n – 3p)2 + 4mn – 6np + 12pm
Using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc +2ca we get
(2m – n – 3p)2 + 4mn – 6np + 12pm
= (2m)2 + (–n)2 + (-3p)2 + 2.2m(–n) + 2(n)(–3p) + 2 (-3p)(2m) + 4mn –
6np + 12pm
= 4m2 + n2 + 9p2 – 4mn – 6np – 12pm + 4mn – 6np + 12pm
= 4m2 + n2 + 9p2
(iv) (x + 2y + 3z + r)2 + (x + 2y + 3z – r)2
Using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc +2ca we get
a = (x + 2y) b = 3z c = r
(x + 2y + 3z + r)2 + (x + 2y + 3z – r)2
= (x + 2y)2 + (3z)2 + r2 + 2 (x + 2y)3z + 2.3z.r + 2.r.(x + 2y) + (x + 2y)2 +
(3z)2 – (r)2 + 2 (x + 2y) 3z + 2.3z (-r) + 2 (-r)(x + 2y)
= 2(x + 2y)2 + 9z2 + r2 + 6 (x + 2y)z + 6zr + 2r (x + 2y) + 9z2 + r2 +
6 (x + 2y) z – 6zr – 2r (x + 2y)
= 2(x2 + 2.x.2y +4y2) + 18z2 + 2r2 + 12 (x + 2y)z
= 2x2 + 8xy + 8y2 + 18z2 + 2r2 +12xz + 24 yz
= 2x2 + 8y2 + 18z2 + 2r2 + 8xy +12xy + 24 yz
1. If a + b + c = 12 and a2 + b2 + c2 = 50, find ab + bc + ca.
Given a + b + c = 12 squaring both sides
(a + b + c)2 = 122
a2 + b2 + c2 + 2ab + 2bc + 2ca = 144
Given a2 + b2 + c2 = 50
50 + 2ab + 2bc + 2ca = 144
2(ab + bc + ca) = 144 – 50
2(ab + bc + ca) = 94
ab + bc + ca = 94/2
ab + bc + ca = 47
1. If a2 + b2 + c2 = 35 and ab + bc + ca = 23, find all possible values of a + b +c.
Given a2 + b2 + c2 = 35 and ab + bc + ca = 23,
(a + b +c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
= a2 + b2 + c2 + 2(ab + bc + ca)
= 35 + 2 (23)
= 35 + 46
(a + b + c)2 = 81
(a + b +c) = ± √81 = ±9
1. Express 4x + 9y + 16z + 12xy – 24yz – 16zx as the square of a trinomial
Using and comparing the coefficient of
(a +b+c)2 = a2 + b2 + c2 + 2ab + 2bc +2ca we get
4x + 9y + 16z + 12xy – 24yz – 16zx
= (2x)2 + (3y)2 + (–4z2) + 2.2x.3y.(–4z) + 2.(–4z) + 2.(–4z).2x
= (2x + 3y – 4z)2
1. If x and y are real numbers and satisfy the equation
(2x + 3y – 4z)2 + (5x – y – 4)2 = 0 find x, y.
[Hint: If a, b are real numbers such that a2 + b2 = 0, then a = b = 0.]
Given if a2 + b2 = 0, then a = b = 0
(2x + 3y – 4z)2 = 0 and (5x – y – 4)2 = 0
(2x + 3y = 5)x1 i.e., 2x + 3y = 5————-(1)
(5x – y = 4)x3 i.e., 15x – 3y = 12—————-(2)
Therefore,
2x + 3y = 5
15x – 3y = 12
—————
17x = 17
—————
x = 1
substitute the value of x in (1)
2(1)+3y = 5
2+ 3y = 5
3y = 5 – 2 = 3
y = 3/3 = 1
Thus, x = 1 and y = 1
## 3.1.5 Conditional Identity – Multiplication of Polynomials
Example :
If x + y = 3, prove that x3 + y3 + 3ab(a + b)
Solution:
We know that (x + y)3 = x3 + y3 + 3xy(x + y) . Substitute x + y = 3 in this identity. We get,
27 = x3 + y3 + 3xy(3) = x3 + y3 + 9xy
## Multiplication of Polynomials – Exercise 3.1.5
1. If a + b + c = 0, prove the following:
(i) (b + c) (b – c) + a (a + 2b) = 0
Given a + b + c = 0 the a + b = -c, b + c = -a, c + a = -b we have
L.H.S = (b + c) (b – c) + a (a + 2b)
= (-a) (b – c) + a (a + b + b)
= -ab + ac + a (-c + b)
= -ab + ab – ac + ac
= 0= R.H.S
(ii) a (a2 – bc) + b (b2 – c) + c (c2 – ab) = 0
L.H.S = a (a2 – bc) + b (b2 – c) + c (c2 – ab)
= a3 – abc + b3 – abc + c3 – abc
= a3 + b3 + c3 – 3abc
We know that if a + b + c = 0 then
a3 + b3 + c3 = 3abc
Hence we have
= 3abc – 3abc
= 0 = R.H.S
(iii) a (b2 + c2) + b (c2 + a2) + c (a2 + b2) = –3abc
L.H.S = a (b2 + c2) + b (c2 + a2) + c (a2 + b2)
= ab2 + ac2+ bc2 + ba2 + ca2 +cb2
= ab2 + ba2 + b2c + bc2 + ac2 + a2c
= ab (a + b) + bc (b + c) + ac (a + c)
= ab (–c) + bc (–a) + ac (–b)
= –abc – abc – abc
= –3abc = R.H.S
[a + b + c =0, a + b = -c, b + c = -a, c + a = -b]
(iv) (ab + bc + ca)2 = a2b2 + b2c2 + c2a2
L.H.S = (ab + bc + ca)2
= (ab)2 + (bc)2 + (ca)2 +2ab.bc + 2bc. ca + 2ca.ab
= a2b2 + b2c2 + c2a2 + 2ab2c + 2bc2a + 2ca2b
= a2b2 + b2c2 + c2a2 + 2abc + (0)
= a2b2 + b2c2 + c2a2 = R.H.S
(v) a2 – bc = b2 – ca = c2 – ab = – (ab + bc + ca)
a.a – bc = a(-b-c)-bc = -ab-ac-ba = -(ab+bc+ca)——(1)
b2 – ca = b.b – ca = b(-c-a)-ca = -bc-ab-ca = -(ab+bc+ca) ———-(2)
c2 – ab = c.c – ab = c(-a-b) – ab = -ac-bc-ab = -(ab+bc+ca) ———-(3)
From equation (1) (2) and (3)
a2 – bc = b2 – ca = c2 – ab = – (ab + bc + ca)
(vi) 2a2 + bc = (a – b) (a –c)
L.H.S = 2a2 + bc
= a2 + a2 + bc
= a2 + a x a + bc
= a2 + a (- b – c) + bc
= a2 – ab – ac + bc
= a (a – b) – c (a – b)
= (a – b) (a –c) = R.H.S
(vii) (a + b) (a – b) + ca – cb = 0
We have a + b + c = 0
a + b = c
L.H.S = (a + b) (a – b) + ac – cb
= –c (a – b) + ac – cd
= – ca + bc + ac – cb
= 0 = R.H.S
(viii) a2 + b2 + c2 = -2(ab + bc + ca)
We have a + b + c = 0
Squaring we get
(a + b + c)2 = 0
a2 + b2 + c2 + 2ab + 2bc +2ca = 0
a2 + b2 + c2 = – 2ab – 2bc – 2ca
a2 + b2 + c2 = – 2(ab + bc + ca)
Hence the proof
1. Suppose a, b, c are non-zero real numbers such that a + b + c = 0,
Prove the following:
(i) 𝐚𝟐/𝐛𝐜 + 𝐛𝟐/𝐜𝐚 + 𝐜𝟐/𝐚𝐛 = 3
L.H.S = a2/bc + b2/ca + c2/ab
= (a2.a+b2.b+c2.c)/abc
= (a3+b3+c3)/abc ………. (1)
We have a + b + c = 0, a + b = –c
Cubing we get
(a + b)3 = (–c)3
a3 + b3 + 3ab + (a + b) = –c3
a3 + b3 – 3ab = –c3
a3 + b3 + c3 = 3abc ………. (2)
Substituting (2) in (1)
L.H.S = 3abc/abc = 3
(ii) ( +𝐛/𝐜 + 𝐛+𝐜/a + 𝐜+𝐚/𝐛 ) ( 𝐛/𝐜+𝐚 + 𝐜/𝐚+𝐛 + 𝐚/𝐛+𝐜 )
Whenever b + c ≠ 0, c + a ≠ 0, a + b ≠ 0
We have a + b + c =0
a + b = –c
b + c = –a
c + a = –b
L.H.S = (𝐚+𝐛/𝐜 + 𝐛+𝐜/a + 𝐜+𝐚/ ) ( 𝐛/𝐜+𝐚 + 𝐜/𝐚+𝐛 + 𝐚/𝐛+𝐜 )
= ( −c/c + −a/a + −a/b ) ( b/−b + c/−c + a/−a )
= (-1-1-1) (-1-1-1)
= (-3) (-3)
= 9 = R.H.S
(iii) 𝐚𝟐/𝟐𝐚𝟐+𝐛𝐜 + 𝐛𝟐/𝟐𝐛𝟐𝐜𝐚 + 𝐜𝟐/𝟐𝐜𝟐𝐚𝐛 = 1, provided the denominators do not become 0.
L.H.S = 𝐚𝟐/𝟐𝐚𝟐+𝐛𝐜 + 𝐛𝟐/𝟐𝐛𝟐𝐜𝐚 + 𝐜𝟐/𝟐𝐜𝟐𝐚𝐛
= 𝐚𝟐/(a-b)(a-c) + 𝐛𝟐/(b-c)(b-a) + 𝐜𝟐/(c-a)(c-b)
= 𝐚𝟐/(a-b)(a-c) – 𝐛𝟐/(a-b)(a-c + 𝐜𝟐/(a-c)(b-c)
= [a2(b−c)– b2(a−c) + c2(a−b)] /(a−b)(b−c)(a−c)
= [a2b− a2c − b2a + b2c + c2a− c2b] /(ab− b2− ac +bc) (a−c)
= [a2b− a2c − b2a + b2c + c2a− c2b]/[a2b − ab2− a2c + abc − abc + b2c + ac2]
= 1 = R.H.S
1. If a + b + c = 0, prove that b2 – 4ac is a square.
We have a + b + c = 0
b = – (a + c)
Squaring on both sides
b2 = [- (a + c)]2
= (a + c)2
b2 = a2 + c2 + 2ac
Subtracting 4ac on both sides
b2 – 4ac = a2 + c2 + 2ac – 4ac
= a2 – 2ac + c2
b2 – 4ac = (a – c)2
We find that b2 – 4ac is the square of (a – c)
1. If a, b, c are real numbers such that a + b + c = 2s, prove the following:
(i) s (s – a) + s (s – b) + s (s – c) = s2
L.H.S. = s (s – a) + s (s – b) + s (s – c)
= s2 – as + s2 – bs + s2 – cs
= 3s2 – as – bs – cs
= 3s2 – s (a + b + c)
= 3s2 – s (2s) (a + b + c = 2s)
= 3s2 – 2s2
= s2 = R.H.S.
(ii) s2 (s – a)2 + s (s – b)2 + s (s – c)2 = a2 + b2 + c2
L.H.S. = s2 (s – a)2 + s (s – b)2 + s (s – c)2
= s2 + s2 + a22sa + s2 + b2 + s2 + c2 – 2as – 2bs – 2cs
= 4s2 + a2 + b2 + c2 – 2s – 2bs – 2cs
= 4s2 + a2 + b2 + c2 – 2s (a + b + c)
= 4s2 + a2 + b2 + c2 – 2as (2s) (a + b + c = 2s)
= 4s2 + a2 + b2 + c2 – 4s2
= a2 + b2 + c2
= R.H.S.
(iii) (s – a) (s – b) + (s – b) (s – c) + (s – c) (s – a) + s2 = ab + bc + ca
L.H.S. = (s – a) (s – b) + (s – b) (s – c) + (s – c) (s – a) + s2
= s2 – as – bs + ab + s2 – bs – cs + bc + s2 – cs – as + ac + s2
= 4s2 – 2 as – 2bs – 2cs + ab + bc +ca
= 4s2 – 2s (a + b + c) + ab + bc +ca
= 4s2 – 2s (2s) + ab + bc +ca
= 4s2 – 4s2 + ab + bc +ca (a + b + c = 2s)
= ab + bc +ca
= R.H.S.
(iv) a2 – b2 – c2 + 2bc = 4 (s – b) ( s – c)
L.H.S = a2 – b2 – c2 + 2bc
= a2 – (b2 + c2 – 2bc)
= a2 – (b – c)2
= (a + b – c) [a – (b – c)]
= (a + b – c) (a + b – c)
= (2s – c – c) (2s – b – b)
= (2s – 2c) (2s – 2b)
= 2(s – c) (2) (s – b)
= 4 (s – c) (s – b)
= R.H.S.
1. If a, b, c are real numbers, a + b + c =2s and s – a ≠ 0, s – b ≠ 0, s – c ≠ 0,
Prove that 𝐚/(𝐬𝐚) + 𝐛/(𝐬𝐛) + 𝐜/(𝐬𝐜) + 2 = 𝐚𝐛𝐜/(𝐬𝐜)( 𝐬𝐛) (𝐬𝐜)
L.H.S = a/(s−a) + b/(s−b) + c/(s−c) + 2
=a(s−b)( s−c) + b(s−a)(s−c) + c(s−a)( s−b) + 2(s−a)(s−b)(s−c)/ (s−a)(s−b)(s−c)
= [a(s-b)(s-c)+b(s-a)(s-c) +c(s-a)(s-b)+2(s-a)(s-b)(s-c)] /(s−a )(s−b)(S−c)
=[a(s2−bs−cs+bc) +b(s2−as−cs+ac) +c(s2−as−bs+ab) +2(s2−as−bs+ab)(s−c)] /(s−a)(s−b)(S−c)
= [as2−abs−acs+abc+bs2−abc−bcs+abc) +cs2−acs−bcs+abc+2(s3−as2−bs2+abs −cs2+acs+bcs−abc] /(s−a)(s−b)(s−c)
= [s2(a+b+c) − 2abc −2acs − 2bcs − 3abc + 2s3−2as2− 2bs2 + 2abs−2cs2+ 2acs+2bcs−2abc] /(s−a)(s−b)(s−c)
= [s2(2s) + abc+2s3−2s2 (a+b+c)]/(s−a)(s−b)(S−c)
=[2s3+ abc+2s3−2s2(2s)]/(s−a) s−b (S−c)
=[4s3+ abc−4s3]/(s−a)(s−b)(S−c)
= abc/(s−a)(s−b)(S−c)
= R.H.S.
1. If a + b + c = 0, prove that a2 – bc = b2 – ca = c2 – ab = (𝐚𝟐 + 𝐛𝟐+𝐜𝟐)/𝟐
We have a + b + c = 0
Squaring will get
(a + b + c)2 = 0
a2 + b2 + c2 + 2ab + 2bc + 2ca = 0
a2 + b2 + c2 + 2b (a + c) + 2ca = 0
a2 + b2 + c2 + 2b (-b) + 2ca = 0
a2 + b2 + c2 = 2b2 – 2ca (hint: a + c = -b)
a2 + b2 + c2 = 2 (b2 – ca)
a2 + b2 + c2 = b2 – ca ……… (1)
IIIly (a + b + c)2 = 0
a2 + b2 + c2 + 2ab + 2bc + 2ca = 0
a2 + b2 + c2 + 2ab + 2c (b + a) = 0
a2 + b2 + c2 + 2ab + 2c (-c) = 0 (a + c = -c)
a2 + b2 + c2 = 2 (c2 – ab)
[a2 + b2 + c2]/2 = (c2 – ab)……………(2)
Also a2 + b2 + c2 + 2ab + 2bc + 2ca = 0
a2 + b2 + c2 + 2a (b + c) + 2bc = 0
a2 + b2 + c2 + 2a (-a) + 2bc = 0
a2 + b2 + c2 = 2 (a2 – bc) ……..(3)
From (1), (2) and (3) we get,
a2 – bc = b2 – ca = c2 – ab = [a2 + b2 + c2]/2
1. If 2(a2 + b2) = (a + b)2, prove that a = b.
2a2 + 2b2 = a2 + b2 + 2ab
2a2 + 2b2 – a2 – b2 – 2ab = 0
a2 + b2 – 2ab = 0
(a – b)2 = 0
a – b =0
a = b
1. If x2 – 3x + 1 = 0, prove that x2 + 𝟏/𝐱𝟐 = 7.
Given x2 – 3x + 1 = 0
x2 + 1 = 3x
x + 1/x = 3 (dividing both sides by x)
Squaring both sides we get
(x + 1/x )2 = 32 = x2 + 1/x 2 + 2x. 1/x = 9
x2 + 1/x 2 + 2 = 9
x2 + 1/x 2 = 9 – 2 = 7
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# Sequence and series questions and solutions pdf
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## Chapter 4 : Series and Sequences
### NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series
This book aims to dispel the mystery and fear experienced by students surrounding sequences, series, convergence, and their applications. The author, an accomplished female mathematician, achieves this by taking a problem solving approach, starting with fascinating problems and solving them step by step with clear explanations and illuminating diagrams. The reader will find the problems interesting, unusual, and fun, yet solved with the rigor expected in a competition. Some problems are taken directly from mathematics competitions, with the name and year of the exam provided for reference. Proof techniques are emphasized, with a variety of methods presented.
Note that some sections will have more problems than others and some will have more or less of a variety of problems. Most sections should have a range of difficulty levels in the problems although this will vary from section to section. Here is a list of all the sections for which practice problems have been written as well as a brief description of the material covered in the notes for that particular section. Sequences — In this section we define just what we mean by sequence in a math class and give the basic notation we will use with them. We will focus on the basic terminology, limits of sequences and convergence of sequences in this section. More on Sequences — In this section we will continue examining sequences. We will determine if a sequence in an increasing sequence or a decreasing sequence and hence if it is a monotonic sequence.
## Sequences And Series Class 11 NCERT Solutions
Write the first five terms of each of the sequences in Exercises 1 to 6 whose nth terms are:. Find the indicated terms in each of the sequences in Exercises 7 to 10 whose nth terms are:. Find the sum of all natural numbers lying between and , which are multiples of 5. In an A. Show that 20th term is —11 2.
So, you can estimate the limit to be 2. The materials are organized by chapter and lesson, with one Word Problem Practice worksheetfor every lesson in Glencoe Math Connects, Course 2. Exercise tests were performed up to maximal effort. Developed on a foundation of problem-solving and assessment, this differentiated course stretches and challenges students of all levels. These questions make suitable bridging material for students with single A-level Mathematics as they begin university - the material is partly revision, partly new material. Resolving these problems The resolution of these problems is accomplished by the use of limits.
Notice that the nth term of the sequence is ar n−1. In the chessboard problem the solution involved adding up the first 64 terms. The sum of the first n terms of a.
### 50+ Sequence and Series Questions with Solution Free PDF For SSC, RRB, FCI Exams – Download Now
Thus, a sequence is a function f: N! See full list on byjus. Infinite Sequence: A sequence, which is not finite, is an infinite sequence. To take just one hair-raising example, suppose the unknown sequence is 2, 3, 5, 7, 11, 13, 17, 19,,wherea nis the nth prime number.
Find the missing number in the following series. Look at this series: 3, 4, 7, 8, 11, 12,. What number should come next? This alternating addition series begins with 3; then 1 is added to give 4; then 3 is added to give 7; then 1 is added, and so on.
Most of the students primarily focus on getting full marks on Mathematics since there is a scope for that if one can master the core concepts.
### Chapter 4 : Series and Sequences
Where is the first term and d is the common difference. Step 2 : The initial term is and common difference is. The n th term of an arithmetic sequence is. Substitute and in above equation. Step 3 : Substitute in
Each page includes appropriate definitions and formulas followed by solved problems listed in order of increasing difficulty. Studying and solving these problems helps you increase problem-solving skills and achieve your personal best on calculus exams! Necessary cookies are absolutely essential for the website to function properly.
#### jee mains Maths chapter Sequences and Series questions with solutions
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# Solutions of a Linear Equation
Go back to 'Linear-Equations'
Consider the following linear equation:
$x + 2y = 4$
Note that $$x = 2$$ and $$y = 1$$ (together) satisfy this equation. We state this fact succinctly by saying that $$\left( {2,1} \right)$$ is a solution of the equation. In general, if $$\left( {p,q} \right)$$ is a solution of the equation $$ax + by + c = 0$$, this means that $$x = p$$ and $$y = q$$ satisfy the equation, that is, $$ap + bq + c = 0$$.
Example 1: Which of the following are solutions of the equation $$3x - 4y + 1 = 0$$?
(A) $$\left( {1,2} \right)$$
(B) $$\left( {5,4} \right)$$
(C) $$\left( { - 3, - 2} \right)$$
(D) $$\left( {2,7} \right)$$
(E) $$\left( {1,1} \right)$$
Solution: Let’s consider option (A). If we use $$x = 1,\,\,y = 2$$ in the given equation, we see that it is not satisfied:
$3\left( 1 \right) - 4\left( 2 \right) + 1 \ne 0$
Thus, $$\left( {1,2} \right)$$ is not a solution for the given equation.
Now, consider option (B). We use $$x = 5,\,\,y = 4$$ and find that
$3\left( 5 \right) - 4\left( 4 \right) + 1 = 0$
Thus, $$\left( {5,4} \right)$$ is a solution for the given equation.
Similarly, you can verify that options (C) and (E) are valid solutions for the given equation, but option (D) is not.
Example 2: Write three solutions of the equation $$x - 2y + 3 = 0$$.
Solution: If we substitute $$x = 0$$, we obtain $$y = \frac{3}{2}$$. Thus, $$\left( {0,\frac{3}{2}} \right)$$ is a solution of this equation.
Substituting $$y = 0$$ gives $$x = - 3$$, which means that $$\left( { - 3,0} \right)$$ is another possible solution.
Substituting $$x = 1$$ gives $$y = 2$$, so that $$\left( {1,2} \right)$$ is yet another solution.
By now, it should be obvious to you that a linear equation in two variables will have infinitely many solutions. Substituting a particular value for one variable gives us a particular value for the other variable, and hence a possible solution.
Example 3: Write two solutions of the equation $$2x - 3y + 1 = 0$$ in which the value of x is a negative non-integer rational number.
Solution: We can use any two negative rational values for x. Substituting $$x = - \frac{1}{2}$$ gives:
\begin{align}&2\left( { - \frac{1}{2}} \right) - 3y + 1 = 0\\& \Rightarrow \,\,\, - 1 - 3y + 1 = 0\\& \Rightarrow \,\,\,y = 0\end{align}
Thus, $$\left( { - \frac{1}{2},0} \right)$$ is a solution. Substituting $$x = - \frac{2}{5}$$ gives:
\begin{align}&2\left( { - \frac{2}{5}} \right) - 3y + 1 = 0\\ &\Rightarrow \,\,\, - \frac{4}{5} - 3y + 1 = 0\\ &\Rightarrow \,\,\,y = \frac{1}{{15}}\end{align}
Thus, $$\left( { - \frac{2}{5},\frac{1}{{15}}} \right)$$ is another possible solution.
Example 4: Write two solutions of the equation $$- \frac{1}{2}x + \sqrt 2 y = \sqrt 3$$ in which the value of y is an irrational number.
Solution: We can take any two irrational values for y. Substituting $$y = \sqrt 2$$ gives:
\begin{align}& - \frac{1}{2}x + \sqrt 2 \left( {\sqrt 2 } \right) = \sqrt 3 \\& \Rightarrow \,\,\, - \frac{1}{2}x + 2 = \sqrt 3 \\& \Rightarrow \,\,\,x = 2\left( {2 - \sqrt 3 } \right)\end{align}
Thus, $$\left( {2\left( {2 - \sqrt 3 } \right),\sqrt 2 } \right)$$ is one such solution. Next, substituting $$y = \sqrt 3$$ gives:
\begin{align}& - \frac{1}{2}x + \sqrt 2 \left( {\sqrt 3 } \right) = \sqrt 3 \\ &\Rightarrow \,\,\, - \frac{1}{2}x + \sqrt 6 = \sqrt 3 \\& \Rightarrow \,\,\,x = 2\left( {\sqrt 6 - \sqrt 3 } \right)\end{align}
Thus, $$\left( {2\left( {\sqrt 6 - \sqrt 3 } \right),\sqrt 3 } \right)$$ is another such solution.
Example 5. If $$\left( { - 2,3} \right)$$ is a solution of the equation $$4x - ky = 1$$, what is the value of k?
Solution: $$x = - 2,\,\,y = 3$$ must satisfy the given equation:
$\begin{array}{l}4\left( { - 2} \right) - k\left( 3 \right) = 1\\ \Rightarrow \,\,\, - 8 - 3k = 1\\ \Rightarrow \,\,\,k = - 3\end{array}$
Example 6: If $$\left( {2, - t} \right)$$ is a solution of $$ax + k = by$$, what is the value of t?
Solution: $$x = 2,\,\,y = - t$$ must satisfy the given equation:
\begin{align}&a\left( 2 \right) + k = b\left( { - t} \right)\\& \Rightarrow \,\,\,t = - \left( {\frac{{2a + k}}{b}} \right)\end{align}
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# Solve three simultaneous equations with three unknowns
$0 = 9a + 3b + c$
$2 = 25a + 5b + c$
$6 = 49a + 7b + c$
These are my 3 equations that have 3 unknowns. How do I solve for these unknowns?
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What does education/quadtric-forms/problem-solving/quadratics have anything to do with this? – Inquest Oct 18 '12 at 23:53
@Inquest See my answer. – Bill Dubuque Oct 19 '12 at 3:08
(1) $9a+3b+c=0$
(2) $25a+5b+c=2$
(3) $49a+7b+c=6$
Using (2)-(1) we have:
(4) $16a+2b=2$
Using (3)-(2) we have:
(5) $24a+2b=4$
Using (5)-(4) we have:
(6) $8a=2$
It follows that $a=\frac{1}{4}$, $b=-1$ and $c=\frac{3}{4}$.
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Since you had tagged your question "quadratic" and "quadratic-forms" (removed by another user) I will interpret that as a hint provided to you to exploit the quadratic structure of the problem. Notice that the problem can be viewed as interpolating a quadratic through $3$ points, as follows.
Let $\rm\:f(x) = a\,x^2 + b\,x + c$
$\rm 0 = f(3)\:\Rightarrow\: f(x) = (x-3)(ax-d),\$ where $\rm\,\ 3d = c = f(0)$
$\rm 2 = f(5)\, =\, 2(5a-d)\:\Rightarrow\:5a-d = 1$
$\rm 6 = f(7)\, =\, 4(7a-d)\:\Rightarrow\:7a-d = 3/2\:\Rightarrow\:2a = \frac{1}2\:\Rightarrow\:a=\frac{1}4$
So $\rm\: d = 5a\!-\!1 = \frac{1}4,\:$ $\rm\:c = 3d = \frac{3}4,\:$ $\rm\: b = -3a\!-\!d = -\frac{3}4-\frac{1}4 = -1$
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# RD Sharma Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.6
The linear pair of equations in two variables has got many applications in science and statistics. This is to say the applications to word problems are enormous and extensively used in everyday life. The RD Sharma Solutions Class 10 are vital tools for students to acquire strong conceptual knowledge and boost their confidence in taking tests. Students can refer to the RD Sharma Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.6 PDF given below for any help regarding this exercise.
## RD Sharma Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.6
### Access RD Sharma Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.6
1. 5 pens and 6 pencils together cost 9, and 3 pens and 2 pencils cost 5. Find the cost of 1 pen and 1 pencil.
Solution:
Let’s assume the cost of a pen and pencil is ₹ x and ₹ y, respectively.
Then, forming equations according to the question,
5x + 6y = 9 … (i)
3x + 2y = 5 … (ii)
On multiplying equation (i) by 2 and equation (ii) by 6, we get
10x + 12y = 18 … (iii)
18x + 12y = 30 … (iv)
Now by subtracting equation (iii) from equation (iv), we get
18x – 10x + 12y – 12y = 30 – 18
8x = 12
x = 3/2 = 1.5
Putting x = 1.5 in equation (i), we find y.
5(1.5) + 6y = 9
6y = 9 – 7.5
y = (1.5)/ 6 = 0.25
Therefore, the cost of one pen = ₹ 1.50, and so the cost of one pencil = ₹ 0.25
2. 7 audio cassettes and 3 videocassettes cost 1110, while 5 audio cassettes and 4 videocassettes cost 1350. Find the cost of audio cassettes and a video cassette.
Solution:
Let’s assume the cost of an audio cassette, and that of a video cassette be ₹ x and ₹ y, respectively. Then, forming equations according to the question, we have
7x + 3y = 1110 … (i)
5x + 4y = 1350 … (ii)
On multiplying equation (i) by 4 and equation (ii) by 3,
We get
28x + 12y = 4440 … (iii)
15x + 4y = 4050 … (iv)
Subtracting equation (iv) from equation (iii),
28x – 13x + 12y – 12y = 4440 – 4050
13x = 390
⇒ x = 30
On substituting x = 30 in equation (I),
7(30) + 3y = 1110
3y = 1110 – 210
y = 900/ 3
⇒ y = 300
Therefore, it’s found that the cost of one audio cassette = ₹ 30
And the cost of one video cassette = ₹ 300
3. Reena has pens and pencils, which together are 40 in number. If she has 5 more pencils and 5 less pens, then the number of pencils would become 4 times the number of pens. Find the original number of pens and pencils.
Solution:
Let’s assume the number of pens and pencils is x and y, respectively.
Forming equations according to the question, we have
x + y = 40 … (i)
(y+5) = 4(x-5)
y + 5 = 4x – 20
5 + 20 = 4x – y
4x – y = 25 … (ii)
Adding equations (i) and (ii),
We get
x + 4x = 40 + 25
5x = 65
⇒ x = 13
Putting x=13 in equation (i), we get
13 + y = 40
⇒ y = 40 – 13 = 27
Therefore, it’s found that the number of pens Reena has is 13.
And the number of pencils Reena has is 27.
4. 4 tables and 3 chairs, together, cost 2250, and 3 tables and 4 chairs cost 1950. Find the cost of 2 chairs and 1 table.
Solution:
Let’s assume the cost of 1 table is ₹ x, and the cost of 1 chair is ₹ y.
Then, according to the question,
4x + 3y = 2250 … (i)
3x + 4y = 1950 … (ii)
On multiplying (i) with 3 and (ii) with 4,
We get,
12x + 9y = 6750 … (iii)
12x + 16y = 7800 … (iv)
Now, subtracting equation (iv) from (iii),
We get,
-7y = -1050
y = 150
Using y = 150 in (i), we find x
4x + 3(150) = 2250
4x = 2250 – 450
x = 1800/ 4
⇒ x = 450
From the question, it’s required to find the value of (x + 2y) ⇒ 450 + 2(150) = 750
Therefore, the total cost of 2 chairs and 1 table is ₹ 750.
5. 3 bags and 4 pens together cost 257, whereas 4 bags and 3 pens together cost 324. Find the total cost of 1 bag and 10 pens.
Solution:
Let the cost of a bag and a pen be ₹ x and ₹ y, respectively.
Then, according to the question,
3x + 4y = 257 … (i)
4x + 3y = 324 … (ii)
On multiplying equation (i) by 3 and (ii) by 4,
We get,
9x + 12y = 770 … (iii)
16x + 12y = 1296 … (iv)
Subtracting equation (iii) from (iv), we get
16x – 9x = 1296 – 771
7x = 525
x = 525/7 = 75
Hence, the cost of a bag = ₹ 75
Substituting x = 75 in equation (i),
We get,
3 x 75 + 4y = 257
225 + 4y = 257
4y = 257 – 225
4y = 32
y = 32/4 = 8
Hence, the cost of a pen = ₹ 8
From the question, it’s required to find the value of (x + 10y) ⇒ 75 +10(8) = 20
Therefore, the total cost of 1 bag and 10 pens = 75 + 80 = ₹ 155
6. 5 books and 7 pens together cost 79, whereas 7 books and 5 pens together cost 77. Find the total cost of 1 book and 2 pens.
Solution:
Let’s assume the cost of a book and a pen be ₹ x and ₹ y, respectively.
Then, according to the question,
5x + 7y = 79 … (i)
7x + 5y = 77 … (ii)
On multiplying equation (i) by 5 and (ii) by 7,
We get,
25x + 35y = 395 … (iii)
49x + 35y = 539 … (iv)
Subtracting equation (iii) from (iv),
We have,
49x – 25x = 539 – 395
24x = 144
x = 144/24 = 6
Hence, the cost of a book = ₹ 6
Substituting x= 6 in equation (i),
We get
5 (6) + 7y = 79
30 + 7y = 79
7y = 79 – 30
7y = 49
y = 49/ 7 = 7
Hence, the cost of a pen = ₹ 7
From the question, it’s required to find the value of (x + 2y) ⇒ 6 + 2(7) = 20
Therefore, the total cost of 1 book and 2 pens = 6 + 14= ₹ 20
7. Jamila sold a table and a chair for 1050, thereby making a profit of 10% on the table and 25% on the chair. If she had taken a profit of 25% on the table and 10% on the chair, she would have got 1065. Find the cost price of each.
Solution:
Let the cost price of one table and one chair be ₹ x and ₹ y, respectively.
So,
The selling price of the table, when it’s sold at a profit of 10% = ₹ x + 10x/100 = ₹ 110x / 100
The selling price of the chair, when it’s sold at a profit of 25% = ₹ y + 25y/100 = ₹ 125y / 100
Hence, according to the question,
110x / 100 + 125y / 100 = 1050 … (i)
Similarly,
The selling price of the table, when it’s sold at a profit of 25% = ₹ (x + 25x/100) = ₹ 125x/ 100
The selling price of the chair, when it’s sold at a profit of 10% = ₹ (y + 10y/100) = ₹ 110y / 100
Hence, again from the question
125x / 100 + 110y / 100 = 1065 … (ii)
Re-written (i) and (ii) with their simplest coefficients,
11x/10 + 5y/4 = 1050…….. (iii)
5x/4 + 11y/10 = 1065…….. (iv)
Adding (iii) and (iv), we get
(11/ 10 + 5/ 4)x + (5/ 4 + 11/ 10)y = 2115
47/ 20x + 47/ 20y = 2115
x + y = 2115(20/ 47) = 900
⇒ x = 900 – y ……. (v)
Using (v) in (iii),
11(900 – y)/10 + 5y/4 = 1050
2(9900 -11y) +25y = 1050 x 20 [After taking LCM]
19800 – 22y + 25y = 21000
3y = 1200
⇒ y = 400
Putting y = 400 in (v), we get
x = 900 – 400 = 500
Therefore, the cost price of the table is ₹ 500, and that of the chair is ₹ 400.
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Centers of a Triangle
by Brant Chesser
We want to review some basic geometry before we go into more depth about it in a future assignment. Today, we want to explore the Orthocenter of a triangle. The Orthocenter of a triangle is the common intersection of the three lines containing the altitudes. An altitude is a perpendicular segment from a vertex to the line of the opposite side. The Orthocenter(H) lies on the lines extended from the altitudes. We want to explore the location of the Orthocenter for different kinds of triangles. We will see where it is for an acute triangle, a right triangle, and an obtuse triangle. Do you have any idea where the Orthocenter might be when we have a right triangle? Well lets explore and find out exactly where it will be.
Lets first construct the Orthocenter by finding the common intersection of the three lines containing the altitudes for an acute triangle. It will look like this when we construct it:
We find that the Orthocenter (H) is inside of the triangle for an acute triangle. So lets explore where it will be if all of the angles are close to sixty degrees. It will then look like the following:
This gives us an Orthocenter location of close to the center of the triangle. We asked earlier where it would be located for a right triangle, so lets now explore this option. The triangle then will look like this:
So the location for the Orthocenter for a right triangle is on the vertex. So what happens to the Orthocenter if we have an obtuse triangle? It has moved from the center of the triangle to the vertex as we have gone from an acute triangle to a right triangle. So this is what it will look like for an obtuse triangle:
Now we observe that for an obtuse triangle, the Orthocenter will lie outside of the triangle. This will hold for the triangle whenever it is an obtuse triangle. Lets look at one further example. So when the top vertex of the triangle is a right angle we get the following triangle:
As we would expect from our earlier exploration the Orthocenter ( H) lies on the vertex. So the Orthocenter should go outside of the triangle if we make the angle obtuse. So when we make it an obtuse angle this is what the result turns out to be:
As we saw before the Orthocenter is outside of the triangle when the angle is obtuse. When the angle is acute, the Orthocenter lies inside of the triangle. A right triangle remember will have the Orthocenter on the vertex like so:
This will give us a better understanding of the Orthocenter and where it lies depending on the shape of the triangle that we are working with on the given assignment. We can explore futher options involving the Orthocenter and other elements of Geometry as we move through the semester.
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
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# Expressions with One or More Variables
## Evaluate expressions given values for variables.
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Expressions with One or More Variables
What if the paycheck for your summer job were represented by the algebraic expression \begin{align*}10h + 25\end{align*}, where h is the number of hours you work? If you worked 20 hours last week, how could you find the value of your paycheck? After completing this Concept, you'll be able to evaluate algebraic expressions like this one.
### Guidance
When we are given an algebraic expression, one of the most common things we might have to do with it is evaluate it for some given value of the variable. The following example illustrates this process.
#### Example A
Let \begin{align*}x = 12\end{align*}. Find the value of \begin{align*}2x - 7\end{align*}.
Solution:
To find the solution, we substitute 12 for \begin{align*}x\end{align*} in the given expression. Every time we see \begin{align*}x\end{align*}, we replace it with 12.
Note: At this stage of the problem, we place the substituted value in parentheses. We do this to make the written-out problem easier to follow, and to avoid mistakes. (If we didn’t use parentheses and also forgot to add a multiplication sign, we would end up turning \begin{align*}2x\end{align*} into 212 instead of 2 times 12!)
#### Example B
Let \begin{align*}y = -2. \end{align*} Find the value of \begin{align*} \frac {7} {y} - 11 y + 2 \end{align*}.
Solution
Many expressions have more than one variable in them. For example, the formula for the perimeter of a rectangle in the introduction has two variables: length \begin{align*}(l)\end{align*} and width \begin{align*}(w)\end{align*}. In these cases, be careful to substitute the appropriate value in the appropriate place.
#### Example C
The area of a trapezoid is given by the equation \begin{align*} A = \frac{ h } { 2 } (a + b)\end{align*}. Find the area of a trapezoid with bases \begin{align*}a = 10 \ cm\end{align*} and \begin{align*}b = 15 \ cm\end{align*} and height \begin{align*}h = 8 \ cm\end{align*}.
Solution:
To find the solution to this problem, we simply take the values given for the variables \begin{align*}a, \ b,\end{align*} and \begin{align*}h\end{align*}, and plug them in to the expression for \begin{align*}A\end{align*}:
The area of the trapezoid is 100 square centimeters.
Watch this video for help with the Examples above.
### Vocabulary
• When given an algebraic expression, one of the most common things we might have to do with it is evaluate it for some given value of the variable. We substitute the value in for the variable and simplify the expression.
### Guided Practice
Let \begin{align*}x= 3\end{align*} and \begin{align*}y = -2. \end{align*} Find the value of \begin{align*} 3xy + \frac{6}{y}-2x \end{align*}.
Solution
### Practice
Evaluate 1-8 using \begin{align*}a = -3, \ b = 2, \ c = 5,\end{align*} and \begin{align*}d = -4\end{align*}.
1. \begin{align*}2a + 3b\end{align*}
2. \begin{align*}4c + d\end{align*}
3. \begin{align*}5ac - 2b\end{align*}
4. \begin{align*} \frac { 2a } { c - d }\end{align*}
5. \begin{align*} \frac { 3b } { d }\end{align*}
6. \begin{align*} \frac { a - 4b } { 3c + 2d }\end{align*}
7. \begin{align*} \frac { 1 } { a + b }\end{align*}
8. \begin{align*} \frac { ab } {cd }\end{align*}
For 9-11, the weekly cost \begin{align*}C\end{align*} of manufacturing \begin{align*}x\end{align*} remote controls is given by the formula \begin{align*}C = 2000 + 3x\end{align*}, where the cost is given in dollars.
1. What is the cost of producing 1000 remote controls?
2. What is the cost of producing 2000 remote controls?
3. What is the cost of producing 2500 remote controls?
### Vocabulary Language: English
algebraic
algebraic
The word algebraic indicates that a given expression or equation includes variables.
Algebraic Expression
Algebraic Expression
An expression that has numbers, operations and variables, but no equals sign.
Evaluate
Evaluate
To evaluate an expression or equation means to perform the included operations, commonly in order to find a specific value.
Exponent
Exponent
Exponents are used to describe the number of times that a term is multiplied by itself.
Expression
Expression
An expression is a mathematical phrase containing variables, operations and/or numbers. Expressions do not include comparative operators such as equal signs or inequality symbols.
Order of Operations
Order of Operations
The order of operations specifies the order in which to perform each of multiple operations in an expression or equation. The order of operations is: P - parentheses, E - exponents, M/D - multiplication and division in order from left to right, A/S - addition and subtraction in order from left to right.
Parentheses
Parentheses
Parentheses "(" and ")" are used in algebraic expressions as grouping symbols.
substitute
substitute
In algebra, to substitute means to replace a variable or term with a specific value.
Trapezoid
Trapezoid
A trapezoid is a quadrilateral with exactly one pair of parallel opposite sides.
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# FAQ: What Is A Limit In Math?
## What does limit in math mean?
A limit tells us the value that a function approaches as that function’s inputs get closer and closer to some number. The idea of a limit is the basis of all calculus.
## What is meant by limit of a function?
The limit of a function at a point a in its domain (if it exists) is the value that the function approaches as its argument approaches. Informally, a function is said to have a limit L at a if it is possible to make the function arbitrarily close to L by choosing values closer and closer to a.
## What does limit exist mean?
In order for a limit to exist, the function has to approach a particular value. In the case shown above, the arrows on the function indicate that the the function becomes infinitely large. Since the function doesn’t approach a particular value, the limit does not exist.
## How do you find the limit?
Find the limit by rationalizing the numerator In this situation, if you multiply the numerator and denominator by the conjugate of the numerator, the term in the denominator that was a problem cancels out, and you’ll be able to find the limit: Multiply the top and bottom of the fraction by the conjugate.
You might be interested: Question: What Is In Math?
## Can 0 be a limit?
When simply evaluating an equation 0 / 0 is undefined. However, in take the limit, if we get 0 / 0 we can get a variety of answers and the only way to know which on is correct is to actually compute the limit. Once again however note that we get the indeterminate form 0 / 0 if we try to just evaluate the limit.
## What are the limit laws?
The limit of a constant times a function is equal to the constant times the limit of the function. The limit of a product is equal to the product of the limits. The limit of a quotient is equal to the quotient of the limits. The limit of a linear function is equal to the number x is approaching.
## What is the limit formula?
Limits formula:- Let y = f(x) as a function of x. If at a point x = a, f(x) takes indeterminate form, then we can consider the values of the function which is very near to a. If these values tend to some definite unique number as x tends to a, then that obtained a unique number is called the limit of f(x) at x = a.
## What is the formal definition of a limit?
Definition: One-Sided Limits ( Formal ) Limit from the Right: Let f(x) be defined over an open interval of the form (a,b) where a 0, there exists a δ>0, such that if 0Who invented limits?
Englishman Sir Issac Newton and German Gottfried Wilhelm von Leibniz independently developed the general principles of calculus (of which the theory of limits is an important part) in the seventeenth century.
You might be interested: Often asked: What Is The Meaning Of Probability In Math?
## Where does a limit not exist?
If the graph is approaching the same value from opposite directions, there is a limit. If the limit the graph is approaching is infinity, the limit is unbounded. A limit does not exist if the graph is approaching a different value from opposite directions.
## Is Infinity a limit?
When we say in calculus that something is “infinite,” we simply mean that there is no limit to its values. We say that as x approaches 0, the limit of f(x) is infinity. Now a limit is a number—a boundary. So when we say that the limit is infinity, we mean that there is no number that we can name.
## How do you prove a limit does not exist?
To prove a limit does not exist, you need to prove the opposite proposition, i.e. We write limx→2f(x)=a if for any ϵ>0, there exists δ, possibly depending on ϵ, such that |f(x)−a|<ϵ for all x such that |x−2|<δ.
## What is a limit on a graph?
A limit is the value that a function approaches as the input approaches a given value.
## How do you approach a limit problem?
Evaluating Limits
1. Just Put The Value In. The first thing to try is just putting the value of the limit in, and see if it works (in other words substitution).
2. Factors. We can try factoring.
3. Conjugate.
4. Infinite Limits and Rational Functions.
5. L’Hôpital’s Rule.
6. Formal Method.
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###### Carl Horowitz
University of Michigan
Runs his own tutoring company
Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!
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# Set Operation: Intersection - Problem 1
Carl Horowitz
###### Carl Horowitz
University of Michigan
Runs his own tutoring company
Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!
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Intersection of two linear inequalities. Here we have the intersection of these two statements. What we really want to do, is figure out what each of these statements are saying independently, and then piece them together. This just a linear inequality, we solve it as we would anything else.
To solve this out, subtract 2, we end up with x is less than or equal to 5.Still dealing with the union, solve this one out, subtract 2 on the other side. X is greater than or equal to -6.
Whenever I solve a union or an intersection, either one, I always make a number line. So make a number line and then plot both of these regions independently. X is less than or equal to 5. We have the number 5, and we have a closed circle because we are including it and it’s everything less than. I always do above the number line, so I can see exactly what’s happening and I don’t have anything just on top of each other. X is greater than, equal to -6. So here a -6, closed circle and that’s everything greater.
Now we’re looking for the intersection; remember intersection is where they both exist. Where they overlap. We want to see where both regions are represented. Looking over at this region over here, -6 and down, that only has one of these statements. It only has the x is less than or equal to, -5 so both aren’t included, so this region doesn’t count.
Between -6 and 5, they’re both represented so this is part of our intersection. And greater than 5, again we only have one element. So the intersection is where these two overlap so it’s really just this middle region. We’re dealing with from -6 to 5; both end points are included so we include both.
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# Find the sum of to n terms of the sequence $\left( {x + \dfrac{1}{{{x^2}}}} \right),\,\,{\left( {{x^2} + \,\,\dfrac{1}{{{x^2}}}} \right)^2},\,{\left( {{x^2} + \,\,\dfrac{1}{{{x^3}}}} \right)^2},.....$.
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Hint: An itemized collection of terms in which repetition of any sort of series are allowed is known as sequence.
Sum of n terms of such sequence means adding all the ‘n’ terms.
It could be done with various algebraic expressions
Here, ${(a + b)^2} = {a^2} + {b^2} + 2ab$
Further the concept of A.P & G.P series are used to solve such algebraic expressions
An arithmetic progression is a sequence of numbers such that the difference of any two successive members is a constant. If the initial term of A.P is a, and common difference is ‘d’, then nth terms of sequence is
${a^n} = a + (n - 1)d,$ where n $=$ no. of terms.
Sum of A.P is given by ${S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right],$ where n $=$ no. of terms, a $=$ first term, d$=$common difference.
Geometric sequence has a common ratio.
The formula for the nth term is:
${a_n} = a{r^{n - 1}}$
Where, ${a_n} =$ nth term of the sequence
a$=$first term of the sequence
r$=$common ratio.
Sum of n terms of a G.P is denoted by ${S_n}$i.e.
${S_n} = \dfrac{{a\left( {{r^n} - 1} \right)}}{{(r - 1)}}$
Where, a $=$first term of the sequence.
r$\, =$common ratio.
n $=$ no. of terms in the sequence.
Let ${S_n}$ denote the sum to n terms of the given sequence
Then,
${S_n} = {\left( {x + \dfrac{1}{x}} \right)^2} + {\left( {{x^2} + \dfrac{1}{{{x^2}}}} \right)^2} + {\left( {{x^3} + \dfrac{1}{{{x^2}}}} \right)^2} + ....$
First, we need to find the general term of the sequence i.e. ${\left( {{x^n} + \dfrac{1}{{{x^n}}}} \right)^2}$
Now, ${S_n} = {\left( {x + \dfrac{1}{x}} \right)^2} + {\left( {{x^2} + \dfrac{1}{{{x^2}}}} \right)^2} + {\left( {{x^3} + \dfrac{1}{{{x^2}}}} \right)^2} + ....{\left( {{x^n} + \dfrac{1}{{{x^n}}}} \right)^2}$_____ (1).
Using identity ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$ throughout the sequence (1), we have:
${S_n} = \left[ {{x^2} + \dfrac{1}{{{x^2}}} + 2x \times x} \right] + \left[ {{x^4} + \dfrac{1}{{{x^4}}} + 2{x^2} \times \dfrac{1}{{{x^2}}}} \right] + \left[ {{x^6} + \dfrac{1}{{{x^6}}} + 2{x^6} \times \dfrac{1}{{{x^6}}}} \right] + ......\left[ {{x^{2n}} + \dfrac{1}{{{x^{2n}}}} + 2{x^n} \times \dfrac{1}{{{x^n}}}} \right]$
Simplify the series ${x^2}$ is canceled out.
${S_n} = \left[ {{x^2} + \dfrac{1}{{{x^2}}} + 2} \right] + \left[ {{x^4} + \dfrac{1}{{{x^4}}} + 2} \right] + \left[ {{x^6} + \dfrac{1}{{{x^6}}} + 2} \right] + ......\left[ {{x^{2n}} + \dfrac{1}{{{x^{2n}}}} + 2} \right]$
Combining term, we get:
${S_n} = \left[ {{x^2} + {x^4} + {x^6} + ..... + {x^{2n}}} \right] + \left[ {\dfrac{1}{{{x^2}}} + \dfrac{1}{{{x^4}}} + \dfrac{1}{{x{}^6}} + ......\dfrac{1}{{{x^{2n}}}}} \right] + \left[ {2 + 2 + .........2n} \right]$. ______ (2).
We see that $\left[ {{x^2} + {x^4} + {x^6} + ..... + {x^{2n}}} \right]$ is a G.P with common ratio ${x^2}$and a$= {x^2}$
Sum of finite. G.P $= \dfrac{{a\left( {{r^n} - 1} \right)}}{{r - 1}}$$= \dfrac{{{x^2}\left( {{{\left( {{x^2}} \right)}^n} - 1} \right)}}{{{x^2} - 1}}$
$\dfrac{{{x^2}\left[ {{x^{2n}} - 1} \right]}}{{{x^2} - 1}}$ _______(3)
Now,
$\dfrac{1}{{{x^2}}} + \dfrac{1}{{{x^4}}} + ........ + \dfrac{1}{{{x^{2n}}}}$ is a G.P with a $= \dfrac{1}{{{x^2}}}$ and r $= \dfrac{1}{{{x^2}}}$
Sum of finite G.P $= \dfrac{{a\left( {{r^n} - 1} \right)}}{{r - 1}} = \dfrac{{\dfrac{1}{{{x^2}}}\left( {{{\left( {\dfrac{1}{{{x^2}}}} \right)}^n} - 1} \right)}}{{\left( {\dfrac{1}{{{x^2}}}} \right) - 1}}$ ____________(4).
Substituting equation (3) & (4) in (2), we have:
${S_n} = {x^2}\left[ {\dfrac{{{x^{2n}} - 1}}{{{x^2} - 1}}} \right] + \dfrac{1}{{{x^2}}}\left[ {\dfrac{{{{\left( {\dfrac{1}{{{x^2}}}} \right)}^n} - 1}}{{\left( {\dfrac{1}{{{x^2}}}} \right) - 1}}} \right] + \left( {2 + .... + 2} \right)$n times
n times $2$ will be $2n$
${S_n} = {x^2}\left[ {\dfrac{{{x^{2n}} - 1}}{{{x^2} - 1}}} \right] + \dfrac{1}{{{x^2}}}\left[ {\dfrac{{{{\left( {\dfrac{1}{{{x^2}}}} \right)}^n} - 1}}{{\left( {\dfrac{1}{{{x^2}}}} \right) - 1}}} \right] + 2n$
Simplify
${S_n} = {x^2}\left[ {\dfrac{{{x^{2n}} - 1}}{{{x^2} - 1}}} \right] + \dfrac{1}{{{x^2}}}\left[ {\dfrac{{\dfrac{{1 - {x^{2n}}}}{{{x^{2n}}}}}}{{\dfrac{{1 - {x^2}}}{{{x^2}}}}}} \right] + 2n$
Common denominator is cancelled out.
${S_n} = {x^2}\left[ {\dfrac{{{x^{2n}} - 1}}{{{x^2} - 1}}} \right] + \dfrac{1}{{{x^{2n}}}}\left[ {\dfrac{{1 - {x^{2n}}}}{{1 - {x^2}}}} \right] + 2n$
${S_n} = {x^2}\left[ {\dfrac{{{x^{2n}} - 1}}{{{x^2} - 1}}} \right] + \dfrac{1}{{{x^{2n}}}}\left[ {\dfrac{{ - \left( {{x^{2n}} - 1} \right)}}{{ - \left( {{x^2} - 1} \right)}}} \right] + 2n$
Taking $\left[ {\dfrac{{{x^{2n}} - 1}}{{{x^2} - 1}}} \right]$ common from first $2$terms
${S_n} = \left[ {\dfrac{{{x^{2n}} - 1}}{{{x^2} - 1}}} \right] + \left[ {{x^2} + \dfrac{1}{{{x^{2n}}}}} \right] + 2n$
Hence, it is the required sum.
Note: In such a question all we need is to recognize the sequence whether it is a G.P or an A.P.
Also, as it clearly stated we need to find the sum of n terms i.e. we need to find the nth terms first. Arithmetic progression is of the form:- a-d, a, a+d, a+2d,.....
where a is the first term and
d is the common difference.
|
# Matrix Determinant Calculator
Created by Luis Hoyos
Last updated: Feb 18, 2023
If you want to calculate matrix determinants, you're in the right place. This determinant solver calculates the determinant of 4x4, 3x3, and 2x2 matrices.
But what is the importance of determinants? Determinants have many applications, which we'll mention in the following section. For example, solving a 3x3 system of equations is the same as calculating the determinant of a 3x3 matrix. Keep reading to learn more about it!
## Why do we need to calculate matrix determinants?
These are some of the applications of determinants:
• For instance, we can describe systems of linear equations using matrices. The use of Cramer's rule is an example in which we use determinants to solve systems of linear equations.
• When using matrices to describe a linear transformation, it's often better to diagonalize them. We do that by calculating matrix determinants, of course.
• The determinant tells us whether the matrix has an inverse and whether we can approximate that inverse with the Moore-Penrose pseudoinverse.
• We usually need the eigenvalues of the previously mentioned transformation. To obtain them, we also need to calculate matrix determinants.
And why do we need matrices? Well, matrices describe many physical quantities, such as stress, strain, turbulence, or the Mohr's circle.
Well, determinants are important, that's clear. Now, let's see how to calculate them.
## Calculating the determinant of 4x4, 3x3, and 2x2 matrices
The following are the formulas to calculate matrix determinants.
### Determinant of a 2x2 matrix
If
$\scriptsize A = \begin{bmatrix} a_1 & b_1 \\ a_2 & b_2 \end{bmatrix}$
then the determinant of $A$ is
$\footnotesize |A| = a_1b_2 - a_2b_1$
### Determinant of a 3x3 matrix
If
$\scriptsize B = \begin{bmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{bmatrix}$
then, to calculate the determinant of such a 3x3 matrix:
$\scriptsize |B| = a_1b_2c_3 + a_2b_3c_1 + a_3b_1c_2 \\\ \ \ \ \ \ \ \ \ - a_3b_2c_1 - a_1b_3c_2 - a_2b_1c_3$
### Determinant of a 4x4 matrix
Finally:
$\scriptsize C = \begin{bmatrix} a_1 & b_1 & c_1 & d_1 \\ a_2 & b_2 & c_2 & d_2 \\ a_3 & b_3 & c_3 & d_3 \\ a_4 & b_4 & c_4 & d_4 \end{bmatrix}$
then, to calculate the determinant of such a 4x4 matrix:
$\scriptsize |C| = a_1b_2c_3d_4 - a_2b_1c_3d_4 + a_3b_1c_2d_4 - \\\ a_1b_3c_2d_4 + a_2b_3c_1d_4 - a_3b_2c_1d_4 + a_3b_2c_4d_1 - \\\ a_2b_3c_4d_1 + a_4b_3c_2d_1 - a_3b_4c_2d_1 + a_2b_4c_3d_1 - \\\ a_4b_2c_3d_1 + a_4b_1c_3d_2 - a_1b_4c_3d_2 + a_3b_4c_1d_2 - \\\ a_4b_3c_1d_2 + a_1b_3c_4d_2 - a_3b_1c_4d_2 + a_2b_1c_4d_3 - \\\ a_1b_2c_4d_3 + a_4b_2c_1d_3 - a_2b_4c_1d_3 + a_1b_4c_2d_3 - \\\ a_4b_1c_2d_3$
As you can see, finding the determinant of a 3x3 and a 2x2 matrix is relatively easy, but calculating the determinant of a 4x4 matrix is an uphill task. The best option is, undoubtedly, using our determinant solver.
After that, you should visit our other math tools! The vector addition calculator is convenient if you need to deal with vectors.
Luis Hoyos
Matrix size
2x2
A=
⌈ a₁ b₁ ⌉ ⌊ a₂ b₂ ⌋
First row
a₁
b₁
Second row
a₂
b₂
Result
Determinant |A|
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# Shanghai Trip: Primary take-aways
January 14, 2020
### This is the second blog in a series about the recent trip that Ruth, Katie and Jo made to Shanghai, representing the Cambridge Maths Hub. This post focuses on three primary ideas.
Idea 1: Planning for a misconception and how to address it
In a Year 4 lesson students were asked to find how many lines of symmetry a rectangle, square, equilateral triangle and circle have. Students were given pre-cut shapes (manipulatives) and an answer sheet. Many students chose to not use the shapes. Some students made mistakes, but these were not addressed until the teacher went over the answers.
During the discussion about how many lines of symmetry appear on a non-square rectangle, the teacher identified the lines of symmetry one at a time. When the diagonal was chosen as a line of symmetry, she made the students fold a rectangle to see that it did not work. This was an example of students self-correcting their work.
On some occasions, teachers try to avoid misconceptions in case students subsequently remember the wrong answer. In this situation the pupils already had the misconception and they were given a concrete example to help them to correct it.
This image appears in the textbook. The statement under the circle explains that it has many lines of symmetry.
Idea 2: Relationship between 3,6 and 9 times tables.
By Grade 2 (Year 3), the children in Shanghai are fluent in their multiplication facts up to 9 x 9. We found this interesting, considering that we are teaching up to 12 x 12 by the end of Year 4. Their thinking is, like our thinking has been previously, that if you learn up to the ten times table you can use this knowledge to calculate any other multiplication fact (as long as you have a deep understanding of the relationships that are involved).
After Shanghai’s children are fluent with recalling multiplication facts, teaching moves on to expose the relationship between fact families with the ultimate goal of being able to use this knowledge to solve problems. For example, “My house number is a multiple of 3,6 and 9. The number is between 30 and 40. What is my house number?”
After the lesson, in a Teacher Research Group (TRG) discussion, we explained that while we do look at the relationship between the multiples in England, this is often to support children to learn seemingly harder multiplication facts (e.g. doubling a multiple of 3 to find a multiple of 6).
Idea 3: Classifying triangles
We observed a lesson in Grade 3 (Year 4) that focused on classifying triangles. After recapping the previous learning, that a triangle had 3 angles and 3 sides, and having already classified triangles using their angles, the children were given plenty of sticks (of 4 colours/lengths and with plastic joiners) to explore and make different triangles. After a few minutes of exploration, the teacher began to collect triangles in groups. She displayed them on the board in a very specific arrangement (equilateral followed by isosceles, with a space left for scalene triangles). The isosceles triangles were collected in such a way that the children were able to identify the missing triangles, based on the stick pattern that was missing from the display.
After collecting all triangles, the teacher asked the pupils to circle the isosceles triangles. This included the equilateral triangle because isosceles triangles are classified as having AT LEAST two equal sides. This generated discussion amongst the English teachers at the back of the room, as many primary teachers in England classifying triangles into 3 distinct groups (equilateral or isosceles or scalene). This lesson therefore exposed the equilateral triangle as a special type of isosceles triangle in the same way as we see a square as being a special type of rectangle. We referred back to the NCETM website and confirmed that this is also the case in the English curriculum, so it was interesting that not all of the English teachers were aware of this.
The layout of this collection of triangles allowed the pupils to come out and circle the triangles to create what we saw as a Venn diagram.
During the TRG after the lesson, we were able to share the NCETM resource referred to earlier with the Chinese teachers, which they found interesting and were keen to use in future to develop challenge activities. (Go NCETM!) It was great to be involved in such a collaborative discussion, with both groups of teachers learning from each other.
Read other Cambridge Maths Hub blogs here
Katrina White Jo Mills Ruth Colenso
Exning Primary Burrowmoor Primary Saffron Walden County High School
Ruth Colenso Katrina White Jo Mills
Saffron Walden County High School
###### RECENT POSTS
January 21, 2020
December 2, 2019
November 15, 2019
October 1, 2019
September 22, 2019
June 18, 2019
April 30, 2019
March 13, 2019
|
# Dividing Fraction by a Fraction: Worksheets, Online Test and Calculators
Dividing fraction by a fraction worksheets, solved worksheet problems or exercises with step-by-step work, formative assessment as online test, calculator and more learning resources to learn, practice, assess and master the basic math skills of fraction.
Dividing Fraction by a Fraction
Name
Date
1
15/7 ÷ 8/7 =
2
19/5 ÷ 14/7 =
3
7/6 ÷ 18/3 =
4
20/5 ÷ 14/5 =
5
18/17 ÷ 20/17 =
6
20/7 ÷ 19/17 =
7
14/11 ÷ 17/12 =
8
5/12 ÷ 4/12 =
9
19/14 ÷ 9/2 =
10
16/3 ÷ 12/6 =
11
6/5 ÷ 5/4 =
12
15/9 ÷ 7/9 =
13
18/9 ÷ 16/8 =
14
20/3 ÷ 11/9 =
15
5/18 ÷ 19/18 =
16
18/17 ÷ 6/17 =
17
10/3 ÷ 15/9 =
18
12/6 ÷ 20/6 =
19
11/5 ÷ 10/5 =
20
9/5 ÷ 19/14 =
Solved Worksheet Problems
Show All Workout
1
15/7 ÷ 8/7 =
15/8
Steps
157
÷
87
=
157
x
78
15 x 77 x 8
=
10556
simplify the fraction
GCF(105, 56) = 7
10556
=
105 ÷ 756 ÷ 7
=
158
157
÷
87
=
158
2
19/5 ÷ 14/7 =
19/10
Steps
195
÷
147
=
195
x
714
19 x 75 x 14
=
13370
simplify the fraction
GCF(133, 70) = 7
13370
=
133 ÷ 770 ÷ 7
=
1910
195
÷
147
=
1910
3
7/6 ÷ 18/3 =
7/36
Steps
76
÷
183
=
76
x
318
7 x 36 x 18
=
21108
simplify the fraction
GCF(21, 108) = 3
21108
=
21 ÷ 3108 ÷ 3
=
736
76
÷
183
=
736
4
20/5 ÷ 14/5 =
10/7
Steps
205
÷
145
=
205
x
514
20 x 55 x 14
=
10070
simplify the fraction
GCF(100, 70) = 10
10070
=
100 ÷ 1070 ÷ 10
=
107
205
÷
145
=
107
5
18/17 ÷ 20/17 =
9/10
Steps
1817
÷
2017
=
1817
x
1720
18 x 1717 x 20
=
306340
simplify the fraction
GCF(306, 340) = 34
306340
=
306 ÷ 34340 ÷ 34
=
910
1817
÷
2017
=
910
6
20/7 ÷ 19/17 =
340/133
Steps
207
÷
1917
=
207
x
1719
20 x 177 x 19
=
340133
207
÷
1917
=
340133
7
14/11 ÷ 17/12 =
168/187
Steps
1411
÷
1712
=
1411
x
1217
14 x 1211 x 17
=
168187
1411
÷
1712
=
168187
8
5/12 ÷ 4/12 =
5/4
Steps
512
÷
412
=
512
x
124
5 x 1212 x 4
=
6048
simplify the fraction
GCF(60, 48) = 12
6048
=
60 ÷ 1248 ÷ 12
=
54
512
÷
412
=
54
9
19/14 ÷ 9/2 =
19/63
Steps
1914
÷
92
=
1914
x
29
19 x 214 x 9
=
38126
simplify the fraction
GCF(38, 126) = 2
38126
=
38 ÷ 2126 ÷ 2
=
1963
1914
÷
92
=
1963
10
16/3 ÷ 12/6 =
8/3
Steps
163
÷
126
=
163
x
612
16 x 63 x 12
=
9636
simplify the fraction
GCF(96, 36) = 12
9636
=
96 ÷ 1236 ÷ 12
=
83
163
÷
126
=
83
11
6/5 ÷ 5/4 =
24/25
Steps
65
÷
54
=
65
x
45
6 x 45 x 5
=
2425
65
÷
54
=
2425
12
15/9 ÷ 7/9 =
15/7
Steps
159
÷
79
=
159
x
97
15 x 99 x 7
=
13563
simplify the fraction
GCF(135, 63) = 9
13563
=
135 ÷ 963 ÷ 9
=
157
159
÷
79
=
157
13
18/9 ÷ 16/8 =
1
Steps
189
÷
168
=
189
x
816
18 x 89 x 16
=
144144
simplify the fraction
GCF(144, 144) = 144
144144
=
144 ÷ 144144 ÷ 144
=
11
= 1
189
÷
168
=
1
14
20/3 ÷ 11/9 =
60/11
Steps
203
÷
119
=
203
x
911
20 x 93 x 11
=
18033
simplify the fraction
GCF(180, 33) = 3
18033
=
180 ÷ 333 ÷ 3
=
6011
203
÷
119
=
6011
15
5/18 ÷ 19/18 =
5/19
Steps
518
÷
1918
=
518
x
1819
5 x 1818 x 19
=
90342
simplify the fraction
GCF(90, 342) = 18
90342
=
90 ÷ 18342 ÷ 18
=
519
518
÷
1918
=
519
16
18/17 ÷ 6/17 =
3
Steps
1817
÷
617
=
1817
x
176
18 x 1717 x 6
=
306102
simplify the fraction
GCF(306, 102) = 102
306102
=
306 ÷ 102102 ÷ 102
=
31
= 3
1817
÷
617
=
3
17
10/3 ÷ 15/9 =
2
Steps
103
÷
159
=
103
x
915
10 x 93 x 15
=
9045
simplify the fraction
GCF(90, 45) = 45
9045
=
90 ÷ 4545 ÷ 45
=
21
= 2
103
÷
159
=
2
18
12/6 ÷ 20/6 =
3/5
Steps
126
÷
206
=
126
x
620
12 x 66 x 20
=
72120
simplify the fraction
GCF(72, 120) = 24
72120
=
72 ÷ 24120 ÷ 24
=
35
126
÷
206
=
35
19
11/5 ÷ 10/5 =
11/10
Steps
115
÷
105
=
115
x
510
11 x 55 x 10
=
5550
simplify the fraction
GCF(55, 50) = 5
5550
=
55 ÷ 550 ÷ 5
=
1110
115
÷
105
=
1110
20
9/5 ÷ 19/14 =
126/95
Steps
95
÷
1914
=
95
x
1419
9 x 145 x 19
=
12695
95
÷
1914
=
12695
## Worksheet: Dividing Fraction by a Fraction
Dividing fraction by a fraction worksheet is the largest collection of practice problems and solved exercises which can be served as homework, classwork or assignment problems for the students to learn, practice, assess, iterate and master the skills of how to solve such fraction problems in basic mathematics.
Online Test
00:00
Start my Challenge
16/20 ÷ 15/14 = ?
17/30 ÷ 9/4 = ?
20/28 ÷ 21/23 = ?
10/24 ÷ 22/26 = ?
22/23 ÷ 12/8 = ?
27/10 ÷ 15/28 = ?
14/7 ÷ 4/8 = ?
4/8 ÷ 23/16 = ?
22/15 ÷ 18/6 = ?
20/2 ÷ 14/12 = ?
## Online Test: Dividing Fraction by a Fraction
This online test to find the dividing fraction by a fraction is a formative assessment which can be used as homework, classwork or assignment problems to assess and improve the learner's math skills on fraction.
### Fraction Division Calculator
Fraction A
Fraction B
5/8 ÷ 2/3 = 15/16
show workout
## Work with steps - Fraction Division
Question:
5/8 divided by 2/3 = ?
|
# Factor a binomial
We'll provide some tips to help you choose the best Factor a binomial for your needs. Math can be a challenging subject for many students.
## The Best Factor a binomial
There are a lot of Factor a binomial that are available online. The best math experts are those who can teach math with clarity and ease. They have a good grasp of the subject matter and can explain it in an organized and insightful way. As well, they have the patience to work with students who might not be up to snuff when it comes to understanding certain concepts. Finally, they have a passion for the subject that shows in their teaching. Clearly, no two people are alike, so finding someone that you click with is key if you want your lessons to be memorable and effective. However, there are some qualities that all great math teachers should share. First and foremost is enthusiasm—it’s one of the most important factors in any lesson. You need to be engaged in order to make sure your students are too; otherwise, you’re just wasting your time. Next is patience—you need to be able to sit back, listen to what your students have to say, and take their feedback seriously. Lastly, you need to know the subject matter inside and out since this is one area where mistakes will happen more often than not.
It is pretty simple to solve a geometric sequence. If we have a sequence A, B, C... of numbers and it looks like AB, then we can simply start at A and work our way down the list. Once we reach C, we are done. In this example, we can easily see AB = BC = AC ... Therefore once we reach C, the solution is complete. Let's try some other examples: A = 1, B = 2, C = 4 AB = BC = AC = ACB ACAB = ABC ==> ABC + AC ==> AC + AB ==> AC + B CABACCA ==> CA + AB ==> CA + B + A ==> CA + (B+A) ==> CABABABABABA The solutions are CABABABABABA and finally ABC.
Trinomial factor or trinomial model is a statistical model that uses the coefficients of the three main terms in a formula. The coefficients describe the relationship between each variable in the formula and the function value (the dependent variable). Since there are three variables in a formula, it follows that each variable’s coefficient is expressed as a ratio. For example, if the coefficients for “age”, “x” and “y” are expressed as “a”:“b”:“c”, where “a” is the coefficient for “age” and “c” for “y”, then it can be inferred that humanity evolved at a rate of 1:1.25:0.25 = 0.61 = 1/3 per 1000 years. In statistics, a factor is an observation that represents one unit of an independent variable. A factor is often thought of as being observable; in other words, it is directly observable by an observer. However, a factor can also be unobservable (e.g., time-dependent); in this case, it can be thought of as being observable given certain assumptions about its underlying structure and behavior. Factors are sometimes referred to as determinants or causes. Factors can arise from the measured variable itself (e.g.,
In order to solve inequality equations, you have to first make sure that every variable is listed. This will ensure that you are accounting for all of the relevant information. Once you have accounted for all variables, you can start to solve the equation. When solving inequality equations, keep in mind that multiplication and division are not commutative operations. For example, if you want to find the value of x in an inequality equation, you should not just divide both sides by x. Instead, you should multiply both sides by the reciprocal of x: To solve inequality equations, it is best to use graphing calculators because they can handle more complex mathematics than simple hand-held calculators can. Graphing calculators can also be used to graph inequalities and other functions such as t and ln(x).
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# Math Expressions Grade 5 Unit 7 Lesson 2 Answer Key Simplify Expressions
## Math Expressions Common Core Grade 5 Unit 7 Lesson 2 Answer Key Simplify Expressions
Math Expressions Grade 5 Unit 7 Lesson 2 Homework
Unit 7 Lesson 2 Simplify Expressions Question 1.
Follow the Order of Operations to simplify 27 ÷ (3 • 3) + 17
Step I Perform operations inside parentheses. ___________
Step 2 Multiply and divide from left to right. ___________
Step 3 Add and subtract from left to right. __________________
Step 1: 27 ÷ (9) + 17
Step 2: 3 + 17
Step 3: 20
Simplify. Follow the Order of Operations.
Math Expressions Grade 5 Simplify Question 2.
54 – 200 ÷ 4
54 – 200 ÷ 4
54 – 50
4
Unit 7 Answer Key Math Expressions Lesson 2 Simplify Question 3.
0.8 ÷ (0.07 – 0.06)
0.8 ÷ ( 0.01 )
80
Unit 7 Lesson 2 Math Expressions Question 4.
3 • 8 – 6 ÷ 2
3 • 8 – 6 ÷ 2
3 . 8 – 3
24 – 3
21
Unit 7 Lesson 2 Answer Key Math Expressions Question 5.
($$\frac{3}{8}$$ + $$\frac{1}{4}$$) • 16
3/8 + 1/4 . 16
0.625 x 16
= 10
Math Expressions Common Core Grade 5 Simplify Question 6.
64 + 46 – 21 + 29
64 + 46 – 21 + 29
139 – 21
118
Simplify Answers Math Expressions Unit 7 Lesson 2 Question 7.
72 ÷ (7 – 1) • 3
72 ÷ (7 – 1) • 3
72 ÷ 6 . 3
12 x 3
= 36
Question 8.
0.8 – 0.5 ÷ 5 + 0.2
0.8 – 0.5 ÷ 5 + 0.2
0.8 + 0.2 – 0.1
1 – 0.1
= 0.9
Question 9.
$$\frac{5}{6}$$ – 4 • $$\frac{1}{12}$$
5/6 – 4 . 1/12
5/6 – 1/3
= 1/2
Question 10.
5 • 15 ÷ 3
5 • 15 ÷ 3
5. 5
= 25
Question 11.
32 ÷ (2.3 + 1.7) • 3
32 ÷ (2.3 + 1.7) • 3
32 ÷ 4 . 3
8 x 3
= 24
Question 12.
(1$$\frac{1}{2}$$ – $$\frac{3}{4}$$) × $$\frac{1}{4}$$
1 1/2 – 3/4 x 1/4
3/4 x 1/4
3/16
Question 13.
(6.3 – 5.1) • (0.7 + 0.3)
(6.3 – 5.1) • (0.7 + 0.3)
1.2 x 1
= 1.2
Question 14.
12 ÷ 0.1 + 12 ÷ 0.01
120 + 1200
1320
Question 15.
$$\frac{1}{2}$$ • $$\frac{1}{2}$$ ÷ $$\frac{1}{2}$$
1/2 x 1/2 ÷ 1/2
1/2 x 1
= 1/2
Question 16.
10 – 4 + 2 – 1
10 – 4 + 2 – 1
10 – 6 – 1
= 7
Math Expressions Grade 5 Unit 7 Lesson 2 Remembering
Solve.
Question 1.
Question 2.
Question 3.
Write an equation to solve the problem. Draw a model if you need to.
Question 4.
The students of Turner Middle School are going on a field trip. There are 540 students attending. A bus can hold 45 students. How many buses are needed for the field trip?
Given, Total number of students = 540
Number of students a bus can hold = 45
Now, 540/45
= 12
Therefore, 12 busses are needed for the field trip.
Question 5.
The area of a rectangular court is 433.37 square meters, and the length of the court is 28.7 meters. What is the width of the court?
Given,
Area of the rectangular court = 433.37 sq.m
Length = 28.7 sq. m
Width = n
Area = length x width
433.37 = 28.7 x n
n = 433.37/28.7
n = 15.1 meters.
Write the computation in words.
Question 6.
5 ÷ $$\frac{1}{8}$$ _____________________
5 divided by 1/8
Question 7.
2.4 ÷ 0.6 + 0.2 __________________________
Divide the sum of 0.6 and 0.2 by 2.4
Question 8.
Stretch Your Thinking Write step-by-step instructions for simplifying the following expression.
10 • [60 ÷ (11 + 4)] – 3
Step 1 : Add 11 + 4
10. [60 ÷ 15]-3
Step 2: Divide 60 by 15
10. [ 4 ] – 3
Step 3: Multiply 10 by 4
40 – 3
Step 4: Subtract 3 from 40
= 37
|
Courses
Facts that Matter- Arithmetic Progressions Class 10 Notes | EduRev
Class 10 : Facts that Matter- Arithmetic Progressions Class 10 Notes | EduRev
The document Facts that Matter- Arithmetic Progressions Class 10 Notes | EduRev is a part of the Class 10 Course Class 10 Mathematics by VP Classes.
All you need of Class 10 at this link: Class 10
Facts that Matter
• Sequence:
When some numbers are arranged in a definite order, according to a definite rule, they are said to form a sequence.
The number occurring at the 1st place is called the 1st term, denoted by T1.
The number occurring at the nth place is called the nth term, denoted by Tn.
Example: Let us consider a rule
Tn = 3n + 1
Putting n = 1, 2, 3, 4, 5, ..... we get
T1 = 3 (1) + 1 = 4
T2 = 3 (2) + 1 = 7
T3 = 3 (3) + 1 = 10
T4 = 3 (4) + 1 = 13
T5 = 3 (5) + 1 = 16
...... ...... ......
...... ...... ......
...... ...... ......
∴ The numbers 4, 7, 10, 13, 16, ..... ..... form a sequence.
The pattern followed by its terms is:
• Finite Sequence
A sequence containing definite number of terms is called a finite sequence.
• Infinite Sequence
A sequence containing infinite number of terms is called an infinite sequence.
• Progressions The sequences in which each term (other than the first and the last) is related to its succeeding term by a fixed rule, are called progressions.
Note:
There are three types of progressions:
I. Arithmetic Progressions (A.P.)
II. Geometric Progressions (G.P.)
III. Harmonic Progressions (H.P.)
But, here we shall learn about arithmetic progression.
• Arithmetic Progression (A.P.)
An arithmetic progression is a list of numbers in which each term is obtained by adding a fixed number to the preceding term except the first term.
The fixed number is called the common difference, which can be positive, negative or zero.
The common difference of an A.P. can be obtained by subtracting any term from its following term.
Generally the first term is denoted by ‘a’ and the common difference is denoted by ‘d’.
• If three numbers a, b, c are in order, then (b – a) = common difference = (c – d)
• When certain number of terms of an A.P. are required, then we select the terms in the following manner:
Number of Terms, Terms and Common difference: Number of Terms Terms Common difference 3 a – d, a, a + d d 4 a – 3d, a – d, a + d, a + 3d 2d 5 a – 2d, a – d, a, a + d, a + 2d d 6 a – 5d, a – 3d, a – d, a + d, a + 3d, a + 5d 2d
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# What Is 6/22 as a Decimal + Solution With Free Steps
The fraction 6/22 as a decimal is equal to 0.272.
Fractions are numerals of the form p/q where p is the numerator and q is the denominator. They are simply another way of denoting division, and therefore all the rules and procedures for evaluating divisions apply to fractions. Fractions are of several types such as improper, proper, etc. 6/22 is a proper fraction.
Here, we are more interested in the division types that result in a Decimal value, as this can be expressed as a Fraction. We see fractions as a way of showing two numbers having the operation of Division between them that result in a value that lies between two Integers.
Now, we introduce the method used to solve said fraction to decimal conversion, called Long Division, which we will discuss in detail moving forward. So, let’s go through the Solution of fraction 6/22.
## Solution
First, we convert the fraction components, i.e., the numerator and the denominator, and transform them into the division constituents, i.e., the Dividend and the Divisor, respectively.
This can be done as follows:
Dividend = 6
Divisor = 22
Now, we introduce the most important quantity in our division process: the Quotient. The value represents the Solution to our division and can be expressed as having the following relationship with the Division constituents:
Quotient = Dividend $\div$ Divisor = 6 $\div$ 22
This is when we go through the Long Division solution to our problem.
Figure 1
## 6/22 Long Division Method
We start solving a problem using the Long Division Method by first taking apart the division’s components and comparing them. As we have 6 and 22, we can see how 6 is Smaller than 22, and to solve this division, we require that 6 be Bigger than 22.
This is done by multiplying the dividend by 10 and checking whether it is bigger than the divisor or not. If so, we calculate the Multiple of the divisor closest to the dividend and subtract it from the Dividend. This produces the Remainder, which we then use as the dividend later.
Now, we begin solving for our dividend 6, which after getting multiplied by 10 becomes 60.
We take this 60 and divide it by 22; this can be done as follows:
 60 $\div$ 22 $\approx$ 2
Where:
22 x 2 = 44
This will lead to the generation of a Remainder equal to 60 – 44 = 16. Now this means we have to repeat the process by Converting the 16 into 160 and solving for that:
160 $\div$ 22 $\approx$ 7Â
Where:
22 x 7 = 154
This, therefore, produces another Remainder which is equal to 160 – 154 = 6. Now we must solve this problem to Third Decimal Place for accuracy, so we repeat the process with dividend 60.
60 $\div$ 22 $\approx$ 2Â
Where:
22 x 2 = 44
Finally, we have a Quotient generated after combining the three pieces of it as 0.272, with a Remainder equal to 16.
Images/mathematical drawings are created with GeoGebra.
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# The "Cubic Formula"
#### Introduction.
Knowledge of the quadratic formula is older than the Pythagorean Theorem. Solving a cubic equation, on the other hand, was the first major success story of Renaissance mathematics in Italy. The solution was first published by Girolamo Cardano (1501-1576) in his Algebra book Ars Magna.
Our objective is to find a real root of the cubic equation
ax3+bx2+cx+d=0.
The other two roots (real or complex) can then be found by polynomial division and the quadratic formula. The solution proceeds in two steps. First, the cubic equation is "depressed"; then one solves the depressed cubic.
#### Depressing the cubic equation.
This trick, which transforms the general cubic equation into a new cubic equation with missing x2-term is due to Nicoḷ Fontana Tartaglia (1500-1557). We apply the substitution
to the cubic equation, to obtain:
Multiplying out and simplifying, we obtain the "depressed" cubic
Let's try this for the example
2x3-30x2+162x-350=0.
Our substitution will be x=y+5; expanding and simplifying, we obtain the depressed cubic equation
y3+6y-20=0.
#### Solving the depressed cubic.
We are left with solving a depressed cubic equation of the form
y3+Ay=B.
How to do this had been discovered earlier by Scipione del Ferro (1465-1526).
We will find s and t so that
3st = A (1) s3-t3 = B. (2)
It turns out that y=s-t will be a solution of the depressed cubic. Let's check that: Replacing A, B and y as indicated transforms our equation into
(s-t)3+3st (s-t)=s3-t3.
This is true since we can simplify the left side by using the binomial formula to:
(s3-3s2t+3st2-t3)+(3s2t-3st2)=s3-t3.
How can we find s and t satisfying (1) and (2)? Solving the first equation for s and substituting into (2) yields:
Simplifying, this turns into the "tri-quadratic" equation
which using the substitution u=t3 becomes the quadratic equation
From this, we can find a value for u by the quadratic formula, then obtain t, afterwards s and we're done.
Let's do the computation for our example
y3+6y=20.
We need s and t to satisfy
3st = 6 (3) s3-t3 = 20 (4)
Solving for s in (3) and substituting the result into (4) yields:
which multiplied by t3 becomes
t6+20t3-8=0.
Using the quadratic formula, we obtain that
We will discard the negative root, then take the cube root to obtain t:
By Equation (4),
Our solution y for the depressed cubic equation is the difference of s and t:
The solution to our original cubic equation
2x3-30x2+162x-350=0
is given by
#### Concluding remarks.
I will not discuss a slight problem you can encounter, if you follow the route outlined. What problem am I talking about?
Shortly after the discovery of a method to solve the cubic equation, Lodovico Ferrari (1522-1565), a student of Cardano, found a similar method to solve the quartic equation.
This section is loosely based on a chapter in the book Journey Through Genius by William Dunham.
#### Exercise 1.
Show that y=2 is a solution of our depressed cubic
y3+6y-20=0.
Then find the other two roots. Which of the roots equals our solution
#### Exercise 2.
Transform the cubic equation
x3-6x2+14x-15=0
into a depressed cubic.
#### Exercise 3.
Find a real root of the cubic equation in Exercise 2.
(This is for practice purposes only; to make the computations a little less messy, the root will turn out to be an integer, so one could use the Rational Zero test instead.)
[Back] [Next: The Geometry of the Cubic Formula]
[Algebra] [Trigonometry] [Complex Variables]
[Calculus] [Differential Equations] [Matrix Algebra]
|
# How to Graph Single–Variable Inequalities
You need to use a number line to graph single-variable inequalities. Learn how to graph an inequality with one variable.
## Step by step guide to graphing single–variable inequalities
• Inequality is similar to equations and uses symbols for “less than” ($$<$$) and “greater than” ($$>$$).
• To solve inequalities, we need to isolate the variable. (like in equations)
• To graph an inequality, find the value of the inequality on the number line.
• For less than or greater than draw open circle on the value of the variable.
• If there is an equal sign too, then use filled circle.
• Draw a line to the right or to the left for greater or less than.
### Graphing Single–Variable Inequalities – Example 1:
Draw a graph for $$x \ > \ 4$$
Solution:
Since the variable is greater than $$4$$, then we need to find $$4$$ and draw an open circle above it. Then, draw a line to the right.
### Graphing Single–Variable Inequalities – Example 2:
Draw a graph for $$x>2$$
Solution:
Since, the variable is greater than $$2$$, then we need to find $$2$$ and draw an open circle above it. Then, draw a line to the right.
### Graphing Single–Variable Inequalities – Example 3:
Graph this inequality. $$x<5$$
Solution:
## Exercises for Graphing Single–Variable Inequalities
### Draw a graph for each inequality.
1. $$\color{blue}{– 2 > x}$$
2. $$\color{blue}{x ≤ – 5}$$
3. $$\color{blue}{x > 7}$$
4. $$\\ \color{blue}{1.5 > x}$$
1. $$\color{blue}{ – 2 > x }$$
2. $$\color{blue}{ x ≤ -5 }$$
3. $$\color{blue}{ x > 7 }$$
4. $$\color{blue}{ -1.5 > x }$$
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## HOW TO CHECK IF THE GIVEN SEQUENCE IS GEOMETRIC OR NOT
How to check if the given sequence is geometric or not ?
• If the common ratios of the given sequence are equal, then we can decide that the given sequence is a geometric progression.
• If the common ratios are not equal, then the given sequence is not a geometric progression.
To find the common ratio, we use the formula
r = a2/a1 (or) r = a3/a
What is geometric progression ?
A Geometric Progression is a sequence in which each term is obtained by multiplying a fixed non-zero number to the preceding term except the first term.
The fixed number is called common ratio. The common ratio is usually denoted by r.
## Check If the Given Sequence is Geometric Sequence Examples
Example 1 :
Find out which of the following sequences are geometric sequences . For those geometric sequences, find the common ratio.
1/2, 1/3, 2/9, 4/47,...........
Solution :
r = t2/t1r = (1/3) / (1/2)r = 2/3 ----(1) r = t3/t2r = (2/9) / (1/3)r = 2/3 ----(2)
Since the common ratios are same, the given sequence is a geometric sequence. The required common ratio is 2/3.
(ii) 12, 1, 1/12, ............
Solution :
r = t2/t1r = 1/12 ----(1) r = t3/t2r = (1/12) / 1r = 1/12 ----(2)
Since the common ratios are same, the given sequence is a geometric sequence. The required common ratio is 1/12.
(iii) √2, 1/√2, 1/2√2,...........
Solution :
r = t2/t1r = (1/√2)/√2r = 1/2 ----(1) r = t3/t2r = (1/2√2) / (1/√2)r = 1/2 ----(2)
Since the common ratios are same, the given sequence is geometric progression.
The required common ratio is 1/2.
(iv) 0.004, 0.02, 1, ..........
Solution :
t1 = 0.004, t2 = 0.02 and t3 = 1
r = t2/t1r = 0.02 / 0.004r = 20 / 4r = 5 ----(1) r = t3/t2r = 1 / 0.02r = 100/2r = 50 ----(2)
Since the common ratios are not same, the given sequence is not geometric progression.
After having gone through the stuff given above, we hope that the students would have understood how to check if the given sequence is geometric or not.
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|
# Calculate remaining balance on a mortgage
(male narrator)
So a couple purchases a home
with a \$180,000 mortgage at a 4% rate for 30 years
with monthly payments. Uh…what will
the remaining balance in our mortgage be
after 5 years? Now to answer this, we’re gonna have
to do it in two parts. First thing we’re
gonna have to do is figure out what are
their monthly payments? Uh…so for that, we know that
our interest rate is 4%. We know we’re doing
monthly payments, uh…based on a 30-year
repayment schedule, uh…with an initial amount
of \$180,000, and we’re looking
for d–our payment amount. So we pull out our loan formula,
and we set it up. So 1 plus .04, divided by 12
to the negative 12, times 30…
all over .04 over 12. Uh…and since we’ve done
a lot of these, I’m gonna go ahead and skip,
uh…a bunch of steps here. So I pull out my calculator,
and I calculate all of that. It comes out to be
about 209, uh…562, uh…and then divide
by that gives me a payment
of about \$858.93. So the monthly payments
on this loan will be \$858.93. So now that takes us to the
second part of the question. Now we know
the monthly payments, we can determine
the remaining balance. So we’re trying to find
the balance after 5 years. So after 5 years, there are 25 years left
to pay on the loan. So what we’re
really asking is how much loan balance
can be paid off in 25 years with monthly payments
of this size? In other words, keeping the interest rate
the same, if we have a loan for 25 years
with payments of \$858.93, what is the loan balance at the beginning
of those 25 years? That is the current balance
of the loan after 5 years. So our P0 here,
where 0 now… times 0’s now 5 years in,
will be… uh…let’s see,
\$858.93 is my d, times 1, minus 1,
plus .04 over 12, to the negative 12,
times 25– because remember we’re going
for 25 years now– over .04 over 12, and again, skipping
several steps here, uh…we come up
with 155,793, uh…91. So in other words,
after 5 years, uh…the remaining balance
on the loan will be 155,793. Now it’s worth thinking
have we paid, uh…so far? So we’ve paid,
uh…\$858.93 for 5 years, 12 times a year,
gives us a total of, uh… let’s see…51,535.80
that we’ve paid, uh…to the loan company,
in payments. Uh…how much of the loan
have we actually paid off? Well, we started
with a \$183,000 loan. After 5 years, our loan balance
is now 155,793, which means
we’ve paid off, uh…\$24,206.09,
uh…paid off of the loan. Well, if we’ve paid
\$51,535.80 and only \$24,206.09 of it
has gone to the loan, then the other 27,329.71 must have gone
to interest so far. And of course,
the next 25 years, we’ll be adding
more interest, but over those
first 5 years, uh…we’ve paid
\$27,000 in interest.
### Paul Whisler
1. David Lippman
At 2:47, the value for P0 is incorrect. It should have been about \$162,758. This would mean the couple has paid off about \$17242 of the loan balance, and about \$34,293 of what they have paid so far has been interest.
2. Euan Gamble
Thank you very much. I was quite stuck on this until you explained it.
3. Yasmine
That's unprofessional video
4. Vinita
Thanks for the clear explanation.
|
# Fundamental theorem of calculus review
Review your knowledge of the fundamental theorem of calculus and use it to solve problems.
## What is the fundamental theorem of calculus?
The theorem has two versions.
### a) $\dfrac{d}{dx}\displaystyle\int_a^x f(t)\,dt=f(x)$
We start with a continuous function $f$ and we define a new function for the area under the curve $y=f(t)$:
$F(x)=\displaystyle\int_a^x f(t)\,dt$
What this version of the theorem says is that the derivative of $F$ is $f$. In other words, $F$ is an antiderivative of $f$. Thus, the theorem relates differential and integral calculus, and tells us how we can find the area under a curve using antidifferentiation.
### b) $\displaystyle\int_a^b\!\! f(x)dx=F(b)\!-\!\!F(a)$
This version gives more direct instructions to finding the area under the curve $y=f(x)$ between $x=a$ and $x=b$. Simply find an antiderivative $F$ and take $F(b)-F(a)$.
## Practice set 1: Applying the theorem
Problem 1.1
$\displaystyle g(x) = \int_{\,1}^{\,x}\sqrt{2t+7}\,dt\,$
$\displaystyle g\,^\prime(9)\, =$
Want to try more problems like this? Check out this exercise.
## Practice set 2: Applying the theorem with chain rule
We can use the theorem in more hairy situations. Let's find, for example, the expression for $\dfrac{d}{dx}\displaystyle \int_{0}^{x^3}\sin(t) \, dt$. Note that the interval is between $0$ and $x^3$, not $x$.
To help us, we define $\displaystyle F(x) = \int_{0}^{x}\sin(t) \, dt$. According to the fundamental theorem of calculus, $F'(x)=\sin(x)$.
It follows from our definition that $\displaystyle\int_{0}^{x^3}\sin(t) \, dt$ is $F(x^3)$, which means that $\dfrac{d}{dx}\displaystyle \int_{0}^{x^3}\sin(t) \, dt$ is $\dfrac{d}{dx}F(x^3)$. Now we can use the chain rule:
\begin{aligned} &\phantom{=}\dfrac{d}{dx}\displaystyle \int_{0}^{x^3}\sin(t) \, dt \\\\ &=\dfrac{d}{dx}F(x^3) \\\\ &=F'(x^3)\cdot\dfrac{d}{dx}(x^3) \\\\ &=\sin(x^3)\cdot 3x^2 \end{aligned}
Problem 2.1
$\displaystyle F(x) = \int_{0}^{x^4}\cos(t) \, dt$
$F'(x) =$
Want to try more problems like this? Check out this exercise.
|
# How do you verify the chain rule?
## How do you verify the chain rule?
Chain Rule If f(x) and g(x) are both differentiable functions and we define F(x)=(f∘g)(x) F ( x ) = ( f ∘ g ) ( x ) then the derivative of F(x) is F′(x)=f′(g(x))g′(x) F ′ ( x ) = f ′ ( g ( x ) ) g ′ ( x ) .
### What is the rule for chain rule?
The chain rule states that the derivative of f(g(x)) is f'(g(x))⋅g'(x). In other words, it helps us differentiate *composite functions*. For example, sin(x²) is a composite function because it can be constructed as f(g(x)) for f(x)=sin(x) and g(x)=x².
#### How do you solve chain rule problems?
Use the chain rule to calculate h′(x), where h(x)=f(g(x)).
1. Solution: The derivatives of f and g are f′(x)=6g′(x)=−2.
2. Solution: The derivatives of f and g are f′(x)=exg′(x)=6x.
3. The derivatives of the component functions are g′(z)=6ezh′(x)=4×3+2x.
Why is chain rule so hard?
The difficulty in using the chain rule: The problem that many students have trouble with is trying to figure out which parts of the function are within other functions (i.e., in the above example, which part if g(x) and which part is h(x).
What does Rolles Theorem say?
Rolle’s theorem, in analysis, special case of the mean-value theorem of differential calculus. Rolle’s theorem states that if a function f is continuous on the closed interval [a, b] and differentiable on the open interval (a, b) such that f(a) = f(b), then f′(x) = 0 for some x with a ≤ x ≤ b.
## Is the chain rule difficult?
The difficulty in using the chain rule: Implementing the chain rule is usually not difficult. The problem that many students have trouble with is trying to figure out which parts of the function are within other functions (i.e., in the above example, which part if g(x) and which part is h(x).
### What is the chain rule example?
The chain rule is a method for finding the derivative of composite functions, or functions that are made by combining one or more functions. An example of one of these types of functions is f(x)=(1+x)2 which is formed by taking the function 1+x and plugging it into the function x2.
#### Is the chain rule hard?
Who invented chain rule?
If you consider that counting numbers is like reciting the alphabet, test how fluent you are in the language of mathematics in this quiz. The chain rule has been known since Isaac Newton and Leibniz first discovered the calculus at the end of the 17th century.
Can you apply the chain rule twice?
Here we have a more complicated chain of compositions, so we use the chain rule twice. At the outermost “layer” we have the function g(x)=1+√1+√x plugged into f(x)=√x, so applying the chain rule once gives ddx√1+√1+√x=12(1+√1+√x)−1/2ddx(1+√1+√x).
## What is Rolle’s and Lagrange’s theorem?
Rolle’s Theorem is a particular case of the mean value theorem which satisfies certain conditions. At the same time, Lagrange’s mean value theorem is the mean value theorem itself or the first mean value theorem. In general, one can understand mean as the average of the given values.
### What is Rose theorem?
#### What is the chain rule used for in real life?
Real World Applications of the Chain Rule The Chain Rule can also help us deduce rates of change in the real world. From the Chain Rule, we can see how variables like time, speed, distance, volume, and weight are interrelated. A horse is carrying a carriage on a dirt path.
|
Subtracting Integers
Lesson 3
In lesson 3 our warm up is on adding integers. We want to review this concept because adding integers is very important to subtract integers. We are going to begin lesson 3 with our SOLVE problem. The high temperature in Hopedale, Ohio was -2 F. The low temperature was -7 F. What was the difference between the high and low temperature for the day? The first step in S study the problem, is to underline the questions, our question is what was the difference between the high and low temperatures for the day?
The second step in S is to answer the questions what is the problem asking me to find? This problem is asking me to find the difference between the high and low temperatures.
In problem 1 we are going to model the problem 5 minus 2. Subtraction is different than addition, in subtraction we model the first number and we take away the second number. We will represent our 5 with five yellows. Our problem is 5 minus 2 which mean we want to take away a 2 or take away a positive 2, because we started with 5 yellows it is possible for us to take away 2 yellows., and this leaves us with a difference of positive 3. To represent this problem pictorial, we will represent our five with 5 yellows and we will take away the two yellows by crossing them out. This leaves us with a final answer of three yellows or positive three.
In problem 2 negative 5 minus negative 2, we are going to subtract a negative 2. We will begin this problem by representing the negative 5 with 5 red integer chips. To subtract negative 2 or take away negative 2, it is possible because we have 2 reds that we can take away. So we take them away and that leaves us with an answer 3 reds or negative 3. To represent this problem pictorially, we will begin by representing our negative 5 with 5 R’s and we want to take away two reds and we will do this by crossing out 2 R’s, which leaves us with a final answer of 3 R’s or negative 3.
To model problem 5 we have 8 minus negative 3, we will begin by representing our 8 with 8 yellow chips. We want to take away negative three, this is impossible right now because we do not have any negatives to take away, however we can create this possibility by using our zero pairs, our value right now is 8, if we add one zero pair ,one yellow and one red, our value is still 8, because there is 8 here plus zero here, so our value stays at 8. But now we have the possibility of taking away one red. We don’t have enough reds to take away yet so we going to add another zero pair , is it possible to take away three reds not yet, so we add another zero pair and now we have created the possibility to take away negative three, we take away the three reds and our answer is all yellow. 1,2,3,4,5,6,7,8,9,10,11 because their yellow our answer is positive 11. To represent this pictorially we will begin , by representing our positive 8, once again we try to take away negative three, it is impossible to take away negative three, so must add a zero pair we still must take away negative 3, we have not changed our value we still have a value of 8, but now have created the possibility of taking away our negative three, and it leaves us with a final answer of 11 Y’s or positive 11.
To model problem 6, we have negative 8 minus positive 3. We will first represent the negative 8 with 8 red integer chips. We want to take away a positive 3 once again we do not have any positive to take away in order to be able to take away positives, we must create the possibility using zero pairs. If we add one zero pair we are still maintaining our value of negative 8 creating the possibility to take away one positive, if we add a second zero pair are still , but we can now take away two positive, and when we add our third zero pair we still maintaining a value of negative 8, but we have created the possibility to take away three. Our final answer is in all one color and it is 8 plus 3 which is 11, and because they are red our answer is negative 11. To represent this problem pictorially we will first represent our negative 8 with R’s, from here we will represent the 3 zero pairs that we must have in order to take away that positive 3, one yellow one red we still cannot take away positive 3, now we can take away positive 2 and when we add one more zero pair, one yellow one red we can take away positive three, which leaves us with 1,2,3,4,5,6,7,8,9,10,11 reds which is negative 11.
To the problem 4 minus one we will begin by representing our 4 with 4 yellow integer chips. The problem tell us we want to take away 1 we are able to take away one positive, so our answer is positive three. To represent this pictorially we will represent our 4 with 4 Y’s and we will take away 1 by crossing one of the Y’s leaving us with 3 Y’s or positive 3.
To model the problem negative 7 minus negative 5 we will first represent our negative 7 with 7 red integer chips. The problem tell us that we want to take away negative five, five red chips that we can take away 1,2,3,4,5 and when we take them away we are left with negative 2 our answer is negative 2. To represent this problem pictorially we will begin by representing our negative 7 with 7 R’s, we want to take away negative 5 so we will cross out 5 R’s, our final answer is negative 2 because we are left 2 R’s
To model the problem negative 3 minus 4, we will first represent our negative 3 with three reds, the problem tell us we want to take away positive 4 because we start with negative 3, it is impossible to take away negative 4, so we must create the possible by using zero pairs. One yellow one red is a zero pair, our value stays negative three but now it is possible for us to take away, one yellow. One yellow, one red, we can take away two yellows, one yellow one red, we are still maintaining a value of negative 3 but by adding zero pairs we’ve created the possibility to take away a positive 4,.. 1,2,3,4, when we take them away we are left with our answer in all one color, and our answer is seven reds or negative 7. To model this problem pictorially we will begin by representing our negative 3 with 3 reds because we want to take away, a positive 4 we must add zero pairs to create the possibility of taking away four yellows and when we take away our positive four we are left with 7 R’s or negative 7.
To represent the problem 6 minus negative 2 we will first represent our 6 with 6 yellows , the problem tell us we want to take away negative 2 but we do not have any negative that we can take away, so must create this possibility by adding zero pairs. When I add one zero pair my value of positive six does not change, it creates the possibility for me to take away one negative, when I add a second zero pair, my value of 6 is still 6 but we have the have the possibility to take away negative 2. When we take away negative 2 we are left with 8 yellows our answer is positive 8. To represent this problem pictorially we will first represent our six with 6 yellows we want to take away negative two but we do not have negatives to take away, so we will add zero pairs to create the possibility of taking away negative 2, our value is still six, but now we can take away negative 2, our final answer will be 8 Y’s or positive 8.
Once we have developed all of our rules for subtractions we are going back to our graphic organizer. Our rules for subtraction,subtraction is the same as adding the opposite. We make two changes to the problem, change subtraction sign to addition sign, and we change the sign of second number. After these two steps are completed we follow the rules for addition
Now that we know the rules for subtraction we are going to go back and complete the SOLVE problem we began with at the beginning of the lesson. We had already underlined our question, which was , what is the difference in the high and low temperature for the day, we had also found what the problem was asking us to find, which is the difference between the high and low temperatures.
In O we organize the facts, we are now going to identify the facts in the problem. The high temperature in Hopedale, Ohio was -2 F, one fact . The low temperature was -7 F, another fact. We must decide if these two facts are necessary or unnecessary. The first fact the high temperature in Hopedale, Ohio was -2 F is necessary. We list all necessary facts. The second fact the low temperature was –7 degree F, also needs to be listed.
Now we are going to line up our plan. We are finding the difference between the high and low temperature for the day so we are going to use subtraction. Our operation was subtractions, we are going to write in words what your plan of action will be. We’re going take away the low temperature from the high temperature. Change subtraction problem to an addition problem, then follow the rules for addition.
In V verify our plan with actions, we will first estimate, we know our temperature has gone from-2 to -7 we are going to estimate that the difference between the two temperature is about 5. Now we are going to carry out our plan, our plan was to take away the low temperature from the high temperature. We are going to change the subtraction problem to an addition problem and follow the rules for addition. We are going to change our subtraction to addition and we are going to add the opposite. Now our problem is negative 2 plus 7. When we have two integers with different signs we subtract and take the sign of the larger group. 7 minus 2 is 5, and 7 is the larger group so it will be a positive 5.
In E we are going to examine our results, we will first answer the question does your answer make sense, if you go back to what your original question was we were looking for the difference between the high and low temperature. So we ended up with an answer positive 5 F. so our answer does make sense. Is your answer reasonable? We look at our estimate, which was about 5, so yes our answer is reasonable. Is our answer accurate, for accuracy you will have your students rework the problem or use a calculator to check their answer. The difference between the high and low temperature was 5, we will write our answer as a complete sentence.
To close the lesson we will review the essential questions. How do you subtract integers using manipulative, represent the first value with a red or yellow chip and take away the second value. What do you do if you can not take away the second number from the first number? You create the possibility by using zero pairs.
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# The value of sin20.sin40.sin60.sin80 is?
Aug 10, 2018
$\sin {20}^{\circ} \sin {40}^{\circ} \sin {60}^{\circ} \sin {80}^{\circ} = \frac{3}{16}$
#### Explanation:
Let ,
$X = \sin {20}^{\circ} \sin {40}^{\circ} \sin {60}^{\circ} \sin {80}^{\circ}$
$\therefore X = \frac{1}{2} \left(\sin {60}^{\circ} \sin {20}^{\circ}\right) \left(2 \sin {80}^{\circ} \sin {40}^{\circ}\right)$
Using ,$\text{Product Identity}$
$2 \sin x \sin y = \cos \left(x - y\right) - \cos \left(x + y\right) , f \mathmr{and} , x = 80 , y = 40$
$\therefore X = \frac{1}{2} \left(\frac{\sqrt{3}}{2} \sin {20}^{\circ}\right) \left\{\cos \left(80 - 40\right) - \cos \left(80 + 40\right)\right\}$
$\therefore X = \frac{\sqrt{3}}{4} \sin {20}^{\circ} \left\{\cos {40}^{\circ} - \cos {120}^{\circ}\right\}$
$\therefore X = \frac{\sqrt{3}}{4} \sin {20}^{\circ} \left\{\cos {40}^{\circ} + \frac{1}{2}\right\}$
$\therefore X = \frac{\sqrt{3}}{8} \left\{2 \sin {20}^{\circ} \cos 40 + \sin 20\right\}$
Using ,$\text{Product Identity}$
$2 \sin x \cos y = \sin \left(x + y\right) + \sin \left(x - y\right) , f \mathmr{and} , x = 20 , y = 40$
$\therefore X = \frac{\sqrt{3}}{8} \left\{\sin 60 + \sin \left(- 20\right) + \sin 20\right\}$
$\therefore X = \frac{\sqrt{3}}{8} \left\{\frac{\sqrt{3}}{2} - \sin {20}^{\circ} + \sin {20}^{\circ}\right\}$
$\therefore X = \frac{\sqrt{3}}{8} \left\{\frac{\sqrt{3}}{2}\right\}$
$\therefore X = \frac{3}{16}$
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# Linear equations in One variable Class 8 Notes
In this page we will explain the topics for the chapter 8 of Linear equations in One variable Class 8 Maths.We have given quality Linear equations in One variable Class 8 Notes and video to explain various things so that students can benefits from it and learn maths in a fun and easy manner, Hope you like them and do not forget to like , social share and comment at the end of the page.
Table of Content
## What is algebraic expression
Algebraic expression is the expression having constants and variable. It can have multiple variable and multiple power of the variable
Example
$10x$
$2x -3$
$3x + y$
$2xy + 11$
$x^2 +2$
## What is algebraic equation
We already know the below terms from previous class
What is equation
An equation is a condition on a variable.
What is variable
A variable takes on different numerical values; its value is not fixed. Variables are denoted usually by letters of the alphabets, such as x, y, z, l, m, n, p, etc
An algebraic equation is an equality involving variables. It says that the value of the expression on one side of the equality sign is equal to the value of the expression on the other side.
Example
$5x = 25$
$2x - 3 = 9$
$11xy+1 =98$
$x^2 +1=y^2$
Important points to remember
1. These all above equation contains the equality (=) sign.
2. The above equation can have more than one variable
3. The above equation can have highest power of variable > 1
4. The expression on the left of the equality sign is the Left Hand Side (LHS). The expression on the right of the equality sign is the Right Hand Side (RHS)
5. In an equation the values of the expressions on the LHS and RHS are equal. This happens to be true only for certain values of the variable. These values are the solutions of the equation.
6. We assume that the two sides of the equation are balanced. We perform the same mathematical operations on both sides of the equation, so that the balance is not disturbed. We get the solution after generally performing few steps
## What is Linear equation in one Variable
We will restrict the above equation with two conditions
a) algebraic equation in one variable
b) variable will have power 1 only
Example
$5x = 25$
$2x - 3 = 9$
We will focus on Linear equation in One variable in this chapter.
## Solving Equations which have Linear Expressions on one Side and Numbers on the other Side
$2x -3= 5$
$3x-11= 22$
How to solve Linear equation in one Variable
1. Transpose (changing the side of the number) the Numbers to the side where all number are present. We know the sign of the number changes when we transpose it to other side
2. Now you will have a equation have variable on one side and number on other side. Add/subtract on both the side to get single term
3. Now divide or multiply on both the side to get the value of the variable
Example
$2x -3= 5$
Solution
Transposing 3 to other side
$2x=5+3$
$2x=8$
Dividing both the sides by 2
x=4
Watch this tutorial on how to solve Equations which have Linear Expressions on one Side and Numbers on the other Side
## Word Problem on Simple Linear equation in one variable
Linear equation can be used to solve many word Problem. The procedure is simple
1. First read the problem carefully. Write down the unknown and known
2. Assume one of the unknown to x and find the other unknown in term of that
3. Create the linear equation based on the condition given
4. Solve them by using the above method
Example:
The perimeter of a rectangular swimming pool is 154 m. Its length is 2 m more than twice its breadth. What are the length and the breadth of the pool?
Solution
The unknown are length and breadth.
Let the breadth be x m.
Then as per question the length will be (2x + 2) m.
Perimeter of swimming pool = 2(l + b) = 154 m
$2(2x + 2 + x) = 154$
$2(3x + 2) = 154$
Dividing both sides by 2
$3x + 2 = 77$
Transposing 2 to R.H.S, we obtain
$3x = 77 - 2$
$3x = 75$
Dividing 3 on both the sides
$x = 25$
So
Breath is 25 m
Length =2x + 2 = 2 × 25 + 2 = 52m
Hence, the breadth and length of the pool are 25 m and 52 m respectively.
## Solving Equations having the Variable on both Sides
$2x -3= 6-x$
$3x-11= 4x$
How to solve them
1. Here we Transpose (changing the side of the number) both the variable and Numbers to the side so that one side contains only the number and other side contains only the variable. We know the sign of the number changes when we transpose it to other side.Same is the case with Variable
2. Now you will have a equation have variable on one side and number on other side. Add/subtract on both the side to get single term
3. Now divide or multiply on both the side to get the value of the variable
Example
$2x -3= 6-x$
Solution
Transposing 3 to RHS and x to LHS
$2x+x=6+3$
$3x=9$
Dividing both the sides by 3
$x=3$
## Reducing Equations to Simpler Form
You will find many situations where the linear equation may be having number in denominator. We can perform the below steps to simplify them and solve it
1. Take the LCM of the denominator of both the LHS and RHS
2. Multiple the LCM on both the sides, this will reduce the number without denominator and we can solve using the method described above
Example
Solve the linear equation
Solution
L.C.M. of the denominators, 2, 3, 4, and 5, is 60.
So Multiplying both sides by 60, we obtain
$30x - 12 = 20x + 15 +60$
Transposing 20x to R.H.S and 12 to L.H.S, we obtain
$30x- 20x = 15 + 12+60$
$10x= 87$
Dividing both the sides by 10
$x=\frac {[87}{10}=8.7$
## Equations Reducible to the Linear Form
Let’s explain this with an example
The ages of Hari and Harry are in the ratio 5:7. Four years from now the ratio of their ages will be 3:4. Find their present ages.
Solution
Let the common ratio between their ages be x. Therefore, Hari’s age and Harry’s age will be 5x years and 7x years respectively and four years later, their ages will be (5x + 4) years and (7x + 4) years respectively.
According to the situation given in the question,
Now this is not a linear equation as we studied, but if you observe, we can reduce to linear equation.
We multiply both the sides by (7x+4)
Multiplying both the sides by (7x+4)
$(5x+4) =\frac {3(7x+4)}{4}$
or
$20x+16=21x+12$
Transposing 20x to RHS and 12 to LHS
4=x
x=4
Hari’s age = 5x years = (5 × 4) years = 20 years
Harry’s age = 7x years = (7 × 4) years = 28 years
Therefore, Hari’s age and Harry’s age are 20 years and 28 years respectively.
## Practice Questions
1. $6 x + 3 = 3 x + 4$
2. $11 x + 4 = 4 x + 10$
3. $13 x + 1 = x + 9$
4. $3 x + 1 = x + 3$
5. $15 x + 5 = 5 x + 5$
6. $12 x + 9 = 9 x + 7$
7. $7 x + 1 = x + 4$
8. $19 x + 10 = 10 x + 4$
9. $5 x + 4 = 4 x + 6$
10. $9 x + 5 = 5 x + 4$
11. $11 x + 1 = x + 8$
12. $14 x + 10 = 10 x + 6$
1. $x = \frac{1}{3}$
2. $x = \frac{6}{7}$
3. $x = \frac{2}{3}$
4. $x = 1$
5. $x = 0$
6. $x = - \frac{2}{3}$
7. $x = \frac{1}{2}$
8. $x = - \frac{2}{3}$
9. $x = 2$
10. $x = - \frac{1}{4}$
11. $x = \frac{7}{10}$
12. $x = -1$
## Frequently asked Questions on CBSE Class 8 Maths Chapter 2: Linear equations in One Variable
### What is LHS and RHS Class 8??
In an equation ,The expression on the left of the equality sign is the Left Hand Side (LHS). The expression on the right of the equality sign is the Right Hand Side (RHS)
### What is the General Form of a linear equation in One Variable?
The general form of the linear equations in one variable is Ax + B = 0 where x is the variable and A is the coefficient of x and b is the constant term. Example 2x +3=0
### What are the properties of linear equation?
Properties of Linear equation (i) Variable will have Power 1 only (ii) It cannot be denominator of the fraction
## Summary
Here is Linear equations in One variable Class 8 notes Summary
1. Linear equations in One Variable is an algebriac equation where we have one variable and it has power 1 only.
2. The solution of the linear equation can be integers or fractions
3. We can transpose both the constant and linear expression in the linear equations to solve the equation
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# Tan Graph
Here we will learn about tan graphs or tangent graphs, including how to recognise the graph of the tangent function, sketch the tangent curve and label important values, and interpret the tangent graph.
There are also tangent graph worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you’re still stuck.
## What is the tan graph?
The tan graph is a visual representation of the tangent function for a given range of angles.
The horizontal axis of a trigonometric graph represents the angle, usually written as \theta , and the y -axis is the tangent function of that angle.
• repeats every 180^o
• Not a continuous curve
• Vertical asymptotes at 90^o \pm 180^o
We will only look at the graph of the tangent function in this lesson for angles in degrees although this can also be represented in radians.
## How to plot the tan graph
Remember that \tan(\theta) is a relationship between the opposite side and the adjacent side of a right angle triangle:
Let’s look at 3 triangles where we would use the tangent ratio to calculate the size of the angle \theta . For each triangle, the adjacent side is the same but the length of the opposite side and the associated angle change.
Here we can see that as \tan(\theta)=\frac{opp}{adj} , as the angle \theta increases, the length of the side opposite to the angle also increases. So for each triangle we have:
• Triangle 1: \tan(\theta)=\frac{3}{10}=0.3
• Triangle 2: \tan(\theta)=\frac{9}{10}=0.9
• Triangle 3: \tan(\theta)=\frac{15}{10}=1.5
So what would happen if the opposite side to the angle is equal to 10
\tan(\theta)=\frac{10}{10}=1
So when the opposite side is equal to the adjacent side, we get \tan(\theta) = 1 .
What about when the opposite side is equal to 0 ?
\tan(\theta)=\frac{0}{10}=0
So when the opposite side is equal to 0, \; \tan(\theta) = 0.
If we plotted a graph to show the value of \tan(\theta) for each value of \theta between 0^o and 90^o , we get the following graph of the tangent function:
Let us add the values of \tan(\theta) for the three triangles from earlier into the graph to show how they would look:
We can now use the graph to find the angle \theta for triangles 1, 2, and 3 :
This graph shows that when \tan(\theta) = 0.3 , \; \theta = 17^o so we have the triangle
This graph shows that when \tan(\theta) = 0.9 , \; \theta = 42^o so we have the triangle
This graph shows that when \tan(\theta) = 1.5 , \; \theta = 56^o so we have the triangle
This shows us that we can use the graph of the tangent function to find missing angles in a triangle. More on this later as we have a large problem to resolve. How can the opposite side of a right angle triangle be equal to 0 ? What if the adjacent side is equal to 0 ?
Unfortunately there is a limit to the use of trig ratios to find angles between 0 and 90^o . For any larger or smaller angles, we need to look at the unit circle.
### Unit circle
The unit circle is a circle of radius 1 with the centre at the origin. We can label the values where the circle intersects the axes because we know that the radius of the unit circle is 1 unit.
We can still construct a triangle within the unit circle with the angle starting from the positive x axis.
Looking at the trigonometric ratios of sine and cosine, we can say that:
• \sin(\theta)=\frac{a}{1} \; \text{so} \; a=\sin(\theta)
• \cos(\theta)=\frac{b}{1} \; \text{so} \; b=\cos(\theta)
• \tan(\theta)=\frac{a}{b} \; \text{so} \; \tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}
This means that the width of the triangle is equal to \cos(\theta) and the height of the triangle is equal to \sin(\theta) . The point on the unit circle is therefore the coordinate ( \cos(\theta) , \; \sin(\theta) ):
### When do we get positive and negative values for tan(θ)
As \tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)} , we can now calculate when the tangent function is positive and negative using division of negative numbers.
\tan(\theta) is as periodic function as we could continue to turn around the origin from the positive x axis beyond 360^o .
We can also turn anticlockwise which would give us the negative values for the value of \tan(\theta) .
The period of the function is 180^o
### What is an asymptote?
An asymptote is a line that a curve gets closer and closer to but never touches as the curve goes towards infinity.
We know that \tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)} .
As \tan(\theta) can be expressed as a fraction, we have an issue when \cos(\theta) = 0 as you cannot divide by 0 . This means that the value for \tan(\theta) is undefined every time \cos(\theta) = 0 , which occurs every at 90^o \pm 180^o when y=0
Given the angles in the table below, the gradient of the graph increases continuously as the value of \theta tends towards 90^o .
We already know that the graph of the tangent function is periodic and so a vertical asymptote occurs every 180^o before and after the asymptote shown above. See the graph below:
### The graph of the tangent function
Key features include:
• Periodic (repeats every 180^o )
• Not a continuous curve
• Vertical asymptotes at 90^o \pm 180^o
• The curve always has a positive gradient
Unlike the graphs for the sine and cosine functions, the tangent function is not a wave. This is because the graph is not continuous
There is therefore no amplitude to the wave as there is no maximum or minimum value.
The tangent function can change when we transform the function. For more information, see the lesson on transformations of trigonometric functions.
You should be able to interpret the unit circle to determine the value of \theta . Here are a couple of examples to help you.
Now that we have generated the graph of \tan(\theta) , we need to be able to interpret it to find values of \theta and \tan(\theta) for any value between the range -360 \leq \theta \leq 360^o (values for between -360^o and 360^o .
The inverse to \tan(\theta)
The inverse function to \tan(\theta) is given as \tan^{-1}(\theta) in GCSE Mathematics. The power of -1 here is used as the inverse function notation and not the reciprocal. In A level Mathematics we learn that the reciprocal to \tan(\theta) is called the cotangent or \cot(\theta) where \cot(\theta)=\frac{1}{\tan(\theta)}=\frac{\cos(\theta)}{\sin(\theta)} . In order to not get confused with with the inverse function notation, we call the inverse of \tan(\theta) \arctan(\theta). We do the same with the inverse functions of the sine and cosine functions, \arcsin(\theta) and \arccos(\theta).
## How to interpret tangent graphs
In order to interpret tangent graphs:
1. Draw a straight line from the axis of the known value to the tangent curve.
2. Draw a straight, perpendicular line at the intersection point to the other axis.
3. Read the value where the perpendicular line meets the other axis.
## Tangent graph examples
### Example 1: 0° ≤ θ ≤ 90°
Use the graph of \tan(\theta) to estimate the value for \theta when \tan(\theta)=1.5 for 0^o \leq \theta \leq 90^o
1. Draw a straight line from the axis of the known value to the tangent curve.
2Draw a straight, perpendicular line at the intersection point to the other axis.
3Read the value where the perpendicular line meets the other axis.
When \tan(\theta)=1.5, \; \theta=56^o
### Example 2: 90° ≤ θ ≤ 270°
Use the graph of \tan(\theta) to estimate the value for \theta when \tan(\theta)=0.5 for 90^o \leq \theta \leq 270^o
Draw a straight line from the axis of the known value to the tangent curve.
Draw a straight, perpendicular line at the intersection point to the other axis.
Read the value where the perpendicular line meets the other axis.
### Example 3: 270 ≤ θ ≤ 360°
Use the graph of \tan(\theta) to estimate the value for \tan(\theta) when \theta=300^o .
Draw a straight line from the axis of the known value to the tangent curve.
Draw a straight, perpendicular line at the intersection point to the other axis.
Read the value where the perpendicular line meets the other axis.
### Example 4: −180 ≤ θ ≤ 0°
Use the graph of \tan(\theta) to estimate the value for \theta when \tan(\theta)=-0.5 for -180 \leq \theta \leq 0^o
Draw a straight line from the axis of the known value to the tangent curve.
Draw a straight, perpendicular line at the intersection point to the other axis.
Read the value where the perpendicular line meets the other axis.
### Example 5: −360° ≤ θ ≤ 0°
Use the graph of \tan(\theta) to estimate the value for \tan(\theta) when \theta=-255^o .
Draw a straight line from the axis of the known value to the tangent curve.
Draw a straight, perpendicular line at the intersection point to the other axis.
Read the value where the perpendicular line meets the other axis.
### Example 6: −360° ≤ θ ≤ 360°
Use the graph of \tan(\theta) to estimate the value for \theta when \tan(\theta)=0.25 for -360^o \leq \theta \leq 360^o
Draw a straight line from the axis of the known value to the tangent curve.
Draw a straight, perpendicular line at the intersection point to the other axis.
Read the value where the perpendicular line meets the other axis.
### Common misconceptions
• Drawing the horizontal line only to one intersection point
When you are finding the value of \theta using a trigonometric graph, only one value is calculated when there are more points of intersection.
For example, only the value of 65^o is read off the graph.
• Sine and cosine graphs switched
The sine and cosine graphs are very similar and can easily be confused with one another. A tip to remember is that you “sine up” from 0 for the sine graph so the line is increasing whereas you “cosine down” from 1 so the line is decreasing for the cosine graph.
• Asymptotes are drawn incorrectly for the graph of the tangent function
The tangent function has an asymptote at 90^o because this value is undefined. As the curve repeats every 180^o , the next asymptote is at 270^o and so on.
• The graphs are sketched using a ruler
Each trigonometric graph is a curve and therefore the only time you are required to use a ruler is to draw a set of axes. Practice sketching each curve freehand and label important values on each axis.
• Value given out of range
When finding a value of \theta using a trigonometric graph, you must make sure that the value of \theta is within the range specified in the question.
For example, the range of values for \theta is given as 0^o \leq \theta \leq 360^o and only the value of \theta=240^o is written for the solution, whereas the solution \theta=300^o is also correct.
Tan graph is part of our series of lessons to support revision on trigonometry. You may find it helpful to start with the main trigonometry lesson for a summary of what to expect, or use the step by step guides below for further detail on individual topics. Other lessons in this series include:
### Practice tangent graphs questions
1. Use the graph of \tan(\theta) to calculate the value of \theta when \tan(\theta)=3 for 0^o \leq \theta \leq 180^o
\theta=252^o
\theta=-108^o
\theta=-288^o
\theta=72^o
2. Use the graph of \tan(\theta) to calculate the value of \theta when \tan(\theta)=-1.2 for -180^o \leq \theta \leq 90^o
\theta=130^o
\theta=-50^o
\theta=-230^o
\theta=310^o
3. Use the graph of \tan(\theta) to calculate the value of \theta when \tan(\theta)=0 for -180^o \leq \theta \leq 180^o
\theta=0^o
\theta=-180^o
\theta=180^o
\theta=180^o, \; 0^o, \; 180^o
4. Use the graph of \tan(\theta) to calculate the value of \tan(\theta) when \theta=150^o
\tan(150)=-0.8
\tan(150)=0.6
\tan(150)=-0.6
\tan(150)=-0.2
5. Use the graph of \tan(\theta) to calculate the value of \tan(\theta) when \theta=-320^o
\tan(-320)=0.8
\tan(-320)=-0.5
\tan(-320)=-0.8
\tan(-320)=0.5
6. What value for \theta would return the same value for 240^o for -90^o \leq \theta \leq 270^o
\tan(240)= \tan(-120)
\tan(240) = \tan(-240)
\tan(240) = \tan(120)
\tan(240) = \tan(60)
### Tangent graphs GCSE questions
1. Below is a sketch of the graph of y=\tan(\theta) for 0^o \leq \theta \leq 360^o .
If \tan(x)=1.4 , what two values for x would return the same value of \tan(\theta) ?
(3 marks)
x=54.5^o
(1)
x+180
(1)
234.5^o
(1)
2. (a) Write an equation in terms of \theta for the size of the angle ABC .
(b) Use the graph of y=\tan(\theta) to estimate the value of \theta in triangle ABC .
(4 marks)
(a)
\tan(\theta)=\frac{25}{15}=1.6…
(1)
\theta =tan^{-1}(1.67)
(1)
(b)
(1)
tan^{-1}(1.67) = [55-65^o]
(1)
3. Write down the solutions to the equation \tan(\theta)=2 for 360^o \leq \theta \leq 720^o .
(2 marks)
[60-70^o] and [240-250^o] highlighted on graph
(1)
[420-430^o] and [600-610^o]
(1)
4. Below are the graphs of y=\sin(\theta) and y=\tan(\theta). Use the graph to find the solutions for \sin(\theta) = \tan(\theta) for -360^o \leq \theta \leq 360^o .
(3 marks)
\theta =-360^o, \; -180^o
(1)
\theta = 0^o
(1)
\theta=180^o, \; -360^o
(1)
## Learning checklist
You have now learned how to:
• recognise, sketch and interpret graphs of trigonometric functions (with arguments in degrees) y = \tan x for angles of any size
## Still stuck?
Prepare your KS4 students for maths GCSEs success with Third Space Learning. Weekly online one to one GCSE maths revision lessons delivered by expert maths tutors.
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# 7.6: Special Right Triangles
Difficulty Level: At Grade Created by: CK-12
## Introduction
The Square Rock Garden
Ms. Kino’s class decided to do a community service project that everyone could enjoy. They decided to create a meditation garden that would be a rock garden.
Chas and Juanita took charge of the project. They drew a sketch of the rock garden and the rest of the class loved it so much that they instantly agreed to use the sketch that the pair had created. Here is their sketch.
“Let’s put a diagonal path in it,” Frankie suggested looking at the sketch.
“That’s a great idea, how long will the path be?” Chas asked.
The class wants to add a diagonal path. If they do that from one corner to another, how long will the path be?
This lesson will teach you all that you need to know to solve this problem. You will have a chance to solve it at the end of the lesson.
What You Will Learn
By the end of this lesson, you will have an understanding of the following skills.
• Recognize that a \begin{align*}45^\circ-45^\circ-90^\circ\end{align*} triangle is an isosceles triangle and that the length of the hypotenuse is the product of the length of either leg and \begin{align*}\sqrt{2}\end{align*}.
• Recognize that a \begin{align*}30^\circ-60^\circ-90^\circ\end{align*} triangle is half of an equilateral triangle, that the length of the hypotenuse is twice the length of the shorter leg, and that the longer leg is the product of the shorter leg and \begin{align*}\sqrt{3}\end{align*}.
• Find exact values of the indicated variable lengths in special right triangles, and check using the Pythagorean Theorem.
• Solve real – world problems involving applications of special right triangles.
Teaching Time
I. Understanding \begin{align*}\underline{45^\circ-45^\circ-90^\circ}\end{align*} Triangles
There are a few types of right triangles it is particularly important to study. Their sides are always in the same ratio, and it is crucial to study the \begin{align*}45^\circ-45^\circ-90^\circ\end{align*} and the \begin{align*}30^\circ-60^\circ-90^\circ\end{align*} triangles and understand the relationships between the sides. It will save you time and energy as you work through math problems both straight-forward and complicated. Let’s start by learning about the \begin{align*}45^\circ-45^\circ-90^\circ\end{align*}.
First, think about that \begin{align*}45^\circ-45^\circ-90^\circ\end{align*} refers to. Those values refer to the angle measures in the right triangle. We can see that there is one 90 degree angle and that the other two angles have the same measure. This particular triangle is also isosceles. An isosceles triangle has two side lengths that are the same. An isosceles right triangle will always have the same angle measurements: \begin{align*}45^\circ,45^\circ\end{align*}, and \begin{align*}90^\circ\end{align*} and will always have two side lengths that are the same. These characteristics make it a special right triangle.
Because these angles will always remain the same, the sides will always be in proportion. To find the relationship between the sides, use the Pythagorean Theorem.
Example
The isosceles right triangle below has legs measuring 1 centimeter. Use the Pythagorean Theorem to find the length of the hypotenuse.
As the problem states, you can use the Pythagorean Theorem to find the length of the hypotenuse. Since the legs are 1 centimeter each, set both \begin{align*}a\end{align*} and \begin{align*}b\end{align*} equal to 1 and solve for \begin{align*}c\end{align*}.
\begin{align*}a^2+b^2 &=c^2\\ (1)^2+(1)^2 &= c^2\\ 1+1 &= c^2\\ 2 &= c^2\\ \sqrt{2} &= \sqrt{c^2}\\ \sqrt{2} &= c\end{align*}
We can look at this and understand that there is also a 1 in front of the square root of two. This shows that the relationship between one side length and the length of the hypotenuse will always be the same. The hypotenuse of an isosceles right triangle will always equal the product of one leg and \begin{align*}\sqrt{2}\end{align*}.
Write this down in your notebook under \begin{align*}45^\circ-45^\circ-90^\circ\end{align*} special right triangles.
Now let’s use this information to solve the problem in the next example.
Example
What is the length of the hypotenuse in the triangle below?
Since the length of the hypotenuse is the product of one leg and \begin{align*}\sqrt{2}\end{align*}, you can easily calculate this length. It is easy because we know that with any 45/45/90 degree triangle, that the hypotenuse is the product of one of the legs and the square root of 2.
One leg is 3 inches, so the hypotenuse will be \begin{align*}3 \sqrt{2}\end{align*}, or about 4.24 inches.
That is a good question. To get that answer, we took the square root of two on the calculator, 1.414 and then multiplied it times 3.
\begin{align*}3 \times 1.414 = 4.242\end{align*}
We rounded to get the answer.
II. Understanding \begin{align*}\underline{30^\circ-60^\circ-90^\circ}\end{align*} Triangles
Another important type of right triangle has angles measuring \begin{align*}30^\circ,60^\circ\end{align*}, and \begin{align*}90^\circ\end{align*}. These triangles are exactly one half of equilateral triangles. Do you remember what an equilateral triangle is? An equilateral triangle is a triangle with equal angle measures. The equal angle measures of an equilateral triangle are \begin{align*}60^\circ-60^\circ-60^\circ\end{align*}. If we divide an equilateral triangle in half, then we have a \begin{align*}30^\circ-60^\circ-90^\circ\end{align*} triangle.
You can tell in the diagram that since the original triangle was equilateral, the short leg will be one-half the length of the hypotenuse.
Example
Find the length of the missing leg in the triangle below. Use the Pythagorean theorem to find your answer.
Just like you did for \begin{align*}45^\circ-45^\circ-90^\circ\end{align*} triangles, use the Pythagorean Theorem to find the missing side. In this diagram, you are given two measurements. The hypotenuse \begin{align*}(c)\end{align*} is 2 feet and the shorter leg \begin{align*}(a)\end{align*} is 1 foot. Find the length of the missing leg \begin{align*}(b)\end{align*}.
\begin{align*}a^2+b^2 &= c^2\\ (1)^2+b^2 &= (2)^2\\ 1+b^2 &=4\\ 1+b^2-1 &= 4-1\\ b^2 &= 3\\ \sqrt{b^2} &= \sqrt{3}\\ b &= \sqrt{3}\end{align*}
You can leave the answer as the radical as shown, or use your calculator to find the approximate value of 1.732 feet.
So, just as there is a constant proportion relating the sides of the \begin{align*}45^\circ-45^\circ-90^\circ\end{align*} triangle, there is also one relating the sides of the \begin{align*}30^\circ-60^\circ-90^\circ\end{align*} triangle. The hypotenuse will always be twice the length of the shorter leg, and the longer leg is always the product of the shorter leg and \begin{align*}\sqrt{3}\end{align*}.
Write this information down under 30/60/90 degree right triangles.
Example
What is the length of the missing leg in the a 30/60/90 degree right triangle with a short leg of 5 and a hypotenuse of 10?
Since the length of the longer leg is the product of the shorter leg and \begin{align*}\sqrt{3}\end{align*}, you can easily calculate this length. The short leg is 5 inches, so the hypotenuse will be \begin{align*}5 \sqrt{3}\end{align*}, or about 8.66 inches.
III. Find Missing Values in Special Right Triangles
In this section you can find exact values of the indicated variable lengths in special right triangles, and check using the Pythagorean Theorem. You now know the special proportions of the side lengths of certain right triangles, now you can use this information to solve problems involving these special right triangles. If you use these relationships correctly, you can solve a lot of problems about right triangles without too much time on calculation.
Example
What is the length of one leg in the triangle below?
The first step in this problem is to identify the type of right triangle depicted. Since the length of both sides is \begin{align*}k\end{align*}, this is an isosceles right triangle. This makes it a 45/45/90 degree triangle.
So, the hypotenuse is the product of one leg and \begin{align*}\sqrt{2}\end{align*}. Set up an equation to find the length of \begin{align*}k\end{align*}.
\begin{align*}k \times \sqrt{2} &= 9 \sqrt{2}\\ k \times \sqrt{2} \div \sqrt{2} &= 9 \sqrt{2} \div \sqrt{2}\\ k &= 9\end{align*}
The length of one leg in this triangle is 9 yards.
Example
What is the length of one leg in the triangle below?
The first step in this problem is to identify the type of right triangle depicted. The angles show that this is a \begin{align*}30^\circ-60^\circ-90^\circ\end{align*} triangle. So, the longer leg is the product of one leg and \begin{align*}\sqrt{3}\end{align*}. The hypotenuse is twice the length of the shorter leg. The shorter leg is 4 meters.
The hypotenuse will be \begin{align*}4 \times 2\end{align*}, or 8 meters long.
Now we can look at how special right triangles appear in real – world situations.
IV. Solve Real – World Problems Involving Applications of Special Right Triangles
Just like any other mathematical concept, you may be tested on this material in a real-world context. Just translate the information into mathematical data and proceed as usual. No new tools or skills are required to solve these problems.
Example
The diagram below shows the shadow a flagpole casts at a certain time of day. If the height of the flagpole is \begin{align*}7 \sqrt{3} \ feet\end{align*}, what is the length of the hypotenuse of the triangle formed by the flagpole and its shadow?
The wording in this problem is complicated, but you only need to notice a few things. You can tell in the picture that this triangle has angles of \begin{align*}30^\circ,60^\circ\end{align*}, and \begin{align*}90^\circ\end{align*}. The actual flagpole is the longer leg in the triangle, so use the ratios in the diagrams above to find the length of the hypotenuse.
The longer leg is the product of the shorter leg and \begin{align*}\sqrt{3}\end{align*}. So, the length of the shorter leg will be 7 feet.
The hypotenuse in a \begin{align*}30^\circ-60^\circ-90^\circ\end{align*} triangle will always be twice the length of the shorter leg, so it will equal \begin{align*}7 \times 2\end{align*}, or 14 feet.
Now let’s apply what we have learned to the problem from the introduction.
## Real-Life Example Completed
The Square Rock Garden
Here is the original problem once again. Reread it and then answer the question about the length of the diagonal path.
Ms. Kino’s class decided to do a community service project that everyone could enjoy. They decided to create a meditation garden that would be a rock garden.
Chas and Juanita took charge of the project. They drew a sketch of the rock garden and the rest of the class loved it so much that they instantly agreed to use the sketch that the pair had created. Here is their sketch.
“Let’s put a diagonal path in it,” Frankie suggested looking at the sketch.
“That’s a great idea, how long will the path be?” Chas asked.
The class wants to add a diagonal path. If they do that from one corner to another, how long will the path be?
Now use what you have learned to figure out the length of the path.
Solution to Real – Life Example
The first step in a word problem of this nature is to add important information to the drawing. Because the problem asks you to find the length of a path from one corner to another, you should draw that path in.
Once you draw the diagonal path, you can tell that this is a triangle question. Because both legs of the triangle have the same measurement (10 feet), this is an isosceles right triangle. The angles in an isosceles right triangle are \begin{align*}45^\circ,45^\circ\end{align*}, and \begin{align*}90^\circ\end{align*}.
In an isosceles right triangle, the hypotenuse is always equal to the product of the length of one leg and \begin{align*}\sqrt{2}\end{align*}. So, the length of the path will be the product of 10 and \begin{align*}\sqrt{2}\end{align*}, or \begin{align*}10 \sqrt{2} \ feet\end{align*}. This value is approximately equal to 14.14 feet.
## Vocabulary
Here are the vocabulary words that are found in this lesson.
Isosceles Triangle
a triangle with two sides the same length.
45/45/90 Triangle
a special right isosceles triangle.
Equilateral Triangle
a triangle with all three angles \begin{align*}60^\circ\end{align*}.
30/60/90 Triangle
a special right triangle that is created when an equilateral triangle is divided in half.
## Time to Practice
Directions: Find the missing hypotenuse in each \begin{align*}45^\circ-45^\circ-90^\circ\end{align*} triangle.
1. Length of each leg = 5
2. Length of each leg = 4
3. Length of each leg = 6
4. Length of each leg = 3
5. Length of each leg = 7
Directions: Now use a calculator to figure out the approximate value of each hypotenuse. You may round to the nearest hundredth.
1. \begin{align*}5 \sqrt{2}\end{align*}
2. \begin{align*}4 \sqrt{2}\end{align*}
3. \begin{align*}6 \sqrt{2}\end{align*}
4. \begin{align*}3 \sqrt{2}\end{align*}
5. \begin{align*}7 \sqrt{2}\end{align*}
Directions: Find the missing length of the longer leg in each \begin{align*}30^\circ-60^\circ-90^\circ\end{align*} triangle.
1. short leg = 3
2. short leg = 4
3. short leg = 2
4. short leg = 8
5. short leg = 10
Directions: Now use a calculator to figure out the approximate value of each longer leg. You may round your answer when necessary.
1. \begin{align*}3 \sqrt{3}\end{align*}
2. \begin{align*}4 \sqrt{3}\end{align*}
3. \begin{align*}2 \sqrt{3}\end{align*}
4. \begin{align*}8 \sqrt{3}\end{align*}
5. \begin{align*}10 \sqrt{3}\end{align*}
Directions: Use what you have learned to solve each problem.
1. Janie had construction paper cut into and equilateral triangle. She wants to cut it into two smaller congruent triangles. What will be the angle measurement of the triangles that result?
2. Madeleine has poster board in the shape of a square. She wants to cut two congruent triangles out of the poster board without leaving any leftovers. What will be the angle measurements of the triangles that result?
3. A square window has a diagonal of \begin{align*}2 \sqrt{2} \ feet\end{align*}. What is the length of one of its sides?
4. A square block of cheese is cut into two congruent wedges. If a side of the original block was 9 inches, how long is the diagonal cut?
5. Jerry wants to find the area of an equilateral triangle but only knows that the length of one side is 4 centimeters. What is the height of Jerry’s triangle?
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Lesson Objectives
• Demonstrate an understanding of how to Factor Trinomials
• Learn how to write a Quadratic Equation in Standard Form
• Learn about the Zero-Product Property
• Learn how to solve a Quadratic Equation using Factoring
## How to Solve Quadratic Equations Using Factoring
Over the course of the last few lessons, we have learned to factor quadratic expressions. A quadratic expression contains a squared variable and no term with a higher degree. We will expand on this knowledge and learn how to solve a quadratic equation using factoring. A quadratic equation is an equation which contains a squared variable and no other term with a higher degree. Generally, we think about a quadratic equation in standard form:
ax2 + bx + c = 0
a ≠ 0 (since we must have a variable squared)
a, b, and c are any real numbers (a can't be zero)
Some examples of a quadratic equation are:
4x2 + 7x - 15 = 0
5x2 + 18x + 9 = 0
### Zero-Product Property
Up to this point, we have not attempted to solve an equation in which the exponent on a variable was not 1. For these types of problems, obtaining a solution can be a bit more work than what we have seen so far. When a quadratic equation is in standard form and the left side can be factored, we can solve the quadratic equation using factoring. This works based on the zero-product property (also known as the zero-factor property). The zero product property tells us if the product of two numbers is zero, then at least one of them must be zero:
xy = 0
x could be 0, y could be a non-zero number
y could be 0, x could be a non-zero number
x and y could both be zero
We can apply this to more advanced examples. Suppose we saw the following:
(x - 2)(x + 3) = 0
In this case, we have a quantity (x - 2) multiplied by another quantity (x + 3). The result of this multiplication is zero. This means we can use our zero-product property. To do this, we set each factor equal to zero and solve:
(x - 2) = 0
(x + 3) = 0
x - 2 = 0
x = 2
x + 3 = 0
x = -3
Essentially, x could be 2 or x could be -3. In either scenario, the equation would be true:
Let's check x = 2:
(x - 2)(x + 3) = 0
(2 - 2)(2 + 3) = 0
0(5) = 0
0 = 0
Let's check x = -3:
(-3 - 2)(-3 + 3) = 0
(-5)(0) = 0
0 = 0
### Solving a Quadratic Equation using Factoring
• Place the quadratic equation in standard form
• Factor the left side
• Use the zero-product property and set each factor with a variable equal to zero
• Check the result
Let's look at a few examples.
Example 1: Solve each quadratic equation using factoring
x2 + 3x = 18
Step 1) Write the quadratic equation in standard form. We want to subtract 18 away from each side of the equation:
x2 + 3x - 18 = 0
Step 2) Factor the left side:
x2 + 3x - 18 » (x + 6)(x - 3)
(x + 6)(x - 3) = 0
Step 3) Use the zero-product property and set each factor with a variable equal to zero
x + 6 = 0
x - 3 = 0
x = -6
x = 3
We can say that x = -6, 3
This means x can be -6 or x can be 3. Either will work as a solution.
Step 4) Check the result:
Plug in a -6 for x:
(-6)2 + 3(-6) - 18 = 0
36 - 18 - 18 = 0
36 - 36 = 0
0 = 0
Plug in a 3 for x:
(3)2 + 3(3) - 18 = 0
9 + 9 - 18 = 0
18 - 18 = 0
0 = 0
Example 2: Solve each quadratic equation using factoring
3x2 - 5 = -14x
Step 1) Write the quadratic equation in standard form. We want to add 14x to both sides of the equation:
3x2 + 14x - 5 = 0
Step 2) Factor the left side:
3x2 + 14x - 5 » (3x - 1)(x + 5)
(3x - 1)(x + 5) = 0
Step 3) Use the zero-product property and set each factor with a variable equal to zero
3x - 1 = 0
x + 5 = 0
x = 1/3
x = -5
We can say that x = -5, 1/3
This means x can be -5 or x can be 1/3. Either will work as a solution.
Step 4) Check the result:
Plug in a -5 for x:
3(-5)2 + 14(-5) - 5 = 0
3(25) - 70 - 5 = 0
75 - 75 = 0
0 = 0
Plug in a 1/3 for x:
3(1/3)2 + 14(1/3) - 5 = 0
3(1/9) + (14/3) - 5 = 0
1/3 + 14/3 - 5 = 0
15/3 - 5 = 0
5 - 5 = 0
0 = 0
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Home » NCERT Solutions » NCERT Solutions Class 5 Maths Chapter 12 Smart Charts – Free Download
# NCERT Solutions Class 5 Maths Chapter 12 Smart Charts – Free Download
NCERT Solutions Class 5 Maths Chapter 12- Smart Charts, provided here are very helpful for students who are preparing for their final exams. These learning materials are prepared by our experts with respect to the 5th standard CBSE syllabus(2020-2021).
Mathematics has always been at the forefront of almost every field – academic or non-academic. Hence, it is quite important to be aware of certain concepts as they are applied in the real world too. Students in the 5th class can get the study materials from here such as notes, textbooks, solutions, etc.
## NCERT Solutions for Class 5 Maths Chapter 12 Smart Charts:-
Solution:
1. Look at the tally marks and write the number for each animal in the table. How many children did Yamini talk to?
Solution:
Yamini talked to 131 children.
2. Which is the most favourite pet animal in this table?
Solution:
Dog is the most favourite pet animal in this table.
3. Which pet will you like to have? What will you name it? Which other animals can be kept at home? Discuss.
Solution:
I would like to have a dog. I will name it Brownie. The other animals that can be kept at home are cats, rabbits, parrots, etc.
Making Tally Marks on the Road:
Sumita stood on the road for half an hour and counted the number of vehicles passing by. She made a tally mark for each vehicle. This helped her in counting quickly the total number of vehicles in each group.
1. Write the number of each vehicle in the table.
Solution:
2. How many vehicles in all did Sumita see on the road in half an hour?
Solution:
Sumita saw 103 vehicles in half an hour.
3. Auto rickshaws are thrice the number of trucks — true/false?
Solution:
False.
4. Make tally marks for 7 more buses, and 2 more trucks.
Solution:
Tally marks for 7 more buses:
Tally marks for 2 more trucks:
Helping Hands:
In the EVS period, the teacher asked children whether they help their parents at home. There were different answers. Children named the work in which they help their parents the most. The teacher collected their answers and made a table.
Now you can fill the chapati chart to show the numbers given in the table.
1) Look and find out:
Children who help in making or serving food are
a) One-third of the total children
b) Half of the total children
c) One-fourth of the total children
Solution:
c) One-fourth of the total children
2) Practice time:
Solution:
Hot and Cold:
Have you seen the weather report on TV or in a newspaper? These are two bar charts. These show the highest temperature (in degrees Celsius) in four cities, on two different days. The cities are Delhi, Shimla, Bangalore and Jaisalmer.
Yes, I have seen the weather report on TV or in a newspaper.
Find out from the bar chart —
1. Which city is hottest on 1 June?
Solution:
Jaisalmer is hottest on 1 June.
2. Which city is coldest n 1 December?
Solution:
Shimla is coldest n 1 December.
3. Which city shows little change in temperature on the two days —-1 June and 1 December?
Solution:
Bangalore shows little change in temperature on the two days.
Lowest change = 28 – 24 = 4o C.
Rabbits in Australia:
Earlier there were no rabbits in Australia. Rabbits were brought to Australia around the year 1780. At that time there were no animals in Australia which ate rabbits. So the rabbits began to multiply at a very fast rate. Imagine what they did to the crops!
The table shows how rabbits grew every year.
1. After each year the number of rabbits was —
a) A little less than double the number of rabbits in the last year.
b) Double the number in the last year.
c) 8 more than the number in the last year.
d) More than double the number of rabbits in the last year.
Solution:
a) A little less than double the number of rabbits in the last year.
2. At the end of year 6, the number of rabbits was close to
a) 400
b) 600
c) 800
Solution:
a) 400
3. After which year did the number of rabbits cross 1000?
Solution:
After the year 1788 the number of rabbits cross 1000.
Family Tree:
Madhav went to a wedding along with his parents. He met many relatives there. But he didn’t know everyone. He met his mother s grandfather, but found that her grandmother is not alive. He also found that her mother (grandmother’s mother) is still alive, and is more than a hundred years old.
Madhav got confused. He couldn’t imagine his mother’s grandmother’s mother! So, Madhav’s mother made a family tree for him —
Madhav’s mother helped him understand her family with the help of this drawing. You can also find out about your older generations using such a family tree.
1. How many grandparents in all does Shobna have?
Solution:
Shobna has four grandparents.
2. How many great, great grandparents in all does Madhav have?
Solution:
Madhav has eight great, great grandparents in all.
3. How many elders will be in the VII generation of his family?
Solution:
32 elders will be in the VII generation of his family.
VI generation: 8 × 2 = 16
VII generation: 16 × 2 = 32
4. If he takes his family tree forward in which generation will he find 128 elders?
Solution:
Number of members in VII generation of Madhav’s family = 32
Number of members in VIII generation of Madhav’s family = 32 × 2 = 64
Number of members in IX generation of Madhav’s family = 64 × 2 = 128
Thus, in the IX generation, number of elders in Madhav’s family will be 128.
Growth Chart of a Plant:
Amit sowed a few seeds of moong dal in the ground. The height of the plant grew to 1.4 cm in the first four days. After that it started growing faster.
Amit measured the height of the plant after every four days and put a dot on the chart. For example if you look at the dot marked on the fourth day, you can see on the left side scale that it is 1.4 cm high.
Now look at the height of each dot in cm and check from the table if he has marked the dots correctly.
Find out from the growth chart:
a) Between which days did the length of the plant change the most?
i) 0-4 ii) 4-8 iii) 8-12 iv) 12-16 v) 16-20
Solution:
iii) 8-12
b) What could be the length of this plant on the 14th day? Guess.
i) 8.7 cm ii) 9.9 cm iii) 10.2 cm iv) 10.5 cm
Solution:
ii) 9.9 cm
c) Will the plant keep growing all the time? What will be its length on the 100th day? Make a guess!
No, I think its length would be about 17-18 cm on 100th day.
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How do you solve 5| - 9n + 5| = 70?
Aug 5, 2017
$n = - 1 \text{ or } n = \frac{19}{9}$
Explanation:
$\text{the expression "-9n+5" inside the absolute value can be}$
$\text{positive or negative}$
$\text{divide both sides by 5}$
$\frac{\cancel{5}}{\cancel{5}} | - 9 n + 5 | = \frac{70}{5}$
$\Rightarrow | - 9 n + 5 | = 14$
$\Rightarrow - 9 n + 5 = 14 \text{ or } - \left(- 9 n + 5\right) = 14$
•color(white)(x)-9n+5=14
$\text{subtract 5 from both sides}$
$\Rightarrow - 9 n = 9$
$\Rightarrow n = - 1$
•color(white)(x)-(-9n+5)=14
$\Rightarrow 9 n - 5 = 14$
$\text{add 5 to both sides}$
$\Rightarrow 9 n = 19$
$\Rightarrow n = \frac{19}{9}$
$\Rightarrow n = - 1 \text{ or "n=19/9" are the solutions}$
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# The mean of 12 numbers is 40. If each number is divided by 8,
Question:
The mean of 12 numbers is 40. If each number is divided by 8, what will be the mean of the new numbers?
Solution:
Let the numbers be x1x2,...x12.
We know:
Mean $=\frac{\text { Sum of observations }}{\text { Number of observations }}$
Thus, we have:
$40=\frac{x_{1}+x_{2}+\ldots+x_{12}}{12}$
$\Rightarrow x_{1}+x_{2}+\ldots+x_{12}=480 \quad \ldots \ldots(\mathrm{i})$
After division, the numbers become:
$\frac{x_{1}}{8}, \frac{x_{2}}{8}, \ldots, \frac{x_{12}}{8}$
$\therefore$ New mean $=\frac{\left(x_{1} / 8\right)+\left(x_{2} / 8\right)+\ldots .+(\sqrt{12} / 8)}{12}$
$=\frac{x_{1}+x_{2}+\ldots+x_{12}}{12 \times 8}$
$=\frac{x_{1}+x_{2}+\ldots+x_{12}}{96}$
$=\frac{480}{96} \quad[$ From (i) $]$
$=5$
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Logic puzzles, riddles, math puzzles and brainteasers - pzzls.com
Vandaag is het 28 May 2024
# Brick tower - science puzzle
Difficulty:
Rating: 2.6/5.0
Suppose you have an infinite number of bricks and no cement or another material to connect the bricks to each other. You would like to build a tower like the one in the figure. How many bricks do you need in order to build a tower in which the highest brick is shifted three bricks with respect to the lowest brick?
# Explanation
If you build a tower from two bricks, the highest brick can maximally be shifted a half brick with respect to the lowest one. This is because if you shift the upper brick somewhat more, the center of mass of the upper brick is not longer right above the lowest brick but next to the lowest brick. Therefore in that case the tower will fall. The center of mass of this tower of two bricks lies at (distance till center of mass lowest brick + distance to center of mass upper brick) / (total mass) = (1/2 + 3/2) / 2 = 3/4 brick lengths from one side of the lowest brick.
You can put the tower of two bricks on top of a third brick. Then the center of mass of the tower of two bricks should be just along the corner of the lowest brick. That implies that the second lowest brick can be shifted maximally 1/4 brick with respect to the lowest brick. The maximal shift possible with a three brick tower is hence 1/2 + 1/4 = 3/4. The center of mass of the tower with three bricks lies at (1/2 + 3/4 + 5/4) / 3 = 5 / 6.
If would like to build a tower from four bricks, you put the three brick tower on a new lower brick. Since the center of mass of the three brick tower lies at 5/6 brick lengths, the three brick tower can be shifted at most 1/6. The total shift will be (1/6 + 1/4 + 1/2) = 11/12.
Continuing this reasoning shows that the maximal shift for a tower consisting of N bricks is equal to
[ 1 + 1/2 + 1/3 + 1/4 + 1/5 + ...... + 1/(N-1) ] / 2.
The question is to determine how many brick you need in order to build a tower with a shift of three bricks. You will get this shift if [1 + 1/2 + 1/3 + ... 1/(N-1) ] / 2 is larger than 3.
Using the formula gives that
if the number of bricks N = 227 then the maximal shift is 2.999981 and
if the number of bricks N = 228 then the maximal shift is 3.002183.
So you need at least 228 bricks in order to get a shift of three bricks.
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Converting a Fraction to a Percentage Denominator of 20, 25, or 50 Online Quiz
Following quiz provides Multiple Choice Questions (MCQs) related to Converting a Fraction to a Percentage Denominator of 20, 25, or 50. You will have to read all the given answers and click over the correct answer. If you are not sure about the answer then you can check the answer using Show Answer button. You can use Next Quiz button to check new set of questions in the quiz.
Q 1 - Write the following fraction as a percentage:
$\mathbf{\frac{9}{20}}$
Explanation
Step 1:
$\frac{9}{20}$ is made into a fraction with denominator of 100.
Step 2:
Multiply and divide the fraction by 5
$\frac{9}{20} = (9 \times 5) \div (20 \times 5) = \frac{45}{100}$
Step 3:
Writing this fraction as a percentage
By definition, $\frac{9}{20} = \frac{45}{100}$ = 45%
Q 2 - Write the following fraction as a percentage:
$\mathbf{\frac{12}{25}}$
Explanation
Step 1:
$\frac{12}{25}$ is made into a fraction with denominator of 100.
Step 2:
Multiply and divide the fraction by 4
$\frac{12}{25} = (12 \times 4) \div (25 \times 4) = \frac{48}{100}$
Step 3:
Writing this fraction as a percentage
By definition, $\frac{12}{25} = \frac{48}{100}$ = 48%
Q 3 - Write the following fraction as a percentage:
$\mathbf{\frac{23}{50}}$
Explanation
Step 1:
$\frac{23}{50}$ is made into a fraction with denominator of 100.
Step 2:
Multiply and divide the fraction by 2
$\frac{23}{50} = (23 \times 2) \div (50 \times 2) = \frac{46}{100}$
Step 3:
Writing this fraction as a percentage
By definition, $\frac{23}{50} = \frac{46}{100}$ = 46%
Q 4 - Write the following fraction as a percentage:
$\mathbf{\frac{11}{20}}$
Explanation
Step 1:
$\frac{11}{20}$ is made into a fraction with denominator of 100.
Step 2:
Multiply and divide the fraction by 5
$\frac{11}{20} = (11 \times 5) \div (20 \times 5) = \frac{55}{100}$
Step 3:
Writing this fraction as a percentage
By definition, $\frac{11}{20} = \frac{55}{100}$ = 55%
Q 5 - Write the following fraction as a percentage:
$\mathbf{\frac{17}{25}}$
Explanation
Step 1:
$\frac{17}{25}$ is made into a fraction with denominator of 100.
Step 2:
Multiply and divide the fraction by 4
$\frac{17}{25} = (17 \times 4) \div (25 \times 4) = \frac{68}{100}$
Step 3:
Writing this fraction as a percentage
By definition, $\frac{17}{25} = \frac{68}{100} = 68%$
Q 6 - Write the following fraction as a percentage:
$\mathbf{\frac{33}{50}}$
Explanation
Step 1:
$\frac{33}{50}$ is made into a fraction with denominator of 100.
Step 2:
Multiply and divide the fraction by 2
$\frac{33}{50} = (33 \times 2) \div (50 \times 2) = \frac{66}{100}$
Step 3:
Writing this fraction as a percentage
By definition, $\frac{33}{50} = \frac{66}{100}$ = 66%
Q 7 - Write the following fraction as a percentage:
$\mathbf{\frac{19}{20}}$
Explanation
Step 1:
$\frac{19}{20}$ is made into a fraction with denominator of 100.
Step 2:
Multiply and divide the fraction by 5
$\frac{19}{20} = (19 \times 5) \div (20 \times 5) = \frac{95}{100}$
Step 3:
Writing this fraction as a percentage
By definition, $\frac{19}{20} = \frac{95}{100}$ = 95%
Q 8 - Write the following fraction as a percentage:
$\mathbf{\frac{18}{25}}$
Explanation
Step 1:
$\frac{18}{25}$ is made into a fraction with denominator of 100.
Step 2:
Multiply and divide the fraction by 4
$\frac{18}{25} = (18 \times 4) \div (25 \times 4) = \frac{72}{100}$
Step 3:
Writing this fraction as a percentage
By definition, $\frac{18}{25} = \frac{72}{100}$ = 72%
Q 9 - Write the following fraction as a percentage:
$\mathbf{\frac{41}{50}}$
Explanation
Step 1:
$\frac{41}{50}$ is made into a fraction with denominator of 100.
Step 2:
Multiply and divide the fraction by 2
$\frac{41}{50} = (41 \times 2) \div (50 \times 2) = \frac{82}{100}$
Step 3:
Writing this fraction as a percentage
By definition, $\frac{41}{50} = \frac{82}{100}$ = 82%
Q 10 - Write the following fraction as a percentage:
$\mathbf{\frac{17}{20}}$
Explanation
Step 1:
$\frac{17}{20}$ is made into a fraction with denominator of 100.
Step 2:
Multiply and divide the fraction by 5
$\frac{17}{20} = (17 \times 5) \div (20 \times 5) = \frac{85}{100}$
Step 3:
Writing this fraction as a percentage
By definition, $\frac{17}{20} = \frac{85}{100}$ = 85%
converting_fraction_to_percentage_denominator_20_25_50.htm
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# Difference between revisions of "1985 AIME Problems/Problem 12"
## Problem
Let $A$, $B$, $C$ and $D$ be the vertices of a regular tetrahedron each of whose edges measures 1 meter. A bug, starting from vertex $A$, observes the following rule: at each vertex it chooses one of the three edges meeting at that vertex, each edge being equally likely to be chosen, and crawls along that edge to the vertex at its opposite end. Let $p = \frac n{729}$ be the probability that the bug is at vertex $A$ when it has crawled exactly 7 meters. Find the value of $n$.
## Solutions
### Solution 1
Let $P(n)$ denote the probability that the bug is at $A$ after it has crawled $n$ meters. Since the bug can only be at vertex $A$ if it just left a vertex which is not $A$, we have $P(n + 1) = \frac13 (1 - P(n))$. We also know $P(0) = 1$, so we can quickly compute $P(1)=0$, $P(2) = \frac 13$, $P(3) = \frac29$, $P(4) = \frac7{27}$, $P(5) = \frac{20}{81}$, $P(6) = \frac{61}{243}$ and $P(7) = \frac{182}{729}$, so the answer is $\boxed{182}$. One can solve this recursion fairly easily to determine a closed-form expression for $P(n)$.
### Solution 2
We can find the number of different times the bug reaches vertex $A$ before the 7th move, and use these smaller cycles to calculate the number of different ways the bug can end up back at $A$.
Define $f(x)$ to be the number of paths of length $x$ which start and end at $A$ but do not pass through $A$ otherwise. Obviously $f(1) = 0$. In general for $f(x)$, the bug has three initial edges to pick from. From there, since the bug cannot return to $A$ by definition, the bug has exactly two choices. This continues from the 2nd move up to the $(x-1)$th move. The last move must be a return to $A$, so this move is determined. So $f(x) = 2^{x-2}3$.
Now we need to find the number of cycles by which the bug can reach $A$ at the end. Since $f(1) = 0$, $f(6)$ cannot be used since on the 7th move the bug cannot move from $A$ to $A$. So we need to find the number of partitions of 7 using only 2,3,4,5, and 7. These are $f(2)f(2)f(3)$, $f(2)f(5)$, $f(3)f(4)$, and $f(7)$. We can calculate these and sum them up using our formula. Also, order matters, so we need to find the number of ways to arrange each partition.
${3\choose1}f(2)f(2)f(3) + {2\choose1}f(2)f(5) + {2\choose1}f(3)f(4) + f(7)$
$= 3(3)(3)(2\cdot3) + 2(3)(2^33) + 2(2\cdot3)(2^23) + (2^53)$
$= 546$
Finally, this is a probability question, so we divide by $3^7$: $\frac{546}{3^7} = \frac{182}{3^6}$.
### Solution 3
There exists a simple heuristic method to arrive at the answer to this question, due to Simon Rubinstein-Salzedo, as follows: after a couple of moves, the randomness of movement of the bug and smallness of the system ensures that we should expect its probability distribution to be very close to uniform. In particular, we would expect $P(n)$ to be very close to $\frac 14$ for decently-sized $n$, for example $n = 7$. (In fact, from looking at the previous solution we can see that it is already close when $n = 3$, and in fact, the earlier values are also the best possible approximations given the restraints on where the bug can be.) Since we know the answer is of the form $\frac n{729}$, we realize that $n$ must be very close to $\frac{729}{4} = 182.25$, as indeed it is.
### Solution 4 (Cheap Solution)
Only do this if you don't know how to solve the problem and want to make a good guess. Since there are 4 vertices of a tetrahedron, there is approximately a 1/4 probability of coming back to A after 7 moves. Dividing 729 by 4 gives a number between 182 and 183. If the ant continuously alternates his location from A to some other vertice, in the end, it will not be at A. Therefore we chose the smaller number, 182. - Rocket123
### Solution 5
Let $a_n$ denote the number of ways that the bug arrives at $A$ after crawling $n$ meters, then we have $a_1=0$.
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# 180 Days of Math for Fourth Grade Day 64 Answers Key
By accessing our 180 Days of Math for Fourth Grade Answers Key Day 64 regularly, students can get better problem-solving skills.
## 180 Days of Math for Fourth Grade Answers Key Day 64
Directions Solve each problem
Question 1.
Subtract 19 from 36.
______________
36 – 19 = 17
Explanation:
Subtract 19 from 36 is 17
Question 2.
Is 0.4 less than 0.04?
______________
Explanation:
0.4 = $$\frac{4}{10}$$
0.04 = $$\frac{0.4}{10}$$
So $$\frac{4}{10}$$ > $$\frac{0.4}{10}$$
Question 3.
81 ÷ 9 = ___
Explanation:
The first number, 81, is called the dividend.
The second number, 9 is called the divisor.
The remainder is 0
9 x 9 = 81
Question 4.
Divide 27 into 3 equal groups.
______________
27 ÷ 3 =9
Explanation:
Explanation:
The first number, 27, is called the dividend.
The second number, 3 is called the divisor.
The remainder is 0
3 x 9 = 27
Question 5.
Order the numbers from smallest to largest.
1,624; 1,264; 1,426
___ ____ ____
1,264 ; 1,426 ; 1,624
Explanation:
Question 6.
8 × = 64
8 × 8= 64
Explanation:
Multiplication and division are inversely proportional. So
8 X C = 64
C = 64 ÷ 8
C= 8
Question 7.
Circle the best estimate for the weight of the object.
100 g 2 kg 5 kg 10 kg
100 g
Explanation:
grams are basic unit to measure the weight. in given units, 100g is the basic unit.
Question 8.
Which would be the best measure for the width of your pointer finger: a foot, a degree, or a centimeter?
Question 9.
Does an angle get bigger if you make its lines longer?
______________
No. Scaling does not change angles .
Explanation:
Regardless of the construction, we then have that segment BC′ is a scaled version of BC with a scale factor of two and, similarly, BA is the scaled version of BA with scale factor two. Scaling does not change angles and we can see this in the picture as ∠A′BC′ is the same as ∠ABC.
Question 10.
Complete the chart below to represent the shaded part of the hundred grid.
shaded part of the hundred grid in fraction is $$\frac{46}{100}$$
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# Math Insight
### Forming planes
#### Plane determined by point and normal vector
Given a point $P$, there are many planes that contain $P$. However, assuming that we are living in three-dimensional space ($\R^3$), a plane is uniquely determined if we also specify a normal vector $\vc{n}$ (i.e., a vector that is perpendicular to the plane).
The following applet illustrates this fact. You can experiment with changing the point $P$ and the normal vector $\vc{n}$ and see how the plane changes.
Plane from point and normal vector. You can change the point $\color{red}{P}$ (in red) and the normal vector $\color{green}{\vc{n}}$ (in green) by dragging the colored balls. Dragging the mouse elsewhere rotates the whole figure.
Note that the plane doesn't care about the length of $\vc{n}$, only its direction. In fact, if you change $\vc{n}$ to point in the opposite direction, you still get the same plane back. Of course, changing $P$ to any other point on the plane, wouldn't change the plane either. (The plane should really extend out to infinity in all directions, so sliding it from side to side wouldn't actually change the plane.)
We can infer the equation for the plane from these properties. Let $\vc{a}$ be the vector representing the point $P$ (i.e., the vector from the origin to $P$). Let $\vc{x}=(x,y,z)$ be the vector representing another point $Q$ on the plane. Since we know $Q$ is in the plane, can you think of any condition on the vector from $P$ to $Q$, i.e., the vector $\vc{x} - \vc{a}$? In below applet, see if you can infer the relationship between the vector $\vc{x} - \vc{a}$ and the vector $\vc{n}$.
Plane from point and normal vector with additional point in the plane. In additional to the point $\color{red}{P}$ (in red) and the normal vector $\color{green}{\vc{n}}$ (in green) that determine the plane, there is another point $Q$ (in yellow) that is constrained to lie in the plane. The vector from $P$ to $Q$ is shown in cyan. You can move the points $P$ and $Q$ and the vector $\vc{n}$ by dragging them with the mouse. Dragging the mouse elsewhere rotates the whole figure.
If the point represented by $\vc{x}$ is in the plane, the vector $\vc{x}-\vc{a}$ must be parallel to the plane, hence perpendicular to the normal vector $\vc{n}$. Two vectors are perpendicular if their dot product is zero. We conclude that for any point represented by $\vc{x}$ that is in the plane, the following equation must be satisfied: \begin{align*} \vc{n} \cdot (\vc{x}-\vc{a})=0. \end{align*} This is the equation for the plane perpendicular to $\vc{n}$ that goes through the point represented by $\vc{a}$.
#### Plane determined by three points
We just determined that to write the equation for a plane, we want a point $P$ in the plane and a normal vector $\vc{n}$. But most of us know that three points determine a plane (as long as they aren't collinear, i.e., lie in straight line). Here is a plane determined by three such points.
Plane determined from three points. The plane is determined by the points $\color{red}{P}$ (in red), $\color{green}{Q}$ (in green), and $\color{blue}{R}$ (in blue), which you can move by dragging with the mouse. The vectors from $\color{red}{P}$ to both $\color{green}{Q}$ and $\color{blue}{R}$ are drawn in the corresponding colors. The normal vector (in cyan) is the cross product of the green and blue vectors.
Since a plane is given by a point (say $\color{red}{P}$) and normal vector, somehow the addition of two points (say, $\color{green}{Q}$ and $\color{blue}{R}$ must determine the normal vector. Following the above logic, the normal vector must be perpendicular to the vector from $\color{red}{P}$ to $\color{green}{Q}$ and the vector from $\color{red}{P}$ to $\color{blue}{R}$. One way to obtain a vector perpendicular to two vectors is take their cross product. The normal vector in the above applet is indeed the cross product of those two vectors.
In summary, if you are given three points, you can take the cross product of the vectors between two pairs of points to determine a normal vector $\vc{n}$. Pick one of the three points, and let $\vc{a}$ be the vector representing that point. Then, the same equation described above, \begin{align*} \vc{n} \cdot (\vc{x}-\vc{a})=0. \end{align*} is the equation for the plane going through the three points.
If you find yourself in a position where you want to find the equation for a plane, look for a way to determine both a normal vector $\vc{n}$ and a point $\vc{a}$ through the plane. Then, you can simply use the above equation. You can see some examples for finding the equation of a plane.
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# How to Find Factors of a Positive Or Negative Number
How to Find Factors of a Positive or Negative Number? The first thing to know is that the factors of a positive number must multiply in order to get the corresponding negative number. For example, the factors of -30 are -1, -2, 3, -5, -6, -10, 15 and so on. Then, we should be able to multiply all these factors to get the corresponding negative number.
Prime factorization
Prime factorization is the process of finding the prime factors of a number. The prime factors of a number are smaller than the original number. Therefore, the number must be divided into smaller parts. This process will help you determine if the number is prime. If it is, then you should write it as a prime factor. You can also use a factor tree to solve factorization problems. Start by drawing two “branches” dividing the original number. At the end of each branch, write the two prime factors. Repeat this process until you find the prime factorization of the number.
Prime factors are the products of two tcn micro sites or more prime factors. Besides, these prime factors are always the same. If a number is divisible by two, then it is prime. Therefore, a prime factorization of a number would have two prime factors and two non-prime factors. This way, you would avoid accidental over-duplication or omission of factors. It will also help you solve a number problem more quickly.
Prime factorization of a number is the process of finding the prime number that divides a given number. It is very simple to perform when working with divisibility or fractions. However, it is more challenging for non-divisible numbers, because you would have to find the prime factors of the quotient. If you can’t find a number with two prime factors, you can use a prime factorization calculator to do the job for you.
One of the best tools to determine the prime factorization of a number is a prime factor tree. This picture shows the prime factorization process. The top of the prime factor tree is the integer N, and from there, you draw branches to the two positive factors of the number. Repeat this process for each number on the end of each branch, until you reach the end of the tree. Eventually, each leaf of the tree will be a prime factor.
Divisibility rule of 7
The Divisibility Rule of Seven can be applied to numbers, including the odd numbers. Simply double the number that is divisible by 7 and take the difference. If the result is a multiple of seven, the number is divisible by ten. If not, the number is not divisible by seven. For example, a 6-digit number is divisible by seven only if the result is 5a+4b+6c+2d+3e+f.
The divisibility rule of seven also applies to prime numbers. Prime numbers have two forms: a divisor that is divisible by seven, and a divisor that is not divisible by seven. Using the divisibility rule of 7 can help you with quick calculations, including problem-solving in exams. In this article, we’ll discuss the rule of seven and other divisibility rules. If you don’t know any of these rules, read on.
The Divisibility Rule of Seven is a fundamental principle in mathematics. The last three digits of a number must be divisible by nine or eight. For example, 34 and 442 are not divisible by seven. Likewise, 1899 is not divisible by nine. The rule applies to both even and odd numbers. In mathematics, the rule of seven is based on dividing the last digit of a number by seven.
In other words, a number divisible by seven is 798. The first digit of the divisor of a number is zero and the remainder is a whole number. The second digit must be a multiple of seven. The last digit of the number must be an even number. If it is an odd number, it must be divisible by two. Likewise, a number divisible by six is a multiple of two.
Divisibility rule of 2
The Divisibility Rule of Two is a mathematical rule for determining whether a number is divisible by two. By this rule, the last digit of a number must be an even number, such as two, four, or six. This is especially important when dealing with odd numbers, such as those greater than eight. However, even numbers are still divisible by two. So, when you’re working with odd numbers, you can apply this rule to them too.
The divisibility rule of two applies to both long and short numbers. Integers should have last digits that are divisible by either 3 or four. Therefore, a number with a last digit of two is divisible by both three and five. If the number is divisible by two, the resulting number should be a multiple of seventeen or twenty-four. The prime factor table may be useful to determine if a number is divisible by two.
The divisibility rule of two is especially useful when reducing fractions with large numbers to their lowest terms. It helps you determine the actual divisor of any number, even if it’s one that isn’t completely divisible by any other number. This rule applies to all numbers between two and nine. You don’t have to use this rule when dividing a large number into small ones, but it will certainly help you get around fractions of large numbers.
Using the divisibility rule of two can be beneficial in a variety of different situations, including math. In many cases, it will help you identify the prime factorization of a number. In general, it’s a good idea to check the last digit of any number to determine if it is even or odd. Once you’ve identified which numbers are divisible by two, you can use the divisibility rule of two to determine whether the number is prime-factorized.
Divisibility rule of 3
The Divisibility Rule of 3 is a useful tool for finding factors of a number. It tells you which numbers divide into the given number exactly without leaving a remainder. The prime factors are 2, 3, 5, 7, 11 and 13, but not 1. You can also use the Rule of Divisibility to determine the factorization of a number. For example, 120 can be divided into two by using the rule of divisibility. The result of this process is a square number.
The simplest way to understand the Rule of Divisibility is to use it to divide a number into a sum of its digits. For example, if a number is 1377, the quotient will be 459 and the remainder will be zero. In the same way, the quotient of a number that ends in 2 is 2, and so is a number like 212.
The Rule of Divisibility of a Number is a basic math concept and can be used to solve problems. For example, if you have the number 74, you can use the rule of 2 to find the factors. Even numbers, on the other hand, end with 6 and are divisible by two. Likewise, if a number is in the ones and tens place, it must be divisible by four.
The Rule of 3 is also helpful in determining the divisibility of a number. For example, if a number is three-digits long, it will not be divisible by two. The number 66, however, is divisible by three. The same goes for 177. If a number is divisible by three, the sum of its digits must be three-digits long.
Total number of factors
The total number of factors of a given numbers is a mathematical concept that is used to find a fraction. For example, the factors of a number 30 are 1, 2, and 3. Each of these prime factors can be counted one time, resulting in a total of eight factors. This is called the prime number system. In addition, number theory can be applied to the problem of finding the total number of factors of a number.
A number can have as many as four factors if the factors are equal to the number. However, it cannot be larger than the number. In other words, an integer must have two factors, and an odd number cannot have a factor that is larger than the number itself. A number’s factors can be a sum or product of other numbers. It can also be an even or odd number, perfect square or cube, and so on. Factoring is a common way to simplify algebraic expressions.
In the mathematical world, the factors of a number are the numbers that divide that number into exactly the same amount. For example, if a number is 7 in base two, then it is a prime factor. For example, if you divide a number by seven, you get a prime factor. You can also use a factor calculator to find the factors of a number. You can even try using a factor calculator to find the factors of any positive integer. You can use a step-by-step guide to find the factors of any number.
The prime factorization method can be used to find the factors of 12 in a number. A number with two prime factors is called a prime factor. It is used to find the total number of factors of a number. Therefore, the prime factorization method is used to find a prime number. There are also many common factors of twelve. In general, a number with two prime factors is a factor of three other prime numbers.
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### What is a Dot Product?
The dot product is a way of multiplying two vectors together. It is calculated by multiplying the corresponding elements of the two vectors and then adding the products together. The result of the dot product is a single number.
### Interpreting the Dot Product
The result of the dot product is a single number. This number represents the amount of overlap between the two vectors. In this case, the two vectors have a lot of overlap, so the dot product is a large number.
|a| indicates the magnitude (length) of vector a, |b| the magnitude of vector b, and theta the angle between the two vectors. The magnitudes of the vectors and the cosine of the angle between them are multiplied to get the dot product.
$$\vec{a} \cdot \vec{b} = |a| |b| \cos \theta$$
The dot product is calculated by multiplying the respective vector components and adding the results.
$$\vec{a} \cdot \vec{b} = a_x b_x + a_y b_y + a_z b_z + \ldots$$
If the two vectors are perpendicular to each other, the dot product will be zero. This is because the product of any two perpendicular vectors will always be zero.
If the two vectors are in opposite directions, the dot product will be negative. This is because the product of any two vectors in opposite directions will always be negative.
The dot product is a mathematical operation that takes two vectors and produces a scalar.
A vector represents a quantity characterized by both its magnitude and direction.
The dot product is usefull in tasks such as text classification and recommendation systems, where it helps in quantifying the similarity between vectors. Additionally, it can assist in measuring the distance between vectors, enabling applications such as clustering and dimensionality reduction.
### Practical Examples
Suppose we are developing a movie recommendation system in which movies are described on a scale of 0 to 10 based on two features: action and romance.
We have three films:
• Movie A: Action = 8, Romance = 3
• Movie B: Action = 2, Romance = 7
• Movie C: Action = 5, Romance = 5
A user watches Movie A and gives it a high rating. This indicates that the user prefers movies with a high level of action and a moderate level of romance. We may describe the user’s preference as a vector, which is the same as the Movie A vector in this case:
user = np.array([8, 3])
Now we’d like to recommend another film to this user. The dot product can assist us in determining which movie best matches the user’s preferences.
• The dot product of user and movieB is 29, indicating a moderate overlap.
• The dot product of user and movieC is 55, suggesting a higher degree of overlap than between user and movieB.
Based on these dot products, we would recommend Movie C to this user because it has a higher dot product, indicating that it is more in accordance with the customer’s preferences.
### Visualizing the Dot Product
The following visualization will help you grasp the concepts more intuitively and deepen your comprehension of this essential mathematical operation.
NoteBook: Dot Product
### Cons
• Limited to Euclidean Space: The standard dot product is applicable only in Euclidean space.
• Dependent on Magnitude: The dot product is influenced not only by the angle between two vectors, but also by their magnitude. As a result, two vector pairs with the same orientation but different lengths will have different dot products. This can be a disadvantage in some situations, such as when you’re only concerned with the vectors’ direction.
• Limited Contextual Information: In text analysis, two documents may have a high dot product because they use similar words, but those words might be used in entirely different contexts, leading to potential misinterpretation of the similarity.
• No Cross-Dimensional Information: In 3D space, the dot product won’t tell you if two vectors are twisting around each other. This information can only be obtained using a different operation known as the cross product.
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# Delaware - Grade 1 - Math - Operations and Algebraic Thinking - Number Sentences - 1.OA.7 , 1.OA.8
### Description
1.OA.7 Understand the meaning of the equal sign, and determine if equations involving addition and subtraction are true or false. For example, which of the following equations are true and which are false? 6 = 6, 7 = 8 – 1, 5 + 2 = 2 + 5, 4 + 1 = 5 + 2. 1.OA.8 Determine the unknown whole number in an addition or subtraction equation relating three whole numbers. For example, determine the unknown number that makes the equation true in each of the equations 8 + ? = 11, 5 = _ – 3, 6 + 6 = _.
• State - Delaware
• Standard ID - 1.OA.7 , 1.OA.8
• Subjects - Math Common Core
### Keywords
• Math
• Operations and Algebraic Thinking
## More Delaware Topics
Apply properties of operations as strategies to add and subtract.2 Examples: If 8 + 3 = 11 is known, then 3 + 8 = 11 is also known. (Commutative property of addition.) To add 2 + 6 + 4, the second two numbers can be added to make a ten, so 2 + 6 + 4 = 2 + 10 = 12. (Associative property of addition.)
Understand that the two digits of a two-digit number represent amounts of tens and ones. Understand the following as special cases: A. 10 can be thought of as a bundle of ten ones — called a “ten.” B. The numbers from 11 to 19 are composed of a ten and one, two, three, four, five, six, seven, eight, or nine ones. C. The numbers 10, 20, 30, 40, 50, 60, 70, 80, 90 refer to one, two, three, four, five, six, seven, eight, or nine tens (and 0 ones).
Express the length of an object as a whole number of length units, by laying multiple copies of a shorter object (the length unit) end to end; understand that the length measurement of an object is the number of same-size length units that span it with no gaps or overlaps. Limit to contexts where the object being measured is spanned by a whole number of length units with no gaps or overlaps.
Distinguish between defining attributes (e.g., triangles are closed and three-sided) versus non-defining attributes (e.g., color, orientation, overall size) ; build and draw shapes to possess defining attributes.
Compose two-dimensional shapes (rectangles, squares, trapezoids, triangles, half-circles, and quarter-circles) or three-dimensional shapes (cubes, right rectangular prisms, right circular cones, and right circular cylinders) to create a composite shape, and compose new shapes from the composite shape.1
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## Thursday, March 8, 2012
### Area of an Irregular Rectangular Figure
Finding Area for
An Irregular Rectangular Shape
When you first take a look at the irregular rectangular shape below, you might be unsure how to find the area for the modified shape. However, finding the area for an irregular shape like the one below takes some steps in a process, but there are two different techniques to use to help find the area. First, let's take a glance at our irregular shape below.
IMPORTANT PRECAUTIONS: When you see the irregular shape, be sure to label the missing measurements so you can go on with either of the two techniques. For the width, we know that one part of the width is 5 units, and another unlabeled width line. To find the length of this line, we must subtract 7-5. So that answer is two units. So that line is 2 units. For the length, one part of the length on the right is 3 units. So we must subtract 5-3 to find the length. So that line is 2 units long. Now, our irregular shape has all the lines labeled. This can be done because opposite sides of a rectangle are equal. Since this shape can fit inside a rectangle and all lines are horizontal and vertical and create right angles, you can determine missing side lengths by subtracting the known lengths.
So now that all the lines are labeled, it's time to use either of the two ways to find our area.
WAY #1: The first way to find the area of an irregular shape is to divide it into multiple rectangles. This will make finding the area much simpler, since you will be finding the area of each divided rectangle and adding up all the areas of each of those rectangles to find the total area.
Now that the shape has been divided into smaller rectangles, let's find the area of each of the rectangles. For Shape 1: 5 X 5 = 25 units squared. For Shape 2: 2 X 3 = 6 units squared. So 25 + 6 units squared gives us our total area of 31 square units. Be sure you are using the appropriate lengths when you are finding area. Notice to find the area of Rectangle 1 we are multiplying 5 X 5 not 5 X 7 even though the length of the whole side is 7 units. That length is not the length of the rectangle we broke out of our entire figure.
Now, for the second way to find the area.
WAY #2: The other way is to make the modified shape appear as a rectangle. So, let's go ahead and make the shape into an actual rectangle.
So now, we multiply 7 X 5 to get 35 units squared for our full rectangle. After that, we will now find the area of the rectangle that was the empty space. Since we know the length and width of the rectangle, we multiply 2 X 2 to get 4 units squared. Now, we subtract the area of the full rectangle from the empty space that was made a rectangle. 35 - 4 = 31 square units, which was the answer we got from the other way to fin d the area! Using either of these processes is a matter of preference, and both will work.
JUST BE SURE TO:
• TO REMEMBER TO LABEL ALL THE LINES IN THE SHAPE BEFORE PROCEEDING WITH THESE STEPS.
• MAKE SURE YOU KNOW THE AREA OF THE DIVIDED RECTANGLE YOU ARE ADDING/SUBTRACTING.
Hope you enjoyed the article (and found it helpful) on finding the area for the irregular rectangular shape.
Post Written by: RC
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# Unit 5 Review Flashcards Unit 5 Review Flashcards ALA: Pre-Algebra Unit 5 Percentages.
## Presentation on theme: "Unit 5 Review Flashcards Unit 5 Review Flashcards ALA: Pre-Algebra Unit 5 Percentages."— Presentation transcript:
Unit 5 Review Flashcards Unit 5 Review Flashcards ALA: Pre-Algebra Unit 5 Percentages
X + X 100% is the WHOLE THING: 100% of 80 is 80
X + X 200% is the twice the whole: 200% of 80 is 160
X + X 50% is half: 50% of 80 is 40
X + X 25% is one-fourth: 25% of 80 is 20
X + X 10% is one-tenth, so move the decimal one spot to the left: 10% of 80 is 8
X + X 20% is twice as much as 10%: 20% of 80 is 16
X + X 15% is halfway between 10% and 20%: 15% of 80 is 12
X + X The meal cost \$12.00 What is a 20% tip?
X + X The meal cost \$12.00 10% of \$12.00 is \$1.20. Double that to get 20%, so the tip would be \$2.40.
X + X The meal cost \$25.00 A coupon gives you 20% off. How much will you pay?
X + X 10% of 25 is \$2.50. So 20% of 25 is double that, or \$5. You’ll save \$5, so you’ll pay \$20.
10 is 50% of what number?
One way: 50% means half. 10 is half of 20. Another way: Set up a proportion 50 = 10 50x = 1000, so x = 20. 100 x Another way: set up an equation 10 = 50% of x 10 =.5x x = 10/.5 = 20
10 is what percent of 50?
X + X One way: set up a proportion x = 10 100 50 50x = 1000, so x = 20. 10 is 20% of 50. Another way: equation 10 = x 50 so x =.2, which is 20%
What is 40% of 150?
In your head: 10% of 150 is 15. 40% is 4 times that much, or 4 x 15 = 60. Another way: proportion 40 = x 100x = 6000 so x = 60. 100 150 Another way: equation 40% of 150 = x, so.4(150) = x, so x = 60.
X + X 1.3% of children are allergic to peanuts. In a pediatric hospital with 2000 patients a year, how many would you expect to have peanut allergies?
X + X You could set up a proportion: 1.3 = x 100 2000 100x = 2600, so x = 26 kids allergic to peanuts.
30% of Americans have hay fever. Of those, 75% find over-the-counter allergy medications helpful. How many people in a town of 1200 would have hay fever that was NOT helped by over-the-counter medication?
X + X 30% of 1200 is 360 people with hay fever. 75% of these 360, or ¾ of them, ARE helped by over-the-counter meds. That’s 270 who ARE helped by the medicine. That leaves 360 – 270, or 90 people, who are NOT helped.
Change the percent to a decimal and a fraction: 43%
Change the percent to a decimal and a fraction: 43% =.43 = 43 100
Change the percent to a decimal and a fraction: 135%
Change the percent to a decimal and a fraction: 135% = 1.35 = 135 = 27 100 20
Change the percent to a decimal and a fraction:.6%
Change the percent to a decimal and a fraction:.6% =.006 = 6 = 3 1000 500
Change the decimal to a fraction and a percent:.6
Change the decimal to a fraction and a percent:.6 = 6 = 3 = 60% 10 5
Change the decimal to a fraction and a percent:.07
Change the decimal to a fraction and a percent:.07 = 7 = 7% 100
Change the fraction to a decimal and a percent: 3 8
Change the fraction to a decimal and a percent: 3 =.375 = 37.5% 8
You are done with the ALA Unit 5 Review
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Class 9 Definitions, Intersecting Lines and Non-intersecting Lines & Pairs of Angles
### Topic covered
color{red} ♦ Introduction
color{red} ♦Basic Terms and Definitions
color{red} ♦Intersecting Lines and Non-intersecting Lines
color{red} ♦Pairs of Angles
### Introduction
As we know that a minimum of two points are required to draw a line. You have also studied some axioms and, with the help of these axioms, you proved some other statements.
Here,we will study the properties of the angles formed when two lines intersect each other, and also the properties of the angles formed when a line intersects two or more parallel lines at distinct points.
In your daily life, you see different types of angles formed between the edges of plane surfaces. For making a similar kind of model using the plane surfaces, you need to have a thorough knowledge of angles.
For instance, suppose you want to make a model of a hut to keep in the school exhibition using bamboo sticks. Imagine how you would make it?
You would keep some of the sticks parallel to each other, and some sticks would be kept slanted. Whenever an architect has to draw a plan for a multistory building, she has to draw intersecting lines and parallel lines at different angles.
In science, you study the properties of light by drawing the ray diagrams. For example, to study the refraction property of light when it enters from one medium to the other medium, you use the properties of intersecting lines and parallel lines.
When two or more forces act on a body, you draw the diagram in which forces are represented by directed line segments to study the net effect of the forces on the body.
At that time, you need to know the relation between the angles when the rays (or line segments) are parallel to or intersect each other. To find the height of a tower or to find the distance of a ship from the light house, one needs to know the angle formed between the horizontal and the line of sight.
Plenty of other examples can be given where lines and angles are used. In the subsequent chapters of geometry, you will be using these properties of lines and angles to deduce more and more useful properties.
### Basic Terms and Definitions
Recall that a part (or portion) of a line with two end points is called a "line-segment" and a part of a line with one end point is called a "ray."
"Note" that the line segment AB is denoted by bar(AB) , and its length is denoted by AB. The ray AB is denoted by bar(AB) , and a line is denoted by bar(AB) .
However, we will not use these symbols, and will denote the line segment AB, ray AB, length AB and line AB by the same symbol, AB. The meaning will be clear from the context. Sometimes small letters l, m, n, etc. will be used to denote lines.
If three or more points lie on the same line, they are called collinear points; otherwise they are called non-collinear points.
Recall that an angle is formed when two rays originate from the same end point. The rays making an angle are called the arms of the angle and the end point is called the vertex of the angle.
You have studied different types of angles, such as acute angle, right angle, obtuse angle, straight angle and reflex angle in earlier classes (see Fig. 6.1).
An acute angle measures between 0° and 90°, whereas a right angle is exactly equal to 90°. An angle greater than 90° but less than 180° is called an obtuse angle.
Also, recall that a straight angle is equal to 180°. An angle which is greater than 180° but less than 360° is called a reflex angle. Further, two angles whose sum is 90° are called complementary angles, and two angles whose sum is 180° are called supplementary angles.
You have also studied about adjacent angles in the earlier classes (see Fig. 6.2). Two angles are adjacent, if they have a common vertex, a common arm and their non-common arms are on different sides of the common arm.
In Fig. 6.2, angle ABD and angle DBC are adjacent angles. Ray BD is their common arm and point B is their common vertex. Ray BA and ray BC are non common arms.
Moreover, when two angles are adjacent, then their sum is always equal to the angle formed by the two noncommon arms. So, we can write
angle ABC = angle ABD + angle DBC.
Note that angle ABC and angle ABD are not adjacent angles. Why? Because their noncommon arms BD and BC lie on the same side of the common arm BA.
If the non-common arms BA and BC in Fig. 6.2, form a line then it will look like Fig. 6.3. In this case, angle ABD and angle DBC are called linear pair of angles.
You may also recall the vertically opposite angles formed when two lines, say AB and CD, intersect each other, say at the point O (see Fig. 6.4). There are two pairs of vertically opposite angles.
One pair is angle AOD and angle BOC. Can you find the other pair?
### Intersecting Lines and Non-intersecting Lines
Draw two different lines PQ and RS on a paper. You will see that you can draw them in two different ways as shown in Fig. 6.5 (i) and Fig. 6.5 (ii).
Recall the notion of a line, that it extends indefinitely in both directions. Lines PQ and RS in Fig. 6.5 (i) are intersecting lines and in Fig. 6.5 (ii) are parallel lines.
Note that the lengths of the common perpendiculars at different points on these parallel lines is the same. This equal length is called the distance between two parallel lines.
### Pairs of Angles
As we've learnt the definitions of some of the pairs of angles such as complementary angles, supplementary angles, adjacent angles, linear pair of angles, etc.
Now, let us find out the relation between the angles formed when a ray stands on a line. Draw a figure in which a ray stands on a line as shown in Fig. 6.6.
Name the line as AB and the ray as OC. What are the angles formed at the point O? They are angle AOC, angle BOC and angle AOB. Can we write angle AOC + angle BOC = angle AOB ? (1) Yes! (Why? Refer to adjacent angles in Section 6.2 ) What is the measure of angle AOB ? It is 180°.
From (1) and (2), can you say that angle AOC + angle BOC = 180° ? Yes ! (Why ?) From the above discussion, we can state the following Axiom :
color {blue} text(Axiom 6.1 : If a ray stands on a line, then the sum of two adjacent angles so formed is 180°).
Recall that when the sum of two adjacent angles is 180°, then they are called a linear pair of angles.
In Axiom 6.1, it is given that ‘a ray stands on a line’. From this ‘given’, we have concluded that ‘the sum of two adjacent angles so formed is 180°’. Can we write Axiom 6.1 the other way? That is, take the ‘conclusion’ of Axiom 6.1 as ‘given’ and the ‘given’ as the ‘conclusion’. So it becomes:
(A) If the sum of two adjacent angles is 180°, then a ray stands on a line (that is, the non-common arms form a line).
Now you see that the Axiom 6.1 and statement (A) are in a sense the reverse of each others. We call each as converse of the other. We do not know whether the statement (A) is true or not.
Let us check. Draw adjacent angles of different measures as shown in Fig. 6.7. Keep the ruler along one of the non-common arms in each case. Does the other non-common arm also lie along the ruler?
You will find that only in Fig. 6.7 (iii), both the non-common arms lie along the ruler, that is, points A, O and B lie on the same line and ray OC stands on it. Also see that angle AOC + angle COB = 125° + 55° = 180°. From this, you may conclude that statement (A) is true. So, you can state in the form of an axiom as follows:
Axiom 6.2 : If the sum of two adjacent angles is 180°, then the non-common arms of the angles form a line.
For obvious reasons, the two axioms above together is called the Linear Pair Axiom.
Let us now examine the case when two lines intersect each other.
Recall, from earlier classes, that when two lines intersect, the vertically opposite angles are equal. Let us prove this result now. See Appendix 1 for the ingredients of a proof, and keep those in mind while studying the proof given below.
Theorem 6.1 : If two lines intersect each other, then the vertically opposite angles are equal.
Proof : In the statement above, it is given that ‘two lines intersect each other’. So, let AB and CD be two lines intersecting at O as shown in Fig. 6.8. They lead to two pairs of vertically opposite angles, namely,
(i) angle AOC and angle BOD (ii) angle AOD and
angle BOC.
We need to prove that angle AOC = angle BOD
and angle AOD = angle BOC.
Now, ray OA stands on line CD.
Therefore, angle AOC + angle AOD = 180° (Linear pair axiom) (1)
Can we write angle AOD + angle BOD = 180°? Yes! (Why?) (2)
From (1) and (2), we can write
angle AOC + angle AOD = angle AOD + angle BOD
This implies that angle AOC = angle BOD (Refer Section 5.2, Axiom 3)
Similarly, it can be proved that angle AOD = angle BOC
Now, let us do some examples based on Linear Pair Axiom and Theorem 6.1.
Q 3200367218
In Fig. 6.9, lines PQ and RS intersect each other at point O. If angle POR : angle ROQ = 5 : 7, find all the angles.
Class 9 Chapter 6 Example 1
Solution:
angle POR + angle ROQ = 180°
(Linear pair of angles)
But angle POR : angle ROQ = 5 : 7
(Given)
Therefore, angle POR = 5/(12) xx 180° = 75°
Similarly, angle ROQ = 7/(12) xx 180° = 105°
Now, angle POS = angle ROQ = 105° (Vertically opposite angles)
and angle SOQ = angle POR = 75°
Q 3210367219
In Fig. 6.10, ray OS stands on a line angle POQ. Ray OR and ray OT are angle bisectors of angle POS and angle SOQ, respectively. If angle POS = x, find angle ROT.
Class 9 Chapter 6 Example 2
Solution:
Ray OS stands on the line POQ.
Therefore, angle POS + angle SOQ = 180°
But, angle POS = x
Therefore, x + angle SOQ = 180°
So, angle SOQ = 180° – x
Now, ray OR bisects angle POS, therefore,
angle ROS = 1/2 xx angle POS
= 1/2 xx x = x/2
angle SOT = 1/2 xx angle SOQ
= 1/2 xx ( 180^o - x)
= 90^0 - x/2
Now, angle ROT = angle ROS + angle SOT
= x/2 + 90^0 - x/2
= 90^o
Q 3200467318
In Fig. 6.11, OP, OQ, OR and OS are four rays. Prove that angle POQ + angle QOR + angle SOR + angle POS = 360°.
Class 9 Chapter 6 Example 3
Solution:
In Fig. 6.11, you need to produce any of
the rays OP, OQ, OR or OS backwards to a point.
Let us produce ray OQ backwards to a point T so
that TOQ is a line (see Fig. 6.12).
Now, ray OP stands on line TOQ.
Therefore, angle TOP + angle POQ = 180° (1)
(Linear pair axiom)
Similarly, ray OS stands on line TOQ.
Therefore, angle TOS + angle SOQ = 180° (2)
But angle SOQ = angle SOR + angle QOR
So, (2) becomes
angle TOS + angle SOR + angle QOR = 180° (3)
Now, adding (1) and (3), you get
angle TOP + angle POQ + angle TOS + angle SOR + angle QOR = 360° (4)
But angle TOP + angle TOS = angle POS
Therefore, (4) becomes
angle POQ + angle QOR + angle SOR + angle POS = 360°
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absolute value
The abs function calculates online the absolute value of a number.
abs(-5) returns 5
Abs, online calculus
Summary :
The abs function calculates online the absolute value of a number.
abs online
Absolute value function
The absolute value of a real number is equal to that number if it is positive, the opposite of this number if it is negative. The absolute value note abs. With this notation :
• If x>=0 abs(x)=abs(x)=x
• If x<0 abs(x)=abs(x)=-x
Calculation of the absolute value
With the absolute value calculator, the function abs can calculate the absolute value online of a number.
To calculate the absolute value of a number, just enter the number and to apply the function abs. Thus, for calculating the absolute value of the number -5, you must enter abs(-5) or directly -5, if the button abs already appears, the result 5 is returned.
Derivative of absolute value
The derivative of the absolute value is equal to :
• 1 if x>=0,
• -1 if x<0
Antiderivative of absolute value
The antiderivative of the absolute value is equal to abs(x).
• intabs(x)=x^2/2 if x>=0,
• intabs(x)=-x^2/2 if x<0
Limits of absolute value
The limits of the absolute value exist at -oo and +oo:
• The absolute value function has a limit in -oo which is +oo .
• lim_(x->-oo)abs(x)=+oo
• The absolute value function has a limit in +oo which is +oo.
• lim_(x->+oo)abs(x)=+oo
Equation with absolute value
The calculator has a solver that allows him to solve a equation with absolute value . The calculations for obtaining the results are detailed, so it will be possible to solve equations like |x|=2 or like |2*x+4|=3 or like |(x^2-1)|=1 with the steps calculations.
Properties of the absolute value function
The absolute value function is an even function, for every real x, abs(-x)=abs(x). The consequence for the curve representative of the absolute value function is that it admits the axis of the ordinates as axis of symmetry.
Syntax :
abs(x), where x is a number
Examples :
abs(-5) returns 5
Derivative absolute value :
To differentiate function absolute value online, it is possible to use the derivative calculator which allows the calculation of the derivative of the absolute value function
The derivative of abs(x) is derivative(abs(x))=1
Antiderivative absolute value :
Antiderivative calculator allows to calculate an antiderivative of absolute value function.
An antiderivative of abs(x) is antiderivative(abs(x))=(x)^2/2
Limit absolute value :
The limit calculator allows the calculation of limits of the absolute value function.
The limit of abs(x) is limit(abs(x))
Graphic absolute value :
The graphing calculator is able to plot absolute value function in its definition interval.
Property of the function absolute value :
The absolute value function is an even function.
Calculate online with abs (absolute value)
List of related calculators :
• Absolute value : abs. The abs function calculates online the absolute value of a number.
• Arccosine : arccos. The arccos function allows the calculation of the arc cosine of a number. The arccos function is the inverse functions of the cosine function.
• Arcsine : arcsin. The arcsin function allows the calculation of the arc sine of a number. The arcsin function is the inverse function of the sine function.
• Arctangent : arctan. The arctan function allows the calculation of the arctan of a number. The arctan function is the inverse functions of the tangent function.
• Hyperbolic cosine : ch. The function ch calculates online the hyperbolic cosine of a number.
• Cosine : cos. The cos trigonometric function calculates the cos of an angle in radians, degrees or gradians.
• Cosecant : cosec. The trigonometric function sec allows to calculate the secant of an angle expressed in radians, degrees, or grades.
• Cotangent : cotan. The cotan trigonometric function to calculate the cotan of an angle in radians, degrees or gradians.
• Hyperbolic cotangent : coth. The coth function calculates online the hyperbolic cotangent of a number.
• Exponential : exp. The function exp calculates online the exponential of a number.
• Napierian logarithm : ln. The ln calculator allows to calculate online the natural logarithm of a number.
• Logarithm : log. The log function calculates the logarithm of a number online.
• Cube root : cube_root. The cube_root function calculates online the cube root of a number.
• Secant : sec. The trigonometric function sec allows to calculate the secant of an angle expressed in radians, degrees, or grades.
• Hyperbolic sine : sh. The sh function allows to calculate online the hyperbolic sine of a number.
• Sine : sin. The sin trigonometric function to calculate the sin of an angle in radians, degrees or gradians.
• Square root : sqrt. The sqrt function allows to calculate the square root of a number in exact form.
• Tangent : tan. The tan trigonometric function to calculate the tan of an angle in radians, degrees or gradians.
• Hyperbolic tangent : th. The function th allows to calculate online the hyperbolic tangent of a number.
Other resources
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# Three coins are tossed once. Find the probability of getting
Question:
Three coins are tossed once. Find the probability of getting
(vi) 3 tails
(vii) exactly two tails
(viii) no tail
(ix) at most two tails.
Solution:
When three coins are tossed once, the sample space is given by
S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
$\therefore$ Accordingly,$n(S)=8$
It is known that the probability of an event A is given by
$\mathrm{P}(\mathrm{A})=\frac{\text { Number of outcomes favourable to } \mathrm{A}}{\text { Total number of possible outcomes }}=\frac{n(\mathrm{~A})}{n(\mathrm{~S})}$
(i) Let B be the event of the occurrence of 3 heads. Accordingly, B = {HHH}
$\therefore \mathrm{P}(\mathrm{B})=\frac{n(\mathrm{~B})}{n(\mathrm{~S})}=\frac{1}{8}$
(ii) Let C be the event of the occurrence of 2 heads. Accordingly, C = {HHT, HTH, THH}
$\therefore \mathrm{P}(\mathrm{C})=\frac{n(\mathrm{C})}{n(\mathrm{~S})}=\frac{3}{8}$
(iii) Let D be the event of the occurrence of at least 2 heads.
Accordingly, D = {HHH, HHT, HTH, THH}
$\therefore \mathrm{P}(\mathrm{D})=\frac{n(\mathrm{D})}{n(\mathrm{~S})}=\frac{4}{8}=\frac{1}{2}$
(iv) Let E be the event of the occurrence of at most 2 heads.
Accordingly, E = {HHT, HTH, THH, HTT, THT, TTH, TTT}
$\therefore \mathrm{P}(\mathrm{E})=\frac{n(\mathrm{E})}{n(\mathrm{~S})}=\frac{7}{8}$
(v) Let F be the event of the occurrence of no head.
Accordingly, F = {TTT}
$\therefore \mathrm{P}(\mathrm{G})=\frac{n(\mathrm{G})}{n(\mathrm{~S})}=\frac{1}{8}$
(vii) Let H be the event of the occurrence of exactly 2 tails.
Accordingly, H = {HTT, THT, TTH}
$\therefore \mathrm{P}(\mathrm{H})=\frac{n(\mathrm{H})}{n(\mathrm{~S})}=\frac{3}{8}$
(viii) Let I be the event of the occurrence of no tail.
Accordingly, I = {HHH}
$\therefore \mathrm{P}(\mathrm{I})=\frac{n(\mathrm{I})}{n(\mathrm{~S})}=\frac{1}{8}$
(ix) Let J be the event of the occurrence of at most 2 tails.
Accordingly, I = {HHH, HHT, HTH, THH, HTT, THT, TTH}
$\therefore \mathrm{P}(\mathrm{J})=\frac{n(\mathrm{~J})}{n(\mathrm{~S})}=\frac{7}{8}$
|
$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$
# 2.1: Relations, Graphs, and Functions
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Skills to Develop
• State the domain and range of a relation.
• Identify a function.
• Use function notation
## Graphs, Relations, Domain, and Range
The rectangular coordinate system1 consists of two real number lines that intersect at a right angle. The horizontal number line is called the x-axis2, and the vertical number line is called the y-axis3. These two number lines define a flat surface called a plane4, and each point on this plane is associated with an ordered pair5 of real numbers $$(x, y)$$. The first number is called the $$x$$-coordinate, and the second number is called the $$y$$-coordinate. The intersection of the two axes is known as the origin6, which corresponds sto the point $$(0, 0)$$.
The $$x$$- and $$y$$-axes break the plane into four regions called quadrants7, named using roman numerals I, II, III, and IV, as pictured. The ordered pair $$(x, y)$$ represents the position of points relative to the origin. For example, the ordered pair $$(−4, 3)$$ represents the position $$4$$ units to the left of the origin, and $$3$$ units above in the second quadrant.
This system is often called the Cartesian coordinate system8, named after the French mathematician René Descartes (1596–1650).
Next, we define a relation9 as any set of ordered pairs. In the context of algebra, the relations of interest are sets of ordered pairs $$(x, y)$$ in the rectangular coordinate plane. Typically, the coordinates are related by a rule expressed using an algebraic equation. For example, both the algebraic equations $$y = |x| − 2$$ and $$x = |y| + 1$$ define relationships between $$x$$ and $$y$$. Following are some integers that satisfy both equations:
Here two relations consisting of seven ordered pair solutions are obtained:
$$y=|x|−2 has solutions \{(−3,1),(−2,0),(−1,−1),(0,−2),(1,−1),(2,0),(3,1)\}$$
$$and$$
$$x=|y|+1 has solutions \{(4,−3),(3,−2),(2,−1),(1,0),(2,1),(3,2),(4,3)\}$$
We can visually display any relation of this type on a coordinate plane by plotting the points.
The solution sets of each equation will form a relation consisting of infinitely many ordered pairs. We can use the given ordered pair solutions to estimate all of the other ordered pairs by drawing a line through the given points. Here we put an arrow on the ends of our lines to indicate that this set of ordered pairs continues without bounds.
The representation of a relation on a rectangular coordinate plane, as illustrated above, is called a graph10. Any curve graphed on a rectangular coordinate plane represents a set of ordered pairs and thus defines a relation.
The set consisting of all of the first components of a relation, in this case the x-values, is called the domain11. And the set consisting of all second components of a relation, in this case the y-values, is called the range12 (or codomain13). Often, we can determine the domain and range of a relation if we are given its graph.
Here we can see that the graph of $$y=|x|−2$$ has a domain consisting of all real numbers, $$ℝ=(−∞,∞)$$, and a range of all y-values greater than or equal to $$−2, [−2,∞)$$. The domain of the graph of $$x=|y|+1$$ consists of all x-values greater than or equal to $$1, [1,∞)$$, and the range consists of all real numbers, $$ℝ=(−∞,∞)$$.
Example $$\PageIndex{1}$$:
Determine the domain and range of the following relation:
Solution
The minimum $$x$$-value represented on the graph is $$−8$$ all others are larger. Therefore, the domain consists of all $$x$$-values in the interval $$[−8,∞)$$. The minimum $$y$$-value represented on the graph is $$0$$; thus, the range is $$[0,∞)$$.
Domain: $$[−8,∞)$$; range: $$[0,∞)$$
## Functions
Of special interest are relations where every $$x$$-value corresponds to exactly one $$y$$-value. A relation with this property is called a function14.
Example $$\PageIndex{2}$$:
Determine the domain and range of the following relation and state whether it is a function or not: $$\{(−1, 4), (0, 7), (2, 3), (3, 3), (4, −2)\}$$
Solution
Here we separate the domain (x-values), and the range (y-values), and depict the correspondence between the values with arrows.
The relation is a function because each x-value corresponds to exactly one y-value.
The domain is $$\{−1, 0, 2, 3, 4\}$$ and the range is $$\{−2, 3, 4, 7\}$$. The relation is a function.
Example $$\PageIndex{3}$$:
Determine the domain and range of the following relation and state whether it is a function or not: $$\{(−4, −3), (−2, 6), (0, 3), (3, 5), (3, 7)\}$$
Solution
The given relation is not a function because the $$x$$-value $$3$$ corresponds to two $$y$$-values. We can also recognize functions as relations where no $$x$$-values are repeated.
The domain is $$\{−4, −2, 0, 3\}$$ and the range is $$\{−3, 3, 5, 6, 7\}$$. This relation is not a function.
Consider the relations consisting of the seven ordered pair solutions to $$y = |x| − 2$$and $$x = |y| + 1$$. The correspondence between the domain and range of each can be pictured as follows:
Notice that every element in the domain of the solution set of $$y = |x| − 2$$ corresponds to only one element in the range; it is a function. The solutions to $$x = |y| + 1$$, on the other hand, have values in the domain that correspond to two elements in the range. In particular, the x-value $$4$$ corresponds to two y-values $$−3$$ and $$3$$. Therefore, $$x = |y| + 1$$ does not define a function.
We can visually identify functions by their graphs using the vertical line test15. If any vertical line intersects the graph more than once, then the graph does not represent a function.
The vertical line represents a value in the domain, and the number of intersections with the graph represent the number of values to which it corresponds. As we can see, any vertical line will intersect the graph of $$y=|x|−2$$ only once; therefore, it is a function. A vertical line can cross the graph of $$x=|y|+1$$ more than once; therefore, it is not a function. As pictured, the $$x$$-value $$3$$ corresponds to more than one $$y$$-value.
Example $$\PageIndex{1}$$:
Given the graph, state the domain and range and determine whether or not it represents a function:
Solution
From the graph we can see that the minimum $$x$$-value is $$−1$$ and the maximum $$x$$-value is $$5$$. Hence, the domain consists of all the real numbers in the set from $$[−1,5]$$. The maximum $$y$$-value is $$3$$ and the minimum is $$−3$$; hence, the range consists of $$y$$-values in the interval $$[−3,3]$$.
In addition, since we can find a vertical line that intersects the graph more than once, we conclude that the graph is not a function. There are many $$x$$-values in the domain that correspond to two $$y$$-values.
Domain: $$[−1,5]$$; range: $$[−3,3]$$; function: no
Exercise $$\PageIndex{1}$$
Given the graph, determine the domain and range and state whether or not it is a function:
Domain: $$(−∞,15]$$; range:$$ℝ$$; function: no
## Function Notation
With the definition of a function comes special notation. If we consider each x-value to be the input that produces exactly one output, then we can use function notation16:
$$f ( x ) = y$$
The notation $$f(x)$$ reads, “f of x” and should not be confused with multiplication. Algebra frequently involves functions, and so the notation becomes useful when performing common tasks. Here $$f$$ is the function name, and $$f (x)$$ denotes the value in the range associated with the value x in the domain. Functions are often named with different letters; some common names for functions are $$f, g, h, C$$, and $$R$$. We have determined that the set of solutions to $$y = |x| − 2$$ is a function; therefore, using function notation we can write:
\begin{aligned} y & = | x | - 2 \\ \color{Cerulean}{\downarrow} & \\ f ( x ) & = | x | - 2 \end{aligned}
It is important to note that $$y$$ and $$f(x)$$ are used interchangeably. This notation is used as follows:
$$\begin{array} { l } { f ( x )\:\: =\:\:\: | x | - 2 } \\ { \:\:\:\:\:\downarrow \quad\quad\quad \downarrow } \\ { f ( \color{Cerulean}{- 5}\color{Black}{ )} = | \color{Cerulean}{- 5}\color{Black}{ |} - 2 = 5 - 2 = 3 } \end{array}$$
Here the compact notation $$f(−5) = 3$$ indicates that where $$x = −5$$ (the input), the function results in $$y = 3$$ (the output). In other words, replace the variable with the value given inside the parentheses.
Functions are compactly defined by an algebraic equation, such as $$f (x) = |x| − 2$$. Given values for $$x$$ in the domain, we can quickly calculate the corresponding values in the range. As we have seen, functions are also expressed using graphs. In this case, we interpret $$f(−5) = 3$$ as follows:
Function notation streamlines the task of evaluating. For example, use the function $$h$$ defined by $$h (x) = \frac{1}{2} x − 3$$to evaluate for $$x$$-values in the set $$\{−2, 0, 7\}$$.
$$\begin{array} { c } { h ( \color{Cerulean}{- 2}\color{Black}{ )} = \frac { 1 } { 2 } ( \color{Cerulean}{- 2}\color{Black}{ )} - 3 = - 1 - 3 = - 4 } \\ { h ( \color{Cerulean}{0}\color{Black}{ )} = \frac { 1 } { 2 } ( \color{Cerulean}{0}\color{Black}{ )} - 3 = 0 - 3 = - 3 } \\ { h ( \color{Cerulean}{7}\color{Black}{ )} = \frac { 1 } { 2 } ( \color{Cerulean}{7}\color{Black}{ )} - 3 = \frac { 7 } { 2 } - 3 = \frac { 1 } { 2 } } \end{array}$$
Given any function defined by $$h(x) = y$$, the value $$x$$ is called the argument of the function17. The argument can be any algebraic expression. For example:
\begin{aligned} h ( \color{Cerulean}{4 a ^ { 3 }}\color{Black}{ )} & = \frac { 1 } { 2 } ( \color{Cerulean}{4 a ^ { 3 } }\color{Black}{)} - 3 = 2 a ^ { 3 } - 3 \\ h ( \color{Cerulean}{2 x - 1}\color{Black}{ )} & = \frac { 1 } { 2 } ( \color{Cerulean}{2 x - 1}\color{Black}{ )} - 3 = x - \frac { 1 } { 2 } - 3 = x - \frac { 7 } { 2 } \end{aligned}
Example $$\PageIndex{5}$$:
Given $$g ( x ) = x ^ { 2 }$$ , find $$g (−2), g ( \frac{1}{2} )$$, and $$g (x + h)$$ .
Solution
Recall that when evaluating, it is a best practice to begin by replacing the variables with parentheses and then substitute the appropriate values. This helps with the order of operations when simplifying expressions.
\begin{aligned} g ( \color{Cerulean}{- 2}\color{Black}{ )} & = ( \color{Cerulean}{- 2}\color{Black}{ )} ^ { 2 } = 4 \\ g ( \color{Cerulean}{\frac { 1 } { 2 }}\color{Black}{)} & = ( \color{Cerulean}{\frac { 1 } { 2 }} \color{Black}{)} ^ { 2 } = \frac { 1 } { 4 } \\ g (\color{Cerulean}{ x + h}\color{Black}{ )} & = ( \color{Cerulean}{x + h}\color{Black}{ )} ^ { 2 } = x ^ { 2 } + 2 x h + h ^ { 2 } \end{aligned}
$$g (−2) = 4,\: g ( \frac{1}{2} ) = \frac{1}{4} ,\: g (x + h) = x^{2} + 2xh + h^{2}$$
At this point, it is important to note that, in general, $$f (x + h) ≠ f (x) + f (h)$$. The previous example, where $$g (x) = x^{2}$$ , illustrates this nicely
$$\begin{array} { l } { g ( x + h ) \neq g ( x ) + g ( h ) } \\ { ( x + h ) ^ { 2 } \neq x ^ { 2 } + h ^ { 2 } } \end{array}$$
Example $$\PageIndex{6}$$:
Given $$f ( x ) = \sqrt { 2 x + 4 }$$ , find $$f (−2), f (0)$$, and $$f \left( \frac { 1 } { 2 } a ^ { 2 } - 2 \right)$$.
Solution
\begin{aligned} f ( \color{Cerulean}{- 2}\color{Black}{ )} & = \sqrt { 2 ( \color{Cerulean}{- 2}\color{Black}{ )} + 4 } = \sqrt { - 4 + 4 } = \sqrt { 0 } = 0 \\ f ( \color{Cerulean}{0}\color{Black}{ )} & = \sqrt { 2 ( \color{Cerulean}{0}\color{Black}{ )} + 4 } = \sqrt { 0 + 4 } = \sqrt { 4 } = 2 \\ f ( \color{Cerulean}{\frac { 1 } { 2 } a ^ { 2 } - 2}\color{Black}{ )} & = \sqrt { 2( \color{Cerulean}{ \frac { 1 } { 2 } a ^ { 2 } - 2}\color{Black}{)} + 4 } = \sqrt { a ^ { 2 } - 4 + 4 } = \sqrt { a ^ { 2 } } = | a | \end{aligned}
$$f (−2) = 0,\: f (0) = 2,\: f \left( \frac { 1 } { 2 } a ^ { 2 } - 2 \right)= |a|$$
Example $$\PageIndex{7}$$:
Given the graph of $$g(x)$$, find $$g(−8), g(0)$$, and $$g(8)$$.
Solution
Use the graph to find the corresponding $$y$$-values where $$x = −8, 0$$, and $$8$$.
$$g(−8)=−2,\: g(0)=0,\: g(8)=2$$
Sometimes the output is given and we are asked to find the input.
Example $$\PageIndex{8}$$:
Given $$f (x) = 5x + 7$$, find $$x$$ where $$f (x) = 27$$.
Solution
In this example, the output is given and we are asked to find the input. Substitute $$f (x)$$ with $$27$$ and solve.
$$\begin{array} { c } { f ( x ) = 5 x + 7 }\\\color{Cerulean}{\downarrow}\quad\quad\quad\:\:\: \\ { 27 = 5 x + 7 } \\ { 20 = 5 x } \\ { 4 = x } \end{array}$$
Therefore, $$f (4) = 27$$. As a check, we can evaluate $$f (4) = 5 (4) + 7 = 27$$.
$$x = 4$$
Example $$\PageIndex{9}$$:
Given the graph of $$g$$, find $$x$$ where $$g(x)=2$$.
Solution
Here we are asked to find the x-value given a particular y-value. We begin with 2 on the y-axis and then read the corresponding x-value.
We can see that $$g(x)=2$$ where $$x=−5$$; in other words, $$g(−5)=2$$.
$$x=−5$$
Exercise $$\PageIndex{2}$$
Given the graph of $$h$$, find $$x$$ where $$h(x)=-4$$
$$x=-15$$ and $$x=15$$
## Key Takeaways
• A relation is any set of ordered pairs. However, in this course, we will be working with sets of ordered pairs $$(x, y)$$ in the rectangular coordinate system. The set of $$x$$-values defines the domain and the set of $$y$$-values defines the range.
• Special relations where every $$x$$-value (input) corresponds to exactly one $$y$$-value (output) are called functions.
• We can easily determine whether or not an equation represents a function by performing the vertical line test on its graph. If any vertical line intersects the graph more than once, then the graph does not represent a function.
• If an algebraic equation defines a function, then we can use the notation $$f (x) = y$$. The notation $$f (x)$$ is read “f of x” and should not be confused with multiplication. When working with functions, it is important to remember that $$y$$ and $$f (x)$$ are used interchangeably.
• If asked to find $$f(a)$$, we substitute the argument $$a$$ in for the variable and then simplify. The argument could be an algebraic expression.
• If asked to find $$x$$ where $$f(x) = a$$, we set the function equal to $$a$$ and then solve for $$x$$.
Exercise $$\PageIndex{3}$$
Determine the domain and range and state whether the relation is a function or not.
1. $$\{ ( 3,1 ) , ( 5,2 ) , ( 7,3 ) , ( 9,4 ) , ( 12,4 ) \}$$
2. $$\{ ( 2,0 ) , ( 4,3 ) , ( 6,6 ) , ( 8,6 ) , ( 10,9 ) \}$$
3. $$\{ ( 7,5 ) , ( 8,6 ) , ( 10,7 ) , ( 10,8 ) , ( 15,9 ) \}$$
4. $$\{ ( 1,1 ) , ( 2,1 ) , ( 3,1 ) , ( 4,1 ) , ( 5,1 ) \}$$
5. $$\{ ( 5,0 ) , ( 5,2 ) , ( 5,4 ) , ( 5,6 ) , ( 5,8 ) \}$$
6. $$\{ ( - 3,1 ) , ( - 2,2 ) , ( - 1,3 ) , ( 0,4 ) , ( 0,5 ) \}$$
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
27.
28.
29.
30.
31.
32.
33.
34.
1. Domain: $$\{ 3,5,7,9,12 \}$$; range: $$\{ 1,2,3,4 \}$$; function: yes
3. Domain: $$\{ 7,8,10,15 \}$$; range: $$\{ 5,6,7,8,9 \}$$; function: no
5. Domain: $$\{5\}$$; range: $$\{ 0,2,4,6,8 \}$$; function: no
7. Domain: $$\{ - 4 , - 1,0,2,3 \}$$; range: $$\{ 1,2,3 \}$$; function: yes
9. Domain: $$\{−1, 0, 1, 2\}$$; range: $$\{0, 1, 2, 3, 4\}$$; function: no
11. Domain: $$\{−2\}$$; range: $$\{−4, −2, 0, 2, 4\}$$; function: no
13. Domain: $$ℝ$$; range: $$[−2, ∞)$$; function: yes
15. Domain: $$(−∞, −1]$$; range: $$ℝ$$; function: no
17. Domain: $$(−∞, 0]$$; range: $$[−1, ∞)$$; function: yes
19. Domain: $$ℝ$$; range: $$(−∞, 3]$$; function: yes
21. Domain: $$ℝ$$; range: $$ℝ$$; function: yes
23. Domain: $$[−5, −1]$$; range: $$[−2, 2]$$; function: no
25. Domain: $$ℝ$$; range: $$[0, ∞]$$; function: yes
27. Domain: $$ℝ$$; range: $$ℝ$$; function: yes
29. Domain: $$ℝ$$; range: $$[−1, 1]$$; function: yes
31. Domain: $$[−8, 8]$$; range: $$[−3, 3]$$; function: no
33. Domain: $$ℝ$$; range: $$[−8, ∞]$$; function: yes
Exercise $$\PageIndex{4}$$
Evaluate.
1. $$g ( x ) = | x - 5 | \text { find } g ( - 5 ) , g ( 0 ) , \text { and } g ( 5 )$$
2. $$g ( x ) = | x | - 5 ; \text { find } g ( - 5 ) , g ( 0 ) , \text { and } g ( 5 )$$
3. $$g ( x ) = | 2 x - 3 | ; \text { find } g ( - 1 ) , g ( 0 ) , \text { and } g \left( \frac { 3 } { 2 } \right)$$
4. $$g ( x ) = 3 - | 2 x | ; \text { find } g ( - 3 ) , g ( 0 ) , \text { and } g ( 3 )$$
5. $$f ( x ) = 2 x - 3 ; \text { find } f ( - 2 ) , f ( 0 ) , \text { and } f ( x - 3 )$$
6. $$f ( x ) = 5 x - 1 ; \text { find } f ( - 2 ) , f ( 0 ) , \text { and } f ( x + 1 )$$
7. $$g ( x ) = \frac { 2 } { 3 } x + 1 ; \text { find } g ( - 3 ) , g ( 0 ) , \text { and } f ( 9 x + 6 )$$
8. $$g ( x ) = - \frac { 3 } { 4 } x - \frac { 1 } { 2 } ; \text { find } g ( - 4 ) , g ( 0 ) , \text { and } g ( 6 x - 2 )$$
9. $$g ( x ) = x ^ { 2 } ; \text { find } g ( - 5 ) , g ( \sqrt { 3 } ) , \text { and } g ( x - 5 )$$
10. $$g ( x ) = x ^ { 2 } + 1 ; \text { find } g ( - 1 ) , g ( \sqrt { 6 } ) , \text { and } g ( 2 x - 1 )$$
11. $$f ( x ) = x ^ { 2 } - x - 2 ; \text { find } f ( 0 ) , f ( 2 ) , \text { and } f ( x + 2 )$$
12. $$f ( x ) = - 2 x ^ { 2 } + x - 4 ; \text { find } f ( - 2 ) , f \left( \frac { 1 } { 2 } \right) , \text { and } f ( x - 3 )$$
13. $$h ( t ) = - 16 t ^ { 2 } + 32 ; \text { find } h \left( \frac { 1 } { 4 } \right) , h \left( \frac { 1 } { 2 } \right) , \text { and } h ( 2 a - 1 )$$
14. $$h ( t ) = - 16 t ^ { 2 } + 32 ; \text { find } h ( 0 ) , h ( \sqrt { 2 } ) , h ( 2 a + 1 )$$
15. $$f ( x ) = \sqrt { x + 1 } - 2 \text { find } f ( - 1 ) , f ( 0 ) , f ( x - 1 )$$
16. $$f ( x ) = \sqrt { x - 3 } + 1 ; \text { find } f ( 12 ) , f ( 3 ) , f ( x + 3 )$$
17. $$g ( x ) = \sqrt { x + 8 } ; \text { find } g ( 0 ) , g ( - 8 ) , \text { and } g ( x - 8 )$$
18. $$g ( x ) = \sqrt { 3 x - 1 } ; \text { find } g \left( \frac { 1 } { 3 } \right) , g \left( \frac { 5 } { 3 } \right) , \text { and } g \left( \frac { 1 } { 3 } a ^ { 2 } + \frac { 1 } { 3 } \right)$$
19. $$f ( x ) = x ^ { 3 } + 1 ; \text { find } f ( - 1 ) , f ( 0 ) , f \left( a ^ { 2 } \right)$$
20. $$f ( x ) = x ^ { 3 } - 8 ; \text { find } f ( 2 ) , f ( 0 ) , f \left( a ^ { 3 } \right)$$
1. $$g ( - 5 ) = 10 , g ( 0 ) = 5 , g ( 5 ) = 0$$
3. $$g ( - 1 ) = 5 , g ( 0 ) = 3 , g \left( \frac { 3 } { 2 } \right) = 0$$
5. $$f ( - 2 ) = - 7 , f ( 0 ) = - 3 , f ( x - 3 ) = 2 x - 9$$
7. $$g ( - 3 ) = - 1 , g ( 0 ) = 1 , g ( 9 x + 6 ) = 6 x + 5$$
9. $$g ( - 5 ) = 25 , g ( \sqrt { 3 } ) = 3 , g ( x - 5 ) = x ^ { 2 } - 10 x + 25$$
11. $$f ( 0 ) = - 2 , f ( 2 ) = 0 , f ( x + 2 ) = x ^ { 2 } + 3 x$$
13. $$h \left( \frac { 1 } { 4 } \right) = 31 , h \left( \frac { 1 } { 2 } \right) = 28 , h ( 2 a - 1 ) = - 64 a ^ { 2 } + 64 a + 16$$
15. $$f ( - 1 ) = - 2 , f ( 0 ) = - 1 , f ( x - 1 ) = \sqrt { x } - 2$$
17. $$g ( 0 ) = 2 \sqrt { 2 } , g ( - 8 ) = 0 , g \left( a ^ { 2 } - 8 \right) = | a |$$
19. $$f ( - 1 ) = 0 , f ( 0 ) = 1 , f \left( a ^ { 2 } \right) = a ^ { 6 } + 1$$
Exercise $$\PageIndex{5}$$
Given the function find $$f(x+h)$$.
1. $$f ( x ) = 3 x - 1$$
2. $$f ( x ) = - 5 x + 2$$
3. $$f ( x ) = x ^ { 2 } + x + 1$$
4. $$f ( x ) = 2 x ^ { 2 } - x - 1$$
5. $$f ( x ) = x ^ { 3 }$$
6. $$f ( x ) = 2 x ^ { 3 } - 1$$
1. $$f ( x + h ) = 3 x + 3 h - 1$$
3. $$f ( x + h ) = x ^ { 2 } + 2 x h + h ^ { 2 } + x + h + 1$$
5. $$f ( x + h ) = x ^ { 3 } + 3 h x ^ { 2 } + 3 h ^ { 2 } x + h ^ { 3 }$$
Exercise $$\PageIndex{6}$$
Find $$x$$ given the function.
1. $$f ( x ) = 2 x - 3 ; \text { find } x \text { where } f ( x ) = 25$$
2. $$f ( x ) = 7 - 3 x ; \text { find } x \text { where } f ( x ) = - 27$$
3. $$f ( x ) = 2 x + 5 ; \text { find } x \text { where } f ( x ) = 0$$
4. $$f ( x ) = - 2 x + 1 ; \text { find } x \text { where } f ( x ) = 0$$
5. $$g ( x ) = 6 x + 2 ; \text { find } x \text { where } g ( x ) = 5$$
6. $$g ( x ) = 4 x + 5 ; \text { find } x \text { where } g ( x ) = 2$$
7. $$h ( x ) = \frac { 2 } { 3 } x - \frac { 1 } { 2 } ; \text { find } x \text { where } h ( x ) = \frac { 1 } { 6 }$$
8. $$h ( x ) = \frac { 5 } { 4 } x + \frac { 1 } { 3 } ; \text { find } x \text { where } h ( x ) = \frac { 1 } { 2 }$$
9. The value of a new car in dollars is given by the function $$V(t) = −1,800t + 22,000$$ where $$t$$ represents the age of the car in years. Use the function to determine the value of the car when it is $$4$$ years old. What was the value of the car new?
10. The monthly income in dollars of a commissioned car salesperson is given by the function $$I(n) = 350n + 1,450$$ where $$n$$ represents the number of cars sold in the month. Use the function to determine the salesperson’s income if he sells $$3$$ cars this month. What is his income if he does not sell any cars in one month?
1. $$x=14$$
3. $$x=-\frac{5}{2}$$
5. $$x=\frac{1}{2}$$
7. $$x=1$$
9. New: $$22,000$$; $$4$$ years old: $$14,800$$
Exercise $$\PageIndex{7}$$
Given the graph of the function $$f$$, find the function values.
1. Find $$f ( 0 ) , f ( 2 )$$ , and $$f ( 4 )$$.
2. Find $$f(-1), f(0)$$, and $$f(1)$$.
3. Find $$f (0), f (2)$$, and $$f (4)$$.
4. Find $$f (−3), f (0)$$, and $$f (3)$$.
5. Find $$f (−4), f (0)$$, and $$f (2)$$.\
6. Find $$f (−6), f (0)$$, and $$f (6)$$.
7. Find $$f (−2), f (2)$$, and $$f (7)$$.
8. Find $$f (0), f (5)$$, and $$f (9)$$.
9. Find $$f (−8), f (0)$$, and $$f (8)$$.
10. Find $$f (−12), f (0)$$, and $$f (12)$$.
1. $$f (0) = 5, f (2) = 1, f (4) = 5$$
3. $$f (0) = 0, f (2) = 2, f (4) = 0$$
5. $$f (−4) = 3, f (0) = 3, f (2) = 3$$
7. $$f (−2) = 1, f (2) = 3, f (7) = 4$$
9. $$f (−8) = 10, f (0) = 0, f (8) = 10$$
Exercise $$\PageIndex{8}$$
Given the graph of a function $$g$$, find the $$x$$-values:
1. Find $$x$$ where $$g (x) = 3, g (x) = 0$$, and $$g (x) = −2$$.
2. Find $$x$$ where $$g (x) = 0, g (x) = 1$$, and $$g (x) = 4$$.
3. Find $$x$$ where $$g (x) = −5, g (x) = 0$$, and $$g (x) = 10$$.
4. Find $$x$$ where $$g (x) = 0, g (x) = 10$$, and $$g (x) = 15$$.
5. Find $$x$$ where $$g (x) = −5, g (x) = −4$$, and $$g (x) = 4$$.\
6. Find $$x$$ where $$g (x) = 1, g (x) = 0$$, and $$g (x) = −3$$.
7. Find $$x$$ where $$g (x) = −4, g (x) = 3$$, and $$g (x) = 4$$.
8. Find $$x$$ where $$g (x) = −5, g (x) = −4$$, and $$g (x) = 4$$.
9. Find $$x$$ where $$g (x) = −10$$ and $$g (x) = 5$$.
10. Find $$x$$ where $$g(x)=2$$.
The value of a certain automobile in dollars depends on the number of years since it was purchased in $$1970$$ according to the following function:
11. What was the value of the car when it was new in $$1970$$?
12. In what year was the value of the car at a minimum?
13. What was the value of the car in $$2005$$?
14. In what years was the car valued at $$4,000$$?
1. $$g (−4) = 3, g (2) = 0$$, and $$g (6) = −2$$.
3. $$g (10) = −5, g (5) = 0$$ and $$g (15) = 0 , g (−5) = 10$$ and $$g (25) = 10$$
5. $$g (−2) = −5, g (−3) = −4$$ and $$g (−1) = −4 , g (−5) = 4$$ and $$g (1) = 4$$
7. $$g (−2) = −4, g (−1) = 3, g (0) = 4$$
9. $$g (−10) = −10$$ and $$g (5) = −10 ; g (−5) = 5$$ and $$g (10) = 5$$
11. $$5,000$$
13. $$10,000$$
Exercise $$\PageIndex{9}$$
Given the linear function defined by $$f(x)=2x-5$$, simplify the following.
1. $$f ( 5 ) - f ( 3 )$$
2. $$f ( 0 ) - f ( 7 )$$
3. $$f ( x + 2 ) - f ( 2 )$$
4. $$f ( x + 7 ) - f ( 7 )$$
5. $$f ( x + h ) - f ( x )$$
6. $$\frac { f ( x + h ) - f ( x ) } { h }$$
7. Simplify $$\frac { c ( x + h ) - c ( x ) } { h }$$ given $$c ( x ) = 3 x + 1$$.
8. Simplify $$\frac { p ( x + h ) - p ( x ) } { h }$$ given $$p ( x ) = 7 x - 3$$.
9. Simplify $$\frac { g ( x + h ) - g ( x ) } { h }$$ given $$g ( x ) = m x + b$$.
10. Simplify $$\frac { q ( x + h ) - q ( x ) } { h }$$ given $$q ( x ) = a x$$.
1. $$4$$
3. $$2x$$
5. $$2h$$
7. $$3$$
9. $$m$$
Exercise $$\PageIndex{10}$$
1. Who is credited with the introduction of the notation $$y = f (x)$$? Provide a brief summary of his life and accomplishments.
2. Explain to a beginning algebra student what the vertical line test is and why it works.
3. Research and discuss the life and contributions of René Descartes.
4. Conduct an Internet search for the vertical line test, functions, and evaluating functions. Share a link to a page that you think others may find useful.
## Footnotes
1A system with two number lines at right angles specifying points in a plane using ordered pairs $$(x, y)$$.
2The horizontal number line used as reference in a rectangular coordinate system.
3The vertical number line used as reference in a rectangular coordinate system.
4The flat surface defined by $$x$$- and $$y$$-axes.
5Pairs $$(x, y)$$ that identify position relative to the origin on a rectangular coordinate plane.
6The point where the $$x$$- and $$y$$-axes cross, denoted by $$(0, 0)$$.
7The four regions of a rectangular coordinate plane partly bounded by the $$x$$- and $$y$$-axes and numbered using the Roman numerals I, II, III, and IV.
8Term used in honor of René Descartes when referring to the rectangular coordinate system.
9Any set of ordered pairs.
10A visual representation of a relation on a rectangular coordinate plane.
11The set consisting of all of the first components of a relation. For relations consisting of points in the plane, the domain is the set of all $$x$$-values.
12The set consisting of all of the second components of a relation. For relations consisting of points in the plane, the range is the set of all $$y$$-values.
13Used when referencing the range.
14A relation where each element in the domain corresponds to exactly one element in the range.
15If any vertical line intersects the graph more than once, then the graph does not represent a function.
16The notation $$f (x) = y$$ , which reads “$$f$$ of $$x$$ is equal to $$y$$.” Given a function, $$y$$ and $$f (x)$$ can be used interchangeably.
17The value or algebraic expression used as input when using function notation.
|
## Lesson: Ordered Pairs Developing the Concept
Now that students have seen you graph points on the coordinate plane, it is time for them to try it. Be sure to have students use the proper vocabulary in explaining what they do.
Materials: overhead transparency or poster paper; grid paper and rulers for students
Preparation: Draw a coordinate grid on a transparency or poster paper.
Pass out grid paper with points A to H marked on them as shown on the grid below. (There are two points in each quadrant.) Put a transparency with the same grid on the overhead for the students to see. Point to points A and F as you ask the questions below.
• Ask: What are the coordinates for this point? Explain how you found the coordinates.
After doing this for both points A and F, have the class find the coordinates for the other six points. Have students explain how they found the coordinates.
• Ask: Where is the point (0, -4) on the plane? Explain.
Repeat for several other points found on either the x- or y-axis.
• Say: Plot on your graph the following points: (-3, 2), (3, -1), (2, 2), and (-4, -6). Have a volunteer plot the points on the overhead after students have plotted them at their desks.
• Say: Look at the points in the first quadrant. What can you tell me about the x- and y-coordinates for those points? (They are both positive numbers.) Now look at the points in the second, third, and fourth quadrants. See if you can tell me anything about the x- and y-coordinates for those points.
After students have told you about the x- and y-coordinates in each of the quadrants, ask questions like the following.
• Ask: In what quadrant would you find a positive x-coordinate and a negative y-coordinate? (IV) In what quadrant would you find a point with a negative x-coordinate and a negative y-coordinate? (III) Where would you find a point with a zero x-coordinate and a negative y-coordinate? (lower half of the y-axis) What ordered pair corresponds to the origin? (0, 0)
• Ask: Who can summarize the information about the signs of the coordinates in the first quadrant? (They are both positive.) Second quadrant? (xis negative; yis positive.) Third quadrant? (They are both negative.)Fourth quadrant? (x is positive; yis negative.)What about the points on the two axes? (Zero coordinates are neither positive nor negative.)
Wrap-Up and Assessment Hints
Provide students with coordinate grids (Learning Tool 20 in the Learning Tools Folder) and have them work with a partner to locate points on the grid. Encourage them to challenge each other by providing pairs of x- and y- coordinates that alternate between positive and negative signs — for example (3, -5) and (-3, 5). Have students label the points on their grids. Circulate to spot-check their answers.
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# Lesson Explainer: Power and Exponents over the Real Numbers Mathematics • 9th Grade
In this explainer, we will learn how to evaluate real numbers raised to positive and negative integer and zero powers and solve simple exponential equations.
Recall that to evaluate a power, we multiply the base the number of times indicated by the exponent. Letβs recap the definition below.
### Definition: Evaluating a Power
For a power with base , , and exponent , , then
Using this definition, we can evaluate not only powers where the base is an integer or a rational number, but also where the base is a real number.
One such type of real number we might evaluate is a square root. Recall that a square root is a number written in the form , . We can use the inverse operation, squaring, to derive two key results.
The first is that if you square , then you get , since squaring and square rooting are inverse operations (i.e., , ).
The second is that if you square root , you either get or , depending on whether is negative or positive. This is because when we square a negative number, say , we get a positive number, 9. However, if we square the additive inverse of , which is 3, we also get 9. This means when we square root a number squared, the result is positive, but the original number might have been negative. To avoid confusion, we then say that . However, if we know , then we can say .
Expanding on these definitions and using the definition of a power, we can say that and . Therefore, we get the following definition.
### Definition: Simplifying Expressions with Squares and Square Roots
For an expression with a base , where ,
• ,
• .
In our first example, we will apply what we know about evaluating powers and calculating squares of radicals to determine the value of an unknown.
### Example 1: Raising a Real Number to a Positive Integer Power
Determine the value of given that .
To find the value of , we need to substitute into . Doing so gives us
We know from the definition of a power that is the same as saying . Therefore, we get
To calculate this, we need to use , giving us
Therefore, the value of is 640.
As well as raising real numbers to the power of a positive integer, we can also raise them to the power of zero or to the power of a negative integer.
If we consider the pattern of powers of 2, we can see that as we increase the power by 1, we multiply by 2, as follows:
If we move backward in the sequence, we can therefore deduce that must be divided by 2. This then gives us . So, .
Similarly, if we move backward again, we can deduce the must be divided by 2. This then gives us . So, .
Again, if we continue moving backward, we can deduce that and and so on. If we write out the sequence again, we get the following:
Notice, that if we rewrite the denominators in the first three terms as powers of 2, we get the following:
We can then say that , , and . Continuing the sequence in the negative direction, we can generalize that , .
These results do not only hold for a base of 2 but also for any real number except zero. We can generalize these in the laws below.
### Laws: Rules for Zero and Negative Exponents
• The law for zero exponents is where .
• The law for negative exponents is where and .
In the following example, we will use the law for zero exponents to determine the value of an expression involving powers of radicals.
### Example 2: Raising a Real Number to a Zero Power
If , find .
To determine the value of , we can substitute , giving us
We know that , where ; therefore,
Therefore, the value of is 1.
In the next example, we will use the law for negative exponents to determine the value of an expression involving powers of radicals.
### Example 3: Raising a Real Number to a Negative Integer Power
If , find .
To find the value of when , we first need to substitute. This gives us
Next, to simplify , we use the law for negative exponents, which states that where and . This then gives us
We can use the rule , , to evaluate the denominator by expanding the power, giving us
Substituting this into the denominator, we get
Therefore, the value of when is .
In the next example, we will discuss how to evaluate an expression with positive and negative exponents.
### Example 4: Evaluating an Expression by Substituting Surds
If , , and , then find the value of in its simplest form.
To find the value of , we start by substituting , , and . Doing so gives us
We can simplify this further by expanding the brackets of the second term, giving us
Canceling common factors of in gives us
Substituting this back into the expression, we get
Next, we can use to evaluate the squared terms, giving us for the first, second, and fourth terms, respectively,
Substituting back into the expression, we get
Since is , then we can use to simplify this, giving us
Substituting back into the expression, we get
To simplify further, we need to find a common factor for the denominators of the fractions. For the first four terms, a common factor is 36, giving us
Finally, a common factor of 36 and 81 is 324, giving us
So far, we have considered how to simplify expressions with integer exponents. Next, we will discuss how to find an unknown exponent.
When comparing two powers with the same base, we can deduce that if the powers are equal, then their exponents must be equal as well. For example, if , then since the bases are the same, the exponents must be the same too, meaning .
Similarly, if we compare two powers with the same exponent, then if they are equal and odd, the bases must be equal. For example, if , then since the exponents are the same, then the bases must be the same too, meaning . However, if we have an even exponent, then it is possible that the bases are not equal, since one could be the negative of the other. For example, , so if the exponents are equal, then the base could be or 2 in order for them to be equal. This means that if the even exponents are equal, then we can say the moduli of the bases are equal. For example, if , then , so or . Letβs summarize these points in the rule below.
### Rule: Equating Powers with the Same Base or the Same Exponent
• If , where , then .
• If , then when and when .
We can use these rules to solve problems where there are unknowns in the exponent. Letβs discuss how to do this in the following example.
### Example 5: Solving a Simple Exponential Equation
Find the value of in the equation .
To find the value of in the equation , we can use the rule that states that if , where , then .
This rule will only work if both sides of the equation have the same base. As such, since the base of the power on the left-hand side of the equation is 3, then we want to write the right-hand with a base of 3 also.
We know that 81 is a power of 3, since . Therefore, . We can then say that the right-hand side is
Next, we can use the law of negative exponents, which states where and , meaning that for , we get
Putting this into the right-hand side of the original equation, we then get
Since the bases are now the same, we can equate the exponents, giving us
Solving for , we get
Therefore, the value of is for the equation .
In this explainer, we have learned how to evaluate real numbers raised to positive, zero, and negative exponents and how to solve simple exponential equations. Letβs recap the key points.
### Key Points
• We can use the property to simplify radicals raised to positive integers.
• We can use the law for zero exponents to simplify real numbers raised to the power of zero. The law states that where .
• We can use the law for negative exponents to simplify real numbers raised to negative powers. The law states that where and .
• We can use the rules for when bases or exponents are equal for equal powers to find unknowns in simple exponential equations. The rules state that
• if , where , then ,
• if , then when and when .
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# Domain solver
In this blog post, we discuss how Domain solver can help students learn Algebra. Our website can solving math problem.
## The Best Domain solver
Best of all, Domain solver is free to use, so there's no sense not to give it a try! One way to solve a problem is by using the process of elimination. This involves looking at all of the possible options and eliminating the ones that are not possible. For example, if you are trying to find out how many books are in a library, you would start by eliminating the options that are not possible. If there are only two books in the library, then you know that the answer is not three or four. You would continue this process until you are left with only one option. This can be a very effective way to solve problems, but it can also be time-consuming.
For example, consider the equation x2 + 6x + 9 = 0. To solve this equation by completing the square, we would first add a constant to both sides so that the left side becomes a perfect square: x2 + 6x + 9 + 4 = 4. Next, we would factor the trinomial on the left side to get (x + 3)2 = 4. Finally, we would take the square root of both sides to get x + 3 = ±2, which means that x = -3 ± 2 or x = 1 ± 2. In other words, the solutions to the original equation are x = -1, x = 3, and x = 5.
Substitution is a method of solving equations that involves replacing one variable with an expression in terms of the other variables. For example, suppose we want to solve the equation x+y=5 for y. We can do this by substituting x=5-y into the equation and solving for y. This give us the equation 5-y+y=5, which simplifies to 5=5 and thus y=0. So, the solution to the original equation is x=5 and y=0. In general, substitution is a useful tool for solving equations that contain multiple variables. It can also be used to solve systems of linear equations. To use substitution to solve a system of equations, we simply substitute the value of one variable in terms of the other variables into all of the other equations in the system and solve for the remaining variable. For example, suppose we want to solve the system of equations x+2y=5 and 3x+6y=15 for x and y. We can do this by substituting x=5-2y into the second equation and solving for y. This gives us the equation 3(5-2y)+6y=15, which simplifies to 15-6y+6y=15 and thus y=3/4. So, the solution to the original system of equations is x=5-2(3/4)=11/4 and y=3/4. Substitution can be a helpful tool for solving equations and systems of linear equations. However, it is important to be careful when using substitution, as it can sometimes lead to incorrect results if not used properly.
Trigonometry is used in a wide variety of fields, including architecture, engineering, and even astronomy. While the concepts behind trigonometry can be challenging, there are a number of resources that can help students to understand and master this important subject. Trigonometry textbooks often include worked examples and practice problems, while online resources can provide interactive lessons and quizzes. In addition, many math tutors offer trigonometry help specifically designed to address the needs of individual students. With a little effort, anyone can learn the basics of trigonometry and unlock its power to solve complex problems.
A calculus solver can be a helpful tool for these students. By enterinng the equation they are trying to solve, the solver will provide step-by-step instructions on how to solve it. This can be a valuable resource for students who are struggling to understand the material or simply need extra practice. With the help of a calculus solver, students can improve their grades and get a better understanding of the subject.
## We cover all types of math issues
Fun and an easy-to-use tool to work out math questions. The app has fun and cool animations which makes it look more professional and not boring. Doesn't have the full feature to work out more complicated questions and word problems but overall, best app to use for working out math questions.
Fiorella Scott
It's a really good app for students to learn and finish the Homework, and it is also a good ap for teachers to grade tests or Assignments 5 stars Works perfectly for all types of problems basic or hard this is a must have app if you’re in middle or high school
Olva Cook
How to do this math problem step by step Answers to word problems Solving matrices calculator with steps Solving radicals Math anwers
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# Resistive Circuit Analysis
1,459
18
3
## Introduction: Resistive Circuit Analysis
Circuit analysis is the process of determining
how the current and voltage supplied to a particular system affects it and its individual components. The completion of this Instructable will teach you a few of the fundamental tools that are used during the analysis of a circuit. Specifically you will learn about Ohm’s Law and resistance equivalencies. You will then apply these tools to determine the current and voltages of components on a circuit board.
Although there are many different components that can be present in any given circuit, for the purposes of this Instructable, only two types of components will be used, batteries and resistors. The function of a battery is common knowledge and does not need explaining. However, the function of the resistor may not be as common as the battery’s. The purpose of the resistor is explained by its name. It resists the flow of current. Resistors are typically used to control the amount of current flowing into a specific component of a circuit that would otherwise be destroyed by a higher current.
There is no previous experience needed for this project. But basic algebra is required if you are not using a scientific calculator. The only tools needed for this project are a pencil and paper.
### Teacher Notes
Teachers! Did you use this instructable in your classroom?
Add a Teacher Note to share how you incorporated it into your lesson.
## Step 1: Understanding Ohm’s Law
Due to their resistance (R), each resistor
maintains a relationship between its voltage (V) and the current (I) flowing through it. This relationship is described by Ohm’s Law which has the mathematical expression, V=IR. This shows that the voltage of a component is the product of its current and its resistance. Given that you are provided with at least two of the three variables you can determine the third. The units of measurement of voltage, current and resistance are volts, amps and ohms respectively. Ohms are typically depicted as the Greek letter Omega (Ω).
## Step 2: Understanding Component Arrangements
A very useful tool for circuit analysis is learning how to reduce a circuit to its equivalent resistance. The equivalent resistance of a circuit is the value of the algebraic summation of multiple resistors in an electric system. In some schematics of circuits that include multiple resistors, the schematic can be redrawn that only has a single resistor that is equivalent to the resistors that it replaced. This, in turn, makes the current flowing through the system easier to calculate. Before this is done, it is better that you have an understanding of the two basic component arrangements and how each of them affect the equivalent resistance (Req).
The first type of arrangement is similar to how the cars of a train are connected to each other. When the head of one resistor is connected to the tail of another, these two resistors are said to be in series. When two or more components share a connection, this connection is called a node. Components that are in series all share the same current.
The other type of component arrangement is formed when the head of one resistor is connected to the head of another resistor and the tail of both of those resistors are also connected to each other to form a loop. In other words, components in parallel share 2 nodes, as contrasted with sharing a single node in a series connection. Components that are in parallel have the same voltage.
In order for these arrangements to be true, the connection that each component shares has to be exclusive: The joint that the two components share cannot be shared by a third.
NOTE: Just because two components are not in series does not automatically mean that they are in parallel and vice versa. Some schematics contain components that are in neither parallel nor series and will use alternate methods that will not be discussed here.
## Step 3: Reducing the Circuit
When resistors are in series, the equivalent
resistance can be calculated by simply summing all of the values of the individual resistors together. Using this method, the equivalent resistance should always yield an equivalent resistance that is larger than the largest individual resistor.
Req = R1 + R2 + R3 + … (Series Equivalent Resistance)
When resistors are in parallel, the equivalent resistance is a bit more complex. The reciprocal of the equivalent resistor is equal to the summation of the individual reciprocal resistor values. Using this method, the equivalent resistance should always have a smaller value than the smallest individual resistor.
1/Req= 1/R1 + 1/R2 + 1/R3 + … (Parallel Equivalent Resistance)
Once the equivalent resistors have been calculated, additional equivalent resistors can also be calculated from those.
The following images will show you examples of resistors that are in series, parallel or neither. They will also show you how the components can be reduced.
## Step 4:
The first image shows that the only resistors in
series are resistors R3 and R4. Because of this, those resistors can be combined into a single resistor.
## Step 5:
Resistors R3 and R4 have now been reduced to form the equivalent resistor Req1. Notice how this new resistor is now in parallel with resistor R2. These resistors can also be reduced to a single component.
## Step 6:
Resistors R2 and Req1 have
now been combined to form the equivalent resistor Req2. Now all of the remaining resistors are in series and can be reduced to a single component.
## Step 7: Calculating Current
After a circuit has been reduced it is often easier to find the current flowing through it. Once you have simplified the circuit, you can then go on to apply Ohm’s Law, V=IR, to determine the current.
## Step 8: Practice
The best way to retain what you have learned here is to practice it. Start by going over the guided example problem below. After that, try the next problem on your own. The answers will be provided at the end.
## Step 9: Example Problem
First, find the final equivalent resistance of the circuit (reduce the system to one resistor and the battery). Then, find the value of the current I.
The black values are the initial values of the resistors and battery.
## Step 10:
The values that are boxed in red are the values
of the resistors that are to be reduced into a single component in the next step. In this is instance, the 15 ohm and the 30 ohm resistors are the ones to be combined using the Series Equivalent Resistance equation.
NOTE: It may appear that the 15 ohm and 45 ohm resistors are in parallel. Even though they do share a loop, the connection they share is also shared by the 30 ohm resistor, making the connection invalid. Also, don’t mistake the 30 ohm and the 60 ohm resistors to be in series. Those resistors do share a single node. However, the node is also shared by the 45 ohm resistor.
## Step 11:
The red value is the new equivalent resistance of
the resistors in the precious step. Now this new resistor is in parallel with the 45 ohm resistor. Calculate the next equivalent resistance of these two resistors using the Parallel Resistance Equivalence equation.
TIP: When calculating the equivalent resistance of ONLY TWO resistors IN PARALLEL you can use this equation:
Req = R1R2/(R1+R2)
## Step 12:
Now there are three resistors that are in series
and will be combined using Series Equivalent Resistance equation.
## Step 13:
These 70 ohm and 162.5 ohm resistors will be reduced using the Parallel Resistance Equivalence equation.
## Step 14:
For the final equivalent resistance, the 48.9 ohm and 55 ohm resistors will be reduced using the Series Equivalent Resistance equation.
Now that you know the final equivalent resistance of the circuit, you can use that to find the current of the system using Ohm’s Law, V=IR.
Solving for I you get, I=V/R. The current, I, flowing through the circuit is determined to be about .3849 amps or 384.9 milliamps .
## Step 15:
Now try the next problem on your own. Remember: When reducing the circuit, it always helps to redraw the circuit with the new equivalent resistors in between each step that you take.
First, find the final equivalent resistance of the circuit (reduce the system to one resistor and the battery). Then, find the value of the current I.
Answers: Req = 75 ohms, I = 1/3 A
115
9 1.6K
14 3.2K
## 3 Discussions
For a powerful way of calculating these values see: https://www.instructables.com/id/Incredibility-Powerful-Resistance-Calculator/
Also a note to the images: the current-arrows in the images are the PHYSICAL direction of the current: From negative to positive. In most textbooks, current and voltage flow from positive to negative which is the TECHNICAL current-direction. Just as a heads up. :)
As an electronics engineer, i say BRAVO for this simple but important message to all you DIY-Electrotechnicians out there!
BUT dont forget that if you have a decent schematic, you may have to use a 64.3kOhm-Resistor. Since in the E12-series such a resistor doesnt exist (It may exist in a E-series bigger than 100 but NOBODY on earth uses them for SURE) you may HAVE to use 2 resistors to get decently close. You may then use 2 in series like 8.2k + 56k to get 64.2k (Quite close, but still 0.16% too low) or you may use 100k and 180k in parallel to get 64.2857k which is 0.02% short. Since most resistors have 1% accuracy both solutions are considered viable.
How did i find out those combinations? I wrote a little tool for that ages back which quickly (12ms) crunches all the combinations of the E12-series and compares the results to the needed value It then presents the best 2 solutions for series and parallel and points out the better of the two.
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# Permutations and Combinations
## Factorial
In probability, a system of counting occurs frequently which involves a series of multiplications of the form:
\$\$n! = n ⋅ (n - 1) ⋅ (n - 2) ⋅ (n - 3) .... 2 ⋅ 1\$\$
This function has the name factorial, and has the symbol '!'.
For example, 6! = 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 ( = 720)
1! = 1, and 0! = 1 (careful: 0! is NOT equal to zero)
Factorials can be divided by other factorials: \${6!}/{4!} = {6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1}/{4 ⋅ 3 ⋅ 2 ⋅ 1} = {6 ⋅ 5} = 30\$
## Permutations
A permutation of a group of symbols is this group in any arrangement in a definite order.
e.g the numbers 1, 2, 3 may be organised in the orders: 123, 132, 213, 231, 312, 321.
There are 6 (3!) possible permutations for 3 symbols. There are 24 (4!) possible permutations for 4 symbols, and 120 possible permutations for 5 (5!) symbols.
Rule: n symbols may be arranged n! ways.
### Permutations of sub-sets
The number of ways 5 items may be arranged, taking 3 at a time is: \${5!}/{2!} = 5 ⋅ 4 = 20\$
Rule: n symbols taken r at a time may be arranged in \${n!}/{(n - r)!}\$ ways.
6 cards taken two at a time form \${n!}/{(n - r)!} = {6!}/{(6 - 2)!} = {6!}/{4!} = 6 ⋅ 5 = 30\$ pairs, where the order is important: (1,2) is different to (2,1).
This can be written as:
\$\$_nP_r = {n!}/{(n - r)!}\$\$
## Combinations
In \$_nP_r\$ the order of the sub-set of r items is important. In many cases, the order is not relevant. For example, in dealing hands of cards, the player can rearrange the cards he is dealt.
Therefore, if the order of the sub-set is not important, there are fewer combinations possible. This quantity is given the symbol C.
\$\$_nC_r = {n!}/{{(n - r)!}r!}\$\$
Content © Renewable.Media. All rights reserved. Created : January 1, 2015
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#### Sydney Brenner
1927 -
Sydney Brenner, born 1927, is a South African molecular biologist. He is a leading figure in biology, having won the Nobel Prize, and authored or co-authored numerous seminal papers on fields related to genome and protein sequencing.
### Quote of the day...
"I knew Descartes," said Isaac Barrow. "René', I used to say. 'Have you got des cartes?' Then after a few tankards, he would say: 'Don't try to cheat me, Wheelie-boy. Cogito ergo sum...' - that's classical pidgin for 'I can think so I can add up as well as the next man'."
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# Area of Shapes
Reviewed by:
Last updated date: 17th Sep 2024
Total views: 413.4k
Views today: 5.13k
## Shapes
In our everyday life, we come across a lot of objects of different shapes and sizes. Some of the shapes are squares, rectangles, circles, cones, rhombus. Area and perimeter give us knowledge that helps in dealing with these different forms of object.
In geometry, a shape is defined as an enclosed figure by the boundaries. The enclosed figure consists of lines, angles, points, and curves. There are basically two types of shapes namely:
1. 2 Dimensional
2. 3 Dimensional
Every shape has formulas to determine its area, volume, edges, angles and corners, surface area, perimeter, and so on. The formulas are different for both 2D and 3D shapes. Let us discuss further how area and other formulas are calculated.
### What is an Area?
An area is a quantity that expresses the closed area in a plane. It is calculated in 2D shapes. The shapes can be drawn on paper. An example includes circle, square, rectangle, trapezium, rhombus, etc. The area will be defined by the surface captured in an enclosed shape. Areas of different shapes are calculated by applying different shape formulas.
Polygon Shapes: It is a 2D shape formed by straight lines. Its example includes triangles, hexagons, and pentagons. The shape is easily identified by its name.
For example, a triangle will consist of 3 lines, a rectangle will consist of 4 lines.
Any shape that consists of 4 different lines is termed a quadrilateral. The area is the enclosed region within that boundary or perimeter of the shape.
### 2D Shapes
A 2-dimensional shape or also known as flat shapes. They are drawn on the plane and have length and breadth. It does not consist of thickness. The different methods used for the measurement of the area of 2D shapes are area and perimeter. Examples of 2D shapes are square, rectangle, triangle, circle, etc.
The area of a shape is simply calculated as the amount of paint required to cover that enclosed area. Each shape has a different method to calculate its area and perimeter. Let us discuss the area of 2D shapes in a simple tabular form:
## Area of 2D Shapes
Shape Area Triangle ½ * base * height Square Side2 Rectangle Length * Breadth Parallelogram Base * vertical height Trapezium ½ * ( a+b )* h Circle π * radius2 Elipse π * ½ minor axis * ½ major axis
### 3D Shapes
These are shapes that have 3 dimensions namely length, breadth, and thickness. Other measures used here are volume and surface area. 3D shapes are formed by the rotation of 2D shapes.
3D solid shapes include a cube, cuboid, sphere, cylinder, and cone. The area of the shape is upgraded to the concept of the surface area of 3D shapes. Each enclosed shape has its area. As discussed earlier, each shape has a different method to calculate the area. Let us discuss the 3D shapes formulae in a tabular form:
## Area of the 3D Shape
The area formula list is easier understood in tabular form and hence easy to memorize as well.
### 3D Shapes Formula - Volume And Surface Area
In practical life, we come across a lot of solids which are simply a combination of basic shapes. Its common example includes tents, capsules, and ice-cream cones. These shapes are similar to the basic shapes that we study. They are the combination of simple basic shapes used in our daily life.
3D shapes are a combination of two or more shapes formed from the fusion of basic shapes. They all come together and form a new shape.
While calculating the surface area and volume of 3D shapes, we need to look at the number of solid shapes that form. For example, a cube is formed when six square shapes are joined adjacent to each other. When measuring the surface area and volume of these shapes, all three dimensions such as length, breadth, and height are taken into account.
When finding the surface area of the cuboid, we add the area of each rectangle constituting the cuboid.
## FAQs on Area of Shapes
Q1. What Real-Life Situations Require us to Use the Area?
Ans: To determine the area of all shapes is quite easy to determine. For a square or rectangle, an area is the number of square units inside a figure. The area is also calculated by multiplying the length and its breadth.
The concept of the area has many practical applications. Some of them are listed below:
1. If looking to sod the lawn, you would need to know the area of the lawn so that it is easier to purchase the right measurement sod for effective and efficient use. This also reduces the wastage.
2. If you wish to apply carpet in the living area, you need to calculate the area covered in the living area.
3. I want to lay tiles on the floor and determine how many tiles will be used in a particular area.
These all are the real-life experiences that would need the requirement to calculate area. This is done to get a basic idea and to also ensure that materials are purchased and applied without any wastage or loss of material.
Q2. How do we Find the Area and Perimeter of an Irregular Shape?
Ans: An irregular shape is any shape that has sides and angles of any length and size.
The following method is used to determine the perimeter and area of irregular shape:
1. Perimeter: To calculate the perimeter, all sides are added. Their addition is referred to as the perimeter of the irregular shape. We name irregular shapes based on how many sides they have.
2. Area: To calculate the area of an irregular shape, simply divide the shape into small regular shapes. This makes the calculation of area more precise and easy to calculate. For any 4 sided or 5 sided figure, bifurcate the shape into small regular triangles and calculate the area of the triangle using heron’s formula.
After calculation, the area of each triangle, add all of them to find the area of irregular triangles.
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How do you find the derivative of g(z)=(z^2+3)e^z?
Sep 9, 2017
$g ' \left(z\right) = \left({z}^{2} + 2 z + 3\right) {e}^{z}$
Explanation:
we will use the product rule
$g \left(z\right) = u \left(z\right) v \left(z\right) \implies g ' \left(z\right) = \textcolor{red}{u ' \left(z\right)} v \left(z\right) + u \left(z\right) \textcolor{b l u e}{v ' \left(z\right)}$
we have
$g \left(z\right) = \left({z}^{2} + 3\right) {e}^{z}$
$u \left(z\right) = {z}^{2} + 3 \implies \textcolor{red}{u ' \left(z\right) = 2 z}$
$v \left(z\right) = {e}^{z} \implies v ' \left(z\right) = \textcolor{b l u e}{{e}^{z}}$
$\therefore g ' \left(z\right) = \textcolor{red}{2 z} {e}^{z} + \left({z}^{2} + 3\right) \textcolor{b l u e}{{e}^{z}}$
simplifying and tidying up
$g ' \left(z\right) = {e}^{z} \left(2 z + {z}^{2} + 3\right)$
$g ' \left(z\right) = \left({z}^{2} + 2 z + 3\right) {e}^{z}$
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# Lesson Video: Using Systematic Trial and Improvement to Find a Root of a Quadratic Equation Mathematics
A detailed explanation and walk-through of examples showing how to use the method of trial and improvement in a tabular systematic way to find a solution to quadratic equations to a given level of accuracy, such as 1 or 2 decimal places.
16:17
### Video Transcript
In this video, we’re gonna use the systematic method of trial and improvement to find a root of a quadratic equation. In these sort of questions, you’re usually given some conditions. So we want a solution between zero and one. You’re given an equation to solve, and you’re given a required level of accuracy to work to. Then you’re expected to produce a table, where you show all your calculations, to justify your conclusions. Now lots of people lose marks really easily on these sort of questions, and so we’re gonna go very carefully through the method. And also we’re gonna look at using diagrams like these to help us to make good decisions about the calculations that we make.
So let’s go ahead and try out an example.
Using the method of systematic trial and improvement and giving your answer correct to one decimal place, find the solution between 𝑥 equals zero and 𝑥 equals one to two 𝑥 squared minus five 𝑥 plus one equals zero.
So they’re asking for the particular method, systematic trial and improvement. They’ve given you the level of accuracy that we need, correct to one decimal place. And they’ve given us some starting conditions here, we want to find the solution between 𝑥 equals zero and 𝑥 equals one. And they’ve given us an equation to solve. Now normally, the first thing you do with these questions is to draw a table. But what we’re gonna do is just have a quick look at the graph. Now obviously you wouldn’t normally see the graph, but that’s gonna help us to see how this method is gonna work. So here’s the graph of 𝑦 equals two 𝑥 squared minus five 𝑥 plus one. Now we’re looking for the specific solution where that is equal to zero. In other words, where the 𝑦-coordinate here is equal to zero. And when the 𝑦-coordinate is zero, we’re talking about any point on the 𝑥-axis. Now this graph cuts the 𝑥-axis in two places, here and here. And the question has said that we’re looking for this solution here between 𝑥 equals zero and 𝑥 equals one. So this is the one that we’re looking for.
So we’re trying to find the 𝑥-coordinate of this point: the value of 𝑥 which generates a 𝑦-coordinate of zero. So zooming in on that curve between 𝑥 is zero and 𝑥 is one, what we’re gonna do is use the 𝑥-coordinate of zero and plug that into the equation and see what value the expression has. So two times zero squared minus five times zero plus one. And that hopefully is gonna give us an answer of one. We’re then gonna do the same for 𝑥 equals one to see what expression value is generated. And we’ll see that one is too big and one is too small. So our solution that generates an answer of zero must be somewhere between 𝑥 equals zero and 𝑥 equals one. Here, it’s too big; it’s getting smaller and smaller and smaller. Somewhere it becomes zero and then it gets smaller and smaller; it’s negative and goes down here. Now obviously we won’t have the graph to look at. But what we’ll be able to do is, once we know that one is too small and one is too big, we will be able to pick a point somewhere in the middle here. And we’ll be able to calculate the new expression value. And we’ll have a new value that gives us an answer that’s too small and a new value that gives us an answer that’s too big. And now- then we’ll try to look between those two points and we’ll keep moving that 𝑥-value in either from the right or from the left. And we’ll gradually zoom in on our actual 𝑥-value that we’re looking for here. So let’s go ahead and try that.
So we’ll construct the table that’s got these four columns. First of all, we’ve got the 𝑥-values that we’re gonna try. And we’ve already been told what values to use for our first two attempts, zero and one. The second column is where we’re gonna plug those 𝑥 values into the expression we were given in the equation and evaluate that expression. So with zero then, we’ve got two lots of zero squared minus five lots of zero plus one. And two times zero squared is zero, take away five lots of zero, that’s zero, plus one leaves us one. Then we’re gonna try plugging in 𝑥 equals one. So two times one squared minus five times one plus one. And I give this an answer of negative two. So now, we can fill out the comments. And when 𝑥 was zero, the result was one; that’s too big. We wanted our result to be zero, remember, because that’s what we had in the original equation. And in the second case, the result was too small because it was less than zero.
So our quadratic curve is gonna be a smooth curve. So somehow it’s gonna go between this point and this point; there must be an 𝑥-value in between zero and one which will generate a 𝑦-coordinate of zero when I plug it in. So I can now fill in the “Range” column. And I know from those two results — one being too big one being too small — that somewhere between the two there must be a value of 𝑥 which generates an 𝑥- a 𝑦-coordinate of zero. Now it’s up to you which particular value you choose. I’m gonna go and try an 𝑥-value of nought point three. And when I do that, I try plugging nought point three into that expression. So two times nought point three squared minus five times nought point three plus one. And that generates a result of nought point three two, a negative nought point three two. So that’s below zero. So that’s too small as well. So just drawing that on my little sketch over at the right-hand side, when I had an 𝑥-coordinate of zero, it gave me a positive result when I put it into that expression. When I have an 𝑥-coordinate of nought point three, I plug that into the expression, I get a negative result. So somewhere between 𝑥 is zero and 𝑥 is nought point three, there must be a value of 𝑥 which generates a 𝑦-coordinate of zero.
So now I know that none of this area over here is any use. So what I’m gonna do now is move the left-hand in a little bit. And I’m gonna go up to nought point two. I’m gonna try nought point two as an 𝑥-value and see whether that gives me a good answer or not. So plugging nought point two into our expression, I’ve got two times nought point two squared minus five times nought point two plus one. And when I evaluate that, I get an answer of nought point nought eight. So an 𝑥-coordinate of nought point two generates a 𝑦-coordinate of nought point nought eight. Now that’s-that’s above zero, so that’s too big. Now nought point two generates an answer which is pretty close to zero, so we’re quite close to the right answer. But unfortunately, that’s not quite good enough a guess. So looking at our graph, we’ve updated it; we’ve zoomed in. So when we had an 𝑥-value of zero, the 𝑦-coordinate was one. An 𝑥 value of nought point two, gave us a 𝑦-coordinate of nought point eight. So the value of 𝑥 which generates a 𝑦-coordinate of zero can’t be in this region over here. Now the value of negative nought point three two was generated when I put in an 𝑥-coordinate of nought point three. So if we look at this, the value of 𝑥 which generates a 𝑦-coordinate of zero must be between zero point two and zero point three. The answer goes from being too big when we put in 𝑥 equals nought point two to being too small when we put in 𝑥 equals nought point three.
So, I’ve got two consecutive answers to one decimal place, which is what they asked for in the question. The answer’s somewhere between nought point two and nought point three. We’ve gotta find out which one is it closer to. Is it closer to nought point two or is it closer to nought point three? So with nought point two generating an answer that’s too big and nought point three generating an answer that’s too small, the solution for 𝑥 must be somewhere between the two. So we can update the range there. Now it’s very tempting at this stage to say, “Well nought point two gives us an answer of nought point nought eight, which is very close to zero. Nought point three gives us an answer of negative nought point three two, which is a bit further away from zero. So the answer must be nought point two, because it generates an answer closer to the answer we’re looking for.” But you can’t really do that.
Now there’s some sort of a curve representing the 𝑦-values of this expression in between nought point two and nought point three. Now that curve could be more or less a straight line coming down like this, it could be a curve that goes in this direction and comes across here, or it could be a curve that goes across here and down. We don’t really know. Now if it was this sort of a curve, actually the answer, the 𝑥 value that generates the 𝑦-coordinate of zero would be closer to nought point three than it would be to nought point two. If it’s either of these two scenarios, then the 𝑥-coordinate here is going to be closer to nought point two than it is to nought point three. So given that we don’t know what the curve looks like, we have to be a bit clever about this.
So what we have to do is look at the 𝑥-coordinate that’s midway between nought point two and nought point three. Now if that generates an answer which is bigger than zero, then we know that the change from being too big to too small is happening in this range over here. If it generates an answer which is too small, then we know the change from being too big to being too small is happening in this range over here. So plugging in the 𝑥-value nought point two five to our expression, we’ve got two times nought point two five squared minus five times nought point two five plus one. And that equals negative nought point one two five, which is too small. So in between nought point two five and nought point three, we suspect all of the results are gonna be too small. But the change from being too big to too small happens in this range here, between nought point two and nought point two five. So being to the left of nought point two five means that it’s definitely closer to nought point two than it was to nought point three. So we’ve narrowed down the range to be in between nought point two and nought point two five, which means our answer, correct to one decimal place, is nought point two.
Now to get full marks on these sorts of questions, we need to see some of these values, a few of these values, correctly evaluated for different values of 𝑥 within the range specified in the question. We also need to see that you’ve narrowed it down to — So if we’re going for one decimal place, we’d need two consecutive answers to one decimal place like nought point two and nought point three, and they need to be correctly evaluated as well. And then finally, you need to show that you’ve correctly proved whether we’re to the left or to the right of that midpoint of those two consecutive answers, nought point two and nought point three. You won’t get the credit for getting a correct answer, nought point two, unless you’ve done this extra stage here where we’ve gone to one more decimal place, two decimal places in this case, to prove whether we’re closer to nought point two or nought point three.
Now, the next part of the question is to refine that answer, correct to two decimal places. So the expression we’re trying to solve here is two 𝑥 squared minus five 𝑥 plus one equals zero. And we’d narrowed it down to 𝑥 was between nought point two and nought point two five. So given that we want two decimal places, we’ve - we’ll change that nought point two to nought point two zero and we’ll do another table. Right. So we just filled in what we know already. The result when 𝑥 is nought point two gives us an answer of nought point nought eight when we evaluate that expression, and that’s too big. And when we plug in 𝑥 equals nought point two five, that gives us a value for the expression which is below zero, so it’s too small. So just quickly sketching out our graph, we know that somewhere between those two 𝑥 values, we’re gonna find a coordinate which generates a 𝑦-coordinate of zero: a value for that expression of zero.
Now, I’m gonna try an 𝑥-value of nought point two two. So you’ve got a range, you could try nought point two one, two two, two three, or two four. So we’ll go with nought point two two. And plugging that into the expression, we’ve got two times nought point two two squared minus five times nought point two two plus one. And when we evaluate that, we get an answer of negative nought point nought nought three two. So it’s quite close to zero, but it’s below the 𝑥-axis. It’s a negative number, so it’s too small. So this means that the 𝑥-coordinate has to be somewhere between nought point two zero and nought point two two. And that will generate a 𝑦-coordinate of zero. So let’s try nought point two one. And two times nought point two one squared minus five times nought point two one plus one gives us a value of nought point nought three eight two. So that’s above zeros, so that’s too big. So this means nought point two one is to the left of our answer, and nought point two two is to the right of our answer. So we need to try the value that’s exactly halfway between those and see what result we get. So plugging in the next value of nought point two one five gives us two times nought point two one five squared minus five times nought point two one five plus one, which is nought point nought one seven four five. It’s positive. It’s too big.
So just re-sketching our little graph there, nought point two one generated a 𝑦-value which was too big. Nought point two one five has also generated a 𝑦-value which is too big. So all of those 𝑥-values between nought point two one and nought point two one five are gonna generate 𝑦-values that are too big. So the change from being too big to being too small happens between nought point two one five and nought point two two. This means that it’s to the right of nought point two one five and therefore it’s closer to nought point two two than it is to nought point two one. So our answer is 𝑥 is nought point two two correct to two decimal places.
So let’s just check we got everything we need. We wanted the answer correct to two decimal places. We’ve got two consecutive answers, one that’s too small and one that’s too big. So nought point two one and nought point two two, they’re consecutive answers to two decimal places. We checked the value in the middle, and that was too big. And that told us that the change from being too big to being too small happens between nought point two one five and nought point two two, which means that the 𝑥-value must be closer to nought point two two than it is to nought point two one. That’s how we got our answer. So that’s full marks.
Again, it would’ve been tempting to say, “Well at nought point two one, the 𝑦-value generated is nought point nought three eight two. But at nought point two two, the 𝑦-value generated is minus nought point nought nought three two. Now that is closer to zero than that is.” That argument on its own is not good enough to get you those last two marks for the question. You have to test the value in the middle and prove whether it’s gonna be to the left or the right of that value.
So just before we go, let’s do a slow motion action replay of what we’ve just done on the graph. We started off evaluating the expression for 𝑥 equals zero and 𝑥 equals one and found out that one was too big and one was too small. So the answer must be somewhere between zero and one. So we then tried an 𝑥-value of nought point three and the-the 𝑦-value that that generated was too small. So at this point, we knew that our 𝑥-coordinate must be between zero and zero point three. We then tried an 𝑥-coordinate of nought point two. That generated an answer that was too big. So now, we knew the answer was between zero point two and zero point three. So we checked out the mid 𝑥-value between nought point two and nought point three, that’s at nought point two five, and found that that gave us an answer which was too small. And this told us that the answer must be between nought point two and nought point two five because that’s where the answer goes from being too big to being too small. We wanted our answer to one decimal place, so it’s either gonna nought point two or nought point three. And by doing this midpoint value and working our way to the left of that, we know that the final answer must be nought point two.
So that pretty much sums up systematic trial and improvement. Good luck.
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# The Math of Sunrise and Sunset
I knew since I was ten (we had quite a comprehensive curriculum at school) that the shortest day of the year falls on December 22nd. What I didn’t ponder until very recently was whether it was also the day of the latest sunrise (and, consequently, the earliest sunset).
While it may seem like a natural consequence of December 22nd being the shortest day, it doesn’t necessarily have to be true. If we model the time of sunrise and of sunset as two sine waves, $a(t) = A \text{sin}(t+α)+K$ and $b(t) = B \text{sin}(t+β)+L$ such that $t_0$ minimizes the difference (we can drop the constants):
$B \text{sin}(t+β) - A \text{sin}(t+α)$
This means that
$\frac{d(b(t)-a(t)}{dt} = 0 \text{at} t_0 \Rightarrow B \text{cos}(t_0+β) - A \text{cos}(t_0+α) = 0$
We need to show that this equality may hold (for some values of $α$, $β$, $A$ and $B$) even if one of the waves is not minimized at $t_0$. Let
\begin{align}\frac{da(t)}{dt} = P \neq 0 \text{at} t_0 \Rightarrow A \text{cos}(t_0+α) & = P\\ B \text{cos}(t_0+β) = A \text{cos}(t_0+α) &= P\\ \text{cos}(t_0+β) = \frac{P}{B}, \text{where} \frac{A}{B} \leq \frac{P}{B} \leq \frac{A}{B}\end{align}
We can always find some values of $A$ and $B$ such that $\frac{P}{B}$ is between -1 and 1, and hence the equation will be satisfied for some values of $t_0$ and $β$.
We can also take a short route and recall that a linear combination of two sine waves of the same frequency but not necessarily the same phase is still a sine wave. Its phase is a function of the difference in phases of the two waves. The value of $t_0$ that minimizes the resulting wave is not necessarily going to minimize any of the two input waves because the three phases are different (and not different by a multiple of π).
In fact, if you look at the sunrise and sunset times in Connecticut around this December, the shortest day, unsurprisingly, falls on December 22nd, but the latest sunrise is on January 4th 2010 and the earliest sunset is on December 8th.
This is great news–it means that starting on December 8th (and not the 22nd), it will finally start getting darker later and later!
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Grade 10 Math Module 1 searching for patterns, sequence and series
Math Module for Grade 10
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Grade 10 Math Module 1 searching for patterns, sequence and series
1. 1. Module 1 Searching for Patterns in Sequences, Arithmetic, Geometric and Others What this module is all about This module will teach you how to deal with a lot of number patterns. These number patterns are called sequences. Go over the lessons and have fun in working with the exercises. What you are expected to learn It is expected that you will be able to demonstrate knowledge and skill related to sequences and apply these in solving problems. Specifically, you should be able to: a) list the next few terms of a sequence given several consecutive terms. b) derive by pattern searching, a mathematical expression (rule) for generating the sequences. c) generate the next few terms of sequences defined recursively. d) describe an arithmetic sequence by any of the following: • giving the first few terms • giving the formula for the nth term • drawing the graph How much do you know
2. 2. A. Write the first five terms of the sequence. 1. an = n n 3+ 3. an = 4n – 3 2. an = 2n - 1 B. Find the indicated term for the sequence. 4. an = -9n + 2; a8 6. an = 52 73 − + n n ; a14 5. an = (n + 1)(2n + 3); a5 C. Find the general term, an, for the given terms of the sequence. 7. 4, 8, 12, 16,… 9. ,... 12 1 , 6 1 , 2 1 8. -10, -20, -30, -40,… D. Find the first four terms of the sequence defined recursively. 10. a1= -1, an = 3an – 1 11. a1 = 5, an = 1 1 − − n an E. Find the common difference and the next three terms of the given arithmetic sequence. 12. 2, 8, 14, 20,… 14. 1, 2, 3, 4,… 13. 7, 8.5, 10, 11.5,… 15. 24, 21, 18, … What you will do Lesson 1 Sequences and its Kinds It is a common experience to be confronted with a set of numbers arranged in some order. The order and arrangement may be given to you or you have to discover a rule for it from some data. For example, the milkman comes every other day. He came on July 17; will he come on Aug 12? Consider that you are given the set of dates 17, 19, 21,… 2
3. 3. arranged from left to right in the order of increasing time. Continuing the set we have 17, 19, 21, …, 29, 31, 2, 4, ….,28, 30… so that the answer to our question is yes. Any such ordered arrangement of a set of numbers is called a SEQUENCE. Look at this second example. Lorna, a 2nd year student in a certain public school, is able to save the money her ninongs and ninangs gave her last Christmas. She then deposits her savings of P1,000 in an account that earns 10% simple interest. The total amount of interest she earned in each of the 1st 4 years of her saving is shown below: Year 1 2 3 4 Total amount 10 20 30 40 The list of numbers 10, 20, 30, 40 is called a sequence. The list 10,20,30,40 is ordered because the position in this list indicates the year in which that total amount of interest is earned. Now, each of the numbers of a sequence is called a term of the sequence. The first term in the sequence 10, 20, 30, 40 is 10, the second term is 20, while the third term is 30 and the fourth term is 40. It is also good to point out that the preceding term of a given term is the term immediately before that given term. For example, in the given sequence 20 is the term that precedes 30. Examples of other sequences are shown below. These sequences are separated into two groups. A finite sequence contains a finite number of terms. An infinite sequence contains an infinite number of terms. Finite sequence Infinite sequence 1, 1, 2, 3, 5, 8 1, 3, 5, 7, … 1, 2, 3, 4, 5, …, 8 1, , 8 1 , 4 1 , 2 1 … 1, -1, 1, -1 1, 1, 2, 3, 5, 8, … In general: A sequence is a set of numbers written in a specific order: a1, a2, a3, a4, a5, a6,………, an 3
4. 4. The number a1 is called the 1st term, a2 is the 2nd term, and in general, an is the nth term. Note that each term of the sequence is paired with a natural number. Given at least the first 3 terms of a sequence, you can easily find the next term in that sequence by simply discovering a pattern as to how the 3rd term is derived from the 2nd term, and the 2nd from the 1st term. You will find that either a constant number is added or subtracted or multiplied or divided to get the next term or a certain series of operations is performed to get the next term. This may seem hard at first but with practice and patience in getting them, you will find that it’s very exciting. Examples: Find the next term in each sequence. 1. 17, 22, 27, 32, … 2. 11 1 , 8 1 , 5 1 , 2 1 … 3. 5, 10, 20, 40,… 4. 3, -3, 3, -3,… Solutions: 1. Notice that 5 is added to 17 to get 22, the same is added to 22 to get 27, and the same (5) is added to 27 to get 32. So to get the next term add 5 to the preceding term, that is, 32 + 5 = 37. The next term is 37. 2. Notice that 1 is the numerator of all the fractions in the sequence while the denominators- 2, 5, 8, 11 form a sequence. 3 is added to 2 to get 5, 3 is also added to 5 to get 8. So that 3 is added to 11 to get 14. The next term is therefore 1/14. 3. For this example, 2 is multiplied to 5 to get 10, 2 is multiplied to 10 to get 20 and 2 is also multiplied to 20 to get 40. So the next term is 80, the result of multiplying 40 by 2. 4. It is easy to just say that the next term is 3 since the terms in the sequence is alternately positive and negative 3. Actually the first, second, and third terms were multiplied by -1 to get the second, third and fourth terms respectively. Try this out A. Write F if the sequence is finite or I if the sequence is infinite. 1. 2, 3, 4, 5, ….., 10 2. 7, 10, 13, 16, 19, 22, 25 4
5. 5. 3. 4, 9, 14, 19, … 4. 2, 6, 18, 54 5. 3, 9, 27, 81, …., 729, … 6. -2, 4, -8, 16, ….. 7. 100, 97, 94, 91, …, -2 8. 64 1 , 32 1 , 16 1 , 8 1 , 4 1 9. 1, 4, 9, 16, 25, …., 144 10. 25 4 , 16 3 , 9 2 , 4 1 B. Answer the puzzle. Why are Policemen Strong? Find the next number in the sequences and exchange it for the letter which corresponds each sequence with numbers inside the box to decode the answer to the puzzle. A 2, 5, 11, 23, __ N 2, 6, 18, 54, __ B 2, 4, 16, __ O 20, 19, 17, __ C 7, 13, 19, __ P 2, 3, 5, 7, 9, 11, 13, 15, __ D 19, 16, 13, __ R 13, 26, 39, __ E 4, 8, 20, 56, __ S 5, 7, 13, 31, __ F 2, 2, 4, 6, 10, 16, __ T 1, 1, 2, 4, 7, 13, 24, __ H 1, 1, 2, 4, 7, 13, __ U 1, 1, 1, 2, 3, 4, 6, 9, 13, __ I 3, 6, 12, 24, __ Y 1, 2, 2, 4, 3, 6, 4, 8, 5, 10, __ L 10, 11, 9, 12, 8, __ Lesson 2 Finding the Terms of a Sequence Frequently, a sequence has a definite pattern that can be expressed by a rule or formula. In the simple sequence 24 14 13 10 19 17 44 52 47 26 26 48 25 256 164 25 47 19 85 164 44 24 164 6 25 47 162 5
6. 6. 2, 4, 6, 8, 10, …. each term is paired with a natural number by the rule an = 2n. Hence the sequence can be written as 2, 4, 6, 8,… 2n,… 1st term 2nd term 3rd term 4th term nth term a1 a2 a3 a4 an Notice how the formula an = 2n gives all the terms of the sequence. For instance, substituting 1, 2, 3, and 4 for n gives the 1st four terms: a1 = 2(1) = 2 a3 = 2(3) = 6 a2 = 2(2) = 4 a4 = 2(4) = 8 To find the 103rd term of this sequence, use n=103 to get a103 = 2(103) = 206. Examples: 1. Find the 1st four terms of the sequence whose general term is given by an = 2n – 1. Solution: To find the first, second, third and fourth terms of this sequence, simply substitute 1, 2, 3, 4 for n in the formula an = 2n-1. If the general term is an = 2n – 1, then the 1st term is a1 = 2(1) – 1 = 1 2nd term is a2 = 2(2) – 1 = 3 3rd term is a3 = 2(3) – 1 = 5 4th term is a4 = 2(4) – 1 = 7. The 1st four terms of this sequence are the odd numbers 1, 3, 5, and 7. The whole sequence can be written as 1, 3, 5, …, 2n – 1 Since each term in this sequence is larger than the preceding term, we say that the sequence is an increasing sequence. A sequence is increasing if an + 1 > an for all n. 6
7. 7. 2. Write the 1st 4 terms of the sequence defined by an = 1 1 +n . Solution: Replacing n with 1, 2, 3, and 4, respectively the 1st four terms are: 1st term = a1 = 11 1 + = 2 1 2nd term = a2 = 12 1 + = 3 1 3rd term = a3 = 13 1 + = 4 1 4th term = a4 = 14 1 + = 5 1 The sequence defined by an = 1 1 +n can be written as 2 1 , 3 1 , 4 1 , 5 1 , ……, 1 1 +n Since each term in the sequence is smaller than the term preceding it, the sequence is said to be a decreasing sequence. A sequence is decreasing if an + 1 < an for all n. 3. Find the 1st 5 terms of the sequence defined by an = n n 2 )1(− . Solution: Again by simple substitution, 1st term = a1 = 1 1 2 )1(− = - 2 1 2nd term = a2 = 2 2 2 )1(− = 4 1 3rd term = a3 = 3 3 2 )1(− = - 8 1 4th term = a4 = 4 4 2 )1(− = 16 1 7
8. 8. 5th term = a5 = 5 5 2 )1(− = - 32 1 The sequence defined by an = n n 2 )1(− can be written as - 2 1 , 4 1 , - 8 1 , 16 1 , - 32 1 ,…, n n 2 )1(− Notice that the presence of (-1) in the sequence has the effect of making successive terms alternately negative and positive. It is often useful to picture a sequence by sketching its graph w/ the n values as the x- coordinates and the an values as the y- coordinates. The corresponding graphs of Examples 1 - 3 are given below Example 1 0 1 2 3 4 5 6 7 8 0 1 2 3 4 5 n a n 8
9. 9. You can also find a specific term, given a rule for the sequence, as seen in the following example. 4. Find the 13th and 100th terms of the sequence whose general term is given by An = 2 )1( n n − Solution: For the 13th term, replace n with 13 and for the 100th term, replace n w/ 100: 13th term = a13 = 2 13 13 )1(− = 169 1− 100th term = a100 = 2 100 100 )1(− = 10000 1 Example 2 0 0.1 0.2 0.3 0.4 0.5 0.6 0 1 2 3 4 5 n a n Example 3 -0.6 -0.4 -0.2 0 0.2 0.4 0 1 2 3 4 5 6 n a n 9
10. 10. Try this out A. Write the 1st 4 terms of the sequence whose nth term is given by the formula. 1. an = n + 1 2. an = 2 – 2n 3. an = n - 1 4. an = 2 n 5. an = 2n + 1 6. an = n² + 1 7. an = 3n - 1 8. an = 1+n n 9. an = 1 - 2n 10. an = n – n 1 11. an = 3n 12. an = (-1) n+1 n 13. an = n² - 1 14. an = (-1) n 2 n 15. an = n² - n 1 16. an = n n 12 − 17. an = 1 )1( 1 + − + n n 18. an = 1 )1( 2 1 + − + n n 19. an = 2 1 3 1 + n 20. an = 3 1 n³ + 1 21. an = (1)n (n²+2n+1) B. Find the indicated term of the sequence whose nth term is given by the formula. 22. an = 3n + 4 a12 23. an= n(n -1) a11 24. an= (-1) n - 1 n² a15 25. an= ( 2 1 )n a8 26. an = 2n - 5 a10 27. an = 1+n n a12 28. an = (-1) n - 1 (n - 1) a25 29. an = ( 3 2 ) n a5 30. an = (n + 2)(n + 3) a17 10
11. 11. 31. an = (n + 4)(n + 1) a7 32. an = 2 12 )1( n n+ − a6 33. an = 4 )1( 2 + − n n a16 34. an = 2 3 n² - 2 a8 35. an = 3 1 n + n² a6 Lesson 3 Finding the nth Term of a Sequence In Lesson 2, some terms of a sequence were found after being given the general term. In this lesson, the reverse is done. That is, given some terms of the sequence, try to find the formula for the general term. Examples: 1. Find a formula for the nth term of the sequence 2, 8, 18, 32,… Solution: Solving a problem like this involve some guessing. Looking over the first 4 terms, see that each is twice a perfect square: 2 = 2(1) 8 = 2(4) 18 = 2(9) 32 = 2(16) By writing each sequence with an exponent of 2, the formula for the nth term becomes obvious: a1 = 2 = 2(1)² a2 = 8 = 2(2)² a3 = 18 = 2(3)² a4 = 32 = 2(4)² . . . an = 2(n)² = 2n² 11
12. 12. The general term of the sequence 2, 8, 18, 32,…. is an = 2n². 2. Find the general term for the sequence 2, 8 3 , 27 4 , 14 5 ,…. Solution: The first term can be written as 1 2 . The denominators are all perfect cubes while the numerators are all 1 more than the base of the cubes of the denominators: a1 = 2/1 = 3 1 11 + a2 = 3/8 = 3 2 12 + a3 = 4/27 = 2 3 13 + a4 = 5/64 = 3 4 14 + Observing this pattern, recognize the general term to be an = 3 1 n n + 3. Find the nth term of a sequence whose first several terms are given 2 1 , 4 3 , 6 5 , 8 7 , . . . Solution: Notice that the numerators of these fractions are the odd numbers and the denominators are the even numbers. Even numbers are in the form usually written in the form 2n, and odd numbers are written in the form 2n – 1 (an odd number differs form an even number by 1). So, a sequence that has these numbers for its first four terms is given by an = n n 2 12 − . 4. Find the nth term of a sequence whose first several terms are given -2, 4, -8, 16, -32,… Solution: 12
13. 13. These numbers are powers of 2 and they alternate in sign, so a sequence that agrees with these terms is given by an = (-1)n 2n . Note: Finding the nth term of a sequence from the 1st few terms is not always automatic. That is, it sometimes takes a while to recognize the pattern. Don’t be afraid to guess the formula for the general term. Many times an incorrect guess leads to the correct formula. Some pointers on how to find the general term of a sequence is given below. Pointers on How to Find the General Term of a Sequence 1. Study each term of the sequence as it compares to its term number. Then answer the following questions: a. Is it a multiple of the term number? b. Is it a multiple of the square or cube of the term number? If each term is a multiple of the term number, there will be a common number. 2. Examine the sequence. Does it increase or decrease? a. If it increases slowly, consider expressions that involve the term number plus or minus a constant like: n + 2 or n – 3. b. If it increases moderately, think about multiples of the term number plus or minus a constant like: 2n or 3n – 1. c. If the sequence increases very rapidly, try powers of the term number plus or minus a constant like: n2 or n2 + 1. 3. If the sequence consists of fractions, examine how the denominator and numerator change as separate sequences. For example: an = 2 1 n n + yields ,... 25 6 , 16 5 , 9 4 , 4 3 , 1 2 Also, though not all sequences can be defined by a formula, like for the sequence of prime numbers, be assured that the sequences discussed or given here are all obvious sequences that one can find a formula or rule for them. 13
14. 14. Try this out A. Write the formula for the nth term of the sequence: 1. The sequence of the natural numbers. 2. The sequence of the negative even integers. 3. The sequence of the odd natural numbers. 4. The sequence of the negative odd numbers. 5. The sequence of the multiples of 7. 6. The sequence of the positive even integers that are divisible by 4. 7. The sequence of the negative integers less than -5. 8. The sequence of the positive odd integers greater than 9. 9. The sequence for which a1 is 8 and each term is 6 more than the preceding term. 10. The sequence for which a1 is 3 and each term is 10 less than the preceding term. B. Determine the general term for each of the following sequences: 1. 2, 3, 4, 5,… 2. 3, 6, 9, 12,… 3. 4, 8, 12, 16, 20,… 4. 3, 4, 5, 6,… 5. 7, 10, 13, 16,… 6. 4, 9, 14, 19,… 7. 3, 12, 27, 48,… 8. 2, 16, 54, 128,… 9. 4, 8, 16, 32,… 10. 1, 8,27, 64,… 11. 1, 4, 9, 16,… Lesson 4 Recursively Defined Sequences Some sequences do not have simple defining formulas like those in Lesson 3. The nth term of a sequence may depend on some or all of the terms preceding it. A sequence defined in this way is called recursive. One has to be sure that the notations: an and an – 1 is understood first before going to the examples below. To illustrate: if the 24th term is an, then it can be written as a24. So that an – 1 means a24 – 1 = a23 or the 23rd term. 14 12. 3, 9, 27, 81,… 13. -2, 4, -8, 16,… 14. -3, 9, -27, 81,… 15. 32 1 , 16 1 , 8 1 , 4 1 ,… 16. 81 1 , 27 1 , 9 1 , 3 1 ,… 17. 25 4 , 16 3 , 9 2 , 4 1 ,… 18. 32 1 , 28 3 , 10 2 , 4 1 ,…
15. 15. Examples: 1. Find the first five terms of the sequence defined recursively by a1 = 1 and an = 3(an – 1 + 2). Solution: The defining formula asks that the preceding term, an – 1, be identified first before one can get the nth term, an. Thus, one can find the second term from the first term, the third term from the second term, the fourth term from the third term, and so on. Since the first term is given as a1 = 1, proceed as follows: a2 = 3(a1 + 2) = 3(1 + 2) = 9 a3 = 3(a2 + 2) = 3(9 + 2) = 33 a4 = 3(a3 + 2) = 3(33 + 2) = 105 a5 = 3(a4 + 2) = 3(105 + 2) = 321 Thus, the first five terms of this sequence are 1, 9, 33, 105, 321,… 2. Find the first 11 terms of the sequence defined recursively by F1 = 1, F2 = 1 and Fn = Fn – 1 + Fn – 2 Solution: To find Fn, the preceding terms, Fn – 1 and Fn – 2 have to be found first. Since the first two terms are given, then F3 = F2 + F1 = 1 + 1 = 2 F4 = F3 + F2 = 2 + 1 = 3 F5 = F4 + F3 = 3 + 2 = 5 By this time it should be clear as to what is happening here. Each term is simply the sum of the first two terms that precede it, so that its easy to write down as many terms as one pleases. Here are the first 11 terms: 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89,… The sequence in Example 2 is called the Fibonacci sequence, named after the 13th century Italian mathematician who used it to solve a problem about the breeding of rabbits. The sequence also occurs in a lot of other applications in nature. 15
16. 16. 8 5 3 2 1 1 The Fibonacci sequence In the branching of a tree To better understand the difference between recursively defined sequences and the general sequences encountered in the last three lessons, study the examples below. 3. Find the 120th term of the sequence defined by an = 2n + 3. Solution: Replacing n with 120, then 120th term = a120 = 2(120) + 3 = 243 The 120th term is 243. Notice that the 120th term from the example above is found without finding first the preceding term, that is, the 119th term! Whereas with the recursively defined sequence one has to find all the terms preceding the required term, 4. Find the 4th term of the recursively defined sequence an = 2 1−na where a1 = -3. Solution: First term = a1 = -3 Second term = a2 = 2 12−a = 2 1a = 2 3− Third term = a3 = 2 13−a = 2 2a = 2 2 3− = 4 3− Fourth term = a4 = 2 14−a = 2 3a = 2 4 3− = 8 3− The fourth term is - 8 3 . 16
17. 17. Try this out Find the first five terms of the given recursively defined sequences: 1. an = 2(an – 1 – 2) and a1 = 3 2. an = 2 1−na and a1 = -8 3. an = 2an – 1 + 1 and a1 = 1 4. an = 11 1 −+ na and a1 = 1 5. an = an – 1 + an – 2 and a1 = 1, a2 = 2 6. an = an – 1 + an – 2 + an – 3 and a1 = a2 = a3 = 1 Lesson 5 Arithmetic Sequences There are two main types of sequences. These are the arithmetic sequences and the geometric sequences. This lesson will show what arithmetic sequences are. The more detailed lesson on arithmetic sequences will be discussed in the next module. Look at the following sequences. 1. 4, 7, 10, 13,… 2. 33, 38, 43, 48,… 3. -2, -6, -10, -14,… 4. 100, 98, 96, 94,… 5. 2 1 , 1, 1 2 1 , 2, … Can you give the next two terms of the above sequences? How did you get the next terms? If you get 16 and 19 for a, then you are correct. Notice that a constant number , 3, is added to each term to get the next term. In b, 5 is added to the preceding term after the first, while in c, -4 is added to get the next term, in d, -2 is added to the preceding term and in e, ½ is added to get the next term. Notice that a constant or common number is added to the preceding term to get the next term in each of the sequences above. All these sequences are called 17
18. 18. arithmetic sequences. The constant number is called the common difference and is represented as d. To find the common difference, d, simply subtract the first term from the second term, a2 – a1, or the second term from the third term, a3 – a2, or the third term from the fourth term, a4 – a3; or in general, d = an – an – 1 Examples: 1. Determine if the sequence is arithmetic or not. If it is, find the common difference and the next three terms. Then graph. -11, -4, 3, 10,… Solution: To find out if the sequence is arithmetic, there must be a common difference between any two terms in the sequence. So that d = a2 – a1 = -4 – (-11) = 7 = a3 – a2 = 3 – (-4) = 7 = a4 – a3 = 10 – 3 = 7 The sequence is arithmetic and the common difference is 7. The next three terms are obtained by adding 7 to the preceding term, so that a5 = a4 + 7 = 10 + 7 = 17 a6 = a5 + 7 = 17 + 7 = 24 a7 = a6 + 7 = 24 + 7 = 31 Example 1 -20 -10 0 10 20 30 40 0 2 4 6 8 n an 18
19. 19. 2. Write the first five terms of the arithmetic sequence with first term 5 and common difference -2. Solution: The second term is found by adding -2 to the first term 5, getting 3. For the next term, add -2 to 3, and so on. The first five terms are 5, 3, 1, -1, -3. Remark: There is another way of finding the specified term of an arithmetic sequence but it will be discussed in the next module. The same thing is true for the general term of any arithmetic sequence. Try this out Determine whether the sequence is arithmetic or not. If it is, find the common difference and the next three terms. 1. 2, 5, 8, 11,… 2. 2, -4, 6, -8, 10,… 3. -6, -10, -14, -18,… 4. 40, 42, 44, 46,… 5. 1.2, 1.8, 2.4,… 6. 1, 5, 9, 13,… 7. ,... 5 1 , 4 1 , 3 1 , 2 1 8. ,...8,7,6,5 9. 98, 95, 92, 89,… 10. 1, 3 5 , 3 4 , 2,… Let’s Summarize 1. A sequence is a list of numbers in which order is important. a1, a2, a3, a4, …, an, … Each number in the list corresponds to each natural number. 2. A sequence may either be finite or infinite. A finite sequence has a specific number of terms. An infinite sequence has an endless number of terms. 19
20. 20. 3. To find the terms of a sequence given its rule, simply replace n with the number of the specific term needed to be found. 4. Most sequences have a general term or rule that describes all the terms in the sequence. There is no specific way of finding the general term of a given sequence. 5. A sequence is defined recursively when the nth term can be found only when the preceding term is found. 6. An arithmetic sequence is a sequence where each term is found by adding a constant number, called the common difference, to the preceding term. 7. The Fibonacci sequence is a special recursively defined sequence. What have you learned A. Write the first five terms of the sequence. 1. an = n n 7+ 3. an = 5n – 2 2. an = 3n - 1 B. Find the indicated term for the sequence. 4. an = -7n + 3; a8 6. an = 53 72 − + n n ; a14 5. an = (n + 2)(2n - 3); a5 C. Find the general term, an, for the given terms of the sequence. 7. 3, 7, 11, 15,… 9. ,... 16 1 , 8 1 , 4 1 , 2 1 8. 0, -4, -8, -12,… E. Find the first four terms of the sequence defined recursively. 10. a1= -1, an = 5an – 1 11. a1 = 6, an = 13 1 − − n an F. Find the common difference and the next three terms of the given arithmetic sequence. 12. 1, 10, 19, 28,… 14. 1, 3, 5, 7,… 13. 5.5, 7, 8.5, 10,… 15. 43, 39, 35, … 20
21. 21. 21
22. 22. Answer Key How much do you know 1 4, 5 8 , 4 7 ,2, 2 5 9. an = nn +2 1 2 1, 2, 4, 8, 16 10 -1, -3, -9, -27 3 1, 5, 9, 13, 17 11 5, 5, 6 5 , 2 5 4 -70 12 d = 6; 26, 32, 38 5 78 13 d = 1.5; 13, 14.5, 16 6 49/23 14 d = 1; 5, 6, 7 7 an = 4n 15 d = -3; 15, 12, 9 8 an = -10n Lesson 1 A. B. Because they can hold up traffic. Lesson 2 A. 1 2, 3, 4, 5 12 1, -2, 3, -4 2 0, -2, -4, -6 13 0, 3, 8, 15 3 0, 1, 2, 3 14 -2, 4, -8, 16 4 2, 4, 8, 16 15 0, 4 63 , 3 26 , 2 7 5 3, 5, 7, 9 16 0, 4 15 , 3 8 , 2 3 6 2, 5, 10, 17 17 5 1 , 4 1 , 3 1 , 2 1 −− 7 2, 5, 8, 11 18 17 1 , 10 1 , 5 1 , 2 1 −− 1 F 3 I 5 I 7 F 9 F 2 F 4 F 6 I 8 F 10 F 22
23. 23. 8 5 4 , 4 3 , 3 2 , 2 1 19 243 2 , 81 2 , 27 2 , 9 2 9 -1, -3, -5, -7 20 3 67 ,10, 3 11 , 3 4 10 0, 4 15 , 3 8 , 2 3 21 4, 9, 16, 25 11 3, 9, 27, 81 B. 22 40 29 243 32 23 110 30 380 24 225 31 88 25 256 1 32 36 1− 26 15 33 20 1 27 13 12 34 94 28 24 35 38 Lesson 3 A. B. 1 an = n + 1 10 an = n3 2 an = 3n 11 an = n2 3 an = 4n 12 an = 3n 4 an = n + 2 13 an = (-2)n 5 an = 3n + 4 14 an = (-3)n 6 an = 5n – 1 15 an = 1 2 1 +n 7 an = 3n2 16 an = n 3 1 8 an = 2n3 17 an = 2 )1( +n n 1 an = n 6 an = 4n 2 an = -2n 7 an = -(n + 5) 3 an = 2n – 1 8 an = 2n + 9 4 an = -2n + 1 9 an = 6n + 2 5 an = 7n 10 an = -10(n + 1)+ 3 23
24. 24. 9 an = 2n + 1 18 an = 13 +n n Lesson 4 1 3, 2, 0, -4, -12 4 1, 8 5 , 5 3 , 3 2 , 2 1 2 -8, -4, -2, -1, - 2 1 5 1, 2, 3, 5, 8 3 1, 3, 7, 15, 31 6 1, 1, 1, 3, 5 Lesson 5 1 Arithmetic d = 3 14, 17, 20 2 No 3 Arithmetic d = -4 -22, -26, -30 4 Arithmetic d = 2 48, 50, 52 5 Arithmetic d = 0.6 3, 3.6, 4.2 6 Arithmetic d = 4 17, 21, 25 7 No 8 No 9 Arithmetic d = 3 86, 83, 80 10 Arithmetic d = 1/3 3, 3 8 , 3 7 What have you learned 1 8, 5 12 , 4 11 , 3 10 , 2 9 9. an = n 2 1 2 1, 3, 9, 27, 81 10 -1, -5, -25, -125 3 3, 8, 13, 18, 23 11 6, 220 3 , 20 3 , 5 6 4 -53 12 d = 9; 37, 46, 55 5 49 13 d = 1.5; 11.5, 13, 14.5 6 35/37 14 d = 2; 9, 11, 13 7 an = 4n – 1 15 d = -4; 31, 27, 23 8 an = 4(1 – n) 24
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# 30 60 90 Triangle – Definition with Examples
The 30 60 90 triangle, a key concept in geometry for young learners at Brighterly, is distinguished by its angles: 30 degrees, 60 degrees, and 90 degrees. Its sides have a fixed ratio, essential for solving various geometric problems.
This triangle’s unique side relationships provide a straightforward approach to understanding geometry. In this article, we’ll explore these properties, explain side length calculations, and demonstrate practical examples. This concise overview is aimed at helping students grasp the fundamentals of the 30 60 90 triangle and apply them in mathematical contexts.
## What Is a 30-60-90 Triangle?
A 30-60-90 triangle is a specific type of right triangle characterized by its angle measurements. The angles in this triangle are 30 degrees, 60 degrees, and 90 degrees. The 90-degree angle makes it a right triangle. This triangle is significant due to its unique properties and ratios between its sides.
## 30 60 90 Triangle Sides
The sides of a 30-60-90 triangle are in a unique ratio that depends on the angles. The side opposite the 30-degree angle is the shortest, let’s call it . The side opposite the 60-degree angle is , and the side opposite the 90-degree angle, which is the hypotenuse, is . This ratio 1:√3:2 remains constant for all 30-60-90 triangles.
## 30 60 90 Triangle Rules
There are specific rules associated with 30-60-90 triangles:
1. Side Ratios: As mentioned, the sides are in the ratio 1:√3:2.
2. Angles: The angles are always 30 degrees, 60 degrees, and 90 degrees.
3. Pythagorean Theorem: This theorem applies as it is a right triangle, so 2.
### Some other important 30-60-90 rules:
• Scaling: Multiplying all sides of a 30-60-90 triangle by the same factor results in another 30-60-90 triangle.
• Angle Bisector: Bisecting the 60-degree angle creates two smaller 30-60-90 triangles.
30-60-90 Triangles Worksheet
## Perimeter of the 30-60-90 Triangle
The perimeter of a 30-60-90 triangle is the sum of its sides: .
## Area of the 30-60-90 Triangle
The area of the triangle is . For our triangle, it would be .
## 30-60-90 Triangle Theorem
The 30-60-90 Triangle Theorem states that in a 30-60-90 triangle, the sides are in the ratio 1:√3:2, and the angles are fixed at 30, 60, and 90 degrees.
## Solved Examples on 30-60-90 Triangle
Here are some 30-60-90 triangle examples:
Example 1: If the shortest side of a 30-60-90 triangle is 4 cm, find the lengths of the other sides. Solution: Other sides are 4√3 cm and 8 cm.
Example 2: Find the area of a 30-60-90 triangle with a hypotenuse of 10 cm. Solution: Shortest side cm, so area is 12.5√ cm².
### 30 60 90 Triangle Worksheets Answers PDF
30 60 90 Triangle Worksheets Answers
### 30-60-90 Triangle Worksheet PDF
30-60-90 Triangle Worksheet
We recommend using Brighterly Worksheets for the topic “30-60-90 triangles”. These worksheets provide you with clear visual aids and practice problems that will help you better understand this topic and improve your problem-solving skills.
## Frequently Asked Questions on 30-60-90 Triangle
### Why is the 30-60-90 triangle special in geometry?
The 30-60-90 triangle is special due to its consistent angle measurements and side ratio of 1:√3:2. This makes calculations predictable and allows for easy application of trigonometric principles.
### Can the side lengths of a 30-60-90 triangle be any numbers?
The side lengths must adhere to the ratio 1:√3:2. Any set of numbers maintaining this ratio can represent the sides of a 30-60-90 triangle.
### How do you find the perimeter of a 30-60-90 triangle?
To find the perimeter, add the lengths of all three sides. If the shortest side is ‘a’, the perimeter is .
### Is it possible to construct a 30-60-90 triangle with a compass and straightedge?
Yes, by drawing a 60-degree angle using a compass and then bisecting it, you can construct a 30-60-90 triangle.
### How does the Pythagorean theorem apply to a 30-60-90 triangle?
The Pythagorean theorem applies as it is a right triangle, verifying the side lengths’ ratio.
### Can a 30-60-90 triangle be scalene?
No, a 30-60-90 triangle cannot be scalene as two of its angles are always the same in every instance, making it an isosceles triangle.
### What is the significance of the 30-60-90 triangle in trigonometry?
This triangle is significant in trigonometry for studying sine, cosine, and tangent functions, as it provides straightforward ratios.
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## Making Change from a Dollar
I was trying to figure out this problem, “How many different ways can we make change of \$ 1.00, given half-dollars, quarters, dimes, nickels, and pennies?” This problem is featured in this Ramblings of a Geek blog post and is also in section 1.2.2 of the book Structure and Interpretation of Computer Programs (PDF version).
I think it involves the pigeonhole principle when trying to figure out how the recursive formula works. Here’s part of the set up from the Ramblings of a Geek blog post:
Let An be the number of ways to make n cents with pennies.
Let Bn be the number of ways to make n cents with pennies and nickels.
Let Cn be the number of ways to make n cents with pennies, nickels, and dimes.
Let Dn be the number of ways to make n cents with pennies, nickels, dimes, and quarters.
The question is asking us to find D100
So if you’re trying to figure out Dn=100 as in the post, where n=the amount, and D is a combination of quarters, dimes, nickels and pennies, you use the “pigeonhole principle” and split the choices into combinations using quarters and combinations without using quarters first.
So, if you use 0 quarters, that could be C100 combinations, where Cn is the number of combinations using dimes, nickels, and pennies but no quarters. That’s your “no quarters” box. Then your other box, to begin with, for a combination to belong in the “using quarters” box, you know automatically that the combination must be using at least one quarter to belong in that box. So since the any combination is for sure using at least one quarter (25 cents), you will have a remaining 75cents (the beginning 100 cents minus the quarter you’re sure exists in the combination) that you’re unsure of the composition of, that is, you have 75 cents that will be made up of a combination of 0-3 quarters, 0-7 dimes, 0-15 nickels, and 0-75 pennies. Since it could be a combination of quarters, nickels, and pennies, that can fit in category “D”, and since the amount is 75 cents, you can label that as D75.
So, D100=C100 + D75. So technically, like in the SICP book and the Ramblings of a Geek blog post, you can think of a recursive formula:
The number of ways to change amount a using n kinds of coins equals the number of ways to change amount a using all but the first kind of coin, plus the number of ways to change amount a – d using all n kinds of coins, where d is the denomination of the first kind of coin.
So we’ve determined above, that D100=C100 + D75. You can use the same logic as above on the D75 to reduce that to a C. For example, for D75, you might get 75 cents by using zero quarters, so you could categorize that as C75…and if you were using at least one quarter, you would know for sure what 25 cents of that 75 cents was made out of (that one quarter), that would leave you with 50 cents you were not sure about the composition of, that might be made up of quarters, dimes, nickels, and pennies, which could be categorized as D50, so D75=C75 + D50. You can do that with the recursive formula, you can then examine D75, that would be equal to D75=C75+D50. And on down the chain until the Ds disappear completely, replaced by all Cs, then replace the Cs with Bs and As.
And with As, there will always only be one way to make anything out of pennies; \$1 is 100 pennies, 50 cents is 50 pennies, etc. So when you examine the Bs, the formula is n/5 + 1. The 1 is for one way you can make anything using all pennies, and the n/5 is how many nickels fit in the amount n. So, say n=15. Here n/5 is 15/5=3. B15=4, using the formula n/5 + 1. That corresponds here to these four situations: you use one nickel and ten pennies, or two nickels and five pennies, or three nickels and no pennies, or no nickels and 15 pennies!
So starting at Dn, you reduce down to Cs, then to Bs and As. Here’s the rest of the Rambling Geek’s example:
We want D100, let’s use some substitution here:
D100 = C100+D75
D100 = C100+C75+D50
D100 = C100+C75+C50+D25
D100 = C100+C75+C50+C25+1
Similarly:
C100 = B100+B90+B80+B70+B60+B50+B40+B30+B20+B10+1
C75 = B75+B65+B55+B45+B35+B25+B15+B5 (note no +1, since C5 = B5)
C50 = B50+B40+B30+B20+B10+1
C25 = B25+B15+B5
Substituting with the values of Bn in the formula above we see:
C100 = 21+19+17+15+13+11+9+7+5+3+1 = 121
C75 = 16+14+12+10+8+6+4+2 = 72
C50 = 11+9+7+5+3+1 = 36
C25 = 6+4+2 = 12
Finally:
D100 = C100+C75+C50+C25+1 = 121+72+36+12+1 = 242
So there are 242 ways to uniquely make change for \$1 using pennies, nickels, dimes, and quarters.
Each time you subtract the denomination of the type of coin being examined from the amount, that’s because you’re setting up the case where you are using that type of coin, such as quarters; so C100 is all the ways to make 100 cents made of 0 quarters and 0-10 dimes, 0-20 nickels, and 0-100 pennies; C75 is all the ways to make 100 cents using 1 quarter, 0-7 dimes, 0-15 nickels, and 0-75 pennies; C50 is all the ways to make 100 cents using two quarters, 0-5 dimes, 0-10 nickels, and 0-50 pennies; C25 is all the ways to make 100 cents using three quarters, 0-2 dimes, 0-5 nickels, and 0-25 pennies; C0 is all the ways to make 100 cents using four quarters, namely, 1.
Disclaimer, I hope that helps explain the logic behind the problem and the solution and hope that my explanation is correct, it might not be! 😉
internal tag: math
(Why doesn’t the search function on wordpress.com blogs search post tags too? I can’t believe that when I search my blog with the term “math,” only posts with “math” in the actual post text are retrieved; if the post is tagged “math” but doesn’t contain the term “math” in the post text, the post won’t be retrieved…hence the “internal tag” I had to add. Bah.)
See these other math related posts including:
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# How to find period of this periodic function?
How can I find a period of this function?
$$2\sin{3x} + 3\sin{2x}$$
Is here any way how to sum both sinuses?
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You might want to take a look at this answer. – JohnD Dec 26 '12 at 20:45
One has period $\pi$ and the other has period $2\pi /3$. What you want now is to see when they "match up". This is obtained in $2\pi$. Basically, this is $3\times 2\pi/3$ and $2\times \pi$. We're just cross multiplying periods.
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You can always go the hard way analyzing, for arbitrary $T$: $$2 \sin(3 (x+T) ) + 3 \sin(2(x+T)) = \\ 2 \sin(3x) \cos(3T) + 2 \cos(3x) \sin(3T) + 3 \sin(2x) \cos(2T) + 2 \cos(2x) \sin(2T)$$ When $T$ equals a period you should have $$\sin(3T) = 0, \quad \sin(2T) = 0, \quad \cos(3T) = 1, \quad \cos(2T) = 1$$ Since $\sin(3T) = \sin(T+2T) = \sin(T) \underbrace{\cos(2T)}_1 + \cos(T) \underbrace{\sin(2T)}_0 = \sin(T)$, and $\cos(3T) = \cos(T) \cos(2T) - \sin(T) \sin(2T) = \cos(T)$. We have: $$\sin(T) = 0, \quad \sin(2T) = 0, \quad \cos(T) = 1, \quad \cos(2T) = 1$$ Repeating the exercise we see that $\sin(2T) = 0, cos(2T)=1$ is implied by $\sin(T)=0$ and $\cos(T) = 1$. Solving $\sin(T)=0$ and $\cos(T)=1$ is easy. There are infinitely many solutions: $$T = 2 \pi n, \quad n \in \mathbb{Z}$$ The minimal solution is $T = 2\pi$.
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To find a period is no problem, note that $2\pi$ is a period of both parts.
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The (smallest) period can be shown to be $2\pi$ by yet another method. Note that $$f(x)=2\sin(3x)+3\sin(2x)=2\sin x(\cos x+1)(4 \cos x-1).$$ So the zeros of $f(x)$ in $[0,2\pi]$ occur at $0, \pi, 2\pi$ and at two other points obtained from $4 \cos x-1=0$ which are $\cos^{-1}(1/4)$ and $2\pi-\cos^{-1}(1/4).$
In any period of $f(x)$ this pattern of zeros would have to repeat. However the numeric estimates $$0.00,\ 1.318,\ 3.141,\ 4.965,\ 6,28$$ are not evenly spaced, in fact the spacings are $$1.318,\ 1.823,\ 1.823,\ 1.318.$$ It seems clear this uneven spacing of the zeros makes it impossible for $f(x)$ to repeat before a full $2\pi$ goes by.
EDIT: It is because this spacing pattern is of the form ABBA(where lengths A,B different) that the period must be $2\pi$ and not smaller. A different function with zero spacing pattern ABAB might have period $\pi$. If extended, the ABBA pattern is ABBAABBAABBA.., and there are no periods of less than four letters length in this pattern. The repeated pattern ABABABAB.. has a period of two letters, hence zero spacing of that form might have period $\pi$ rather than $2\pi$.
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The fundamental period of sin function is $2\pi$. Now,Add the period of $2\pi$ in each sin function of given equation.
$y=2\sin(3x) + 3\sin(2x)$
$y=2\sin(3x+2\pi) + 3\sin(2x+2\pi)$
taking $3$ common from 1st sin fun. and $2$ from 2nd sin fun.
$y=2\sin[3(x+2\pi/3)] + 3\sin[2(x+2\pi/2)]$
so, the period of 1st sin fun. is $\frac{2\pi}{3}$
and the peiod of 2nd sin fun. is ${\pi}{}$
Finally,add these two periods as: (2pi/3 + 2pi/2)
which is = $\frac{5\pi}{3}$ Answer.
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Please format your answer using $\LaTeX$ to be more readable. Some help is here. I'll do some of it to be an example. – Ross Millikan May 1 '13 at 15:45
thanks dear.... – engr.haseeb May 1 '13 at 18:24
The period of the first is $2 \pi/3$, that of the second is $\pi$. When we take the sum, the resulting period is the least common multiple of the two, which is $2 \pi$.
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the correct ans will be $2\pi$, as the period of first is $2\pi$ by $3$ and second is $2\pi$ by $4$.
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# Find derivative of w.r.t x where $y=\left[ \sin x \right]\ln x$\begin{align} & \text{(A)-cosx-}\dfrac{\text{lnx}}{\text{lnx}} \\ & \text{(B)}\dfrac{\text{1}}{\text{2}}\text{cosx+lnx}\text{.cosx}\text{.sinx} \\ & \text{(C)}\left| \text{x} \right|\text{cosx+lnx}\text{.cosx}\text{.Insin} \\ & \text{(D)lnxcosx+}\dfrac{\text{sinx}}{\text{x}} \\ \end{align}
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Hint: The given function has two independent functions of x in product form. Since both are x dependent thus they can’t be directly differentiated with respect to x. These type of problems are solved by a particular method using product rule, given by $\dfrac{\text{dy}}{\text{dx}}=\text{u}\times \dfrac{\text{d}}{\text{dx}}(v)+\text{v}\times \dfrac{d}{\text{dx}}\left( u \right)$, where function is $\text{y=u}\times \text{v}$. We will also be using standard derivatives like $\dfrac{\text{d}}{\text{dx}}\left( \text{sinx} \right)=\text{cosx},\text{ }\dfrac{\text{d}}{\text{dx}}\left( \text{cosx} \right)=\text{-sinx}$, $\dfrac{\text{d}}{\text{dx}}(\text{lnx)=}\dfrac{1}{\text{x}}$ to solve the question.
Complete step-by-step solution:
Let us learn about the product rule first. In product rule, we have two functions of x, assigning them as first and second functions respectively. We keep the first function constant and differentiate the second function with respect to x then, keeping the second function constant and differentiate the first function with respect to x.
Suppose, $\text{y=u}\times \text{v}$; u = first function of x and v = second function of x.
Then, product rule is given by $\dfrac{\text{dy}}{\text{dx}}=\text{u}\times \dfrac{\text{d}}{\text{dx}}(v)+\text{v}\times \dfrac{d}{\text{dx}}\left( u \right)$.
Here we have the function $y=\text{(sinx)}\text{.lnx}$.
The given function has two independent functions $\sin x$ and $\ln x$. So, y cannot be differentiated directly.
As we have already discussed we use chain rule for solving these types of particular questions. Here, we have the first function as $\sin x$ and the second function as $\ln x$.
Therefore, we can write it as $\dfrac{\text{dy}}{\text{dx}}=\dfrac{\text{d}}{\text{dx}}\left( \left( \text{sinx} \right).\text{lnx} \right)$
Now, applying the product rule, we get
$\Rightarrow \dfrac{\text{dy}}{\text{dx}}=\text{lnx}\text{.}\dfrac{\text{d}}{\text{dx}}\left( \text{sinx} \right)+\text{sinx}\text{.}\dfrac{\text{d}}{\text{dx}}\left( \ln x \right)$
Since, we know that $\dfrac{\text{d}}{\text{dx}}\left( \text{sinx} \right)=\text{cosx and }\dfrac{\text{d}}{\text{dx}}\left( \text{lnx} \right)=\dfrac{1}{\text{x}}$
On putting these derivatives, we get:
\begin{align} & \Rightarrow \dfrac{\text{dy}}{\text{dx}}=\text{lnx}\text{.cosx+sinx}\text{.}\dfrac{1}{\text{x}} \\ & \Rightarrow \dfrac{\text{dy}}{\text{dx}}=\dfrac{1}{\text{x}}\left( \text{x}\text{.lnc}\text{.cosx+sinx} \right) \\ \end{align}
Therefore, option D is correct.
Note: Students may directly differentiate the y and give the answer as $\dfrac{\text{d}}{\text{dx}}\left( \left( \text{sinx} \right)\text{lnx} \right)=\dfrac{\text{cosx}}{\text{x}}$ which is totally wrong. So this kind of silly mistake must be avoided. Also take the derivative of functions sinx and cosx properly, without making sign changes.
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### Consecutive Numbers
An investigation involving adding and subtracting sets of consecutive numbers. Lots to find out, lots to explore.
### Roll These Dice
Roll two red dice and a green dice. Add the two numbers on the red dice and take away the number on the green. What are all the different possible answers?
### Domino Square
Use the 'double-3 down' dominoes to make a square so that each side has eight dots.
# Sealed Solution
## Sealed Solution
Here is a set of ten cards, each showing one of the digits from 0 to 9:
The ten cards are divided up between five envelopes so that there are two cards in each envelope.
The sum of the two numbers inside it is written on each envelope:
What numbers could be inside the "8" envelope?
Thank you to Alan Parr who inspired this task.
### Why do this problem?
Sealed Solution offers the chance for children to work in a systematic way and is a great context in which to encourage them to explain and justify their reasoning.
### Possible approach
Begin by familiarising children with the context: Using digit cards 0 to 5, invite the class to watch as you put 0 and 1 in one envelope and write their total on the outside (or on a 'post-it' note stuck to the envelope). Put 3 and 5 in another envelope, again writing their total on the envelope. Explain that the other two cards will go in the last envelope. What will the total be? How do they know?
Try this again, this time putting 0 and 5 in one envelope and recording the total. But then put two cards in another envelope without showing them to the children. Write the total on the outside of the envelope. Repeat this for the third envelope. (For example you could have 1 and 3 in the first and 2 and 4 in the second.) What numbers are in the two envelopes? How do they know?
Try again, this time keeping 0 and 5 in the first envelope but suggest that you want to put the other cards in pairs into the envelopes, so that the totals on the other two are the same. What could you do? How do they know? At each stage, children can be working in pairs, perhaps using mini-whiteboards and digit cards to try out their ideas.
As a final 'warm-up' challenge with cards 0 to 5, show three envelopes with the totals 4, 5 and 6 written on them. Invite learners to find out which pair of cards could be inside each envelope. In this case, there are two possible solutions, which will prepare them well for the challenge as written in the problem.
Use the images in the problem to explain that we now have digit cards 0 to 9, and that they have been put in pairs into five envelopes with the totals of 3, 7, 8, 13 and 14. Allow time for learners to work in their pairs on the task, bringing everyone together for a mini plenary/plenaries as appropriate, to share approaches and ideas so far.
In the final plenary, focus on ways in which different pairs have worked systematically. You may decide to invite one pair to explain how they worked systematically but without giving any solutions and instead use that way as a whole group to find all the solutions. It is important to remember that there are lots of ways to work systematically and different pairs will have different approaches, which may be quite different from the way you would have chosen!
### Key questions
Which envelope shall we try first? Why?
What could be in this envelope?
Are there any numbers which you know definitely aren't in this envelope? Why?
Are there any other solutions?
### Possible support
Having digit cards available for children to use will free up their thinking and will make it easier to try out different ideas without worrying about crossing ideas out on paper.
### Possible extension
Children could make up their own problem along these lines.
Alan Parr, the creator of this task, wrote to tell us:
'I've recently returned to this for the first time in ages, working with some Year 6s [10 and 11 year olds]. They found it so accessible and involving that we took it to places I'd never previously dreamt of.' You can read what they did in the first April 2015 post on Alan's blog, and he writes about the task again in two March 2017 posts.
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10-Solved problems for 3×3 matrix by Crout’s LU-option 1.
Solved problems for 3×3 matrix by Crout’s LU-option -1.
We have two solved problems for the 3×3 matrix by Crout’s LU decomposition. for which we will apply the same technique used to derive the elements of the lower matrix and the upper matrix.
The video for first solved problem-Option-1.
The video used is for the illustration of the first solved problem of Crout’s LU decomposition for 3 x3 matrices. The solution is based on option-1, for which the matrix A is converted to a lower matrix to get the values for the different elements of the two matrices L and U.
The first solved problem for 3×3 matrix by Crout’s LU-Option-1.
The first problem for the 3×3 matrix is Crout’s LU decomposition. It is required to find both the Lower and upper matrix decomposition LU by using Crout’s method through a given matrix A. The given matrix is 3×3.
We will apply the previous post technique to get the L1 and U1 matrix components by using the following steps.
Step-1 Convert matrix A to a lower matrix.
First, we have matrix A as (1 1 1, 4 3 -1, 3 5 3 ). We will estimate the determinant value of this matrix A, it is estimated as equal to +10. Since the determinant value is not equal to zero, then the matrix is invertible and we can proceed to find L and U matrices based on Crout’s method.
The three elements in the first column vector of the A matrix which are 1 4 3 will be set as L11 &L21 and L31. L11=1, L21=4,L31=3. While for U12, we will divide a21/a11=1/1=1. While U13=a13/a11=1.
As for the upper matrix U1, the u11=1, U12=a21/a11=1/1=1 and u13=a13/a11=1/1=1. The detailed procedures are shown in the slide image.
Step-2-Derive the values for the L22, L23, and U23-option-1.
We will use a11 as a pivot, we multiply (-1)*(a12/a11)*column 1 and add the result to the 2nd column, similarly, we multiply (-1)*(a13/a11)*column 1 and add the result to the 3rd column. If we call the new matrix L1, the elements of the matrix are as follows:
The first row is (1 0 0). The second row is (4 -1 -5). The third row is (3 2 0). If we compare the first row of L1 to the next matrix, we have (L11 0 0), then L11=1. If we compare the second row of L1 to the next matrix, we have (L21 L22 L22*U23), then L21=4.
If we compare the third row of L1 to the next matrix, we have (L31 L32 L32*U23+L33), then L31=3 and L32 =2.
Step-3-Derive the value for the L33.
We want to proceed to get the final l matrix, we want to let -5 equal zero. We multiply (-1)*(-5/-1)*column 2 and add the result to the 3rd column. The final elements of the Lower matrix l will be as follows:
The first row is (1 0 0). The second row is ( 4 -1 -5*-1+(-5)). The third row is ( 3 2 -5*2+(0)).
Step-4-Check the product of(L*U)=A-matrix for a 3×3 matrix.
In the final slide, we have all the elements in the first column of the matrix identical to the first column of the lower matrix. The diagonals of the U matrix U11=U22=U33=1. L22=-1. U23=+5 while L33=-10. U12=1 and u13=1.
We can write both the lower matrix and the upper matrix and check whether the multiplication of the lower matrix by the upper matrix will give us the final A matrix or not. Using the row-by-column multiplication, we can see that L*U multiplication will give the matrix A.
The second solved problem-Crout’s LU-Option-1.
First, we have matrix A as ( 10 3 4, 2 – 10 3, 3 2 -10 ). We will estimate the determinant value of this matrix A, which is estimated as equal to +1163 Since the determinant value is not equal to zero, then the matrix is invertible and we can proceed to find L and U matrices based on Crout’s method. The second solved problem, the detailed procedure to estimate the determinant is shown in the next slide image.
Step-1 Convert matrix A to a lower matrix.
The three elements in the first column vector of the A matrix which are 10 2 3 will be set as L11 &L21 and L31. L11=10, L21=2,L31=3. While for U12, we will divide a21/a11=3/10=3/10. While U13=a13/a11=4/10.
Step-2-Derive the values for the L22, L23, and U23.
We will use a11 as a pivot, we multiply (-1)*(a12/a11)*column 1 and add the result to the 2nd column, similarly, we multiply (-1)*(a13/a11)*column 1 and add the result to the 3rd column. If we call the new matrix L1, the elements of the matrix are as follows:
The first row is (10 3 4). The second row is (2 -10 3). The third row is (3 2 -10). If we compare the first row of L1 to the next matrix, we have (L11 0 0), then L11=1. If we compare the second row of L1 to the next matrix, we have (L21 L22 L22*U23), then L21=2.
If we compare the third row of L1 to the next matrix, we have (L31 L32 L32*U23+L33), then L31=3 and L32=1.1. To get U23 divide ((2.2)/(-10.6)=-11/53.
Step-3-Derive the value for the L33-Crout’s LU decomposition.
We want to proceed to get the final l matrix, we want to let 2.2 to be equal zero. We multiply (-1)*(2.2/-10.60)*column 2 and add the result to the 3rd column. The final elements of the Lower matrix l will be as follows:
The first row is (10 0 0). The second row is ( 2 -10.60 0). The third row is ( 3 1.1 -1163/106). As we can see L33 will be equal to -1163/106.
Step-4-Check the product of(L*U)=A-matrix for a 3×3 matrix.
Using the row-by-column multiplication, we can see that L*U multiplication will give the matrix A.
The full detailed steps are shown in the previous slide images, at the end we will check the product of L*U against the Matrix A value.
This is a link to the next post is the introduction to the permutation matrix.
References
This is the Omni calculator for estimating various items of linear algebra -LU Decomposition Calculator.
Another calculator to use is a Calculator for matrices.
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# How to Calculate Percent Difference – Definition, Formula, Examples
Home » Math Vocabulary » How to Calculate Percent Difference – Definition, Formula, Examples
## What Is the Percent Difference in Math?
Percent difference is the ratio of the absolute difference between two values to the average of the two values, calculated as a percentage.
Percent Difference $= \frac{Absolute \; Difference}{Average} \times 100$
In simple words, the percent difference is the percentage of the relative difference. The relative difference between two quantities compares the difference between the two values in relation to their average.
Relative Difference $= \frac{Absolute \; Difference}{Average}$
The percent difference is frequently used when comparing data sets or estimating the degree of change between two numbers.
Note that the “percent difference” and the “percent change” are two different concepts. We will compare them later in the article. Also, do not confuse this formula with the percent decrease or increase formulas.
## What Is the Percent Difference Formula?
The percentage difference is the ratio of the difference in their values to their average, multiplied by 100.
Percent difference formula for the two values A and B can be given as
Percent difference $= \frac{A – B}{ \frac{(A + B)}{2}} \times 100$
## How to Find Percent Difference
Follow the steps given below to learn how to calculate the percent difference between two values A and B.
Step 1: Find the absolute difference.
Find the absolute value of the difference between A and B. The order does not matter since we will be taking the absolute value.
Absolute Difference $= |A \;-\; B|$
Step 2: Find the average of A and B. Simply add the values together and divide the sum by 2.
Average $= \frac{(A + B)}{2}$
Step 3: Divide the absolute difference by the average and multiply by 100 in order to calculate the percent difference.
Percent Difference $= \frac{Absolute \; Difference}{Average} \times 100$
Percent difference $= \frac{|A \;-\; B|}{ \frac{(A + B)}{2}} \times 100$
## Percent Difference between Two Numbers
The percent difference between the two numbers is expressed as follows:
Percent Difference $= \frac{|A – B|}{ \frac{(A + B)}{2}} \times 100$
Example: Alex has 10 years of working experience. Paxton has 20 years of working experience. What is the percent difference?
Here, $A = 10, B = 20$
Note that these values represent the working experience of Alex and Paxton in years.
Can you decide which value is more important here? No, both values hold the same importance!
In such cases, we can use the percent difference formula.
We choose the average value as the reference point. We take absolute difference to treat both values equally. So, the order does not matter when you subtract.
$|A \;-\; B| = |10 \;-\; 20| = | \;-\;10| = 10$
$\frac{(A + B)}{2} = \frac{10 + 20}{2} = 15$
Percent Difference $= \frac{|A – B|}{\frac{(A + B)}{2}} \times 100$
Percent Difference $= \frac{10}{15} \times 100$
Percent Difference $≈ 66.67\%$
## Percent Difference vs. Percent Change
• The context of the data being analyzed should be regarded when interpreting percent differences.
• The percent difference just measures the relative difference between two values and does not reveal the pattern or direction of change. It does not distinguish between positive and negative changes.
• When there is an old value and a new value, we should use the percentage change formula.
• When there is an approximate value and an exact value, we should use the percentage error formula.
• When both the values have equal importance or when it is not possible to determine the reference value, we use the percent difference formula.
## Conclusion
You know the percent difference, definition, concept, formula, etc. Let’s move toward the numerical section of a percent difference, including solved examples and MCQs, for a clearer understanding.
## Solved Examples on Percent Difference
1. Mr. Dunphy’s garden has 78 rose plants. Mr. Prichet’s garden has 106 rose plants. Find the percent difference.
Solution:
Firstly, we will calculate the difference between the two numbers.
Difference $= 78\;-\;106 = \;-\; 28$
We can remove the negative sign by giving the absolute value of the difference.
Absolute Difference $= |Difference| = |\;-\;28| = 28$
Average $= \frac{78 + 106}{2} = \frac{184}{2} = 92$
Now, find the percentage difference.
Percentage Difference $= (\frac{Absolute \;Difference}{Average}) \times 100$
Percentage Difference $= \frac{28}{92} \times 100$
Percentage Difference $≈ 30.43\%$
2. In a video game tournament, Team A scored 980 and Team B scored 1220 points. What is the percentage difference between the scores?
Solution:
On calculating the difference between the two scores, we get
Difference $=$ Score of Team B – Score of Team A $= 1220\;-\;980 = 240$
Absolute difference $= | 240 | = 240$
Average $= \frac{1220 + 980}{2} = 1100$
Now, we can calculate the percentage difference between the teams’ scores.
Percentage Difference $= (\frac{Difference}{Average})\times100$
Percentage Difference $= (\frac{240}{1100}) \times 100$
Percentage Difference $≈ 21.81\%$
3. What is the percentage difference between 7.2 and 8.5?
Solution:
Here, we have to calculate the difference between two numbers.
Difference $= 7.2 \;-\; 8.5 = \;-\; 1.3$
We will remove the negative sign by taking the absolute value of the difference.
Absolute Difference $= |Difference| = |\;-\;1.3| = 1.3$
Now, you can calculate the percentage difference.
Average $= \frac{7.2 + 8.5}{2} = 7.85$
Percentage Difference $= (\frac{Absolute \;Difference}{Average}) \times 100$
Percentage Difference $= (\frac{1.3}{7.85}) \times 100$
Percentage Difference $≈ 16.56\%$
## Practice Problems on Percent Difference
1
### The population of two cities A and B is 50,000 and 55,000 respectively. What is the percentage difference?
$5\%$
$5.8\%$
$920\%$
$9.52\%$
CorrectIncorrect
Correct answer is: $9.52\%$
Population difference $= 55,000 \;-\; 50,000 = 5,000$
Average $= \frac{(50,000 + 55,000)}{2} = 52,500$
Percent difference $= \frac{5,000}{52,500} \times 100 ≈ 9.52\%$
2
### The weight of an object A is 200 grams. The weight of object B is 180 grams. What is the percent difference?
$8\%$
$10\%$
$14\%$
$25\%$
CorrectIncorrect
Correct answer is: $10\%$
Percent Difference $= \frac{|Weight \;difference|}{(Average\; weight)} \times 100$
Percent Difference $= (\frac{20}{190}) \times 100$
Percent Difference $≈ 10.53\%$
3
### The population of town A is 10,000 and that of B is 12,500. What is the percent difference?
$22\%$
$25\%$
$45\%$
$52\%$
CorrectIncorrect
Correct answer is: $22\%$
Difference $= 12,500 \;-\; 10,000 = 2,500$
Average$= 11,250$
Percent difference $=(\frac{2,500}{11.250}) \times 100 ≈ 22.22\%$
4
### The price of a watch is $\$80$. The price of a clock is$\$58$. Choose the correct percent difference in the price.
$25\%$
$22\%$
$32\%$
$35\%$
CorrectIncorrect
Correct answer is: $32\%$
Difference $= \$80 \;-\; \$58 = \$22$Average$= 69$Percent difference$=(\frac{16}{72}) \times 100 ≈ 32\%\$.
## Frequently Asked Questions on Percent Difference
Since there is no way to decide which value is the reference value, we consider the average of two values as the reference value. It helps us treat both the values equally.
Since we focus on the magnitude of the change (regardless if it is positive or negative) and not on the direction of the change, we don’t take the minus sign into account. Since both the values are equally important, the use of the negative sign is not useful.
When we want to compare the exact value with the estimated value (or measurement), we prefer the percent error formula. It helps us quantify the accuracy or precision of a measurement.
Percent change generally indicates growth or decline, the percent error measures accuracy in the measurement, and the percent difference helps us quantify variation or inconsistency in the two values.
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