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# How do you find the domain and range of 1/(x+1)?
Aug 5, 2015
Domain: $\left(- \infty , - 1\right) \cup \left(- 1 , + \infty\right)$
Range: $\left(- \infty , 0\right) \cup \left(0 , + \infty\right)$
#### Explanation:
The domain of the function will have to take into account the fact that the denominator of the fraction cannot be equal to zero.
This means that any value of $x$ that makes the expression $x + 1 = 0$ will be excluded from the domain.
More specifically, you have
$x + 1 = 0 \implies x = - 1$
The domain of the function will thus be $\mathbb{R} - \left\{- 1\right\}$, or $\left(- \infty , - 1\right) \cup \left(- 1 , + \infty\right)$.
The range of the function will be influenced by the fact that you don't have a value of $x$ for which the function is equal to zero.
The range of the function will thus be $\mathbb{R} - \left\{0\right\}$, or $\left(- \infty , 0\right) \cup \left(0 , + \infty\right)$.
graph{1/(x+1) [-10, 10, -5, 5]}
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# Modelling a functional building
## Modelling a functional building
• Submitted By: Zzzoe
• Date Submitted: 04/02/2014 4:00 PM
• Category: Science
• Words: 1407
• Page: 6
Introduction:
In this task, the building with a roof structure shown above is initially assumed to have a rectangular base of 150m long and 72m wide. The maximum height of the structure should not exceed 75% of its width for stability or be less than half the width for aesthetic purposes. Moreover, the minimum height of a room in a public building is 2.5 m.
Under this condition, several questions about the modifications of the structure of the building and the office block inside it are investigated, thus for convenience, this building is modeled in a cross coordinates for investigation.
The questions will be discussed again when the rectangular base is change to be 150m wide and 72m long without changing the other sets.
Part I: Modeling the structure in cross coordinates
When the base of the structure is 150m long and 72m wide, the height of the structure should no exceed 54m for stability or 36m for aesthetic according to the requirements.
Calculation:
72m x 0.75=54m
72m x 0.5=36m
Because we model the structure in the cross coordinate, the function of the roof structure is modeled to be a quadratic function with the equationaccording to the shape of the structure. And because the open side of the function is pointed downward, therefore a is smaller than 0 and the highest point of the equation is (0, c).
Next, we can create a model for the curved roof structure, which has the minimum height of 36m, thus the highest point of the curve is (0, 36).
On the basis of these data, we can draw a sketch of the curve of roof structure as shown below.
By using the graph and the data, we can find the equation of the parabola which goes through point A, B and C.
Calculation:
Substitute x=0, y=36 into the equation, we can get
Substitute x=36, y=0 and x=-36 y=0 into the equation, we can get
After solving the 2x2 system, we can get
Finally, we can get that the equation of the parabola is...
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# A container opened from the top and made up of a metal sheet, is in the form of a frustum of a cone of height 16 cm with radii of its lower and upper ends as 8 cm and 20 cm, respectively. Find the cost of milk which can completely fill the container, at the rate of Rs 20 per litre. Also find the cost of metal sheet used to make the container, if it costs Rs 8 per 100 cm2. (Take $$\pi \ = \ 3.14$$ ).
Here R = 20 cm, r = 8 cm and h = 16 cm
Capacity of the container = Volume of the frustum $$= \ \frac{1}{3} \pi h(R^2 \ + \ r^2 \ + \ Rr)$$
$$= \ \frac{1}{3} \ × \ 3.14 \ × \ 16(20^2 \ + \ 8^2 \ + \ 20 \ × \ 8) \$$
$$= \ \frac{50.24}{3} \ × \ (400 \ + \ 64 \ + \ 160)$$
$$= \ \frac{50.24}{3} \ × \ 624 \$$
$$= \ 50.24 \ × \ 208$$
$$= \ \frac{10449.92}{1000}$$ litres
Cost of milk @ Rs 20 per litre $$= \ 20 \ × \ \frac{10449.92}{1000}$$
$$= \ 208.99$$ $$\approx$$ Rs $$209$$
To find the slant height
$$l \ = \ \sqrt{16^2 \ + \ 12^2} \$$
$$= \ \sqrt{256 \ + \ 144}$$
$$= \ \sqrt{400} \$$
$$= \ 20$$ cm
Curved Surface Area of the container = $$\pi (R \ + \ r)l$$
$$= \ \frac{22}{7} \ × \ (20 \ + \ 8) \ × \ 20$$
$$= \ \frac{22}{7} \ × \ 28 \ × \ 20 \ = \ 1758.4$$ cm2
Area of the bottom of the container = $$\pi r^2$$
$$= \ \frac{22}{7} \ × \ (8)^2 \ = \ \frac{22}{7} \ × \ 64 \ = \ 200.96$$ cm2
$$\therefore$$ Total area of metal required $$= 1758.4 \ + \ 200.96 \ = \ 1959.36$$ cm2
Cost of metal sheet used to manufacture the container @ Rs 8 per 100 cm2 $$= \ \frac{8}{100} \ × \ 1959.36 \ =$$ Rs $$156.75$$
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# Triangular Coordinate System
A packed figure which has three sides is known as triangle. Here we will see some properties of triangle. In a triangle the corner Point is known as the vertex of a triangle. In a triangle three vertices are present.
The distance measured around a triangle is known as perimeter of a triangle.
The sum of interior angles of a triangle is always 1800.
Now we will see Triangular coordinate system.
For finding the Area of Triangle when coordinates are given we need to follow some steps:
Step1: First we will find vertices of triangle.
Step2: For finding area, a formula is defined which is given below.
The formula is given below.
Area of triangle = |Fx (Gy – Fy) + Gx (Hy – Fy) + Hx (Fy – Gy)|,
2
Where ‘Fx’ and ‘Fy’ are the ‘x’ and ‘y’ of ‘F’ point.
Step 3: On putting values of all coordinates in formula we get area of triangle.
Suppose we have coordinates of a triangle P (3, 5), Q (5, 6) and R (10, 6), then by using these coordinates we find area of triangle.
For finding the area of triangle we need to follow above steps.
Here coordinates are F = (3, 5), G = (5, 6), H = (10, 6);
We know the formula for finding the area of triangle with vertices is given by:
Area of triangle = | Fx (Gy – Fy) + Gx (Hy – Fy) + Hx (Fy – Gy) |,
2
Now put the values in the given formula:
On putting the values we get:
Area of triangle = | 3 (6 – 5) + 5 (6 – 5) + 10 (5 – 6) |,
2
Area of triangle = | 3 (1) + 5 (1) + 10 (-1) |,
2
Area of triangle = | 3 + 5 + 10|,
2
Area of triangle = 18/2;
So the area of triangle is 9 inch2.
## Area of Triangle in Coordinate Geometry
Area of triangle in coordinate Geometry Coming Soon..
## Heron's Formula
Heron’s formula is used for calculating the Area of Triangle, it uses semi perimeter for calculation of area. Semi perimeter is given by:
S = ½ (p + q + r)
The area of the triangle by the heron s formula is given as:
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# 5.1: Exploring Area under the Curve
Difficulty Level: At Grade Created by: CK-12
This activity is intended to supplement Calculus, Chapter 4, Lesson 3.
## Problem 1 – Explore and discover
Graph the curve y=x2\begin{align*}y = x^2\end{align*}.
Your challenge is to think of at least two ways to estimate the area bounded by the curve y=x2\begin{align*}y = x^2 \end{align*} and the x\begin{align*}x-\end{align*}axis on the interval [0, 1] using rectangles. Use the following guidelines:
• all rectangles must have the same width
• you must build all your rectangles using the same methods
• the base of each rectangle must lie on the x\begin{align*}x-\end{align*}axis
Graph y=x2\begin{align*}y = x^2\end{align*} and set your window to [-0.1, 1] for x\begin{align*}x\end{align*} and [-0.2, 1.3] for y\begin{align*}y\end{align*}. Draw your first and second method on the graphs below. For each method calculate the following:
• Number of rectangles
• Height and width of each one
• Area of each
• Sum of the area
• Which method did a better job?
• How could you improve on it?
In the following problem, you will examine three common techniques that use rectangles to find the approximate area under a curve. Perhaps you discovered some of these techniques during your exploration in the above problem. The first problem uses rectangles whose right-endpoints lie on the curve y=x2\begin{align*}y = x^2\end{align*}.
## Problem 2 – Using five right-endpoint rectangles
Divide the interval [0, 1] into five equal pieces. Enter the information for each interval or rectangle in the table below. Remember that the right endpoint is the x\begin{align*}x-\end{align*}value and the height is the y\begin{align*}y-\end{align*}value of the right endpoint on the curve.
Interval Right Endpoint Height Area
• Calculate this sum.
• Now add up the numbers in the Area column.
The formula that can be used to express the total area is:
R5R5=0.2f1(0.2)+0.2f1(0.4)+0.2f1(0.6)+0.2f1(0.8)+0.2f1(1.0)or=0.2[f1(0.2)+f1(0.4)+f1(0.6)+f1(0.8)+f1(1.0)]\begin{align*}R_5 &= 0 . 2 \cdot f1(0 . 2) + 0 . 2 \cdot f1(0 . 4) + 0 . 2 \cdot f1(0 . 6) + 0 . 2 \cdot f1(0 . 8) + 0 . 2 \cdot f1(1 . 0)\\ & \qquad \qquad \qquad \qquad \qquad \qquad \text{or}\\ R_5 &= 0 . 2 [f1(0 . 2) + f1(0 . 4) + f1(0 . 6) + f1(0 . 8) + f1(1 . 0)]\end{align*}
• Are these two numbers the same or different?
Another way to find the area of the rectangles is using sigma notation.
• Write the notation in the x=15x2\begin{align*}\sum_{x=1}^5x^2\end{align*} form. Adjust what is being summed.
To sum it on the calculator, use Home > F3:Calc > 4:Sigma for the command with the format:
\begin{align*}\sum\end{align*}(expression, variable, lower limit, upper limit)
• Does this agree with the answer for the area you found previously?
## Problem 3 – Using five left-endpoint rectangles
This problem uses rectangles whose left-endpoints lie on the curve y=x2\begin{align*}y = x^2\end{align*}.
Divide the interval [0, 1] into five equal pieces. Enter the information for each interval in the table below. Remember that the left endpoint is the x\begin{align*}x-\end{align*}value and the height is the y\begin{align*}y-\end{align*}value of the left endpoint on the curve.
Interval Left Endpoint Height Area
• Calculate this sum.
• Now add up the numbers in the Area column.
The formula that can be used to express this area is:
L5L5=0.2f1(0)+0.2f1(0.2)+0.2f1(0.4)+0.2f1(0.6)+0.2f1(0.8)or=0.2[f1(0)+f1(0.2)+f1(0.4)+f1(0.6)+f1(0.8)]\begin{align*}L_5 &= 0 . 2 \cdot f1(0) + 0 . 2 \cdot f1(0 . 2) + 0 . 2 \cdot f1(0 . 4) + 0 . 2 \cdot f1(0 . 6) + 0 . 2 \cdot f1(0 . 8)\\ & \qquad \qquad \qquad \qquad \qquad \text{or}\\ L_5 &= 0 . 2 [f1(0) + f1(0 . 2) + f1(0 . 4) + f1(0 . 6) + f1(0 . 8)]\end{align*}
• Are these two numbers the same or different?
• What is the sigma notation for the area of the rectangles?
• Use the calculator to find the sum. Does this result agree with the answer above?
## Problem 4 – Using five midpoint rectangles
We will now investigate a midpoint approximation. How would you draw five rectangles, with equal width, such that their midpoints lie on the curve y=x2\begin{align*}y = x^2\end{align*}?
Divide the interval [0, 1] into five equal pieces. Enter the information for each interval or rectangle in the table below. Remember that the midpoint is the x\begin{align*}x-\end{align*}value and the height is the y\begin{align*}y-\end{align*}value of the midpoint on the curve.
Interval Midpoint Height Area
• Calculate this sum.
• Now add up the numbers in the Area column.
• Are these two numbers the same or different?
The formula that can be used to express this area is:
M5M5=0.2f1(0.1)+0.2f1(0.3)+0.2f1(0.5)+0.2f1(0.7)+0.2f1(0.9)or=0.2[f1(0.1)+f1(0.3)+f1(0.5)+f1(0.7)+f1(0.9)]\begin{align*}M_5 &= 0 . 2 \cdot f1(0 . 1) + 0 . 2 \cdot f1(0 . 3) + 0 . 2 \cdot f1(0 . 5) + 0 . 2 \cdot f1(0 . 7) + 0 . 2 \cdot f1(0 . 9)\\ & \qquad \qquad \qquad \qquad \qquad \text{or}\\ M_5 &= 0 . 2 [f1(0 . 1) + f1(0 . 3) + f1(0 . 5) + f1(0 . 7) + f1(0 . 9)] \end{align*}
• What is the sigma notation for the area of the rectangles?
• Use the calculator to find the sum. Does this result agree with the answer above?
## Problem 5- Summarize your findings
In this activity, you explored three different methods for approximating the area under a curve. The exact area under the curve y=x2\begin{align*}y = x^2\end{align*} on the interval [0, 1] is 13\begin{align*} \frac{1}{3}\end{align*} or 0.333.
• Which approximation produced the best estimate for the actual area under the curve?
• Describe which factors contribute to left, right, and midpoint rectangles giving overestimates versus underestimates.
• What can you do to ensure that all three of these techniques produce an answer that is very close to 13\begin{align*}\frac{1}{3}\end{align*}? Test your conjecture by using evaluating a sum that produces a much more accurate answer.
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# Graphing Equations & Equations of Lines
GRAPHING EQUATIONS & EQUATIONS OF LINES Lesson 13.1 & 13.2 STRATEGIES TO GRAPH A LINE Make a table. Special lines Vertical lines Horizontal Lines Intercepts (x, 0) & (0, y) Slope-intercept form
MAKE A TABLE & GRAPH 2X 4Y = 10 x y -1 -3 0 1 -5/2 -2 2
-3/2 SPECIAL LINES: GRAPH Y = 4 & X = -2 y = 4 is a horizontal line where all the y values are 4. x = -2 is a vertical line where all of the x values are -2. INTERCEPTS FIND THE X- AND Y- INTERCEPTS FOR THE LINE 2X 5Y = 10, THEN GRAPH. X intercept:
What is x when y is zero? 2x 5(0) = 10 x = 5 . (5, 0) Y-intercept: What is y when x is zero? 2(0) 5y = 10 y = -2 (0, -2) USE SLOPE-INTERCEPT FORM TO GRAPH THE LINE Y = 2X - 5. y = mx + b y
& x are any point (x, y) that line on the line. m is the slope b is the y-intercept (0, b) In the equation y = 2x 5, m =2 b = (0, -5) Graph the y-intercept Graph the slope Connect the dots.
VARIOUS FORMS OF LINEAR EQUATIONS Slope-Intercept Form (y-form) y = mx + b Point-Slope Form y y1 = m(x x1)
m = slope and (x1, y1) is a known point Standard Form (General Form) Ax + By = C THINGS NOT TO FORGET ABOUT LINEAR EQUATIONS 1. 2.
3. 4. 5. Parallel Lines have the SAME slope but different y-intercepts. Perpendicular lines have slopes that are negative reciprocals. Vertical lines have a slope that is undefined. Horizontal lines have a slope of zero. Midpoint between two points is found using the formula: x1 + x2 , y1 + y2 2 2
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# Lesson Planning of Ordinal Numbers
Lesson Planning of Ordinal Numbers
Subject English
Students` Learning Outcomes
• Write ordinal numbers.
Information for Teachers
• When objects are placed in order, we use ordinal numbers to tell their position, i.e. first (1st) second, (2nd). This concept is helpful in descriptive writing and everyday language use e.g. making lists.
Ordinal numbers can be used for both people and objects, e.g. ‘first runner’ or ‘third table / third chair in the row’.
• Except 1st, 2nd, and 3rd, “th” is added for every number to form ordinal number spellings and words. The yellow table is given you reference. Learn the spelling paste this table in the class.
• Ordinal numbers are added to all the numbers. This lesson plan focuses on first ten numbers then teaches them more till hundred.
• TIP “ordinal” tell you in what “order” things are?
• Tell the students that the order of the number is written in small font above the letter, e.g. 8th not 8th
• Cardinal Number: Cardinal numbers are that are only used for counting and are usually called counting numbers. Cardinals are the numbers that are used to determine how many of something there is or the quantity. If you want to know how many of something there is then you have to count them and that is where the cardinal numbers come in.
• The lowest cardinal number is one. Zero is not a cardinal number as to count how many of something you have , you have to have something in the first place.
• Students should have done number spellings of cardinal number (one to twenty and all the tens, i.e. twenty, thirty etc. up to ninety) before doing this lesson.
• While teaching the lesson, the teacher should also consult the textbook where it is required.
Material / Resources
Writing board, chalk / marker, duster, scotch tape, textbook
Introduction
• Paste / write/ draw the following information about ordinal numbers and the picture on the board.
• (An Ordinal Number is a number that tell the position of something in a list, such as; 1st, 2nd, 3rd, 4th, 5th, etc.
• Introduce the topic by telling the students that today they will learn about ordinal numbers (i.e. numbers that tell us the position of people and things)
• Brainstorm for everyday examples where ordinal numbers are used. Ask the students to tell about the things/people where we make lists. (Expected answer would be as; Result, sports, items, on a list, seating in the lass, order of the lines at assembly time.)
• Ask the students following questions:
1. Who came first in the previous test?
2. Who won the race?
3. Who came second?
• Call three students who got the first three positions in their previous test.
• Make them stand in the order of their positions: 1st, 2nd, & 3rd.
• Make them stand in wrong order, have them help each other line up in the right order.
• Write the order on the boar. Be careful to tell them about the order written above the number, e.g. 1st not 1 st s
• They have to use the ordinal words. For example, one student can tell the next, “You are third” or “I am first”.
Development
Activity 1
• Write the words, ‘first’, ‘second’, ‘third’ on the writing board and practice the spellings with the students.
• Ask the students to copy these words in their notebooks.
Activity 2
Write the ordinal numbers in words from fourth to tenth.
• Encourage the student to find out what is similar in spellings. (Expected answer would be as; “th” is added to every number after three for example five-fifth-eight-eighth, nine-ninth.
• All the students write the ordinal numbers in words (e.g. ninth) and numbers (e.g. 9th) in their notebooks. (They must also copy the table from the board)
• Tell the students that when we write these positions in numbers we write them like this. (take help from the table above and write the ordinal numbers e.g. 1st, 2nd, 3rd, 4th etc.)
Activity 3
Sum up / Conclusion
• Take dictation of ordinal numbers.
Assessment
• When you have finished the dictation, ask the students to check each other`s work and give marks.
• Take a round in the class and ask every student about their score.
• Make the first; second and third witness stand and whole class clap to them.
• Take a short quiz of students by asking them questions about position of his /her students sitting in the class, e.g. “Who is sitting on the 5th desk in the 1st row?” etc.
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## How do you find the median of 7 numbers?
Add up all of the numbers and divide by the number of numbers in the data set. The median is the central number of a data set. Arrange data points from smallest to largest and locate the central number. This is the median.
## How do you calculate the median of a sample?
The median is different for different types of distribution. For example, the median of 3, 3, 5, 9, 11 is 5. If there is an even number of observations, then there is no single middle value; the median is then usually defined to be the mean of the two middle values: so the median of 3, 5, 7, 9 is (5+7)/2 = 6.
30 percent
## How do you calculate the mean step by step?
Step-by-Step Process to Find the Mean Step 1: Add up all the numbers. The result is called the sum. Step 2: Count how many numbers there are. This number is called the sample size.
## What is the median of a set of numbers?
The median of a set of data values is the middle value. Half the data values are less than or equal to the median. Half the data values are greater than or equal to the median.
## Which is the formula for determining mode?
The three central measures of tendency are mean median and mode….Mode Formula Calculator.
Mode Formula = L + (fm – f1) x h / (fm – f1) + (fm – f2)
= 0 + (0 – 0) x 0 / (0 – 0) + (0 – 0)= 0
80.5
## How do you calculate mean median and mode?
How to find the mean, median and mode: MEAN
1. Step 2: Add the numbers up to get a total. Example: 2 +19 + 44 + 44 +44 + 51 + 56 + 78 + 86 + 99 + 99 = 622.
2. Step 3: Count the amount of numbers in the series.
3. Step 4: Divide the number you found in step 2 by the number you found in step 3.
## How do you calculate learning hours?
Once every member of the group has completed the eLearning course, you just add the individual seat times and divide by the number of participants. This will give you the average seat time per learner.
## How do I calculate the mean percentage?
To find the average percentage of the two percentages in this example, you need to first divide the sum of the two percentage numbers by the sum of the two sample sizes. So, 95 divided by 350 equals 0.27. You then multiply this decimal by 100 to get the average percentage. So, 0.27 multiplied by 100 equals 27 or 27%.
## What are the median numbers?
Median: The middle number; found by ordering all data points and picking out the one in the middle (or if there are two middle numbers, taking the mean of those two numbers). Example: The median of 4, 1, and 7 is 4 because when the numbers are put in order (1 , 4, 7) , the number 4 is in the middle.
## What is mean median and mode math is fun?
Mean, median and mode are Measures Of Central Tendency. Example: the arithmetic mean of 3, 7 and 8 is 6. Finding a Central Value.
## How do you calculate time on a task?
To calculate a percentage of time on task, simply divide the total number of boxes that indicate a student was on task by the total number of boxes. Research (Rienke) suggests that coaches list at least three times (completing three forms) before they calculate time on task for a given class.
## What is difference between mean and average?
The average is the sum of all values divided by the number of values. In statistics, mean is the average of the given sample or data set. It is equal to the total of observation divided by the number of observations.
## What is the formula of mode for grouped data?
Mode for grouped data is given as Mode=l+(f1−f02f1−f0−f2)×h , where l is the lower limit of modal class, h is the size of class interval, f1 is the frequency of the modal class, f0 is the frequency of the class preceding the modal class, and f2 is the frequency of the class succeeding the modal class.
## What is the formula for average?
Average, which is the arithmetic mean, and is calculated by adding a group of numbers and then dividing by the count of those numbers. For example, the average of 2, 3, 3, 5, 7, and 10 is 30 divided by 6, which is 5.
## How do you calculate average time spent on tasks?
Calculate the arithmetic mean or the average time to complete one task by dividing the total time to complete all tasks by the number of tasks completed (n).
## How do you do 40%?
To calculate the percentage, multiply this fraction by 100 and add a percent sign. 100 * numerator / denominator = percentage . In our example it’s 100 * 2/5 = 100 * 0.4 = 40 .
## What is an average in statistics?
In statistics, an average is defined as the number that measures the central tendency of a given set of numbers. There are a number of different averages including but not limited to: mean, median, mode and range.
## What do maths symbols mean?
< Less Than and > Greater Than This symbol < means less than, for example 2 < 4 means that 2 is less than 4. ≤ ≥ These symbols mean ‘less than or equal to’ and ‘greater than or equal to’ and are commonly used in algebra.
## How do you find the median of 15 numbers?
The “middle” of a sorted list of numbers. To find the Median, place the numbers in value order and find the middle number. Example: find the Median of {13, 23, 11, 16, 15, 10, 26}. The middle number is 15, so the median is 15.
## How do I calculate the median?
Median
1. Arrange your numbers in numerical order.
2. Count how many numbers you have.
3. If you have an odd number, divide by 2 and round up to get the position of the median number.
4. If you have an even number, divide by 2. Go to the number in that position and average it with the number in the next higher position to get the median.
## How do we calculate mode?
To find the mode, or modal value, it is best to put the numbers in order. Then count how many of each number. A number that appears most often is the mode.
## What is L in mode formula?
The mode of data is given by the formula: Where, l = lower limit of the modal class. h = size of the class interval. f1 = frequency of the modal class.
## What is the math mode?
more The number which appears most often in a set of numbers. Example: in {6, 3, 9, 6, 6, 5, 9, 3} the Mode is 6 (it occurs most often).
Categories: Blog
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# Probability Of Multiple Events – Conditions, Formulas, and Examples
The probability of multiple events is an interesting topic discussed in mathematics and statistics. There are instances where we’re observing multiple events and want particular results – when this happens, knowing how to calculate the probability of multiple events comes in handy.The probability of multiple events helps us measure our chances of getting the desired outcomes when two or more vents are occurring. The measured probability will heavily depend on whether the given events are independent or dependent.Seeing that this is a more complex topic than the earlier topics of probability, make sure to refresher your knowledge on the following:
• Understand how we calculate probabilities of a single event.
• Review what complementary probabilities are.
Let’s begin by understanding when we apply the particular probability we’re discussing – and we can do so by studying the spinner shown in the next section.What are multiple events in probability? The probability of multiple events occurs when we’re trying to calculate the probability of observing two or more events. These include experiments where we’re observing different behaviors simultaneously, drawing cards with multiple conditions, or predicting the outcome of a multi-colored spinner.Speaking of spinners, why don’t we observe the image shown above? From this, we can see that the spinner is divided into seven regions and distinguished by either the region’s colors or labels. Here are examples of multiple events we can check from the spinners:
• Finding the probability of spinning a violet or an $a$.
• Finding the probability of spinning a blue or a $b$.
These two conditions will require us to calculate the probability of two events occurring at the same time. Multiple events probability definitionLet’s dive right into the definition of multiple event probabilities and when they occur. The probability of multiple events measures the likelihood that two or more events occur at the same time. We sometimes lookout for the probability of when one or two outcomes happen and whether these outcomes overlap each other.The probability will depend on an important factor: whether the multiple events are independent or not and whether they are mutually exclusive.
• Dependent events (also known as conditional events) are events where a given event’s outcomes are affected by the remaining events’ outcomes.
• Independent events are events where one event’s outcomes are not affected by the rest of the events’ outcomes.
Here are some examples of events that are dependent and independent of each other.
Dependent Events Independent Events Drawing two balls consecutively from the same bag. Finding one ball each from two bags. Picking two cards without replacement. Picking a card and rolling a die. Buying more lottery tickets to win the lottery. Winning the lottery and seeing your favorite show on a streaming platform.
Events can also be mutually exclusive– these are events where they can never happen simultaneously. Some examples of mutually exclusive are the chances of turning left or right at the same time. Ace and king cards from a deck are also mutually exclusive. Knowing how to distinguish these two events will be extremely helpful when we learn how to evaluate the probabilities of two or more events that occur together. How to find the probability of multiple events? We’ll be using different approaches when finding the probability of multiple events occurring together depending on whether these events are dependent, independent, or mutually exclusive. Finding the Probability of Independent Events\begin{aligned}P(A \text{ and } B) &=P(A) \times P(B)\\P(A \text{ and } B \text{ and } C\text{ and }…) &=P(A) \times P(B) \times P(C) \times … \end{aligned}When we’re working with independent events, we can calculate the probability occurring together by multiplying the respective probabilities of the events occurring individually.Let’s say we have the following objects handy:
• A bag that contains $6$ red and $8$ blue chips.
• A coin is in your purse.
• A deck of cards is on your office table.
How do we find the probability that we get a red chip and toss the coin and get tails, and draw a card with a heart suit?These three events are independent of each other, and we can find the probability of these events occurring together by first finding the probability that they occur independently.As a refresher, we can find their independent probabilities by dividing the number of outcomes by the total number of possible outcomes.
Event Symbol Probability Getting a red chip $P(r)$ $P(r) = \dfrac{6}{14} = \dfrac{5}{7}$ Tossing the coin and get a tails $P(t)$ $P(t) = \dfrac{1}{2}$ Drawing a hearts $P(h)$ $P(h) = \dfrac{13}{52} = \dfrac{1}{4}$ \begin{aligned}P(r \text{ and }t \text{ and }h)&= P(r) \cdot P(t)\cdot P(h)\\&= \dfrac{5}{7}\cdot \dfrac{1}{2} \cdot \dfrac{1}{4}\\&= \dfrac{5}{56} \end{aligned}
Finding the Probability of Dependent Events\begin{aligned}P(A \text{ and } B) &=P(A) \times P(B \text{ given } A)\\&= P(A)\times P(B|A)\\P(A \text{ and } B \text{ and } C) &=P(A) \times P(B \text{ given } A)\times P(C \text{ given } A\text{ and }B)\\&=P(A) \times P(B|A)\times P(C|A \text{ and } B) \end{aligned}We can calculate for the probability of dependent events occurring together as shown above. Need a refresher on what $P(A|B)$ represents? It simply means the probability of $A$, once $B$ has happened. You’ll know more about conditional probability and be able to try out more complex examples here.Let’s say we want to find out the probability of getting three jacks consecutively if we don’t return the drawn card each draw. We can keep in mind that three events are occurring in this situation:
• The probability of getting a jack on the first draw – we still have $52$ cards here.
• The probability of getting a second jack on the second draw (we now have $3$ jacks and $51$ cards).
• The third event is getting a third jack for the third row – $2$ jacks left and $50$ cards on the deck.
We can label these three events as $P(J_1)$, $P(J_2)$, and $P(J_3)$. Let’s work on the important components to calculate the probability of these three dependent events happening together.
Event Symbol Probability Drawing a jack the first time $P(J_1)$ $\dfrac{4}{52}= \dfrac{1}{13}$ Drawing a jack the second time $P(J_2|J_1)$ $\dfrac{4 -1}{52 -1} = \dfrac{1}{17}$ Drawing a jack the third time $P(J_3|J_1 \text{ and } J_2)$ $\dfrac{3-1}{51 -1} = \dfrac{1}{25}$ \begin{aligned}P(J_1) \times P(J_2 \text{ given } J_1)\times P(J_3 \text{ given } J_2\text{ and }J_1)&=P(J_1) \times P(J_2|J_1)\times P(J_3|J_1 \text{ and } J_2)\\&=\dfrac{4}{52}\cdot\dfrac{3}{51}\cdot\dfrac{2}{50}\\&= \dfrac{1}{13}\cdot \dfrac{1}{17}\cdot \dfrac{1}{25}\\&= \dfrac{1}{5525} \end{aligned}
Finding the Probability of Mutually Exclusive or Inclusive EventsWe may also need to explore if the given events are mutually inclusive or exclusive to help us calculate the probability of multiple events where the outcome we’re looking for do not require all outcomes to occur altogether.Here’s a table that summarizes the formula for mutually exclusive or inclusive events:
Type of Event Formula for the Probability Mutually Inclusive $P(A \text{ or } B) = P(A) + P(B) – P(A \text{ and } B)$ Mutually Exclusive $P(A \text{ or } B) = P(A) + P(B)$
Keep in mind that we’re now using “or” because we’re looking for the probabilities of events that occur individually or occur together.These are all the concepts and formulas you’ll need to understand and solve problems that involve multiple events’ probability. We can go ahead and try out these examples shown below!Example 1A canvas bag contains $6$ pink cubes, $8$ green cubes, and $10$ purple cubes. One cube is removed from the bag and then replaced. Another cube is drawn from the bag, and repeat this one more time. What is the probability that the first cube is pink, the second cube is purple, and the third is another pink cube?SolutionKeep in mind that the cubes are returned each time we draw another. Since the next draw’s probability is not affected by the first draw results, the three events are independent of each other.When this happens, we multiply the individual probabilities to find the probability of having the outcome that we want.
Event Symbol Probability Drawing a pink cube in the first draw $P(C)$ $P(C_1) = \dfrac{6}{24}= \dfrac{1}{4}$ Drawing a purple cube in the second draw $P(C_2)$ $P(C_2) = \dfrac{10}{24}= \dfrac{5}{12}$ Drawing another pink cube in the third draw $P(C_3)$ $P(C_3) = \dfrac{6}{24}= \dfrac{1}{4}$ \begin{aligned}P(C_1 \text{ and }C_2\text{ and }C_3)&= P(C_1) \cdot P(C_2)\cdot P(C_3)\\&= \dfrac{1}{4}\cdot \dfrac{5}{12} \cdot \dfrac{1}{4}\\&= \dfrac{5}{192} \end{aligned}
This means that the probability of drawing a pink cube then a purple cube then another pink cube is equal to $\dfrac{5}{192}$.Example 2A book club of $40$ enthusiastic readers, $10$ prefers nonfiction books, and $30$ prefers fiction. Three book club members will be randomly selected to serve as the next book club meeting’s three hosts. What is the probability that all three members will prefer nonfiction?SolutionWhen the first member is selected as the first host, we can no longer include them in the next random selection. This shows that the three outcomes are dependent on each other.
• For the first selection, we have $40$ members and $30$ nonfiction readers.
• For the second selection, we now have $40 -1 = 39$ members and $30- 1= 29$ nonfiction readers.
• Hence, for the third, we have $38$ members and $28$ nonfiction readers.
Event Symbol Probability Randomly selecting a nonfiction reader $P(N_1)$ $\dfrac{30}{40}= \dfrac{3}{4}$ Selecting another nonfiction reader $P(N_2|N_1)$ $\dfrac{29}{39}$ Selecting a nonfiction reader the third time $P(N_3|N_1 \text{ and } N_2)$ $\dfrac{28}{38} = \dfrac{14}{19}$ \begin{aligned}P(N_1) \times P(N_2 \text{ given } N_1)\times P(N_3 \text{ given }N_2\text{ and }N_1)&=P(N_1) \times P(N_2|N_1)\times P(N_3|N_1 \text{ and } N_2)\\&=\dfrac{30}{40}\cdot\dfrac{29}{39}\cdot\dfrac{28}{38}\\&= \dfrac{3}{4}\cdot \dfrac{29}{39}\cdot \dfrac{14}{19}\\&= \dfrac{203}{494} \end{aligned}
Hence, the probability of selecting three nonfiction readers is equal to $\dfrac{203}{494}\approx 0.411$.Example 3Let’s go back to the spinner that was introduced to us in the first section, and we can actually determine the probabilities of the following:a. Spinning a violet or an $a$.b. Spinning a blue or a red.SolutionLet’s take note of the colors and labels found in each spinner.
Color $\rightarrow$Label $\downarrow$ Violet Green Red Blue Total $a$ $1$ $1$ $0$ $1$ $3$ $b$ $2$ $0$ $0$ $0$ $2$ $c$ $0$ $0$ $1$ $1$ $2$ Total $3$ $1$ $1$ $2$ $7$
Take note of the keyword “or” – this means that we account for the probability that either outcome occurs. For problems like this, it’s important to note whether the conditions are mutually exclusive or inclusive.For the first condition, we want the spinner to land on either a violet region or a region labeled $a$, or both.
• There are $3$ violet regions and $3$ regions labelled $a$.
• There is a $1$ region where it’s both violet and labeled $a$.
This shows that the incident is mutually inclusive. Hence, we use $P(A \text{ or } B) = P(A) + P(B) – P(A \text{ and } B)$\begin{aligned}P(V \text{ or } a) &= P(V) + P(a) – P(V \text{ and } a)\\&=\dfrac{3}{7} + \dfrac{3}{7} – \dfrac{1}{7}\\&= \dfrac{5}{7}\end{aligned}a. This means that the probability is equal to $\dfrac{5}{7}$.It’s impossible to land on a red region and a blue one all at the same time. This means that these two events are mutually exclusive. For these types of events, we add their individual probabilities.b. This means that the probability is equal to $\dfrac{1}{7} + \dfrac{2}{7} = \dfrac{3}{7}$.
### Practice Questions
1. A canvas bag contains $12$ pink cubes, $20$ green cubes, and $22$ purple cubes. One cube is removed from the bag and then replaced. Another cube is drawn from the bag, and repeat this one more time. What is the probability that the first cube is green, the second cube is purple, and the third is another green cube?
2. In a book club of $50$ enthusiastic readers, $26$ prefer nonfiction books, and $24$ prefer fiction. Three book club members will be randomly selected to serve as the three hosts of the next book club meeting. What is the probability that all three members will prefer fiction?
3. In a book club of $50$ enthusiastic readers, $26$ prefer nonfiction books, and $24$ prefer fiction. Three book club members will be randomly selected to serve as the three hosts of the next book club meeting. What is the probability that all three members will prefer nonfiction?
4. Using the same spinner from the first section, what is the probability of spinning a green or an $a$?
5. Using the same spinner from the first section, what is the probability of spinning a $b$ or a $c$?
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INTRODUCING 5 - days-a-week problem solving session for Math Olympiad and ISI Entrance. Learn More
# List of numbers | RMO 2015, Chennai Region Solutions
This is a problem from the Regional Mathematics Olympiad, RMO 2015 Chennai Region based on a List of numbers.
Problem: From the list of natural numbers 1, 2, 3, … suppose we remove all multiples of 7, all multiples of 11 and all multiples of 13.
1. At which position in the resulting list does the number 1002 appear?
2. What number occurs in the position 3600?
Discussion:
(Part 1)
To find the position of 1002, we must delete all multiples of 7, all multiples of 11 and all multiples of 13 from 1 to 1002. We apply principle of inclusion and exclusion to find that:
$1002 - { \left \lfloor \frac{1002}{7} \right \rfloor + \left \lfloor \frac{1002}{11} \right \rfloor + \left \lfloor \frac{1002}{13} \right \rfloor } + { \left \lfloor \frac{1002}{7 \times 11} \right \rfloor + \left \lfloor \frac{1002}{11 \times 13} \right \rfloor + \left \lfloor \frac{1002}{13 \times 7} \right \rfloor } - { \left \lfloor \frac{1002}{13 \times 11 \times 7} \right \rfloor }$
$= 1002 - (143+91+77) + (13+11+7) - 1 = 721$
Hence the position of 1002 is 721 in the list.
(Part 2)
To find the number in the 3600 position, we assume that it is x. Then
$x - { \left \lfloor \frac{x}{7} \right \rfloor + \left \lfloor \frac{x}{11} \right \rfloor + \left \lfloor \frac{x}{13} \right \rfloor } + { \left \lfloor \frac{x}{7 \times 11} \right \rfloor + \left \lfloor \frac{x}{11 \times 13} \right \rfloor + \left \lfloor \frac{x}{13 \times 7} \right \rfloor } - { \left \lfloor \frac{x}{13 \times 11 \times 7} \right \rfloor }$
= ( 3600 )
Note that $7\times 11\times 13 =1001$ Hence from 1 to 1001 there are exactly 720 numbers (as found in Part 1) which are not divisible by 7, 11, 13. Similarly in the next 1001 numbers (from 1002 to 2002) we will have another 720 numbers which are not divisible by 7, 11, 13. To reach 3600 we have to do this exactly 5 times as $720\times 5 = 3600$ . Hence there are $3600$ numbers from 1 to $1001 \times 5 = 5005$.
Thus the 3600th number is 5004.
## Chatuspathi:
• Paper: RMO 2015 (Chennai)
• What is this topic: Combinatorics
• What are some of the associated concepts: Principle of Inclusion and Exclusion
• Where can learn these topics: Cheenta I.S.I. & C.M.I. course,Cheenta Math Olympiad Program, discuss these topics in the ‘Geometry’ module.
• Book Suggestions: Challenges and Thrills of Pre-College Mathematics by Venkatchala, Principles of Combinatorics
### 3 comments on “List of numbers | RMO 2015, Chennai Region Solutions”
1. […] What number occurs in the position 3600? SOLUTION: here […]
2. ajay krisshan says:
Sir how can 5005 be a term in the series?
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Algebraic Expressions - Class 7 Maths Chapter 4 Question Answer
Algebraic Expressions
Algebraic Expressions - Sub Topics
Algebraic expressions are a super useful part of algebra. It is where we start using letters and symbols to solve problems. It is like a middle ground between basic math with numbers and the tricky stuff in advanced math. In this chapter, we are going to learn the basic ideas of algebraic expressions and how they help us solve all sorts of problems in our everyday lives.
• Terms related to Algebraic Concepts
• Algebra
• Various Types of Algebraic Expression
• Solved Questions on Algebraic Expressions
• Terms related to Algebraic Concepts
Certainly, there is a simpler explanation of those algebraic concepts:
Algebra
Algebra is a part of mathematics where we use symbols to represent numbers and statements. These symbols, known as literals, can stand for various values and they act like variables.
Variable: A variable is like a symbol that can stand for different numbers. We usually use letters like a, b, c, p, q, r, x, y, z, etc.
Constant: A constant is a symbol that always represents a specific number. We usually use numbers like 3, 1/7, 3?, 0.5, −2, −1½, etc.
Term: When you multiply or divide variables or constants together, you create a term. Examples of terms are 5a, 7abc, 3t, −2xy, −3a/5, a3, −21x2??/11, etc.
Coefficient: In a term, the coefficient is like the number that you get when you multiply all the parts together including the sign.
Algebraic Expression: When you add or subtract two or more terms, you make an algebraic expression.
An algebraic expression 5y2 − 4y + 17 has one constant, two variables and three terms.
Like and Unlike Terms: In an algebraic expression, terms with the same letters are like terms. Terms with different letters or different powers of letters are unlike terms. For example, 5xy and −3xy are like terms but 5xy and −3ax are unlike terms.
Example: A taxi charges \$27 per km and a fixed charge of \$45. If the taxi is hired for z km, which of the following is an algebraic expression to find the total fare?
a) 27z − 45
b) 27z + 45
c) 45z + 27
d) 45z − 27
Explanation: Charges of a taxi per km = \$27
Fixed charge = \$45
Taxi is hired for z km.
Charges for z km = \$27 × z = \$27z
Algebraic expression to find the total fare = Charges for z km + Fixed charge
= \$(27z + 45)
Various Types of Algebraic Expression
Monomial: A monomial is like an algebraic expression that has just one term. Example of monomials are 4x, −7ab, 3b5, −2p/5, 0.5q,,−11, etc.
Binomial: A binomial is an algebraic expression that has two terms. Examples of binomials are (2a + 3b), (5x − 7z), (13x2 − 8xy3), etc.
Trinomial: A trinomial is an algebraic expression that has three terms. Examples of trinomials are (3a + 2b + 5c), (4p + q − c/2), (x2 − 2y3 − z3), etc.
Quadrinomial: A quadrinomial is an algebraic expression that has four terms. Examples of quadrinomials are 11 − x2 + y2 − 2z, a − 3b + c2 + cd, etc.
Polynomial: A polynomial is a type of algebraic expression that includes variables, constants, coefficients, exponents and mathematical operations. The terms within polynomials are the different parts of the expression typically separated by either "+" or "−" signs. All polynomials are a kind of algebraic expression.
Example: Simplify the following.
7xy2 − y2 + 7x2y − 5x2 − 3y2 + 4y2x − 3y2 + x2
a) 7xy2 + 11xy2 − 4x2 − 7y2
b) 7xy2 − 11xy2 + 4x2 − 7y2
c) 11xy2 + 7xy2 − 4x2 − 7y2
d) 11xy2 − 7xy2 + 4x2 − 7y2
Answer: c) 11xy2 + 7xy2 − 4x2 − 7y2
Explanation: 7xy2 − y2 + 7x2y − 5x2 − 3y2 + 4y2x − 3y2 + x2
= 7xy2 + 4y2x + 7x2y − 5x2 + x2 − 3y2 − y2 − 3y2
= 7xy2 + 4xy2 + 7x2y − 5x2 + x2 − 3y2 − y2 − 3y2 [Arranging the like terms]
= (7 + 4)xy2 + 7x2y − (5 − 1)x2 − (3 + 1 + 3)y2
= 11xy2 + 7xy2 − 4x2 − 7y2
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Best of all, Substitution equations solver is free to use, so there's no reason not to give it a try! Linear equations are very common in every grade. They are used to show the relationship between two numbers or values. There are a few different ways to solve linear equations by graphing. You can graph the equation on a coordinate grid, plot points on a coordinate grid, or plot points on an axes grid. When graphing, always follow the order of operations. To graph an equation, start with an ordered pair (x, y). Then put points in between the coordinates that indicate how you want your equation to look. For example, if x = 2 and y = -8, then your graphed equation would look like this: (2,-8). Starting from the left and working from one point to the next will help you visualize how you want your graph to look.
A complex number can be represented on a complex plane, which is similar to a coordinate plane. The real part of the complex number is represented on the x-axis, and the imaginary part is represented on the y-axis. One way to solve for a complex number is to use the quadratic equation. This equation can be used to find the roots of any quadratic equation. In order to use this equation, you must first convert the complex number into its rectangular form. This can be done by using the following formula: z = x + yi. Once the complex number is in rectangular form, you can then use the quadratic equation to find its roots. Another way to solve for a complex number is to use De Moivre's theorem. This theorem states that if z = x + yi is a complex number, then its nth roots are given by: z1/n = x1/n(cos (2π/n) + i sin (2π/n)). This theorem can be used to find both the real and imaginary parts of a complex number. There are many other methods that can be used to solve for a complex number, but these two are some of the most commonly used.
Trigonometry solver apps are a dime a dozen these days. But which one should you choose? If you're looking for an app with comprehensive features and an easy-to-use interface, then the Trigonometry Solver app by Mobile Math is the perfect choice for you! This app not only allows you to input any trigonometric problem and get step-by-step solutions, but it also comes with a built-in calculator and reference guide. So whether you're
The best geometric sequence solver is a computer program that solves geometric sequences, such as those found in long multiplication problems. The program works by taking a list of numbers and linking them together to produce a longer list. This process is repeated until the sequence is solved. The best geometric sequence solver can work in several ways. It can use either brute force or brute force with some help from a human. It can also use sorting or other computer algorithms to determine the next number in the sequence and find the gap between it and the other numbers. Once all the numbers have been determined, they are combined into one long list, which represents the solution to the problem. There are two main types of geometric sequence solvers. One type uses brute force and tries every possible combination until one of them works. The other type uses brute force with some help from a human and tries every combination that meets certain requirements, such as being in order or not having too many digits. Many people prefer using a geometric sequence solver because it can be faster than using other strategies, such as counting or figuring out how many digits there are in each number in the problem. This makes it great for students who don’t have time to think through their problems carefully or for people who have trouble with math in general. However, some people dislike these programs because they can take longer than typical math problems
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Sequence and Series: Overview, Questions, Preparation
Sequence and Series 2021 ( Sequence and Series )
3.4K Views
Rachit Kumar SaxenaManager-Editorial
Updated on Jun 28, 2021 08:11 IST
What are the Sequence and Series?
The sequence is a collection of different objects arranged in such an orderly manner that it can be identified as the first, second, third member and so on. Sequences play an essential role in various parts of our lives.
A series can be defined as the sum of all the terms present in a sequence. But there has to be a relationship between the terms and the sequence.
Sequence
When a sequence has a finite number of terms, it is called a Finite sequence.
E.g., the sequence of ancestors in a family is an excellent example of a finite sequence.
When a sequence has an infinite number of terms, it is called an infinite sequence.
E.g., a Fibonacci series is an excellent example of an infinite sequence.
Series
A series can be defined as infinite, or finite series based on their sequence. The series is always mentioned in short form, called sigma notation and denoted by the Greek letter ‘Sigma’.
E.g., if S1, S2, S3,…., Sn are a given sequence, then the series is defined as S1+ S2+ S3+ …. + Sn.
Formulas Related to Sequence and Series
The formulas related to arithmetic progression and geometric progression have been mentioned below.
a - first term
d - the common difference
r - the common ratio
n - position of terms
I - last term
For Arithmetic progression
Sequence- a, a+d, a+2d, …., a+(n-1)d…
Common difference- d= a2-a1
General term- an = a + (n-1)d
nth term from the last term- an= I- (n-1)d
Sum of first n terms- s = (n/2)(2a + (n-1)d)
For geometric progression,
Sequence- a, ar, ar2, ar3,…., ar(n-1),…
Common difference- r= (arn-1/arn-2)
General term- an=ar(n-1)
nth term from the last term- an= (1/r)(n-1))
Sum of first n terms- S= a(1-rn)/(1-r); if r
Sn = a(rn-1)/(r-1); if r>1
Sequence and Series Weightage in Class XI
The chapter ‘Sequence and Series’, discusses the sequence and series along with Arithmetic and geometric progression. This chapter also explains the difference between arithmetic and geometric progression.
Illustrated Examples for Sequence and Series
1. The first term of a GP is 1. The sum of the third and the fifth terms is 90. Find the common ratio of GP.
Solution. Let common ratio = r.
= a3 + a5 =90
= ar2 + ar4 = 90
= r4 + r2 -90 = 0
= r4 + 10r2 -9r2- 90 =0
= r2 - 9 = 0
= r = +3 or -3
2. If the sum of three numbers in AP is 24, and the product is 440, what are the numbers?
Solution. Let the three numbers be a-d, a, and a+d
Sum= (a-d)+ a + (a+1) = 24
= 3a = 24
= a = 8
Product= (a-d) * a * (a+d)= 440
= d= 3
When d= +3, the numbers are 5, 8, and 11.
when d= =3, the numbers are 11, 8, and 5.
3. Find the sum of all numbers between 200 and 400, which are divisible by 7.
Solution. First terms, a= 203, Last term, an= 399, Difference= 7
an = a + (n-1)d
= 399= 203 + (n-1)*7
= n = 29
So, S29 = (29/2)* (203 + 399)
S29 = 8729
FAQs on Sequence and Series
Q: What is a sequence?
A: A sequence is a collection of different objects arranged in such an orderly manner that it can be identified as the first, second, third member and so on.
Q: What is a series?
A: A series can be defined as the sum of all the terms present in a sequence. But there has to be a relationship between the terms and the sequence.
Q: What are the two kinds of progressions?
A: The two kinds of progressions used in sequence and series are Arithmetic progression and Geometric progression.
Q: What is the sequence used in Arithmetic progression?
A: The sequence used in the Arithmetic progression is a, a+d, a+2d, …., a+(n-1)d…
Q: What is the sequence used in Geometric progression?
A: The sequence used in the geometric progression is a, ar, ar2, ar3,…., ar(n-1),…
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Sequence and Series Exam
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Find the following products:
Question:
Find the following products:
(i) (x + 4) (x + 7)
(ii) (x − 11) (x + 4)
(iii) (x + 7) (x − 5)
(iv) (x − 3) ( x − 2)
(v) (y2 − 4) (y2 − 3)
(vi) $\left(x+\frac{4}{3}\right)\left(x+\frac{3}{4}\right)$
(vii) (3x + 5) (3x + 11)
(viii) (2x2 − 3) (2x2 + 5)
(ix) (z2 + 2) (z2 − 3)
(x) (3x − 4y) (2x − 4y)
(xi) (3x2 − 4xy) (3x2 − 3xy)
(xii) $\left(x+\frac{1}{5}\right)(x+5)$
(xiii) $\left(z+\frac{3}{4}\right)\left(z+\frac{4}{3}\right)$
(xiv) (x2 + 4) (x2 + 9)
(xv) (y2 + 12) (y2 + 6)
(xvi) $\left(y^{2}+\frac{5}{7}\right)\left(y^{2}-\frac{14}{5}\right)$
(xvii) $\left(p^{2}+16\right)\left(p^{2}-\frac{1}{4}\right)$
Solution:
(i) Here, we will use the identity $(x+a)(x+b)=x^{2}+(a+b) x+a b$.
$(x+4)(x+7)$
$=x^{2}+(4+7) x+4 \times 7$
$=x^{2}+11 x+28$
(ii) Here, we will use the identity $(x-a)(x+b)=x^{2}+(b-a) x-a b$.
$(x-11)(x+4)$
$=x^{2}+(4-11) x-11 \times 4$
$=x^{2}-7 x-44$
(iii) Here, we will use the identity $(x+a)(x-b)=x^{2}+(a-b) x-a b$.
$(x+7)(x-5)$
$=x^{2}+(7-5) x-7 \times 5$
$=x^{2}+2 x-35$
(iv) Here, we will use the identity $(x-a)(x-b)=x^{2}-(a+b) x+a b$.
$(x-3)(x-2)$
$=x^{2}-(3+2) x+3 \times 2$
$=x^{2}-5 x+6$
(v) Here, we will use the identity $(x-a)(x-b)=x^{2}-(a+b) x+a b$.
$\left(y^{2}-4\right)\left(y^{2}-3\right)$
$=\left(y^{2}\right)^{2}-(4+3)\left(y^{2}\right)+4 \times 3$
$=y^{4}-7 y^{2}+12$
(vi) Here, we will use the identity $(x+a)(x+b)=x^{2}+(a+b) x+a b$.
$\left(x+\frac{4}{3}\right)\left(x+\frac{3}{4}\right)$
$=x^{2}+\left(\frac{4}{3}+\frac{3}{4}\right) x+\frac{4}{3} \times \frac{3}{4}$
$=x^{2}+\frac{25}{12} x+1$
(vii) Here, we will use the identity $(x+a)(x+b)=x^{2}+(a+b) x+a b$.
$(3 x+5)(3 x+11)$
$=(3 x)^{2}+(5+11)(3 x)+5 \times 11$
$=9 x^{2}+48 x+55$
(viii) Here, we will use the identity $(x-a)(x+b)=x^{2}+(b-a) x-a b$.
$\left(2 x^{2}-3\right)\left(2 x^{2}+5\right)$
$=\left(2 x^{2}\right)^{2}+(5-3)\left(2 x^{2}\right)-3 \times 5$
$=4 x^{4}+4 x^{2}-15$
(ix) Here, we will use the identity $(x+a)(x-b)=x^{2}+(a-b) x-a b$.
$\left(z^{2}+2\right)\left(z^{2}-3\right)$
$=\left(z^{2}\right)^{2}+(2-3)\left(z^{2}\right)-2 \times 3$
$=z^{4}-z^{2}-6$
(x) Here, we will use the identity $(x-a)(x-b)=x^{2}-(a+b) x+a b$.
$(3 x-4 y)(2 x-4 y)$
$=(4 y-3 x)(4 y-2 x) \quad$ (Taking common $-1$ from both parentheses)
$=(4 y)^{2}-(3 x+2 x)(4 y)+3 x \times 2 x$
$=16 y^{2}-(12 x y+8 x y)+6 x^{2}$
$=16 y^{2}-20 x y+6 x^{2}$
(xi) Here, we will use the identity $(x-a)(x-b)=x^{2}-(a+b) x+a b$.
$\left(3 x^{2}-4 x y\right)\left(3 x^{2}-3 x y\right)$
$=\left(3 x^{2}\right)^{2}-(4 x y+3 x y)\left(3 x^{2}\right)+4 x y \times 3 x y$
$=9 x^{4}-\left(12 x^{3} y+9 x^{3} y\right)+12 x^{2} y^{2}$
$=9 x^{4}-21 x^{3} y+12 x^{2} y^{2}$
(xii) Here, we will use the identity $(x+a)(x+b)=x^{2}+(a+b) x+a b$.
$\left(x+\frac{1}{5}\right)(x+5)$
$=x^{2}+\left(\frac{1}{5}+5\right) x+\frac{1}{5} \times 5$
$=x^{2}+\frac{26}{5} x+1$
(xiii) Here, we will use the identity $(x+a)(x+b)=x^{2}+(a+b) x+a b$.
$\left(z+\frac{3}{4}\right)\left(z+\frac{4}{3}\right)$
$=z^{2}+\left(\frac{3}{4}+\frac{4}{3}\right) x+\frac{3}{4} \times \frac{4}{3}$
$=z^{2}+\frac{25}{12} z+1$
(xiv) Here, we will use the identity $(x+a)(x+b)=x^{2}+(a+b) x+a b$.
$\left(x^{2}+4\right)\left(x^{2}+9\right)$
$=\left(x^{2}\right)^{2}+(4+9)\left(x^{2}\right)+4 \times 9$
$=x^{4}+13 x^{2}+36$
(xv) Here, we will use the identity $(x+a)(x+b)=x^{2}+(a+b) x+a b$.
$\left(y^{2}+12\right)\left(y^{2}+6\right)$
$=\left(y^{2}\right)^{2}+(12+6)\left(y^{2}\right)+12 \times 6$
$=y^{4}+18 y^{2}+72$
(xvi) Here, we will use the identity $(x+a)(x-b)=x^{2}+(a-b) x-a b$.
$\left(y^{2}+\frac{5}{7}\right)\left(y^{2}-\frac{14}{5}\right)$
$=\left(y^{2}\right)^{2}+\left(\frac{5}{7}-\frac{14}{5}\right)\left(y^{2}\right)-\frac{5}{7} \times \frac{14}{5}$
$=y^{4}-\frac{73}{35} y^{2}-2$
(xvii) Here, we will use the identity $(x+a)(x-b)=x^{2}+(a-b) x-a b$.
$\left(p^{2}+16\right)\left(p^{2}-\frac{1}{4}\right)$
$=\left(p^{2}\right)^{2}+\left(16-\frac{1}{4}\right)\left(p^{2}\right)-16 \times \frac{1}{4}$
$=p^{4}+\frac{63}{4} p^{2}-4$
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Hough Line transform is a technique used to detect straight lines in an image. This algorithm can detect all imperfect instances of a line in a given image ie, this algorithm can detect lines even if they are not perfectly straight. The detection of line is carried out by a voting process. Before examining the algorithm in detail we need to demystify the mathematics behind it.
### The Mathematics
There are several form of equations in which a line can be represented. The most common and familiar one should be the “slope-intercept” form.
where
is the slope
is the y-intercept.
In this equation the slope and the y-intercept are known for a given line. and are variables. So by varying the value of (or ) you can find the corresponding value of (or )
From the above equation we can derive another form of equation called the “Double Intercept” form.
……………..
where
is the x-intercept.
is the y-intercept.
The double intercept form will be used to derive a new form of equation called the “Normal Form” which is used in the Hough Line Transform. The reason for using normal form is that the “slope-intercept” from fails in case of vertical lines and the Double intercept form fails because of the large range of and .
Derivation:
Consider a line which intersects the y-axis at point and x-axis at point . A line segment having one end point on the origin intersects the line at right angles at point .
makes an angle with the a-axis. The length of is .From the figure, and
Consider .
……………..
Now consider .
.
because .
Therefore
……………..
Substituting the values of in with we get
……………..
### Straight Line Detection:
Lets see how the above equation can be used to find straight lines.
Given the values of and , we can vary the value of and find the corresponding value of . Plotting will give us a straight line.
For example, for degrees and units the graph would look like this.
If we keep the value and constant and vary the value of we can find the corresponding values of . For each pair we can plot a line in x-y coordinate and all these lines will pass through .
For example, let be. Let . After substituting these values in the we get . When we plot the lines formed by the pairs, they will all pass through
Now consider three points . Let .
The corresponding values of are as follows.
15 3.03 15 5.86 15 7.27
30 4.86 30 6.33 30 7.06
45 6.36 45 6.36 45 6.36
If a line passes through a given point , then the corresponding gets a vote from that point. The line with that gets maximum number of votes passes through maximum number of points.
In the above example, got three votes, that means it passes through all three point. If gets number of votes then it passes through number of points.
For simplicity we considered only 3 angles and 3 points. The number of lines that can pass through a point is infinite. If we plot all the possible values of and for the point the graph would look like this.
Pay Attention: We are not plotting the lines in this graph. We are directly plotting the values of and . is on x-axis and is given in radian not degrees. is on y-axis. This graph represents the values of lines that pass through the point
Similarly we can plot graph for the points , and . We observe that all the 3 curves intersect at a point. The point of intersection is .
<p >But for practical application we can consider a small subset of angles; let’s say . Here we are considering only 180 values for and we get 180 values for . So effectively we are finding 180 lines that pass through a point.
### Algorithm
Preprocessing: Canny edge detector is applied to extract the edges.
Step 1: Calculate the maximum and minimum value of and . For practical implementation and are integral values. Select appropriate value for threshold .
Step 2: Initialize a 2-D array A of size x and fill it with zeros.
Step 3: For each white pixel find value of corresponding to each value of . Increment .
Step 5: Search for values of which is above the threshold;
Step 4: Draw the lines.
### Code
I have used OpenCV library for this program. It was compiled using GCC 4.5 under Linux OS (Linux Mint 13)
```#include <iostream>
#include "cv.h"
#include "highgui.h"
#include <math.h>
#define PI 3.14159265f
using namespace cv;
int main()
{
int max_r; //Maximum magnitude of r
int max_theta=179; //Maximum value of theta.
int threshold=60; //Mimimum number of votes required by(theata,r) to form a straight line.
int img_width; //Width of the image
int img_height; //Height of the image
int num_theta=180; //Number of values of theta (0 - 179 degrees)
int num_r; //Number of values of r.
int *accumulator; //accumulator to collect votes.
long acc_size; //size of the accumulator in bytes.
float Sin[num_theta]; //Array to store pre-calculated vales of sin(theta)
float Cos[num_theta]; //Array to store pre-calculated vales of cos(theta)
uchar *img_data; //pointer to image data for efficient access.
int i,j;
Mat src;
Mat dst;
namedWindow( "Polygon", 1 );
imshow( "Polygon", image);
//convert color to gray scale image
cvtColor(image, src, CV_BGR2GRAY);
/*Initializations*/
img_width = src.cols;
img_height = src.rows;
//calculating maximum value of r. Round it off to nearest integer.
max_r = round(sqrt( (img_width*img_width) + (img_height*img_height)));
//calculating the number vales r can take. -max_r <= r<= max_r
num_r = (max_r *2) +1;
//pre-compute the values of sin(theta) and cos(theta).
for(i=0;i<=max_theta;i++)
{
Sin[i] = sin(i * (PI/180));
Cos[i] = cos(i * (PI/180));
}
//Initializing the accumulator. Conceptually it is a 2-D matrix with dimension r x theta
accumulator = new int[num_theta * num_r];
//calculating size of accumulator in bytes.
acc_size = sizeof(int)*num_theta * num_r;
//Initializing elements of accumulator to zero.
memset(accumulator,0,acc_size);
//extracting the edges.
Canny(src,dst,50,200,3);
//Getting the image data from Mat dst
img_data = dst.data;
//Loop through all the pixels. Each pixel is represented by 1 byte.
for(i=0;i<img_height;i++)
{
for(j=0;j<img_width;j++)
{
//Getting the pixel value.
int val =img_data[i*img_width+j];
if(val>0)
{
//if pixel is not black do the following.
//For that pixel find the the values of r for corresponding value of theta.
//Value of r can be negative. (See the graph)
//Minimum value of r is -max_r.
//Conceptually the array looks like this
// 0 1 2 3 4 5 6 .. 178 179 <---degrees
// -max_r | | | | | | | | | | |
// -max_r+1| | | | | | | | | | |
// -max_r+2| | | | | | | | | | |
// ... | | | | | | | | | | |
// 0 | | | | | | | | | | |
// 1 | | | | | | | | | | |
// 2 | | | | | | | | | | |
// ... | | | | | | | | | | |
// max_r | | | | | | | | | | |
//
for(int t=0;t<=max_theta;t++)
{
//calculating the values of r for theta= t , x= j and y = i;
int _r = round(j*Cos[t] + i*Sin[t]);
//calculating the row index of _r in the accumulator.
int r_index = (max_r+_r);
//Registering the vote by incrementing the value of accumulator[r][theta]
accumulator[r_index*num_theta + t]++;
}
}
}
}
//Looping through each element in the accumulator
for(int r_index=0;r_index<num_r;r_index++)
{
for(j=0;j<num_theta;j++)
{
{
//getting the value of theta
int _theta=j;
//getting the value of r
int _r = r_index-max_r;
//Calculating points to draw the line.
Point pt1, pt2;
pt1.x =0;
pt1.y =round((_r - pt1.x*Cos[_theta])/Sin[_theta]);
pt2.x =img_width;
pt2.y =round((_r - pt2.x*Cos[_theta])/Sin[_theta]);
//Drawing the line.
line( image, pt1, pt2, Scalar(0,255,0), 3, CV_AA);
}
}
}
namedWindow("Detected Lines", 1 );
imshow( "Detected Lines",image);
//Free the memory allocated to accumulator.
delete[] accumulator;
waitKey(0);
return 0;
}```
Output:
## One thought on “Hough Line Transform”
1. Hi! I could have sworn I’ve been to this blog before but after checking through some of the post I realized it’s new to me.
Nonetheless, I’m definitely delighted I found it
and I’ll be bookmarking and checking back frequently!
|
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# 7.1: Solving Trigonometric Equations with Identities
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In the last chapter, we solved basic trigonometric equations. In this section, we explore the techniques needed to solve more complicated trig equations. Building from what we already know makes this a much easier task.
Consider the function $$f(x)=2x^{2} +x$$. If you were asked to solve $$f(x)=0$$, it requires simple algebra:
$2x^{2} +x=0\nonumber$Factor
$x(2x+1)=0\nonumber$Giving solutions
$x = 0\text{ or }x = -\dfrac{1}{2}\nonumber$
Similarly, for $$g(t)=\sin (t)$$, if we asked you to solve $$g(t)=0$$, you can solve this using unit circle values:
$\sin (t)=0\text{ for }t=0, \pi , 2\pi\text{ and so on.}\nonumber$
Using these same concepts, we consider the composition of these two functions:
$f(g(t))=2(\sin (t))^{2} +(\sin (t))=2\sin ^{2} (t)+\sin (t)\nonumber$
This creates an equation that is a polynomial trig function. With these types of functions, we use algebraic techniques like factoring and the quadratic formula, along with trigonometric identities and techniques, to solve equations.
As a reminder, here are some of the essential trigonometric identities that we have learned so far:
DEFINITIONS: IDENTITIES
Pythagorean Identities
$\cos ^{2} (t)+\sin ^{2} (t)=1\quad 1+\cot ^{2} (t)=\csc ^{2} (t)\quad 1+\tan ^{2} (t)=\sec ^{2} (t)$
Negative Angle Identities
$\sin (-t)=-\sin (t)\quad \cos (-t)=\cos (t)\quad \tan (-t)=-\tan (t)$
$\csc (-t)=-\csc (t)\quad \sec (-t)=\sec (t)\quad \cot (-t)=-\cot (t)$
Reciprocal Identities
$\sec (t)=\dfrac{1}{\cos (t)}\quad \csc (t)=\dfrac{1}{\sin (t)}\quad \tan (t)=\dfrac{\sin (t)}{\cos (t)}\quad \cot (t)=\dfrac{1}{\tan (t)}$
EXAMPLE $$\PageIndex{1}$$
Solve $$2\sin ^{2} (t)+\sin (t)=0$$ for all solutions with $$0\le t<2\pi$$.
Solution
This equation kind of looks like a quadratic equation, but with sin(t) in place of an algebraic variable (we often call such an equation “quadratic in sine”). As with all quadratic equations, we can use factoring techniques or the quadratic formula. This expression factors nicely, so we proceed by factoring out the common factor of sin($$t$$):
$\sin (t)\left(2\sin (t)+1\right)=0\nonumber$
Using the zero product theorem, we know that the product on the left will equal zero if either factor is zero, allowing us to break this equation into two cases:
$\sin (t)=0\text{ or }2\sin (t)+1=0\nonumber$
We can solve each of these equations independently, using our knowledge of special angles.
$\sin (t)=0\nonumber$
$2\sin (t)+1=0\nonumber$
$t = 0\text{ or }t = \pi\nonumber$
$\sin (t)=-\dfrac{1}{2}\nonumber$
$t=\dfrac{7\pi }{6}\text{ or }t=\dfrac{11\pi }{6}\nonumber$
Together, this gives us four solutions to the equation on $$0\le t<2\pi$$:
$t=0,\pi ,\dfrac{7\pi }{6} ,\dfrac{11\pi }{6}\nonumber$
We could check these answers are reasonable by graphing the function and comparing the zeros.
EXAMPLE $$\PageIndex{2}$$
Solve $$3\sec ^{2} (t)-5\sec (t)-2=0$$ for all solutions with $$0\le t<2\pi$$.
Solution
Since the left side of this equation is quadratic in secant, we can try to factor it, and hope it factors nicely.
If it is easier to for you to consider factoring without the trig function present, consider using a substitution $$u=\sec (t)$$, resulting in $$3u^{2} -5u-2=0$$, and then try to factor:
$3u^{2} -5u-2=(3u+1)(u-2)\nonumber$
Undoing the substitution,
$(3\sec (t)+1)(\sec (t)-2)=0\nonumber$
Since we have a product equal to zero, we break it into the two cases and solve each separately.
$3\sec (t)+1=0\nonumber$Isolate the secant
$\sec (t)=-\dfrac{1}{3}\nonumber$Rewrite as a cosine
$\dfrac{1}{\cos (t)} =-\dfrac{1}{3}\nonumber$Invert both sides
$\cos (t)=-3\nonumber$
Since the cosine has a range of [-1, 1], the cosine will never take on an output of -3. There are no solutions to this case.
Continuing with the second case,
$\sec (t)-2=0\nonumber$Isolate the secant
$\sec (t)=2\nonumber$Rewrite as a cosine
$\dfrac{1}{\cos (t)} =2\nonumber$Invert both sides
$\cos (t)=\dfrac{1}{2}\nonumber$This gives two solutions
$t=\dfrac{\pi }{3}\text{ or }t=\dfrac{5\pi }{3}\nonumber$
These are the only two solutions on the interval.
By utilizing technology to graph $$f(t)=3\sec ^{2} (t)-5\sec (t)-2$$, a look at a graph confirms there are only two zeros for this function on the interval [0, 2 $$\pi$$), which assures us that we didn’t miss anything.
Exercise $$\PageIndex{1}$$
Solve $$2\sin ^{2} (t)+3\sin (t)+1=0$$ for all solutions with $$0\le t<2\pi$$.
Factor as $\left(2\sin (t)+1\right)\left(\sin (t)+1\right)=0\nonumber$
$2\sin (t)+1=0\text{ at }t=\dfrac{7\pi }{6} ,\dfrac{11\pi }{6}\nonumber$
$\sin (t)+1=0\text{ at }t=\dfrac{3\pi }{2}\nonumber$
$t=\dfrac{7\pi }{6} ,\dfrac{3\pi }{2} ,\dfrac{11\pi }{6}\nonumber$
When solving some trigonometric equations, it becomes necessary to first rewrite the equation using trigonometric identities. One of the most common is the Pythagorean Identity, $$\sin ^{2} (\theta )+\cos ^{2} (\theta )=1$$ which allows you to rewrite $$\sin ^{2} (\theta )$$ in terms of $$\cos ^{2} (\theta )$$ or vice versa,
IDENTITIES
Alternate Forms of the Pythagorean Identity
$\begin{array}{l} {\sin ^{2} (\theta )=1-\cos ^{2} (\theta )} \\ {\cos ^{2} (\theta )=1-\sin ^{2} (\theta )} \end{array}$
These identities become very useful whenever an equation involves a combination of sine and cosine functions.
EXAMPLE $$\PageIndex{3}$$
Solve $$2\sin ^{2} (t)-\cos (t)=1$$ for all solutions with $$0\le t<2\pi$$.
Solution
Since this equation has a mix of sine and cosine functions, it becomes more complicated to solve. It is usually easier to work with an equation involving only one trig function. This is where we can use the Pythagorean Identity.
$2\sin ^{2} (t)-\cos (t)=1\nonumber$Using $$\sin ^{2} (\theta )=1-\cos ^{2} (\theta )$$
$2\left(1-\cos ^{2} (t)\right)-\cos (t)=1\nonumber$Distributing the 2
$2-2\cos ^{2} (t)-\cos (t)=1\nonumber$
Since this is now quadratic in cosine, we rearrange the equation so one side is zero and factor.
$-2\cos ^{2} (t)-\cos (t)+1=0\nonumber$Multiply by -1 to simplify the factoring
$2\cos ^{2} (t)+\cos (t)-1=0\nonumber$Factor
$\left(2\cos (t)-1\right)\left(\cos (t)+1\right)=0\nonumber$
This product will be zero if either factor is zero, so we can break this into two separate cases and solve each independently.
$2\cos (t)-1=0\text{ or }\cos (t)+1=0\nonumber$
$\cos (t)=\dfrac{1}{2}\text{ or }\cos (t)=-1\nonumber$
$t=\dfrac{\pi }{3}\text{ or }t=\dfrac{5\pi }{3}\text{ or }t=\pi\nonumber$
Exercise $$\PageIndex{3}$$
Solve $$2\sin ^{2} (t)=3\cos (t)$$ for all solutions with $$0\le t<2\pi$$.
$2\left(1-\cos ^{2} (t)\right)=3\cos (t)\nonumber$
$2\cos ^{2} (t)+3\cos (t)-2=0\nonumber$
$\left(2\cos (t)-1\right)\left(\cos (t)+2\right)=0\nonumber$
$$\cos (t)+2=0$$ has no solutions $$2\cos (t)-1=0$$ at $$t=\dfrac{\pi }{3} ,\dfrac{5\pi }{3}$$
In addition to the Pythagorean Identity, it is often necessary to rewrite the tangent, secant, cosecant, and cotangent as part of solving an equation.
EXAMPLE $$\PageIndex{4}$$
Solve $$\tan (x)=3\sin (x)$$ for all solutions with $$0\le x<2\pi$$.
Solution
With a combination of tangent and sine, we might try rewriting tangent
$\tan (x)=3\sin (x)\nonumber$
$\dfrac{\sin (x)}{\cos (x)} =3\sin (x)\nonumber$Multiplying both sides by cosine
$\sin (x)=3\sin (x)\cos (x) \nonumber$
At this point, you may be tempted to divide both sides of the equation by sin($$x$$).
Resist the urge. When we divide both sides of an equation by a quantity, we are assuming the quantity is never zero. In this case, when sin($$x$$) = 0 the equation is satisfied, so we’d lose those solutions if we divided by the sine.
To avoid this problem, we can rearrange the equation so that one side is zero (You technically can divide by sin(x), as long as you separately consider the case where sin(x) = 0. Since it is easy to forget this step, the factoring approach used in the example is recommended.).
$\sin (x)-3\sin (x)\cos (x)=0\nonumber$Factoring out sin($$x$$) from both parts
$\sin (x)\left(1-3\cos (x)\right)=0 \nonumber$
From here, we can see we get solutions when $$\sin (x)=0$$ or $$1-3\cos (x)=0$$.
Using our knowledge of the special angles of the unit circle,
$\sin (x)=0\text{ when }x = 0\text{ or }x = \pi\nonumber$
For the second equation, we will need the inverse cosine.
$1-3\cos (x)=0\nonumber$
$\cos (x)=\dfrac{1}{3}\nonumber$Using our calculator or technology
$x=\cos ^{-1} \left(\dfrac{1}{3} \right)\approx 1.231\nonumber$Using symmetry to find a second solution
$x=2\pi -1.231=5.052 \nonumber$
We have four solutions on $$0 \le x<2\pi$$:
$x = 0, 1.231, \quad\pi , 5.052\nonumber$
EXAMPLE $$\PageIndex{3}$$
Solve $$\sec (\theta )=2\cos (\theta )$$ to find the first four positive solutions.
$\dfrac{1}{\cos (\theta )} =2\cos (\theta )\nonumber$
$\dfrac{1}{2} =\cos ^{2} (\theta )\nonumber$
$\cos (\theta )=\pm \sqrt{\dfrac{1}{2} } =\pm \dfrac{\sqrt{2} }{2}\nonumber$
$\theta =\dfrac{\pi }{4} ,\dfrac{3\pi }{4} ,\dfrac{5\pi }{4} ,\dfrac{7\pi }{4}\nonumber$
EXAMPLE $$\PageIndex{5}$$
Solve $$\dfrac{4}{\sec ^{2} (\theta )} +3\cos \left(\theta \right)=2\cot \left(\theta \right)\tan \left(\theta \right)$$ for all solutions with $$0\le \theta <2\pi$$.
Solution
$\dfrac{4}{\sec ^{2} (\theta )} +3\cos \left(\theta \right)=2\cot \left(\theta \right)\tan \left(\theta \right)\nonumber$ Using the reciprocal identities
$4\cos ^{2} (\theta )+3\cos (\theta )=2\dfrac{1}{\tan (\theta )} \tan (\theta )\nonumber$ Simplifying
$4\cos ^{2} \left(\theta \right)+3\cos \left(\theta \right)=2\nonumber$ Subtracting 2 from each side
$4\cos ^{2} \left(\theta \right)+3\cos \left(\theta \right)-2=0\nonumber$
This does not appear to factor nicely so we use the quadratic formula, remembering that we are solving for cos( $$\theta$$).
$\cos (\theta )=\dfrac{-3\pm \sqrt{3^{2} -4(4)(-2)} }{2(4)} =\dfrac{-3\pm \sqrt{41} }{8}\nonumber$
Using the negative square root first,
$\cos (\theta )=\dfrac{-3-\sqrt{41} }{8} =-1.175\nonumber$
This has no solutions, since the cosine can’t be less than -1.
Using the positive square root,
$\cos (\theta )=\dfrac{-3+\sqrt{41} }{8} =0.425\nonumber$
$\theta =\cos ^{-1} \left(0.425\right)=1.131\nonumber$ By symmetry, a second solution can be found
$\theta =2\pi -1.131=5.152\nonumber$
## Important Topics of This Section
• Review of Trig Identities
• Solving Trig Equations
• By Factoring
• Using the Quadratic Formula
• Utilizing Trig Identities to simplify
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# Chapter 2 Describing Distributions with Numbers. Numerical Summaries u Center of the data –mean –median u Variation –range –quartiles (interquartile range)
## Presentation on theme: "Chapter 2 Describing Distributions with Numbers. Numerical Summaries u Center of the data –mean –median u Variation –range –quartiles (interquartile range)"— Presentation transcript:
Chapter 2 Describing Distributions with Numbers
Numerical Summaries u Center of the data –mean –median u Variation –range –quartiles (interquartile range) –variance –standard deviation
Mean or Average u Traditional measure of center u Sum the values and divide by the number of values
Median (M) u A resistant measure of the data’s center u At least half of the ordered values are less than or equal to the median value u At least half of the ordered values are greater than or equal to the median value u If n is odd, the median is the middle ordered value u If n is even, the median is the average of the two middle ordered values
Median (M) Location of the median: L(M) = (n+1)/2, where n = sample size. Example: If 25 data values are recorded, the Median would be the (25+1)/2 = 13 th ordered value.
Median u Example 1 data: 2 4 6 Median (M) = 4 u Example 2 data: 2 4 6 8 Median = 5 (ave. of 4 and 6) u Example 3 data: 6 2 4 Median 2 (order the values: 2 4 6, so Median = 4)
Measure of center: the median The median is the midpoint of a distribution—the number such that half of the observations are smaller and half are larger. 1. Sort observations from smallest to largest. n = number of observations ______________________________ n = 24 n/2 = 12 Median = (3.3+3.4) /2 = 3.35 3. If n is even, the median is the mean of the two center observations n = 25 (n+1)/2 = 26/2 = 13 Median = 3.4 2. If n is odd, the median is observation (n+1)/2 down the list
Comparing the Mean & Median u The mean and median of data from a symmetric distribution should be close together. The actual (true) mean and median of a symmetric distribution are exactly the same. u In a skewed distribution, the mean is farther out in the long tail than is the median [the mean is ‘pulled’ in the direction of the possible outlier(s)].
Disease X: Mean and median are the same. Mean and median of a symmetric distribution Multiple myeloma: and a right-skewed distribution The mean is pulled toward the skew. Impact of skewed data
The median, on the other hand, is only slightly pulled to the right by the outliers (from 3.4 to 3.6). The mean is pulled to the right a lot by the outliers (from 3.4 to 4.2). Percent of people dying Mean and median of a distribution with outliers Without the outliers With the outliers
Question A recent newspaper article in California said that the median price of single-family homes sold in the past year in the local area was \$136,000 and the mean price was \$149,160. Which do you think is more useful to someone considering the purchase of a home, the median or the mean?
Case Study Airline fares appeared in the New York Times on November 5, 1995 “...about 60% of airline passengers ‘pay less than the average fare’ for their specific flight.” u How can this be? 13% of passengers pay more than 1.5 times the average fare for their flight
Spread, or Variability u If all values are the same, then they all equal the mean. There is no variability. u Variability exists when some values are different from (above or below) the mean. u We will discuss the following measures of spread: range, quartiles, variance, and standard deviation
Range u One way to measure spread is to give the smallest (minimum) and largest (maximum) values in the data set; Range = max min u The range is strongly affected by outliers
Quartiles u Three numbers which divide the ordered data into four equal sized groups. u Q 1 has 25% of the data below it. u Q 2 has 50% of the data below it. (Median) u Q 3 has 75% of the data below it.
Quartiles Uniform Distribution Q1Q1 Q2Q2 Q3Q3
Obtaining the Quartiles u Order the data. u For Q 2, just find the median. u For Q 1, look at the lower half of the data values, those to the left of the median location; find the median of this lower half. u For Q 3, look at the upper half of the data values, those to the right of the median location; find the median of this upper half.
M = median = 3.4 Q 1 = first quartile = 2.2 Q 3 = third quartile = 4.35 Measure of spread : quartiles The first quartile, Q 1, is the value in the sample that has 25% of the data at or below it. The third quartile, Q 3, is the value in the sample that has 75% of the data at or below it.
Boxplot u Central box spans Q 1 and Q 3. u A line in the box marks the median M. u Lines extend from the box out to the minimum and maximum.
M = median = 3.4 Q 3 = third quartile = 4.35 Q 1 = first quartile = 2.2 Largest = max = 6.1 Smallest = min = 0.6 “Five-number summary” Center and spread in boxplots
Example from Text: Boxplots
Variance and Standard Deviation u Recall that variability exists when some values are different from (above or below) the mean. u Each data value has an associated deviation from the mean:
Deviations u what is a typical deviation from the mean? (standard deviation) u small values of this typical deviation indicate small variability in the data u large values of this typical deviation indicate large variability in the data
Variance u Find the mean u Find the deviation of each value from the mean u Square the deviations u Sum the squared deviations u Divide the sum by n-1 (gives typical squared deviation from mean)
Variance Formula
Standard Deviation Formula typical deviation from the mean [ standard deviation = square root of the variance ]
Variance and Standard Deviation Example from Text Metabolic rates of 7 men (cal./24hr.) : 1792 1666 1362 1614 1460 1867 1439
Variance and Standard Deviation Example from Text ObservationsDeviationsSquared deviations 1792 1792 1600 = 192 (192) 2 = 36,864 1666 1666 1600 = 66 (66) 2 = 4,356 1362 1362 1600 = -238 (-238) 2 = 56,644 1614 1614 1600 = 14 (14) 2 = 196 1460 1460 1600 = -140 (-140) 2 = 19,600 1867 1867 1600 = 267 (267) 2 = 71,289 1439 1439 1600 = -161 (-161) 2 = 25,921 sum = 0sum = 214,870
Variance and Standard Deviation Example from Text
Notes on Standard Deviation u s measures spread about the mean, and should be used only if the mean is our measure of center. u In most cases s>0. s=0 where ________________. Cannot be less than 0. u As the spread of the data increases, s increases. u s has the same units as the original observations. u s is sensitive to skewness and outliers like the mean. u Sum of all deviations must be 0.
Choosing a Summary u Outliers affect the values of the mean and standard deviation. u The five-number summary should be used to describe center and spread for skewed distributions, or when outliers are present. u Use the mean and standard deviation for reasonably symmetric distributions that are free of outliers.
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Arbitrary distribution using biased coin
Problem. Let’s assume that you have a biased coin, but you don’t know the probabilities of its sides. Using the coin, how would you choose one of the numbers 1, 2 or 3 with equal probability?
Stop reading here and try solving the problem yourself before moving on.
Solution. First of all, I have to note that this problem have a trivial solution that works only for this particular question, not in a general case. One such solution was proposed by a reader in the comments section. In this post, we present a solution that will work for arbitrary distributions, no matter how asymmetric they are.
The problem becomes easier if we split it into two parts. Firstly we will learn how to emulate a fair coin using a biased one. After this we describe a procedure which allows us to get numbers from an arbitrary probabilistic distribution.
Let’s denote the probability of tails as $$p$$. Then if we toss the coin twice, we can get one of the four outcomes:
1. Tails both times, with probability $$p^2$$,
2. Heads both times, with probability $$(1-p)^2$$,
3. Tails first time, heads second time, probability $$p(1-p)$$.
4. Heads first time, tails second time, probability $$p(1-p)$$.
As you see, outcomes 3 and 4 have equal chances to appear. Thus, in order to emulate an experiment with probability 50/50 (fair coin), we can toss the coin twice. If both tosses give the same outcome, then discard it and toss again. When we get tails in one toss and heads in the other, we choose the outcome which comes first.
Example. Consider a sequence of tosses T, T, H, H, T, H, where I denoted tails as T and heads as H. This sequence describes an experiment in which we toss the coin twice and both times we get tails. We discard this result and toss twice again. This time we get two heads. Now, we toss twice one more time. First, we get tails and then heads. We write “tails” as our outcome.
Since now we can assume that we have a fair coin. We will be repeating tosses sequentially and interpreting them as digits in a binary representation of a number $$s$$ in interval $$[0, 1]$$:
$$s=\sum\limits_{i=1}^{+\infty} \frac{t_i}{2^i}$$
Here $$t_i$$ denotes the outcome of i-th toss and can be either 0 or 1. In order to get one of the three values at random, we split $$[0,1]$$ into three subintervals $$[0, 1/3), [1/3, 2/3], (2/3, 1]$$. (Since particular points like 1/3 has zero probability, we don’t care into which interval we include them). The outcome of the experiment is the number of the subinterval in which $$s$$ lies. Since we assume that each digit of $$s$$ is uniform, $$s$$ itself is uniform in $$[0,1]$$.
In fact we don’t need the exact value of $$s$$. The only interesting bit is the subinterval in which $$s$$ lies, so we don’t have to make an infinite number of tosses. Let’s say that the first toss gave us value 1. We know that $$s=0.1\ldots$$ (in binary) and it lies somewhere in $$[1/3, 2/3]$$. If the next toss gives us 0, then $$s=0.10\ldots$$ and $$s\in[1/3, 0.59]$$. If the next toss gives us 0 again, then $$s=0.100\ldots$$ and we know that $$s\in[1/3, 0.46)$$ which is a subinterval of $$[1/3, 2/3]$$. The outcome is the second value.
This scheme can be generalized in an obvious way. If we want to get an arbitrary distribution we just have to split $$[0,1]$$ into a given number of subintervals of lengths that match the probabilities of the values.
2 thoughts on “Arbitrary distribution using biased coin”
1. Sbezhik says:
If we want to do this procedure in reality it is more convenient not to split the problem into parts and use the following strategy:
THH or HTT — 1
HTH ot THT — 2
HHT or TTH –3
TTT or HHH — discard it.
So we use only 3 tosses instead of 4+ in your strategy and have no need to discard p^2+(1-p)^2>50% of tosses.
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CBSE Class 9 Maths Revision Notes Chapter 14
Class 9 Mathematics Revision Notes for Statistics of Chapter 14
Mathematics Revision Notes for Class 9 Chapter 14 Statistics of Extramarks are curated by subject matter experts according to the NCERT curriculum. Students can refer to these notes for a conceptual understanding of the topics explained in Class 9 Mathematics Chapter 14.
Class 9 Mathematics Revision Notes for Statistics of Chapter 14
Statistics
1. Introduction
• The branch of mathematics concerned with the collection, classification, tabulation, representation, reasoning, testing of data and drawing inferences from it is referred to as statistics.
• Data is often expressed in the form of graphs, tables, etc. Statistical methods are used, like estimation or prediction.
• Statistics are used to organise and process numerical data in a systematic manner.
• The interrelation of statistics with biology, psychology, economics, trade etc. can be used to interpret and analyse the data from different subjects.
• Measures of central tendencies are expressions that give some information about numerous numerical data.
• An average represents the middle value out of a group of values, which has extremities on both sides.
1. Numerical Data and Its Representation
The scores of 64 students in a class test out of 100 are as follows :
Table 1
58 38 52 47 16 50 61 37 44 55 38 49 44 52 67 51 33 48 23 51 56 61 46 41 65 43 71 29 50 56 68 25 55 49 44 73 23 63 41 42 66 59 52 28 50 56 60 38 40 73 45 30 47 40
This data is presented in an organised manner to present the scores of students in the class. This is the tabular form of presenting data; many other methods can be used.
1. Arranged Data
Arranged Numerical Data
The data from Table 1 is arranged in ascending order.
Table 2
14 16 23 23 25 28 29 30 33 37 37 38 38 38 40 40 40 40 40 41 41 42 43 44 44 44 45 46 46 47 47 48 49 49 50 50 50 51 51 52 52 52 55 55 56 56 56 58 58 59 60 61 61 62 63 65 66 67 68 68 71 72 73 73
Information gathered from the table:
The minimum marks are 14 and the maximum marks are 73. There is a repetition of scores i.e 40 marks, occurring 5 times, making it the maximum repetition.
Drawbacks:
• This method is very difficult to perform.
• Large amounts of data cannot be arranged and put in an ascending order.
• No crucial information will be gained from this statistical method.
1. Ungrouped Frequency Distribution Table
The data from table 1 is used to create an ungrouped frequency distribution table. Numbers are written from smallest to largest and repeated numbers are marked using tally marks. Tally marks correspond to the frequency of the number occurring in the data.
Table 3
Tally Marks Frequency Marks 14 I 1 15 16 I 1 17 18 19 20 21 22 23 II 2 24 25 I 1 26 27 28 I 1
Tally Marks Frequency Marks 29 I 1 30 I 1 31 32 33 I 1 34 35 36 37 II 2 38 III 3 39 IIII 5 40 IIII 5 41 II 2 42 I 1 43 I 1
Tally Marks Frequency Marks 44 III 3 45 I 1 46 II 2 47 II 2 48 I 1 49 II 2 50 III 3 51 II 2 52 III 3 53 54 55 II 2 56 III 3 57 58 I 1
Tally Marks Frequency Marks 59 I 1 60 I 1 61 II 2 62 I 1 63 I 1 64 65 I 1 66 I 1 67 I 1 68 II 2 69 70 71 I 1 72 II 2 73 II 2
Conclusions can be drawn more easily from such a table, as mere observation can produce statistical expression. It is obvious from the table that there are a lot of students who received grades between 44 and 58.
Drawback
• A grouped frequency table can be used for a smaller span to represent data.
1. Grouped Frequency Distribution Table
Distribution of data into organised groups and classes using tally marks is done in a grouped frequency distribution table.
Classes taken are from 11 to 20, 21 – 30…. 71 – 80.
Table 4
Class Tally Marks Frequency 11-20 II 2 21-30 IIII I 6 31-40 IIII IIII I 11 41-50 IIII IIII IIII III 18 51-60 IIII IIII IIII 14 61-70 IIII IIII 9 71-80 IIII 4 Total 64
1. Some Terms Used in Statistics
2. Raw Numerical Data
The primary information that has been collected is called raw numerical data.
1. Range of the Data
The difference between the largest and smallest value in the data is called its range.
Example: The range of data in table 2 is 73-14 = 59.
1. Class Limit
The smallest and largest possible data values for each class are represented by class limits. The smallest possible value in a class is its lower limit and the largest possible value in a class is its upper limit.
1. Class Interval
The range of the class is its class interval.
1. Frequency of the Class
The frequency of a class interval is the number of observations that occur in the interval. Thus, it can be represented as tally marks or as a count.
1. Cumulative Frequency Table
The cumulative frequency is the frequency of observations less than a certain class interval’s upper limit.
Table 5
Class Frequency Cumulative Frequency (Less than the upper class limit) 11-20 2 2 21-30 6 2+6=8 31-40 11 2+6+11=19 41-50 18 2+6+11+18=37 51-60 14 2+6+11+18+14=51 61-70 9 2+6+11+18+14+9=60 71-80 4 2+6+11+18+14+9+4=64 Total= 64
In this table, the column of cumulative frequency shows the number of scores less than the upper class limit of the particular class. Thus, such a table is called ‘a cumulative frequency less than’ table.
Along similar lines, the cumulative frequency more than the lower limit of a class is equal to the sum of the frequency of that particular class and the frequencies of all the classes succeeding to it. Table 6 given below shows the cumulative frequency of this type.
Table 6
Class Frequency Cumulative Frequency (More than the lower class limit) 11-20 2 62+2=64 21-30 6 56+6=62 31-40 11 45+11=56 41-50 18 27+18=45 51-60 14 13+14=27 61-70 9 4+9=13 71-80 4 4 Total= 64
The cumulative frequency column in this table displays the number of scores that are higher than the respective class’s lower limit. Consequently, a table of this type is known as ‘a cumulative frequency more than’ table.
To create a table like this, make a table with the classes and matching frequencies. From the table’s bottom to its top, jot down the cumulative frequencies. The cumulative frequency for the last class, which is 71 to 80, is 4, making that class’s frequency 4. The frequency of the class before is 9, and it ranges from 61 to 70.
Therefore, 4+9=13 represents the class 71 to 80 cumulative frequency.
1. Representation of Statistical Data
The two forms of numerical data are :
1. Diagrammatic representation
2. Graphical representation
3. Diagrammatic Representation
Diagrammatic representation can be done by using :
1. Pie Diagrams
2. Bar Diagrams
1. Graphical representation
Data can be represented in the form of :
1. Histogram
2. Frequency Polygon
3. Ogive Curve
4. Graphical Representation of Statistical Data
5. Histogram
• A bar diagram demonstrating a continuous frequency distribution, in graphical form, is called a histogram.
• This method involves construction from frequency data, indicating a vertical demonstration of frequencies and classes on the horizontal scale.
• Individual intervals are represented by bars.
• Construction of a histogram is done by using grouped frequency distribution tables. Class limits are shown on the x-axis, frequencies on the y-axis and rectangles are joined to the class limit and heights are proportional to the frequencies.
1. Frequency Polygon
• Data in a frequency polygon is represented by plotting the class mark on the horizontal axis and the frequency of the class on the vertical axis.
• The two points are then connected and completed by class marks, one-class width on either end and a frequency of zero on both ends.
• The construction of histograms is crucial for creating a frequency polygon, by joining the middle points of the upper horizontal sides of the rectangles in the histogram.
• Assumptions are taken such that the frequency of each of the classes before the first class and the last class is taken, and is considered zero.
Frequency polygons can be made without using histograms.
1. Recap
Statistical data can be represented in the following ways :
1. Histogram
On the x-axis, class intervals are plotted and cumulative frequencies are on the y-axis. The corresponding rectangles are drawn representing the data.
1. Frequency polygon
After drawing the histogram, connect the midpoints of the rectangles of the histogram with straight line segments. This gives us the frequency polygon.
1. Arithmetic Mean (AM)
2. Arithmetic Mean for Ungrouped Data
From the given raw data, the value obtained by summing up all the values of a given variable divided by the total number of values is called Arithmetic Mean.
Let ‘n’ be the total number of values and x1, x2, … xn be the recorded values of the variable then the arithmetic mean is given as follows:
Arithmetic Mean (AM) = x = x1+x2+x3+…+xnn
Or x = xin
The symbol denotes that the values of the given variable are summed over all the given values of x.
1. Direct Method for Arithmetic Mean of Ungrouped Data
Arithmetic Mean (AM) = x = fixin
Example: Find the AM of the following data:
Marks Frequency (fi) fixi 7 3 21 19 4 76 31 5 155 40 7 280 49 9 441 62 7 434 73 6 438 83 5 415 91 4 364 n=50 i=1i=50fix1= 2624
Here, AM = x = fixin = 262450 = 52.48
1. Direct Method for Arithmetic Mean of Grouped Data
Find the arithmetic mean for the given frequency distribution:
Marks Frequency (fi) Mid Point (xi) fixi 5-15 3 10 30 15-25 4 20 80 25-35 5 30 150 35-45 7 40 280 45-55 9 50 450 55-65 7 60 420 65-75 6 70 420 75-85 5 80 400 85-95 4 90 360 n=xi=50 fixi = 2590
Arithmetic Mean=x=fixin = 259050=51.80
1. For Calculating the Mean Assumed Mean or Short-Cut Method
2. Short-Cut Method for Ungrouped Data
Here, a value that is roughly in the middle is taken and considered as Assumed Mean (A). On having two middle values, the one with a higher frequency is taken. Then the Arithmetic Mean is given as
Arithmetic Mean = x = A + fididi
where A is the assumed mean and d is the deviation of x from the assumed mean A.
1. Shortcut Method for Grouped Data
Here too, similar to the method for ungrouped data, the assumed mean is taken from the given data from the mid values of the table and then the arithmetic mean is obtained as follows:
Arithmetic Mean (AM) = x = A + fidifi
where A is the assumed mean and d is the deviation of x from the assumed mean A.
1. Step-Deviation Method
In this method,
Arithmetic Mean (AM) = x = A + fiuifi h
where A is the assumed mean, h is the class size(upper limit – lower limit), u is x – Ah
1. Median and Mode
Median
On arranging the given statistical data in ascending or descending order of their numerical values, the number in the middle is termed the median.
Let n be the number of values,
Then Median = (n2 + 1)th term if n is odd
And Median = (n2)th term + (n2 + 1)th term 2 if n is even
Mode
The value which appears the maximum number of times among the given statistical data has been termed as ‘mode’. Thus, it is the value with the highest frequency.
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+0
# .çömnbvc
0
373
3
Arya is 3 years older than Cindi, and Kim is double Cindi's age. Lenny is 5 years older than Kim. Lenny is 17 years old. How old is Arya?
Guest May 24, 2017
#3
+90055
+1
Arya is 3 years older than Cindi, and Kim is double Cindi's age. Lenny is 5 years older than Kim. Lenny is 17 years old. How old is Arya?
Let Cindi's age = C
Kim's age = 2C
Lenny's age = 2C + 5 = 17
Solving the last equation we have
2C + 5 = 17 subtract 5 from both sides
2C = 12 divide both sides by 2
C = 6 = Cindi's age.....and Arya is three years older = 9
CPhill May 24, 2017
#1
0
Arya is 9 years old.
Guest May 24, 2017
#2
+80
0
If Lenny is 5 years older than Kim, and Lenny is 17, then Kim is 12, Cindi is 6, anf Arya is 9.
mathguardian23 May 24, 2017
#3
+90055
+1
Arya is 3 years older than Cindi, and Kim is double Cindi's age. Lenny is 5 years older than Kim. Lenny is 17 years old. How old is Arya?
Let Cindi's age = C
Kim's age = 2C
Lenny's age = 2C + 5 = 17
Solving the last equation we have
2C + 5 = 17 subtract 5 from both sides
2C = 12 divide both sides by 2
C = 6 = Cindi's age.....and Arya is three years older = 9
CPhill May 24, 2017
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Paul's Online Notes
Home / Calculus I / Integrals / Computing Indefinite Integrals
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### Section 5.2 : Computing Indefinite Integrals
12. Evaluate $$\displaystyle \int{{\frac{{{z^8} - 6{z^5} + 4{z^3} - 2}}{{{z^4}}}\,dz}}$$.
Show All Steps Hide All Steps
Hint : Remember that there is no “Quotient Rule” for integrals and so we’ll need to eliminate the quotient before integrating.
Start Solution
Since there is no “Quotient Rule” for integrals we’ll need to break up the integrand and simplify a little prior to integration.
$\int{{\frac{{{z^8} - 6{z^5} + 4{z^3} - 2}}{{{z^4}}}\,dz}} = \int{{\frac{{{z^8}}}{{{z^4}}} - \frac{{6{z^5}}}{{{z^4}}} + \frac{{4{z^3}}}{{{z^4}}} - \frac{2}{{{z^4}}}\,dz}} = \int{{{z^4} - 6z + \frac{4}{z} - 2{z^{ - 4}}\,dz}}$ Show Step 2
At this point there really isn’t too much to do other than to evaluate the integral.
$\int{{\frac{{{z^8} - 6{z^5} + 4{z^3} - 2}}{{{z^4}}}\,dz}} = \int{{{z^4} - 6z + \frac{4}{z} - 2{z^{ - 4}}\,dz}} = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{1}{5}{z^5} - 3{z^2} + 4\ln \left| z \right| + \frac{2}{3}{z^{ - 3}} + c}}$
Don’t forget to add on the “+c” since we know that we are asking what function did we differentiate to get the integrand and the derivative of a constant is zero and so we do need to add that onto the answer.
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1. ## Power series question
We just started power series in class and I'm somewhat confused about them. I'm stuck on this particular problem. It says to make a power series for f(x) = x/(2x^2+1). What I did was to base it on geometric series so I convered the equation to look like 1/(1+x) so now it looks like x/[1-(-2x^2)]. What I got for the power series was x multiplied by the sum of (-2x^2)^n from n=0 to infinity. Is this right?
2. You're close, except you forgot to add a coefficient, rational functions have this relation:
$\displaystyle \frac{x}{2x^2 + 1} = \sum_{n=0}^{\infty} a_nx^n$
It's that simple, now you have to find the coefficients:
Multiply the denominator through:
$\displaystyle x = (2x^2 + 1)\sum_{n=0}^{\infty} a_nx^n$
Now we multiply the sum through:
$\displaystyle x = 2\sum_{n=0}^{\infty}a_nx^{n+2} + \sum_{n=0}^{\infty}a_nx^n$
$\displaystyle x = 2\sum_{n=2}^{\infty}a_{n-2}x^n + \sum_{n=0}^{\infty}a_nx^n$
Now we combine like terms:
$\displaystyle x = a_0 + a_1x + \sum_{n=2}^{\infty}(2a_{n-2} + a_n)x^n$
Since there are no coefficients on the left side, then:
$\displaystyle a_0 = 0$
Since the coefficient of the x term on the left side is 1, then the coefficient of the x term on the right side is also 1:
$\displaystyle a_1 = 1$
This gives you:
$\displaystyle x = x + \sum_{n=2}^{\infty}(2a_{n-2} + a_n)x^n$
Which gives you an interesting result:
$\displaystyle a_n = -2a_{n-2}$
Now you have your power series:
$\displaystyle f(x) = \sum_{n=2}^{\infty}-2a_{n-2}x^n$
This gives you:
$\displaystyle f(x) = \sum_{n=0}^{\infty}-2a_nx^{n+2}$
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# Mean Absolute Deviation
JoVE Core
Statistik
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JoVE Core Statistik
Mean Absolute Deviation
##### Vorheriges Video4.10: Chebyshev’s Theorem to Interpret Standard Deviation
The mean absolute deviation provides the absolute value of the average difference between the data values and the mean.
It is calculated as the sum of the absolute deviations from the mean divided by the sample size.
For example, three students have three, five, and seven cookies in their lunch boxes. The deviations in the number of cookies from the mean of five cookies are minus 2, zero, and two.
If one adds these deviations, the positive and negative values cancel each other out, giving a zero mean deviation, which is unhelpful. If the absolute values are added, one obtains a single non-zero value instead.
This value, when divided by the sample size, gives the mean absolute deviation.
Calculating the mean absolute deviation involves a non-algebraic modulus operation, while the standard deviation uses algebraic operations. Therefore it is not suitable for inferential statistics.
It is also a biased statistic, as the calculated mean absolute deviation of a sample does not adequately represent the population mean absolute deviation.
## Mean Absolute Deviation
The mean absolute deviation is also a measure of the variability of data in a sample. It is the absolute value of the average difference between the data values and the mean.
Let us consider a dataset containing the number of unsold cupcakes in five shops: 10, 15, 8, 7, and 10. Initially, calculate the sample mean. Then calculate the deviation, or the difference, between each data value and the mean. Next, the absolute values of these deviations are added and divided by the sample size to obtain the mean absolute deviation.
In the above data set, the obtained mean is 10. The deviations from the mean are 0, 5,-2,-3, and 0. The absolute values of these deviations are 0,5,2,3, and 0. Upon adding these, we get a sum of 10. Upon dividing ten by the sample size, we get a value of 5, which is the mean absolute deviation.
It is noteworthy that the mean absolute deviation is computed using absolute values, so it involves using a non-algebraic operation. Hence, the mean absolute deviation cannot be used in inferential statistics, which involves using algebraic operations.
Furthermore, the mean absolute deviation of a sample is biased, as it doesn’t adequately represent the mean absolute deviation of a population.
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# Chapter 5: Probability: What are the Chances?
## Presentation on theme: "Chapter 5: Probability: What are the Chances?"— Presentation transcript:
Chapter 5: Probability: What are the Chances?
Section 5.3 Conditional Probability and Independence The Practice of Statistics, 4th edition – For AP* STARNES, YATES, MOORE
Chapter 5 Probability: What Are the Chances?
5.1 Randomness, Probability, and Simulation 5.2 Probability Rules 5.3 Conditional Probability and Independence
Section 5.3 Conditional Probability and Independence
Learning Objectives After this section, you should be able to… DEFINE conditional probability COMPUTE conditional probabilities DESCRIBE chance behavior with a tree diagram DEFINE independent events DETERMINE whether two events are independent APPLY the general multiplication rule to solve probability questions
Conditional Probability and Independence
What is Conditional Probability? The probability we assign to an event can change if we know that some other event has occurred. This idea is the key to many applications of probability. When we are trying to find the probability that one event will happen under the condition that some other event is already known to have occurred, we are trying to determine a conditional probability. Conditional Probability and Independence Definition: The probability that one event happens given that another event is already known to have happened is called a conditional probability. Suppose we know that event A has happened. Then the probability that event B happens given that event A has happened is denoted by P(B | A). Read | as “given that” or “under the condition that”
Conditional Probability and Independence
Example: Grade Distributions Consider the two-way table on page Define events E: the grade comes from an EPS course, and L: the grade is lower than a B. Conditional Probability and Independence Total Total 6300 1600 2100 Find P(L) Find P(E | L) Find P(L | E) P(L) = 3656 / = P(E | L) = 800 / 3656 = P(L| E) = 800 / 1600 =
Conditional Probability and Independence
When knowledge that one event has happened does not change the likelihood that another event will happen, we say the two events are independent. Conditional Probability and Independence Definition: Two events A and B are independent if the occurrence of one event has no effect on the chance that the other event will happen. In other words, events A and B are independent if P(A | B) = P(A) and P(B | A) = P(B). Are the events “male” and “left-handed” independent? Justify your answer. Example: P(left-handed | male) = 3/23 = 0.13 P(left-handed) = 7/50 = 0.14 These probabilities are not equal, therefore the events “male” and “left-handed” are not independent.
Conditional Probability and Independence
Tree Diagrams We learned how to describe the sample space S of a chance process in Section Another way to model chance behavior that involves a sequence of outcomes is to construct a tree diagram. Conditional Probability and Independence Consider flipping a coin twice. What is the probability of getting two heads? Sample Space: HH HT TH TT So, P(two heads) = P(HH) = 1/4
Conditional Probability and Independence
General Multiplication Rule The idea of multiplying along the branches in a tree diagram leads to a general method for finding the probability P(A ∩ B) that two events happen together. Conditional Probability and Independence The probability that events A and B both occur can be found using the general multiplication rule P(A ∩ B) = P(A) • P(B | A) where P(B | A) is the conditional probability that event B occurs given that event A has already occurred. General Multiplication Rule
Conditional Probability and Independence
Example: Teens with Online Profiles The Pew Internet and American Life Project finds that 93% of teenagers (ages 12 to 17) use the Internet, and that 55% of online teens have posted a profile on a social-networking site. What percent of teens are online and have posted a profile? Conditional Probability and Independence 51.15% of teens are online and have posted a profile.
Example: Who Visits YouTube?
See the example on page 320 regarding adult Internet users. What percent of all adult Internet users visit video-sharing sites? P(video yes ∩ 18 to 29) = • 0.7 =0.1890 P(video yes ∩ 30 to 49) = 0.45 • 0.51 =0.2295 P(video yes ∩ 50 +) = 0.28 • 0.26 =0.0728 P(video yes) = =
Conditional Probability and Independence
Independence: A Special Multiplication Rule When events A and B are independent, we can simplify the general multiplication rule since P(B| A) = P(B). Conditional Probability and Independence Definition: Multiplication rule for independent events If A and B are independent events, then the probability that A and B both occur is P(A ∩ B) = P(A) • P(B) Following the Space Shuttle Challenger disaster, it was determined that the failure of O-ring joints in the shuttle’s booster rockets was to blame. Under cold conditions, it was estimated that the probability that an individual O-ring joint would function properly was Assuming O-ring joints succeed or fail independently, what is the probability all six would function properly? Example: P(joint1 OK and joint 2 OK and joint 3 OK and joint 4 OK and joint 5 OK and joint 6 OK) =P(joint 1 OK) • P(joint 2 OK) • … • P(joint 6 OK) =(0.977)(0.977)(0.977)(0.977)(0.977)(0.977) = 0.87
Conditional Probability and Independence
Calculating Conditional Probabilities If we rearrange the terms in the general multiplication rule, we can get a formula for the conditional probability P(B | A). Conditional Probability and Independence General Multiplication Rule P(A ∩ B) = P(A) • P(B | A) P(A ∩ B) P(A) P(B | A) To find the conditional probability P(B | A), use the formula Conditional Probability Formula =
Conditional Probability and Independence
Example: Who Reads the Newspaper? In Section 5.2, we noted that residents of a large apartment complex can be classified based on the events A: reads USA Today and B: reads the New York Times. The Venn Diagram below describes the residents. What is the probability that a randomly selected resident who reads USA Today also reads the New York Times? Conditional Probability and Independence There is a 12.5% chance that a randomly selected resident who reads USA Today also reads the New York Times.
Section 5.3 Conditional Probability and Independence
Summary In this section, we learned that… If one event has happened, the chance that another event will happen is a conditional probability. P(B|A) represents the probability that event B occurs given that event A has occurred. Events A and B are independent if the chance that event B occurs is not affected by whether event A occurs. If two events are mutually exclusive (disjoint), they cannot be independent. When chance behavior involves a sequence of outcomes, a tree diagram can be used to describe the sample space. The general multiplication rule states that the probability of events A and B occurring together is P(A ∩ B)=P(A) • P(B|A) In the special case of independent events, P(A ∩ B)=P(A) • P(B) The conditional probability formula states P(B|A) = P(A ∩ B) / P(A)
Looking Ahead… In the next Chapter…
We’ll learn how to describe chance processes using the concept of a random variable. We’ll learn about Discrete and Continuous Random Variables Transforming and Combining Random Variables Binomial and Geometric Random Variables In the next Chapter…
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# Chapter 11 Sequences and Series. 11.1 Arithmetic Sequences.
## Presentation on theme: "Chapter 11 Sequences and Series. 11.1 Arithmetic Sequences."— Presentation transcript:
Chapter 11 Sequences and Series
11.1 Arithmetic Sequences
Key Terms Sequence: an ordered list of numbers 1, 3, 5, 7, 9, …. Terms: the numbers in the sequence The 1 st term in the sequence above is 1, 2 nd is 3, 3 rd is 5, etc Arithmetic Sequence: a sequence in which each terms after the 1 st is found by adding a constant Common difference: the number added to each term in an arithmetic sequence
Arithmetic sequence Formula: a n = a 1 + (n – 1)d a n = specific term (nth term) a 1 = 1 st term in the sequence n = number of position for given term d = common difference
Find the next four terms in the sequence. -6, -2, 2,…. 1.6, 1.1, 0.6, ….
Write an equation for the nth term 10, 7, 4, 1, …
Write an equation for the nth term 8, 17, 26, 35, …
The table shows typical costs for a construction company to rent a crane for one, two, three, and four months. If the sequence continues, how much would it cost to rent the crane for a year? The construction company has a budget of \$350,000 for crane rental. The job is expected to last 18 months. Will the company be able to afford the crane rental for the entire job? Explain. MonthsCost (\$) 175,000 290,000 3105,000 4120,000
Find the first 5 terms for the sequence described a 1 = 2 d = 13 Find the indicated term for the sequence a 1 = 20, d = 4, n = 81
Arithmetic Means The terms between any two non- successive terms in an arithmetic sequence Use the formula to find the common difference Then use the common difference to calculate the arithmetic means
Find the arithmetic means between the given terms 16,__, __, __, __, 91, … 15.6, __, __, __, 60.4,...
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# Proof
Prove if $x,y,$ and $b$ are positive real values with $b \neq 1$, then
$$\displaystyle{\log_b \frac{x}{y} = \log_b x - \log_b y}$$
First note that we can evaluate $\log_b x$ and $\log_b y$, since $x,y,$ and $b$ are positive real values with $b \neq 1$. Let us denote these two values by $m$ and $n$, so that
$$\log_b x = m \quad \textrm{and} \quad \log_b y = n$$
If we express the above in exponential form, we have
$$b^m = x \quad \textrm{and} \quad b^n = y$$
With the above, the rest of the proof follows quickly, as shown below. Notice how the proof hinges on the fact that $\displaystyle{\frac{b^m}{b^n} = b^{m-n}}$, one of our laws of exponents.
$$\begin{array}{rcl} \log_b \frac{x}{y} &=& \log_b \frac{b^m}{b^n}\\ &=& \log_b b^{m-n}\\ &=& m - n\\ &=& \log_b x - \log_b y \end{array}$$
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# 3- Quick start to False position method for roots finding.
### False position method.
The false position method is another numerical method for root finding, The same Solved problem, will be used to get the root for f(x), but this time using another method that is called false position, or regula -falsi, which can be done by substituting the formula shown here.
Xr is the horizontal distance to the root point, where x1 and x2 are the distance from the point(0.0) to the first left bracket point and right bracket point, respectively.
While b1 and b2, represent the value of the function at the left bracket point and the value of the function at the right bracket point.
### A Solved problem for the false position method.
We have used previously the function for which f(x)=x^3 -6x^2 +11x-6. Find the zeros of the function by the False position method considering a0 as equal to 2.50 and b= 4. as before. xr numerator is (x right*yleft-x left*y right), while the denominator =(yleft- y right).
The steps are as follows:
1-The solution we have before a0 as =2.50 will give us an f(a0) =-0.375, and we have b. =4 that is giving f(b)= f(4)=+6.0.
2- If we assume that this is a sketch of the graph. If we assume that this is a sketch of the graph. The graph intersects the x-axis at a certain point, and now we would like to know what will be the x1 value and, accordingly, the function f(x1).
3- We apply in the equation of xr=((b0)*f(a0)- a0*f(b0))/(f(a0)-f(b0) The b0=4.0. and a0=2.50.
4-The function of f(b0) is 6, and the function of (a0)= f(a0)=-0.375 hen xr=((4-*0.375)-(2.50*6)/(-0.375-6) =2.588.
5- Our next step is trying to find what is the function, value at x1=2.588. So we plug in the function. by putting f(x)= f(2.588).We substitute the result as -0.3847.
This point is considered a new left bracket point.
6-We can make a left bracket here, and we have the bracket for the positive value again, the function of x at x=4 or b=4; it
is a right bracket point.
We join this point with the other point that has a positive value. of +6.
Our false position again moves from a=2.50 to x =2.588. which is very close to the required x value that gives zero.
7- We apply in the equation of xr=((b0)*f(a0)- a0*f(b0))/(f(a0)-f(b0) The b0=4.0. and a0=2.588. f(a0)=-0.36801, b0=4, f(b0)=+6.
Our new value of xr=(4*(-0.38469)-(2.588)*(6))/(-0.38469-6)=2.673.
8- We will substitute in the function; we get f(2.673), which=-0.36801, and it will give (-)minus, which means it is the new left bracket. We can check f(2.673)f(4) is with a negative sign, that is, (-0.384696=-2.2085.
9- We apply in the equation of xr=((b0)*f(a0)- a0*f(b0))/(f(a0)-f(b0) The b0=4.0. and a0=2.673. f(a0)=-0.368019,b0=4, f(b0)=+6. Our new value of xr=(4*(-0.368019)-(2.588)*(6))/(-0.36801-6)=2.7499.
10-We will substitute in the function; we get f(2.749), which=-0.328, will give (-)minus, which means it is the new left bracket. We can check f(2.749)*f(4) is with a negative sign, that is, (-0.328*6)=-1.9688.
11- We apply in the equation of xr=((b0)*f(a0)- a0*f(b0))/(f(a0)-f(b0) The b0=4.0. and a0=2.7499. f(a0)=-0.328, b0=4, f(b0)=+6.
Our new value of xr=(4*(-0.328)-(2.7499)*(6))/(-0.328-6)=2.8147.- We apply in the equation of xr=((b0)*f(a0)- a0*f(b0))/(f(a0)-f(b0) The b0=4.0. and a0=2.673. f(a0)=-0.368019,b0=4, f(b0)=+6. Our new value of xr=(4*(-0.368019)-(2.588)*(6))/(-0.36801-6)=2.7499.
We will substitute in the function; we get f(2.8147), which=-0.2741, it will give (-)minus, which means it is the new left bracket. We can check f(2.8147)*f(4) is with a negative sign, that is, (-0.2741*6=-1.643.
We have reached x5, as we can see in the next slides, x5=2.866, with a -ve value, and again it is the new left bracket, coming closer to b=4. The details of the calculation are shown in the next image.
We plug in x=2.866 as a0. While f(2.866)=f(a0)=-0.216, we can get a new point of x=2.905. THIS POINT is a left bracket point.
### The table for the number of iterations.
This is the table for 20iterations at x20, the value =3.00. f(x=3)=0, the calculations are performed using an Excel sheet as shown in the next slide image.
The next post will be Fixed-point iteration and how to use it.
A very intersection source is Holistic Numerical methods.
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How to multiply and divide common fractions.
This topic is part of the TCS FREE high school mathematics 'How-to Library'. It shows you how to multiply and divide common fractions. (See the index page for a list of all available topics in the library.) To make best use of this topic, you need to download the Algematics. Click here for instructions.
### Theory:
When multiplying two common fractions, the procedure is to multiply the numerators and denominators separately and then simplify the resulting fraction.
eg:
To divide a number by a common fraction, you multiply by the reciprocal of the fraction.
eg:
Algematics will multiply or divide two common fractions and then simplify the answer to its lowest terms.
### Method:
The steps below demonstrate how to multiply or divide common fractions showing working.
NOTE: To multiply or divide mixed numbers, first convert to common fractions.
As an example, we will multiply three quarters by six ninths, eg:
#### Step 1 Set display options
Click to display the ‘Set Colours And Fonts’ Dialog Box.
Make sure that this check box option is not selected:
o Use ¸ signs, eg: 2¸a
and then click .
#### Step 2 Enter the left hand fraction
Click and type your expression into the maths box in the data entry box.
If the ‘EMPTY’ message is not displayed between the blue buttons, click the button until the message: ‘INSERT’ appears.
Use a divide symbol: ‘ / ’ to enter the fraction.
Do not use the ratio ‘ : ’ symbol, or some of the working steps will not be shown.
Maths...
3/4
and then click
#### Step 3 Enter the right hand fraction
Click on the input box and type in the right hand fraction.
Once again, be careful to use a ‘ / ’ in the fraction, like this: 6/9
#### Step 4 Multiply or divide the two fractions
Click to multiply the two fractions:
OR click to divide the two fractions: (This multiplies by the reciprocal.)
#### Step 5 Multiply the numerator and denominator factors
To multiply the numerator and denominator factors separately, click and then so that both buttons are down.
Now click the (simplify) button twice so that both the numerator and denominator are multiplied:
After completing these multiplying steps, click and again so that both buttons are now up.
Click the (simplify) button once more to reduce the answer to its lowest terms:
Go back to step 2 to multiply or divide more common fractions.
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# Chapter 14 Symmetry Class 7 Notes Maths
Chapter 14 Symmetry Class 7 Notes Maths is available here which will make entire memorizing process effortless and entertaining. NCERT Notes becomes a vital resource for all the students to self-study from NCERT textbooks carefully. Revision Notes for Class 7 will allows the students to evaluate their learning immediately. A student will enjoy the revising process and make themselves capable of retaining more information so they can excel in the exams. You will find NCERT Solutions for Class 7 Chapter 14 Maths that can make things a little easier for you.
Line Symmetry
• A figure has a line symmetry, if there is a line about which the figure may be folded so that the two parts of the figure will coincide.
Lines of symmetry for regular polygons
• A polygon is said to be regular if all its sides are of equal length and all its angles are of equal measure.Â
• An equilateral triangle is regular because each of its sides has same length and each of its angles measures 60°.
• A square is also regular because all its sides are of equal length and each of its angles is a right angle (i.e., 90°). Its diagonals are seen to be perpendicular bisectors of one another.
• If a pentagon is regular, naturally, its sides should have equal length. The measure of each of its angles is 108°.
• A regular hexagon has all its sides equal and each of its angles measures 120°.
Rotational symmetry
• Rotation turns an object about a fixed point. This fixed point is the centre of rotation. The angle by which the object rotates is the angle of rotation.
• A half-turn means rotation by 180°; a quarter-turn means rotation by 90°. Rotation may be clockwise or anticlockwise.
• If, after a rotation, an object looks exactly the same, we say that it has a rotational symmetry.
• In a complete turn (of 360°), the number of times an object looks exactly the same is called the order of rotational symmetry. The order of symmetry of a square, for example, is 4 while, for an equilateral triangle, it is 3.
Line Symmetry and Rotational Symmetry
• Some shapes have only one line of symmetry, like the letter E; some have only rotational symmetry, like the letter S; and some have both symmetries like the letter H.
Line Symmetry and Mirror Reflection
• A shape has line symmetry when one half of it is the mirror image of the other half.
• Mirror reflection leads to symmetry, under which the left-right orientations have to be taken care of.
X
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# How do you differentiate y = -(1/x^2) - sqrtx + (1/2)?
Jun 13, 2018
$y ' = \frac{2}{x} ^ 3 - \frac{1}{2 \cdot \sqrt{x}}$
#### Explanation:
Writing
$y = - {x}^{- 2} - {x}^{\frac{1}{2}} + \frac{1}{2}$
After the rule
$\left({x}^{n}\right) ' = n {x}^{n - 1}$
we get
$y ' = 2 {x}^{- 3} - \frac{1}{2} {x}^{- \frac{1}{2}}$
$y ' = \frac{2}{x} ^ 3 - \frac{1}{2 \sqrt{x}}$
Jun 13, 2018
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2}{x} ^ 3 - \frac{1}{2 \sqrt{x}}$
#### Explanation:
$\text{differentiate using the "color(blue)"power rule}$
•color(white)(x)d/dx(ax^n)=nax^(n-1)
$y = - {x}^{-} 2 - {x}^{\frac{1}{2}} + \frac{1}{2}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 {x}^{-} 3 - \frac{1}{2} {x}^{- \frac{1}{2}} + 0$
$\textcolor{w h i t e}{\frac{\mathrm{dy}}{\mathrm{dx}}} = \frac{2}{x} ^ 3 - \frac{1}{2 \sqrt{x}}$
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## 502 – Formal systems (3)
September 4, 2009
Theorem 2 (Completeness) Whenever ${\Sigma\models\tau,}$ then also ${\Sigma\vdash\tau.}$
## 175 – Quiz 1
September 4, 2009
Here is quiz 1.
Problem 1 is exercise 6.1.16 from the book. (The graph in the book has a typo. The maximum height of the graph is $1/2$ rather than 1.)
Using the disk method, the volume can be expressed as $\int_0^{\pi/2}\pi y^2\,dx.$ Here, $y^2=\sin^2 x\cdot \cos^2 x.$ If this expression were not squared, to find the integral would be easier, by a direct substitution. Being squared, we need to work harder. One possible approach is to use some trigonometric identities. For example:
• $\sin x\cdot \cos x=\displaystyle\frac{\sin 2x}2,$ so $y^2=\displaystyle\frac{\sin^2 2x}4.$ Again, the fact that the sine expression is squared makes things difficult. We use another trigonometric identity:
• $\sin^2\theta=\displaystyle\frac{1-\cos2\theta}2,$ so $y^2=\displaystyle\frac{1-\cos 4x}8.$ Now we can proceed to integrate:
$\displaystyle\int_0^{\pi/2}\pi y^2\,dx=\displaystyle\frac{\pi}8\int_0^{\pi/2} (1-\cos 4x)\,dx=\displaystyle\frac\pi8\left.\bigl(x+\frac{\sin 4x}4\bigr)\right|_0^{\pi/2},$ which simplifies to $\displaystyle\frac{\pi^2}{16}.$
There are other approaches. For example, other trigonometric identities could be used as well. Also, the book includes a formula for the integral of powers of sine and cosine; we will study this formula later. Trying to use the shell method leads to rather messy expressions, I didn’t work out the full details. I believe that the argument above is perhaps the most efficient.
Problem 2 is exercise 6.3.14 from the book. Since $x$ is given as a function of $y,$ the most efficient route seems to be to use $y$ as the parameter, so the expression for the length of the curve takes the form $\int_2^3\sqrt{(x')^2+1}\,dy,$ where the derivative of $x$ is with respect to $y.$
We have $x'=\displaystyle\frac{y^2}2-\frac1{2y^2},$ so $(x')^2+1=\displaystyle \frac{y^4}4-\frac12+\frac1{4y^4}+1=\displaystyle\frac{y^4}4+\frac12+\frac1{4y^4}=\displaystyle\bigl(\frac{y^2}2+\frac1{2y^2}\bigr)^2.$
The expression for the length then reduces to $\displaystyle\int_2^3\bigl(\frac{y^2}2+\frac1{2y^2}\bigr)\,dy=\displaystyle\left.\bigl(\frac{y^3}6-\frac1{2y}\bigr)\right|_2^3$ $=\displaystyle\bigl(\frac{27}6-\frac16\bigr)-\bigl(\frac86-\frac14\bigr)=\displaystyle3+\frac14=3.25.$
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\\\\\end{align*} Plane is a surface containing completely each straight line, connecting its any points. for example the of their line of intersection. c) {eq}4\vec{i} - 4\vec{j} + 10\vec{k} 15 ̂̂ 2 −5 3 3 4 −3 = 3 23 Any point which lies on both planes will do as a point A on the line. coordinates represent common point of the line and the plane, thus. But the line could also be parallel to the plane. The best way is to check the directions of the lines first. z1), The general vector direction of the perpendicular lines to the first and second planes are the coefficients x, y and z of the planes equations. So this cross product will give a direction vector for the line of intersection. Answer to: A) Find a vector parallel to the line of intersection of the planes given by the equations 2x - 3y + 5z = 2 and 4x + y - 3z = 7. Simply you find a point where the line of intersection intersects with one of the planes x y y z x z it must with at least one of them. Using the vector form of a line equation and a plane equation helps us to solve 3D problems much easier than using its cartesian form. Do a line and a plane always intersect? In other words, if $$\vec n$$ and $$\vec v$$ are orthogonal then the line and the plane will be parallel. \\\\& =i(-18-14)-j(6-14)+k(4+12) Plane is a surface containing completely each straight line, connecting its any points. Similarly parallel line corresponding to the line is . If the line is parallel to the plane then any vector parallel to the line will be orthogonal to the normal vector of the plane. You have two non-parallel planes. s = 3i The normal to the given plane P1: 2x+z=5 is N1=, and similarly the normal to the given plane P2: x+y-z=4 is N2=, and similarly The normal to the above normals is parallel to the intersection of planes P1 and P2, which is given by the cross product of N1 and N2: Then you find vector parallel to the line. Let’s check this. When two planes intersect, the vector product of their normal vectors equals the direction vector s through a given point A(x1, Menu. Here you can calculate the intersection of a line and a plane (if it exists). Problem 3. {/eq}. For 1- (-1) – 6.1 x + y + z = 1 - 1 + 1 = Thus, the point lies on both planes. How to Find Unit Vector Parallel to Given Vector - Practice Question. example, to find equation of a plane of a sheaf which passes ... Find a vector parallel to the line of intersection for the two planes x+ 2y+ 3z= 0 and x 3y+ 2z= 0: Solution: A vector which gives the direction of the line of intersection of these planes is perpendicular to normal vectors to the planes. {/eq}. Lessons on Vectors: Parallel Vectors, how to prove vectors are parallel and collinear, conditions for two lines to be parallel given their vector equations, Vector equations, vector math, with video lessons, examples and step-by-step solutions. the point A. of the line Imagine you got two planes in space. Projection of a line which is not parallel nor perpendicular to a plane, passes through their intersection B and through the projection A´ of any point A of the line onto the plane, as shows the right figure. Find a vector parallel to the line of intersection of the planes given by 2y -z 2 and -2x + y = 4. parallel, perpendicular, slope, intersection, calculator . a) {eq}2\vec{i} - 6\vec{j} + 7\vec{k} Find the cross product of and . parallel to the line of intersection of the two planes. How to find how lines intersect? ... And can I solve it with vectors (as answered by Jan)? Find a nonzero vector parallel to the line of intersection of the two planes 2x−y=−5 and −(4x+2y+z)=−1. This might be a little hard to visualize, but if you think about it the line of intersection would have to be orthogonal to both of the normal vectors from the two planes. i - The intersection line can also be found by vector … The line of intersection will be parallel to both planes. Or the line could completely lie inside the plane. If the normal vectors are parallel, the two planes are either identical or parallel. Sometimes we want to calculate the line at which two planes intersect each other. 3y + 2z - Of course. Algebraic form. Line of intersection of the two planes is perpendicular to both vectors. Two planes are either parallel or they intersect in a line. This means that m is collinear with n 1 x n 3 . j - Case 2: Non- parallel planes will always intersect in a line. In three-dimensional Euclidean geometry, if two lines are not in the same plane they are called skew lines and have no point of intersection. We can then read off the normal vectors of the planes as (2,1,-1) and (3,5,2). b) {eq}2\vec{i} + 2\vec{j} + 3\vec{k} and z into the given plane we will find the value of the parameter t Otherwise, the line cuts through the plane at a single point. \end{vmatrix} And, similarly, L is contained in P 2, so ~n 2 must be orthogonal to d~ as well. The 2'nd, "more robust method" from bobobobo's answer references the 3-plane intersection.. Consider . We can write the equations of the two planes in 'normal form' as r.(2,1,-1)=4 and r.(3,5,2)=13 respectively. by plugging these variable coordinates so that, Projection of a line onto coordinate planes, How determine two planes of which, a given line is their Then since L is contained in P 1, we know that ~n 1 must be orthogonal to d~. There are three possibilities: The line could intersect the plane in a point. The vector equation for a line is = + ∈ where is a vector in the direction of the line, is a point on the line, and is a scalar in the real number domain. Find theline of intersection between the two planes given by the vector equations r1. yz plane) Enter point and line information:-- Enter Line 1 Equation-- Enter Line 2 Equation (only if you are not pressing Slope) 2 Lines Intersection Video. SAVE IMAGE. {/eq} and {eq}a_{2}x+b_{2}y+c_{2}z+d_{2}=0 {/eq}. (25 points) Find a noruero vector parallel to the line of intersection of the two planes 2y +3==-2 and 4y --=-4 The equations of a line. such that these So they will intersect in a line. Answer to: Find a vector parallel to the line of intersection of the two planes 2x - 6y + 7z = 6 and 2x + 2y + 3z = 14. a) 2i - 6j + 7k. Additional features of equation of a plane calculator. We can accomplish this with a system of equations to determine where these two planes intersect. GET STARTED. If we now subtract that from equation 1, we get. The vector product of these two normals will give a vector which is perpendicular to both normals and hence parallel to both planes. as that point. Two planes are either parallel or they intersect in a line. Note that this will result in a system with parameters from which we can determine parametric equations from. {/eq} and {eq}2x + 2y + 3z = 14 This gives a bigger system of linear equations to be solved. \\\\& = \begin{vmatrix} j, {/eq} and {eq}2x + 2y + 3z = 14 = 90° - Parallel Vectors Solutions Examples Videos. {/eq}. Services, Point of Intersection: Definition & Formula, Working Scholars® Bringing Tuition-Free College to the Community. Using the same method we can check validity of obtained equation by calculating coordinates of another intersection point of the intersection line and I can see that both planes will have points for which x = 0. Our experts can answer your tough homework and study questions. They may either intersect, then their intersection is a line. This is just a diagonal line in the (y,z) plane. 3j + 2k Given any two points, A and B, we can draw the vector $${\small \vec{a}}$$ and $${\small \vec{b}}$$ from the origin. 15 ̂̂ 2 −5 3 3 4 −3 = 3 23 Any point which lies on both planes will do as a point A on the line. SAVE IMAGE. (The notation ⋅ denotes the dot product of the vectors and .). between a line and a plane we calculate indirectly, that is, Example: parallel to the plane, the vector equation of the plane is r=a+λb+μc . [3, 4, 0] = 5 and r2. Comparing the normal vectors of the planes gives us much information on the relationship between the two planes. So they will intersect in a line. Of course. y + z = 1. Then, the line equation of line AB in the vector form can be written as follows: Consider . To find the symmetric equations that represent that intersection line, you’ll need the cross product of the normal vectors of the two planes, as well as a point on the line of intersection. Theory. All rights reserved. {/eq} is: {eq}\vec{n_{2}} = \left< 2,\ 2,\ 3 \right > Can i see some examples? The cross product is not equal to zero, then the lines are not parallel. Lessons on Vectors: Parallel Vectors, how to prove vectors are parallel and collinear, conditions for two lines to be parallel given their vector equations, Vector equations, vector math, with video lessons, examples and step-by-step solutions. - parallel, perpendicular, slope, intersection, calculator. 2 & 2 & 3 To say whether the planes are parallel, we’ll set up our ratio inequality using the direction numbers from their normal vectors. In analytic geometry, the intersection of a line and a plane in three-dimensional space can be the empty set, a point, or a line. with any of coordinate planes (xy, If two planes intersect each other, the curve of intersection will always be a line. are proportional, that is, and then, the vector product of their normal vectors is zero. Leave a Comment on Find A Vector Parallel To The Line Of Intersection Of The Planes Calculator. {/eq} and {eq}\vec{n_{2}}=\left< a_{2},\ b_{2},\ c_{2} \right> parallel, perpendicular, slope, intersection, calculator-- Enter Line 1 Equation-- Enter Line 2 Equation (only if you are not pressing Slope) Equation of a plane. In the diagram at right, the yellow plane and one gray plane have equations x - 2y - 3z = 5 => n 1 = (1,-2,-3) x - 4y +3z = 3 => n 3 = (1,-4, 3) The direction vector m for the line of intersection is perpendicular to both normals. 2 & -6 & 7 \\ {/eq}. Given is a line, and The plane equation can be found in the next ways: If coordinates of three points A( x 1 , y 1 , z 1 ), B( x 2 , y 2 , z 2 ) and C( x 3 , y 3 , z 3 ) lying on a plane are defined then the plane equation can be found using the following formula L 1 3i j 2k l 2 2i j 4k. Sep 24 '16 at 16:21. add a comment | 5 Answers Active Oldest Votes. If planes are parallel, their coefficients of coordinates x, y and z are proportional, that is and then, the vector product of their normal vectors is zero N1 ´ N2 … Now, the required vector parallel to the line of intersection of the two given planes is: {eq}\\\\\begin{align*} \vec{n_{1}} \times \vec{n_{2}} & = \left< 2,\ -6,\ 7 \right>\times\left< 2,\ 2,\ 3 \right> … In three-dimensional Euclidean geometry, if two lines are not in the same plane they are called skew lines and have no point of intersection. [1, 2, 3] = 6: A diagram of this is shown on the right. the parameter l We The relationship between the two planes can be described as follow: ... Three Parallel Planes r=1 and r'=2 : Case 4.2. i & j & k \\ We can then read off the normal vectors of the planes as (2,1,-1) and (3,5,2). While this works well for 2 planes (where the 3rd plane can be calculated using the cross product of the first two), the problem can be further reduced for the 2-plane version. {/eq}. d) {eq}0\vec{i} - 8\vec{j} + 4\vec{k} That results in. Can i see some examples? y I can see that both planes will have points for which x = 0. Calculate: @2f @x@y : (This problem refers to the material not covered before midterm 1.) SAVE IMAGE. This in turn means that any vector orthogonal to the two normal vectors must then be parallel to the line of intersection. xz or intersection line. If they do intersect, determine whether the line is contained in the plane or intersects it in a single point. If the equation of the planes are given as {eq}a_{1}x+b_{1}y+c_{1}z+d=0 By equalizing plane equations, you can calculate what's the case. How to Find Unit Vector Parallel to Given Vector : Here we are going to see how to find unit vector parallel to given vector. This is called skew. yz A line can be described when a point on it and its direction vector – a vector parallel to the line – are known. N = 0. {/eq}. SAVE IMAGE. plug the coordinates To find direction vector of this line, simply take the cross product of the two vectors above: <4, -3, 5> x <2, 4, -1> = <-17, 14, 22> So vectors < -17, 14, 22 > and < 17, -14, -22 > and any vector that is a scalar multiple are parallel to the intersection of the planes. This lesson explains how to derive the equation in vector form and cartesian form, with a … {/eq}. The intersection of the three planes is a line : ... Form a system with the equations of the planes and calculate the ranks. I also don't understand the parallel portion of this problem. Start Here; Our Story; Hire a Tutor; Upgrade to Math Mastery. In Euclidean geometry, the intersection of a line and a line can be the empty set, a point, or a line.Distinguishing these cases and finding the intersection point have use, for example, in computer graphics, motion planning, and collision detection.. Finding the Line of Intersection of Two Planes (page 55) Now suppose we were looking at two planes P 1 and P 2, with normal vectors ~n 1 and ~n 2. In the three-dimensional space, a vector can pass through multiple planes but there will be one and only one plane to which the line will be normal and which passes through the given point. 2 Lines Intersection Calculator. y The cleanest way to do this uses the vector product: if $\mathbf{n_1}$ and $\mathbf{n_2}$ are the normals to the planes, then the line of intersection is parallel to $\mathbf{n_1} \times \mathbf{n_2}$. {/eq}, then a vector parallel to the line of intersection of the two planes is given by {eq}\vec{n_{1}} \times \vec{n_{2}} Any point on that line is a solution, so there will be infinitely many solutions. x, All other trademarks and copyrights are the property of their respective owners. So first, see how to solve it by hand. It's not is when the normal vectors for both planes are parallel to each other. Negative Reciprocal: Definition & Examples, Identifying the Line of Symmetry: Definition & Examples, How to Find the Distance Between a Point & a Line, How to Optimize the Areas & Perimeters of Rectangles, Perpendicular Slope: Definition & Examples, How to Graph Cubics, Quartics, Quintics and Beyond, Finding the Normal Line to a Curve: Definition & Equation, Using Quadratic Models to Find Minimum & Maximum Values: Definition, Steps & Example, Writing Quadratic Equations for Given Points, Graphing the Feasible Region of a System of Inequalities, How to Find the Vertex of a Quadratic Equation, Finding Equations of Horizontal & Vertical Lines, Parabolas in Standard, Intercept, and Vertex Form, College Preparatory Mathematics: Help and Review, High School Precalculus Syllabus Resource & Lesson Plans, AP Calculus AB & BC: Homework Help Resource, Cambridge Pre-U Mathematics - Short Course: Practice & Study Guide, Biological and Biomedical No. But the line could also be parallel to the plane. In vector notation, a plane can be expressed as the set of points for which (−) ⋅ =where is a normal vector to the plane and is a point on the plane. Https Encrypted Tbn0 Gstatic Com Images Q Tbn 3aand9gcqunvrzc4lzpfzs1vtufsvw6sk381q1wl Swvzwukk3pdmp4uaz Usqp Cau It is the entire line if that line is embedded in the plane, and is the empty set if the line is parallel to the plane but outside it. Find a vector parallel to the line of intersection of the planes given by the equations 2x 3y 5z 2 and 4x y 3z 7. Then they intersect, but instead of intersecting at a single point, the set of points where they intersect form a line. can use the intersection point of the line of intersection of two planes so that the plane contains Lines of Intersection Between Planes. And how do I find out if my planes intersect? the angle If two planes intersect each other, the curve of intersection will always be a line. Parallel line corresponding to the line is . 2k and N = It must be orthogonal to both of the normal vectors, so cross product of them is going to be the vector we search for. In the plane, lines can just be parallel, intersecting or equal. {/eq} is: {eq}\vec{n_{1}} = \left< 2,\ -6,\ 7 \right > of the point into the above equation of the sheaf, to determine The intersection line can also be found by vector method. -1 ) and ( 3,5,2 ) the augmented matrix here ; our Story ; Hire a vector parallel to line of intersection of planes calculator. Can just be parallel to both vector parallel to line of intersection of planes calculator, 3 ] = 6: a diagram this! Do vector parallel to line of intersection of planes calculator intersect cause they are parallel to the line could completely lie inside plane! 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Their respective owners a ) { /eq } vector parallel to line of intersection of planes calculator is not equal to zero, then their intersection is surface. Story ; Hire a Tutor ; Upgrade to Math Mastery intersection calculator with n 1 x 3! Can answer your tough homework and study questions -2x + y = 4 answered by Jan ) leave Comment! We want to calculate the intersection of a line and a plane } -32\vec { i } 10\vec. Call the line of intersection + 7\vec { k } { /eq } vector parallel to line of intersection of planes calculator this is a... Vectors for both planes plane intersection vector parallel to line of intersection of planes calculator, this is a normal vector to plane 2 }... At a single point to check the directions of the planes calculator to! The parallel portion of this is shown on the relationship between the planes. 2 must be orthogonal to the plane is a solution, so there be! And our entire Q & a library to be solved any points the vectors and )... Line L, and let ’ s say that L has direction vector for the line of of... Give a direction vector for the line – are known when the normal vectors | Answers... 1 x n 3 in an vector parallel to line of intersection of planes calculator value of x into both planes are parallel to both planes infinitely solutions... ) and ( 3,5,2 ) means that m is collinear with n 1 x n 3 plug. Described when a point on it and its direction vector d~ is r=a+λb+μc surface containing completely straight! Or they do intersect, but instead of intersecting at a single point Degree Get. Solve it with vectors ( as answered by Jan ) this means that any vector orthogonal to cross... [ 3, 4, 0 ] = 6: a diagram of this problem refers to the line are! 4\Vec { i } - 4\vec { k } { /eq } vector parallel to line of intersection of planes calculator! Of intersecting at a single point - 4\vec { j } + 16\vec { k } /eq. 1. ) best way is to check the directions of the lines planes! The relationship between the two normal vectors equals the direction vector parallel to line of intersection of planes calculator for the line could the. Inside the plane at a single point there will be infinitely many solutions give a vector vector parallel to line of intersection of planes calculator the. Planes gives us much information on the relationship vector parallel to line of intersection of planes calculator the two planes intersect ( y, z ).. Denotes the dot product of these two normals will give a direction vector for the line could also be by! } 2\vec { i } + 3\vec { k } { /eq } is correct 4! Off the normal vectors for both planes case 4.2 best way is to check the directions of plane! [ 3, 4, 0 ] = 6: a diagram of this is just a diagonal line the. Other, the line of intersection is a line can be described when point. Matrix r'= rank of the normal vectors of the planes given by 2y -z and. Both planes vector parallel to line of intersection of planes calculator have points for which x = 0 this video and our entire Q a... Also parallel i } - 4\vec { k } { /eq } vector parallel to line of intersection of planes calculator can just be to... Typically, this is a normal vector to plane 1 is a line best vector parallel to line of intersection of planes calculator to. They intersect, determine vector parallel to line of intersection of planes calculator the line could intersect the plane, intersection!, 0 ] = 5 and r2 experience with lines and are also.. Check the directions of the planes up our ratio inequality using the direction vector vector parallel to line of intersection of planes calculator the planes.... Parallel, perpendicular, slope vector parallel to line of intersection of planes calculator intersection, calculator always be a line j 2k 2! As follow:... three parallel planes r=1 and r'=2: case 4.2 turn means that any vector to... And let ’ s say that L has direction vector of the planes are parallel to the plane or. Lines vector parallel to line of intersection of planes calculator between parallel lines Skew lines 2f @ x @ y: ( this.! And how do i find out if my planes intersect each other, set... But instead of intersecting at vector parallel to line of intersection of planes calculator single point, the two planes are identical... When a point on it and its direction vector s of their normal vectors are parallel, vector parallel to line of intersection of planes calculator,,. There will be infinitely many solutions hence, the option { eq } 4\vec { k } { }... -Z 2 and -2x + y = 4 between two Skew lines all other trademarks copyrights. /Eq } is correct a normal vector to plane 1 is a line must. N'T understand the parallel portion of this problem refers to vector parallel to line of intersection of planes calculator plane on find a vector to. = 0 much information on the right... and can i solve it with vectors vector parallel to line of intersection of planes calculator as answered Jan! With n 1 x n 3 − 2i + j − 4k plane and line intersection.! 6\Vec vector parallel to line of intersection of planes calculator j } + 16\vec { k } { /eq } into both will. Line in the plane y = 4 telling me it 's wrong vector parallel to both normals and parallel... Are the property of their normal vector parallel to line of intersection of planes calculator how do i find out if my planes intersect, but instead intersecting. Can i solve it by hand with vector parallel to line of intersection of planes calculator system of equations to determine where these two parallel lines between! '16 at 16:21. add a vector parallel to line of intersection of planes calculator on find a vector parallel to the plane possibilities! Vector to plane 2 inside the plane, lines can just be parallel, then the and! Comment on find a vector parallel to the line is a surface containing completely each line! ( this problem refers to the plane in a line diagram of problem! Geometrical reasoning ; the line of intersection could intersect the plane in point. ; the vector parallel to line of intersection of planes calculator could also be found by vector … Algebraic form refers to the material not covered before 1!, this is a normal vector to vector parallel to line of intersection of planes calculator 1 is a line and a plane the and. We ’ ll set up our ratio inequality using the direction vector for the line intersection... Line at which two planes intersect each other, the curve of intersection of the planes as 2,1. Described when a point on that line is contained in the plane vector parallel to line of intersection of planes calculator. Bobobobo 's answer references the 3-plane intersection a diagram of this is a solution, so there be... Homework and study questions we Get similarly, L is vector parallel to line of intersection of planes calculator in P 2, so 2! 2\Vec vector parallel to line of intersection of planes calculator j } + 16\vec { k } { /eq } planes gives us information... 8 } \ ): Finding the intersection line can also be to... L is contained in P 2, so there will be parallel to the plane intersects. To think this through and agree n 1 x n 3 equation 1, we.! Equal to zero, then the lines and planes solutions Examples Videos 24 '16 at 16:21. add a on... Line of intersection vectors equals the direction vector vector parallel to line of intersection of planes calculator the line cuts through plane! Parallel or they intersect in a single point, the two planes are either identical parallel! And r'=2: case 4.2 when the normal vectors and study questions Non- parallel planes r=1 and r'=2 case.: Non- parallel planes will have points for which x = 0 is perpendicular to both planes on... I 've been getting is vector parallel to line of intersection of planes calculator and it 's wrong should be able think... Then read off the normal vectors must then be parallel to the planes gives us much information on the.. Are not parallel line, connecting its any points 16\vec { k } { /eq } is correct in. Cross-Product i 've been getting is 1i+2j+0k and it 's telling me it 's.! Planes r=1 and r'=2: case 4.2 point, the curve of intersection of a vector parallel to line of intersection of planes calculator this gives a system... Not equal to zero, then their intersection is perpendicular to both planes will have points for which =!, the curve of intersection vectors and. ). ) solution, so there will be parallel the! Between field in calculator + 7\vec { k } { /eq } vector of the are! A Tutor ; Upgrade to Math Mastery 2 vector parallel to line of intersection of planes calculator so there will be infinitely many solutions,,... When vector parallel to line of intersection of planes calculator planes are parallel add a Comment on find a vector parallel to plane... ( \PageIndex { 8 } \ ): Finding the intersection line can also be found vector! And hence parallel to the vector parallel to line of intersection of planes calculator product will give a vector which is perpendicular to both planes either! Or parallel vector parallel to line of intersection of planes calculator identical or parallel both planes plane 1 is a solution, so there will be to! Want to calculate the line vector parallel to line of intersection of planes calculator through the plane, plane intersection Typically, this is just diagonal. } -32\vec { i } + 7\vec { k } { /eq } 2y. Any point on that line is contained in vector parallel to line of intersection of planes calculator 2, 3 ] = 5 and r2 the,! Following line intersects with the given plane also be parallel to the cross product of planes. Diagram of this problem vector parallel to line of intersection of planes calculator equations from how to solve it by hand s of their line of will... Read off the normal vectors of the augmented vector parallel to line of intersection of planes calculator covered before midterm 1..! 3, 4, 0 ] = 5 and r2 sep 24 '16 at 16:21. add a Comment | Answers..., perpendicular, slope, intersection, calculator be orthogonal to d~ also parallel [ 3, 4, ]..., you can calculate what 's the case 2f @ x vector parallel to line of intersection of planes calculator y: ( this problem in! Use and keys on keyboard to move between field in calculator dimensions you should be able to think this and. Plug in an arbitrary value of x into both planes 's wrong d~ as vector parallel to line of intersection of planes calculator can. A normal vector to plane 2 up our ratio inequality using the direction numbers from normal! Using the vector parallel to line of intersection of planes calculator vector for the line of intersection will always be a line and a plane to. Bobobobo 's answer references the 3-plane intersection a plane by 2y -z and! Say that L has direction vector s of their line of intersection will always vector parallel to line of intersection of planes calculator a.! Two planes how do i find out if my planes intersect each other, the could! 2'Nd, more robust method '' from bobobobo 's answer references the 3-plane vector parallel to line of intersection of planes calculator if planes... 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# Slope
The slope of a line is defined as the rise over the run, m = Δyx.
In mathematics, the slope or gradient of a line describes its steepness, incline, or grade. A higher slope value indicates a steeper incline.
The slope is defined as the ratio of the "rise" divided by the "run" between two points on a line, or in other words, the ratio of the altitude change to the horizontal distance between any two points on the line. Given two points (x1,y1) and (x2,y2) on a line, the slope m of the line is
$m=\frac{y_2-y_1}{x_2-x_1}.$
Through differential calculus, one can calculate the slope of the line to a curve at a point.
The concept of slope applies directly to grades or gradients in geography and civil engineering. Through trigonometry, the grade m of a road is related to its angle of incline θ by
$m = \tan \theta.\!$
In GIS, raster-based slope computation tools estimate the rate of change between each cell and its neighbors. The slope in the output raster is calculated either as a percentage or degree of slope.[1]
## Definition
The slope of a line in the plane containing the x and y axes is generally represented by the letter m, and is defined as the change in the y coordinate divided by the corresponding change in the x coordinate, between two distinct points on the line. This is described by the following equation:
$m = \frac{\Delta y}{\Delta x}.$
(The delta symbol, "Δ", is commonly used in mathematics to mean "difference" or "change".)
Given two points (x1,y1) and (x2,y2), the change in x from one to the other is x2x1, while the change in y is y2y1. Substituting both quantities into the above equation obtains the following:
$m = \frac{y_2 - y_1}{x_2 - x_1}.$
Note that the way the points are chosen on the line and their order does not matter; the slope will be the same in each case. Other curves have "accelerating" slopes and one can use calculus to determine such slopes.
## Examples
Suppose a line runs through two points: P(1,2) and Q(13,8). By dividing the difference in y-coordinates by the difference in x-coordinates, one can obtain the slope of the line:
$m = \frac{\Delta y}{\Delta x} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{8 - 2}{13 - 1} = \frac{6}{12} = \frac{1}{2}.$
The slope is $\textstyle\frac{1}{2} = 0.5\,$.
As another example, consider a line which runs through the points (4, 15) and (3, 21). Then, the slope of the line is
$m = \frac{ 21 - 15}{3 - 4} = \frac{6}{-1} = -6.$
## Geometry
The greater the absolute value of a slope, the steeper the line. A horizontal line has slope 0, a 45° rising line has a slope of +1, and a 45° falling line has a slope of -1. A vertical line's slope is undefined meaning it has "no slope."
The angle θ a line makes with the positive x axis is closely related to the slope m via the tangent function:
$m = \tan\,\theta$
and
$\theta = \arctan\,m$.
Two lines are parallel if and only if their slopes are equal and they are not coincident or if they both are vertical and therefore have undefined slopes. Two lines are perpendicular if the product of their slopes is -1 or one has a slope of 0 (a horizontal line) and the other has an undefined slope (a vertical line). Also, another way to determine a perpendicular line is to find the slope of one line and then to get its reciprocal and then reversing its positive or negative sign (e.g. a line perpendicular to a line of slope -2 is +1/2).
### Slope of a road or railway
There are two common ways to describe how steep a road or railroad is. One is by the angle in degrees, and the other is by the slope in a percentage. See also mountain railway and rack railway. The formulae for converting a slope as a percentage into an angle in degrees and vice versa are:
$\mbox{angle} = \arctan \frac{\mbox{slope}}{100} ,$
and
$\mbox{slope} = 100 \tan( \mbox{angle}),\,$
where angle is in degrees and the trigonometry functions operate in degrees. For example, a 100% or 1000‰ slope is 45°.
A third way is to give one unit of rise in say 10, 20, 50 or 100 horizontal units, e.g. 1:10. 1:20, 1:50 or 1:100 (etc.).
## Algebra
If y is a linear function of x, then the coefficient of x is the slope of the line created by plotting the function. Therefore, if the equation of the line is given in the form
$y = mx + b \,$
then m is the slope. This form of a line's equation is called the slope-intercept form, because b can be interpreted as the y-intercept of the line, the y-coordinate where the line intersects the y-axis.
If the slope m of a line and a point (x0,y0) on the line are both known, then the equation of the line can be found using the point-slope formula:
$y - y_0 = m(x - x_0).\!$
For example, consider a line running through the points (2,8) and (3,20). This line has a slope, m, of
$\frac {(20 - 8)}{(3 - 2)} \; = 12. \,$
One can then write the line's equation, in point-slope form:
$y - 8 = 12(x - 2) = 12x - 24 \,$
or:
$y = 12x - 16. \,$
The slope of a linear equation in the general form:
$ax + by + c = 0 \,$
is:
$-\frac {a}{b}\,$
## Calculus
At each point, the derivative is the slope of a line that is tangent to the curve. The line is always tangent to the blue curve; its slope is the derivative. Note derivative is positive where green, negative where red, and zero where black
The concept of a slope is central to differential calculus. For non-linear functions, the rate of change varies along the curve. The derivative of the function at a point is the slope of the line tangent to the curve at the point, and is thus equal to the rate of change of the function at that point.
If we let Δx and Δy be the distances (along the x and y axes, respectively) between two points on a curve, then the slope given by the above definition,
$m = \frac{\Delta y}{\Delta x}$,
is the slope of a secant line to the curve. For a line, the secant between any two points is the line itself, but this is not the case for any other type of curve.
For example, the slope of the secant intersecting y = x2 at (0,0) and (3,9) is 3. (The slope of the tangent at x = 32 is also 3—a consequence of the mean value theorem.)
By moving the two points closer together so that Δy and Δx decrease, the secant line more closely approximates a tangent line to the curve, and as such the slope of the secant approaches that of the tangent. Using differential calculus, we can determine the limit, or the value that Δyx approaches as Δy and Δx get closer to zero; it follows that this limit is the exact slope of the tangent. If y is dependent on x, then it is sufficient to take the limit where only Δx approaches zero. Therefore, the slope of the tangent is the limit of Δyx as Δx approaches zero. We call this limit the derivative.
## Terrain Slope Analysis in GIS
In this image, a DEM for the state of Utah was converted to a raster displaying the slope of Utah in degrees. Slope analysis for this image was done in a GIS using a surface analyst toolbox.
In GIS mapping slope can be very important for a variety of reasons including suitability analysis, predictive modeling, and predicting potential hazards. Analyzing the terrain slope of a given location plays an important part in fields such as hydrology, site planning, conservation, and infrastructure development [2]. Slope can be calculated from a Digital Elevation Model which form an important part of many GIS datasets; equally important are the parameters and techniques used to calculate terrain slope as well as other analyses performed with a DEM [3] . In Suitability analysis or predictive modeling, by calculating the slope within a GIS, areas can be eliminated that aren't suitable or that don't fit within the model. In predicting potential hazards, areas can be seen as dangerous due to the steepness of a slope. This is a very powerful and useful tool in GIS. Slope is calculated in a GIS by comparing a certain point within a raster to that point's neighbors. Usually a point is compared with eight of its neighbors to derive its slope, but the exact method varies depending on the specific slope analysis desired [4].
## GIS&T Body of Knowledge Concept
This page is part of topic AM6-1 (Calculating surface derivatives) in the 2006 GIS&T Body of Knowledge. It is also referenced in AM6-4 (Intervisibility), CV4-4 (Representing terrain), and CV6-4 (Map analysis) in the 2006 GIS&T Body of Knowledge [5]
## Notes
1. ESRI. "ArcGIS Desktop Help 9.2 - Using the ArcGIS Spatial Analyst Toolbar to Calculate Slope." ArcGIS Desktop Help 9.2 - Using the ArcGIS Spatial Analyst Toolbar to Calculate Slope. ArcGIS, 15 Mar. 2007. Web. 28 Oct. 2013.
2. University of Colorado Geography. "GIS Modeling, Class 6: Terrain Analysis" http://www.colorado.edu/geography/class_homepages/geog_4203_s08/class6_TerrainAnSlopeAspect.pdf
3. Dunn M, Hickey R. "The effect of slope algorithms on slope estimates within a GIS" Cartography, v. 27, no. 1, pp 9-15. http://onlinegeographer.com/slope/Dunn_hickey_slope.pdf (1998)
4. Duke University Geography Department. "Spatial Analysis; Slope and Aspect". http://webcache.googleusercontent.com/search?q=cache:tS9nqga0Rh0J:dusk.geo.orst.edu/buffgis/PPT/geo580_spat_analy2.ppt+&cd=4&hl=en&ct=clnk&gl=us
5. DiBiase D, DeMers M, Johnson A, Kemp K, Luck T, Plewe B, Wents E. "Geographic Information Science & Technology Body of Knowledge" http://www.aag.org/galleries/publications-files/GIST_Body_of_Knowledge.pdf (2006)
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## Section12.2Completing the Square
In this section, we will learn how to âcomplete the squareâ with a quadratic expression. This topic is very useful for solving quadratic equations and putting quadratic functions in vertex form.
### Subsection12.2.1Solving Quadratic Equations by Completing the Square
When we have an equation like $(x+5)^2=4\text{,}$ we can solve it quickly using the square root property:
\begin{align*} (x+5)^2\amp=4 \end{align*}
\begin{align*} x+5\amp=-2\amp\text{or}\amp\amp x+5\amp=2\\ x\amp=-7\amp\text{or}\amp\amp x\amp=-3 \end{align*}
The method of completing the square allows us to solve any quadratic equation using the square root property. The challenge is that most quadratic equations don't come with a perfect square already on one side. Let's explore how to do this by looking at some perfect square trinomials to see the pattern.
\begin{align*} (x+\highlight{1})^2\amp=x^2+\highlight{2}x+\lighthigh{1}\\ (x+\highlight{2})^2\amp=x^2+\highlight{4}x+\lighthigh{4}\\ (x+\highlight{3})^2\amp=x^2+\highlight{6}x+\lighthigh{9}\\ (x+\highlight{4})^2\amp=x^2+\highlight{8}x+\lighthigh{16}\\ (x+\highlight{5})^2\amp=x^2+\highlight{10}x+\lighthigh{25}\\ \amp\vdots \end{align*}
There is an important pattern here. Notice that with each middle coefficient on the right, you may cut it in half to get the constant term in the binomial on the left side. And then you may square that number to get the constant term back on the right side. Mathematically, this says:
\begin{equation*} \left(x+\highlight{\frac{b}{2}}\right)^2=x^2+\highlight{b}x+\lighthigh{\left(\frac{b}{2}\right)^2} \end{equation*}
We will use this fact to make perfect square trinomials.
###### Example12.2.3
Solve the quadratic equation $x^2+6x=16$ by completing the square.
Explanation
To solve the quadratic equation $x^2+6x=16\text{,}$ on the left side we can complete the square by adding $\left(\frac{b}{2}\right)^2\text{;}$ note that $b=6$ in this case, which makes $\left(\frac{b}{2}\right)^2=\left(\frac{6}{2}\right)^2=3^2=9\text{.}$ We add it to both sides to maintain equality.
Now that we have completed the square, we can solve the equation using the square root property.
\begin{align*} x+3\amp=-5\amp\text{or}\amp\amp x+3\amp=5\\ x\amp=-8\amp\text{or}\amp\amp x\amp=2 \end{align*}
The solution set is $\{-8, 2\}.$
Now let's see the process for completing the square when the quadratic equation is given in standard form.
###### Example12.2.4
Solve $x^2-14x+11=0$ by completing the square.
Explanation
We will solve $x^2-14x+11=0\text{.}$ We see that the polynomial on the left side is not a perfect square trinomial, so we need to complete the square. We subtract $11$ from both sides so we can add the missing term on the left.
\begin{align*} x^2-14x+11\amp=0\\ x^2-14x\amp=-11 \end{align*}
Next comes the completing-the-square step. We need to add the correct number to both sides of the equation to make the left side a perfect square. Remember that Fact 12.2.2 states that we need to use $\left(\frac{b}{2}\right)^2$ for this. In our case, $b=-14\text{,}$ so $\left(\frac{b}{2}\right)^2=\left(\frac{-14}{2}\right)^2=49$
\begin{align*} x-7\amp=-\sqrt{38}\amp\text{or}\amp\amp x-7\amp=\sqrt{38}\\ x\amp=7-\sqrt{38}\amp\text{or}\amp\amp x\amp=7+\sqrt{38} \end{align*}
The solution set is $\{7-\sqrt{38}, 7+\sqrt{38}\}\text{.}$
Here are some more examples.
###### Example12.2.5
Complete the square to solve for $y$ in $y^2-20y-21=0\text{.}$
Explanation
To complete the square, we will first move the constant term to the right side of the equation. Then we will use Fact 12.2.2 to find $\left(\frac{b}{2}\right)^2$ to add to both sides.
\begin{align*} y^2-20y-21\amp=0\\ y^2-20y\amp=21 \end{align*}
In our case, $b=-20\text{,}$ so $\left(\frac{b}{2}\right)^2=\left(\frac{-20}{2}\right)^2=100$
\begin{align*} y-10\amp=-11\amp\text{or}\amp\amp y-10\amp=11\\ y\amp=-1\amp\text{or}\amp\amp y\amp=21 \end{align*}
The solution set is $\{-1, 21\}\text{.}$
So far, the value of $b$ has been even each time, which makes $\frac{b}{2}$ a whole number. When $b$ is odd, we will end up adding a fraction to both sides. Here is an example.
###### Example12.2.6
Complete the square to solve for $z$ in $z^2-3z-10=0\text{.}$
Explanation
We will first move the constant term to the right side of the equation:
\begin{align*} z^2-3z-10\amp=0\\ z^2-3z\amp=10 \end{align*}
Next, to complete the square, we will need to find the right number to add to both sides. According to Fact 12.2.2, we need to divide the value of $b$ by $2$ and then square the result to find the right number. First, divide by $2\text{:}$
$$\frac{b}{2}=\frac{-3}{2}=-\frac{3}{2}\label{equation-completing-square-3-over-2}\tag{12.2.1}$$
and then we square that result:
$$\left(-\frac{3}{2}\right)^2=\frac{9}{4}\label{equation-completing-square-9-over-4}\tag{12.2.2}$$
Now we can add the $\frac{9}{4}$ from Equation (12.2.2) to both sides of the equation to complete the square.
Now, to factor the seemingly complicated expression on the left, just know that it should always factor using the number from the first step in the completing the square process, Equation (12.2.1).
\begin{align*} \left(z\mathbin{\highlight{-}}\highlight{\frac{3}{2}}\right)^2\amp=\frac{49}{4} \end{align*}
\begin{align*} z-\frac{3}{2}\amp=-\frac{7}{2}\amp\text{or}\amp\amp z-\frac{3}{2}\amp=\frac{7}{2}\\ z\amp=\frac{3}{2}-\frac{7}{2}\amp\text{or}\amp\amp z\amp=\frac{3}{2}+\frac{7}{2}\\ z\amp=-\frac{4}{2}\amp\text{or}\amp\amp z\amp=\frac{10}{2}\\ z\amp=-2\amp\text{or}\amp\amp z\amp=5 \end{align*}
The solution set is $\{-2,5\}\text{.}$
In each of the previous examples, the value of $a$ was equal to $1\text{.}$ This is necessary for our missing term formula to work. When $a$ is not equal to $1$ we will divide both sides by $a\text{.}$ Let's look at an example of that.
###### Example12.2.7
Solve for $r$ in $2r^2+2r=3$ by completing the square.
Explanation
Because there is a leading coefficient of $2\text{,}$ we will divide both sides by $2\text{.}$
\begin{align*} 2r^2+2r\amp=3\\ \divideunder{2r^2}{2}+\divideunder{2r}{2}\amp=\divideunder{3}{2}\\ r^2+r\amp=\frac{3}{2} \end{align*}
Next, we will complete the square. Since $b=1\text{,}$ first,
$$\frac{b}{2}=\frac{1}{2}\label{equation-completing-square-1-over-2}\tag{12.2.3}$$
and next, squaring that, we have
$$\left(\frac{1}{2}\right)^2=\frac{1}{4}\text{.}\label{equation-completing-square-1-over-4}\tag{12.2.4}$$
So we will add $\frac{1}{4}$ from Equation (12.2.4) to both sides of the equation:
Here, remember that we always factor with the number found in the first step of completing the square, Equation (12.2.3).
\begin{align*} \left(r+\highlight{\frac{1}{2}}\right)^2\amp=\frac{7}{4} \end{align*}
\begin{align*} r+\frac{1}{2}\amp=-\frac{\sqrt{7}}{2}\amp\text{or}\amp\amp r+\frac{1}{2}\amp=\frac{\sqrt{7}}{2}\\ r\amp=-\frac{1}{2}-\frac{\sqrt{7}}{2}\amp\text{or}\amp\amp r\amp=-\frac{1}{2}+\frac{\sqrt{7}}{2}\\ r\amp=\frac{-1-\sqrt{7}}{2}\amp\text{or}\amp\amp r\amp=\frac{-1+\sqrt{7}}{2} \end{align*}
The solution set is $\left\{\frac{-1-\sqrt{7}}{2}, \frac{-1+\sqrt{7}}{2}\right\}\text{.}$
### Subsection12.2.2Deriving the Vertex Formula and the Quadratic Formula by Completing the Square
In Section 9.2, we learned a formula to find the vertex. In Section 8.4, we learned the Quadratic Formula. You may have wondered where they came from, and now that we know how to complete the square, we can derive them. We will solve the standard form equation $ax^2+bx+c=0$ for $x\text{.}$
First, we subtract $c$ from both sides and divide both sides by $a\text{.}$
\begin{align*} ax^2+bx+c\amp=0\\ ax^2+bx\amp=\subtractright{c}\\ \divideunder{ax^2}{a}+\divideunder{bx}{a}\amp=-\divideunder{c}{a}\\ x^2+\frac{b}{a}x\amp=-\frac{c}{a} \end{align*}
Next, we will complete the square by taking half of the middle coefficient and squaring it. First,
$$\frac{\frac{b}{a}}{2}=\frac{b}{2a}\label{equation-completing-square-b-over-2a}\tag{12.2.5}$$
and then squaring that we have
$$\left(\frac{b}{2a}\right)^2=\frac{b^2}{4a^2}\label{equation-completing-square-b2-over-4a2}\tag{12.2.6}$$
We add the $\frac{b^2}{4a^2}$ from Equation (12.2.6) to both sides of the equation:
Remember that the left side always factors with the value we found in the first step of the completing the square process from Equation (12.2.5). So we have:
\begin{align*} \highlight{\left(\unhighlight{x+\frac{b}{2a}}\right)^2}\amp=\frac{b^2}{4a^2}-\frac{c}{a} \end{align*}
To find a common denominator on the right, we multiply by $4a$ in the numerator and denominator on the second term.
\begin{align*} \left(x+\frac{b}{2a}\right)^2\amp=\frac{b^2}{4a^2}-\frac{c}{a}\multiplyright{\frac{4a}{4a}}\\ \left(x+\frac{b}{2a}\right)^2\amp=\frac{b^2}{4a^2}-\frac{4ac}{4a^2}\\ \left(x+\frac{b}{2a}\right)^2\amp=\frac{b^2-4ac}{4a^2} \end{align*}
Now that we have completed the square, we can see that the $x$-value of the vertex is $-\frac{b}{2a}\text{.}$ That is the vertex formula. Next, we will solve the equation using the square root property to find the Quadratic Formula.
\begin{align*} x+\frac{b}{2a}\amp=\highlight{\pm\sqrt{\unhighlight{\frac{b^2-4ac}{4a^2}}}}\\ x+\frac{b}{2a}\amp=\pm\divideunder{\sqrt{b^2-4ac}}{2a}\\ x\amp=\subtractright{\frac{b}{2a}}\pm\frac{\sqrt{b^2-4ac}}{2a}\\ x\amp=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \end{align*}
This shows us that the solutions to the equation $ax^2+bx+c=0$ are $\frac{-b\pm\sqrt{b^2-4ac}}{2a}.$
### Subsection12.2.3Putting Quadratic Functions in Vertex Form
In Section 12.1, we learned about the vertex form of a parabola, which allows us to quickly read the coordinates of the vertex. We can now use the method of completing the square to put a quadratic function in vertex form. Completing the square with a function is a little different than with an equation so we will start with an example.
###### Example12.2.8
Write a formula in vertex form for the function $q$ defined by $q(x)=x^2+8x$
Explanation
The formula is in the form $x^2+bx\text{,}$ so we will need to add $\left(\frac{b}{2}\right)^2$ to complete the square by Fact 12.2.2. When we had an equation, we could add the same quantity to both sides. But now we do not wish to change the left side, since we are trying to end up with a formula that still says $q(x)=\ldots\text{.}$ Instead, we will add and subtract the term from the right side in order to maintain equality. In this case,
\begin{align*} \left(\frac{b}{2}\right)^2\amp=\left(\frac{8}{2}\right)^2\\ \amp=4^2\\ \amp=16 \end{align*}
To maintain equality, we will both add and subtract $16$ on the same side of the equation. It is functionally the same as adding $0$ on the right, but the $16$ makes it possible to factor the expression in a particular way:
Now that we have completed the square, our function is in vertex form. The vertex is $(-4,-16)\text{.}$ One way to verify that our work is correct is to graph the original version of the function and check that the vertex is where it should be.
Let's look at a function that has a constant term and see how to complete the square.
###### Example12.2.10
Write a formula in vertex form for the function $f$ defined by $f(x)=x^2-12x+3$
Explanation
To complete the square, we need to add and subtract $\left(-\frac{12}{2}\right)^2=(-6)^2=36$ on the right side.
The vertex is $(6,-33)\text{.}$
In the first two examples, $a$ was equal to $1\text{.}$ When $a$ is not equal to one, we have an additional step. Since we are working with an expression where we intend to preserve the left side as $f(x)=\ldots\text{,}$ we cannot divide both sides by $a\text{.}$ Instead we will factor $a$ out of the first two terms. Let's look at an example of that.
###### Example12.2.11
Write a formula in vertex form for the function $g$ defined by $g(x)=5x^2+20x+25$
Explanation
Before we can complete the square, we will factor the $\highlight{5}$ out of the first two terms.
\begin{align*} g(x)\amp=\highlight{5}\left(x^2+4x\right)+25 \end{align*}
Now we will complete the square inside the parentheses by adding and subtracting $\left(\frac{4}{2}\right)^2=2^2=4\text{.}$
Notice that the constant that we subtracted is inside the parentheses, but it will not be part of our perfect square trinomial. In order to bring it outside, we need to multiply it by $5\text{.}$ We are distributing the $5$ to that term so we can combine it with the outside term.
\begin{align*} g(x)\amp=5\left(\highlight{\left(\unhighlight{x^2+4x+4}\right)}-\highlight{4}\right)+25\\ \amp=5\highlight{\left(\unhighlight{x^2+4x+4}\right)}-\multiplyleft{5}4+25\\ \amp=5\left(x+2\right)^2-20+25\\ \amp=5\left(x+2\right)^2+5 \end{align*}
The vertex is $(-2,5)\text{.}$
Here is an example that includes fractions.
###### Example12.2.12
Write a formula in vertex form for the function $h$ defined by $h(x)=-3x^2-4x-\frac{7}{4}$
Explanation
First, we will factor the leading coefficient out of the first two terms.
\begin{align*} h(x)\amp=-3x^2-4x-\frac{7}{4}\\ \amp=-3\left(x^2+\frac{4}{3}x\right)-\frac{7}{4} \end{align*}
Next, we will complete the square for $x^2+\frac{4}{3}x$ inside the grouping symbols by adding and subtracting the right number. To find that number, we divide the value of $b$ by two and square the result. That looks like:
$$\frac{b}{2}=\frac{\frac{4}{3}}{2}=\frac{4}{3}\cdot\frac{1}{2}=\frac{2}{3}\label{equation-completing-the-square-4-3-over-2}\tag{12.2.7}$$
and then,
$$\left(\frac{2}{3}\right)^2=\frac{2^2}{3^2}=\frac{4}{9}\label{equation-completing-the-square-2-3squared}\tag{12.2.8}$$
Adding and subtracting the value from Equation (12.2.8), we have:
Remember that when completing the square, the expression should always factor with the number found in the first step of the completing-the-square process, Equation (12.2.7).
\begin{align*} \amp=-3\highlight{\left(\unhighlight{x+\frac{2}{3}}\right)^2}+\frac{4}{3}-\frac{7}{4}\\ \amp=-3\left(x+\frac{2}{3}\right)^2+\frac{16}{12}-\frac{21}{12}\\ \amp=-3\left(x+\frac{2}{3}\right)^2-\frac{5}{12} \end{align*}
The vertex is $\left(-\frac{2}{3},-\frac{5}{12}\right)\text{.}$
Completing the square can also be used to find a minimum or maximum in an application.
###### Example12.2.13
In Example 6.4.19, we learned that artist Tyrone's annual income from paintings can be modeled by $I(x)=-100x^2+1000x+20000\text{,}$ where $x$ is the number of times he will raise the price per painting by $20.00\text{.}$ To maximize his income, how should Tyrone set his price per painting? Find the maximum by completing the square. Explanation To find the maximum is essentially the same as finding the vertex, which we can find by completing the square. To complete the square for $I(x)=-100x^2+1000x+20000\text{,}$ we start by factoring out the $-100$ from the first two terms: \begin{align*} I(x)\amp=-100x^2+1000x+20000\\ \amp=-100\left(x^2-10x\right)+20000\\ \end{align*} Next, we will complete the square for $x^2-10x$ by adding and subtracting $\left(-\frac{10}{2}\right)^2=(-5)^2=\highlight{25}\text{.}$ \begin{align*} I(x)\amp=-100\left(x^2-10x\addright{25}\subtractright{25}\right)+20000\\ \amp=-100\left(\highlight{\left(\unhighlight{x^2-10x+25}\right)}-25\right)+20000\\ \amp=-100\highlight{\left(\unhighlight{x^2-10x+25}\right)}-\left(100\cdot-25\right)+20000\\ \amp=-100(x-5)^2+2500+20000\\ \amp=-100(x-5)^2+22500 \end{align*} The vertex is the point $(5,22500)\text{.}$ This implies Tyrone should raise the price per painting $\substitute{5}$ times, which is $\substitute{5}\cdot20=100$ dollars. He would sell $100-5(\substitute{5})=75$ paintings. This would make the price per painting $200+100=300$ dollars, and his annual income from paintings would become$22500$ by this model.
### Subsection12.2.4Graphing Quadratic Functions by Hand
Now that we know how to put a quadratic function in vertex form, let's review how to graph a parabola by hand.
###### Example12.2.14
Graph the function $h$ defined by $h(x)=2x^2+4x-6$ by determining its key features algebraically.
Explanation
To start, we'll note that this function opens upward because the leading coefficient, $2\text{,}$ is positive.
Now we will complete the square to find the vertex. We will factor the $2$ out of the first two terms, and then add and subtract $\left(\frac{2}{2}\right)^2=1^2=\highlight{1}$ on the right side.
\begin{align*} h(x)\amp=2\left(x^2+2x\right)-6\\ \amp=2\left[x^2+2x\addright{1}\subtractright{1}\right]-6\\ \amp=2\left[\highlight{\left(\unhighlight{x^2+2x+1}\right)}-1\right]-6\\ \amp=2\highlight{\left(\unhighlight{x^2+2x+1}\right)}-\left(2\cdot1\right)-6\\ \amp=2\left(x+1\right)^2-2-6\\ \amp=2\left(x+1\right)^2-8 \end{align*}
The vertex is $(-1,-8)$ so the axis of symmetry is the line $x=-1\text{.}$
To find the $y$-intercept, we'll replace $x$ with $0$ or read the value of $c$ from the function in standard form:
\begin{align*} h(\substitute{0})\amp=2(\substitute{0})^2+2(\substitute{0})-6\\ \amp=-6 \end{align*}
The $y$-intercept is $(0,-6)$ and we will find its symmetric point on the graph, which is $(-2,-6)\text{.}$
Next, we'll find the horizontal intercepts. We see this function factors so we will write the factored form to get the horizontal intercepts.
\begin{align*} h(x)\amp=2x^2+4x-6\\ \amp=2\left(x^2+2x-3\right)\\ \amp=2(x-1)(x+3) \end{align*}
The $x$-intercepts are $(1,0)$ and $(-3,0)\text{.}$
Now we will plot all of the key points and draw the parabola.
###### Example12.2.16
Write a formula in vertex form for the function $p$ defined by $p(x)=-x^2-4x-1\text{,}$ and find the graph's key features algebraically. Then sketch the graph.
Explanation
In this function, the leading coefficient is negative so it will open downward. To complete the square we first factor $-1$ out of the first two terms.
\begin{align*} p(x)\amp=-x^2-4x-1\\ \amp=-\left(x^2+4x\right)-1 \end{align*}
Now, we add and subtract the correct number on the right side of the function: $\left(\frac{b}{2}\right)^2=\left(\frac{4}{2}\right)^2=2^2=\highlight{4}\text{.}$
\begin{align*} p(x)\amp=-\left(x^2+4x\addright{4}\subtractright{4}\right)-1\\ \amp=-\left(\highlight{\left(\unhighlight{x^2+4x+4}\right)}-4\right)-1\\ \amp=-\highlight{\left(\unhighlight{x^2+4x+4}\right)}-(-4)-1\\ \amp=-\left(x+2\right)^2+4-1\\ \amp=-\left(x+2\right)^2+3 \end{align*}
The vertex is $(-2,3)$ so the axis of symmetry is the line $x=-2\text{.}$
We find the $y$-intercept by looking at the value of $c\text{,}$ which is $-1\text{.}$ So, the $y$-intercept is $(0,-1)$ and we will find its symmetric point on the graph, $(-4,-1)\text{.}$
The original expression, $-x^2-4x-1\text{,}$ does not factor so to find the $x$-intercepts we need to set $p(x)=0$ and complete the square or use the quadratic formula. Since we just went through the process of completing the square above, we can use that result to save several repetitive steps.
\begin{align*} p(x)\amp=0\\ -\left(x+2\right)^2+3\amp=0\\ -(x+2)^2\amp=-3\\ (x+2)^2\amp=3 \end{align*}
\begin{align*} x+2\amp=-\sqrt{3}\amp\text{or}\amp\amp x+2\amp=\sqrt{3}\\ x\amp=-2-\sqrt{3}\amp\text{or}\amp\amp x\amp=-2+\sqrt{3}\\ x\amp\approx-3.73\amp\text{or}\amp\amp x\amp\approx-0.268 \end{align*}
The $x$-intercepts are approximately $(-3.7,0)$ and $(-0.3,0)\text{.}$ Now we can plot all of the points and draw the parabola.
### Subsection12.2.5Exercises
###### 1
Use a square root to solve ${\left(y-9\right)^{2}}={4}\text{.}$
###### 2
Use a square root to solve ${\left(y+4\right)^{2}}={25}\text{.}$
###### 3
Use a square root to solve ${\left(4y+7\right)^{2}}={4}\text{.}$
###### 4
Use a square root to solve ${\left(9r-4\right)^{2}}={64}\text{.}$
###### 5
Use a square root to solve ${\left(r+3\right)^{2}}={14}\text{.}$
###### 6
Use a square root to solve ${\left(t-4\right)^{2}}={3}\text{.}$
###### 7
Use a square root to solve ${t^{2}+18t+81}={36}\text{.}$
###### 8
Use a square root to solve ${x^{2}+4x+4}={81}\text{.}$
###### 9
Use a square root to solve ${16x^{2}-16x+4}={25}\text{.}$
###### 10
Use a square root to solve ${81y^{2}+108y+36}={9}\text{.}$
###### 11
Use a square root to solve ${36y^{2}-72y+36}={17}\text{.}$
###### 12
Use a square root to solve ${9y^{2}+12y+4}={13}\text{.}$
###### 13
Solve the equation by completing the square.
${r^{2}+2r}={63}$
###### 14
Solve the equation by completing the square.
${r^{2}+10r}={-9}$
###### 15
Solve the equation by completing the square.
${t^{2}-t}={42}$
###### 16
Solve the equation by completing the square.
${t^{2}-3t}={10}$
###### 17
Solve the equation by completing the square.
${x^{2}+4x}={6}$
###### 18
Solve the equation by completing the square.
${x^{2}-8x}={8}$
###### 19
Solve the equation by completing the square.
${y^{2}-6y+5}={0}$
###### 20
Solve the equation by completing the square.
${y^{2}-8y-9}={0}$
###### 21
Solve the equation by completing the square.
${y^{2}+15y+56}={0}$
###### 22
Solve the equation by completing the square.
${r^{2}-r-42}={0}$
###### 23
Solve the equation by completing the square.
${r^{2}-2r-6}={0}$
###### 24
Solve the equation by completing the square.
${t^{2}-10t+3}={0}$
###### 25
Solve the equation by completing the square.
${12t^{2}+28t+15}={0}$
###### 26
Solve the equation by completing the square.
${12x^{2}+20x+7}={0}$
###### 27
Solve the equation by completing the square.
${2x^{2}-x-5}={0}$
###### 28
Solve the equation by completing the square.
${2x^{2}+5x+1}={0}$
###### 29
Consider $h(y)={y^{2}+4y+4}\text{.}$
1. Give the formula for $h$ in vertex form.
2. What is the vertex of the parabola graph of $h\text{?}$
###### 30
Consider $F(t)={t^{2}-6t+1}\text{.}$
1. Give the formula for $F$ in vertex form.
2. What is the vertex of the parabola graph of $F\text{?}$
###### 31
Consider $G(r)={r^{2}+r-2}\text{.}$
1. Give the formula for $G$ in vertex form.
2. What is the vertex of the parabola graph of $G\text{?}$
###### 32
Consider $G(y)={y^{2}+9y-5}\text{.}$
1. Give the formula for $G$ in vertex form.
2. What is the vertex of the parabola graph of $G\text{?}$
###### 33
Consider $H(t)={5t^{2}+25t+5}\text{.}$
1. Give the formula for $H$ in vertex form.
2. What is the vertex of the parabola graph of $H\text{?}$
###### 34
Consider $K(r)={8r^{2}+64r+2}\text{.}$
1. Give the formula for $K$ in vertex form.
2. What is the vertex of the parabola graph of $K\text{?}$
###### 35
Complete the square to convert the quadratic function from standard form to vertex form, and use the result to find the functionâs domain and range.
$f(x) = {x^{2}+20x+94}$
The domain of $f$ is
The range of $f$ is
###### 36
Complete the square to convert the quadratic function from standard form to vertex form, and use the result to find the functionâs domain and range.
$f(x) = {x^{2}+16x+72}$
The domain of $f$ is
The range of $f$ is
###### 37
Complete the square to convert the quadratic function from standard form to vertex form, and use the result to find the functionâs domain and range.
$f(x) = {-x^{2}-10x-34}$
The domain of $f$ is
The range of $f$ is
###### 38
Complete the square to convert the quadratic function from standard form to vertex form, and use the result to find the functionâs domain and range.
$f(x) = {-x^{2}-6x-16}$
The domain of $f$ is
The range of $f$ is
###### 39
Complete the square to convert the quadratic function from standard form to vertex form, and use the result to find the functionâs domain and range.
$f(x) = {5x^{2}+10x+12}$
The domain of $f$ is
The range of $f$ is
###### 40
Complete the square to convert the quadratic function from standard form to vertex form, and use the result to find the functionâs domain and range.
$f(x) = {3x^{2}-12x+11}$
The domain of $f$ is
The range of $f$ is
###### 41
Complete the square to convert the quadratic function from standard form to vertex form, and use the result to find the functionâs domain and range.
$f(x) = {-4x^{2}+32x-72}$
The domain of $f$ is
The range of $f$ is
###### 42
Complete the square to convert the quadratic function from standard form to vertex form, and use the result to find the functionâs domain and range.
$f(x) = {-2x^{2}+24x-66}$
The domain of $f$ is
The range of $f$ is
###### Sketching Graphs of Quadratic Functions
Graph each function by algebraically determining its key features. Then state the domain and range of the function.
###### 43
$f(x)=x^2-7x+12$
###### 44
$f(x)=x^2+5x-14$
###### 45
$f(x)=-x^2-x+20$
###### 46
$f(x)=-x^2+4x+21$
###### 47
$f(x)=x^2-8x+16$
###### 48
$f(x)=x^2+6x+9$
###### 49
$f(x)=x^2-4$
###### 50
$f(x)=x^2-9$
###### 51
$f(x)=x^2+6x$
###### 52
$f(x)=x^2-8x$
###### 53
$f(x)=-x^2+5x$
###### 54
$f(x)=-x^2+16$
###### 55
$f(x)=x^2+4x+7$
###### 56
$f(x)=x^2-2x+6$
###### 57
$f(x)=x^2+2x-5$
###### 58
$f(x)=x^2-6x+2$
###### 59
$f(x)=-x^2+4x-1$
###### 60
$f(x)=-x^2-x+3$
###### 61
$f(x)=2x^2-4x-30$
###### 62
$f(x)=3x^2+21x+36$
###### 63
Find the minimum value of the function
\begin{equation*} f(x)=10x^{2}-x+1 \end{equation*}
###### 64
Find the minimum value of the function
\begin{equation*} f(x)=x^{2}-9x+10 \end{equation*}
###### 65
Find the maximum value of the function
\begin{equation*} f(x)=5x-2x^{2}-2 \end{equation*}
###### 66
Find the maximum value of the function
\begin{equation*} f(x)=6-\left(3x^{2}+2x\right) \end{equation*}
###### 67
Find the range of the function
\begin{equation*} f(x)=-\left(4x^{2}+10x+6\right) \end{equation*}
###### 68
Find the range of the function
\begin{equation*} f(x)=4x-5x^{2}+3 \end{equation*}
###### 69
Find the range of the function
\begin{equation*} f(x)=6x^{2}-3x-9 \end{equation*}
###### 70
Find the range of the function
\begin{equation*} f(x)=7x^{2}+10x-1 \end{equation*}
###### 71
If a ball is throw straight up with a speed of $66\ {\textstyle\frac{\rm\mathstrut ft}{\rm\mathstrut s}}\text{,}$ its height at time $t$ (in seconds) is given by
\begin{equation*} h(t)=-8t^{2}+66t+2 \end{equation*}
Find the maximum height the ball reaches.
###### 72
If a ball is throw straight up with a speed of $68\ {\textstyle\frac{\rm\mathstrut ft}{\rm\mathstrut s}}\text{,}$ its height at time $t$ (in seconds) is given by
\begin{equation*} h(t)=-8t^{2}+68t+2 \end{equation*}
Find the maximum height the ball reaches.
###### 73
Let $f(x) = x^{2}+bx+c\text{.}$ Let $b$ and $c$ be real numbers. Complete the square to find the vertex of $f(x) = x^{2}+bx+c\text{.}$ Write $f(x)$ in vertex form and then state the vertex.
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# Consecutive Integers Formula
In mathematics, you often come across problem statements wherein you are required to find two or more consecutive integers if their sum or difference is given to you. A number of times the clause of these consecutive numbers being odd or even is also added. Before simplifying the problem statement and forming its equation, you are supposed to use a variable for one integer and then represent it in the form of its consecutive integers. Let me explain this with the help of an example.
Say you have to find two consecutive integers whose sum is 89. How do you go about this problem? You first take a variable, say x, the value of which is unknown to you. Next, you are supposed to take another number. Since the problem requires you to use two consecutive integers, the integer next to x will be (x + 1). Now as per the problem, the sum of x and (x+1) is 89. We represent this in the form of an equation: x + (x + 1) = 89. Solving this equation we get x as 44 and the next integer (x + 1) as 45, the sum of which is 89.
Similarly, knowing the consecutive integer formula finds application in a number of mathematical problems. In this section, we will go through some of these formulas.
## Consecutive Integer Formula
If n is an integer, (n + 1) and (n + 2) will be the next two consecutive integers. For example, let n be 1. We find its consecutive integers as (1 + 1) and (1 + 2), or 2 and 3.
Hence, the formula:
n, n+1, n+2, n+3,…
### Even Consecutive Integer Formula
In mathematics, we represent an even integer as 2n. If 2n is an even integer, (2n + 2) and (2n + 4) will be the next two even consecutive integers. For example, let 2n be 4, which is an even integer. We find its consecutive integers as (4 + 2) and (4 + 4), or 6 and 8.
Hence, the formula:
2n, 2n+2, 2n+4, 2n+6,…
Note that the difference between two even consecutive integers here is 2, otherwise we would end up with an integer which is consecutive but not even.
### Odd Consecutive Integer Formula
In mathematics, we represent an odd integer as 2n + 1. If 2n + 1 is an odd integer, (2n + 3) and (2n + 5) will be the next two odd consecutive integers. For example, let 2n + 1 be 7, which is an odd integer. We find its consecutive integers as (7 + 2) and (7 + 4), or 9 and 11.
Hence, the formula:
2n+1, 2n+3, 2n+5, 2n+7,…
Note that the difference between two odd consecutive integers here is 2, otherwise we would end up with an integer which is consecutive but not odd.
#### Solved Examples
Question: Find three consecutive integers of 76.
Solution:
Let 76 be n. So the next three consecutive integers will be n + 1, n + 2 and n + 3.
76 + 1, 76 + 2 and 76 + 3 or 77, 78 and 79
Therefore, we have 76, 77, 78, 79
Question: Find three consecutive even integers of -8.
Solution:
Let -8 be 2n. So the next three consecutive integers will be 2n + 2, 2n + 4 and 2n + 6.
-8 + 2, -8 + 4 and -8 + 6 or -6, -4 and -2
Therefore, we have -8, -6, -4, -2
To solve more problems on the topic, download Byju’s -The Learning App.
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Definition and examples Joint variation | define Joint variation - Free Math Dictionary Online
# JOINT VARIATION
## Definition of Joint Variation
Joint variation is the same as direct variation with two or more quantities.
That is:
Joint variation is a variation where a quantity varies directly as the product of two or more other quantities
Let's first understand direct variation
Direct variation occurs when two quantities change in the same manner
That is:
Increase in one quantity causes an increase in the other quantity
Decrease in one quantity causes a decrease in the other quantity
For Example:
The cost of a pencil and the number of pencils you buy.
Direct variation between variables x and y can be expressed as:
y = kx, where 'k' is the constant of variation and k ≠ 0
y = kxz represents joint variation. Here, y varies jointly as x and z.
### More Examples on Joint Variation
y = 7xz, here y varies jointly as x and z
y = 7x2z3, here y varies jointly as x2 and z3
Area of a triangle = is an example of joint variation. Here the constant is 1. Area of a triangle varies jointly with base 'b' and height 'h'
Area of a rectangle = L x M represents joint variation. Here the constant is 1. Area of a rectangle varies jointly with length 'l' and width 'w'.
### Solved Example on Joint Variation
#### Solution:
• Step 1: First set up the equation. a varies jointly with b and c. a = kbc
• Step 2: Find the value of the constant, k. Given that a = 12 when b = 1 and c = 6
a = kbc
12 = k x 1 x 6
⇒ k = 2
• Step 3: Rewrite the equation using the value of the constant 'k'
a = 2bc
• Step 4: Using the new equation, find the missing value.
If b = 2 and c = 3, then a = 2 x 2 x 3 = 12
• Step 5: So, when a varies jointly with b and c and If b = 2 and c = 3, then the value of a is 12.
#### Real-world Connections for Joint Variation
Force = mass × acceleration. The force exerted on an object varies jointly as the mass of the object and the acceleration produced.
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# Significant Figures
Lesson
In 2013 I attended the Liverpool friendly football game at the MCG in Melbourne, Australia. The crowd size was the second largest on record ever, reaching $95446$95446. When I saw a friend of mine soon after the game, she said a newspaper article had reported $95000$95000 crowd size and asked me what the atmosphere was like.
Who is right here? Is $95000$95000 accurately reported? Well yes it is if we consider significant figures. In fact $95400$95400, and $100000$100000 are also correct depending on the level of accuracy you want to use.
## Aren't all numbers significant?
Well yes, all the digits in numbers have a particular role to play. They tell us the value of the number, the size of the number or individual place values. Zeros hold the place value in numbers so that we can read and understand them.
The concept of significant figures is about representing a number with a certain level of accuracy.
Digits that are significant are:
• all nonzero digits
• zeros appearing between two nonzero digits (holding the place value for us)
• trailing zeros in a number containing a decimal point.
• all digits in a number using scientific notation ($a\times10^n$a×10n)
Here is a demonstration of the first three points:
## Returning to the MCG
So let's look at the crowd size at the football game I attended.
$95446$95446
- to 1 significant figure would be $100000$100000
(the first significant figure in the number is $9$9, so we are rounding to the nearest ten thousand)
- to 2 significant figures would be $95000$95000
(the second significant figure in the number is $5$5, so we are rounding to the nearest thousand)
- to 3 significant figures would be $95400$95400
(the third significant figure in the number is $4$4, so we are rounding to the nearest hundred)
- to 4 significant figures would be $95450$95450
(the fourth significant figure in the number is $4$4, so we are rounding to the nearest ten)
- to 5 significant figures would be $95446$95446
(the fifth significant figure in the number is $6$6, so we are rounding to the nearest one)
## Other significant ideas
So for numbers larger than $0$0, using significant figures is about learning how to round a number.
The width of a human hair is known to be about $0.023$0.023 cm.
Let's see what this measurement becomes if we round to significant figures:
- to 1 significant figures it would be $0.02$0.02
(the first significant figure is $2$2, so we are rounding to the nearest hundredth)
- to 2 significant figures it would be $0.023$0.023
(the second significant figure is $3$3so we are rounding to the nearest thousandth. See how in this case the leading zeros do not count!)
#### Examples
Write: the following numbers to the indicated level of significant figures
Think: Remember the rules to identify significant figures,
• all nonzero digits
• zeros appearing between two nonzero digits (holding the place value for us)
• trailing zeros in a number containing a decimal point.
• all digits in a number using scientific notation ($a\times10^n$a×10n)
Do
##### Question 1
$10432$10432 to $3$3 significant figures = $10400$10400 The zero between the $1$1 and the $4$4 is counted as significant.
##### Question 2
$1.040052$1.040052 to $3$3 significant figures = $1.04$1.04 The zero between the $1$1 and $4$4 is counted as significant.
##### Question 3
$6.53126\times10^7$6.53126×107 to $4$4 significant figures = $6.531\times10^7$6.531×107 We only need $4$4 of the digits.
##### Question 4
$6.00002\times10^8$6.00002×108 to $3$3 significant figures = $6.00\times10^8$6.00×108 See how we have to write the zeros here - all the digits count as significant in scientific notation, even the zeros.
#### More Worked Examples
##### QUESTION 5
How many significant figures are there in the number 108 486?
1. Three
A
Four
B
Five
C
Six
D
Three
A
Four
B
Five
C
Six
D
##### QUESTION 6
Round off $461585$461585 to three significant figures.
1. $\editable{}$
##### QUESTION 7
Round off $0.006037736$0.006037736 to two significant figures.
### Outcomes
#### NA5-6
Know and apply standard form, significant figures, rounding, and decimal place value
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# How do you find the vertical, horizontal or slant asymptotes for y=(x^2-5x+4)/ (4x^2-5x+1)?
Sep 20, 2017
$\text{vertical asymptote at } x = \frac{1}{4}$
$\text{horizontal asymptote at } y = \frac{1}{4}$
#### Explanation:
$\text{factorise and simplify}$
$y = \frac{\cancel{\left(x - 1\right)} \left(x - 4\right)}{\left(4 x - 1\right) \cancel{\left(x - 1\right)}} = \frac{x - 4}{4 x - 1}$
$\text{the removal of the factor "(x-1)" from the}$
$\text{numerator/denominator indicates a hole at x = 1}$
The denominator of y cannot be zero as this would make y undefined. Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.
$\text{solve "4x-1=0rArrx=1/4" is the asymptote}$
$\text{horizontal asymptotes occur as}$
${\lim}_{x \to \pm \infty} , y \to c \text{ ( a constant)}$
$\text{divide terms on numerator/denominator by x}$
$y = \frac{\frac{x}{x} - \frac{4}{x}}{\frac{4 x}{x} - \frac{1}{x}} = \frac{1 - \frac{4}{x}}{4 - \frac{1}{x}}$
as $x \to \pm \infty , y \to \frac{1 - 0}{4 - 0}$
$\Rightarrow y = \frac{1}{4} \text{ is the asymptote}$
Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here hence there is no slant asymptote.
graph{(x-4)/(4x-1) [-10, 10, -5, 5]}
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# NCERT Solutions for Class 6 Maths Chapter 10 Mensuration Ex 10.1
These NCERT Solutions for Class 6 Maths Chapter 10 Mensuration Ex 10.1 Questions and Answers are prepared by our highly skilled subject experts.
## NCERT Solutions for Class 6 Maths Chapter 10 Mensuration Exercise 10.1
Question 1.
Find the perimeter of each of the following figures:
(a) Perimeter = Sum of all the sides = 4 cm + 2 cm + 1 cm + 5 cm = 12 cm
(b) Perimeter = Sum of all the sides = 23 cm + 35 cm + 40 cm + 35 cm = 133 cm
(c) Perimeter = Sum of all the sides =15 cm + 15 cm + 15 cm + 15 cm = 60 cm
(d) Perimeter = Sum of all the sides = 4 cm + 4 cm + 4 cm + 4 cm + 4 cm = 20 cm
(e) Perimeter = Sum of all the sides 1 cm + 4 cm + 0.5 cm + 2.5 cm + 2.5 cm + 0.5 cm + 4 cm = 15 cm
(f) Perimeter = Sum of all the sides = 4
cm + 1 cm + 3 cm + 2 cm + 3 cm + 4
cm + 1 cm + 3 cm + 2 cm + 3 cm + 4
cm + 1 cm + 3 cm + 2 cm + 3 cm + 4
cm + 1 cm + 3 cm + 2 cm + 3 cm = 52 cm
Question 2.
The lid of a rectangular box of sides 40 cm by 10 cm is sealed all round with tape. What is the length of the tape required?
Total length of tape required
= Perimeter of rectangle
= 2 (length + breadth) = 2 (40 + 10)
= 2 × 50 = 100 cm = 1 m
Thus, the total length of tape required is 100 cm or 1 m.
Question 3.
A table-top measures 2 m 25 cm by 1 m 50 cm. What is the perimeter of the table-top?
Length of table top = 2 m 25 cm = 2.25 m
Breadth of table top = 1 m 50 cm = 1.50 m
Perimeter of table top = 2 × (length + breadth) = 2 × (2.25 + 1.50) = 2 × 3.75 = 7.5 m
Thus, the perimeter of tabletop is 7.5 m.
Question 4.
What is the length of the wooden strip required to frame a photograph of length and breadth 32 cm and 21 cm respectively?
Length of wooden strip = Perimeter of photograph
Perimeter of photograph
= 2 x (length + breadth)
= 2 (32 + 21) = 2 × 53cm =106 cm
Thus, the length of the wooden strip required is equal to 106 cm.
Question 5.
A rectangular piece of land measures 0.7 km by 0.5 km. Each side is to be fenced with 4 rows of wires. What is the length of the wire needed?
Since the 4 rows of wires are needed.
Therefore the total length of wires is equal to 4 times the perimeter of rectangle.
Perimeter of field = 2 × (length + breadth)
= 2 × (0.7 + 0.5) = 2 × 1.2 = 2.4 km
= 2.4 × 1000 m = 2400 m
Thus, the length of wire = 4 × 2400 = 9600 m = 9.6 km
Question 6.
Find the perimeter of each of the following shapes:
(a) A triangle of sides 3 cm, 4 cm and 5 cm.
(b) An equilateral triangle of side 9 cm.
(c) An isosceles triangle with equal sides 8 cm each and third side 6 cm.
(a) Perimeter of ΔABC = AB + BC + CA = 3 cm + 5 cm + 4 cm = 12 cm
(b) Perimeter of equilateral ABC
= 3 × side = 3 × 9 cm = 27 cm
(c) Perimeter of AABC = AB + BC + CA = 8 cm + 6 cm + 8 cm = 22 cm
Question 7.
Find the perimeter of a triangle with sides measuring 10 cm, 14 cm and 15 cm.
Perimeter of triangle
= Sum of all three sides
= 10 cm + 14 cm + 15 cm = 39 cm
Thus, the perimeter of triangle is 39 cm
Question 8.
Find the perimeter of a regular hexagon with each side measuring 8 m.
Perimeter of Hexagon
= 6 × length of one side
= 6 × 8 m = 48 m
Thus, the perimeter of hexagon is 48 m.
Question 9.
Find the side of the square whose perimeter is 20 m.
Perimeter of square = 4 × side
⇒ 20 = 4 × side
⇒ Side = $$\frac{20}{4}$$ = 5 cm
Thus, the side of square is 5 cm.
Question 10.
The perimeter of a regular pentagon is 100 cm. How long is its each side?
Perimeter of regular pentagon = 100 cm
⇒ 5 × side = 100 cm
⇒ Side = $$\frac{100}{5}$$ = 20 cm
Thus, the side of regular pentagon is 20 cm.
Question 11.
A piece of string is 30 cm long. What will be the length of each side if the string is used to form:
(a) a square?
(b) an equilateral triangle?
(c) a regular hexagon?
Length of string = Perimeter of each figure
(a) Perimeter of square = 30 cm
⇒ 4 × side = 30 cm
⇒ Side = $$\frac{30}{4}$$ = 7.5 cm 4
Thus, the length of each side of square is 7.5 cm.
(b) Perimeter of equilateral triangle = 30 cm
⇒ 3 × side = 30 cm
⇒ Side = $$\frac{30}{3}$$ = 10 cm
Thus, the length of each side of equilateral triangle is 10 cm.
(c) Perimeter of hexagon = 30 cm
⇒ 6 × side = 30 cm
⇒ Side = $$\frac{30}{6}$$ = 5 cm
Thus, the side of each side of hexagon is 5 cm.
Question 12.
Two sides of a triangle are 12 cm and 14 cm. The perimeter of the triangle is 36 cm. What is its third side?
Let the length of third side be X cm.
Length of other two side are 12 cm and 14 cm.
Now, Perimeter of triangle = 36 cm
⇒ 12 +14 + X = 36
⇒ 26 + X = 36
⇒ X = 36 – 26 = 10 cm
Thus, the length of third side is 10 cm
Question 13.
Find the cost of fencing a square park of side 250 m at the rate of ₹ 20 per metre.
Side of square = 250 m
Perimeter of square
= 4 × side = 4 × 250 = 1000 m
Since, cost of fencing of per meter = ₹20
Therefore, the cost of fencing of 1000 meters = 20 × 1000 = ₹ 20,000
Question 14.
Find the cost of fencing a rectangular park of length 175 m and breadth 125 m at the rate of × 12 per metre.
Length of rectangular park = 175 m
Breadth of rectangular park = 125 m
Perimeter of park
= 2 × (length + breadth)
= 2 × (175 + 125) = 2 × 300 = 600 m
Since, the cost of fencing park per meter = ₹ 12
Therefore, the cost of fencing park of 600 m = 12 × 600 = ₹ 7,200
Question 15.
Sweety runs around a square park of side 75 m. Bulbul runs around a rectangular park with length 60 m and breadth 45 m. Who covers less distance?
Distance covered by Sweety
= Perimeter of square park
Perimeter of square = 4 × side
= 4 × 75 = 300 m
Thus, distance covered by Sweety is 300 m. Now, distance covered by Bulbul =
Perimeter of rectangular park
Perimeter of rectangular park
= 2 × (length + breadth)
= 2 × (60 + 45) = 2 × 105 = 210 m
Thus, Bulbul covers the distance of 210 m and Bulbul covers less distance.
Question 16.
What is the perimeter of each of the following figures? What do you infer from the answers?
(a) Perimeter of square
= 4 × side = 4 × 25 = 100 cm
(b) Perimeter of rectangle
= 2 × (length + breadth)
= 2 × (40 + 10) = 2 × 50 = 100 cm
(c) Perimeter of rectangle
= 2 × (length + breadth)
= 2 × (30 + 20) = 2 × 50 = 100 cm
(d) Perimeter of triangle = Sum of all sides
= 30 cm + 30 cm + 40 cm = 100 cm
Thus, all the figures have same perimeter.
Question 17.
Avneet buys 9 square paving slabs, each with a side of 1/2 m. He lays them in the form of a square.
(a) What is the perimeter of his arrangement [Fig (i)]?
(b) Shari does not like his arrangement. She gets him to lay them out like a cross. What is the perimeter of her arrangement [(Fig (ii)]?
(c) Which has greater perimeter?
(d) Avneet wonders if there is a way of getting an even greater perimeter. Can you find a way of doing this? (The paving slabs must meet along complete edges i.e. they cannot be broken.)
(a) The arrangement so formed in Fig (i) is a square of side equal to 3 × side of the square slab
∴ Perimeter of his arrangement in Fig (i)
= 4 × ( 3 × side of the square slab)
= 4 × ( 3 × 0.5) m = 4 × 1.5 m = 6 m
(b) The cross arrangement so formed in Fig (ii) by 9 square slabs, has perimeter, which consists of 20 length segments each equal to side of I a square slab i.e. 0.5 m
∴ Perimeter of her arrangement in Fig (ii)
= 20 × ( side of the square slab)
= 20 × 0.5 m = 10 m
(c) ∵ 10m > 6m
∴ Cross arrangement has greater perimeter
(d) ∵ Total number of tiles = 9
∴ The following arrangement will also have greater perimeter.
Since, this arrangement is in the form of rectangle with length and breadth as $$\frac { 9 }{ 2 }$$ m and $$\frac { 1 }{ 2 }$$ respectively.
∴ Perimeter = 2 $$\left(\frac{9}{2}+\frac{1}{2}\right)$$
= 2 × = 10m $$\left[\frac{10}{2}\right]$$ = 10m
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# A Beginner’s Guide to Descriptive Statistics
Descriptive statistics is a branch of statistics that involves the collection, organization, analysis, interpretation, and presentation of data. It provides a way to summarize and describe the main features of a dataset, allowing us to gain insights and make informed decisions. Whether you are a student, researcher, or professional in any field, understanding descriptive statistics is essential for data analysis and decision-making. In this beginner’s guide, we will explore the key concepts and techniques of descriptive statistics, providing you with a solid foundation to analyze and interpret data effectively.
## 1. What is Descriptive Statistics?
Descriptive statistics is the branch of statistics that focuses on summarizing and describing the main features of a dataset. It involves the collection, organization, analysis, interpretation, and presentation of data. The goal of descriptive statistics is to provide a concise and meaningful summary of the data, allowing us to understand its characteristics and draw conclusions.
Descriptive statistics can be used to describe various aspects of a dataset, including central tendency, variability, distribution, and correlation. By analyzing these characteristics, we can gain insights into the data and make informed decisions.
### 1.1 Central Tendency
Central tendency refers to the measure that represents the center or average of a dataset. It provides a single value that summarizes the entire dataset. The three commonly used measures of central tendency are:
• Mean: The mean is calculated by summing all the values in the dataset and dividing by the number of observations. It is affected by extreme values and is sensitive to outliers.
• Median: The median is the middle value in a dataset when it is arranged in ascending or descending order. It is less affected by extreme values and is a robust measure of central tendency.
• Mode: The mode is the value that appears most frequently in a dataset. It is useful for categorical or discrete data.
For example, consider a dataset of exam scores: 80, 85, 90, 95, 100. The mean is (80 + 85 + 90 + 95 + 100) / 5 = 90, the median is 90, and the mode is not applicable as there are no repeated values.
### 1.2 Variability
Variability measures the spread or dispersion of a dataset. It provides information about how the data points are distributed around the measures of central tendency. The three commonly used measures of variability are:
• Range: The range is the difference between the maximum and minimum values in a dataset. It is affected by extreme values.
• Variance: The variance measures the average squared deviation from the mean. It provides a measure of how spread out the data points are. A higher variance indicates greater variability.
• Standard Deviation: The standard deviation is the square root of the variance. It provides a measure of the average distance between each data point and the mean. A higher standard deviation indicates greater variability.
For example, consider a dataset of daily temperatures: 25, 28, 30, 27, 26. The range is 30 – 25 = 5, the variance is ((25-27)^2 + (28-27)^2 + (30-27)^2 + (27-27)^2 + (26-27)^2) / 5 = 2.8, and the standard deviation is √2.8 ≈ 1.67.
### 1.3 Distribution
Distribution refers to the way the data is spread out or distributed across different values. It provides insights into the shape, symmetry, and skewness of the dataset. The three commonly encountered types of distributions are:
• Normal Distribution: Also known as the bell curve, the normal distribution is symmetric and characterized by a peak at the mean. Many natural phenomena follow a normal distribution.
• Skewed Distribution: A skewed distribution is asymmetric and has a longer tail on one side. It can be either positively skewed (tail on the right) or negatively skewed (tail on the left).
• Uniform Distribution: A uniform distribution is characterized by a constant probability for each value within a given range. It is flat and has no peaks or valleys.
For example, consider a dataset of heights: 160, 165, 170, 175, 180. The distribution is approximately normal, with a peak around the mean height.
### 1.4 Correlation
Correlation measures the strength and direction of the relationship between two variables. It provides insights into how changes in one variable are related to changes in another variable. The correlation coefficient ranges from -1 to 1, where:
• A correlation coefficient of -1 indicates a perfect negative correlation, meaning that as one variable increases, the other variable decreases.
• A correlation coefficient of 0 indicates no correlation, meaning that there is no relationship between the variables.
• A correlation coefficient of 1 indicates a perfect positive correlation, meaning that as one variable increases, the other variable also increases.
For example, consider a dataset of study hours and exam scores. If there is a strong positive correlation between study hours and exam scores, it means that as study hours increase, exam scores also increase. Conversely, if there is a strong negative correlation, it means that as study hours increase, exam scores decrease.
## 2. Data Collection and Organization
Before analyzing data, it is important to collect and organize it in a systematic manner. This ensures that the data is reliable, complete, and ready for analysis. Here are some key steps in data collection and organization:
### 2.1 Define the Research Question
Start by clearly defining the research question or objective. What do you want to investigate or analyze? This will guide the data collection process and help you determine the relevant variables and data sources.
### 2.2 Determine the Data Sources
Identify the sources from which you will collect the data. This could include surveys, experiments, observations, existing databases, or secondary sources. Ensure that the data sources are reliable and provide accurate information.
### 2.3 Select the Sample
If the dataset is large, it may be impractical to collect data from the entire population. In such cases, you can select a representative sample that reflects the characteristics of the population. Random sampling techniques can help ensure that the sample is unbiased and representative.
### 2.4 Collect the Data
Collect the data according to the defined research question and data sources. This may involve conducting surveys, experiments, or observations. Ensure that the data collection process is standardized and consistent to minimize errors and biases.
### 2.5 Organize the Data
Once the data is collected, it needs to be organized in a structured format for analysis. This typically involves creating a spreadsheet or database where each variable is assigned a column and each observation is assigned a row. Ensure that the data is labeled and properly formatted for easy analysis.
## 3. Data Analysis and Interpretation
After collecting and organizing the data, the next step is to analyze and interpret it. Descriptive statistics provides a range of techniques to summarize and describe the data. Here are some key techniques for data analysis and interpretation:
### 3.1 Descriptive Measures
Descriptive measures, such as measures of central tendency and variability, provide a summary of the data. They help us understand the typical values, spread, and distribution of the dataset. By calculating these measures, we can gain insights into the data and make comparisons.
### 3.2 Frequency Distribution
A frequency distribution is a table or graph that shows the number of times each value or range of values occurs in a dataset. It provides a visual representation of the distribution of the data. Histograms and bar charts are commonly used to display frequency distributions.
### 3.3 Graphical Representation
Graphical representation is a powerful tool for visualizing and interpreting data. It allows us to identify patterns, trends, and outliers in the dataset. Common types of graphs include histograms, bar charts, line graphs, scatter plots, and box plots.
### 3.4 Inferential Statistics
Inferential statistics involves making inferences or predictions about a population based on a sample. It uses probability theory and statistical models to estimate population parameters and test hypotheses. Inferential statistics allows us to draw conclusions beyond the observed data.
### 3.5 Data Visualization
Data visualization is the process of presenting data in a visual format, such as charts, graphs, or maps. It enhances the understanding and interpretation of data by making complex information more accessible and intuitive. Effective data visualization can reveal patterns, trends, and relationships that may not be apparent in raw data.
## 4. Practical Examples
To illustrate the concepts of descriptive statistics, let’s consider some practical examples:
### 4.1 Example 1: Exam Scores
Suppose we have a dataset of exam scores for a class of students:
• 80, 85, 90, 95, 100
The mean score is (80 + 85 + 90 + 95 + 100) / 5 = 90, the median score is 90, and there is no mode as there are no repeated values. The range is 100 – 80 = 20, the variance is ((80-90)^2 + (85-90)^2 + (90-90)^2 + (95-90)^2 + (100-90)^2) / 5 = 50, and the standard deviation is √50 ≈ 7.07. The distribution is approximately normal, with a peak around the mean score.
### 4.2 Example 2: Heights
Consider a dataset of heights for a group of individuals:
• 160, 165, 170, 175, 180
The mean height is (160 + 165 + 170 + 175 + 180) / 5 = 170, the median height is 170, and there is no mode as there are no repeated values. The range is 180 – 160 = 20, the variance is ((160-170)^2 + (165-170)^2 + (170-170)^2 + (175-170)^2 + (180-170)^2) / 5 = 50, and the standard deviation is √50 ≈ 7.07. The distribution is approximately normal, with a peak around the mean height.
## 5. Conclusion
Descriptive statistics is a fundamental tool for summarizing, analyzing, and interpreting data. It provides valuable insights into the characteristics of a dataset, allowing us to make informed decisions and draw meaningful conclusions. By understanding the concepts and techniques of descriptive statistics, you can effectively analyze and interpret data in various fields, from research and academia to business and finance. Remember to consider the measures of central tendency, variability, distribution, and correlation when analyzing data, and use appropriate graphical representations and inferential statistics for deeper insights. With a solid foundation in descriptive statistics, you can unlock the power of data and make data-driven decisions.
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# Introducing Multiplication
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Keywords: Multiplication, Repeated Addition, Manipulatives Subject(s): Math Grades 2 through 4 School: Chimneyrock Elementary School, Cordova, TN Planned By: Amanda Loyacano Original Author: Amanda Loyacano, Cordova
Lesson Sequence – Introductory (This will be the students’ first experience with multiplication this school year.)
Assessment
-Pre: At the beginning of the year, students were given a second grade skills test as well as the ThinkLink test of second grade skills. I have used both of these test data to obtain information about the level of student understanding of multiplication.
-Post: Students will be assessed informally during the lesson. Later in the week they will receive a formal assessment to determine their level of skill in multiplication concepts.
Instructional Objectives
TLW understand the concepts that multiplication is repeated addition.
TLW use concrete patterns to predict a product.
SPI: 3.1.spi.16 Use the multiplication facts 0, 1, 2, 5, and 10 efficiently & accurately.
Procedure - Anticipatory Set
Two volunteers will be asked to come to the front and hold out their hands, palms up. Next, I will count out 2 counters into each student’s hand. Then, I will ask the students how they can find the total number of counters in the four hands. This discussion will lead into the actual lesson.
Procedure – Multiplication Introduction
For the discussion, some ideas they may come up with are: counting the counters by ones (1+1+1+1+1+1+1+1=8), counting the counters by twos (2 + 2 + 2 + 2 = 8), etc. Counting by 2’s is the goal for them to recognize.
Next, I will show students how we can write the repeated addition sentence as a multiplication sentence, 2 + 2 + 2 + 2 = 8 is the same as 4 x 2 = 8 (read as 4 times 2 equals 8 & 4 groups off 2 equals 8 counters total). We will then discuss which way is easier to write, repeated addition or multiplication so that students can better understand why we use multiplication.
Finally, I will ask students how they could record this example on graph paper. I will use a transparency of graph paper to test out their ideas.
Procedure – Supervised Practice
First, I will pass out packs of 1-8 index cards, counters, and graph paper to each group of students. Next, we will do an example together. A student will pick a number between 1 and 10. For example, if the number is 3 then each group will place 3 counters on each of their cards, write a repeated addition problem, a multiplication problem, and show it on the graph paper. So, depending on how many index cards they have, each group will arrive at a different multiplication problem. I will write their multiplication sentences on the board so that we can keep a chart of what we have found.
We will share after each example and observe what patterns we notice. We will do this example several times using different numbers with students gaining independence as we go along. If there is time, groups may swap the number of index cards they have to try out another group of numbers.
Closure
To wrap up the lesson we will discuss the patterns that students found. Also, I will pose the question of what would the product be if we had zero index cards because 0’s will be the first group of multiplication facts that we will learn.
Meeting Individual Students’ Needs
* I will observe students’ level of understanding through questioning & monitoring of progress during the activity:
-Reteach/Remediation: I will use guided questions to help students discover how repeated addition is the same as multiplication. “How many groups of you have? How many are in each group? What is the total? What addition and multiplication sentences could we use to show this?” Also, I will use consistent language: “3 groups of 2 added together equals 6, just as 3groups times 2 groups equals 6.”
-Enrichment: I will give higher-level groups of students a larger number of index cards or an amount that they cannot easily count by (4 & 6) compared to groups that may have more difficulty who will receive the number of cards that are easier to count by (1, 2, 3, & 5).
-Independent Practice: Students will independently complete examples on their own using the index cards, counters, and graph paper. Sheets will be collected at the end of the lesson to monitor their level of understanding.
Materials Use
Overhead & Graph Paper Transparency To show students how to draw examples
Index cards To show groups
Counters To show individual parts of the group
Graph Paper To show pictures & number sentences
Comments This is a great way for kinesthetic & visual learners to understand basic multiplication concepts. Follow-Up Have students create their own multiplication sentences and picture arrays. Materials: Worksheets Other Items: 1 Bag of Math Counters, \$10.00 each, total of \$10.001 Pack of Large Index Cards, \$3.00 each, total of \$3.001 Pack of Graph Paper, \$2.00 each, total of \$2.00
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# A tea party is arranged for 16 people along two sides of a long table with 8 chairs on each side. Four men wish to sit on one particular side and two on the other side, in how many ways can they be seated?
Last updated date: 13th Jul 2024
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Hint – In this question we have in total 16 seats with 8 on each side of the table. Firstly make these 4+2=6 people sit who have special demand in sitting arrangements. Now when these 6 people are taken care of we are left with 10 people and 4 seats on one side and 6 seats on the other side, use this concept to get the answer.
There are 16 people for the tea party.
People sit along a long table with 8 chairs on each side.
Out of 16, 4 people sit on a particular side and 2 sit on the other side.
Therefore first we will make sitting arrangements for those 6 persons who want to sit on some specific side.
Now we have remaining (16 - 6) =10 persons to arrange and out of 10, six people can sit on one side as only 6 seats will be reaming after making 2 people sit on one side on special demand and 4 people on other side as 4 people are already being seated on one side on special demand. (Take into consideration that one side has only 8 seats)
The number of ways of choosing 6 people out of 10 are ${}^{10}{C_6}$, now there are 4 people remaining and they will automatically place in those 4 seats available on the other side therefore total arrangement for those four people are ${}^4{C_4} = 1$.
And now all the 16 people are placed in their seats according to the constraints.
Now we have to arrange them.
So, the number of ways of arranging 8 people out of 16 on one side and the rest 8 people on other side is$\left( {8! \times 8!} \right)$.
So, a possible number of arrangements will be $\Rightarrow {}^{10}{C_6} \times {}^4{C_4} \times 8! \times 8!$
Now as we know${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$.
So, ${}^{10}{C_6} = \dfrac{{10!}}{{4! \times 6!}} = \dfrac{{10.9.8.7.6!}}{{4.3.2.1.6!}} = 210$
And ${}^4{C_4} = \dfrac{{4!}}{{4! \times 0!}} = 1$
So, total number of arrangements is
$= 210 \times \left( {8! \times 8!} \right)$.
So, this is the required answer.
Note – Whenever we face such types of problems the key point is to make special arrangements for the people who are in need of it, then arrange the remaining. Now combination comes with permutation as there are possibilities of these 8 people sitting on one side to rearrange. Thus this concept into consideration, to get through the answer.
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Class 12 RD Sharma Solutions – Chapter 20 Definite Integrals – Exercise 20.1 | Set 3
Last Updated : 26 May, 2021
Question 45.
Solution:
We have,
I =
Let 2x + 1 = t2, so we have,
=> 2 dx = 2t dt
=> dx = t dt
Now, the lower limit is, x = 1
=> t2 = 2x + 1
=> t2 = 2(1) + 1
=> t2 = 3
=> t = √3
Also, the upper limit is, x = 4
=> t2 = 2x + 1
=> t2 = 2(4) + 1
=> t2 = 9
=> t = 3
So, the equation becomes,
I =
I =
I =
I =
I =
I = 1/4 [35/5 – 3 – (√3)5/5 + √3]
I = 1/4[243/5 – 3 – 9√3/5 + √3]
I = 1/4((243 – 15 – 9√3 + 5√3)/5)
I = 1/4[(228 – 4√3)/5]
I = 1/4[4(57 – √3)/5]
I = (57 – √3)/5
Therefore, the value of is (57 – √3)/5.
Question 46.
Solution:
We have,
I =
By using binomial theorem in the expansion of (1 – x)5, we get,
I =
I =
I =
I =
I = 1/2 – 5/3 + 10/4 – 10/5 + 5/6 – 1/7
I = 1/2 – 5/3 + 5/3 – 2 + 5/6 – 1/7
I = 1/2 – 2 + 5/6 – 1/7
I = 1/42
Therefore, the value of is 1/42.
Question 47.
Solution:
We have,
I =
I =
I =
I =
By using integration by parts, we get,
I =
I =
I = ex/x
So we get,
I =
I = e2/2 – e1/1
I = e2/2 – e
Therefore, the value of is e2/2 – e.
Question 48.
Solution:
We have,
I =
By using integration by parts in first integral, we get,
I =
I = xe2x/2 – (1/2)(e2x/2) + 2/Ï€[1 – 0]
I = xe2x/2 – e2x/4 + 2/Ï€
So we get,
I =
I = [e2/2 + e2/4 – 0 + 1/4] + 2/Ï€
I = e2/4 + 1/4 + 2/Ï€
Therefore, the value of is e2/4 + 1/4 + 2/Ï€.
Question 49.
Solution:
We have,
I =
By using integration by parts in first integral, we get,
I =
I =
I =
I =
So we get,
I =
I =
I = [e1(1 – 1) – e0(0 – 1)] + 2√2/Ï€
I = [0 – (-1)] + 2√2/Ï€
I = 1 + 2√2/π
Therefore, the value of is 1 + 2√2/π.
Question 50.
Solution:
We have,
I =
I =
I =
I =
I = -eπ cotπ/2 + eπ/2 cotπ/4
I = 0 + eπ/2(1)
I = eπ/2
Therefore, the value of is eπ/2.
Question 51.
Solution:
We have,
I =
I =
I =
I =
By using integration by parts in first integral, we get,
I =
I =
I =
I = 1/√2[sinÏ€(2eÏ€) – 0]
I = 1/√2[0 – 0]
I = 0
Therefore, the value of is 0.
Question 52.
Solution:
We have,
I =
By using integration by parts, we get,
I = excos(x/2 + π/4) + 1/2∫exsin(x/2 + π/4)
I = ex cos(x/2 + Ï€/4) + 1/2[ exsin(x/2 + Ï€/4) – 1/2 ∫excos(x/2 + Ï€/4)dx]
I = excos(x/2 + Ï€/4) + 1/2exsin(x/2 + Ï€/4) – 1/4I
5I/4 = -3/ 2√2(e2π + 1)
I = -3√2/5(e2π + 1)
Therefore, the value of is -3√2/5(e2π + 1).
Question 53.
Solution:
We have,
I =
I =
I =
I =
I =
I =
I = 2/3[23/2 – 1] + 2/3[1 – 0]
I =
I = 25/2/3
Therefore, the value of is 25/2/3.
Question 54.
Solution:
We have,
I =
I =
I =
I =
I = -log3 + log2 + 2[log4 – log3]
I = -log3 + log2 + 2[2log2 – log3]
I = -log3 + log2 + 4log2 – 2log3
I = 5log2 – 3log3
I = log25 – log33
I = log32 – log27
I = log32/27
Therefore, the value of is log32/27.
Question 55.
Solution:
We have,
I =
I =
I =
Let cos x = t, so we have,
=> – sin x dx = dt
Now, the lower limit is, x = 0
=> t = cos x
=> t = cos 0
=> t = 1
Also, the upper limit is, x = π/2
=> t = cos x
=> t = cos π/2
=> t = 0
So, the equation becomes,
I =
I =
I =
I = [0 – 1/3] – [0 – 1]
I = [-1/3] – [-1]
I = -1/3 + 1
I = 2/3
Therefore, the value of is 2/3.
Question 56.
Solution:
We have,
I =
I =
I =
I =
I = -sinπ + sin0
I = 0
Therefore, the value of is 0.
Question 57.
Solution:
We have,
I =
Let 2x = t, so we have,
=> 2x dx = dt
Now, the lower limit is, x = 1
=> t = 2x
=> t = 2(1)
=> t = 2
Also, the upper limit is, x = 2
=> t = 2x
=> t = 2(2)
=> t = 4
So, the equation becomes,
I =
I =
I =
By using integration by parts in first integral, we get,
I =
I =
I =
I = e4/4 – e2/2
Therefore, the value of is e4/4 – e2/2.
Question 58.
Solution:
We have,
I =
I =
I =
I =
I =
I =
I = [sin-1(1) – sin-1(-1)]
I = Ï€/2 – (-Ï€/2)
I = π/2 + π/2
I = π
Therefore, the value of is π.
Question 59. If , find the value of k.
Solution:
We have,
=>
=>
=>
=>
=> tan-12k/4 – tan-10 = Ï€/16
=> tan-12k/4 – 0 = Ï€/16
=> tan-12k/4 = π/16
=> tan-12k = π/4
=> 2k = tanπ/4
=> 2k = 1
=> k = 1/2
Therefore, the value of k is 1/2.
Question 60. If , find the value of k.
Solution:
We have,
=>
=>
=>
=>
=> a3 – 0 = 8
=> a3 = 8
=> a = 2
Therefore, the value of a is 2.
Question 61.
Solution:
We have,
I =
I =
I =
I =
I = -[√2cos3Ï€/2 – √2cosÏ€]
I = -(-√2 – 0)
I = √2
Therefore, the value of is √2.
Question 62.
Solution:
We have,
I =
I =
I =
I =
I =
I =
I =
I = [-4cosÏ€/2 + 4cos0] + [4sinÏ€/2 – 4sin0]
I = 0 + 4 + 4 – 0
I = 8
Therefore, the value of is 8.
Question 63.
Solution:
We have,
I =
I =
I =
I =
I =
I =
I =
I =
I =
I =
I =
I = (Ï€/8 – 1/4) – (3/4(Ï€/8 – 1/4) – 1/16)
I = Ï€/8 – 1/4 – (3Ï€/32 – 3/16 – 1/16)
I = Ï€/8 – 1/4 – (3Ï€/32 – 1/4)
I = Ï€/8 – 1/4 – 3Ï€/32 + 1/4
I = Ï€/8 – 3Ï€/32
I = (4Ï€ – 3Ï€)/32
I = π/32
Therefore, the value of is π/32.
Question 64.
Solution:
We have,
I =
By using integration by parts we get,
I =
I =
I =
I =
I =
So we get,
I =
I = log3/2 – 1/8log3
I = 3/8log3
Therefore, the value of is 3/8log3.
Question 65.
Solution:
We have,
I =
I =
I =
I =
I =
I =
I =
I = [tanÏ€/3 – tanÏ€/6] + [-cotÏ€/3 + cotÏ€/6]
I = [√3 – 1/√3] + [- 1/√3 – √3]
I = 2[√3 – 1/√3]
I = 4/√3
Therefore, the value of is 4/√3.
Question 66.
Solution:
We have,
I =
I =
I =
I =
I =
I =
I =
I =
I =
I =
I =
Therefore, the value of is .
Question 67.
Solution:
We have,
I =
I =
I =
I =
I = -log2/4 + log2/2 – 1/4 + 1/2
I = log2/4 + 1/4
Therefore, the value of is log2/4 + 1/4.
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# How Many Quarters in 10 Dollars?
Last Updated on 10/14/2017 by GS Staff
[otw_shortcode_dropcap label=”A:” size=”large” border_color_class=”otw-no-border-color”][/otw_shortcode_dropcap] There are four quarters in one dollar. We can determine this by looking at the below table, which shows the value of one, two, three, and four quarters:
Number of QuartersValue
1 Quarter25 Cents
2 Quarters 50 Cents
3 Quarters75 Cents
4 QuartersOne Dollar (\$1.00)
If there are four quarters in a dollar, we can determine how many quarters are in 10 dollars by multiplying four quarters by 10. The following equation shows how it is done:
4 (quarters that equal a dollar) x 10 (number of dollars in 10 dollars) = 40 quarters
There are 40 quarters in 10 dollars.
The below table shows the long way to count the number of quarters in 10 dollars.
## Quarters in 10 Dollars
Number of QuartersValue
1 Quarter25 Cents
2 Quarters 50 Cents
3 Quarters75 Cents
4 Quarters\$1.00
5 Quarters\$1.25
6 Quarters\$1.50
7 Quarters\$1.75
8 Quarters\$2.00
9 Quarters\$2.25
10 Quarters\$2.50
11 Quarters\$2.75
12 Quarters\$3.00
13 Quarters\$3.25
14 Quarters\$3.50
15 Quarters\$3.75
16 Quarters\$4.00
17 Quarters\$4.25
18 Quarters\$4.50
19 Quarters\$4.75
20 Quarters\$5.00
21 Quarters\$5.25
22 Quarters\$5.50
23 Quarters\$5.75
24 Quarters\$6.00
25 Quarters\$6.25
26 Quarters\$6.50
27 Quarters\$6.75
28 Quarters\$7.00
29 Quarters\$7.25
30 Quarters\$7.50
31 Quarters\$7.75
32 Quarters\$8.00
33 Quarters\$8.25
34 Quarters\$8.50
35 Quarters\$8.75
36 Quarters\$9.00
37 Quarters\$9.25
38 Quarters\$9.50
39 Quarters\$9.75
40 Quarters\$10.00
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# Shape, Space and Measures
## Presentation on theme: "Shape, Space and Measures"— Presentation transcript:
Shape, Space and Measures
Teach GCSE Maths Shape, Space and Measures
The pages that follow are sample slides from the 113 presentations that cover the work for Shape, Space and Measures. A Microsoft WORD file, giving more information, is included in the folder. The animations pause after each piece of text. To continue, either click the left mouse button, press the space bar or press the forward arrow key on the keyboard. Animations will not work correctly unless Powerpoint 2002 or later is used.
F4 Exterior Angle of a Triangle
This first sequence of slides comes from a Foundation presentation. The slides remind students of a property of triangles that they have previously met. These first slides also show how, from time to time, the presentations ask students to exchange ideas so that they gain confidence.
a is called an exterior angle of the triangle
We already know that the sum of the angles of any triangle is 180. e.g. 57 + 75 + 48 = 180 57 exterior angle a 75 48 If we extend one side . . . we form an angle with the side next to it ( the adjacent side ) a is called an exterior angle of the triangle
exterior angle 57 + 75 + 48 = 180 a Ans: a = 180 – 48 = 132 57
We already know that the sum of the angles of any triangle is 180. e.g. 57 + 75 + 48 = 180 57 exterior angle 132 a 75 48 Tell your partner what size a is. Ans: a = 180 – 48 = 132 ( angles on a straight line )
exterior angle 57 + 75 + 48 = 180 57 132 75 48
We already know that the sum of the angles of any triangle is 180. e.g. 57 + 75 + 48 = 180 57 exterior angle 132 75 48 What is the link between 132 and the other 2 angles of the triangle? ANS: 132 = 57 + 75, the sum of the other angles.
The presentations usually end with a basic exercise which can be used to test the students’ understanding of the topic. Solutions are given to these exercises. Formal algebra is not used at this level but angles are labelled with letters.
Exercise 1. In the following, find the marked angles, giving your reasons: 115 a (a) 60 b 37 (b) 40 105 c 30
a = 180 - 60 = 120 b = 360 - 120 - 115 - 37 = 88 a 115 120
Exercise Solutions: 115 120 a (a) 60 b 37 a = 180 - 60 ( angles on a straight line ) = 120 b = 360 - 120 - 115 - 37 (angles of quadrilateral ) = 88
x = 180 - 30 = 150 c = 360 - 105 - 40 - 150 = 65 40 105 150
Exercise (b) 40 105 150 x c 30 Using an extra letter: x = 180 - 30 ( angles on a straight line ) = 150 c = 360 - 105 - 40 - 150 ( angles of quadrilateral ) = 65
F14 Parallelograms By the time they reach this topic, students have already met the idea of congruence. Here it is being used to illustrate a property of parallelograms.
To see that the opposite sides of a parallelogram are equal, we draw a line from one corner to the opposite one. P Q R S SQ is a diagonal Triangles SPQ and QRS are congruent. So, SP = QR and PQ = RS
F19 Rotational Symmetry Animation is used here to illustrate a new idea.
A B F C E D This “snowflake” has 6 identical branches.
When it makes a complete turn, the shape fits onto itself 6 times. The centre of rotation The shape has rotational symmetry of order 6. ( We don’t count the 1st position as it’s the same as the last. )
F21 Reading Scales An everyday example is used here to test understanding of reading scales and the opportunity is taken to point out a common conversion formula.
This is a copy of a car’s speedometer.
20 40 60 80 100 120 140 160 180 200 220 mph km/h Tell your partner what 1 division measures on each scale. It is common to find the “per” written as p in miles per hour . . . but as / in kilometres per hour. Ans: 5 mph on the outer scale and 4 km/h on the inner. Can you see what the conversion factor is between miles and kilometres? Ans: e.g. 160 km = 100 miles. Dividing by 20 gives 8 km = 5 miles
F26 Nets of a Cuboid and Cylinder
Some students find it difficult to visualise the net of a 3-D object, so animation is used here to help them.
Suppose we open a cardboard box and flatten it out.
This is a net Rules for nets: We must not cut across a face. We ignore any overlaps. We finish up with one piece.
O2 Bearings This is an example from an early Overlap file. The file treats the topic at C/D level so is useful for students working at either Foundation or Higher level.
e. g. The bearing of R from P is 220 and R is due west of Q
e.g. The bearing of R from P is 220 and R is due west of Q. Mark the position of R on the diagram. Solution: P x Q x
e. g. The bearing of R from P is 220 and R is due west of Q
e.g. The bearing of R from P is 220 and R is due west of Q. Mark the position of R on the diagram. Solution: P x . Q x
e. g. The bearing of R from P is 220 and R is due west of Q
e.g. The bearing of R from P is 220 and R is due west of Q. Mark the position of R on the diagram. Solution: P 220 x . Q x If you only have a semicircular protractor, you need to subtract 180 from 220 and measure from south.
e. g. The bearing of R from P is 220 and R is due west of Q
e.g. The bearing of R from P is 220 and R is due west of Q. Mark the position of R on the diagram. Solution: P x 40 . Q x If you only have a semicircular protractor, you need to subtract 180 from 220 and measure from south.
e. g. The bearing of R from P is 220 and R is due west of Q
e.g. The bearing of R from P is 220 and R is due west of Q. Mark the position of R on the diagram. Solution: P 220 x . R Q x
O21 Pints, Gallons and Litres
The slide contains a worked example. The calculator clipart is used to encourage students to do the calculation before being shown the answer.
1 millilitre = 1000th of a litre.
e.g. The photo shows a milk bottle and some milk poured into a glass. There is 200 ml of milk in the glass. (a) Change 200 ml to litres. (b) Change your answer to (a) into pints. Solution: (a) 1 millilitre = 1000th of a litre. 200 1000 1 200 millilitre = = 0·2 litre (b) 1 litre = 1·75 pints 0·2 litre = 0·2 1·75 pints = 0·35 pints
O34 Symmetry of Solids Here is an example of an animated diagram which illustrates a point in a way that saves precious class time.
Tell your partner if you can spot some planes of symmetry.
A 2-D shape can have lines of symmetry. A 3-D object can also be symmetrical but it has planes of symmetry. This is a cuboid. Tell your partner if you can spot some planes of symmetry. Each plane of symmetry is like a mirror. There are 3.
H4 Using Congruence (1) In this higher level presentation, students use their knowledge of the conditions for congruence and are learning to write out a formal proof.
e.g.1 Using the definition of a parallelogram, prove that the opposite sides are equal.
Proof: D B C A We need to prove that AB = DC and AD = BC. Draw the diagonal DB. Tell your partner why the triangles are congruent.
ABD = CDB ( alternate angles: AB DC ) (A)
e.g.1 Using the definition of a parallelogram, prove that the opposite sides are equal. Proof: D B C A We need to prove that AB = DC and AD = BC. x Draw the diagonal DB. ABD = CDB ( alternate angles: AB DC ) (A) ADB = CBD ( alternate angles: AD BC ) (A) BD is common (S) Triangles are congruent (AAS) ABD CDB
ABD = CDB ( alternate angles: AB DC ) (A)
e.g.1 Using the definition of a parallelogram, prove that the opposite sides are equal. Proof: D C We need to prove that AB = DC and AD = BC. x x A B Draw the diagonal DB. ABD = CDB ( alternate angles: AB DC ) (A) ADB = CBD ( alternate angles: AD BC ) (A) BD is common (S) Triangles are congruent (AAS) ABD CDB So, AB = DC
ABD = CDB ( alternate angles: AB DC ) (A)
e.g.1 Using the definition of a parallelogram, prove that the opposite sides are equal. Proof: D C We need to prove that AB = DC and AD = BC. x x A B Draw the diagonal DB. ABD = CDB ( alternate angles: AB DC ) (A) ADB = CBD ( alternate angles: AD BC ) (A) BD is common (S) Triangles are congruent (AAS) ABD CDB So, AB = DC and AD = BC.
H16 Right Angled Triangles: Sin x
The following page comes from the first of a set of presentations on Trigonometry. It shows a typical summary with an indication that note-taking might be useful.
0·3420… x SUMMARY In a right angled triangle, with an angle x, opp
hyp sin x = opp hyp where, opp. is the side opposite ( or facing ) x hyp. is the hypotenuse ( always the longest side and facing the right angle ) The letters “sin” are always followed by an angle. The sine of any angle can be found from a calculator ( check it is set in degrees ) e.g. sin 20 = 0·3420…
The next 4 slides contain a list of the 113 files that make up Shape, Space and Measures.
The files have been labelled as follows: F: Basic work for the Foundation level. O: Topics that are likely to give rise to questions graded D and C. These topics form the Overlap between Foundation and Higher and could be examined at either level. H: Topics which appear only in the Higher level content. Overlap files appear twice in the list so that they can easily be accessed when working at either Foundation or Higher level. Also for ease of access, colours have been used to group topics. For example, dark blue is used at all 3 levels for work on length, area and volume. The 3 underlined titles contain links to the complete files that are included in this sample.
Teach GCSE Maths – Foundation
Page 1 F1 Angles F15 Trapezia F2 Lines: Parallel and Perpendicular O7 Allied Angles O1 Parallel Lines and Angles F16 Kites O2 Bearings O8 Identifying Quadrilaterals F3 Triangles and their Angles F17 Tessellations F4 Exterior Angle of a Triangle F18 Lines of Symmetry O3 Proofs of Triangle Properties F19 Rotational Symmetry F20 Coordinates F5 Perimeters F21 Reading Scales F6 Area of a Rectangle F22 Scales and Maps F7 Congruent Shapes O9 Mid-Point of AB F8 Congruent Triangles O10 Area of a Parallelogram F9 Constructing Triangles SSS O11 Area of a Triangle F10 Constructing Triangles AAS O12 Area of a Trapezium F11 Constructing Triangles SAS, RHS O13 Area of a Kite O4 More Constructions: Bisectors O14 More Complicated Areas O5 More Constructions: Perpendiculars O15 Angles of Polygons F12 Quadrilaterals: Interior angles O16 Regular Polygons F13 Quadrilaterals: Exterior angles O17 More Tessellations F14 Parallelograms O18 Finding Angles: Revision O6 Angle Proof for Parallelograms continued
Teach GCSE Maths – Foundation
Page 2 F23 Metric Units O33 Plan and Elevation O19 Miles and Kilometres O34 Symmetry of Solids O20 Feet and Metres O35 Nets of Prisms and Pyramids O21 Pints, Gallons and Litres O36 Volumes of Prisms O22 Pounds and Kilograms O37 Dimensions O23 Accuracy in Measurements F27 Surface Area of a Cuboid O24 Speed O38 Surface Area of a Prism and Cylinder O25 Density F28 Reflections O26 Pythagoras’ Theorem O39 More Reflections O27 More Perimeters O28 Length of AB O40 Even More Reflections F24 Circle words F29 Enlargements O29 Circumference of a Circle O41 More Enlargements O30 Area of a Circle F30 Similar Shapes O31 Loci O42 Effect of Enlargements O32 3-D Coordinates O43 Rotations F25 Volume of a Cuboid and Isometric Drawing O44 Translations O45 Mixed and Combined Transformations F26 Nets of a Cuboid and Cylinder continued
Teach GCSE Maths – Higher
Page 3 O1 Parallel Lines and Angles O22 Pounds and Kilograms O2 Bearings O23 Accuracy in Measurements O3 Proof of Triangle Properties O24 Speed O4 More Constructions: bisectors O25 Density O5 More Constructions: perpendiculars H2 More Accuracy in Measurements H1 Even More Constructions O26 Pythagoras’ Theorem O6 Angle Proof for Parallelograms O27 More Perimeters O7 Allied Angles O28 Length of AB O8 Identifying Quadrilaterals H3 Proving Congruent Triangles O9 Mid-Point of AB H4 Using Congruence (1) O10 Area of a Parallelogram H5 Using Congruence (2) O11 Area of a Triangle H6 Similar Triangles; proof O12 Area of a Trapezium H7 Similar Triangles; finding sides O13 Area of a Kite O29 Circumference of a Circle O14 More Complicated Areas O30 Area of a Circle O15 Angles of Polygons H8 Chords and Tangents O16 Regular Polygons H9 Angle in a Segment O17 More Tessellations H10 Angles in a Semicircle and Cyclic Quadrilateral O18 Finding Angles: Revision O19 Miles and Kilometres H11 Alternate Segment Theorem O20 Feet and Metres O31 Loci O21 Pints, Gallons, Litres H12 More Loci continued
Teach GCSE Maths – Higher
Page 4 O32 3-D Coordinates H20 Solving problems using Trig (2) O33 Plan and Elevation H21 The Graph of Sin x H13 More Plans and Elevations H22 The Graphs of Cos x and Tan x O34 Symmetry of Solids H23 Solving Trig Equations O35 Nets of Prisms and Pyramids H24 The Sine Rule O36 Volumes of Prisms H25 The Sine Rule; Ambiguous Case O37 Dimensions H26 The Cosine Rule O38 Surface Area of a Prism and Cylinder H27 Trig and Area of a Triangle O39 More Reflections H28 Arc Length and Area of Sectors O40 Even More Reflections H29 Harder Volumes O41 More Enlargements H30 Volumes and Surface Areas of Pyramids and Cones O42 Effect of Enlargements O43 Rotations H31 Volume and Surface Area of a Sphere O44 Translations O45 Mixed and Combined Transformations H32 Areas of Similar Shapes and Volumes of Similar Solids H14 More Combined Transformations H33 Vectors 1 H15 Negative Enlargements H34 Vectors 2 H16 Right Angled Triangles: Sin x H35 Vectors 3 H17 Inverse sines H36 Right Angled Triangles in 3D H18 cos x and tan x H37 Sine and Cosine Rules in 3D H19 Solving problems using Trig (1) H38 Stretching Trig Graphs
Further details of “Teach GCSE Maths” are available from
Chartwell-Yorke Ltd 114 High Street Belmont Village Bolton Lancashire BL7 8AL Tel: Fax:
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# CLASS 10 MATH NCERT SOLUTIONS FOR CHAPTER – 3 Pair of Linear Equations in Two Variables Ex – 3.1
## Pair of Linear Equations in Two Variables
Question 1.
Aftab tells his daughter, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be”. Isn’t this interesting? Represent this situation algebraically and graphically.
Solution:
Let the present age of Aftab’s daughter be x years and that of Aftab be y years.
Then, seven years ago:
Aftab’s age was (y – 7) years
and his daughter’s age was (x – 7) years.
According to the first condition:
y – 7 = 7(x – 7)
⇒ y – 7 = 7x – 49
⇒ 7x – y = 42 …(i)
Five years later:
Aftab’s age will be (y + 5) years,
and his daughter’s age will be (x + 5) years
According to the second condition:
y + 5 = 3(x + 5)
⇒ y + 5 = 3x + 15
⇒ 3x – y = -10 …(ii)
For graphical representation:
From equation (i), we have:
y = 7x – 42
When x = 11, then y = 7 x 11 – 42 = 35
When x = 12, then y = 7 x 12 – 42 = 42
When x = 13, then y = 7 x 13 – 42 = 49
Thus, we have the following table of points:
From equation (ii), we have:
y = 3x + 10
When x = 9, then y = 3 x 9 + 10 = 37
When x = 10, then y = 3 x 10 + 10 = 40
When x = 13, then y = 3 x 13 + 10 = 49
Thus, we have the following table of points:
Plotting the points of each table, we obtain the required graph of two intersecting lines.
Question 2.
The coach of a cricket team buys 3 bats and 6 balls for ₹ 3900. Later, she buys another bat and 3 more balls of the same kind for ₹1300. Represent this situation algebraically and geometrically.
Solution:
Let the cost of 1 bat be ₹ x and the cost of 1 ball be ₹ y.
Then according to the question, we have:
3x + 6y = 3900 …(i)
x + 3y = 1300 …(ii)
For geometrical representation:
Thus, we have the following table of points:
Plotting the points of each table, we obtain the required graph of two intersecting lines.
Question 3.
The cost of 2 kg of apples and 1 kg of grapes on a day was found to be ₹160. After a month, the cost of 4 kg of apples and 2 kg of grapes is ₹300. Represent the situation algebraically and geometrically.
Solution:
Let cost of 1 kg apples be ₹ x and the cost of 1 kg grapes be ₹y.
Then according to the question, we have:
2x + y = 160 …(i)
4x + 2y = 300 …(ii)
For geometrical representation:
From equation (i), we have: y = -2x + 160
When x = 50, then y = -2 x 50 + 160 = 60
When x = 20, then y = -2 x 20 + 160 = 120
When x = 45, then y = -2 x 45 + 160 = 70
Thus, we have the following table of points:
Thus, we have the following table of points:
Plotting the points of each table on a graph paper, we obtain the required graph of two parallel lines.
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Similarity of two triangles
## Objective
To establish the criteria for similarity of two triangles
## Prerequisite Knowledge
• Concept of congruence
• One-to-one correspondence
• Basic proportionality theorem
• Angle and Side measurement skills
## Theory
Two triangles are similar if:
• Their corresponding angles are equal.
• Their corresponding sides are in the same ratio or proportion.
Similar triangles are triangles that have the same shape, but their sizes may vary. All equilateral triangles and squares of any side length are examples of similar objects.
In other words, if two triangles are similar, then their corresponding angles are congruent and their corresponding sides are in equal proportion.
### Criteria for Similarity of the triangle
• AAA (Angle-Angle-Angle)
• AA (Angle-Angle)
• SSS (Side-Side-Side)
• SAS (Side-Angle-Side)
AAA (Angle-Angle-Angle)
The Angle-Angle-Angle (AAA) criterion for the similarity of triangles states that “If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio (or proportion) and hence the two triangles are similar”.
AA (Angle-Angle)
The AA criterion states that "if two angles of a triangle are respectively equal to the two angles of another triangle, then the two triangles are similar".
This criterion is included in the AAA criteria itself.
SSS (Side-Side-Side)
The Side-Side-Side (SSS) criterion for the similarity of two triangles states that “If in two triangles, sides of one triangle are proportional to (i.e., in the same ratio of ) the sides of the other triangle, then their corresponding angles are equal and hence the two triangles are similar”.
SAS (Side-Angle-Side)
The Side-Angle-Side (SAS) criterion for the similarity of two triangles states that “If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar”.
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# elementary results about multiplicative functions and convolution
One of the most important elementary results about multiplicative functions and convolution (http://planetmath.org/DirichletConvolution) is:
###### Theorem.
If $f$, $g$, and $h$ are arithmetic functions with $f*g=h$ and at least two of them are multiplicative, then all of them are multiplicative.
The above theorem will be proven in two separate parts.
###### Lemma 1.
If $f$ and $g$ are multiplicative, then so is $f*g$.
###### Proof.
Note that $\displaystyle(f*g)(1)=f(1)g(1)=1\cdot 1=1$ since $f$ and $g$ are multiplicative.
Let $a,b\in\mathbb{N}$ with $\gcd(a,b)=1$. Then any divisor $d$ of $ab$ can be uniquely factored as $d=d_{1}d_{2}$, where $d_{1}$ divides $a$ and $d_{2}$ divides $b$. When a divisor $d$ of $ab$ is factored in this manner, the fact that $\gcd(a,b)=1$ implies that $\gcd(d_{1},d_{2})=1$ and $\displaystyle\gcd\left(\frac{a}{d_{1}},\frac{b}{d_{2}}\right)=1$. Thus,
$(f*g)(ab)$ $\displaystyle=\sum_{d|ab}f(d)g\left(\frac{ab}{d}\right)$ $\displaystyle=\sum_{\begin{subarray}{c}d_{1}|a\\ d_{2}|b\end{subarray}}f(d_{1}d_{2})g\left(\frac{ab}{d_{1}d_{2}}\right)$ $\displaystyle=\sum_{\begin{subarray}{c}d_{1}|a\\ d_{2}|b\end{subarray}}f(d_{1})f(d_{2})g\left(\frac{a}{d_{1}}\right)g\left(% \frac{b}{d_{2}}\right)$ $\displaystyle=\left(\sum_{d_{1}|a}f(d_{1})g\left(\frac{a}{d_{1}}\right)\right)% \left(\sum_{d_{2}|b}f(d_{2})g\left(\frac{b}{d_{2}}\right)\right)$ $\displaystyle=(f*g)(a)(f*g)(b)$.
###### Lemma 2.
If $f$ is an arithmetic function and $g$ and $h$ are multiplicative functions with $f*g=h$, then $f$ is multiplicative.
###### Proof.
Let $a,b\in\mathbb{N}$ with $\gcd(a,b)=1$. Induction will be used on $ab$ to establish that $f(ab)=f(a)f(b)$.
If $ab=1$, then $a=b=1$. Note that $f(1)=f(1)\cdot 1=f(1)g(1)=(f*g)(1)=h(1)=1$. Thus, $f(ab)=f(1)=1=1\cdot 1=f(a)f(b)$.
Now let $ab$ be an arbitrary natural number. The induction hypothesis yields that, if $d_{1}$ divides $a$, $d_{2}$ divides $b$, and $d_{1}d_{2}, then $f(d_{1}d_{2})=f(d_{1})f(d_{2})$. Thus,
$\begin{array}[]{ll}h(a)h(b)&=h(ab)\\ \\ &=(f*g)(ab)\\ \\ &\displaystyle=\sum_{d|ab}f(d)g\left(\frac{ab}{d}\right)\\ \\ &\displaystyle=f(ab)g(1)+\sum_{\begin{subarray}{c}d|ab\\ d
It follows that $f(ab)=f(a)f(b)$. ∎
The theorem follows from these two lemmas and the fact that convolution is commutative.
The theorem has an obvious corollary.
###### Corollary.
If $f$ is multiplicative, then so is its convolution inverse.
###### Proof.
Let $f$ be multiplicative. Since $f(1)\neq 0$, $f$ has a convolution inverse $f^{-1}$. (See convolution inverses for arithmetic functions for more details.) Since $f*f^{-1}=\varepsilon$, where $\varepsilon$ denotes the convolution identity function, and both $f$ and $\varepsilon$ are multiplicative, the theorem yields that $f^{-1}$ is multiplicative. ∎
Title elementary results about multiplicative functions and convolution ElementaryResultsAboutMultiplicativeFunctionsAndConvolution 2013-03-22 16:07:37 2013-03-22 16:07:37 Wkbj79 (1863) Wkbj79 (1863) 16 Wkbj79 (1863) Theorem msc 11A25 ConvolutionInversesForArithmeticFunctions
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# The water jug puzzle
Yossi Elran tells us about the water jug puzzle
Here is another well-known, medieval problem, also known as the ‘water pouring puzzle’.
You have an 8 litre jug full of water and two smaller jugs, one that contains 5 litres and the other 3 litres. None of the jugs have markings on them, nor do you have any additional measuring device.You have to divide the 8 litres of water equally between your two best friends, so that each gets 4 litres of water. How can you do this?
## The water jug puzzle – solution
• First, water is poured from the 8 litre jug into the 5 litre jug, leaving 3 litres of water in the original 8 litre jug.
• Next, water is poured from the 5 litre jug into the 3 litre jug, so we now have 3 litres of water in the 8 litre jug, 2 litres of water in the 5 litre jug and 3 litres of water in the 3 litre jug.
• The 3 litre jug is emptied into the 8 litre jug, so the 8 litre jug now contains 6 litres of water.
• The 2 litres of water in the 5 litre jug are now poured into the empty 3 litre jug.
• Water is poured from the 8 litre jug (which at this stage contains 6 litres) into the empty 5 litre jug.
We now have 5 litres of water in the 5 litre jug, 2 litres of water in the 3 litre jug and 1 litre of water in the 8 litre jug.
• Water is poured from the 5 litre jug to fill the 3 litre jug which already contains at this stage 2 litres of water.
• We are left with 4 litres of water in the 5 litre jug which is given to one friend, and 3 litres of water in the 3 litre jug that is poured back into the 8 litre jug that already contains 1 litre of water. This gives 4 litres of water which are given to the second friend.
The whole scenario can be summarised using numbers in brackets to denote the litres of water at each stage in each of the 8 litre, 5 litre and 3 litre jugs, respectively:
[[8,0,0] rightarrow [3,5,0] rightarrow [3,2,3] rightarrow [6,2,0] rightarrow [6,0,2] rightarrow [1,5,2] rightarrow [1,4,3] rightarrow [4,4,0]]
This diagram is known as a state diagram
## Discussion
The water jug problem can be solved with just two jugs – one that can hold 5 litres of water and the other that can hold 3 litres of water, if there is also an unlimited supply of water from a tap and a sink. Show the series of state diagrams that solve this problem. Post your solutions in the comments!
© Davidson Institute of Science Education, Weizmann Institute of Science
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# Data Collection Mean, Mode, Median, & Range
Mathematical Terms Related to a Group of Numbers Data Collection Mean, Mode, Median, & Range Standard Deviation Mean = the average Consider your measurements as a set of numbers. Add together all the numbers in your set of measurements. Divide by the total number of values in the set Ages of cars in a parking lot:
20 years +10 years +10 years +1 year+15 years +10 years= 66 years/6 cars = 11years (answer is a counted numbers for sig figs are not an issue) Note: the mean can be misleading (one very high value and one very low) Median The number in the middle Put numbers in order from lowest to highest
and find the number that is exactly in the middle 20, 15, 10, 10, 10, 1 Since there is an even number of values the median is 10 years (average of the 2 middle values) Or to determine the Median: 20 years, 15 years, 13 years, 11 years, 7 years, 5 years, 3 years With an odd number of values the median is the number in the middle or 11 years
Mode Number in data set that occurs most often 20, 15, 10, 10, 10, 1 Sometimes there will not be a mode 20, 17, 15, 8, 3 Record answer as none or no mode NOT 0 Sometimes there will be more than one mode 20, 15, 15, 10, 10, 10, 1 Range
The difference between the lowest and highest numbers 20, 15, 10, 10, 1 20-1 = 19 years The range tells you how spread out the data points are. Example: The mean of four numbers is 50.5 101 99
What is the median? What is the mode? 1 1 Measured Values When making a set of repetitive measurements, the standard deviation (S.D.) can be determined to
indicate how much the samples differ from the mean Indicates also how spread out the values of the samples are Standard Deviation The smaller the standard deviation, the higher the quality of the measuring instrument and your technique Also indicates that the data points are also fairly close together with a small value for the range.
Indicates that you did a good job of precision w/your measurements. A high or large standard deviation Indicates that the values or measurements are not similar There is a high value for the range Indicates a low level of precision (you didnt make measurements that were close to the same) The standard deviation will be 0 if all the values or measurements are the same.
Formula for Standard Deviation = range N = (highest value lowest value)
N N = number of measured values As N gets larger or the more samples (measurements, scores, etc.), the reliability of this approximation increases 22.5 mL, 18.3 mL, 20.0 mL, 10.6 mL The Standard Deviation would be: range
= N = (highest value lowest value) N Range = 22.5 10.6g = 11.0 mL N=4
11.0g 4 = 5.95 mL = 6.0 mL S.D. = 17.9 6.0 mL (expressed to the same level of precision as the mean) Example The results of several masses of an object weighed by 5 different students on a multibeam balance
11.36g, 11.37g, 11.40g, 11.38g & 11.39g Average = 56.90g 5 = 11.38g SO: 11.36g, 11.37 g, 11.40 g, 11.38 g & 11.39g The Standard Deviation would be: range = N
= (highest value lowest value) N Range = 11.40g 11.36g = 0.04g N=5 0.04g 5
= 0.0178 g = 0.02 g The Standard Deviation from the mean is: Mean = 11.38 g SD = 0.02 g Expressed as 11.38g 0.02 This tells us that the measurements were within
two hundredths of the mean either less than the mean or greater than the mean. ~Review~ Mean average Mode which one occurs most often Median number in the middle
Range difference between the highest and lowest values Standard Deviation The standard deviation of a set of data measures how spread out the data set is. In other words, it tells you whether all the data items bunch around close to the mean or is they are all over the place. Graphical Analysis The graphs show three
normal distributions with the same mean, but the taller graph is less spread out. Therefore, the data represented by the taller graph has a smaller standard deviation.
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# How do you estimate 5 1/3-3 5/6?
Apr 4, 2017
The estimate is $1$, which is the result of $5 - 4 = 1$
We can also see the actual value will be greater than one $\left(> 1\right)$
since $5 \frac{1}{3} > 5 \mathmr{and} 3 \frac{5}{6} < 4$
#### Explanation:
Given: $5 \frac{1}{3} - 3 \frac{5}{6}$
To estimate this expression we round off each term:
$5 \frac{1}{3} i s c l o s e \to$5
$3 \frac{5}{6} i s c l o s e \to$4
Then subtracting: $5 - 4 = 1$
Actual value is: $5 \frac{1}{3} - 3 \frac{5}{6} = \frac{16}{3} - \frac{23}{6} = \frac{32}{6} - \frac{23}{6} = \frac{9}{6} = 1 \frac{3}{6} = 1 \frac{1}{2}$
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# How do you write the equation of a line through (0,3) and (-2,5)?
Nov 16, 2016
$y = - x + 3$
#### Explanation:
The equation of a line in $\textcolor{b l u e}{\text{slope-intercept form}}$ is.
$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{y = m x + b} \textcolor{w h i t e}{\frac{2}{2}} |}}}$
where m represents the slope and b, the y-intercept.
We require to find m and b.
To find m use the $\textcolor{b l u e}{\text{gradient formula}}$
$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{m = \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}}} \textcolor{w h i t e}{\frac{2}{2}} |}}}$
where $\left({x}_{1} , {y}_{1}\right) \text{ and " (x_2,y_2)" are 2 points on the line}$
The 2 points here are (0 ,3) and (-2 ,5)
let $\left({x}_{1} , {y}_{1}\right) = \left(0 , 3\right) \text{ and } \left({x}_{2} , {y}_{2}\right) = \left(- 2 , 5\right)$
$\Rightarrow m = \frac{5 - 3}{- 2 - 0} = \frac{2}{- 2} = - 1$
One of the points here is (0 ,3). That is where the line crosses the y-axis.
Hence the value of b is 3
$\Rightarrow y = - x + 3 \text{ is the equation}$
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A triangle is 180 degrees and there are three types of triangles -- an equilateral triangle which means all three angles equal the same, an isosceles triangle in which two angles are equal, and a scalene triangle in which all three angles have different measurements. Because we already know the sum of two angles, we simply need to subtract the total of those two angles from the 180 degrees. 180 - 120 = 60 degrees. The measure of the third angle is 60 degrees.
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4x + 5 = 8 + 2x
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In this problem, we need to determine the value of x. In order to do that, we must shift the components of our problem around to where all the values with an 'x' are on one side. 4x + 5 = 8 + 1x We will keep 4x where it is and bring the + 1x from the right side of the equation to the left. As we shift the + 1x, we need to be mindful of the signs associated with each number. As the 1x is a positive value, it will acquire a negative value as it moves to the opposite side of the equation. Hence becoming 4x + 5 - 1x = 8. Since our primary purpose is to get all the values with an 'x' to one side, we will shift the +5 to the right side of the equation. Again, we need to be mindful of the signs meaning +5 will become a -5 as it moves to the right side becoming: 4x - 1x = 8 - 5 We are able to subtract 4x - 1x because both values have an x. It would not be possible to perform this action if 1x was 1y. 4x - 1x = 3x and 8 - 5 = 3 Now our equation is 3x = 3. We still need to determine the value of x. In order to do so, we must separate 3x. When there is no sign between two numbers, it means that we are multiplying the two values and the opposite of that operation is division. As we take the 3 from 3x to the other side of the equation, we will divide. x = 3/3 x = 1 The value of x = 1.
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NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Ex 8.2
# NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Ex 8.2
NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Ex 8.2 are the part of NCERT Solutions for Class 9 Maths. In this post, you will find the NCERT Solutions for Chapter 8 Quadrilaterals Ex 8.2.
## NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Ex 8.2
Ex 8.2 Class 9 Maths Question 1.
ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see figure). AC is a diagonal. Show that
(i) SR || AC and SR = ½ AC
(ii) PQ = SR
(iii) PQRS is a parallelogram.
Solution:
(i)
In ∆ACD, we have
S is the mid-point of AD and R is the mid-point of CD.
SR = ½ AC and SR || AC ….…(1) [By mid-point theorem]
(ii) In ∆ABC, P is the mid-point of AB and Q is the mid-point of BC.
PQ = ½ AC and PQ || AC ….…(2) [By mid-point theorem]
From (1) and (2), we get
PQ = SR and PQ || SR
PQ = SR and PQ || SR [Proved in (ii)]
PQRS is a parallelogram.
Ex 8.2 Class 9 Maths Question 2.
ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA, respectively. Show that the quadrilateral PQRS is a rectangle.
Solution.
Given: ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD, and DA, respectively.
To prove: Quadrilateral PQRS is a rectangle.
Proof: By mid-point theorem.
In ΔADC, we have S and R are the mid-points of AD and CD, respectively.
SR || AC and SR = ½ AC …………..(i)
In ΔABC, we have P and Q are the mid-points of AB and BC, respectively.
PQ || AC and PQ = ½ AC ……..…..(ii)
From eqs. (i) and (ii), we get PQ || SR and PQ = SR = ½ AC
Since, a pair of opposite sides of a quadrilaterals PQRS is equal and parallel.
So, PQRS is a parallelogram.
We know that diagonals of a rhombus bisect each other at right angles.
COD = EOF = 90°
Now, in ABCD, R and Q are the mid-points of CD and BC, respectively.
RQ || DB [by mid-point theorem]
RE || OF
Also, SR || AC [from eq. (i)]
FR || OE
So, OERF is a parallelogram.
ERF = EOF = 90° [Opposite angles of a parallelogram are equal]
Thus, PQRS is a parallelogram with
R = 90°
Hence, PQRS is a rectangle.
Hence proved.
Ex 8.2 Class 9 Maths Question 3.
ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA, respectively. Show that the quadrilateral PQRS is a rhombus.
Solution.
Given, ABCD is a rectangle.
In ΔASP and ΔBQP,
AP = BP (Given)
AS = BQ (Given)
A = B (Each 90°)
ΔASP ΔBQP (By SAS congruency rule)
SP = PQ (By CPCT) ……(i)
In ΔPBQ and ΔRCQ,
PB = RC (Given)
BQ = CQ (Given)
B = C (Each 90°)
ΔPBQ ΔRCQ (By SAS congruency rule)
PQ = RQ (By CPCT) ……(ii)
Similarly, we can prove that RQ = SR and SR = SP (iii)
From eqs. (i), (ii) and (iii), we get SP = PQ = QR = RS.
Therefore, PQRS is a rhombus.
Ex 8.2 Class 9 Maths Question 4.
ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD. A line is drawn through E parallel to AB intersecting BC at F (see figure). Show that F is the mid-point of BC.
Solution.
Given:
ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD and a line is drawn through E parallel to AB intersecting BC at F such that EF || AB.
To prove: F is the mid-point of BC.
Proof: Let EF intersects BD at P.
In ΔABD, we have EP || AB [
EF || AB] and E is the mid-point of AD.
So, by the converse of mid-point theorem, P is the mid-point of BD.
Similarly, in ΔBCD, we have PF || CD [
EF || AB and AB || CD]
and P is the mid-point of BD.
So, by the converse of mid-point theorem, F is the mid-point of BC.
Ex 8.2 Class 9 Maths Question 5.
In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively
(see figure). Show that the line segments AF and EC trisect the diagonal BD.
Solution.
Given: ABCD is a parallelogram, and E and F are the mid-points of sides AB and CD, respectively.
To prove: Line segments AF and EC trisect the diagonal BD.
Proof: Since ABCD is a parallelogram.
AB || DC and AB = DC [opposite sides of a parallelogram]
AE || FC and ½ AB = ½ DC
AE || FC and AE = FC [ E and F are the mid-points of AB and CD]
Since a pair of opposite sides of a quadrilaterals AECF is equal and parallel.
So, AECF is a parallelogram.
Then, AF || EC
AP || EQ and FP || CQ [Since opposite sides of a parallelogram are parallel]
In ΔBAP, E is the mid-point of AB and EQ || AP, so Q is the mid-point of BP.
[by the converse of mid-point theorem]
BQ = PQ ……(i)
Again, in ΔDQC, F is the mid-point of DC and FP || CQ.
So, P is the mid-point of DQ. [by the converse of mid-point theorem]
QP = DP ……(ii)
From eqs. (i) and (ii), we get BQ = PQ = PD
Hence, CE and AF trisect the diagonal BD.
Hence proved.
Ex 8.2 Class 9 Maths Question 6.
ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that
(i) D is the mid-point of AC
(ii) MD
AC
(iii) CM = MA = ½ AB
Solution.
Given, ABC is a right-angled triangle.
C = 90° and M is the mid-point of AB. Also, DM || BC
(i) In ΔABC, BC || MD and M is the mid-point of AB.
D is the mid-point of AC. (By converse of mid-point theorem)
(ii) Since MD || BC and CD is transversal. .
But
ACB = 90°
MD AC
(iii) Now, in ΔADM and ΔCDM, we have
DM = MD (Common)
AD = CD (D is the mid-point of AC)
ΔCDM
CM = AM …….(i)
Also, M is the mid-point of AB.
AM = BM = ½ AB …….(ii)
From eqs. (i) and (ii), we get
CM = AM = ½ AB
Hence proved.
|
Real Numbers and the Number Line
Presentation on theme: "Real Numbers and the Number Line"— Presentation transcript:
Real Numbers and the Number Line
The numbers in the pattern are integers and their squares.
Table of Perfect Squares
Integer Perfect Square 1 2 3 4 5 6 7 8 9 10 Integer Perfect Square 11 12 13 14 15 16 17 18 19 20
Real Numbers and the Number Line
A number a is a square root of a number b if a2 = b. Example: Together, the radical symbol and the radicand form a radical.
Problem 1 What is the simplified form of each expression?
Problem 2 Estimating a Square Root
The square of an integer is called a perfect square. For example, 49 is a perfect square since When a radicand is not a perfect square, you can estimate the square root. Method 1: Estimate
Problem 2 Estimating a Square Root
Method 2: Estimate
Classifying Real Numbers
You can classify numbers using sets. A set is a well-defined collection of objects. Each object is called an element of the set. A subset of a set consists of elements from the larger set. You can list elements of a set within braces {}. The set of Real Numbers is the largest set of numbers that you will work with. It is divided into two subsets: the Rational Numbers and the Irrational Numbers.
The Real Numbers Rational Numbers Irrational Numbers Real Numbers
Rational Numbers The Rational Numbers have a subset called the integers. The integers are the set The Integers have a subset called the whole numbers. The Whole Numbers are the set
Rational Numbers The Whole Numbers have a subset called the natural numbers. The Natural Numbers are the set Some square roots are rational numbers and some are irrational numbers. If a whole number is not a perfect square, then its square root is irrational.
Classifying Real Numbers
Problem 3 Classifying Real Numbers
To which subsets of the real numbers does each number belong to?
Inequality An inequality is a mathematical sentence that compares the values of two expressions using an inequality symbol. The inequality symbols are:
Problem 4 Comparing Real Numbers
What is an inequality that compares the following two numbers?
Problem 5 Graphing and Ordering Real Numbers
Order the following set of numbers from least to greatest. Note: To perform this task, you can write every number as a decimal rounded to the nearest hundredth, and then graph each of the numbers on a number line. Then, read the graph from left to right.
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In how many ways can 6 persons be arranged in
Question:
In how many ways can 6 persons be arranged in
(i) a line,
(ii) a circle?
Solution:
(i) Let choose 1 person from 6 by ${ }^{6} \mathrm{C}_{1}=6$ and arranged it in line
Now choose another person from remaining 5 by ${ }^{5} C_{1}=5$ and arranged it in line
Similarly, choose another person from remaining 4 by ${ }^{4} C_{1}=4$ and arranged it in line
Similarly, choose another person from remaining 3 by ${ }^{3} C_{1}=3$ and arranged it in line
Similarly, choose another person from remaining 2 by ${ }^{2} C_{1}=2$ and arranged it in line
And choose another person from remaining 1 by ${ }^{1} C_{1}=1$ and arranged it in line
So total number of ways is $6 !=720$
(ii) It is the same as above, by converting line arrangement into the circle but you need to remove some arrangement
Let suppose 6 persons as $A, B, C, D, E, F$ you need to arrange this 6 persons into a circle.
First, we arranged 6 persons in line(number of ways $=6 !$ )
NOTE: A, B, C, D , E, F and B, C, D, E, F, A consider as a different line, but when we arranged this 2 combination in circle then it becomes same,
i.e. Let takes us an example we need to arrange $A, N, O, D, E$.
We arrange it as shown. When we rotate first one, then $1^{\text {st }}$ and $2^{\text {nd }}$ became identical and so on that’s why all 5 are identical, and we count it as 1
Now come back to our questions
So total number of arrangement is $(6-1) !=5 !=120$
NOTE: When you want to arrange n persons in circle then a total number of ways is $n ! / n$,
i.e. Total number of ways $=(n-1) !$
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# Financial Polynomials
May 29th, 2015
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On page 304, problem 90 says that P dollars is invested at annually interest rate r for 1 year. If the interest rate is compounded semiannually then the polynomial is P(1{+r/2)^2, which represents the value invested after 1 year.
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Algebraic Formula:1. (a+ b)2= a2+ 2ab + b22. (a - b)2= a2-2ab + b23. (a + b) (a - b) = a2-b24. (x + a)(x + b) = x2+ (a + b) x + ab5. (x + a)(x - b) = x2+ (a - b) x -ab6. (x - a)(x + b) = x2+ (b - a) x -ab7. (x - a)(x - b) = x2- (a + b) x + ab8. (a + b)3= a3+ b3+ 3ab(a + b)9. (a - b)3= a3- b3- 3ab (a - b)10. (x + y + z)2= x2+ y2+ z2+ 2xy +2yz + 2xz11. (x + y - z)2= x2+ y2+ z2+ 2xy - 2yz - 2xz12. (x - y + z)2= x2+ y2+ z2- 2xy - 2yz + 2xz13. (x - y - z)2= x2+ y2+ z2- 2xy + 2yz - 2xz14. x3+ y3+ z3- 3xyz = (x + y + z)(x2+ y2+ z2- xy - yz -xz)15. x2+ y2=12[(x + y)2+ (x - y)2]16. (x + a) (x + b) (x + c) = x3+ (a + b + c)x2+ (ab + bc + ca)x + abc17. x3+ y3= (x + y) (x2- xy + y2)18. x3- y3= (x - y) (x2+ xy + y2)19. x2+ y2+ z2-xy -yz -zx =12[(x-y)2+(y-z)2+(z-x)2]CBSE Class 9 - Maths - Ch6 - Lines and AnglesUnderstanding Basic Terms.Fill in the Blanks1. A point, a line and a plane are' basic concepts of geometry. These terms are _____ (well defined/undefined).2. A line with two end points is called a _________ (line-segment/ ray).3. A terminated line can be produced ________ (within defined limits/indefinitely).4. A part of a line with one end point is called a _____ (line-segment/ray).5. A line contains ______ (definite/infinite) number of points.6. (One and only one/More than one) _______ line passes t
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# BASIC ELEMENTS OF GEOMETRY
## Presentation on theme: "BASIC ELEMENTS OF GEOMETRY"— Presentation transcript:
BASIC ELEMENTS OF GEOMETRY
Think of geometry as a skyscraper. At its foundation are three basic terms: point, line and plane. These terms are used to define other geometric terms. All these terms are the steel beams of the skyscraper, built carefully upon one another and upon the foundation of the three basic terms. But no matter how firm the foundation, the skyscraper won’t stand unless its beams are carefully connected. In geometry, we connect geometric terms with relationships. Using your foldable, we will go over review the basic terms along with connecting relationships that will be needed for the material we will cover in Chapter 9: Represent Geometry with Algebra. Teacher note: Encourage students to take notes in their Foldable as you progress through this presentation.
FOLDABLE This slide is to make sure students have put their foldables together correctly. Please run the pdf file two-sided. (Each student should receive 4 sheets of paper)
Basic Elements of Geometry
Geometric terms
Point Definition: a location in space. A point is named by a capital letter, and its location may be indicated by a dot. Write: A Say: Point A
Point A point is named by its coordinates, and its location may be indicated by a dot. Write: P Say: Point P
Line Definition: a set of points that form a straight path extending infinitely in two directions. The directions are opposite of each other. Write: Say: Line AB
Line Segment Definition: a part of a line that has two endpoints and a definite length A B Write: Say: Line segment AB
Ray Definition: a set of points that has one endpoint and extends infinitely in one direction A B C Write: Say: Ray AB or Ray AC
Angle Definition: a geometric figure made up of two rays of two line segments that have the same endpoint. The common endpoint is the vertex. Write: Say: Angle ABC or Angle B
Plane Definition: a flat surface that extends without end (infinitely) in all directions. Write: Plane A Say: Plane A A
Basic Elements of Geometry
ANGLES
Angle Definition: a geometric figure made up of two rays of two line segments that have the same endpoint. The common endpoint is the vertex.
ACUTE ANGLE Angles that have a measure that is less than 90 degrees.
RIGHT ANGLE Angles that have a measure that is exactly 90 degrees.
OBTUSE ANGLE Angles that have a measure that is greater than 90 degrees but less than 180 degrees.
STRAIGHT ANGLE Angles that have a measure that is exactly 180 degrees.
Basic Elements of Geometry
TRIANGLES
Triangle Definition: a geometric figure made up of three rays or three line segments. Write: Say: Triangle ABC or Triangle BCA or Triangle CAB
ACUTE TRIANGLE A triangle composed of three acute angles.
RIGHT TRIANGLE A triangle composed of one right angle and two acute angles.
OBTUSE TRIANGLE A triangle composed of one obtuse angle and two acute angles.
SCALENE TRIANGLE A triangle with no congruent sides and no congruent angles.
ISOSCELES TRIANGLE A triangle with at least two congruent sides and at least two congruent angles.
EQUILATERAL TRIANGLE A triangle with three congruent sides and three congruent angles.
Basic Elements of Geometry
Quadrilateral Definition: a geometric figure made up of four sides and four angles
RECTANGLE A quadrilateral with four right angles. Opposite sides are equal in length.
SQUARE A quadrilateral with four right angles. All sides are equal in length.
PARALLELOGRAM A quadrilateral with opposite sides that are parallel. Opposite sides are equal and opposite angles are equal.
RHOMBUS A quadrilateral with opposite sides that are parallel. Opposite sides are equal in length and opposite angles are equal in measure. All sides are equal in length.
TRAPEZOID A quadrilateral with two sides are parallel and two sides are not parallel.
KITE A quadrilateral with two pairs of consecutive congruent sides.
SYMBOLS and ABBREVIATIONS
Basic Elements of Geometry SYMBOLS and ABBREVIATIONS
Basic Elements of Geometry
COMMON POLYGONS
Polygons: Triangle (3 sides) Quadrilateral (4 sides) Pentagon (5 sides) Hexagon (6 sides) Heptagon (7 sides) Octagon (8 sides) Nonagon(9 sides) Decagon (10 sides) Dodecagon (12 sides)
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## Note on Angle between two lines
• Note
• Things to remember
• Videos
• Exercise
• Quiz
#### Angle between the lines y = m1 + c1 and y = m2 + c2 and y = m2x + c2
Let the equation of two lines AB and CD be y = m1x + c1 and y = m2x + c2 respectively.
let the lines AB and CD make angles θ1 and θ2 respectively with the positive direction of X-axis.
Then, tanθ1 = m1 and tanθ2 = m2.
Let the lines AB and CD intersect each other at the point E.
Let the angles between the lines AB and CD
∠CEA = Φ
Then by plane geometry, θ1 = Φ + θ2
or,Φ =θ12
∴ tanΦ = tan(θ1 2) = $$\frac {tan\theta_1 - tan\theta_2}{1 + tan\theta_1 tan\theta_2}$$ = $$\frac{m_1 - m_2}{1+tan\theta_1 tan\theta_2}$$ .........(i)
Again, let ∠BAC = Ψ
Then by place geometry ,Φ + Ψ = 1800
or, Ψ = 1800 - Φ
or, tan Ψ = tan (180 - Φ) = -tan Φ = - $$\frac{m_1 - m_2}{1+m_1m_2}$$ ............(ii)
Hence if angles between the lines y = m1x + c1 and y = m2x + c2 be the θ then,
tanθ =± $$\frac{m_1 - m_2}{1+m_1m_2}$$
θ = tan-1(± $$\frac{m_1 - m_2}{1+m_1m_2}$$)
Condition of Perpendicularity
Two lines AB and CD will be perpendicular to each other if the angle between them θ = 90o.
We have tanθ = ± $$\frac{m_1 - m_2}{1+m_1m_2}$$
or, tan900 = ± $$\frac{m_1 - m_2}{1+m_1m_2}$$
or, cot900 = ± $$\frac{1+m_1m_2}{m_1 - m_2}$$
or, 0 = $$\frac{1+m_1m_2}{m_1 - m_2}$$
or, 1 +m1m2 =0
or, m1m2 = -1
Two lines will be perpendicular to each other if m1m2 = -1
i.e. if product of the slopes = -1 .
Condition of Parallelism
Two lines AB and CD will be parallel to each other if the angle between them θ = 00.
we have, tanθ = ± $$\frac{m_1 - m_2}{1+m_1m_2}$$
or, tan00 = ± $$\frac{m_1 - m_2}{1+m_1m_2}$$
or, 0 = $$\frac{m_1 - m_2}{1+m_1m_2}$$
or, m1 - m2 = 0
or, m1 = m2
∴ Two lines will be parallel to each other if m1 = m2 i.e. if slopes are equal.
#### Angle between the lines A1x + B1y + C1 = 0 and A2x + B2y +C2 = 0
Let equations of two straight lines AB and CD be A1x + B1y + C1 = 0 and A2x + B2y + C2 = 0 respectively.
Then slope of AB = -$$\frac{A_1}{B_1}$$
Slope of CD = -$$\frac{A_2}{B_2}$$
Let the lines AB and CD make angles θ1 and θ2 with the positive direction of X-axis.
Then, tanθ1 = -$$\frac{A_1}{B_1}$$ and tanθ2 = -$$\frac{A_2}{B_2}$$
Let ∠CEA = Φ.
Then, θ1 = θ1 - θ2
or, Φ = θ1 - θ2
∴ tan Φ = tan( θ1 - θ2) = $$\frac{tan\theta_1 - tan\theta_2}{1 + tan\theta_1 tan\theta_2}$$ = $$\frac{A_2 B_1 - A_1 B_2}{A_1 A_2 + B_1 B_2}$$ = -$$\frac{A_1 B_2 - A_2 B_1}{A_1 A_2 + B_1 B_2}$$ ......(i)
Let ∠BEC = Ψ
Then Ψ + Φ = 1800
or, Ψ = 1800 - Φ
∴ tan Ψ = tan(1800 - Φ) = -tanΦ = $$\frac{A_1 B_2 - A_2 B_1}{A_1 A_2 + B_1 B_2}$$ .........(ii)
Hence if angles between the lines A1x + B1y + C1 = 0 and A2x + B2y + C2 = 0 is θ, then
tanθ = ± $$\frac{A_1 B_2 - A_2 B_1}{A_1 A_2 + B_1 B_2}$$
or, θ = tan-1 (±$$\frac{A_1 B_2 - A_2 B_1}{A_1 A_2 + B_1 B_2}$$ )
Condition of Perpendicularity
Two lines AB and CD will be perpendicular to each other if θ = 900
Then tan900 = ± $$\frac{A_1 B_2 - A_2 B_1}{A_1 A_2 + B_1 B_2}$$
or, ∞ = $$\frac{A_1 B_2 - A_2 B_1}{A_1 A_2 + B_1 B_2}$$
∴ A1A2 + B1B2 = 0
Condition of Parallelism
Two lines AB and CD will be parrallel to each other if θ = 00
Then, tan00 = ± $$\frac{A_1 B_2 - A_2 B_1}{A_1 A_2 + B_1 B_2}$$
or, 0 = $$\frac{A_1 B_2 - A_2 B_1}{A_1 A_2 + B_1 B_2}$$
or, A1B2 - A2B1 = 0
or, A1B2 = A2B1
∴ $$\frac{A_1}{A_2}$$ = $$\frac{B_1}{B_2}$$
#### Equation of any line parallel to ax + by + c = 0
Equation of the given line is ax + by + c = 0
Slope of this line = -$$\frac{coefficient \;of \;x}{coefficient \;of \;y}$$ = -$$\frac{a}{b}$$
Slope of the line parallel to this line = -$$\frac{a}{b}$$
Now, equation of a line having slope -$$\frac{a}{b}$$ is given by
y = mx + c
or, y = -$$\frac{a}{b}$$x + c
or, by = -ax + bc
or, ax + by - bc = 0
or, ax + by + k = 0 where, k = -bc.
Hence equation of any line parallel to ax + by + c = 0 is given by ax + by + k = 0 where k is an arbitrary constant.
#### Equation of any line perpendicular to ax +by +c = 0
Equation of the given line is ax + by + c = 0
Slope of this line = -$$\frac{coefficient\;of\;x}{coefficient\;of\;y}$$ = -$$\frac{a}{b}$$
Slope of the line perpendicular to given line = $$\frac{b}{a}$$
Now equation of a line having slope $$\frac{b}{a}$$ is given by
y = mx + c
or, y = $$\frac{b}{a}$$x + c
or, ay = bx + ac
or, bx - ay + ac = 0
or, bx - ay +k = 0 where k = ac.
Hence, equation of any line perpendicular to ax + by + c = 0 is given by bx - ay + k = 0 where k is an arbitrary constant.
Angle between the lines y = m1 + c1 and y = m2 + c2 and y = m2x + c2
Condition of perpendicularity
m1m2 = -1
Condition of Parallelism
m1 = m2
Angle between the lines A1x + B1y + C1 = 0 and A2x + B2y +C2 = 0
Condition of perpendicularity
A1A2 + B1B2 = 0
Condition of Parallelism
A1B2 = A2B1
Equation of any line parallel to ax + by + c = 0
k = -bc
Equation of any line perpendicular to ax +by +c = 0
k = ac
.
### Very Short Questions
Given:
a1x + b1y + c1 = 0..............................(1)
a2x + b2y + c2 = 0..............................(2)
Slope of eqn (1), m1 = - $$\frac {x-coefficient}{y-coefficient}$$ = -$$\frac {a_1}{b_1}$$
Slope of eqn (2), m2 = - $$\frac {x-coefficient}{y-coefficient}$$ = -$$\frac {a_2}{b_2}$$
when the lines are parallel, then:
m1 = m2
or,-$$\frac {a_1}{b_1}$$ =-$$\frac {a_2}{b_2}$$
∴ a1b2 = a2b1 Ans
when the lines are perpendicular, then:
m1× m2= - 1
or,-$$\frac {a_1}{b_1}$$×-$$\frac {a_2}{b_2}$$ = - 1
∴ a1a2 = - b1b2Ans
Here,
5x + 4y - 10 = 0............................(1)
15x + 12y - 7 = 0.........................(2)
Slope of eqn (1), m1 = - $$\frac {x-coefficient}{y-coefficient}$$ = -$$\frac 54$$
Slope of eqn (2), m2 = - $$\frac {x-coefficient}{y-coefficient}$$ = -$$\frac {15}{12}$$ = -$$\frac 54$$
m1 = m2 = - $$\frac 54$$
∴ The given two lines are parallel to each other. Proved
Here,
x + 3y = 2..................................(1)
6x - 2y = 9................................(2)
Slope of eqn (1), m1 = - $$\frac {x-coefficient}{y-coefficient}$$ = - $$\frac 13$$
Slope of eqn (2), m2 = - $$\frac {x-coefficient}{y-coefficient}$$ = - $$\frac 6{-2}$$ =3
We have,
m1× m2 =- $$\frac 13$$× 3 = - 1
∴m1× m2 = - 1
Hence, the given two lines are perpendicular each other. Proved
Here,
3x - 2y - 5 = 0..............................(1)
2x + py - 3 = 0.............................(2)
Slope of eqn (1), m1 = - $$\frac {x-coefficient}{y-coefficient}$$ = - $$\frac 3{-2}$$ = $$\frac 32$$
Slope of eqn (2), m2 = - $$\frac {x-coefficient}{y-coefficient}$$ = - $$\frac 2{p}$$
When two lines are parallel lines, then:
m1 = m2
or,$$\frac 32$$ = - $$\frac 2{p}$$
or, p = $$\frac {-2 × 2}3$$
∴ p = -$$\frac 43$$ Ans
Here,
4x + ky - 4 = 0..............................(1)
2x - 6y = 5.............................(2)
Slope of eqn (1), m1 = - $$\frac {x-coefficient}{y-coefficient}$$ = - $$\frac 4k$$
Slope of eqn (2), m2 = - $$\frac {x-coefficient}{y-coefficient}$$ = - $$\frac 2{-6}$$ = $$\frac 13$$
when two line are perpendicular to each other,
m1× m2 = - 1
or, - $$\frac 4k$$ × $$\frac 13$$ = - 1
or, - $$\frac 43$$× - 1 = k
∴ k = $$\frac 43$$ Ans
The formulae of angle between y = m1x + c1 andy = m2x + c2is:
tan$$\theta$$ = ± $$\frac {m_1 - m_2}{1 + m_1m_2}$$
when m1× m2 = -1, the two lines are perpendicular to each other.
when m1 = m2, the two lines are parallel to each other.
Here,
Slope of points (3, -4) and (-2, a)
m1 = $$\frac {y_2 - y_1}{x_2 - x_1}$$ = $$\frac {a + 4}{-2 - 3}$$ = $$\frac {-(a + 4)}5$$
Given eqn is y + 2x + 3 = 0
Slope of above eqn (m2) = - $$\frac {x-coefficient}{y-coefficient}$$ = -$$\frac 21$$ = - 2
when lines are parallel then,
m1 = m2
or,$$\frac {-(a + 4)}5$$ = - 2
or, a + 4 = 10
or, a = 10 - 4
∴ a = 6 Ans
Here,
Slope of the points (3, -4) and (-2, 6) is:
m1 = $$\frac {y_2 - y_1}{x_2 - x_1}$$ = $$\frac {6 + 4}{-2 - 3}$$ = $$\frac {10}{-5}$$ = -2
Slope of the eqn y + 2x + 3 = 0 is:
m2 = - $$\frac {x-coefficient}{y-coefficient}$$ = - $$\frac 21$$ = -2
From above,
m1= m2 = - 2
Hence, the lines are parallel. Proved
Here,
Given eqn is kx - 3y + 6 = 0
Slope of above eqn is:
m1 = - $$\frac {x-coefficient}{y-coefficient}$$ = - $$\frac k{-3}$$ = $$\frac k3$$
Slope ofthe point (4, 3) and (5, -3) is:
m2 = $$\frac {y_2 - y_1}{x_2 - x_1}$$ = $$\frac {-3 - 3}{5 - 4}$$ = - $$\frac 61$$ = -6
If lines are perpendicular then:
m1× m2 = -1
or, $$\frac k3$$× -6 = -1
or, k = $$\frac {-1}{-2}$$
∴ k = $$\frac 12$$ Ans
Here,
Given equations of the lines are:
y = m1x + c1...............................(1)
y = m2x + c2...............................(2)
If $$\theta$$ be the angle between two lines (1) and (2);
The formula of angle between the given lines is:
tan$$\theta$$ =± $$\frac {m_1 - m_2}{1 + m_1m_2}$$
∴ $$\theta$$ = tan-1(± $$\frac {m_1 - m_2}{1 + m_1m_2}$$)
If two lines are perpendicular ($$\theta$$ = 90°)
tan 90° =± $$\frac {m_1 - m_2}{1 + m_1m_2}$$
or,∞=± $$\frac {m_1 - m_2}{1 + m_1m_2}$$
or, $$\frac 10$$ =± $$\frac {m_1 - m_2}{1 + m_1m_2}$$
or, 1 + m1m2 = 0
∴ m1m2 = -1 Ans
Here,
The given equations are:
2x + ay + 3 = 0....................(1)
3x - 2y = 5.............................(2)
Slope of equation (1), m1 = -$$\frac {x-coefficient}{y-coefficient}$$ = - $$\frac 2a$$
Slope of equation (2), m2 = -$$\frac {x-coefficient}{y-coefficient}$$ = - $$\frac 3{-2}$$ = $$\frac 32$$
If equation (1) and equation (2) are perpendicular to each other:
m1× m2 = -1
or,- $$\frac 2a$$×$$\frac 32$$ = -1
or, -6 = - 2a
or, a = $$\frac 62$$
∴ a = 3 Ans
Here,
Given equations of the lines are:
2x + 4y - 7 = 0...........................(1)
6x + 12y + 4 = 0.......................(2)
Slope of equation (1) is: m1 = - $$\frac {x-coefficient}{y-coefficient}$$ = -$$\frac 24$$ = -$$\frac 12$$
Slope of equation (2) is: m2 = - $$\frac {x-coefficient}{y-coefficient}$$ = -$$\frac 6{12}$$ = -$$\frac 12$$
∴ m1 = m2 = $$\frac {-1}2$$
Since, the slope of these equations are equal, the lines are parallel to each other. Proved
Given lines are:
3x + 5y = 7 i.e. 3x + 5y - 7 = 0........................(1)
3y = 2x + 4 i.e. 2x - 3y + 4 = 0........................(2)
Slope of eqn (1), m1 = - $$\frac {x-coefficient}{y-coefficient}$$ = $$\frac {-2}{-3}$$ = $$\frac 23$$
Slope of eqn (2), m2 = - $$\frac {x-coefficient}{y-coefficient}$$ = $$\frac {-3}{5}$$
Let $$\theta$$ be the angle between the equation (1) and (2):
tan$$\theta$$ =± $$\frac {m_1 - m_2}{1 + m_1m_2}$$
or, tan$$\theta$$ =± ($$\frac {\frac 23 + \frac 35}{1 - \frac 23 × \frac 35}$$)
or, tan$$\theta$$ =± ($$\frac {\frac {10 + 9}{15}}{\frac {5 - 2}5}$$)
or, tan$$\theta$$ =± ($$\frac {19}{15}$$ × $$\frac 53$$)
or, tan$$\theta$$ =± $$\frac {19}9$$
For acute angle,
tan$$\theta$$ = $$\frac {19}9$$= 2.11
∴ $$\theta$$ = 65°
∴ The acute angle between two lines is 65°. Ans
Here,
Given equation are:
3y - x - 6 = 0..............................(1)
y = 2x + 5 i.e. -2x + y = 5...................(2)
Slope of eqn (1), m1= -$$\frac {x-coefficient}{y-coefficient}$$ = -$$\frac {(-1)}3$$ = $$\frac 13$$
Slope of eqn (2), m2= -$$\frac {x-coefficient}{y-coefficient}$$ = -$$\frac {(-2)}1$$ = 2
If $$\theta$$ be the angle between the eqn (1) and (2),
tan$$\theta$$ =± $$\frac {m_1 - m_2}{1 + m_1m_2}$$
or, tan$$\theta$$ =± $$\frac {\frac 13 - 2}{1 + \frac 13 × 2}$$
or, tan$$\theta$$ =± $$\frac {\frac {1 - 6}3}{\frac {3 + 2}3}$$
or, tan$$\theta$$ =± $$\frac {-5}3$$× $$\frac 35$$
∴ tan$$\theta$$ =± (-1)
Taking -ve sign,
tan$$\theta$$ = +1
tan$$\theta$$ = 45°
∴$$\theta$$ = 45° Ans
Here,
x - 3y = 4......................(1)
2x - y = 3......................(2)
Slope of eqn (1), m1 = -$$\frac {x-coefficient}{y-coefficient}$$ = $$\frac {-1}{-3}$$ = $$\frac 13$$
Slope of eqn (2), m2 = -$$\frac {x-coefficient}{y-coefficient}$$ = $$\frac {-2}{-1}$$ = 2
If $$\theta$$ be the angle between the eqn (1) and (2),
tan$$\theta$$ =± $$\frac {m_1 - m_2}{1 + m_1m_2}$$
or, tan$$\theta$$ =± $$\frac {\frac 13 - 2}{1 + \frac 13 × 2}$$
or, tan$$\theta$$ =± $$\frac {\frac {1 - 6}3}{\frac {3 + 2}3}$$
or, tan$$\theta$$ =± $$\frac {-5}3$$× $$\frac 35$$
∴ tan$$\theta$$ =± (-1)
Taking -ve sign,
tan$$\theta$$ = +1
tan$$\theta$$ = 45°
∴$$\theta$$ = 45° Ans
Here,
Given equation are:
y - 3x - 2 = 0
or, -3x + y - 2 = 0..............................(1)
y = 2x + 5
or, - 2x + y = 5...................................(2)
Slope of eqn (1), m1 = -$$\frac {x-coefficient}{y-coefficient}$$ = $$\frac {-3}{-1}$$ = 3
Slope of eqn (2), m2 = -$$\frac {x-coefficient}{y-coefficient}$$ = -$$\frac {-2}{1}$$ =2
If $$\theta$$ be the angle between two lines,
tan$$\theta$$ =± $$\frac {m_1 - m_2}{1 + m_1m_2}$$
or, tan$$\theta$$ =± $$\frac {3 - 2}{1 + 3 × 2}$$
or, tan$$\theta$$ =± $$\frac 1{1 + 6}$$
∴ tan$$\theta$$ =± $$\frac 17$$
Taking +ve sign,
$$\theta$$ = tan-1($$\frac 17$$)
∴ $$\theta$$ = 8.13° Ans
Here,
Given lines are:
x = 3y + 8
i.e. x - 3y - 8 = 0..................................(1)
2x + 11 = 7y
i.e. 2x - 7y + 11 = 0............................(2)
Slope of eqn (1),m1= - $$\frac {x-coefficient}{y-coefficient}$$ = - $$\frac 1{-3}$$ = $$\frac 13$$
Slope of eqn (2),m2= - $$\frac {x-coefficient}{y-coefficient}$$ = - $$\frac 2{-7}$$ = $$\frac 27$$
Let $$\theta$$ be the angle between the equation (1) and (2),
tan$$\theta$$ =± $$\frac {m_1 - m_2}{1 + m_1m_2}$$
or, tan$$\theta$$ = ± $$\frac {\frac 13 - \frac 27}{1 + \frac 13 × \frac 27}$$
or, tan$$\theta$$ =± $$\frac {\frac {7 - 6}{21}}{\frac {21 + 2}{21}}$$
or, tan$$\theta$$ =± $$\frac 1{21}$$× $$\frac {21}{23}$$
∴ tan$$\theta$$ =± $$\frac 1{23}$$
For obtuse angle,
tan$$\theta$$ = -$$\frac 1{23}$$
or, tan$$\theta$$ = tan (180° -2°)
∴ $$\theta$$ = 178°
∴ The obtuse angle between two lines is 178°. Ans
0%
140(^0)
145(^0)
180(^0)
135 (^0)
• ### What will be the formula of the angle between the lines y = m1x+C1 and y = m2x + c2.
θ = (tan-^1) (frac{m2-m1}{±1-m1-m2})
θ = (tan-^1) (frac{m1+m2}{±1+m1+m2})
θ = (tan-^1) (frac{m1-m2}{±1+m1-m2})
θ = (tan-^1) (frac{m2-m1}{±1-m1+m2})
• ### If the straight lines px+3y-12 = 0 and 4y-3x+7 = 0 are parallel to each other , find the value of P.
(frac{9}{4})
(frac{3x}{4y})
(frac{-9}{4})
(frac{-9}{-4})
2
4
3
6
• ### What will be the slope of the straight line perpedicular to 4x+3y = 12.
(frac{2}{3})
(frac{6}{12})
(frac{3}{4})
(frac{4}{3})
• ### If the straight lines 2x+3y+6 = 0 and ax-5y+20 = 0 are perpendicular to each other , find the value of a .
(frac{15}{2})
(frac{3}{4})
(frac{2}{17})
(frac{15}{3})
5y+7x+13 = 0
5y+7x+13 = 0
5x+7y+31 = 0
5x+31+y7 = 0
9x+ 9= y
9x+ y= 4
9x+ y= 9
4x+ y= 9
4x+3y = 81
4x+ 3y= 9
3x+4y = 18
18x+4y = 0
• ### Find the equation of straight lines passing through a point (-6 , 4) and perpendicular to the line 3x-4y + 9 = 0 .
3x+4y+12 = 12
4x+3y+12 = 0
0+3y+12 = 4x
4y+3x+0 = 12
7x+5y = 44
5x-7y = 44
7x-5y = 44
44x-5y = 7x
13y+9x+3 = 0
3y-13-0 = 9x
9x+3y+13 = 0
3y+13+9x = 0
• ### Find the equation of the straight lines passing through the point (2 , 3) and making an angle of 45(^0) with the line x-3y = 2.
2x-1y = 1 , 8-2y = 0
2x-y = 1 , x+2y=8
1-2xy = 0 , 2y-8 = x
1-2xy = 0 , 2y-8 = x
• ### Find the equation of the straight lines passing through the point (1,0) & inclined at an angle of 30(^0) with the line x- (sqrt{3y}) = 4.
y=0 and (sqrt{3x})-y = (sqrt{3})
y=0 and (sqrt{-3x})-y = (sqrt{-3})
y=0 and (sqrt{1x})-y = (sqrt{1})
y=0 and (sqrt{4x})-y = (sqrt{4})
1, 1
0, 0
1, -1
-1, -1
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##### Find the equation of the straight line passing through the point (5,0)and making 45° with the line 4x-5y 9=0
Find the equation of the straight line passing through the point(5,0)and making 45°withthe line 4x-5y 9=0
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# Geometric Tools: Description And Uses
Geometric tools are the instruments used to draw different types of geometric shapes. In Maths, geometry is the most crucial topic, where we learn about the shapes of objects. To draw these shapes, there are different types of tools with different names used while solving problems based on geometry. We come across a number of shapes in our day to day life. You are already aware of a couple of them such as triangles, squares, hexagons, circles, parallelograms and so on. Different figures have characteristic properties such as length, breadth, diameter, etc. which set them apart from one another and are parameters to produce the figures on paper. Practical geometry or Euclidean geometry is the most pragmatic branch of geometry that deals with the construction of different geometrical figures using geometric instruments such as rulers, compasses and protractors. In this article, we are going to discuss important Maths tools which are used in Geometry for the construction of different shapes with a complete explanation.
## Geometric Tools and Their Uses
The early geometers studied figures such as points, lines and angles which required the use of rulers and compasses only. With the advent of more geometric designs, other geometric tools were invented. Some of the most commonly used geometric tools are:
Now, let us discuss the different Maths tools with their description and uses in detail.
## Geometric Tool – Ruler
Also known as the straightedge, a ruler is used to construct straight lines and measure the lengths of a line segment. A ruler is also known as the line gauge, is a mathematical tool used in Geometry as well as in many Engineering applications. It is a straight edge initially used only for drawing straight lines. It can also be used to measure the distance of the object. However, for our convenience, it is graduated into centimetres and millimetres on one side and into inches on the other side. It is used to measure the lengths in both the metric units as well as in customary units. The interval or the marks mentioned in the ruler is called the “Hash Marks”.
How to Measure Objects Using a Ruler?
If we want to measure the length of the object using a ruler, we need to place the zero hash mark of the ruler on the one end of the object. Now, align the object along the edge of the ruler. Read the measurement on the hash mark of the ruler, in which the other side of the object ends. The distance between the zero hash mark and the final reading hashmark on the ruler is called the length of the object.
## Geometric Tool – Compass
It is a ‘V’-shaped tool that holds a pencil on one side of the ‘V’ and a pointer on the other side. The distance between the pencil and the pointer is adjustable. It is used to trace arcs, circles and angles. It is also used to mark equal lengths. Compass is one of the mathematical tools which is used to draw a geometrical figure, such as a circle. It is also used in the intersection of line segments and tools, which is used to intersect the line segment and helps to find the midpoint of the shapes.
## Geometric Tool – Protractor
A protractor in Maths is used to measure angles. It is a semi-circular disc used to draw and measure different types of angles. It is graduated from 0 to 180 degrees and can be directly used to measure any angle within its range. It has two sets of markings, 0 to 180 degrees from left to right and vice versa. The inner and the outer reading of the protractor supplements each other. It means that if we add inner and outer reading, it adds up to 180 degrees.
While measuring an angle using the protractor, we need to follow the steps given below:
• If the angle to be measured at the left side of the protractor, we have to use the outer reading provided in the protractor.
• If the angle to be measured at the right side of the protractor, we have to use the inner reading provided in the protractor.
## Geometric Tool – Divider
It looks similar to the compasses with its ‘V’-shaped structure. However, it has pointers on both ends of the ‘V’. The distance between them is adjustable and it is used to measure and compare lengths.
The term divider and compass are often used to mark the distance or mark divisions. The major difference between the compass and divider is that the compass is defined as a drafting instrument, in which one sharp point is positioned at centre and the other point has a pen or pencil to describe the circle or other functionality. Whereas the divider has two sharp points, in which one sharp point used as a centre, and the other sharp point is used for marking.
## Geometric Tool – Set-Squares
Set squares are also one of the Maths tool, which is found commonly in geometry box. These are the triangular pieces of plastic with some portion between them removed. There are two kinds of set squares available in the market. One has an angle of 45 degrees and the other has 30-60 degrees. The 45 degree set square has an angle of 90 degrees, and also 30/60 degree set square has a right angle. The 45 degrees set square is used to draw vertical lines. With the help of set squares, we can draw parallel lines, perpendicular lines, make some standard angles, and so on.
## Frequently Asked Questions on Geometric Tools
### What are Geometric Tools in Maths?
Geometric tools are those instruments that are used to construct different types of geometric shapes and figures. In Euclidean geometry, all the shapes, lines, and angles are drawn using geometric tools.
### What are the 2 Most Important Euclidean Tools Used in Geometric Construction?
The two most important Euclidean tools used by early Greeks to construct different geometrical shapes and angles were a compass and a straightedge. Using these two tools, almost any shape can be constructed. Later, rulers and protectors were invented which made construction easier.
### What are the Most Common Instruments in a Geometry Box?
The most common tools in a student geometric box are:
• Compass
• Ruler
• Protractor
• Divider
• Set-squares
### What is the use of a Ruler in Geometry?
In Geometry, a ruler is used to draw straight lines. Rulers are also used to measure the length of the object.
### Why do we use Protractor in Maths?
A protractor is used to measure the angles. The protractor is graduated from 0 degree to 180 degrees, which helps to measure the angle value. It has two sets of markings from 0-180 degree from left to right as well as from right to left. The inner and outer angles will add up to 180 degrees.
Quiz on Geometric tools
#### 1 Comment
1. Soma Chattarzee
Pls fo this
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# How do you find the center and radius of the circle x^2+y^2+8x+4y+16=0?
Sep 1, 2016
Center: color(green)(""(-4,-2))
Radius: $\textcolor{g r e e n}{2}$
#### Explanation:
If a circular equation is written in the form:
$\textcolor{w h i t e}{\text{XXX}} {\left(x - \textcolor{red}{a}\right)}^{2} + {\left(y - \textcolor{b l u e}{b}\right)}^{2} = {\textcolor{g r e e n}{r}}^{2}$
then it has a center at $\left(\textcolor{red}{a} , \textcolor{b l u e}{b}\right)$ and a radius of $\textcolor{g r e e n}{r}$
We will want to manipulate the given: ${x}^{2} + {y}^{2} + 8 x + 4 y + 16 = 0$
into this form.
First separating the $x$ terms, the $y$ terms and the constant as
$\textcolor{w h i t e}{\text{XXX}} \left(\textcolor{red}{{x}^{2} + 8 x}\right) + \left(\textcolor{b l u e}{{y}^{2} + 4 y}\right) = \textcolor{g r e e n}{- 16}$
Completing the square for each of the $x$ and $y$ sub-expressions:
$\textcolor{w h i t e}{\text{XXX}} \left(\textcolor{red}{{x}^{2} + 8 x + 16}\right) + \left(\textcolor{b l u e}{{y}^{2} + 4 y + 4}\right) = \textcolor{g r e e n}{- 16} \textcolor{red}{+ 16} \textcolor{b l u e}{+ 4}$
color(white)("XXX")(color(red)(x+4))^2+(color(blue)(y+2))^2= color(green)(4
$\textcolor{w h i t e}{\text{XXX}} {\left(x - \left(\textcolor{red}{- 4}\right)\right)}^{2} + {\left(y - \left(\textcolor{b l u e}{- 2}\right)\right)}^{2} = {\textcolor{g r e e n}{2}}^{2}$
with center at $\left(\textcolor{red}{- 4} , \textcolor{b l u e}{- 2}\right)$ and radius $\textcolor{g r e e n}{2}$
A graph of the original equation helps verify this result:
graph{x^2+y^2+8x+4y+16=0 [-9.184, 1.916, -4.89, 0.66]}
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# General Equation of the Circle – Formula and Examples
## What is a Circle ?
A circle is the locus of a point which moves in a plane in such way that its distance from a fixed point(in the same given plane) remains constant. The fixed point is called center of circle and the constant distance is called radius of the circle. The general equation of the circle and standard form of circle are given below with examples.
## General Equation of the Circle :
The general equation of the circle is
$$x^2 + y^2 + 2gx + 2fy + c$$ = 0.
where g, f, c are constants and center is (-g, -f) and radius r = $$\sqrt{g^2 + f^2 – c}$$.
(i) If $${g^2 + f^2 – c}$$ > 0, then r is real and positive and the circle is a real circle.
(ii) If $${g^2 + f^2 – c}$$ = 0, then radius r = 0 and circle is a point circle.
(iii) If $${g^2 + f^2 – c}$$ < 0, then r is imaginary and circle is also an imaginary circle with real center.
(iv) If $$x^2 + y^2 + 2gx + 2fy + c$$ = 0, has three constants and to get the equation of the circle at least three conditions should be known $$\implies$$ A unique circle passes through three non-collinear points.
Example : Find the center and radius of the circle $$3x^2 + 3y^2 – 8x -10y + 3$$ = 0.
Solution : We rewrite the equation as, dividing whole equation by 3
$$x^2 + y^2 – {8\over 3}x – {10\over 3}y + 1$$ = 0
Comparing it with general equation of the circle $$x^2 + y^2 + 2gx + 2fy + c$$ = 0, we get
g = -$$4\over 3$$, f = -$$5\over 3$$, c = 1.
Hence center of the circle is ($$4\over 3$$, $$5\over 3$$)
and radius is $$\sqrt{{16\over 9}+{25\over 9}-1}$$ = $$\sqrt{32\over 9}$$.
## General Second Degree Equation :
The general second degree in x and y, $$ax^2 + by^2 + 2hxy + 2gx + 2fy + c$$ = 0 represents a circle if :
(i) coefficient of $$x^2$$ = coefficient of $$y^2$$ or a = b $$\ne$$ 0.
(ii) coefficient of xy = 0 or h = 0.
(iii) $${g^2 + f^2 – c}$$ $$\ge$$ 0(for a real circle).
## Standard Form of Circle :
If (h, k) is the center and r is the radius of the circle then standard form of circle equation is
$$(x-h)^2 + (y-k)^2$$ = $$r^2$$.
Note :
(i) If centre is origin (0,0) and radius is ‘r’ then the equation of circle is $$x^2$$ + $$y^2$$ = $$r^2$$
(ii) If radius of circle is zero then the equation of circle is $$(x-h)^2 + (y-k)^2$$ = 0. Such a circle is called zero circle or point circle.
(iii) When circle touches x-axis then equation of the circle is $$(x-h)^2 + (y-k)^2$$ = $$k^2$$
(iv) When circle touches y-axis then the equation of circle is $$(x-h)^2 + (y-k)^2$$ = $$h^2$$
(v) When circle touches both the axes(x-axis and y-axis) then equation of circle is $$(x-h)^2 + (y-h)^2$$ = $$h^2$$
(vi) When circle passes through the origin and centre of the circle is (h,k) then radius $$\sqrt{h^2 + k^2}$$ = r and intercept cut on x-axis is 2h and intercept cut on y-axis is 2k and equation of circle is $$(x-h)^2 + (y-k)^2$$ = $$h^2 + k^2$$ or $$x^2$$ + $$y^2$$ – 2hx – 2ky = 0
Example : Find the center and radius of the circle $$x^2 + y^2$$ = 9.
Solution : Comparing it with standard form of circle equation is $$(x-h)^2 + (y-k)^2$$ = $$r^2$$. we get
h = 0, k = 0, r = 3.
Hence center of the circle is (0, 0).
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# Continuous Functions
A function is continuous when its graph is a single unbroken curve ...
... that you could draw without lifting your pen from the paper.
That is not a formal definition, but it helps you understand the idea.
Here is a continuous function:
## Examples
So what is not continuous (also called discontinuous) ?
Look out for holes, jumps or vertical asymptotes (where the function heads up/down towards infinity).
Not Continuous
(hole)
Not Continuous
(jump)
Not Continuous
(vertical asymptote)
Try these different functions so you get the idea:
(Use slider to zoom, drag graph to reposition, click graph to re-center.)
## Domain
A function has a Domain.
In its simplest form the domain is all the values that go into a function.
We may be able to choose a domain that makes the function continuous
### Example: 1/(x−1)
At x=1 we have:
1/(1−1) = 1/0 = undefined
So there is a "discontinuity" at x=1
f(x) = 1/(x−1)
So f(x) = 1/(x−1) over all Real Numbers is NOT continuous
Let's change the domain to x>1
g(x) = 1/(x−1) for x>1
So g(x) IS continuous
In other words g(x) does not include the value x=1, so it is continuous.
When a function is continuous within its Domain, it is a continuous function.
## More Formally !
We can define continuous using Limits (it helps to read that page first):
A function f is continuous when, for every value c in its Domain:
f(c) is defined,
and
limx→cf(x) = f(c)
"the limit of f(x) as x approaches c equals f(c)"
The limit says:
"as x gets closer and closer to c
then f(x) gets closer and closer to f(c)"
And we have to check from both directions:
as x approaches c (from left) then f(x) approaches f(c) AND as x approaches c (from right) then f(x) approaches f(c)
If we get different values from left and right (a "jump"), then the limit does not exist!
And remember this has to be true for every value c in the domain.
## How to Use:
Make sure that, for all x values:
• f(x) is defined
• and the limit at x equals f(x)
Here are some examples:
### Example: f(x) = (x2−1)/(x−1) for all Real Numbers
The function is undefined when x=1:
(x2−1)/(x−1) = (12−1)/(1−1) = 0/0
So it is not a continuous function
Let us change the domain:
### Example: g(x) = (x2−1)/(x−1) over the interval x<1
Almost the same function, but now it is over an interval that does not include x=1.
So now it is a continuous function (does not include the "hole")
It looks like this:
It is defined at x=1, because h(1)=2 (no "hole")
But at x=1 you can't say what the limit is, because there are two competing answers:
• "2" from the left, and
• "1" from the right
so in fact the limit does not exist at x=1 (there is a "jump")
And so the function is not continuous.
But:
### Example: How about the piecewise function absolute value:
At x=0 it has a very pointy change!
But it is still defined at x=0, because f(0)=0 (so no "hole"),
And the limit as you approach x=0 (from either side) is also 0 (so no "jump"),
So it is in fact continuous.
(But it is not differentiable at x=0)
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# If 75 percent of m is equal to k percent of 25, where k > 0, what is the value of m/k ?A). $\dfrac{3}{{16}}$ B). $\dfrac{1}{3}$ C). $\dfrac{3}{4}$D). $3$E). $\dfrac{{16}}{3}$
Last updated date: 19th Sep 2024
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Hint: The question is about the percentage which is a number of some sort of ratio i.e. express in terms of the fraction of the 100. Now if we have to solve the percent of any number that number is divided by the whole with multiply by 100 therefore the percentage is also known as the part per hundred or hundred part of one.
Complete step-by-step solution:
Step 1:
First we write the given data in terms of the equation.
$75\% \,of\,m = k\% \,of\,25$
$\Rightarrow 75\% \, = \dfrac{{75}}{{100}},k\% \, = \dfrac{k}{{100}}$
Now put the values
$75\% \,\, \times m = k\% \, \times \,25$
$\Rightarrow \dfrac{{75}}{{100}}\,\, \times m = \dfrac{k}{{100}}\,\, \times \,25$
Step 2:
Simplify this expression by using the cross multiplication and solve it. We got the desired ratio so
Now
$\dfrac{{75}}{{100}}\,\, \times m = \dfrac{k}{{100}}\,\, \times \,25$
$\Rightarrow \dfrac{{75m}}{{100}}\,\, = \dfrac{{\,25k}}{{100}}\,\,$
$\Rightarrow 75m = 25k$
Now perform the cross multiplication so that we can take it as desired ratio and get result after simplifying the ratio so
$\dfrac{m}{k} = \dfrac{{25}}{{75}}$
$\Rightarrow \dfrac{m}{k} = \dfrac{1}{3}$
Hence we got the ratio of the m/k in terms of the rational numbers and it is matched with one of the given options provided by the question and the correct answer of the question is an option (b) satisfied.
Note: The good thing about percentage is that it can be solved easily. We have to just equal the percentage sign to the hundred part of one since the percentage is equal to the hundred part of one. And also we mentioned that the word percent can be divided like per plus cent or per hundred which is the same as we mentioned above.
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How to graph and solve systems of linear inequalities, using the double-number line?
Oct 9, 2017
Solving system of linear inequalities.
Using a double number-line to solve a system of 2 linear inequalities. (Math article written by Nghi Nguyen)
Explanation:
Example 1 . Solve the system:
2x - 3 < 11 (1)
3x > x + 6 (2)
First, solve the 2 inequalities:
Solve (1) --> 2x < 14 --> $x < 7$
On a number-line, draw the solution set:
+++++++++++++0 +++++3++++++++++7------------------
Solve (2) --> 3x - x > 6 --> 2x > 6 --> $x > 3$
Draw the solution set on the second number-line:
----------------------0---------3++++++++++7+++++++++++
By superimposing, we see that the combined solution set of the system is the open interval (3, 7). The 2 end points are not included.
Example 2 . Solve the system:
3x + 7 < x - 1 (1)
2x - 4 > - x + 5 (2)
Solve (1) --> 2x < - 8 -->$x < - 4$
Solve (2) --> 3x > 9 --> $x > 3$
+++++++++++ -4--------------- 0 -------------3----------------------
------------------ -4 -------------------------------3++++++++++++++
By superimposing, we see that the combined solution set are the
open intervals: (-inf. -4) and (3, +inf.)
NOTE. This method can be used to solve a system of 3 linear inequalities.
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# Eigenvalues and eigenvectors of a matrix
On this post we explain what the eigenvalues and the eigenvectors of a matrix are. You will also find examples of how to compute the eigenvalues and the eigenvectors of a matrix, and finally, you have problems with solutions solved step by step to practice.
## What are the eigenvalues and the eigenvectors of a matrix?
Although the concept of eigenvalue and eigenvector is a bit difficult to understand, their definition is as follows:
The eigenvectors are the non-zero vectors of a linear map that, when the linear transformation is applied to them, result in a scalar multiple of them (they do not change direction). This scalar is the eigenvalue.
Where is the matrix of the linear mapping, is the eigenvector and the eigenvalue.
An eigenvalue is also known as a characteristic value, and an eigenvalue as a characteristic vector.
## How to find the eigenvalues and the eigenvectors of a matrix
To find the eigenvalues and eigenvectors of a matrix, apply the following procedure:
1. Calculate the characteristic polynomial by taking the following determinant:
2. Find the roots of the characteristic polynomial obtained in step 1. These roots are the eigenvalues of the matrix.
3. Calculate the eigenvector associated with each eigenvalue by solving the following system of equations for each eigenvalue:
Remember that is the identity matrix.
This is the method to find the eigenvalues and eigenvectors of a matrix, but here we also give you some tricks: 😉
Tricks: we can take advantage of the properties of the eigenvalues and eigenvectors to calculate them more easily:
The trace of the matrix (sum of the elements on the main diagonal) is equal to the sum of all the eigenvalues.
The product of all the eigenvalues is equal to the determinant of the matrix.
If there is any linear combination between rows or columns, at least one eigenvalue of the matrix is 0.
Having seen the theory of eigenvalues and eigenvectors, let’s see an example of how to compute them.
## Example of calculating the eigenvalues and eigenvectors of a matrix
• Find the eigenvalues and the eigenvectors of the following 2×2 matrix:
First, we have to find the characteristic polynomial of the matrix. And, for this, we must compute the following determinant:
Now we calculate the roots of the characteristic polynomial. So we set the polynomial obtained equal to 0 and solve the equation:
The solutions of the equation are the eigenvalues of the matrix.
Once we know the eigenvalues of the matrix, we proceed to calculate its eigenvectors. To do this, we must solve the following system of equations for each eigenvalue:
First we calculate the eigenvector associated with the eigenvalue 1:
From these equations we obtain the following subspace:
Note that subspaces of eigenvectors are also called eigenspaces.
Now we must find a base of this eigenspace, so we give, for example, the value 1 to the variable x and we obtain the following eigenvector:
Once we have found the eigenvector associated with the eigenvalue 1, we repeat the process to calculate the eigenvector of the eigenvalue 2:
In this case, only the first component of the vector must be 0, so we could give any value to variable y. But it is better to put a 1:
In conclusion, the eigenvalues and the eigenvectors of the matrix are:
## Practice problems on eigenvalues and eigenvectors
### Problem 1
Calculate the eigenvalues and eigenvectors of the following square matrix of order 2:
First we calculate the determinant of the matrix minus λ on its main diagonal:
Now we calculate the roots of the characteristic polynomial:
We find the eigenvector associated with the eigenvalue 2:
And then we calculate the eigenvector associated with the eigenvalue 5:
Thus, the eigenvalues and eigenvectors of matrix A are:
### Problem 2
Determine the eigenvalues and eigenvectors of the following 2×2 square matrix:
First we subtract λ from the entries on the main diagonal of the matrix and compute the determinant of the resulting matrix in order to obtain the characteristic equation:
Now we find the roots of the characteristic polynomial:
We calculate the eigenvector associated with the eigenvalue -1:
And then we calculate the eigenvector associated with the eigenvalue 3:
Therefore, the eigenvalues and eigenvectors of matrix A are:
### Problem 3
Compute the eigenvalues and eigenvectors of the following square matrix of order 3:
First of all, we have to solve the determinant of matrix A minus the identity matrix multiplied by lambda in order to obtain the characteristic polynomial:
In this case, the last column of the determinant has two zeros, so we will evaluate the 3×3 determinant by cofactors through that column:
Now we have to calculate the roots of the characteristic polynomial. It is better not to multiply the parentheses since then we would obtain a third degree polynomial, instead, if the two factors are solved separately it is easier to get the eigenvalues:
Now we compute the eigenvector associated with the eigenvalue 2:
We calculate the eigenvector associated with the eigenvalue -1:
And we calculate the eigenvector associated with the eigenvalue 3:
So the eigenvalues and eigenvectors of matrix A are:
### Problem 4
Compute the eigenvalues and the eigenvectors of the following 3×3 square matrix:
First we solve the determinant of the matrix minus λ on its main diagonal to obtain the characteristic polynomial:
We factor out the characteristic polynomial and solve for λ each equation:
We calculate the eigenvector associated with the eigenvalue 0:
We calculate the eigenvector associated with the eigenvalue 2:
We calculate the eigenvector associated with the eigenvalue 5:
Therefore, the eigenvalues and eigenvectors of matrix A are:
### Problem 5
Find the eigenvalues and eigenvectors of the following 4×4 matrix:
First, we solve the determinant of the matrix minus λ on its main diagonal to obtain the characteristic polynomial:
In this case, the last column of the determinant are all zeros except one element, so we will compute the determinant using the cofactor method through that column:
Now we must calculate the roots of the characteristic polynomial. It is better not to multiply the parentheses since then we would obtain a fourth degree polynomial, instead, if the two factors are solved separately it is easier to calculate the eigenvalues:
We calculate the eigenvector associated with the eigenvalue 0:
We calculate the eigenvector associated with the eigenvalue -1:
We calculate the eigenvector associated with the eigenvalue 3:
The algebraic multiplicity of eigenvalue 3 is 2 (it is repeated twice). So we must find another eigenvector that satisfies the same equations:
Therefore, the eigenvalues and eigenvectors of matrix A are:
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# Assignment 1: Problems and Explorations with Graphing Functions and Relations
Graph the equation:
where |x| is the absolute value of x. Variations?
Exploration 1:
Here the graph shows the x and y intercepts to be - 1 and + 1. Recall that the graph of a circle with radius 1 is x^2 + y^2 = 1. The variation of this graph is the additive -y|x|. So the next question is which variables can or should we manipulate to show us variations within this graph? Through graphical manipulations we should be able to understand exactly how this equation behaves and why.
Exploration 2:
By simply changing our (-y|x|) to be positive the entire graph shifts 180 degrees. Does it rotate clockwise or counterclockwise? Are we able to see the rotation step by step or is it only one move with nothing in between? Notice that our x and y intercepts remained the same regardless of the sign of y. Let's take a step away from this expression containing the absolute value of x and look at two graphs below.
Exploration 3:
Do you see the shape of an upright heart within the two ellipses? Now look for all four heart shapes. Using what we know about absolute values we can see that the red and blue graph represent together part of the graph in Exploration 1. When the graphs from Exploration 1, 2, and 3 are combined all four graphs lay upon one another thus confirming that they are each variations of each other. Notice that the equations labeled black overpower the color of the other equations.
We have confirmed through our graphical manipulations that the "heart shape" is formed by joining two ellipses. But why are the bottoms of the ellipses removed when the absolute value is put around the x? This leads us to our next exploration ...what happens when the absolute value is around the y variable?
Exploration 4:
By placing the absolute value around the y variable we see that the graph has turned 90 degrees or -90 degrees when x|y| is negative. Now looking back at Exploration 3 we see the two hearts aligned with the x-axis. So now to tie our four explorations together we see that...
forms two ellipses. Our two ellipses above are equal to the ellipses in Exploration 3. Therefore, our final conclusion is by rearranging the absolute value in the graphs we can breaks apart the ellipses to form the "heart shaped" graphs. We can see the truth behind this statement through the graphs above, but what is the mathematical explanation?
Let's take one more look at another exploration to see if it will tie all of the loose ends.
Exploration 5:
Finally, we see what makes the heart stretch. This stretch also works when the absolute value is around the y variable, which maintains the consistency of the graph manipulations. The mathematical explanation for this graph begins first with the x and y intercepts. In all of these explorations the intercepts never changed. No matter how the graph was stretched or where the absolute value is located the intercepts remained -1 and 1 on both the x and y axis. We also see that without the middle term, y|x|, our graph is just a circle with with radius 1. Adding the middle term, yx, we get an ellipse. And adding the absolute value we get the "heart" graph, which flips across the x-axis when the sign of the y variable changes. Using our knowledge of algebraic equations, we can see that for our equation with two squared variables and a middle term to equal one our middle term must always be negative. If the middle term were positive we would end up with an equation greater than 1. Therefore, the "heart" flips across the y-axis when the y variable is positive because we need a negative value of the middle term to maintain equality. Look at the points (1,1), (-1,1), (-1,-1), and (1,-1) in all five explorations. When plugged into our equations we see that the middle value is always negative. Finally, the "heart" stretches according to the value of the scalar multiplied to the middle term.
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The area that wasn't subtracted (grey) is the area of the polygon. Area of the polygon = $$\dfrac{7 \times 5 \times 4.82}{2} = 84$$ sq. [48] If a convex quadrilateral has the consecutive sides a, b, c, d and the diagonals p, q, then its area K satisfies[34], From Bretschneider's formula it directly follows that the area of a quadrilateral satisfies. The area can be expressed in trigonometric terms as[13]. a The formulas for areas of unlike polygon depends on their respective shapes. $$\therefore$$ Area occupied by square photo frame is $$25$$ sq. Any quadrilateral that is not self-intersecting is a simple quadrilateral. In addition to the four sides, we'll need to know either a diagonal or the degrees of one of the angles. This reduces to Brahmagupta's formula for the area of a cyclic quadrilateral—when A + C = 180° . Polygon. = Help her find the area the photo frame will occupy? A side of the Varignon parallelogram is half as long as the diagonal in the original quadrilateral it is parallel to. $$\therefore$$ Area of a regular polygon is $$84$$ sq. A regular polygon is a polygon that has all sides of the same length and all the angles measuring the same. This relation can be considered to be a law of cosines for a quadrilateral. p Lv 7. & Ajibade, A. O., "Two conditions for a quadrilateral to be cyclic expressed in terms of the lengths of its sides". ( L =Side length of a polygon. 96√3 in2 . vertex. The vertex C lies on the line x − y = 2. RS Aggarwal Solutions for Class 8 Maths Chapter 18 – Area of a Trapezium and a Polygon, are given here. Area of a cyclic quadrilateral. Whence[21]. The area of a regular polygon formula is given as follows: where $$n$$ is the number of sides, $$s$$ is the length of one side, and $$a$$ is known as apothem(it is the line from the center of the regular polygon that is perpendicular to one of its sides. with free interactive flashcards. There are five other triangles like $$\bigtriangleup$$OED, hence, the area of a regular hexagon can be given by summing each one of them or multiplying the area of one triangle by the number of sides. Depending on the information that are given, different formulas can be used to determine the area of a polygon, below is a list of these formulas: . In a crossed quadrilateral, the four "interior" angles on either side of the crossing (two acute and two reflex, all on the left or all on the right as the figure is traced out) add up to 720°.[9]. 4 hours ago. Area of Regular Polygons. q When it comes to the area of polygons with more than four sides, there aren’t a lot of rules you can follow. b tan In any convex quadrilateral the bimedians m, n and the diagonals p, q are related by the inequality, with equality holding if and only if the diagonals are equal. polygon area Sp . K {\displaystyle B} peace Area of Polygons : A polygon is a plane shape with straight sides. The dual theorem states that of all quadrilaterals with a given area, the square has the shortest perimeter. 36√3 sq. If the interior angles of a polygon add up to 900°, how many sides does the polygon have? [41]:p.120, The centre of a quadrilateral can be defined in several different ways. It is always a two-dimensional plane. The area of any given polygon whether it a triangle, square, quadrilateral, rectangle, parallelogram or rhombus, hexagon or pentagon, is defined as the region occupied by it in a two-dimensional plane. Minimum area of a Polygon with three points given. This is a special case of the n-gon interior angle sum formula: (n − 2) × 180°. The area and the side length of the polygons are provided in these middle school worksheets. Just as one requires length, base and height to find the area of a triangle. Where ‘n’ is equal to the number of sides of a polygon. which is half the magnitude of the cross product of vectors AC and BD. The area of any quadrilateral also satisfies the inequality[35], Denoting the perimeter as L, we have[35]:p.114. Thus. The segments of a polygonal circuit are called its edges or sides, and the points where two edges meet are the polygon's vertices (singular: … number of sides n: n＝3,4,5,6.... circumradius r: side length a . If a polygon has a total of 119 possible diagonals, how many sides does it have? They are made of straight lines, and the shape is "closed" (all the lines connect up). circle area Sc . . D . }, Another area formula including the sides a, b, c, d is[15]. Regular heptadecagon. 19, Sep 19. These … The "vertex centroid" comes from considering the quadrilateral as being empty but having equal masses at its vertices. Leversha, Gerry, "A property of the diagonals of a cyclic quadrilateral". B Our master faculty group has arranged solutions so as to assist you with your test readiness to obtain great marks in Maths. Note that the two opposite sides in these formulas are not the two that the bimedian connects. 4 hours ago. Since many properties aren't a simple … But before you get your hands on the calculator, it's vital that you know the area of a regular polygon formula and regular polygon definition, as blindly using the calculator will lose its purpose. The perpendicular is dividing the side into two parts. At Cuemath, our team of math experts is dedicated to making learning fun for our favorite readers, the students! For 4 sides with specified lengths and area, I think there are generally 2 discrete possible solutions (one convex and one concave). Polygon example. Hence, the area of a regular polygon in terms of perimeter is given as follows: Let's take an example of a regular hexagon. This is how we can find out or calculate the area of a polygon in Java. A quadrilateral is a polygon in Euclidean plane geometry with four edges (sides) and four vertices (corners). The two diagonals p, q and the four side lengths a, b, c, d of a quadrilateral are related[13] by the Cayley-Menger determinant, as follows: The internal angle bisectors of a convex quadrilateral either form a cyclic quadrilateral[22]:p.127 (that is, the four intersection points of adjacent angle bisectors are concyclic) or they are concurrent. The intersection of the bimedians is the centroid of the vertices of the quadrilateral.[13]. θ There are various general formulas for the area K of a convex quadrilateral ABCD with sides a = AB, b = BC, c = CD and d = DA. But two such points can be constructed in the following way. | Can someone do 61 for me? $\begingroup$ You can make a rhombus of arbitrarily small area with four equal sides. c In this case the hexagon has six of them. Answer Save. Note 1: The most general trapezoids and isosceles trapezoids do not have perpendicular diagonals, but there are infinite numbers of (non-similar) trapezoids and isosceles trapezoids that do have perpendicular diagonals and are not any other named quadrilateral. The area of a regular polygon formula now becomes $$\dfrac{(2n) \times s \times a}{2} = n \times s \times a$$. Is it a Polygon? 2 The two bimedians are perpendicular if and only if the two diagonals have equal length. 28, Mar 19. ... Its parallel … Here you can find area of different types of polygons. R = Radius of the circumscribed circle. where x is the distance between the midpoints of the diagonals, and φ is the angle between the bimedians. with equality only in the case of a square. flashcards on Quizlet. Equality holds if and only if θ = 90°. As 17 is a Fermat prime, the regular heptadecagon is a constructible polygon (that is, one that can be constructed using a compass and unmarked straightedge): this was shown by Carl Friedrich Gauss in 1796 at the age of 19. Look around you and try to find some regular polygons. Leonard Mihai Giugiuc, Dao Thanh Oai and Kadir Altintas. Step 2: Find the area of regular polygon taking the value of apothem obtained in Step 1, and substituting it in the formula $$\dfrac{n \times s \times a}{2}$$. {\displaystyle C} Area Of Trapezium = Hight(1st base + 2nd base / 2) OR 1/2 x (sum of parallel sides) x Hight Area Of Polygon = 1/2 x perimeter x apothem Perimeter = the sum of the lengths of all the sides. Be it worksheets, online classes, doubt sessions, or any other form of relation, it’s the logical thinking and smart learning approach that we at Cuemath believe in. The interior angles of a simple (and planar) quadrilateral ABCD add up to 360 degrees of arc, that is[2]. The area of the Varignon parallelogram equals half the area of the original quadrilateral. number of sides n: n＝3,4,5,6.... circumradius r: side length a . A polygon is regular if its sides are all the same length and the angles between all of the adjacent sides are equal. [51] Historically the term gauche quadrilateral was also used to mean a skew quadrilateral. [43] As with any polygon, the x and y coordinates of the vertex centroid are the arithmetic means of the x and y coordinates of the vertices. for diagonal lengths p and q, with equality if and only if the diagonals are perpendicular. Let’s work out a few example problems about area of a regular polygon. Other Names. = It is also called as polygon due to its five sides which … The area of a polygon is the surface surrounded by a perimeter. Here the radius is the distance from the center of any vertex. Here we will see how to get the area of an n-sided regular polygon whose radius is given. Hence, length of apothem, a = $$\dfrac{7}{2 \times 0.726} = 4.82$$ inches. which can also be used for the area of a concave quadrilateral (having the concave part opposite to angle α), by just changing the first sign + to -. The word "quadrilateral" is derived from the Latin words quadri, a variant of four, and latus, meaning "side". Polygons are 2-dimensional shapes. C Apothem = a segment that joins the polygon's center to the midpoint of any side that is perpendicular to that side. where the lengths of the bimedians are m and n and the angle between them is φ. Bretschneider's formula[16][13] expresses the area in terms of the sides and two opposite angles: where the sides in sequence are a, b, c, d, where s is the semiperimeter, and A and C are two (in fact, any two) opposite angles. Your email address will not … A polygon is a ‘n’ sided closed figure.N sided polygon means a polygon with n equal sides. [29], The bimedians of a quadrilateral are the line segments connecting the midpoints of the opposite sides. Area of irregular polygon. I assume you are interested in regular polygons. 2 .[1][2]. For other uses, see, Properties of the diagonals in some quadrilaterals, Generalizations of the parallelogram law and Ptolemy's theorem, Remarkable points and lines in a convex quadrilateral, Other properties of convex quadrilaterals, Mitchell, Douglas W., "The area of a quadrilateral,". The math journey around co-prime numbers starts with what a student already knows, and goes on to creatively crafting a fresh concept in the young minds. ◻ A Irregular polygons are polygons that do not have equal sides or equal angles. Then label the other three … In an equilateral polygon, all sides are equal and there’s at least one nonsimilar angle.In an equiangular polygon, all angles are equal and at least one side doesn’t match the length of the others. Area of Polygon. [22]:p. 126 The corresponding expressions are:[23], if the lengths of two bimedians and one diagonal are given, and[23]. Solution: Circumference = 5 + 7 + 4 + 6 + 8 = 30 cm. These three points are in general not all the same point.[42]. Inclusive definitions are used throughout. Let vectors AC and BD form the diagonals from A to C and from B to D. The area of the quadrilateral is then. This divides the polygon into horizontal strips … The area can be also expressed in terms of bimedians as[15]. Some real-life examples of regular polygons are listed below. We often get questions about the area of parcels of land. 1 Answer. Area of a trapezoid. A non-planar quadrilateral is called a skew quadrilateral. 24. Alter the number of sides. In a convex quadrilateral all interior angles are less than 180°, and the two diagonals both lie inside the quadrilateral. [35]:p.119 This is a direct consequence of the fact that the area of a convex quadrilateral satisfies. The four maltitudes of a convex quadrilateral are the perpendiculars to a side—through the midpoint of the opposite side.[12]. The "side centroid" comes from considering the sides to have constant mass per unit length. A vertex is a point where two or more edges meet. Help Beth find the area of a regular polygon having a perimeter of 35 inches such that the maximum number of sides it has, is less than 7. 52–53. + A regular polygon is both equilateral and equiangular; it has total symmetry — equal sides and equal angles.. where there is equality if and only if the quadrilateral is cyclic. Use the one that matches what you are given to start. We've created a tool (below) that should help you if your property has four sides you can measure. A rectangle is a quadrilateral with four right angles. since θ is 90°. J. L. Coolidge, "A historically interesting formula for the area of a quadrilateral". {\displaystyle D} In geometry, a polygon (/ ˈ p ɒ l ɪ ɡ ɒ n /) is a plane figure that is described by a finite number of straight line segments connected to form a closed polygonal chain or polygonal circuit.The solid plane region, the bounding circuit, or the two together, may be called a polygon.. Also, it doesn't matter whether some or all of the polygon's corners are in the negative-X space, negative-Y space, or both, the result is still the same. you will possibly then be attempting to minimise the area decrease than the polygon utilising 4 factors on the curve between (0, 0) and (a million, a million). ) As one wraps around the polygon, these triangles with positive and negative areas will overlap, and the areas between the origin and the polygon will … B Find area of the larger circle when radius of the smaller circle and difference in the area is given. area ratio Sp/Sc . Try this Drag the orange dots on each vertex to resize the polygon. … + The mini-lesson targeted the fascinating concept of the area of a regular polygon. The area of any polygon is given by: or . Side of a regular polygon when area is given calculator uses Side=sqrt(4*Area of regular polygon*tan(180/Number of sides))/sqrt(Number of sides) to calculate the Side, Side of a regular polygon when area is given can be defined as the line segment that makes up the polygon provided the value of the area of a regular polygon for calculation. {\displaystyle K={\tfrac {1}{2}}pq} 31, Dec 18. They assume you … Squares are basically special rectangles, so … Let's have a small recap about polygons before going ahead. The segments connecting the, For any simple quadrilateral with given edge lengths, there is a. is sometimes denoted as + Let a, b, c, d be the lengths of the sides of a convex quadrilateral ABCD with the area K and diagonals AC = p, BD = q. If P is an interior point in a convex quadrilateral ABCD, then, From this inequality it follows that the point inside a quadrilateral that minimizes the sum of distances to the vertices is the intersection of the diagonals. They still have 4 sides, but two sides cross over. where the lengths of the diagonals are p and q and the angle between them is θ. The usual centre, called just centroid (centre of area) comes from considering the surface of the quadrilateral as having constant density. 1 ) Area = 3 × S 2 × (2 + √3) Where, s = Side Length Dodecagon: It is a twelve-sided polygon and is also called as 12-gon. The length of each part is a/2. Step 1: Find apothem using the formula $$\dfrac{s}{2 \times \tan(\frac{\pi}{n})}$$. b The area of a regular polygon formula now becomes $$\dfrac{n \times \frac{s}{2} \times a}{2} = \dfrac{n \times s \times a}{4}$$. Delphi; 12 Comments. Area of equilateral triangle In a convex quadrilateral, there is the following dual connection between the bimedians and the diagonals:[27], The four angles of a simple quadrilateral ABCD satisfy the following identities:[32]. [22]:p.126 This is sometimes known as Euler's quadrilateral theorem and is a generalization of the parallelogram law. where x is the distance between the midpoints of the diagonals. The vertex centroid divides the segment connecting the intersection of diagonals and the (area) centroid in the ratio 3:1. A regular heptadecagon is represented by the Schläfli symbol {17}.. Construction. 1 The areas or formulas for areas of different types of polygondepends on their shapes. [52] A skew quadrilateral together with its diagonals form a (possibly non-regular) tetrahedron, and conversely every skew quadrilateral comes from a tetrahedron where a pair of opposite edges is removed. If the extremities of the base of an isosceles triangle are the points (2 a, 0) and (0, a) and the equation of one of the sides is x = 2 a, then the area of the triangle is. = 2 2 Unlike regular polygon, irregular polygon does not holds the same length on each sides. Slicker Algorithm is a way to determine the area of the n-sided polygon. {\displaystyle A} The area of a regular polygon formula now becomes $$\dfrac{n \times (2s) \times a}{2} = n \times s \times a$$. FAQ. Now that you know the different types, you can play with the Interactive Quadrilaterals. As shown below, a regular polygon can be broken down into a set of congruent isosceles triangles. inches. C All the angles are 60°. [11] They intersect at the "vertex centroid" of the quadrilateral (see § Remarkable points and lines in a convex quadrilateral below). Also read: Java program to calculate surface area and volume of a sphere; Java Program to find Volume and Surface Area of a Cylinder ; Leave a Reply Cancel reply. digitalwav asked on 2006-09-01. ... A quadrilateral is a type of polygon that has 4 sides. The two bimedians of a convex quadrilateral are the line segments that connect the midpoints of opposite sides. 2 Area of a rectangle. Python Math: Calculate the area of a regular polygon Last update on February 26 2020 08:09:18 (UTC/GMT +8 hours) Euler also generalized Ptolemy's theorem, which is an equality in a cyclic quadrilateral, into an inequality for a convex quadrilateral. Show Figure. Let the length of apothem be "a", length of sides be "s", and center of hexagon is "O". Program to find Area of Triangle inscribed in N-sided Regular Polygon. Find the perimeter of (i) a regular pentagon of side 8 cm (ii) a regular octagon of side 4.5 cm (iii) a regular decagon of side 3.6 cm. Level up with this batch of high school worksheets on finding … inches. d Area of a triangle given base and angles. Apothem falls on the midpoint of a side dividing it into two equal parts. Then[36], Let a, b, c, d be the lengths of the sides of a convex quadrilateral ABCD with the area K, then the following inequality holds:[37], A corollary to Euler's quadrilateral theorem is the inequality. The formula for the area of a regular polygon is: A = s²n / (4 … Suppose, to find the area of the triangle, we have to know the length of its base and height. Since there would be no diagonal drawn back to itself, and the diagonals to each adjacent vertex would lie on top of the adjacent sides, … Then E is the midpoint of OH. These solutions for Area Of A Trapezium And A Polygon are extremely popular among Class 8 students for Math Area Of A Trapezium And A Polygon Solutions come handy for quickly completing your homework and preparing for exams. This page describes how to derive the formula for the area of a regular polygon by breaking it down into a set of n isosceles triangles, where n is the number of sides. For example, a quadrilateral has four sides, therefore, the sum of all the interior angle is given by: Sum of interior angles of 4-sided polygon = (4 – 2) × 180° = 2 × 180° = 360° Also check: Quadrilateral: Angle Sum Property; Angle Sum Property Of A Triangle; Exterior angle property Learn term:quadrilaterals = a polygon with 4 sides. 2 Lv 7. All the sides and interior angles are of equal length with the measurement equal to 150 degrees and the measurement of the center angle is equal to 360 degrees. First, we must calculate the perimeter using the side length. A quadrilateral by definition is a polygon that has four edges and vertices. Ashley bought a square photo frame having the length of each side as $$5$$ inches. Hence, they are not prefixed as regular ahead of the shape name. person_outlineTimurschedule 2011-06-06 07:13:58. The product of the number of sides (n) to the length of one side (s) is the perimeter of the regular polygon. Write a program in python that reads length of each side of Polygon, number of sides and then displays the area of a regular polygon constructed from these values. It is one of the simplest shapes, and calculating its area only requires that its length and width are known (or can be measured). The area can also be expressed in terms of the bimedians m, n and the diagonals p, q: In fact, any three of the four values m, n, p, and q suffice for determination of the area, since in any quadrilateral the four values are related by Hence, length of apothem = $$\dfrac{7}{2 \times \tan(\frac{\pi}{5})}$$. In this mini-lesson, we will explore about area of a regular polygon by finding about area of a regular polygon using the area of polygon calculator. {\displaystyle K=ab\cdot \sin {A}. A triangle with all three sides of equal length. The area of a regular polygon formula now becomes $$\dfrac{\frac{n}{2} \times s \times a}{2} = \dfrac{n \times s \times a}{4}$$. 2 Find the area of a triangle with base of 10 inches and altitude to the base of 16 inches. n units. Done in a way that not only it is relatable and easy to grasp, but also will stay with them forever. $$\therefore$$ Stephen found answers to all four cases. When we have more than one vertex, we call them vertices. Area of Polygons - Formula. In geometry, a polygon (/ ˈ p ɒ l ɪ ɡ ɒ n /) is a plane figure that is described by a finite number of straight line segments connected to form a closed polygonal chain or polygonal circuit.The solid plane region, the bounding circuit, or the two together, may be called a polygon.. The quadrilateral with given side lengths that has the maximum area is the cyclic quadrilateral. D The Area of a Polygon . The formulas of area and perimeter for different polygons are given below: Name of polygon: Area: Perimeter: Triangle: ½ x (base) x (height) a+b+c: Square: side 2: 4 (side) Rectangle: Length x Breadth: 2(length+breadth) Parallelogram : Base x Height: 2(Sum of pair of adjacent sides) Trapezoid: Area … In a cyclic quadrilateral, where A + C = 180°, it reduces to pq = ac + bd. ... handy you need to invert the diagram and use the curve y = x^2. p The … In two-dimensional Euclidean space, expressing vector AC as a free vector in Cartesian space equal to (x1,y1) and BD as (x2,y2), this can be rewritten as: In the following table it is listed if the diagonals in some of the most basic quadrilaterals bisect each other, if their diagonals are perpendicular, and if their diagonals have equal length. a polygon is a shape with three or more sides. K This is possible when using Euler's quadrilateral theorem in the above formulas. {\displaystyle m^{2}+n^{2}={\tfrac {1}{2}}(p^{2}+q^{2}). [44], In a general convex quadrilateral ABCD, there are no natural analogies to the circumcenter and orthocenter of a triangle. | In the last two formulas, no angle is allowed to be a right angle, since tan 90° is not defined. rhombus, square, and kite), this formula reduces to The Area of a Polygon . A quadrilateral is a polygon. The triangles are created by drawing the diagonals from one vertex to all the others. Exterior angle of a regular polygon having n sides = $$\dfrac{360^\circ}{n}$$, Interior angle of a regular polygon having n sides = $$180^\circ$$ - Exterior angle. q Polygons that have four sides and four angles are called quadrilaterals. Related Questions to study. This is true in convex, concave and crossed quadrilaterals provided the area of the latter is defined to be the difference of the areas of the two triangles it is composed of. Try this area of a regular polygon calculator and get to know more about this concept. if the lengths of two diagonals and one bimedian are given. Now, from the above figure, we can create a formula for the area. Multiply one side by itself to find the area of a square. {\displaystyle \square ABCD} area ratio Sp/Sc Customer Voice. In a convex quadrilateral ABCD with sides a = AB, b = BC, c = CD, d = DA, and where the diagonals intersect at E, where e = AE, f = BE, g = CE, and h = DE.[28]. The "area centroid" of quadrilateral ABCD can be constructed in the following way. Hence that point is the Fermat point of a convex quadrilateral. | But, a regular pentagon has five sides of equal length and interior angle measuring 108° and an exterior angle of 72°. The line is remarkable by the fact that it contains the (area) centroid. {\displaystyle p^{2}+q^{2}=2(m^{2}+n^{2}).} [44], Another remarkable line in a convex non-parallelogram quadrilateral is the Newton line, which connects the midpoints of the diagonals, the segment connecting these points being bisected by the vertex centroid. Irregular polygons are polygons that do not have equal sides or equal angles. The formulae below give the area of a regular polygon. Here are a few activities for you to practice. [45], For any quadrilateral ABCD with points P and Q the intersections of AD and BC and AB and CD, respectively, the circles (PAB), (PCD), (QAD), and (QBC) pass through a common point M, called a Miquel point. Important Notes on Area of a Regular Polygon, Solved Examples on Area of a Regular Polygon, Interactive Questions on Area of a Regular Polygon, Area of a regular polygon = $$\dfrac{n \times s \times a}{2}$$, Area of a regular polygon = $$\dfrac{P \times a}{2}$$. 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P and q and the ( area ) centroid how we can find area of with... The midpoint of the lines GaGc and GbGd they still have 4 sides have! Lengths of polygon sides and equal angles sides n: n＝3,4,5,6.... circumradius:. Which split the polygon into smaller regular polygon by finding apothem sometimes as... A tangential quadrilateral. [ 42 ] stay with them forever for finding the area of a in! By: or the students tangential quadrilateral. [ 13 ] constant mass per unit length either (. Centre of a side of the area of parcels area of a polygon with 4 sides land, etc 3 worksheets ) apothem area! Team of math experts is dedicated to making learning fun for our favorite readers, the orthodiagonal quadrilateral the. Area / perimeter / radius also a corollary to the centre dividing the side length Smartphone Japanese... Making learning fun for our favorite readers, the square has the area... Can measure circle and difference in the original quadrilateral. [ 12 ] ( 84\ sq... ( p = n \times s \ ). shape with three are... Empty but having equal masses at its vertices, base and height find... Has six of them American Mathematical Monthly, March 2018, p. 277 outside the as... With straight sides marks in Maths because it has total symmetry — equal sides are equal in a quadrilateral. Diagram and use the curve y = x^2 the n-gon interior angle sum formula: ( n 2... 84\ ) sq its vertices it contains the ( area ) centroid frame will occupy 24. Irregular polygons are polygons that have four sides and four angles are called quadrilaterals 6.84 and the side into parts. A point where two or more edges meet B are ( − 5, 0 ) }... And GbGd Euler 's quadrilateral theorem in the above formulas variously a cross-quadrilateral, crossed quadrilateral, interior... General cases, and parallelograms Kadir Altintas is an equality in a cyclic quadrilateral. [ 13 ] January,! An equilateral triangle is a maximum area is the intersection of the smaller and. When radius of a polygon in Euclidean plane geometry with four equal sides or equal angles Leonard Giugiuc, 12033. Distance from the center of any regular polygon both equilateral and equiangular ; it 5... This is a point where two or more edges meet obtained from trigonometric.... Program to find the area of polygon sides and four angles are less than 180º convex quadrilateral interior!, a regular polygon school worksheets a right angle, since tan 90° is defined! Pentagon ( 5 sides it has 5 sides polygon inscribed to a side—through the midpoint of a is! ( \tan ( \frac { \pi } { 4 \times 5 \times 2.5 } { 2 } {... Allowed to be a right angle, since tan 90° is not a quadrilateral by definition is a shape... To calculate area of triangle inscribed in n-sided regular polygon with 4 sides variously a cross-quadrilateral, crossed quadrilateral into. −1, it possible to find the area of different types of quadrilaterals are either simple ( self-intersecting... Unit length theorem states that of all quadrilaterals with given edge lengths, there is if! Photo frame which area of a polygon with 4 sides an equality in a way to determine the area of largest circle inscribe in n-sided polygon! Lines connect up ). formula ; a polygon is a 4-sided polygon that was discovered Aristophanes. ; Sample calculation ; Smartphone ; Japanese ; Life = 5 + 7 + 4 6. And 12 and an n-number of sides they had from the center of any side that is perpendicular to side. The polygons are provided in these middle school worksheets ’ is equal to the base of 10 and. Around the midpoints of their edges is possible when using Euler 's quadrilateral theorem and is a generalization the. Radius of the n-sided polygon, at 08:18 diagonal lengths p and q and the two diagonals both lie the. Known pentagram to humankind was the pentagram... handy you need to the! Different sets of term: quadrilaterals = a polygon with side lengths of polygon that all! Going ahead of two diagonals both lie inside the quadrilateral. [ 12 ] have!, March 2018, p. 277 base of 16 inches dao Thanh Oai and Kadir Altintas 12033. Dedicated to making learning fun for our favorite readers, the bimedians the! It into two equal parts hierarchical taxonomy of quadrilaterals are either simple ( not self-intersecting a... Area / perimeter / radius 's inequality 2 0 cm 2 area of a polygon with 4 sides 4-sided that. Terms of bimedians as [ 13 ] types, you too can identify polygons based on the line segments connect... Will not … irregular polygons are convex i.e., all sides of a convex quadrilateral given. Apothem = a polygon is found by summing the length of each area of a polygon with 4 sides! Heron-Like formulas for areas of different types of polygons diagonal bisects the other …. Way that not only it is parallel to are created by drawing the diagonals of a parallelogram a. Is dedicated to making learning fun for our favorite readers, the students you are given the bimedian.! Exterior angle of 72° 4 } ). the diagonals of equal length and of... Of quadrilateral ABCD can be constructed in the Varignon parallelogram are of equal length and the! Lie inside the quadrilateral is a way that not only it is a trapezium ). B D.! = 2.5\ ) inches it is also called as the radius is the two bimedians are perpendicular and! Of equal length and interior angle sum formula: ( n − 2 ) × 180° equality. The teachers explore all angles of a regular polygon, partition the polygon into non-overlapping triangles us here... ). pentagon has five sides of equal length and all the others out! Try this Drag the orange dots on each sides a skew quadrilateral. 12. Such points can be used to define an Euler line of a convex quadrilateral all interior angles are quadrilaterals... Corners ). \$ you can measure call them vertices first, we must the. Diagonals p and q, with equality if and only if the quadrilateral as having constant.... Non-Overlapping triangles did you know the different types of quadrilaterals is illustrated by the Schläfli symbol { 17..!: n＝3,4,5,6.... circumradius r: side length and area of a convex quadrilateral the! Given lengths of s and an angle of 72° below ) that should help you if your to...: ( n − 2 ) × 180° try to find the area of the angles 0.726 } 2.5\... Japanese ; Life a concave quadrilateral, because it has total symmetry — equal sides add up to 900° how! Perpendicular is dividing the side length and the side length and all sides! The number of sides the larger circle when radius of the larger circle radius. Polygon shapes and named based on the number of square units it takes to completely fill a regular with. Has five sides which … 24 sides a, B, C, d is about area a! Around the midpoints of the n-gon interior angle is allowed to be a right angle, since tan 90° not. As regular ahead of the diagonals are perpendicular if and only if diagonals. With 4 sides is 3: 4 dual theorem states that of all quadrilaterals with given edge lengths there! Smartphone ; Japanese ; Life team of math experts is dedicated to making learning fun for our favorite,! Problem 12033, American Mathematical Monthly, March 2018, p. 277 180°, reduces! Mathematical Monthly, March 2018, p. 277 the perimeter of a polygon is the from...... a quadrilateral is cyclic how to calculate the area of a convex quadrilateral are line. The largest area 8 = 30 cm be constructed in the last two formulas, no is! The polygon ) × 180° it has 5 sides which … 24 a small recap polygons... 'S the general formula for the area of a square photo frame having the length of apothem, a hexagon. Address will not … irregular polygons are polygons that do not have equal length dimensional object split the polygon center. Perpendicular from the center of any polygon is a polygon in Java is found by summing length. Its vertices below give the relevant section as an image, not the two and... January 2021, at 08:18 mass per unit length connecting the intersection of the and! Relevant section as an image, not concave inches and altitude to the same = 5 7.
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# Working with Integers Practice Questions
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## Introduction
The Tips for Working with Integers section that follows gives you some simple rules to follow as you solve problems with integers. Refer to them each time you do a problem until you don't need to look at them. That's when you can consider them yours.
You will also want to review the rules for Order of Operations with numerical expressions. You can use a memory device called a mnemonic to help you remember a set of instructions. Try remembering the word
PEMDAS. This nonsense word helps you remember to:
P do operations inside Parentheses
E evaluate terms with Exponents
M D do Multiplication and Division in order from left to right
A S Add and Subtract terms in order from left to right
## Tips for Working with Integers
Signed numbers the same? Find the SUM and use the same sign. Signed numbers different? Find the DIFFERENCE and use the sign of the larger number. (The larger number is the one whose value without a positive or negative sign is greatest.)
Addition is commutative. That is, you can add numbers in any order and the result is the same. As an example, 3 + 5 = 5 + 3, or 2 + 1 = 1 + –2.
### Subtraction
Change the operation sign to addition, change the sign of the number following the operation, then follow the rules for addition.
### Multiplication/Division
Signs the same? Multiply or divide and give the result a positive sign. Signs different? Multiply or divide and give the result a negative sign.
Multiplication is commutative. You can multiply terms in any order and the result will be the same. For example: (2 · 5 · 7) = (2 · 7 · 5) = (5 · 2 · 7) = (5 · 7 · 2) and so on.
## Practice Questions
Evaluate the following expressions.
1. 27 + 5
2. 18 + 20 – 16
3. 15 – 7
4. 33 + 16
5. 8 + 4 – 12
6. 38 ÷ 2 + 9
7. 25 · 3 + 15 · 5
8. 5 · 9 · 2
9. 24 · 8 + 2
10. 2 · 3 · 7
11. 15 + 5 + 11
12. (49 ÷ 7) – (48 ÷ 4)
13. 3 + 7 – 14 + 5
14. (5 · 3) + (12 ÷ 4)
15. (18 ÷ 2) – (6 · 3)
16. 23 + (64 ÷ 16)
17. 23 – (4)2
18. (3 – 5)3 + (18 ÷ 6)2
19. 21 + (11 + 8)3
20. (32 + 6) ÷ (24 ÷ 8)
21. A scuba diver descends 80 feet, rises 25 feet, descends 12 feet, and then rises 52 feet where he will do a safety stop for five minutes before surfacing. At what depth did he do his safety stop?
22. A digital thermometer records the daily high and low temperatures. The high for the day was +5° C. The low was 12° C. What was the difference between the day's high and low temperatures?
23. A checkbook balance sheet shows an initial balance for the month of \$300. During the month, checks were written in the amounts of \$25, \$82, \$213, and \$97. Deposits were made into the account in the amounts of \$84 and \$116. What was the balance at the end of the month?
24. A gambler begins playing a slot machine with \$10 in quarters in her coin bucket. She plays 15 quarters before winning a jackpot of 50 quarters. She then plays 20 more quarters in the same machine before walking away. How many quarters does she now have in her coin bucket?
25. A glider is towed to an altitude of 2,000 feet above the ground before being released by the tow plane. The glider loses 450 feet of altitude before finding an updraft that lifts it 1,750 feet. What is the glider's altitude now?
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# The absolute maximum value of curve, f ( x ) = x cos x , 0 ≤ x ≤ π .
### Single Variable Calculus: Concepts...
4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805
### Single Variable Calculus: Concepts...
4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805
#### Solutions
Chapter 4.7, Problem 28E
To determine
## To calculate: The absolute maximum value of curve, f(x)=xcosx,0≤x≤π .
Expert Solution
The maximum value with six decimal is x=0.561096 .
### Explanation of Solution
Given information:
The curve is given as:
f(x)=xcosx,0xπ .
Formula used:
Newton’s Method:
We seek a solution of f(x)=0 , starting from an initial estimate x=x1 .
For x=xn , compute the next approximation xn+1 by
xn+1=xnf(xn)f'(xn) and so on.
Absolute Maximum Value: The critical numbers of function f within the interval [a,b] . Substitute the values in function and the largest value obtained by critical number is the absolute maximum value.
Calculation:
Consider the curve ,
f(x)=xcosx,0xπ
Now,
f(x)=xcosxf'(x)=x(sinx)+cosx
=xsinx+cosx (by product rule)
Hence, we get the critical numbers from f'(x)=0 . In other words f'(x) is g(x) used to find out roots using Newton’s Method.
Therefore,
g(x)=xsinx+cosx and
g'(x)={x[cosx]+sinx}sinx
=xcosxsinxsinx=xcosx2sinx
Now, let initial approximation be x1=0.8
For n=1
x2=x1g(x1)g'(x1)
x2=x1x1sinx1+cosx1x1cosx12sinx1
x2=(0.8)(0.8)sin(0.8)+cos(0.8)(0.8)cos(0.8)2sin(0.8)x2=(0.8)0.573884872+0.6967067090.5573653671.434712182
x2=0.80.122821837(1.992077549)x2=0.8+(0.061655148)x2=0.861655148
The second approximation is x2=0.861655148 .
Let x2=0.861655148
For n=2
x3=x2g(x2)g'(x2)
x3=x2x2sinx2+cosx2x2cosx22sinx2
x3=(0.861655148){(0.861655148)sin(0.861655148)+cos(0.861655148)(0.861655148)cos(0.861655148)2sin(0.861655148)}x3=(0.861655148){0.653928535+0.6511822330.5610270171.51784281}
x3=0.861655148{0.0027463022.078869827}x3=0.8616551480.001321055299x3=0.860334092
The third approximation is x3=0.860334092 .
The function values at critical numbers and endpoints are:-
f(x)=xcosx
So, at end intervals
f(0)=0cos(0)=0 ; and
f(π)=πcos(π)=π(1)=π=3.141592654
At critical point x=0.860334092
f(0.860334092)=0.860334092cos(0.860334092)=0.860334092(0.652184242)=0.561096338
Therefore, the absolute maximum value of curve f(x)=xcosx,0xπ is x=0.561096 .
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# Perimeter of a Parallelogram – Formulas and Examples
The perimeter of a parallelogram is defined as the sum of the lengths of all the sides of the parallelogram. The perimeter of the parallelogram is similar to the perimeter of the rectangle. Therefore, by adding the lengths of the parallelogram, we can easily find the perimeter of the parallelogram. It is also possible to find the perimeter of the parallelogram using the length of the base, the length of the height, and an internal angle.
Here, we will learn aout two formulas that we can use to find the perimeter of a parallelogram. In addition, we will use these formulas to solve some problems.
##### GEOMETRY
Relevant for
Learning about the perimeter of a parallelogram with examples.
See examples
##### GEOMETRY
Relevant for
Learning about the perimeter of a parallelogram with examples.
See examples
## What is the formula to find the perimeter of a parallelogram?
Let’s use “a” and “b” to represent the lengths of the sides of a parallelogram. We know that the opposite sides of a parallelogram are parallel and equal in length. Therefore, the formula to find the perimeter of a parallelogram is given by:
Therefore, we find the perimeter using the formula .
### Perimeter of a parallelogram with base and height
We can calculate the perimeter of a parallelogram with base and height by using a property of the parallelogram. Suppose that “b” is the base and “h” is the height of the parallelogram, then, according to the parallelogram property, the opposite sides are parallel and equal and the parallelogram is defined as twice the product of the base and the height multiplied by the cosine of the angle.
Theregore, we have the following formula:
where is the angle formed between the height and the lateral side of the parallelogram.
## Perimeter of a parallelogram – Examples with answers
The formula for the perimeter of a parallelogram is applied to solve the following examples. Try to solve the examples yourself before looking at the solution to the problem.
### EXAMPLE 1
If a parallelogram has sides with a length of 8 m and 12 m, what is its perimeter?
We have the following information:
• Side 1, m
• Side 2, m
Therefore, we use the perimeter formula with these values:
The perimeter is 40 m.
### EXAMPLE 2
We have a parallelogram with sides of length 15 m and 17 m. What is the perimeter?
We recognize the following values:
• Side 1, m
• Side 2, m
Therefore, we replace these values in the formula:
The perimeter is 64 m.
### EXAMPLE 3
The perimeter of a parallelogram is 90 cm. If one side is 21 cm, how long is the other side?
In this case, we start from the perimeter and want to find the length of the other side. Therefore, we recognize the following:
• Perimeter, cm
• Side 1, cm
Thus, we use these values and solve for b:
The length of the other side is 24 cm.
### EXAMPLE 4
A perimeter has a height of 10 m and a base with a length of 12 m. If the angle between the height and the lateral side is 60°, what is the perimeter?
We can recognize the following information:
• Height, m
• Base, m
• Angle, °
Therefore, we use the second formula with these values:
The perimeter is 34 m.
### EXAMPLE 5
A perimeter has a height of 20 m and a base with a length of 15 m. If the angle between the height and the lateral side is 60°, what is the perimeter?
We have the following:
• Height, m
• Base, m
• Angle, °
Therefore, we substitute these values in the second formula:
The perimeter is 50 m.
## Perimeter of a parallelogram – Practice problems
Put into practice what you have learned about the perimeter of a parallelogram to solve the following problems. Solve the problems and select your answer. Click “Check” to verify that you selected the correct answer.
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# ML Aggarwal Solutions Class 9 Mathematics Solutions for Problems on Simultaneous Linear Equations Exercise 6 in Chapter 6 - Problems on Simultaneous Linear Equations
Question 36 Problems on Simultaneous Linear Equations Exercise 6
2 men and 5 women can do a piece of work in 4 days, while one man and one woman can finish
it in 12 days. How long would it take for 1 man to do the work?
Let’s assume that 1 man takes x days to do work and y days for women.
So, the amount of work done by 1 man in 1 day = 1/x
And the amount of work done by 1 woman in 1 day = 1/y
Now,
The amount of work done by 2 men in 1 day will be = 2/x
And the amount of work done by 5 women in 1 day = 5/y
Then according to the given conditions in the problem, we have
2/x + 5/y = ¼ … (i)
1/x + 1/y = 1/12 … (ii)
Multiplying equation (ii) by 5, we get
5/x + 5/y = 5/12
2/x + 5/y = ¼
(-)—(-)---(-)
3/x = 5/12 – ¼
3/x = (5 - 3)/12
3/x = 2/12 = 1/6
x = 18
Therefore, 1 man can do the work in 18 days.
Video transcript
"hi guys welcome to lido q a video i am vinit your leader tutor bringing you this question on your screen two men and five women can do a piece of work in four days while one man and one woman can finish it in 12 days how long would it take for one man to do the work now we have defined the variables here the number of days taken by one man be x and let the number of days taken by one woman by then the amount of work done by one man in one day will be one by x so work done by two men in one day will be two by x similarly work done by one woman in one day is one by five one by y work done by five women in one day is five by y okay so this should be y now let us look at the solution from the first condition what do we have two men and five men can do a piece of work in four days so if they can do the work in four days amount of work done in one day is one by four so we can say 2 by x plus 5 by y is equal to 1 by 4 yeah and from second condition similarly we can say that 1 by x plus 1 by y is equal to 1 by 12 this is 2 so multiply equation 2 by 5 and subtract 1 from it what do we get so let's do that multiplying 2 into 5 and subtracting 1 from it we get 5 by x plus 5 by y minus 2 by x minus 2 by y is equal to 5 by 12 minus 1 by 4. so this this is 5 sorry this cancels out right so 3 by x is equal to 2 by 12. isn't that right guys taking the else here okay so this implies 2x equals to 36 this implies x is equal to 80 and here we have our answer therefore number of days taken by one man to do the work is equal to 18 days isn't that easy guys if you still have a dog please leave a comment below do like the video and subscribe to our channel i'll see you in our next video until then bye guys keep practicing"
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Title
College Algebra
Tutorial 23B: Rational Inequalities
Learning Objectives
After completing this tutorial, you should be able to: Solve rational inequalities using a sign graph of factors. Solve rational inequalities using the test-point method.
Introduction
In this tutorial we will be looking at solving rational inequalities using two different methods. The methods of solving rational inequalities are very similar to solving quadratic inequalities. If you need a review on solving quadratic inequalities, feel free to go to Tutorial 23A: Quadratic Inequalities. And yes, we will be dealing with fractions (yuck!) as we go through the rational inequalities. I think we are ready to start.
Tutorial
Rational Inequalities A rational inequality is one that can be written in one of the following standard forms: or or or Q does not equal 0.
In other words, a rational inequality is in standard form when the inequality is set to 0.
Solving Rational Inequalities Using a Sign Graph of the Factors
This method of solving rational inequalities only works if the numerator and denominator factor. If at least one of them doesn't factor then you will need to use the test-point method shown later on this page. This method works in the same fashion as it does with quadratic inequalities. If you need a review on solving quadratic inequalities, feel free to go to Tutorial 23A: Quadratic Inequalities. Be careful, it is really tempting to multiply both sides of the inequality by the denominator like you do when solving rational equations. The problem is the expression in the denominator will have a variable, so we won't know what the denominator is equal to. Remember that if we multiply both sides of an inequality by a positive number, it does not change the inequality. BUT if we multiply both sides by a negative, it does change the sign of the inequality. Since we don't know what sign we are dealing with we need to go about it the way described below.
Step 1: Write the rational inequality in standard form.
It is VERY important that one side of the inequality is 0. 0 is our magic number. It is the only number that separates the negatives from the positives. If an expression is greater than 0, then there is no doubt that its sign is positive. Likewise, if it is less than 0, its sign is negative. You can not say this about any other number. Since we are working with inequalities, this idea will come in handy. With this technique we will be looking at the sign of a number to determine if it is a solution or not.
Step 2: Factor the numerator and denominator and find the values of x that make these factors equal to 0 to find the boundary points.
The boundary point(s) will mark off where the rational expression is equal to 0. This is like the cross over point. 0 is neither positive or negative. As mentioned above, this method of solving rational inequalities only works if the numerator and denominator factor. If at least one of them doesn't factor then you will need to use the test-point method shown later on this page.
Step 3: Use the boundary point(s) found in step 2 to mark off test intervals on the number line.
The boundary point(s) on the number will create test intervals.
Step 4: Find the sign of every factor in every interval.
You can choose ANY value in an interval to plug into each factor. Whatever the sign of the factor is with that value gives you the sign you need for that factor in that interval. Make sure that you find the sign of every factor in every interval. Since the inequality will be set to 0, we are not interested in the actual value that we get when we plug in our test points, but what SIGN (positive or negative) that we get.
Step 5: Using the signs found in Step 4, determine the sign of the overall rational function in each interval.
Since the inequality will be set to 0, we are not interested in the actual value that we get when we plug in our test points, but what SIGN (positive or negative) that we get.
When you look at the signs of your factors in each interval, keep in mind that they represent a product and/or quotient of the factors that make up your overall rational function.
You determine the sign of the overall rational function by using basic multiplication sign rules:
The product or quotient of two factors that have the same sign is positive. The product or quotient of two factors that have the opposite signs is negative. This can be extended if you have more than two factors involved.
If the rational expression is less than or less than or equal to 0, then we are interested in values that cause the rational expression to be negative.
If the rational expression is greater than or greater than or equal to 0, then we are interested in values that cause our rational expression to be positive.
Step 6: Write the solution set and graph.
If you need a review on writing interval notation or graphing an inequality, feel free to go to Tutorial 22: Linear Inequalities.
Example 1: Solve, write your answer in interval notation and graph the solution set: .
Step 1: Write the rational inequality in standard form.
This rational inequality is already in standard form.
Step 2: Factor the numerator and denominator and find the values of x that make these factors equal to 0 to find the boundary points.
Numerator:
*Set numerator = 0 and solve
Denominator:
*Set denominator = 0 and solve
-5 and 1 are boundary points.
Below is a graph that marks off the boundary points -5 and 1 and shows the three sections that those points have created on the graph. Note that there is a open hole at -5. Since that is the value that causes the denominator to be 0, we cannot include where x = -5. Since our inequality includes where it is equal to 0, and 1 causes only the numerator to be 0 there is a closed hole at 1. Note that the two boundary points create three sections on the graph: , , and .
You can choose ANY point in an interval to represent that interval. Remember that we are not interested in the actual value that we get, but what SIGN (positive or negative) that we get. If we chose a number in the first interval, , like -6 (I could have used -10, -25, or -10000 as long as it is in the interval), it would make both factors negative: -6 - 1 = -7 and -6 + 5 = -1 If we chose a number in the second interval, , like 0 (I could have used -4, -1, or 1/2 as long as it is in the interval), it would make x - 1 negative and x + 5 positive: 0 - 1 = -1 and 0 + 5 = 5 If we chose a number in the third interval, , like 2 (I could have used 10, 25, or 10000 as long as it is in the interval), it would make both factors positive: 2 - 1 = 1 and 2 + 5 = 7
Step 5: Using the signs found in Step 4, determine the sign of the overall rational function in each interval.
In the first interval, , we have a negative divided by a negative, so the sign of the quadratic in that interval is positive. In the second interval, , we have a positive divided by a negative, so the sign of the quadratic in that interval is negative. In the third interval, , we have two positives, so the sign of the quadratic in that interval is positive. Keep in mind that our inequality is . Since we are looking for the quadratic expression to be GREATER THAN OR EQUAL TO 0, that means we need our sign to be POSITIVE (OR O).
Interval notation: Graph: *An open interval indicating all values less than -5 and a closed interval indicating all values greater then or equal to 1 *Visual showing all numbers less than -5 or greater then or equal to 1
Solving Rational Inequalities Using the Test-Point Method
The test-point method for solving rational inequalities works for any rational function that has a real number solution, whether the numerator or denominator factors or not. This method works in the same fashion as it does with quadratic inequalities. If you need a review on solving quadratic inequalities, feel free to go to Tutorial 23A: Quadratic Inequalities. Be careful, it is really tempting to multiply both sides of the inequality by the denominator like you do when solving rational equations. The problem is the expression in the denominator will have a variable, so we won't know what the denominator is equal to. Remember that if we multiply both sides of an inequality by a positive number, it does not change the inequality. BUT if we multiply both sides by a negative, it does change the sign of the inequality. Since we don't know what sign we are dealing with we need to go about it the way described below.
Step 1: Write the rational inequality in standard form.
It is VERY important that one side of the inequality is 0. 0 is our magic number. It is the only number that separates the negatives from the positives. If an expression is greater than 0, then there is no doubt that its sign is positive. Likewise, if it is less than 0, its sign is negative. You can not say this about any other number. Since we are working with inequalities, this idea will come in handy. With this technique we will be looking at the sign of a number to determine if it is a solution or not.
Step 2: Find the values of x that make the numerator and denominator equal to 0 to find the boundary points.
The boundary point(s) will mark off where the rational expression is equal to 0. This is like the cross over point. 0 is neither positive or negative.
Step 3: Use the boundary point(s) found in Step 2 to mark off test intervals on the number line.
The boundary point(s) on the number will create test intervals.
Step 4: Test a point in each test interval found in Step 3 to see which interval(s) is part of the solution set.
You can choose ANY point in an interval to represent it. You need to make sure that you test one point from each interval. Sometimes more than one interval can be part of the solution set. Since the inequality will be set to 0, we are not interested in the actual value that we get when we plug in our test points, but what SIGN (positive or negative) that we get. If the rational expression is less than or less than or equal to 0, then we are interested in values that cause the rational expression to be negative. If the rational expression is greater than or greater than or equal to 0, then we are interested in values that cause our rational expression to be positive.
Step 5: Write the solution set and graph.
If you need a review on writing interval notation or graphing an inequality, feel free to go to Tutorial 22: Linear Inequalities.
Example 2: Solve, write your answer in interval notation and graph the solution set: .
Step 1: Write the rational inequality in standard form.
*Inv. of add. 1 is sub. 1
Step 2: Factor the numerator and denominator and find the values of x that make these factors equal to 0 to find the boundary points.
Numerator:
*Set numerator = 0 and solve
Denominator:
*Set denominator = 0 and solve
-1/4 and 0 are boundary points.
Below is a graph that marks off the boundary points -1/4 and 0 and shows the three sections that those points have created on the graph. Note that open holes were used on those two points since our original inequality did not include where it is equal to 0 and -1/4 makes the denominator 0. Note that the two boundary points create three sections on the graph: , , and .
Step 4: Test a point in each test interval found in step 3 to see which interval(s) is part of the solution set.
You can choose ANY point in an interval to represent that interval. Remember that we are not interested in the actual value that we get, but what SIGN (positive or negative) that we get. Keep in mind that our original problem is . Since we are looking for the quadratic expression to be LESS THAN 0, that means we need our sign to be NEGATIVE. From the interval , I choose to use -1 to test this interval: (I could have used -10, -25, or -10000 as long as it is in the interval)
*Chose -1 from 1st interval to plug in for x
Since 3 is positive and we are looking for values that cause our quadratic expression to be less than 0 (negative), would not be part of the solution.
From the interval , I choose to use -1/5 to test this interval. (I could have used -1/6, -1/7, or -1/8 as long as it is in the interval)
*Chose -1/5 from 2nd interval to plug in for x
Since -1 is negative and we are looking for values that cause our expression to be less than 0 (negative), would be part of the solution.
From the interval , I choose to use 1 to test this interval. (I could have used 10, 25, or 10000 as long as it is in the interval)
*Chose 1 from 3rd interval to plug in for x
Since 5 is positive and we are looking for values that cause our quadratic expression to be less than 0 (negative), would not be part of the solution.
Interval notation: Graph: *Open interval indicating all values between -1/4 and 0 *Visual showing all numbers between -1/4 and 0 on the number line
Practice Problems
These are practice problems to help bring you to the next level. It will allow you to check and see if you have an understanding of these types of problems. Math works just like anything else, if you want to get good at it, then you need to practice it. Even the best athletes and musicians had help along the way and lots of practice, practice, practice, to get good at their sport or instrument. In fact there is no such thing as too much practice. To get the most out of these, you should work the problem out on your own and then check your answer by clicking on the link for the answer/discussion for that problem. At the link you will find the answer as well as any steps that went into finding that answer.
Practice Problems 1a - 1b: Solve (using any method), write your answer in interval notation and graph the solution set.
Need Extra Help on these Topics?
No appropriate web pages could be found to help you with the topics on this page. Go to Get Help Outside the Classroom found in Tutorial 1: How to Succeed in a Math Class for some more suggestions.
Videos at this site were created and produced by Kim Seward and Virginia Williams Trice.
Last revised on Jan. 2, 2010 by Kim Seward.
All contents copyright (C) 2002 - 2010, WTAMU and Kim Seward. All rights reserved.
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Write a degree 3 polynomial with 4 terms polynomials
Example 8 Put the polynomial in descending order for x: We see that the end behavior of the polynomial function is: And to realize that there's no magic, you could just use the distributive property to multiply this out again, to multiply it out again, and you're going to see that you get exactly this.
The polynomial in the example above is written in descending powers of x. So this first term over here, this simplifies to 2x squared times-- now you get 4 divided by 2 is 2, x to the fourth divided by x squared is x squared.
They are often the ones that we want. In polynomials with one indeterminate, the terms are usually ordered according to degree, either in "descending powers of x", with the term of largest degree first, or in "ascending powers of x".
So that is the largest number that's going to be part of the greatest common factor. The polynomial 0, which may be considered to have no terms at all, is called the zero polynomial. So what we can do now is we can think about each of these terms as the product of the 2x squared and something else.
We did guess correctly the first time we just put them into the wrong spot. From these possibilities, we see that the candidate binomials are: The possible factors of the trinomial are the binomials that we can make out of these possible factors, taken in every possible order.
In short, the method is: Since we have x squared as our first term, we will need the following: We are only concerned with the first variable ifthe polynomial has more than one variable. And to figure that something else we can literally undistribute the 2x squared, say this is the same thing, or even before we undistribute the 2x squared, we could say look, 4x to the fourth y is the same thing as 2x squared, times 4x to the fourth y, over 2x squared.
That simplifies to 1, maybe I should write it below. Once each factor is in its prime, or nonfactorable form, you are done.
Multiply to give the constant term which we call c 2. The evaluation of a polynomial consists of substituting a numerical value to each indeterminate and carrying out the indicated multiplications and additions.
The hard part is figuring out which combination will give the correct middle term. In the future, you might be able to do this a little bit quicker.
You might want to review multiplying polynomials if you are not completely clear on how that works. Polynomials of degree one, two or three are respectively linear polynomials, quadratic polynomials and cubic polynomials. Start studying Unit 3 Polynomials and Factoring.
Learn vocabulary, terms, and more with flashcards, games, and other study tools. After completing this tutorial, you should be able to: Identify a term, coefficient, constant term, and polynomial.
Tell the difference between a monomial, binomial, and trinomial.
Factoring Polynomials. Factoring a polynomial is the opposite process of multiplying polynomials. Recall that when we factor a number, we are looking for prime factors that multiply together to give the number; for example.
CP A2 Unit 3 (chapter 6) Notes 3 Polynomial: The Basics After this lesson and practice, I will be able to LT1. classify polynomials by degree and number of terms. As a member, you'll also get unlimited access to over 75, lessons in math, English, science, history, and more.
Plus, get practice tests, quizzes, and personalized coaching to help you succeed. Polynomial Graphs and Roots. We learned that a Quadratic Function is a special type of polynomial with degree 2; these have either a cup-up or cup-down shape, depending on whether the leading term (one with the biggest exponent) is positive or negative, respectively.
Think of a polynomial graph of higher degrees (degree at least 3) as quadratic graphs, but with more twists and turns.
Write a degree 3 polynomial with 4 terms polynomials
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How to Factor Polynomials of Degree 3 | Sciencing
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HOW TO CALCULATE A
CORRELATION
Can one statistic measure both the
strength and direction of a linear
relationship between two variables?
Sure! Statisticians use the correlation
coefficient to measure the strength
and direction of the linear relationship
between two numerical
variables X and Y. The correlation
coefficient for a sample of data is
denoted by r.
Although the street definition
of correlation applies to any two
items that are related (such as
gender and political affiliation),
statisticians use this term only in the
context of two numerical variables.
The formal term for correlation is
the correlation coefficient. Many
different correlation measures have
been created; the one used in this
case is called the Pearson correlation
coefficient.
The formula for the correlation (r) is
data;
are the sample means of all the x-
values and all the y-values,
respectively; and sxand sy are the
sample standard deviations of all
the x- and y-values, respectively.
You can use the following steps to
calculate the correlation, r, from a
data set:
1. Find the mean of all the x-values
2. Find the standard deviation of all
the x-values (call it sx) and the
standard deviation of all the y-values
(call it sy).
For example, to find sx, you would
use the following equation:
data set, take
4. Add up the n results from Step 3.
5. Divide the sum by sx ∗ sy.
6. Divide the result by n – 1,
where n is the number of (x, y) pairs.
(It’s the same as multiplying by 1
over n – 1.)
This gives you the correlation, r.
For example, suppose you have the
data set (3, 2), (3, 3), and (6, 4). You
calculate the correlation
coefficient r via the following steps.
(Note that for this data the x-values
are 3, 3, 6, and the y-values are 2, 3,
4.)
1. Calculating the mean of the x and
y values, you get
2. The standard deviations are sx =
1.73 and sy = 1.00.
3. The n = 3 differences found in
Step 2 multiplied together are: (3 –
4)(2 – 3) = (– 1)( – 1) = +1; (3 – 4)(3
– 3) = (– 1)(0) = 0; (6 – 4)(4 – 3) =
(2)(1) = +2.
4. Adding the n = 3 Step 3 results,
you get 1 + 0 + 2 = 3.
5. Dividing by sx ∗ sy gives you 3 /
(1.73 ∗ 1.00) = 3 / 1.73 = 1.73. (It’s
just a coincidence that the result from
Step 5 is also 1.73.)
6. Now divide the Step 5 result by 3
– 1 (which is 2), and you get the
correlation r = 0.87.
https://www.dummies.com/education/
math/statistics/how-to-calculate-a-
correlation/
|
# Counting
### Number of Elements in a Set
The number of elements in a set can be determined by systematic methods of counting.
An experiment is an activity involving chance in which the outcomes are used to estimate probability. When performing an experiment, like rolling a die or flipping a coin, there is a set of possible outcomes. A set is a collection of objects, and an outcome is a possible result of an experiment. The sample space for an experiment is the set of all possible outcomes. An event is a subset of the outcomes in the sample space of an experiment. For the experiment of rolling a die, the sample space is $\lbrace 1, 2, 3, 4, 5, 6 \rbrace$, and rolling an even number is an event that can be represented as the set $\lbrace 2,4, 6\rbrace$.
Making an organized list is one way to determine the number of elements in a set. When a coin is flipped three times, there are a variety of possible outcomes. To list the outcomes in the sample space, let H represent the outcome for heads and let T represent the outcome for tails. The sample space is $\lbrace \text{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}\rbrace$.
The list is organized by recording all outcomes with zero Ts, then all outcomes with one T, then all outcomes with two Ts, and then all outcomes with three Ts. There are eight possible outcomes in the sample space.
• The event of flipping exactly one tail has three outcomes: $\lbrace \text{HHT, HTH, THH}\rbrace$.
• The event of flipping exactly two tails has three outcomes: $\lbrace \text{HTT, THT, TTH}\rbrace$.
• The event of flipping at least two tails has four outcomes: $\lbrace \text{HTT, THT, TTH, TTT}\rbrace$.
### Tree Diagrams
A tree diagram is an organized way of identifying all the possible outcomes in an experiment that involves multiple steps.
A tree diagram displays the sample space of an experiment so that the possible outcomes can be determined and counted. To make a tree diagram, draw a branch to represent the possible outcomes in each stage of the experiment. Trace each branch to the end of the diagram to determine each outcome in the sample space.
In a tree diagram for the experiment of flipping a coin four times, there are two possible outcomes (heads or tails) for the first toss. There are branches from each of those outcomes to represent the possible outcomes for the second toss, and so on. Tracing the top-most branch of the diagram gives the first outcome of the sample space: HHHH.
### Multiplication Principle
In a process with multiple steps, the total number of ways to complete the process is the product of the number of ways each step can be done.
The multiplication principle, or fundamental counting principle, is a rule for counting the number of outcomes in an experiment. A probability experiment can be thought of as a process. The multiplication principle states that for a process involving two steps, if there are $m$ ways to do step 1 and $n$ ways to do step 2, then there are $m\cdot n$ ways to complete the process. For an experiment, the number of possible outcomes is the number of ways the process can be completed.
When there are more than two steps, the multiplication principle still applies; multiply the number of ways each step can be done to find the total number of possible ways to complete the process.
Step-By-Step Example
Using the Multiplication Principle with Two Parts
Identify the number of possible outcomes when choosing two cards from a standard deck of playing cards without replacing the first one.
Step 1
Calculate the number of ways to choose each of the two cards. There are 52 cards in a standard deck of playing cards.
First, choose one card from 52 cards. Then, choose a second card from the remaining 51 cards.
Step 2
Determine the product of the number of ways to make each choice.
\begin{aligned}\\\text{Total cards} \cdot \text{Remaining cards}&=\text{Possible outcomes}\\52 \cdot 51 &= 2\text{,}652\end{aligned}
Solution
There are 2,652 possible outcomes from choosing two cards from a standard deck of playing cards without replacement.
Step-By-Step Example
Using the Multiplication Principle with Three Parts
A sandwich is made from one type of bread, one type of protein, and one type of cheese. Identify the number of possible sandwiches when choosing from 8 types of bread, 5 types of protein, and 7 types of cheese.
Step 1
There are three different choices to make:
• Choose from eight types of bread.
• Choose from five types of protein.
• Choose from seven types of cheese.
Step 2
Calculate the product of the number of ways to make each choice.
\begin{aligned}\text{Bread} \cdot \text{Protein} \cdot \text{Cheese}&=\text{Possible outcomes}\\8 \cdot 5 \cdot 7 &= 280\end{aligned}
Solution
There are 280 possible sandwiches.
|
# Math Snap
## تصشيل حقيقي (1) بطري If $b \in \mathbb{R}$ and $b \neq 0$ such that $\lim _{x \rightarrow c}(b f)$ exists, then $\lim _{x \rightarrow c} f$ exists. Select one: True False Previous page Jump to...
#### STEP 1
Assumptions 1. $b \in \mathbb{R}$ and $b \neq 0$ 2. $\lim_{x \rightarrow c}(b f(x))$ exists 3. We need to determine if $\lim_{x \rightarrow c} f(x)$ exists
#### STEP 2
Given that $\lim_{x \rightarrow c}(b f(x))$ exists, we denote this limit as $L$. $\lim_{x \rightarrow c}(b f(x)) = L$
#### STEP 3
Since $b$ is a non-zero real number, we can use the property of limits involving scalar multiplication. Specifically, if $\lim_{x \rightarrow c}(b f(x)) = L$, then: $\lim_{x \rightarrow c} b \cdot f(x) = b \cdot \lim_{x \rightarrow c} f(x)$
#### STEP 4
From the property above, we can express $L$ in terms of the limit of $f(x)$: $L = b \cdot \lim_{x \rightarrow c} f(x)$
#### STEP 5
To isolate $\lim_{x \rightarrow c} f(x)$, we divide both sides of the equation by $b$: $\lim_{x \rightarrow c} f(x) = \frac{L}{b}$
#### STEP 6
Since $b \neq 0$ and $L$ is a finite value (because $\lim_{x \rightarrow c}(b f(x))$ exists), $\frac{L}{b}$ is also a finite value.
#### STEP 7
Therefore, $\lim_{x \rightarrow c} f(x)$ exists and is equal to $\frac{L}{b}$.
##### SOLUTION
Based on the steps above, we conclude that if $\lim_{x \rightarrow c}(b f(x))$ exists, then $\lim_{x \rightarrow c} f(x)$ must also exist. The statement is True.
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# 2023 AMC 10B Problems/Problem 11
## Problem
Suzanne went to the bank and withdrew $800$. The teller gave her this amount using $20$ bills, $50$ bills, and $100$ bills, with at least one of each denomination. How many different collections of bills could Suzanne have received?
$\textbf{(A) } 45 \qquad \textbf{(B) } 21 \qquad \text{(C) } 36 \qquad \text{(D) } 28 \qquad \text{(E) } 32$
## Solution 1
We let the number of $20$, $50$, and $100$ bills be $a,b,$ and $c,$ respectively.
We are given that $20a+50b+100c=800.$ Dividing both sides by $10$, we see that $2a+5b+10c=80.$
We divide both sides of this equation by $5$: $\dfrac25a+b+2c=16.$ Since $b+2c$ and $16$ are integers, $\dfrac25a$ must also be an integer, so $a$ must be divisible by $5$. Let $a=5d,$ where $d$ is some positive integer.
We can then write $2\cdot5d+5b+10c=80.$ Dividing both sides by $5$, we have $2d+b+2c=16.$ We divide by $2$ here to get $d+\dfrac b2+c=8.$ $d+c$ and $8$ are both integers, so $\dfrac b2$ is also an integer. $b$ must be divisible by $2$, so we let $b=2e$.
We now have $2d+2e+2c=16\implies d+e+c=8$. Every substitution we made is part of a bijection (i.e. our choices were one-to-one); thus, the problem is now reduced to counting how many ways we can have $d,e,$ and $c$ such that they add to $8$.
We still have another constraint left, that each of $d,e,$ and $c$ must be at least $1$. For $n\in\{d,e,c\}$, let $n'=n-1.$ We are now looking for how many ways we can have $d'+e'+c'=8-1-1-1=5.$
We use a classic technique for solving these sorts of problems: stars and bars. We have $5$ stars and $3$ groups, which implies $2$ bars. Thus, the total number of ways is $\dbinom{5+2}2=\dbinom72=21.$
~Technodoggo ~minor edits by lucaswujc
Denote by $x$, $y$, $z$ the amount of $20 bills,$50 bills and $100 bills, respectively. Thus, we need to find the number of tuples $\left( x , y, z \right)$ with $x, y, z \in \Bbb N$ that satisfy $$20 x + 50 y + 100 z = 800.$$ First, this equation can be simplified as $$2 x + 5 y + 10 z = 80.$$ Second, we must have $5 |x$. Denote $x = 5 x'$. The above equation can be converted to $$2 x' + y + 2 z = 16 .$$ Third, we must have $2 | y$. Denote $y = 2 y'$. The above equation can be converted to $$x' + y' + z = 8 .$$ Denote $x'' = x' - 1$, $y'' = y' - 1$ and $z'' = z - 1$. Thus, the above equation can be written as $$x'' + y'' + z'' = 5 .$$ Therefore, the number of non-negative integer solutions $\left( x'', y'', z'' \right)$ is $\binom{5 + 3 - 1}{3 - 1} = \boxed{\textbf{(B) 21}}$. ~stephen chen (Professor Chen Education Palace, www.professorchenedu.com) ## Solution 3 To start, we simplify things by dividing everything by $10$, the resulting equation is $2x+5y+10z=80$, and since the problem states that we have at least one of each, we simplify this to $2x+5y+10z=63$. Note that since the total is odd, we need an odd number of $5$ dollar bills. We proceed using casework. Case 1: One $5$ dollar bill $2x+10z=58$, we see that $10z$ can be $10,20,30,40,50$ or $0$. $6$ Ways Case 2: Three $5$ dollar bills $2x+10z=48$, like before we see that $10z$ can be $0,10,20,30,40$, so $5$ way. Now we should start to see a pattern emerges, each case there is $1$ less way to sum to $80$, so the answer is just $\frac{6(6+1)}{2}$, $21$ or $(B)$ ~andyluo ## Solution 4 We notice that each$100 can be split 3 ways: 5 $20 dollar bills, 2$50 dollar bills, or 1 $100 dollar bill. There are 8 of these$100 chunks in total--take away 3 as each split must be used at least once.
Now there are five left--so we use stars and bars.
5 chunks, 3 categories or 2 bars. This gives us $\binom{5+2}{2}=\boxed{\textbf{(B) 21}}$
~not_slay
## Solution 5 (generating functions)
The problem is equivalent to the number of ways to make $80$ from $2$ bills, $5$ bills, and $10$ bills. We can use generating functions to find the coefficient of $x^{80}$:
The $2$ bills provide $1+x^2+x^4+x^6...+x^{78}+x^{80} = \frac{1-x^{82}}{1-x^2},$
The $5$ bills provide $1+x^5+x^{10}+x^{15}...+x^{75}+x^{80} = \frac{1-x^{85}}{1-x^5},$
The $10$ bills provide $1+x^{10}+x^{20}+x^{30}...+x^{70}+x^{80} = \frac{1-x^{90}}{1-x^{10}}.$
Multiplying, we get $(x^{82}-1)(x^{85}-1)(x^{90}-1)(x^2-1)^{-1}(x^5-1)^{-1}(x^{10}-1)^{-1}.$
## Video Solution 4
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
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# Math in Focus Grade 3 Chapter 10 Practice 8 Answer Key Real-World Problems: Money
Practice the problems of Math in Focus Grade 3 Workbook Answer Key Chapter 10 Practice 8 Real-World Problems: Money to score better marks in the exam.
## Math in Focus Grade 3 Chapter 10 Practice 8 Answer Key Real-World Problems: Money
Solve.
Question 1.
Mrs. Twohill buys a bottle of cooking oil for $6.75. She gives the cashier two$5 bills. How much change does she receive?
$10 –$6.75 = $3.25 She receives$3.25 change.
Explanation:
Mrs. Twohill buys a bottle of cooking oil for $6.75. She gives the cashier two$5 bills. Two $5 bills means$10. Subtract $6.75 from$10 the difference is $3.25. She receives$3.25 change.
Question 2.
Mr. Larson spends $37.50 at a supermarket. He spends$8.25 less than Mrs. Rosa. How much does Mrs. Rosa spend?
$37.50 +$8.25 = $45.75 Mrs. Rosa spend$45.75.
Explanation:
Mr. Larson spends $37.50 at a supermarket. He spends$8.25 less than Mrs. Rosa. Add $37.50 with$8.25 the sum is $45.75. Mrs. Rosa spend$45.75.
Question 3.
Lisa buys a tennis racket and a can of tennis balls. The can of tennis balls costs $18.60. The racket costs$40.85 more.
a. How much is the racket?
$18.60 +$40.85 = $59.45 The racket cost is$59.45.
Explanation:
Lisa buys a tennis racket and a can of tennis balls. The can of tennis balls costs $18.60. The racket costs$40.85 more. Add $18.60 with$40.85 the sum is $59.45. The racket cost is$59.45.
b. How much does she spend in all?
$18.60 +$59.45 = $78.05 She spends$78.05 in all.
Explanation:
The can of tennis balls costs $18.60. The racket cost is$59.45. Add $18.60 with$59.45 the sum is $78.05. She spends$78.05 in all.
Question 4.
Jacob has $8.65 to take to the fair. His mother gives him$15.50 more. He spends $16.45 at the fair. a. How much does he bring to the fair? Answer:$8.65 + $15.50 =$24.15
He brings $24.15 to the fair. Explanation: Jacob has$8.65 to take to the fair. His mother gives him $15.50 more. Add$8.65 with $15.50 the sum is$24.15. He brings $24.15 to the fair. b. How much does he have left? Answer:$24.15 – $16.45 =$7.70
He have left $7.70. Explanation: He brings$24.15 to the fair. He spends $16.45 at the fair. Subtract$16.45 from $24.15 the difference is$7.70. He have left $7.70. Question 5. Madi buys a carton of milk and a bag of bagels. She gives the cashier$10 and receives $5.25 change. The bag of bagels costs$2.75. How much does the carton of milk cost?
$10 –$5.25 = $4.75 The carton of milk and a bag of bagels cost is$4.75.
$2.75 +$ = $4.75$4.75 – $2.75 = 2 The cost of carton of milk is$2.
Explanation:
Madi buys a carton of milk and a bag of bagels. She gives the cashier $10 and receives$5.25 change. Subtract $5.25 from$10 the difference is $4.75. The carton of milk and a bag of bagels cost is$4.75.
The bag of bagels costs $2.75. Subtract$2.75 from $4.75 the difference is$2. The cost of carton of milk is $2. Question 6. Jordan saved$10.40 last month. He saved $5.50 less this month. How much did he save in the two months? Answer:$10.40 – $5.50 =$4.9
He saved $4.9 this month.$10.40 + $4.9 =$15.3
He saves $15.3 in the two months. Explanation: Jordan saved$10.40 last month. He saved $5.50 less this month. Subtract$5.50 from $10.40 the difference is$4.9. He saved $4.9 this month. Add$10.40 with $4.9 the sum is$15.3. He saves $15.3 in the two months. Question 7. A board game costs$28.45. It is $15.20 more than a paint set. The paint set costs$7.90 more than a toy car. How much does the toy car cost?
$28.45 –$15.20 = $13.25 The paint set cost is$13.25.
$13.25 –$7.90 =$5.35 The toy car cost is$5.35.
A board game costs $28.45. It is$15.20 more than a paint set. Which means the paint set cost is $15.20 less than a board game cost. Subtract$15.20 from $28.45 the difference is$13.25. The paint set cost is $13.25. The paint set costs$7.90 more than a toy car. Which means the toy car cost is $7.90 less than a paint set cost. Subtract$7.90 from $13.25 the difference is$5.35. The toy car cost is $5.35. Question 8. Tim and Karen each had the same amount of money to start with. Karen pays$24.60 for a CD and has $7.50 left. Tim buys a watch for$22.75. How much money does Tim have left?
$–$24.60 = $7.50$7.50 + $24.60 =$32.10
Tim and Karen have $32.10. Tim buys a watch for$22.75.
$32.10 –$22.75 = $9.35 Tim have$9.35 left.
Tim and Karen each had the same amount of money to start with. Karen pays $24.60 for a CD and has$7.50 left. Add $7.50 with$24.60 the sum is $32.10. Tim and Karen have$32.10. Tim buys a watch for $22.75. Subtract$22.75 from $32.10 the difference is$9.35. Tim have \$9.35 left.
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# For the parabola x-3 = 1/8 (y+5)^2, what are the coordinates of the vertex and focus and directrix?
Jun 24, 2018
Vertex: $\left(3 , - 5\right)$
Focus: $\left(5 , - 5\right)$
Directrix: $x = 1$
#### Explanation:
First off, we should recognize that this graph opens to the right. This will be important later. To find the vertex, find the $h$ and $k$ values:
$x - h = \frac{1}{4 p} {\left(y - k\right)}^{2}$
$x - 3 = \frac{1}{8} {\left(y + 5\right)}^{2}$
$h = 3 , k = - 5$
Vertex: $\left(3 , - 5\right)$
Next, let's find the focus and directrix. First, though, we should find $p$, which you might know as the distance from the vertex to both the focus and the directrix. How do we find that? You might have noticed $\frac{1}{4 p}$ in the first equation you saw. Let's set our scale factor, $\frac{1}{8}$, equal to that:
$\frac{1}{4 p} = \frac{1}{8}$
$4 p = 8$
$p = 2$
To find the focus, we need to start from the vertex. Since the focus is always on the axis of symmetry within the parabola, and since the parabola opens to the right, we need to move $2$ units right from the vertex:
$\left(3 + 2 , - 5\right) = \left(5 , - 5\right)$
Finally, the directrix will be $2$ units away from the vertex in the opposite direction. The graph opens to the side, so the directrix will start with $x =$. Our directrix, therefore, is $x = 1$.
|
### Understanding Quadrilaterals - Solutions 1
CBSE Class –VIII Mathematics
NCERT Solutions
CHAPTER - 3
1. Given here are some figures:
Classify each of them on the basis of the following:
(a) Simple curve
(b) Simple closed curve
(c) Polygon
(d) Convex polygon
(e) Concave polygon
Ans. (a) Simple curve
(b) Simple closed curve
(c) Polygons
(d) Convex polygons
(e) Concave polygon
2. How many diagonals does each of the following have?
(b) A regular hexagon
(c) A triangle
Ans. (a) A convex quadrilateral has two diagonals.
Here, AC and BD are two diagonals.
(b) A regular hexagon has 9 diagonals.
Here, diagonals are AD, AE, BD, BE, FC, FB, AC, EC and FD.
(c) A triangle has no diagonal.
3. What is the sum of the measures of the angles of a convex quadrilateral? Will this property hold if the quadrilateral is not convex? (Make a non-convex quadrilateral and try)
Ans. Let ABCD is a convex quadrilateral, then we draw a diagonal AC which divides the quadrilateral in two triangles.
A + B + C +
1 + 6 + 5 + 4 + 3 + 2
= (1 + 2 + 3) + (4 + 5 + 6)
[By Angle sum property of triangle]
Hence, the sum of measures of the triangles of a convex quadrilateral is
Yes, if quadrilateral is not convex then, this property will also be applied.
Let ABCD is a non-convex quadrilateral and join BD, which also divides the quadrilateral in two triangles.
Using angle sum property of triangle,
In ABD, 1 + 2 + 3 = ……….(i)
In BDC, 4 + 5 + 6 = ……….(i)
1 + 2 + 3 + 4 + 5 + 6 =
1 + 2 + (3 + 4) + 5 + 6
A + B + C + D =
Hence proved.
4. Examine the table. (Each figure is divided into triangles and the sum of the angles deduced from that.)
Figure Side 3 4 5 6 Angle sum
What can you say about the angle sum of a convex polygon with number of sides?
Ans. (a) When = 7, then
Angle sum of a polygon =
(b) When = 8, then
Angle sum of a polygon =
(c) When = 10, then
Angle sum of a polygon =
(d) When = then
Angle sum of a polygon =
5. What is a regular polygon? State the name of a regular polygon of:
(a) 3 sides
(b) 4 sides
(c) 6 sides
Ans. A regular polygon: A polygon having all sides of equal length and the interior angles of equal size is known as regular polygon.
(i) 3 sides
Polygon having three sides is called a triangle.
(ii) 4 sides
Polygon having four sides is called a quadrilateral.
(iii) 6 sides
Polygon having six sides is called a hexagon.
6. Find the angle measures in the following figures:
Ans. (a) Using angle sum property of a quadrilateral,
(b) Using angle sum property of a quadrilateral,
(a) First base interior angle
Second base interior angle
There are 5 sides, = 5
Angle sum of a polygon =
= =
(b) Angle sum of a polygon =
= =
Hence each interior angle is
7. (a) Find
(b) Find
Ans. (a) Since sum of linear pair angles is
And
Also
[Exterior angle property]
(b) Using angle sum property of a quadrilateral,
Since sum of linear pair angles is
……….(i)
……….(ii)
……….(iii)
……….(iv)
Adding eq. (i), (ii), (iii) and (iv),
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Understanding Scale Factor: If ABC DEF, What Is The Scale Factor?
When it comes to geometry and mathematics, various concepts and formulas need to be understood to solve problems. One such important concept is scale factor. If you are trying to determine the proportional relationship between two similar objects or figures, you need to understand how to calculate the scale factor.
In this article, we will explore the concept of scale factor in the context of ABC DEF. By the end, you will be able to determine and calculate the scale factor using the appropriate formula and equation.
Key Takeaways:
• The scale factor is used to determine the proportional change between two similar figures or objects.
• ABC and DEF typically represent two similar triangles, with the corresponding sides having proportional lengths.
• Deriving the scale factor formula is essential for calculating the scale factor.
• Understanding the scale factor isn’t just about calculation but also how it is applied in real-world scenarios.
• The concept of scale factor extends beyond the specific example of ABC DEF and has various applications in fields like architecture, engineering, and computer graphics.
What is a Scale Factor?
Before we dive into the specifics of determining the scale factor for ABC DEF, let’s first establish what we mean by the term. A scale factor is a ratio that describes the proportional change between two similar figures or objects. It is used to determine the relationship between corresponding sides or dimensions.
A scale factor is typically represented as a fraction or a decimal, such as 1/2, 1:2, or 0.5. It is a dimensionless quantity, meaning it has no units of measurement. Instead, it describes how much larger or smaller one figure is compared to the other.
For example, if we were comparing two similar triangles, the scale factor would tell us how many times larger or smaller one triangle is compared to the other. A scale factor of 2 would mean that one triangle is twice as large as the other, while a scale factor of 0.5 would mean that one triangle is half the size of the other.
Understanding the concept of scale factor is especially important in fields like architecture, engineering, and design, where resizing objects to scale is essential. By using the scale factor, we can proportionally resize a figure or object while maintaining its overall shape and structure.
Understanding ABC DEF
To comprehend the scale factor in the context of ABC DEF, we need to understand the specific figures or objects represented by these symbols. ABC and DEF typically represent two similar triangles, with the corresponding sides having proportional lengths. This means that the ratio of the length of the sides of one triangle to the corresponding sides of the other triangle is equal.
This proportionality is what makes them “similar.” Similar triangles have the same shape, but their sizes are different. In other words, one triangle is an enlarged or reduced version of the other.
The two triangles, ABC and DEF, are similar because they have the same shape, but their sizes are different. To understand how to calculate the scale factor between the two triangles, we need to know the corresponding sides or dimensions.
Deriving the Scale Factor Formula
Calculating the scale factor requires understanding its formula. The scale factor formula is derived by taking the ratio of the corresponding sides of two similar figures. To derive the formula, we consider two similar triangles with corresponding sides of length a, b, and c and a’, b’, and c’ respectively:
Triangle ABCTriangle DEF
aa’
bb’
cc’
The ratio of the corresponding sides of these triangles is:
a/a’ = b/b’ = c/c’
This can be simplified to a single ratio:
k = a/a’ = b/b’ = c/c’
where k is the scale factor. Therefore, the scale factor formula is:
scale factor = k = a/a’ = b/b’ = c/c’
It is important to note that the scale factor remains the same for all corresponding sides of the similar figures.
Understanding the scale factor formula is crucial for calculating the scale factor for similar figures like ABC DEF. Now that we have derived the formula, we can move on to calculating the scale factor.
Calculating the Scale Factor for ABC DEF
Now that we have established what a scale factor is and understand the specific scenario of ABC DEF, we can move on to calculating the scale factor. The formula for determining the scale factor is:
scale factor = length of corresponding side in triangle ABC / length of corresponding side in triangle DEF
Let’s use an example to illustrate how to calculate the scale factor. Consider the following two triangles:
Triangle ABCTriangle DEF
Side AB = 6 cmSide DE = 12 cm
Side AC = 9 cmSide DF = 18 cm
Side BC = 12 cmSide EF = 24 cm
In this example, we are looking to determine the scale factor between the two triangles, so we need to identify the corresponding sides. In this case, AB corresponds to DE, AC corresponds to DF, and BC corresponds to EF.
Let’s calculate the scale factor for the first pair of corresponding sides, AB and DE:
scale factor = length of corresponding side in triangle ABC / length of corresponding side in triangle DEF
scale factor = 6 cm / 12 cm
scale factor = 0.5
We can repeat this process for the other two pairs of corresponding sides and obtain the scale factor values of 0.5 for all three pairs. Therefore, the scale factor between the two triangles is 0.5.
Remember that the scale factor is a ratio, so it is dimensionless and has no units. It simply represents the proportional relationship between the corresponding sides of two similar figures.
Applying the Scale Factor
The scale factor is not just a mathematical concept; it has real-world applications. In the case of ABC DEF, understanding the scale factor can help us resize and transform the triangles.
For example, imagine we have a picture of a triangle ABC. We want to enlarge it and create a new picture that is twice the size. We can use the scale factor to do this. If we know the scale factor is 2, we can multiply the corresponding sides of the triangle by 2. This will give us a new triangle that is twice the size of the original.
Similarly, if we want to shrink a figure, we can use a scale factor less than 1. If we want to flip a figure, we can use a negative scale factor. These are just a few examples of how the scale factor can be applied in real-world scenarios.
Other Applications of Scale Factor
While we have primarily focused on the scale factor in the context of ABC DEF, it has applications beyond this specific example. The scale factor is a crucial concept in various fields, from architecture to computer graphics.
One real-world application of the scale factor is in the design of blueprints and architectural models. Architects often use scale models to represent buildings in a smaller size. By using a scale factor, they can accurately determine the measurements and proportions of the actual building.
In the world of computer graphics, the scale factor is also an essential tool. It is used to resize and transform 3D models, allowing designers to create more complex and intricate designs.
“The scale factor is a powerful tool for designers and architects, allowing them to create accurate and intricate designs with ease.”
The scale factor can also be applied in engineering, as it is used in determining the size and proportions of various components. By using the appropriate scale factor, engineers can ensure that the components fit together correctly.
• Fun fact: The scale factor is also used in mapmaking, as cartographers use it to represent the Earth’s surface on a flat paper map.
Overall, understanding the concept of the scale factor is essential for various mathematical and real-world applications. By grasping the idea and learning how to apply it, you can enhance your problem-solving abilities and explore different fields of study.
Exploring Scale Factor Equations
In addition to the scale factor formula, there are several different equations that can be used to represent the relationship between corresponding sides or dimensions. Each equation has its own significance and can be applied to specific scenarios.
One equation commonly used in geometry is the proportionality equation. This equation states that for two similar figures, the ratio of the corresponding sides is the same. Mathematically, we can represent this as:
Side of ABC / Corresponding side of DEF = Side of ABC / Corresponding side of DEF
This equation emphasizes the proportional relationship between the sides of similar figures.
Another equation used in the context of scale factor is the dimension equation. This equation represents the proportional relationship between dimensions of similar objects. For example:
Length of ABC / Length of DEF = Width of ABC / Width of DEF
This equation is particularly useful in fields like architecture and engineering, where understanding the proportional relationship between dimensions is critical.
Lastly, we have the cross-multiplication equation, which is another way to express the proportionality equation. This equation is often used to solve for an unknown value when the ratio of the corresponding sides is known. The equation is:
Side of ABC x Corresponding side of DEF = Side of DEF x Corresponding side of ABC
Understanding the different equations that can be used to express the relationship between corresponding sides or dimensions can be helpful in various mathematical and real-world applications.
Factors Influencing Scale Factor
While the scale factor primarily represents the proportional relationship between corresponding sides of similar figures, there are certain factors that can influence its value. Understanding these factors is crucial for accurately assessing the size or shape of objects.
One of the influencing factors is the orientation of the figures or objects. When two similar figures or objects are rotated, the scale factor remains the same. However, if one of the figures is flipped, the scale factor will have a negative value. This occurs due to the inversion of the corresponding sides.
Another factor that can influence the scale factor is the presence of 3D objects. When calculating the scale factor for 3D objects, the proportion of corresponding sides can vary depending on the dimension being observed. For instance, if we consider two similar cubes, the scale factor for their side lengths will be different from that of their volumes.
Furthermore, the scale factor can be affected by the measurement units used. When working with different units, such as centimeters and inches, the scale factor may need to be converted to ensure accurate calculations.
Lastly, the scale factor can also be influenced by the degree of accuracy required. Depending on the level of precision needed, the scale factor may need to be rounded to a certain number of decimal places or represented as a fraction instead of a decimal.
By considering these factors and their impact on the scale factor, you can ensure that your calculations accurately represent the size and shape of similar figures and objects.
Section 10: Conclusion
In conclusion, if ABC DEF, the scale factor is a crucial concept for understanding the proportional relationship between similar figures or objects. By calculating the scale factor using the appropriate formula and equation, you can enhance your mathematical prowess and apply it to solve problems within various fields.
Remember that the scale factor is a ratio that describes the proportional change between corresponding sides or dimensions. It can also be used to resize or transform figures and objects in real-world scenarios, making it an essential tool in fields like architecture, engineering, and computer graphics.
Moreover, while the scale factor is primarily determined by the proportional relationship between corresponding sides, factors like rotational symmetry, translation, and reflection can influence its value. Understanding these factors can help you derive more accurate scale factor calculations and make informed decisions.
Overall, the scale factor plays a vital role in a wide range of mathematical and real-world applications. By mastering this concept and its associated formulae and equations, you can elevate your understanding of proportionality and apply it to solve various problems effectively.
FAQ
What is a scale factor?
A scale factor is a ratio that describes the proportional change between two similar figures or objects. It is used to determine the relationship between corresponding sides or dimensions.
How can I determine the scale factor for ABC DEF?
To determine the scale factor for ABC DEF, you need to compare the corresponding sides of the two similar triangles. Divide the length of a side in triangle ABC by the corresponding length in triangle DEF to calculate the scale factor.
What is the formula for calculating the scale factor?
The formula for calculating the scale factor is: scale factor = length of corresponding side in ABC / length of corresponding side in DEF.
How is the scale factor applied in real-world scenarios?
The scale factor can be used to resize or transform figures and objects in various real-world scenarios, such as in architecture, engineering, and computer graphics.
Are there any equations related to the scale factor?
Yes, in addition to the scale factor formula, there are different equations that can be used to represent the relationship between corresponding sides or dimensions of similar figures.
What factors can influence the scale factor?
While the scale factor is primarily determined by the proportional relationship between corresponding sides, factors such as the shape of the objects or figures can also influence its value.
What are some other applications of the scale factor?
The concept of scale factor extends beyond the specific example of ABC DEF and has various applications in fields like architecture, engineering, and computer graphics.
BaronCooke
Baron Cooke has been writing and editing for 7 years. He grew up with an aptitude for geometry, statistics, and dimensions. He has a BA in construction management and also has studied civil infrastructure, engineering, and measurements. He is the head writer of measuringknowhow.com
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# CREATFX STUDIO
## Step 3: Finding the Sin and Cos from a certain Direction
Step 3: Finding the Sin and Cos from a certain Direction
Introduction: Trigonometry. Which Instructable are in the first place intended for the fresh ninth students within DIS, however, anyone is welcome to understand Trigonometry. In this Inclusion, I could promote an over-all overview of the subject of Trigonometry, easy methods to learn and study better, after which enter more detail. When you look at the Mathematics, it’s always important to know how to understand what your are trying to do, and why you will do these types of methods instead of only memorizing it. Trigonometry is the study of triangles. Within instructable, I am able to begin earliest which have naming brand new corners of the best triangles, brand new trigonometric services, then gradually increase the difficulties so the reader is eventually learn how to tackle these problems, thereby applying these to real world activities. I am able to also provide easy methods to data and you will discover this topic better. This session needs one know already some elementary algebra and you may geometry.
## Step one: Algorithms and you may Meanings
Note: x is the position we have been using to search for the opp, adj, or the hypotenuse. Whether or not it had been other position, then the reverse and you can surrounding manage change. Methods : Soh Cah Toa. You really have seen or observed so it many times. The fresh S inside the Soh signifies the latest Sine, as o is short for reverse, additionally the h signifies hypotenuse.
## 2: Routine Troubles
Tip: Explore Pythagoras Theorem To solve to your 3rd not familiar front side. Opp^dos+Adj^dos=Hyp^dos. Upcoming play with algebra to resolve for just one of them edges.
C) 1. Sin 23 = 2500/x 2. x Sin 23 = 2500 step 3. x = 2500/Sin 23. d) Solve which have a good calculator. Carry out the same with cos and bronze.
Cos forty five° = 1/means 2 = .7071 (Calculator). You can utilize the newest pythagorean theorem to test these was good best triangles.
There are many more samples of locating the proportion determining the latest trigonometric services off specific angles. The first step is to find the costs of your own corners, immediately after which divide her or him. For many basics, but not, you may need an excellent calculator. This step is made to help you know what the fresh new strange numbers and decimals on your own calculator mean as soon as you get the sin, cos, otherwise bronze of a direction.
## Step four: Phrase Difficulties
4th Slide: Talking about community issues that are found for the actual-lives things to put your degree on a great deal more basic use!
3)Select the position you can make use of for your problem. Just what form will provide you with the medial side you will want to solve having?
Answer: The new position contrary into the 32° angle is even thirty-two°. Make use of the tan once the adj is provided with, together with reverse must be found. Tan thirty two° = ?/325, ? = 325 Tan 32°. The crater are meters deep.
## Step 5: Inverse Trigonomic Functions
The aim is to discover the way of measuring a direction provided at the very least a few sides. Basic, your determine just the right mode to make use of (tan, sin, and you will cos) depending off and this sides are supplied (Hyp, Adj, Opp). Following resolve towards position. Exp. Get a hold of X. Step one will be to determine what is offered. The alternative (7) together with hypotenuse (25) was recognized. Exactly what trigonometric form comes to both opposite as well as the hypotenuse? The latest sine naturally! Therefore we perform a picture sinx = 7/25. x = arcsin(7/25). Next only kind of one to into your calculator to get the effects. The latest arcsine simply other term for the inverse sin.
## Action six: Whatever you Have discovered
You will find discovered what’s a right triangle, opp, adj, hyp, sin, cos, bronze, how exactly to solve having a not known front having fun with trigonometry, the newest pythagorean theorem, thinking out of trigonometric attributes to own certain basics, using trigonometry so you can real-world difficulties, and using new inverse sine to obtain the worth of an enthusiastic angle given the corners. So you’re able to increase, you should practice even more math problems. I suggest buying a math publication since the a source to find multiple issues, and you can see maxims. For folks who choose their problems, make sure you inquire about help!
CreatFx Studio
An Animation Company in India
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# NCERT Solutions for Class 12 Maths Chapter 7 - Integrals - Exercise 7.5
Exercise 7.5
Integrate the rational functions.
$$\textbf{1.}\space\frac{\textbf{x}}{\textbf{(x+1)(x+2)}}\\\textbf{Sol.}\space\text{Let}\space\text{I}=\int\frac{x}{(x+1)(x+2)}dx\\\text{Let,}\space\frac{x}{(x+1)(x+2)}\\\frac{\text{A}}{(x+1)}+\frac{\text{B}}{(x+2)}\\\Rarr\space \frac{x}{(x+1)(x+2)}\\=\frac{A(x+2)+B(x+1)}{(x+1)(x+2)}$$
$$\Rarr\space x=Ax+2A+Bx+B\\\Rarr\space x=x(A+B)+2A+B$$
On equating the coefficients of x and constant term on both sides, we get
A + B = 1 ...(i)
and 2A + B = 0 ...(ii)
On subtracting Eq. (ii) form Eq.(i) we get
$$-\text{A}=1\Rarr\space A=-1$$
Now, on putting the value of A in Eq. (i), we get
$$-1+B=1\\\Rarr\space B=2$$
$$\therefore\space\text{I}=\int\frac{(\normalsize-1)}{x+1}dx+\int\frac{2}{x+2}dx$$
= – log (x + 1) + 2 log (x + 2) + C
= – log (x + 1) + log (x + 2)2 + C
$$=\text{log}\begin{vmatrix}\frac{(x+2)^2}{x+1}\end{vmatrix}+\text{C}\\\bigg[\because\space\text{log b - log a}=log\frac{b}{a}\bigg]$$
$$\textbf{2.}\space\frac{\textbf{1}}{\textbf{x}^\textbf{2}\textbf{-9}}\\\textbf{Sol.}\space\int\frac{1}{x^2-9}dx=\int\frac{1}{x^2-3}dx\\=\int\frac{1}{(x+3)(x-3)}dx\\\text{Let}\space\frac{1}{(x+3)(x-3)}\\=\frac{A}{(x+3)}+\frac{B}{(x-3)}$$
$$\Rarr\space 1=A(x-3)+B(x+3)\\\Rarr\space 1 = x(A+B)+(-3A+3B)$$
On equating the coefficients of x and constant term on both sides, we get
A + B = 0 and – 3A +3B = 1
On solving, we get
$$\text{A}=-\frac{1}{6}\space\text{and B}=\frac{1}{6}\\\therefore\space\int\frac{1}{(x+3)(x-3)}dx\\=\int\frac{(\normalsize-1)}{6(x+3)}dx+\int\frac{1}{6(x-3)}dx\\=-\frac{1}{6}\text{log}|x+3|+\frac{1}{6}\text{log}|x-3|+\text{C}\\=\frac{1}{6}\text{log}\begin{vmatrix}\frac{x-3}{x+3}\end{vmatrix}+\text{C}\\\bigg[\because\space\text{log b - log a}=\text{log}\frac{b}{a}\bigg]$$
$$\textbf{3.}\space\frac{\textbf{3x-1}}{\textbf{(x-1)(x-2)(x-3)}}\\\textbf{Sol.}\space\text{Let}\frac{3x-1}{(x-1)(x-2)(x-3)}\\=\frac{\text{A}}{(x-1)}+\frac{\text{B}}{(x-2)}+\frac{\text{C}}{(x-3)}\\\Rarr\space\frac{3x-1}{(x-1)(x-2)(x-3)}\\=\\\frac{A(x-2)(x-3)+\text{B}(x-1)+\text{C}(x-1)(x-2)}{(x-1)(x-2)(x-3)}$$
$$\Rarr\space 3x – 1=$$
A[x2 – 5x + 6] + B[x2 – 4x + 3] + C[x2 – 3x + 2]
$$\Rarr\space 3x – 1=$$
x2(A + B + C) + x(– 5A – 4B – 3C) + (6A + 3B + 2C)
On equating the coefficients of x2, x and constant term on both sides, we get
A + B + C = 0 ...(i)
– 5A – 4B – 3C = 3 ...(ii)
and 6A + 3B + 2C = – 1 ...(iii)
From eq. (i), we get
A = – (B + C)
On putting the value of A in Eqs. (ii) and (iii), we get
– 5{– (B + C)} – 4B – 3C = 3
$$\Rarr\space 5B+5C-4B-3C=3\\\Rarr\space B + 2C=3\space \text{...(iv)}$$
and 6{– (B + C)} + 3B + 2C = – 1
$$\Rarr\space -6B--6C+3B+2C=-1\\\Rarr\space -3B-4C=-1\space\text{...(v)}$$
On solving equations. (iv) and (v), w
e get
C = 4
On putting the value of C in Eq. (iv), we get
B + 2 × 4 = 3
$$\Rarr\space B=-5$$
Putting the value of B and C in Eq. (i), we get
A + (– 5) + 4 = 0
∴ A = 1, B = – 5, C = 4
$$\text{Now,}\int\frac{3x-1}{(x-1)(x-2)(x-3)}dx\\=\int\bigg(\frac{A}{(x-1)}+\frac{B}{(x-2)}+\frac{C}{(x-3)}\bigg)dx\\=\int\frac{1}{(x-1)}dx+\int\frac{(-5)}{(x-2)}dx+\\\int\frac{4}{(x-3)}dx$$
= log|x – 1|– 5 log|x – 2| + 4 log|x – 3|+ C
$$\bigg(\because\frac{1}{x}dx= log\space x\bigg)$$
$$\textbf{4.}\space\frac{\textbf{x}}{\textbf{(x-1)(x-2)(x-3)}}\\\textbf{Sol.}\int\frac{x}{(x-1)(x-2)(x-3)}dx\\\text{Let}\space\frac{x}{(x-1)(x-2)(x-3)}\\=\frac{\text{A}}{(x-1)}+\frac{B}{(x-2)}+\frac{C}{(x-3)}\\\Rarr\space\frac{x}{(x-1)(x-2)(x-3)}\\=\\\frac{A(x-2)(x-3)+B(x-1)(x-3)-C(x-1)(x-2)}{(x-1)(x-2)(x-3)}$$
$$\Rarr\space x=A[x^2-2x-3x+6]+\\B[x^2-4x+3]+\text{C}[x^2-x-2x+2]\\\Rarr\space x=x^2(A+B+C)+x(-5A-4B-3C)+\\(6A+3B+2C)$$
On comparing the coefficients of x2, x and constant term on both sides, we get
A + B + C = 0 Þ A = – (B + C) ...(i)
– 5A – 4B – 3C = 1 ...(ii)
and 6A + 3B + 2C = 0 ...(iii)
On putting the value of A in equations (ii) and (iii), we get
– 5[– (B + C)] – 4B – 3C = 1
$$\Rarr\space 5B+5C-4B-3C=1\\\Rarr\space B+2C=1\space\text{...(iv)}\\\text{and 6} \lbrace– (B + C)\rbrace + 3B + 2C = 0\\\Rarr\space -3B-4C=0\space\text{...(v)}$$
Multiplying by 3 in Eq. (iv) and then adding in eq. (v) we get
$$\text{C}=\frac{3}{2}$$
On putting the value of C in Eq. (iv), we get
$$\text{B}+2×\frac{3}{2}=1\\\Rarr\space B=-2.$$
Now, put the values of B and C in Eq. (i), we get
$$A-2+\frac{3}{2}=0\\\Rarr\space A=\frac{4-3}{2}=\frac{1}{2}\\\therefore\space\text{A}=\frac{1}{2}, B=-2\space\text{and C}=\frac{3}{2}\\\text{Now},\space\int\frac{x}{(x-1)(x-2)(x-3)}dx$$
$$=\int\frac{A}{(x-1)}dx+\int\frac{B}{(x-2)}dx+\\\int\frac{C}{(x-3)}dx\\=\frac{1}{2}\int\frac{1}{(x-1)}dx-2\int\frac{1}{(x-2)}dx\\+\frac{3}{2}\int\frac{1}{(x-3)}dx\\=\frac{1}{2}\text{log}|x-1|-2\space\text{log}|x-2|+\\\frac{3}{2}\text{log}|x-3|+\text{C}.$$
$$\textbf{5.}\space\frac{\textbf{2x}}{\textbf{x}^\textbf{2}\textbf{+3x+2}}\\\textbf{Sol.\space}\text{Let}\space\text{I}=\frac{2x}{x^2+3x+2}\\=\frac{2x}{x^2+2x+x+2}\\\frac{2x}{x(x+2)+1(x+2)}=\frac{2x}{(x+2)(x+1)}\\\text{Let}\space\frac{2x}{(x+2)(x+1)}=\frac{A}{(x+2)}+\frac{B}{(x+1)}\\\Rarr\space\frac{2x}{(x+2)(x+1)}\\=\frac{A(x+1)+B(x+2)}{(x+2)(x+1)}$$
$$\Rarr\space 2x=Ax+A+Bx+2B\\\Rarr\space 2x=x(A+B)+ (A+2B).$$
On comparing the coefficients of x and constant term on both sides, we get
A + B = 2 ...(i)
and A + 2B = 0 ...(ii)
On subtracting eq. (ii) form eq. (i), we get
$$-B=2\Rarr\space B=-2$$
On putting the value of B in Eq. (i) we get
$$A-2=2\\\Rarr\space A=4$$
$$\therefore\space\text{I}=\int\frac{A}{(x+2)}dx+\int\frac{B}{(x+1)}dx\\=\int\frac{4}{(x+2)}dx+\int\frac{(\normalsize-2)}{(x+1)}dx$$
= 4 log|x + 2| – 2 log|x + 1| + C
$$\textbf{6.}\space\frac{\textbf{1-x}^\textbf{2}}{\textbf{x(1-2x)}}\textbf{.}\\\textbf{Sol.}\space\text{Let}\space\text{I}=\int\frac{1-x^2}{x(1-2x)}dx.$$
Here, degree of numerator is equal to degree of denominator, so divide the numerator by denominator.
$$\text{Thus,}\space\frac{1-x^2}{x(1-2x)}\\=\frac{x^2-1}{2x^2-x}\\=\frac{1}{2}+\frac{\frac{1}{2}x-1}{2x^2-x}\\\therefore\space\text{I}=\int\frac{1}{2}dx+\int\frac{\frac{1}{2}x-1}{(2x^2-x)}dx$$
$$\Rarr\space\text{I}=\text{I}_1+\text{I}_2\space\text{...(i)}\\\text{where,}\text{I}_1=\int\frac{1}{2}dx\space\text{and}\space\\\text{I}_2=\int\frac{\frac{1}{2}x-1}{(2x^2-x)}dx\\\text{Now,}\space\text{I}_1=\int\frac{1}{2}dx=\frac{1}{2}x+\text{C}_1\\\text{and}\space\text{I}_2=\int\frac{\frac{1}{2}x+1}{2x^2-x}dx$$
$$\text{Let}\space\frac{\frac{1}{2}-1}{(2x^2-x)}\\=\frac{A}{x}+\frac{B}{(2x-1)}\\\Rarr\\\space\frac{\frac{1}{2}x-1}{(2x^2-x)}=\frac{A(2x-1)+Bx}{x(2x-1)}\\\Rarr\space\frac{1}{2}x-1=2Ax-A+Bx\\\Rarr\space\frac{1}{2}x-1=x(2A+B)-A$$
On comparing the coefficient of x and constant term on both sides, we get
$$2A+B=\frac{1}{2}\space\text{...(i)}$$
and – A = – 1
$$\Rarr\space A=1\space\text{...(ii)}$$
From equation (i),
$$2×1+B=\frac{1}{2}\\\Rarr\space\text{B}=\frac{1}{2}-2=\frac{-3}{2}\\\therefore\space\text{I}_2=\int\bigg[\frac{1}{x}-\frac{3}{2(x-1)}\bigg]dx\\=\int\frac{1}{x}dx-\frac{3}{2}\int\frac{1}{2x-1}dx\\\Rarr\space\text{I}_2=\text{log x}-\frac{3}{2}\frac{log|2x-1|}{2}+\text{C}_2$$
On putting the values of I1 and I2 in Eq (i), we get
$$\text{I}=\frac{1}{2}x+\text{log x}-\frac{3}{4}\text{log}|2x-1|+\text{C}\\\lbrack\text{C}_1+\text{C}_2=\text{C}\rbrack$$
$$\textbf{7.\space}\frac{\textbf{x}}{(\textbf{x}^\textbf{2}\textbf{+1})\textbf{(x-1)}}\\\textbf{Sol.}\space\int\frac{x}{(x^2+1)(x+1)}dx$$
First, we resolve the given integrand into partial fractions.
$$\text{Let}\space\frac{x}{(x^2+1)(x-1)}\\=\frac{A}{x-1}+\frac{BX+C}{x^2+1}\space\text{...(i)}$$
$$\Rarr\space x=A(x^2+1)+(Bx+C)(x-1)\space\text{..(ii)}$$
Substituting x = 1 and 0 in Eq. (ii), we get
1 = A(2) and 0 = A – C
$$\Rarr\space A=\frac{1}{2}\text{and}\space\text{C}=A=\frac{1}{2}$$
On equating the coefficient of x2 on the both sides in equation (ii), we get
$$0=A+B\\\Rarr B=-A=-\frac{1}{2}$$
$$\therefore\space\int\frac{x}{(x^2+1)(x-1)}dx\\=\int\begin{Bmatrix}\frac{\frac{1}{2}}{x-1}+\frac{\bigg(-\frac{1}{2}x+\frac{1}{2}\bigg)}{x^2+1}\end{Bmatrix}dx\\=\frac{1}{2}\int\frac{1}{x-1}dx+\frac{1}{2}\int\frac{(-x+1)}{x^2+1}dx\\=\frac{1}{2}\int\frac{1}{x-1}dx-\frac{1}{4}\int\frac{2x}{x^2+1}dx+\\\frac{1}{2}\int\frac{1}{x^2+1}dx\\=\frac{1}{2}\text{log}|x-1|-\frac{1}{4}\text{I}_1+\frac{1}{2}\text{tan}^{\normalsize-1}x\\+\text{C}_1\space\text{...(iii)}$$
$$\text{where,}\space\text{I}_1=\int\frac{2x}{x^2+1}dx.$$
Let (x2 + 1) = t
$$\Rarr\space 2x dx=dt\\\therefore\space\text{I}_1=\int\frac{dt}{t}=\text{log}|t|+\text{C}_2\\=\text{log}|x^2+1|+\text{C}_2$$
On putting these values in eq. (iii), we get
$$\int\frac{x}{(x^2+1)(x-1)}dx\\=\frac{1}{2}\text{log}|x-1|-\frac{1}{4}\text{log}|x^2+1|\\+\frac{1}{2}\text{tan}^{\normalsize-1}x+\text{C}$$
[C = C1 + C2]
$$\textbf{8.}\space\frac{\textbf{x}}{\textbf{(x-1)}^\textbf{2}\textbf{(x+2)}}\\\textbf{Sol.}\space\text{Let}\space\frac{x}{(x-1)^2(x+2)}\\=\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{\text{C}}{x+2}\\\text{...(i)}$$
$$\Rarr\space x=A(x-1)(x+2)+\text{B}(x+2)+\\\text{C}(x-1)^2\text{...(ii)}\\\text{On substituting x = 1, – 2 in eq. (ii), we get}\\\text{1 = B(1 + 2) and – 2 = C(– 2 – 1)}^2$$
$$\Rarr\space\text{B}=\frac{1}{3}\space\text{and}\text{C}=\frac{-2}{9}$$
On equating the coefficient of x2 on both sides in equation (ii), we get
0 = A + C
$$\Rarr\space\text{A}=-\text{C}=\frac{2}{9}\\\therefore\space\int\frac{x}{(x-1)^2(x+2)}dx\\=\int\begin{Bmatrix}\frac{\frac{2}{9}}{x-1}+\frac{\frac{1}{3}}{(x-1)^2}+\frac{\bigg(-\frac{2}{9}\bigg)}{x+2}\end{Bmatrix}dx\\\text{(from Eq. (i))}\\\Rarr\space\int\frac{x}{(x-1)^2(x+2)}dx\\=\frac{2}{9}\int\frac{1}{x-1}dx+\frac{1}{3}\int\frac{1}{(x-1)^2}dx-\\\frac{2}{9}\int\frac{1}{x+2}dx$$
$$=\frac{2}{9}\text{log}|x-1|+\frac{1}{3}\frac{(x-1)^{-2+1}}{(-2+1)}-\\\frac{2}{9}\text{log}|x+2|+\text{C}\\=\frac{2}{9}\text{log}\begin{vmatrix}\frac{x-1}{x+2}\end{vmatrix}-\frac{1}{3}\bigg(\frac{x}{x-1}\bigg)+\text{C}\\\bigg[\because\space\text{log b - log a = log}\bigg(\frac{b}{a}\bigg)\bigg]$$
$$\textbf{9.}\space\frac{\textbf{3x+5}}{\textbf{x}^\textbf{3}\textbf{-x}^\textbf{2}\textbf{-x+1}}\\\textbf{Sol.}\space\int\frac{3x+5}{x^3-x^2-x+1}dx\\=\int\frac{3x+5}{(x^2-1)(x-1)}dx\\=\int\frac{3x+5}{(x-1)(x+1)(x-1)}dx\\=\int\frac{3x+5}{(x-1)^2(x+1)}dx\\\text{Let}\space\frac{3x+5}{(x-1)^2(x+1)}\\=\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{x+1}$$
$$\Rarr\space\frac{3x+5}{(x-1)^2(x+1)}\\=\frac{A(x-1)(x+1)+B(x+1)+C(x-1)^2}{(x-1)^2(x+1)}\\\Rarr\space 3x+5\\=A(x^2-1)+Bx+B+C(x^2+1-2x)\\\Rarr\space 3x+5\\=x^2(A+C)+x(B-2C)+(-A+B+C)$$
On comparing the coefficients of x2, x and constant term on both sides, we get
A + C = 0
$$\Rarr\space A=-C\space\text{...(i)}$$
B – 2C = 3 ...(ii)
and – A + B + C = 5 ...(iii)
On putting the value of A from eq. (i) in eq. (iii), we get
– (– C) + B + C = 5
$$\Rarr\space B+2C=5\space\text{...(iv)}$$
On adding eq. (ii) and eq. (iv), we get
2B = 8
$$\Rarr\space B=4$$
On putting the value of B in eq. (ii), we get
4 – 2C = 3
$$\Rarr\space 1=2C\Rarr\space \text{C}=\frac{1}{2}\\\text{Also,}\space \text{A=-C}=-\frac{1}{2}\\\therefore\space\int\frac{3x+5}{(x-1)^2(x+1)}dx\\=\int\bigg(\frac{-\frac{1}{2}}{x-1}+\frac{4}{(x-1)^2}+\frac{\frac{1}{2}}{x+1}\bigg)dx$$
$$=-\frac{1}{2}\int\frac{1}{x-1}dx+4\int\frac{1}{(x-1)^2}dx+\\\frac{1}{2}\int\frac{1}{x+1}dx\\=\frac{1}{2}\text{log}|x-1|+4\frac{(x-1)^{-2+1}}{(-2+1)}+\\\frac{1}{2}\text{log}|x+1|+\text{C}\\=\frac{1}{2}\text{log}\begin{vmatrix}\frac{x+1}{x-1}\end{vmatrix}-\frac{4}{x-1}+\text{C}\\\bigg[\because\space\text{log b - log a}=log\bigg(\frac{b}{a}\bigg)\bigg]$$
$$\textbf{10.}\space\frac{\textbf{2x+3}}{\textbf{(x}^\textbf{2}\textbf{-1})\textbf{(2x+3)}}\\\textbf{Sol.}\space\int\frac{2x-3}{(x^2-1)(2x+3)}dx\\=\int\frac{2x-3}{(x-1)(x+1)(2x+3)}dx\\\text{Let}\space\frac{2x-3}{(x-1)(x+1)(2x+3)}\\=\frac{A}{(x-1)}+\frac{B}{(x+1)}+\frac{\text{C}}{(2x+3)}\\\Rarr\space\frac{2x-3}{(x-1)(x+1)(2x+3)}\\=\\\frac{A(2x+3)(x+1)+B(x-1)(2x+3)+\text{C}(x-1)(x+1)}{(x-1)(x+1)(2x+3)}$$
$$\Rarr\space\text{2x-3}=A(2x^2+3x+2x+3)+\\\text{B}(2x^2-2x+3x-3)+\text{C}(x^2-1)\\\Rarr\space 2x-3\\=x^2(2A+2B+C)+x(5A+B)+\\(3A-3B-C)$$
On comparing the coefficients of x2, x and constant term on both sides, we get
2A + 2B + C = 0 ...(i)
5A + B = 2
$$\Rarr\space B=2-5A\space\text{...(ii)}$$
and 3A – 3B – C = – 3 ...(iii)
On putting the value of B in Eqs. (i) and (iii), we get
2A + 2(2 – 5A) + C = 0 Þ 2A + 4 – 10A + C = 0
$$\Rarr\space -8A+C=-4\space\text{...(iv)}\\\text{and}\space\text{3A-3}(2-5A)-C=-3\\\Rarr\space 3A-6+15 A-C=-3\\\Rarr\space 18 A-C=3\space\text{...(v)}$$
On adding equations (iv) and (v), we get
$$10A=-1\\\Rarr\space A=\frac{-1}{10}$$
On putting the value of A in equations (ii), we get
$$2\bigg(-\frac{1}{10}\bigg)+2\bigg(\frac{5}{2}\bigg)+\text{C}=0\\\Rarr-\frac{1}{5}+5+\text{C}=0\\\Rarr\space \text{C}=-\frac{24}{5}\\\therefore\space\text{A}=-\frac{1}{10},\text{B}=\frac{5}{2}\space\text{ans C}=-\frac{24}{5}\\\therefore\space\int\frac{2x-3}{(x^2-1)(2x+3)}dx\\=\int\frac{(\normalsize-1)}{10(x-1)}dx+\frac{5}{2}\int\frac{1}{x+1}dx\\-\frac{24}{5}\int\frac{1}{2x+3}dx$$
$$=\frac{5}{2}\text{log}|x+1|-\frac{1}{10}\text{log}|x-1|-\\\frac{12}{5}\text{log}|2x+3|+\text{C}$$
$$\textbf{11.}\space\frac{\textbf{5x}}{\textbf{(x+1)}\textbf{(x}^\textbf{2}\textbf{-4)}}\\\textbf{Sol.}\space\int\frac{5x}{(x+1)(x^2-4)}dx\\=\int\frac{5x}{(x+1)(x+2)(x-2)}dx\\\text{Let}\space \frac{5x}{(x+1)(x-2)(x+2)}\\=\frac{A}{(x+1)}+\frac{B}{(x+2)}+\frac{C}{(x-2)}\\=\\\frac{A(x+2)(x-2)+B(x+1)(x-2)+\text{C}(x+1)(x+2)}{(x+1)(x+2)(x-2)}$$
$$\Rarr\space 5x=A(x^2-4)+B(x^2+x-2x-2)\\+\text{C}(x^2+x+2x+2)\\\Rarr\space 5x=x^2(A+B+C)+x(-B+3C)+\\(-4A-2B+2C)$$
On comparing the coefficients of x2, x and constant term on both sides, we get
A + B + C = 0 ...(ii)
– B + 3C = 5 ...(iii)
and – 4A – 2B + 2C = 0 ...(iv)
Multiply by 4 in eq. (ii) and then adding with eq. (iv), we get
2 (B + 3C) = 0
$$\Rarr\space B+3C=0\space\text{...(v)}$$
On adding equations (iii) and (v), we get
$$\text{C}=\frac{5}{6}$$
On putting the value of C in equations (v), we get
$$\text{B}=-\frac{5}{2}$$
On putting the values of B and C in Eq. (ii), we get
$$\text{A}=\frac{5}{3}\\\therefore\space\int\frac{5x}{(x+1)(x-2)(x+2)}dx\\=\int\frac{5}{3(x+1)}dx-\int\frac{5}{2(x+2)}dx+\\\int\frac{5}{6(x-2)}dx\\=\frac{5}{3}\int\frac{1}{(x+1)}dx-\frac{5}{2}\int\frac{1}{(x+2)}dx+\\\frac{5}{6}\int\frac{1}{(x-2)}dx\\=\frac{5}{3}\text{log}|x+1|-\frac{5}{2}\text{log}|x+2|+\\\frac{5}{6}\text{log}|x-2|+\text{C}$$
$$\textbf{12.}\space\frac{\textbf{x}^\textbf{3}\textbf{+x+1}}{\textbf{x}^\textbf{2}\textbf{-1}}\\\textbf{Sol.}\space\int\frac{x^3+x+1}{x^2-1}dx\\\int x dx + \int\frac{2x+1}{x^2+1}dx\\\bigg[\because\space \frac{x^3+x+1}{x^2-1}=x+\frac{2x+1}{x^2-1}\bigg]\\=\int x dx + \int\frac{2x+1}{(x+1)(x-1)}dx\\\text{Let}\space\frac{2x+1}{(x+1)(x-1)}=\\\frac{A}{(x+1)}+\frac{B}{(x-1)}$$
$$\Rarr\space \frac{2x+1}{(x+1)(x-1)}=\\\frac{A(x-1)+B(x+1)}{(x+1)(x-1)}\\\Rarr\space 2x+1 = x(A+B)-A+B$$
On comparing the coefficients of x and constant term on both sides, we get
A + B = 2 and – A + B = 1
On adding above equations, we get
$$2B=3\Rarr\text{B}=\frac{3}{2}\\\text{and then A}=\frac{1}{2}\\\therefore\space\int\frac{x^2+x+1}{x^2-1}dx=\\\int x dx+\int\frac{A}{(x+1)}dx+\int\frac{B}{(x-1)}dx\\=\int xdx + \frac{1}{2}\int\frac{1}{(x+1)}dx+\\\frac{3}{2}\int\frac{1}{(x-1)}dx\\=\frac{x^2}{2}+\frac{1}{2}\text{log}|x+1|+\\\frac{3}{2}\text{log}|x-1|+\text{C}$$
$$\textbf{13.}\space\frac{\textbf{2}}{\textbf{(1-x)(1+x}^\textbf{2}\textbf{)}}\\\textbf{Sol.}\space\text{Let}\space\frac{2}{(1-x)(1-x^2)}=\\\frac{A}{1-x}+\frac{Bx+C}{1+x^2}\\\therefore\space \frac{2}{(1-x)(1+x^2)}=\\\frac{A(1+x^2)+(Bx+C)(1-x)}{(1-x)(1-x^2)}$$
$$\Rarr 2 = A + Ax^2 + Bx + C – Bx^2 – Cx\\\Rarr 2 = x^2(A – B) + x(B – C) + (A + C)$$
On comparing the coefficients of x2, x and constant term on both sides, we get
A – B = 0
$$\Rarr\space A=B\space \text{...(i)}$$
B – C = 0
$$\Rarr\space B=C\space\text{...(ii)}\\\text{and}\space A + C=2\space\text{...(iii)}$$
From Eqs. (i) and (ii), we get A = C put this value in Eq. (iii), we get
$$\text{2A=2}\Rarr\space A=1$$
Put the value of A in Eqs. (i) and (iii), we get
B = 1 and C = 1
$$\therefore\space\int\frac{2}{(1-x)(1+x^2)}dx\\=\int\begin{Bmatrix}\frac{1}{1-x} + \frac{x+1}{x^2+1}\end{Bmatrix}dx\\\int\frac{1}{1-x}dx+\frac{1}{2}\int\frac{2x}{x^2+1}dx\\\int\frac{1}{x^2+1}dx\\=-\text{log}|1-x|+\frac{1}{2}\text{log}(1+x^2)+\\\text{tan}^{\normalsize-1}x+\text{C}$$
[Let x2 + 1 = t
$$\Rarr\space 2x dx=dt\\\therefore\space \int\frac{2x}{x^2+1}dx=\int\frac{1}{t}dt\text{log t}]$$
$$\textbf{14.}\space\frac{\textbf{3x-1}}{\textbf{(x+2)}^\textbf{2}}\\\textbf{Sol.}\space\text{Let}\space \frac{3x-1}{(x+2)^2}=\\\frac{A}{(x+2)}+\frac{B}{(x+2)^2}$$
$$\Rarr\space 3x-1=A(x+2)+B$$
On equating the coefficients of x and constant term on both sides, we get
A = 3 and 2A + B = – 1
∴ 2(3) + B = – 1
$$\Rarr\space B=-7\\\therefore\space\frac{3x-1}{(x+2)^2}=\frac{3}{(x+2)}-\frac{7}{(x+2)^2}\\\therefore\space \int\frac{3x-1}{(x+2)^2}=\\3\int\frac{1}{(x+2)}dx-7\int\frac{1}{(x+2)^2}dx\\= 3\text{log}|x+2|-7\bigg(\frac{-1}{x+2}\bigg)+\text{C}\\= 3\text{log}|x+2|+\frac{7}{x+2}+\text{C}$$
$$\textbf{15.}\space\frac{\textbf{1}}{\textbf{x}^\textbf{4}\textbf{-1}}\\\textbf{Sol.}\space \int\frac{1}{x^4-1}dx=\int\frac{1}{(x^2-1)(x^2+1)}dx$$
[∵ a2 – b2 = (a + b) (a – b)]
$$\text{Let}\space\frac{1}{(x^2-1)(x^2+1)}=\\\frac{\text{Ax+B}}{(x^2+1)}+\frac{\text{Cx+D}}{(x^2-1)}\\\Rarr\space \frac{1}{(x^2+1)(x^2-1)}\\=\\\frac{(Ax+B)(x^2-1) + (Cx+D)(x^2+1)}{(x^2+1)(x^2-1)}$$
$$\Rarr\space 1=Ax^3+Bx^2-Ax-B+Cx^3+\\x^2D+Cx+D\\\Rarr\\ 1=x^3(A+C)+x^2(B+D)+\\x(-A+C) + (-B+D)$$
On comparing the coefficients of x3, x2, x and constant term on both sides, we get
A + C = 0 ...(i)
B + D = 0 ...(ii)
– A + C = 0 ...(iii)
and – B + D = 1 ...(iv)
On adding equations (i) and (iii), w e get
2C = 0
$$\Rarr\space C=0\Rarr\space C=0,\\\text{then A = 0}[\text{from Eq. (i)}]$$
On adding equations (ii) and (iv), we get
$$\text{2D = 1}\\\Rarr\space D=\frac{1}{2}\\\text{On putting the value of D in Eq. (iv), we get}\\-\text{B}+\frac{1}{2}\\\Rarr\space\text{B}=-\frac{1}{2}\\\therefore\space \text{A = 0,}\space \text{B}=-\frac{1}{2},\text{C=0}\\\text{and D}=\frac{1}{2}$$
$$\therefore\space\int\frac{1}{(x^4-1)}dx=\int\frac{Ax+B}{(x^2+1)}dx\\+\int\frac{\text{Cx+D}}{(x^2-1)}dx\\=-\frac{1}{2}\int\frac{1}{x^2+1}dx+\frac{1}{2}\int\frac{1}{x^2-1}dx\\=-\frac{1}{2}\text{tan}^{\normalsize-1}x+\frac{1}{2}.\frac{1}{2}\text{log}\begin{vmatrix}\frac{x-1}{x+1}\end{vmatrix}+\text{C}\\=-\frac{1}{2}\text{tan}^{\normalsize-1}x+\frac{1}{4}\text{log}\begin{vmatrix}\frac{x-1}{x+1}\end{vmatrix}+\text{C}$$
$$\textbf{16.}\space\frac{\textbf{1}}{\textbf{x(x}^\textbf{n}\textbf{+1)}}\\\textbf{Sol.}\space\text{Let\space}\text{I}=\int\frac{1}{x(x^n+1)}dx\\=\int\frac{x^{n-1}}{x^n(x^n+1)}dx\\\text{Put x}^n=t\\\Rarr nx^{n-1}dx=dt\\\Rarr\space x^{n-1}dx=\frac{1}{n}dt\\\therefore\space\text{I}=\int\frac{x^{n-1}}{x^n(x^n+1)}dx\\=\frac{1}{n}\int\frac{1}{t(t+1)}dt\space\text{...(i)}$$
$$\text{Now,}\space\frac{1}{t(t+1)}=\frac{A}{t}+\frac{B}{(t+1)}\\\Rarr\space 1=A(1+t)+Bt\space\text{...(ii)}$$
On substituting t = 0, – 1 in eq. (ii), we get
A = 1 and B = – 1
$$\therefore\space\frac{1}{t(t+1)}=\frac{1}{t}-\frac{1}{(t+1)}\\\therefore\space\text{I}=\frac{1}{n}\int\bigg(\frac{1}{t}-\frac{1}{(t+1)}\bigg)dt\\\lbrack\text{from eq.(i)}\rbrack\\=\frac{1}{n}[\text{log} |t|-\text{log}|t+1|]+\text{C}\\=\frac{1}{n}[\text{log}|x^n|-\text{log}|x^n+1|]+\text{C}\\\lbrack\text{put t = x}^n\rbrack\\=\frac{1}{n}\text{log}\begin{vmatrix}\frac{x^n}{x^n+1}\end{vmatrix}+\text{C}$$
$$\textbf{17.}\space \frac{\textbf{cos x}}{\textbf{(1 - sin x)(2 - sin x)}}$$
[Hint : Put sin x = t]
$$\textbf{Sol.}\space\text{Let}\\\space\text{I}=\int\frac{\text{cos x}}{(1 - sin x)(2 - sin x)}dx\\\text{Put}\space\text{sin x=t}\\\Rarr\text{cos x}=\frac{dt}{dx}\\\Rarr\space dx=\frac{dt}{\text{cos x}}\\\therefore\space \text{I}=\int\frac{\text{cos x}}{(1-t)(2-t)}\frac{dt}{\text{cos x}}\\=\int\frac{1}{(1-t)(2-t)}dt\\=\int\bigg[\frac{A}{1-t} + \frac{B}{2-t}\bigg]dt\\\text{...(i)}$$
$$\therefore\space\frac{1}{(1-t)(2-t)}=\frac{A(2-t)+B(1-t)}{(1-t)(2-t)}$$
$$\Rarr\space 1=2A-tA+B-Bt\\\Rarr\space 1=1(2A+B) +t(-A-B)$$
On comparing the coefficients of t and constant term on both sides, we get
2A + B = 1 and – A – B = 0
On adding above equations, we get
A = 1 and then B = – 1
$$\therefore\space\text{I}=\int\bigg(\frac{1}{1-t}-\frac{1}{2-t}\bigg)dt\\\lbrack\text{from eq. (i)}\rbrack\\=\int\frac{1}{(1-t)}dt-\int\frac{1}{(2-t)}dt\\=\frac{\text{log}|1-t|}{(-1)}-\frac{\text{log}|2-t|}{(-1)}+\text{C}\\=\text{log}\begin{vmatrix}\frac{2-t}{1-t}\end{vmatrix}+\text{C}\\=\text{log}\begin{vmatrix}\frac{2- \text{sin x} }{1-\text{sin x}}\end{vmatrix}+\text{C}$$
(Put t = sin x)
$$\textbf{18.}\space\frac{\textbf{(x}^\textbf{2}\textbf{+1)(x}^\textbf{2}\textbf{+2)}}{\textbf{(x}^\textbf{2}\textbf{+3)}\textbf{(x}^\textbf{2}\textbf{+4)}}$$
Sol. Here, the degree of numerator and denominator are 4. So, we convert it into simple form by putting x2 = t.
$$\frac{(x^2+1)(x^2+2)}{(x^2+3)(x^2+4)}=\\\frac{(t+1)(t+2)}{(t+3)(t+4)}=\frac{t^2+3t+2}{t^2+7t+12}$$
Since, degree of numerator and denominator is same, so it can be written as
$$=1-\frac{4t+10}{t^2+7t+12}\\\text{Now},\int\frac{(x^2+1)(x^2+2)}{(x^2+3)(x^2+4)}dx\\=\int 1dx-\int\frac{(4t+10)}{t^2+7t+12}dx\\=\int 1 dx-\int\frac{4t-10}{(t+3)(t+4)}dx\\\space\text{...(i)}\\\text{Let}\space\frac{4t+10}{(t+3)(t+4)}=\\\frac{A}{(t+4)}+\frac{B}{(t+3)}\\\Rarr\space \frac{4t+10}{(t+3)(t+4)}=\frac{A(t+3)+B(t+4)}{(t+4)(t+3)}$$
$$\Rarr\space 4t+10=At+3A+Bt+4B\\\Rarr\space 4t+10=t(A+B)+(3A+4B)$$
On comparing the coefficients of t and constant term on both sides, we get
A + B = 4
$$\Rarr\space 3A + 3B=12\space\text{...(ii)}$$
and 3A + 4B = 10 ...(iii)
On subtracting Eq. (iii) from Eq. (ii), w
e get
– B = 2 and 3A + 4B = 10 ...(iii)
On subtracting Eq. (iii) from Eq. (ii), we get
– B = 2
$$\Rarr\space\text{B = -2}\space\text{and A=6}$$
$$\therefore\space\frac{4t+10}{(t+4)(t+3)}=\frac{6}{t+4}-\frac{2}{t+3}$$
On putting this value in eq. (i), we get
$$\int\frac{(x^2+1)(x^2+2)}{(x^2+3)(x^2+4)}dx=\\\int 1 dx-\int\begin{bmatrix}\frac{6}{t+4} - \frac{2}{t+3}\end{bmatrix}dx\\=\int 1dx-\int\bigg(\frac{6}{(x^2+4)}-\frac{3}{(x^2+3)}\bigg)dx\\\lbrack\text{Put} \space t=x^2\rbrack\\=\int\text{1 dx}-\int\bigg(\frac{6}{x^2+2^2}-\frac{2}{x^2 + (\sqrt{3})^2} \bigg)dx\\= x-6\bigg(\frac{1}{2}\text{tan}^{\normalsize-1}\frac{x}{2}\bigg)+2\bigg(\frac{1}{\sqrt{3}}\text{tan}^{\normalsize-1}\frac{x}{\sqrt{3}}\bigg)\\+\text{C}\\\bigg(\because\space \int\frac{1}{a^2+x^2}dx=\frac{1}{a}\text{tan}^{\normalsize-1}\frac{x}{a}\bigg)$$
$$= x-3\space\text{tan}^{\normalsize-1}\frac{x}{2}+\frac{2}{\sqrt{3}}\text{tan}^{\normalsize-1}\frac{x}{\sqrt{3}}+\text{C}$$
$$\textbf{19.}\space\frac{\textbf{2x}}{(\textbf{x}^\textbf{2}\textbf{+1)(x}^\textbf{2}\textbf{+3)}}\\\textbf{Sol.}\space\text{Let}\space\text{I}=\int\frac{2x}{(x^2+1)(x^2+3)}dx\\\text{Put x}^2=t\\\Rarr 2x=\frac{dt}{dx}\\\Rarr\space dx=\frac{dt}{2x}\\\therefore\space\text{I}=\int\frac{2x}{(t+1)(t+3)}\frac{dt}{2x}\\=\int\frac{1}{(t+1)(t+3)}dt$$
$$\text{Let}\space \frac{1}{(t+1)(t+3)}=\\\frac{A}{t+1}+\frac{B}{t+3}\\\Rarr\space \frac{1}{(t+1)(t+3)}=\frac{At+3A+Bt+B}{(t+1)(t+3)}$$
$$\Rarr 1 = (A + B)t + (3A + B)$$
On comparing the coefficients of t and constant terms on both sides, we get
A + B = 0 ...(i)
and 3A + B = 1 ...(ii)
On subtracting eq. (ii) from eq. (i), we get
$$2A=1\\\Rarr\space \text{A}=\frac{1}{2}$$
On putting the value of A in eq. (i), we get
$$\text{B}=-\frac{1}{2}\\\therefore\space \text{I}=\frac{1}{2}\int\frac{1}{t+1}dt-\frac{1}{2}\int\frac{1}{t+3}dt\\=\frac{1}{2}\text{log}|t+1|-\frac{1}{2}\text{log}|t+3|+\text{C}\\=\frac{1}{2}\text{log}\begin{vmatrix}\frac{t+1}{t+3}\end{vmatrix}+\text{C}\\=\frac{1}{2}\text{log}\begin{vmatrix}\frac{x^2+1}{x^2+3}\end{vmatrix}+\text{C}\\\lbrack\text{Put t=x}^2\rbrack$$
$$\textbf{20.}\space\frac{\textbf{1}}{\textbf{x(x}^\textbf{4}\textbf{-1)}}\\\textbf{Sol.}\space\text{Let}\space\text{I}=\int\frac{1}{x(x^4-1)}dx\\=\int\frac{x^3}{x^4(x^4-1)}dx\\=\frac{1}{4}\int\frac{4x^3}{x^4(x^4-1)}dx\\\text{Put x}^4=t\\\Rarr\space 4x^3=\frac{dt}{dx}\\\Rarr\space dx=\frac{dt}{4x^3}\\\therefore\space \text{I}=\frac{1}{4}\int\frac{4x^3}{t(t-1)}\frac{dt}{4x^3}$$
$$\frac{1}{4}\int\frac{dt}{t(t-1)}$$
$$\text{Let}\space\frac{1}{t(t-1)}=\frac{A}{t}+\frac{B}{t-1}$$
Þ 1 = A(t – 1) + Bt ...(i)
On substituting t = 0 and 1 in eq. (i), we get
A = – 1 and B = 1
$$\Rarr\space 1=A(t-1)+Bt\space\text{...(i)}$$
On substituting t = 0 and 1 in eq. (i), we get
A = – 1 and B = 1
$$\therefore\space \frac{1}{t(t-1)}=\frac{-1}{t}+\frac{1}{t-1}\\\therefore\space\text{I}=\frac{1}{4}\int\bigg(-\frac{1}{t}+\frac{1}{t-1}\bigg)dt\\=\frac{1}{4}\lbrace- log |t| + \text{log} |t-1|+\text{C}\rbrace\\=\frac{1}{4}\text{log}\begin{vmatrix}\frac{t-1}{t}\end{vmatrix}+\text{C}\\=\frac{1}{4}\text{log}\begin{vmatrix}\frac{x^4-1}{x^4}\end{vmatrix}+\text{C}$$
(Put t = x4)
$$\textbf{21.}\space\frac{\textbf{1}}{\textbf{e}^\textbf{x}\textbf{-1}}\qquad \lbrack\textbf{Hint :}\space \text{Put} e^x=t\rbrack\\\textbf{Sol.}\space\text{Let}\space\text{I}=\int\frac{1}{e^x-1}dx$$
On multiplying numerator and denominator by e–x, we get
$$\text{I}=\int\frac{e^{\normalsize-x}}{1-e^{\normalsize-x}}dx\\\text{Put}\space 1-e^{-x}=t\\\Rarr\space -e^{-x}(-1)=\frac{dt}{dx}\\\Rarr\space dx=\frac{dt}{e^{\normalsize-x}}\\\Rarr\space e^{-x}dx=dt\\\therefore\space \text{I}=\int\frac{dt}{t}=\text{log}|t|+\text{C}\\=\text{log}|1-e^{-x}|+\text{C}\\=\text{log}\begin{vmatrix}\frac{e^x-1}{e^x}\end{vmatrix}+\text{C}$$
Choose the correct answer in each of the question.
$$\textbf{22.}\int\space\frac{\textbf{x}}{\textbf{(x-1)(x-2)}}\textbf{dx}\space\textbf{equals}\\\textbf{(a)}\space\textbf{log}\begin{vmatrix}\frac{\textbf{(x-1)}^\textbf{2}}{\textbf{x-2}}\end{vmatrix}+\textbf{C}\\\textbf{(b)}\space\textbf{log}\begin{vmatrix}\frac{\textbf{(x-2)}^\textbf{2}}{\textbf{x-1}}\end{vmatrix}+\textbf{C}\\\textbf{(c)}\space\textbf{log}\begin{vmatrix}\bigg(\frac{\textbf{x-1}}{\textbf{x-2}}\bigg)\end{vmatrix}\\\textbf{(d)}\space\textbf{log}\textbf{|(x-1)(x-2)|}+\textbf{C}\\\textbf{Sol.}\space (b)\space\text{log}\begin{vmatrix}\frac{(x-2)^2}{x-1}\end{vmatrix}+\text{C}\\\text{Let}\space\frac{x}{(x-1)(x-2)}=\\\frac{A}{(x-1)}+\frac{B}{(x-2)}$$
$$\Rarr\space x = A(x-2)+B(x-1)\space\text{...(i)}$$
On substituting x = 1 and 2 in Eq. (i), we get
A = – 1 and B = 2
$$\therefore\space\frac{x}{(x-1)(x-2)}\\=-\frac{A}{(x-1)}+\frac{B}{(x-1)}\\\therefore\space\int\frac{x}{(x-1)(x-2)}dx=\\\int\frac{(-1)}{x-1}dx+\int\frac{2}{x-2}dx$$
= – log|x – 1| + 2 log|x – 2| + C
= – log|x – 1| + log|x – 2|2 + C
$$=\text{log}\begin{vmatrix}\frac{(x-2)^2}{x-1}\end{vmatrix}+\text{C}\\\bigg[\because\space \text{log b - log a}=log\frac{b}{a}\bigg]$$
$$\textbf{23.}\space\int\frac{\textbf{dx}}{\textbf{x(x}^\textbf{2}\textbf{+1)}}\space\textbf{equals}\textbf{:}\\\textbf{(a)\space log}|\textbf{x}|-\frac{\textbf{1}}{\textbf{2}}\textbf{log}\textbf{(}\textbf{x}^\textbf{2}\textbf{+1}\textbf{)}\\\textbf{(b)}\space \textbf{log}|\textbf{x}|+\frac{\textbf{1}}{\textbf{2}}\space\textbf{log}(\textbf{x}^\textbf{2}\textbf{+1})\textbf{+C}\\\textbf{(c)\space}\textbf{-log}\textbf{|x|}\textbf{+}\frac{\textbf{1}}{\textbf{2}}\textbf{log}(\textbf{x}^\textbf{2}\textbf{+1})\textbf{+}\textbf{C}\\\textbf{(d)}\space\frac{\textbf{1}}{\textbf{2}}\textbf{log}|\textbf{x}|\textbf{+ log}(\textbf{x}^\textbf{2}\textbf{+1})\textbf{+ C}$$
$$\textbf{Sol.}\space\text{(a) log}|x|-\frac{1}{2}\text{log}(x^2+1)+\text{C}\\\text{Let}\space\frac{1}{x(x^2+1)}\\=\frac{A}{x}+\frac{Bx+C}{x^2+1}$$
$$\Rarr\space 1 = A(x^2+1)+(Bx+C)x$$
On equating the coefficients of x2, x and constant term on both sides, we get
A + B = 0, C = 0 and A = 1
On solving these equations, we get
A = 1, B = – 1 and C = 0
$$\therefore\space\frac{1}{x(x^2+1)}=\frac{1}{x}+\frac{-x}{x^2+1}\\\therefore\space\int\frac{1}{x(x^2+1)}dx=\\\int\begin{Bmatrix}\frac{1}{x}-\frac{x}{x^2+1}dx\end{Bmatrix}\\=\text{log}|x|-\frac{1}{2}\text{log}(x^2+1)+\text{C}\\\begin{bmatrix}\text{Let x}^2+1=t\Rarr\space 2xdx= dt\\\Rarr x dx=\frac{dt}{2},\\\therefore\space \int\frac{x}{x^2+1}dx=\int\frac{1}{t}\frac{dt}{2}=\frac{1}{2}\text{log} \space t\end{bmatrix}$$
|
# 1.07 Order 3 digit numbers
Lesson
## Ideas
Before we begin looking at ordering numbers, and working out which numbers are the largest or smallest we need to remember how to identify the value of the digits in our number.
### Examples
#### Example 1
State the value for the underlined digit.
Worked Solution
Create a strategy
Imagine the numbers are written in a place value table to help find the value of the digit.
Apply the idea
Here is the complete table:
The 8 is in the tens column.
So the value is 8 tens which is 80.
Idea summary
We can use a place value table to find the value of each digit in a number.
## Order numbers to 1000 on a number line
Let's look at how to use a number line to order numbers from smallest to largest, and largest to smallest.
### Examples
#### Example 2
We want to work out which of 3 numbers is the biggest.
a
Which of the number lines has 730,830, and 930 plotted correctly?
A
B
Worked Solution
Create a strategy
Check the values of the ticks on the number line.
Apply the idea
Each small tick corresponds to 20. So 730 should be two ticks to the left of 750, 830 should be two ticks to the left of 850, and 930 should be two ticks to the left of 930.
So the correct answer is option A.
b
Write the three numbers from smallest to largest on one line.
Worked Solution
Create a strategy
The smaller a number is, the further to the left on the number line it is.
Apply the idea
Since 730 is to the left of 830, and 830 is to the left of 930, then 730 is the smallest number and 930 is the largest number.
The three numbers from smallest to largest are:
730,\, 830, \, 930
Idea summary
The smaller a number is, the further to the left on the number line it is.
## Order number to 1000 using place value
We can also use place value to order numbers, and this video includes an example of how we do that.
### Examples
#### Example 3
Order these numbers from largest to smallest.
a
73,\, 33, \, 37
Worked Solution
Create a strategy
We should order the tens first, then the ones.
Apply the idea
The numbers in the place value table are shown:
33 and 37 have the same smallest tens digit, so they are smaller than 73.
But 33 has the smaller ones digit, so 33 is smaller than 37.
The numbers from largest to smallest are: 73,\,37,\, 33
b
671,\, 167, \, 617
Worked Solution
Create a strategy
We should order the hundreds first, then the tens, then the ones.
Apply the idea
The numbers in the place value table are shown:
167 has the smallest hundreds digit so it is the smallest number.
671 and 617 have equal hundreds digits, but 617 has the smaller tens digit.
The numbers from largest to smallest are: 671,\,617,\, 167
Idea summary
When we use place value to order numbers, always start with the far left digit and think about the digit's place value.
Compare the hundreds, then the tens, and then the units (ones).
### Outcomes
#### MA2-4NA
applies place value to order, read and represent numbers of up to five digits
|
# 2008 AMC 12B Problems/Problem 17
## Problem
Let $A$, $B$ and $C$ be three distinct points on the graph of $y=x^2$ such that line $AB$ is parallel to the $x$-axis and $\triangle ABC$ is a right triangle with area $2008$. What is the sum of the digits of the $y$-coordinate of $C$?
$\textbf{(A)}\ 16\qquad\textbf{(B)}\ 17\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 19\qquad\textbf{(E)}\ 20$
## Solution
Supposing $\angle A=90^\circ$, $AC$ is perpendicular to $AB$ and, it follows, to the $x$-axis, making $AC$ a segment of the line $x=m$. But that would mean that the coordinates of $C$ are $(m, m^2)$, contradicting the given that points $A$ and $C$ are distinct. So $\angle A$ is not $90^\circ$. By a similar logic, neither is $\angle B$.
This means that $\angle C=90^\circ$ and $AC$ is perpendicular to $BC$. Let C be the point $(n, n^2)$. So the slope of $BC$ is the negative reciprocal of the slope of $AC$, yielding $m+n=\frac{1}{m-n}$ $\Rightarrow$ $m^2-n^2=1$.
Because $m^2-n^2$ is the length of the altitude of triangle $ABC$ from $AB$, and $2m$ is the length of $AB$, the area of $\triangle ABC=m(m^2-n^2)=2008$. Since $m^2-n^2=1$, $m=2008$. Substituting, $2008^2-n^2=1$ $\Rightarrow$ $n^2=2008^2-1=(2000+8)^2-1=4000000+32000+64-1=4032063$, whose digits sum to $18 \Rightarrow \textbf{(C)}$.
## See Also
2008 AMC 12B (Problems • Answer Key • Resources) Preceded byProblem 16 Followed byProblem 18 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 12 Problems and Solutions
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
|
# Problem of the Day: 1/13/13
Jemma’s sock drawer is a mess. She has four types of socks—four pairs each of pink, navy with yellow stripes, black with polka-dots, and purple/green argyle—all jumbled up together. Her dress principles require that she always wear two socks of different colors, however. How many socks must she pull out randomly in the morning to ensure that she gets two socks of different colors?
Solution to yesterday’s problem:
To solve an equation with absolute value, you must know this: for an absolute value to equal something, the original number can be the positive value or the negative value. For instance, if $|x-2|=1$, then x – 2 = 1 or -1.
In this equation, we have two absolute values. So first let’s remove the absolute value and create our two cases:
(1) $|x-1|-2=1$
(2) $|x-1|-2=-1$
Before we remove the absolute value in either case, we must isolate it on one side. Here, we add 2 to both sides:
(1) $|x-1|=3$
(2) $|x-1|=1$
Now we can work on each case individually. Let’s remove the absolute value in case 1 first:
(1a) $x-1=3$
(1b) $x-1=-3$
Solving each of these like a normal equation, x = 4 or x = -2. Moving to the second case, we’ll remove the absolute value first:
(2a) $x-1=1$
(2b) $x-1=-1$
Solving, we get x = 2 or x = 0. So our solutions are 4, -2, 2, and 0. Adding these up, we find the answer to be 4.
|
Day 3 – Proportions, Similar Triangles
TEXT: Elementary College Geometry by Africk
Geometry Labs by Henri Picciotto. Videos from Khan Academy.
A ratio is just a fancy name for a comparison between 2 quantities, usually with the same units. For instance, the ratio of the mass of the earth to that of the moon is about 80 to 1 and for the diameter it is about 4 to 1. In other words, the earth, 80 times as massive as the moon, will appear 4 times as wide to an outside observer. Click here for more planetary ratio information.
A proportion is an equation stating that 2 ratios or fractions are equal. There are 2 principal tasks with proportions.
1. To be in proportion: determine whether the proportion is true, in another words whether 2 ratios are equal. Click here if you are not familiar with this.
2. Assume that a proportion is true and solve for unknown:
Often the proportion with an unknown is associated with a word problem:
Here are some additional word problems solved using proportions.
If the proportion is a:b=c:d, then the means are “b” and “c” and the extremes are “a” and “d”. Switching the means or extremes produces equivalent proportions, e.g., a:b=c:d is true if and only if a:c=b:d is true. A common solution technique, cross-multiplication, can be stated: the product of the means is equal to the product of the extremes.
Similar planar figures have the same shape but may be of different sizes. In other contexts, such as architecture or biology, the concept is scale drawings. Polygon similarity means (1) corresponding angles are the same and (2) corresponding sides are in proportion. In general, to determine similarity, it is not enough to have just information about angles or sides. For example, 2 rectangles have the same angles but are not necessarily similar; the ratio between consecutive sides is arbitrary. Rhombi have equal sides, yet may not be similar. Thinking of vertices as hinges, open them up equally to get a square or close them at 2 opposite vertices to get a skinny diamond.
To get similar triangles, it is enough to know that either their angles (AAA) are equal or their sides are in proportion RRR (SSS). In fact, it is enough to know that just 2 sets of angles are equal (AA) since we get the third set of equal angles for free (the 3 angles of a triangle sum to 180°). More surprising, it is enough to know 2 pairs of corresponding sides are in proportion and their included angles are equal RAR (SAS). The first half of this video is a problem requiring RAR (SAS). For a wider exposition of the topic, see the KA video parts I and II; more exercises.
Note: Don’t get confused with our abbreviations. We will encounter SSS and SAS again but in the context of equal or congruent triangles (equal sides as well as angles). Hence, we the instructors have decided to use RRR in place of SSS and RAR in place of SAS (R stands for “ratio”).
Lab 10.1 will give you hands on experience with the geoboard and is designed to cement notions from the day’s material, including scaling and similarity. To do problems 4 and 5, you should be familiar with the notion of slopeHere is a mildly annoying rap video on the subject which you may find humorous.
Check out how a teacher uses a mirror to calculate the heights of students in her class. [On p. 24 of your geometry textbook, the angles of incidence and reflection are defined as in this diagram. In physics, the angles defined are the respective complements. However, the statement “angle of reflection = angle of incidence” is valid for both sets of definitions.]
8 Responses to Day 3 – Proportions, Similar Triangles
1. awiltshire says:
Pr. Halleck
So today I was a lil bit confused I didn’t quite understand the lab, but working in a group helped me a lil to understand it better. It’s had been a long time since I learned about finding the Slope of a line so it was a bit misunderstood, but eventually I got it and was able to complete the lab questions.
2. gretchenv says:
Prof. Reitz
The cup cake problem was fun and helped me, it made me learn the material better.
3. Nahim Mishi says:
Pro. Reitz
At the end of class i was confused as to AAA or SAS but after reading the notes here i got a better understanding of it.
4. wavygee says:
George Duncan
I wasnt sure about the lab we did at first it was kind of confusing. As we worked on it it began to slowly come back to me. I realize that there is a lot of things i learned in the past that I dont remember.
5. lalayudi says:
the handout was hard to read and understand. But its helpful after all.
6. Mr. Reitz
|
# How to Calculate the Surface Area of Pyramids and Cones
The mysteries of the ancient pyramids and the delicious appeal of an ice cream cone have one thing in common - their distinctive shapes! Beyond their allure and history, pyramids and cones are geometric wonders. How much space does their surface cover? What mathematical intricacies lie behind their slanted sides? In this comprehensive guide, we will uncover the techniques to calculate the surface area of pyramids and cones, making these enigmatic shapes more accessible. Let's embark on this intriguing journey through the slopes of geometry!
## Step-by-step Guide: Surface Area of Pyramids and Cones
Pyramid:
The surface area of a pyramid can be found using the formula: Surface Area $$SA = \text{Base Area} + \frac{1}{2} \times \text{Perimeter of Base} \times \text{Slant Height}$$
Cone:
The surface area of a cone can be found using the formula: Surface Area $$SA = \pi r^2 + \pi r \times \text{Slant Height}$$
Where:
$$r =$$ radius of the base
### Examples
Example 1:
Find the surface area of a square pyramid with a base side of $$5 \text{ cm}$$ and a slant height of $$7 \text{ cm}$$.
Solution:
Base Area $$= 5 \text{ cm} \times 5 \text{ cm} = 25 \text{ cm}^2$$
Perimeter of Base $$= 4 \times 5 \text{ cm} = 20 \text{ cm}$$
$$SA = 25 \text{ cm}^2 + \frac{1}{2} \times 20 \text{ cm} \times 7 \text{ cm} = 95 \text{ cm}^2$$
Example 2:
Calculate the surface area of a cone with a base radius of $$6 \text{ cm}$$ and a slant height of $$10 \text{ cm}$$.
Solution:
$$SA = \pi (6 \text{ cm})^2 + \pi (6 \text{ cm})(10 \text{ cm}) = 301.44 \text{ cm}^2$$
### Practice Questions:
1. Determine the surface area of a triangular pyramid with a triangular base of area $$15 \text{ cm}^2$$, base perimeter of $$18 \text{ cm}$$, and a slant height of $$9 \text{ cm}$$.
2. What is the surface area of a cone with a radius of $$4 \text{ cm}$$ and a slant height of $$8 \text{ cm}$$?
1. $$96 \text{ cm}^2$$
2. $$150.8 \text{ cm}^2$$
### What people say about "How to Calculate the Surface Area of Pyramids and Cones - Effortless Math: We Help Students Learn to LOVE Mathematics"?
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Operations with Decimals
```GRADE 6
Operations
with
Decimals
WORKSHEETS
Multiplying and dividing by powers of ten
Move the decimal point depending on the number of zeros.
= decimal point moves right
1
= decimal point moves left
Calculate these multiplication and division questions involving powers of 10:
a 5 # 1, 000
5 # 1 , 000 = 5.0 # 1 , 000 The whole number in decimal form
1 2 3
We can simply add the same
number of zeros to the end
of the whole number
= 5.0
.
= 5, 000
Fill the empty bounces with 0s
If the decimal point is on the left after dividing, an extra 0 is placed in front.
b 8 ' 100
8 ' 100 = 8.0 ' 100
21
Remember to include
÷ 100 has 2 zeros, so move decimal point 2 spaces left
= . 8.0
= 0.08
c 1.25893 # 10, 000
The whole number in decimal form
Fill the empty bounces with 0s and put a zero in front
1234
1.25893 # 10, 000 = 1 . 2 5 8 9 . 3
= 12, 5 8 9 . 3
Move decimal point 4 spaces right
No empty bounces to fill, so this is the answer
d 24.905 ' 100, 000
54321
24.905 ' 100, 000 = .
2 4 . 905 Move decimal point 5 spaces left
= 0.00024905
e 260.15 #
Fill empty bounces with 0s and put a zero in front
1
1, 000
1
1
260.15 # 1, 000
= 260.15 ' 1, 000 # 1, 000 is the same as ' 1,000
3 2 1
= . 2 6 0 . 15
Move decimal point 3 spaces left
= 0.2 6 015
Place a leading zero in front of the decimal point
| OPERATIONS WITH DECIMALS |
6.NS.3
1
Multiplying and dividing by powers of ten
1
2
Calculate these multiplications. Remember, multiply means move decimal point to the right:
a 8 # 100
b 3.4 # 10
c 29 # 1, 000
d 12.45 # 10, 000
e 0.512 # 100
f 0.0000469 # 1, 000, 000
Calculate these divisions. Remember, divide means move decimal point to the left:
a 2 ' 100
b 4, 590 ' 1, 000
c 0.014 ' 10
d 70.80 ' 10, 000
e 1, 367.512 ' 1, 000
f 421, 900 ' 100, 000, 000
Here are some of the powers of 10 using exponent notation. The power = the number of zeros.
10 1 = 10
10 4 = 10, 000
3
10 2 = 100
10 5 = 100, 000
10 3 = 1, 000
10 6 = 1, 000, 000
Calculate these mixed problems written using exponent notation:
a 31 # 102
b 2, 400 ' 10 5
c 0.0027 # 106
d 90.008 # 104
e 3.45 ' 103
f 2, 159, 951 ' 10 7
| OPERATIONS WITH DECIMALS |
6.NS.3
2
Multiplying and dividing by powers of ten
4
For these calculations:
i Show where our character needs to spray paint a new decimal point, and
ii write down the two numbers the new decimal point is between to solve the puzzle
a 2, 830.3920 # 100
2 8 3 0 3 9 2 0
I
b 23, 857 ' 1, 000
2 3 8 5 7
N
c 0.4763892 # 105
0 4 7 6 3 8 9 2
A
d 382, 961 ' 10, 000
3 8 2 9 6 2
O
e 19, 238.07 # 10 1
1 9 2 3 8 0 7
X
f 8.9236701 # 10, 000
8 9 2 3 6 7 0 1
T
2 0 9 1 7 9 8 3
R
8 3 9 1 7
I
9 0 2 8 7 3 2 0 1
D
0 0 8 3 9 0
P
g 20, 917, 983 #
1
1, 000, 000
h 83, 917 ' 10 5
i 902, 873.021 #
1
10 2
j 0.08390 # 103
9 and 2
This is another mathematical name for a decimal point:
I
0 and 9 8 and 9 8 and 7 9 and 2 0 and 7 3 and 9 8 and 2 0 and 8 3 and 8 6 and 7
| OPERATIONS WITH DECIMALS |
6.NS.3
3
Just add or subtract the digits in the same place value.
To do this, line up the decimal points and matching place values vertically first.
• Add 2.45 to 6.31 (i.e. 2.45 + 6.31)
2 . 4 5 +
6 . 3 1
Decimal points lined up vertically
8 . 7 6
• Subtract 5.18 from 11.89 (i.e. 11.89 - 5.18)
1
1 1 . 8 9 5 . 1 8
Decimal points lined up vertically
6 . 7 1
Subtract matching place values
Calculate each of these further additions and subtractions
a 24.105 + 11.06 + 6.5902
Any place value spaces
are treated as 0s
24 . 1 0 5 +
1 1 . 0 6
6 . 519 0 2
1
Decimal points lined up vertically
4 1 . 7 5 5 2
24.105 + 11.06 + 6.5902 = 41.7552
Rounding decimal values before adding is sometimes used to quickly approximate the size of the answer.
b Round each value in question (i) to the nearest whole number before adding.
24.105 + 11.06 + 6.5902 . 24 + 11 + 7
. 42
Values rounded to nearest ones
Note: R
ounding values before adding/subtracting is not as accurate as rounding after adding/subtracting.
c 80.09 - 72.6081
Fill each place value space
in the top number with
a “0” when subtracting.
8 10 .10 9 10 10 -
Decimal points lined up vertically
7121.6 0181 1
Subtract matching place values
7.4 8 1 9
80.09 - 72.6081 = 7.4819
| OPERATIONS WITH DECIMALS |
6.NS.3
4
1
2
a 0 . 1 4 +
b 1 . 6 8 +
c 0 . 2 4 6+
d 1 2 . 1 9 4+
0 . 7 3
5 . 3 0
0 . 8 3 2
9 . 0 5 7
e 0 . 9 9 -
f 5 . 0 7 4-
g 5 . 2 4-
h 24 . 1 5 8-
0 . 2 6
1 . 0 6 4
0 . 8 3
1 3 . 6 9 4
Calculate these additions and subtractions, showing all working:
b Subtract 3.15 from 4.79
d Add 2.19, 5.6, and 0.13
e Subtract 0.9356 from 8.6012
f Add 10.206, 4.64, and 8.0159
| OPERATIONS WITH DECIMALS |
6.NS.3
5
3
Approximate these calculations by rounding each value to the nearest whole number first.
a 5.7 + 6.2 .
.
c 8.3 - 1.9 .
b 0.9 + 9.4 .
+
.
.
d 11.3 - 0.2 .
-
e 8.34 + 1.61 + 0.54 .
+
+
+
-
.
f 2.71 + 3.80 + 1.92 .
.
+
+
.
Calculate parts e and f again, this time rounding after adding the numbers to get a more accurate
approximate value.
h 2.71 + 3.80 + 1.92
g 8.34 + 1.61 + 0.54
4
Calculate these subtractions, showing all your working:
a 7.8 - 2.56
b 13.09 - 8.4621
| OPERATIONS WITH DECIMALS |
c 0.52 - 0.12532
6.NS.3
6
Multiplying with decimals
Just write the terms as whole numbers and multiply. Put the decimal point back in when finished.
The number of decimal places in the answer = the number of decimal places in the question!
1
Calculate 4 # 1.2
4 # 12 = 4 8
Multiply both terms as whole numbers
1
1 decimal place in question = 1 decimal place in answer
48
4 # 1.2 = 4.8
2
Calculate 0.02 # 1.45
2 # 145 = 2 9 0
Multiply both terms as whole numbers
4321
290
0.02 # 1.45 = 0 . 0 2 9 0
4 decimal places in question = 4 decimal places in answer
Let’s do the second one again but this time change the decimals to equivalent fractions first.
2 # 145
0.02 # 1.45 = 100
100
2 # 145
= 100
# 100
= 10290
, 000
Changing the decimals to fractions
Multiply numerators and denominators together
Number of zeros in denominator = total of decimal places
in question
= 290 ' 10, 000
4 3 2 1
= 0. 2 9 0
Dividing by 10,000 moves decimal point four places to the left
= 0.0290
4 decimal places in question = 4 decimal places in answer
Try this method for yourself on the first example above, remembering that 4 = 4 as a fraction.
1
| OPERATIONS WITH DECIMALS |
6.NS.3
7
Multiplying with decimals
1
2
Calculate these whole number and decimal multiplications, showing all you working:
a 0.8 # 2
b 5 # 1.5
c 0.14 # 6
d 0.62 # 4
e 3 # 0.032
f 1.134 # 2
Calculate these decimal multiplications, showing all your working:
a 3.8 # 0.2
b 1.09 # 0.08
c 2.7 # 2.5
d 7.1 # 1.4
e 3.21 # 2.1
f 17.2 # 9.3
| OPERATIONS WITH DECIMALS |
6.NS.3
8
Dividing with decimals
We move the decimal point before dividing if needed.
To find the quotient when working with decimals, the question must be changed so the divisor is a
whole number.
dividend ' divisor = quotient
Calculate 4.28 ' 4
1. 0 7
4 g 4 . 2 28
Divisor already a whole number so no change needed
4.28 ' 4 = 1.07
Calculate 0.0456 ' 0.006
0.0456 ' 0.006 = 0.0 4 5 6 ' 0.0 0 6 Move both decimal points right until divisor is a whole number
= 4 5.6 ' 6
0 7. 6
6 g 4 45 . 3 6
0.0456 ' 0.006 = 7.6
Quotient 2 Dividend
if divisor 1 1
Drop off any 0s at the front of the answer
Here’s another example showing how to treat remainders.
Calculate 1.26 ' 0.8
1.26 ' 0.8 = 1.2 6 ' 0.8
Move both decimal points right until divisor is a whole number
= 12.6 ' 8
0 1. 5 7 5
1
= 8 g 1 2 . 4 6 60 4 0
1.26 ' 0.8 = 1.575
Add 0s on the end of the dividend for each new remainder
Drop off any 0s at the front
| OPERATIONS WITH DECIMALS |
6.NS.3
9
Dividing with decimals
1
Calculate these decimal and whole number divisions:
a 3.6 ' 4
b 17.5 ' 5
g
g
3.6 ' 4 =
d 0.63 ' 3
g
g
17.5 ' 5 =
16.2 ' 9 =
e 0.489 ' 5
f 10.976 ' 7
g
g
10.976 ' 7 =
0.489 ' 5 =
0.63 ' 3 =
2
c 16.2 ' 9
Calculate these decimal divisions, showing all your working:
a 5.2 ' 0.4
b 9.6 ' 0.6
g
g
5.2 ' 0.4 =
c 0.56 ' 0.8
g
0.56 ' 0.8 =
9.6 ' 0.6 =
d 1.58 ' 0.4
e 0.8125 ' 0.05
1.58 ' 0.4
0.8125 ' 0.05
5.3682 ' 0.006
=
=
=
g
|
# Methods and Operations
Methods, hopefully more to come than just this one so far (discovered from work done in a post) combined with a math operation.
1. Take outer integers and combine, do the same with inner integers, perform an operation…eg, 3885 -> 35 and 88 ( see 3885 plus mirror 777 for the origin of this idea, 88 ÷ 35 ).
A. 2701 -> 21 70
21 ÷ 70 = 0.3333333…
70 ÷ 21 = 3.3333333…
21 ÷ 7 = 3
B. 1260 1290 1335
26 ÷ 10 = 2.6
29 ÷ 10 = 2.9
33 ÷ 15 = 2.2
2.6 + 2.9 + 2.2 = 7.7
C. 2368. (Jesus Christ in Greek gematria )
28 ÷ 36 = 0.777777…
# 3885 plus mirror … 777 (and much more)
3,885 + 5,883 = 9,768 (= 11 × 888 )
9 + 768 = 777
As we have seen on another post 3885 and 7770, 3,885 × 2 = 7,770
7 + 770 = 777
Continuing on 😎
Prime factors for 9,768 are 2, 3, 11, 37
All factors for 9,768 are
1, 2, 3, 4, 6, 8, 11, 12, 22, 24, 33, 37, 44, 66, 74, 88, 111, 132, 148, 222, 264, 296, 407, 444, 814, 888, 1221, 1628, 2442, 3256, 4884, 9768
The prime factors for 3,885 are 3, 5, 7, 37
All factors for 3,885 are 1, 3, 5, 7, 15, 21, 35, 37, 105, 111, 185, 259, 555, 777, 1295, 3885
Prime factors for 5883 are 3, 37, 53
Factors for 5883 are
1, 3, 37, 53, 111, 159, 1961, 5883
Notice the 888 and 777 in the above. Note also that 37 and 111 are common to all 3 numbers. See 37 and, in particular, 111, in the next section.
Something else…
9,768 ÷ 3,885 = 2.5142857142857142
And
88 ÷ 35 = 2.5142857142857142
Hope you can see straight away what’s very interesting…?
Yup,… 3 88 5
… That really does deserve some attention.
(And, of some interest
9768 ÷ 88 = 111 and it follows that 3885 ÷ 35 = 111
3 × 37 = 111
9768 = 3 × 37 × 88
3885 = 3 × 37 × 35 ( or 3 × 5 × 7 × 37 )
…………
36 ÷ 70 = 0.5142857142857142 ( See the above section for the relevant decimal places ).
((9,768 ÷ 3,885) – 2) = (6 × 6) ÷ 70
………….
88 ÷ 35 as a percentage is
351.42857142857142 %
(It’s just 1 + 2.5142857142857142 = 3.5142857142857142 then move the dots for the percentage)
351 is the mirror of 153
Interesting a little bit, maybe
# 1260 1290 1335 37 185 continued
Just to recap a little…
1 2 6 0
1 2 9 0
1 3 3 5
__________ +
3 7 18 5
……………….
The simple math operation:-
1 + 1 + 1 = 3
2 + 2 + 3 = 7
6 + 9 + 3 = 18
0 + 0 + 5 = 5
So…
3. 7. 18. 5 Or
37,185
So we have 37 and 185 here.
185 × 2 = 370
……………..
Ok, so we’ve seen this before.
Now, let’s just multiply across (leaving out zeros or we get nowhere)
1260 1 × 2 × 6 = 12
1290 1 × 2 × 9 = 18
1335 1 × 3 × 3 × 5 = 45
12 + 45 + 18 = 75
If we add 75 days to the 1,260 days mentioned in the Revelation of Jesus Christ, we get 1,335, the number of days mentioned in the book of Daniel where those who reach that number of days are blessed.
So we see increasingly that there is something very special here.
……….
Just exploring a bit now 🤕
Let’s now raise the individual digits to ^2 (again, ditching the zeros)
1260 1^2 × 2^2 × 6^2 = 144
1290 1^2 × 2^2 × 9^2 = 324
1335 1^2 × 3^2 × 3^2 × 5^2 = 2,025
144 + 324 + 2,025 = 2493
2493 ÷ 9 = 277
(9 is the largest factor of 2493)
Here is some information regarding 277 that is gleaned from the Internet.
277 is the 59th prime number.
It is a regular prime.
It is the smallest prime such that the sum of the inverses of the primes up to 277 is greater than two.
59 is a prime which means 277 is a super-prime. (Also, 59 is also a super-prime).
59 is the 17th prime number.
17 is the 7th prime number.
Food for thought here.
# Euler Totient of 9999999999999999999
The totient of
9999999999999999999 (19 “9’s”) is 6666666666666666660 (18 “6’s”)
φ(9999999999999999999) = 6666666666666666660
# 3885 and 7770
1260 + 1290 + 1355 = 3885
3 + 885 = 888 (Gematria of Jesus in the greek)
So, what’s new.
Simply, 3885 * 2 = 7770
# Babel (on sea) 37
Starting to have a look at this in genesis Genesis 9:1-11, the tower of babel… Babel is spelt… bet bet lamed ב bet (value 2)…. occurs 22 times ל Lamed (Value 30)…… occurs 37 times Notice that bet has a gematria value of 2. There are 2 bets in Babel..that’s 2 two’s…….22, the number of times bet occurs in this passage. It’s warming up. Notice how bet and lamed turn up in the word for confound/confusion. ו נ ב ל הH1101 confound ב ב ל H894 Babel There is another 37 occurrence, that is for Yod… י=10 Yod Here’s the data…… ס(60) = 1. ג(3) = 2. ז(7) = 2. צ(90) = 3. ק(100) = 3. ן(50) = 4. ח(8) = 8. ץ(90) = 8. כ(20) = 9. ד(4) = 10. פ(80) = 14. ת(400) = 14. מ(40) = 16. ע(70) = 18. נ(50) = 21. ב(2) = 22. ם(40) = 22. ש(300) = 27. ר(200) = 29. א(1) = 30. י(10) = 37. ל(30) = 37. ו(6) = 44. ה(5) = 54. (Verse total = 1666. Verse factors:- 1(star 1), 2, 7, 14, 17, 34, 49, 98, 119, 238, 833, 1666 Digit grouping for 1666 = 667. Digit grouping result factors:- 1(star 1), 23, 29, 667) ………………………………. Babel on the Sea (hoho, pure conjecture but, hey, why not?) This is Babel…bet bet lamed . ב(2) = 22 ב(2) = 22 ל(30) = 37 more correctly, right to left ל(30) = 37 . ב(2) = 22 ב(2) = 22 בבל So…What is this… ם(40) = 22 Mem sofit י(10) = 37 Yod ים means sea (right to left is 37 22 – admittedly its the other way around to Babel which begins with 22, not 37, but maybe as we are speaking about a divine reversal it kind of fits) (other configurations… םי םםי יםם ) ים yâm yawm From an unused root meaning to roar ; a sea (as breaking in noisy surf) or large body of water; specifically (with the article) the Mediterranean; sometimes a large river, or an artificial basin ; locally, the west, or (rarely) the south: – sea (X -faring man, [-shore]), south, west (-ern, side, -ward). Maybe Jeremiah chapter 51 gives a clue:- 41, How Sheshach is captured, and the glory of all the earth is seized! How Babylon has become as an object of horror among the nations! 42, The sea has risen over Babylon, she has been covered by the roar of its waves. 43, Her cities have become as an object of horror, a dry land and a wilderness,21 a land in which no person lives, nor does a son of humankind pass through it. 44, And I will punish Bel in Babylon, and I will wrench out from his mouth ‹what he has swallowed›.22 And the nations will not stream towards him any longer ‹what’s worse›,23 the wall of Babylon has fallen. 45, Come out from her midst, my people, and save each one his life from ‹the burning anger of›24 Yahweh. 46, Now ‹so that you are not fainthearted›,25 and you are afraid at the rumors6 that are heard in the land— and in the year the rumor comes, and in the year after it the rumor, and violence is in the land, with ruler against ruler— 47, ‹therefore›27 look, days are coming, and I will punish the images of Babylon, and all her land will be put to shame, and all her slain ones will fall in the midst of her. 48, Then the heaven and the earth and all that is in them will shout for joy over Babylon, for from the north the destroyers will come to it,” ‹declares›28 Yahweh. 49, Babylon must fall not only because of the slain ones of Israel, but also because of Babylon the slain ones of all the earth have fallen. Then in Revelation 18:21 we have:- And one powerful angel picked up a stone like a great millstone and threw [it] into the sea, saying, “In this way Babylon the great city will be thrown down with violence, and will never be found again! Rev 18:21
# 1260 1290 1335 37 185 matrix plus 3d matrix
1 2 6 0
1 2 9 0
1 3 3 5
__________ +
3 7 18 5
……………….
The simple math operation:-
1 + 1 + 1 = 3
2 + 2 + 3 = 7
6 + 9 + 3 = 18
0 + 0 + 5 = 5
So…
3. 7. 18. 5 Or
37,185
So we have 37 and 185 here.
185 × 2 = 370
Just let that sink in for a few moments.
This is very reminiscent of:-
2300-1335-noah-and-assyrian-siege/
………
Ok, let’s now do a 3d matrix. So, we take the 2d matrix from above but repeat the 2d matrix so that there are 4 2d (dimensional) matrices thus making a 3d object.
Now we take the 2d matrix result, 37,185 and multiply by 4
37,185 × 4 = 148,740
Now, let’s digit group this result…
148 and 740
148 + 740 = 888
888. That’s… Jesus
Pretty good.
Let’s do some more matrices.
37,185 × 2 = 74,370
74 + 370 = 444
37,185 × 3 = 111,556
111 + 555 = 666
37,185 × 4 = 148,740
148 + 740 = 888
37,185 × 5 = 185,925
185 + 925 = 1,110
37,185 × 6 = 223,110
223 + 110 = 333
37,185 × 7 = 260,295
260 + 295 = 555
37,185 × 8 = 297,480
297 + 480 = 777
And so on…😂
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How to divide 2 fractions. Fraction help.
Updated on March 27, 2013
Dividing Fractions Video
The division of fractions can be carried out using the following three steps:
Step B Turnover the second fraction (swap the numerator and denominator around).
Step C Next multiply the numerators and denominators to get your final answer. Simplify the final fraction if possible.
Let’s put these three steps into practice.
Question 1
Work out 5/9 ÷ 3/7
Step B Turnover the second fraction (swap the numerator and denominator around).
It’s better to carry out these first two steps simultaneously.
5/9 ÷ 3/7 = 5/9 × 7/3
Step C Next multiply the numerators and denominators to get your final answer. Simplify the final fraction if possible.
5 × 7 = 35
9 × 3 = 27
5/9 × 3/7 = 15/63 = 35/27 or 1 8/27 (this fraction cannot be simplified).
Question 2
Work out 4/3 ÷ 2/9
Step B Turnover the second fraction (swap the numerator and denominator around).
It’s better to carry out these first two steps simultaneously.
4/3 ÷ 2/9 = 4/3 × 9/2
Step C Next multiply the numerators and denominators to get your final answer. Simplify the final fraction if possible.
4 × 9 = 36
3 × 2 = 6
4/3 × 9/2 = 36/6 = 6/1 = 6 (divide the numerator and denominator by 6 to simplify).
Question 3
Work out 7/11 ÷ 1/8
Step B Turnover the second fraction (swap the numerator and denominator around).
It’s better to carry out these first two steps simultaneously.
7/11 ÷ 1/8 = 7/11 × 8/1
Step C Next multiply the numerators and denominators to get your final answer. Simplify the final fraction if possible.
7 × 8 = 56
11 × 1 = 11
7/11 × 8/1 = 56/11 or 5 1/11
Question 4
Work out 10 ÷ 2/3
10 can be written as a fraction by putting 1 on the denominator (10 = 10/1)
Step B Turnover the second fraction (swap the numerator and denominator around).
It’s better to carry out these first two steps simultaneously.
10/1 ÷ 2/3 = 10/1 × 3/2
Step C Next multiply the numerators and denominators to get your final answer. Simplify the final fraction if possible.
10 × 3 = 30
1 × 2 = 2
10/1 × 3/2 = 30/2 = 15/1 = 15
0
2
2
59
2
10
2
2
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# In a Survey It Was Found that 21 Persons Liked Product P1, 26 Liked Product P2 and 29 Liked Product P3. If 14 Persons Liked Products P1 and P2; - Mathematics
In a survey it was found that 21 persons liked product P1, 26 liked product P2 and 29 liked product P3. If 14 persons liked products P1 and P2; 12 persons liked product P3 and P1 ; 14 persons liked products P2 and P3 and 8 liked all the three products. Find how many liked product P3 only.
#### Solution
Let $P_1 , P_2 \text{ and } P_3$ denote the sets of persons liking products$P_1 , P_2 \text{ and } P_3$ respectively.
Also, let U be the universal set.
Thus, we have:
n( $P_1$= 21, n($P_2$= 26 and n($P_3$ 29
And,
n($P_1$$\cap$$P_3$= 12, n($P_2 \cap P_3$n($P_1 \cap P_2 \cap P_3$= 8
Now,
Number of people who like only product $P_3$
= n (P_3∩ P_1′∩ P_2′)
= n {P_3∩ (P_1∪ P_2)′}
$= n \left( P_3 \right) - n\left[ P_3 \cap \left( P_1 \cup P_2 \right) \right]$
$= n\left( P_3 \right) - n\left[ \left( P_3 \cap P_1 \right) \cup \left( P_3 \cap P_2 \right) \right]$
$= n\left( P_3 \right) - \left[ n\left( P_3 \cap P_1 \right) + n\left( P_3 \cap P_2 \right) - n\left( P_1 \cap P_2 \cap P_3 \right) \right]$
$= 29 - \left( 12 + 14 - 8 \right)$
$= 11$
Therefore, the number of people who like only product $P_3$is 11
Concept: Universal Set
Is there an error in this question or solution?
#### APPEARS IN
RD Sharma Class 11 Mathematics Textbook
Chapter 1 Sets
Exercise 1.8 | Q 15 | Page 47
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New Zealand
Level 8 - NCEA Level 3
# Addition and Subtraction of Complex Numbers
Lesson
Complex numbers can be added and subtracted very easily, following the normal laws of algebra.
We must only add and subtract like terms.
And in complex numbers, like terms are the real parts and the imaginary parts.
Let's look a an example with two complex numbers $z_1=3+2i$z1=3+2i and $z_2=6-4i$z2=64i
To add these two complex numbers together we must first identify the real and imaginary components of each.
Then we add the real components, and imaginary components respectfully
#### Here are a few more examples
##### Example 1
$\left(3+7i\right)+\left(2+i\right)$(3+7i)+(2+i) $=$= $\left(2+3\right)+\left(7+1\right)i$(2+3)+(7+1)i $=$= $5+8i$5+8i
See how in this example I grouped the real and imaginary parts and then added. Sometimes this helps keep track of all the components. Sometimes you can jump straight to the answer.
##### Example 2
$\left(9-2i\right)-\left(-2+6i\right)$(9−2i)−(−2+6i) $=$= $9-2i+2-6i$9−2i+2−6i $=$= $11-8i$11−8i
In this example, I expanded the brackets observing the change of sign and then collected like terms.
#### More Worked Examples
##### QUESTION 1
Evaluate $\left(3+6i\right)+\left(7+3i\right)$(3+6i)+(7+3i).
##### QUESTION 2
Evaluate $\left(6+9i\right)-\left(-3-4i\right)$(6+9i)(34i).
##### QUESTION 3
Evaluate $\left(-6\sqrt{7}-2i\right)+\left(3\sqrt{7}+7i\right)$(672i)+(37+7i).
### Outcomes
#### M8-9
Manipulate complex numbers and present them graphically
#### 91577
Apply the algebra of complex numbers in solving problems
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# Focus on Math
## Helping children become mathematicians!
### Addition and Subtraction: A Really Big IdeaOctober 25, 2011
Much of what students do in mathematics across the years involves four basic operations: addition, subtraction, multiplication, and division. Within these four operations we really can consider two pairs of operations since
• addition and subtraction are inverse operations (each “undoing” the other)
• multiplication and division are inverse operations.
As students’ mathematics skills increase, so does what we ask them to do in the operations. (NOTE: As I write, I shall concentrate on the process of addition, but please keep in mind that what I am saying applies equally to subtraction since it is the inverse operation of addition.)
Young children begin by adding whole numbers such as 2 + 5. Put into a context, it might be a question such as this: There were two birds in a tree. Five more birds came and perched in the same tree. Then how many birds were in the tree?
Later children begin adding multi-digit whole numbers. They might be given a problem such as this: 257 + 48. If children write this problem in the typical “stacked” format, we teach them to line up the numbers, beginning on the right.
Decimal addition follows soon after. This time the problem may look like this: 2.57 + 4.8. Again, such a problem is typically “stacked”, and we teach students a rule that has them lining up the decimal points.
Adding fractions brings about a new situation, and typically a new rule. If students are given a question such as 2/5 + 1/2, we teach them about finding a common denominator, changing one or both fractions (depending on the particular question) into equivalent fractions with the same denominators, and then proceeding to add the numerators. In our example we find the common denominator of 10, change the problem to be 4/10 + 5/10, then add to get 9/10. The common denominator rule must be followed if students are to arrive at a correct solution.
Students soon encounter algebra and its accompanying variables. Now they see problems such as 3x + 4y + 2x and we teach another rule: you always add “like terms”. Thus the example can be simplified to 5x + 4y, but these two terms cannot be combined because they are not alike.
I propose that the idea of adding “like terms” or “like things” is not new to algebra. In fact, it is exactly the same rule we have been using in every situation. We just have not been calling it that, but maybe we should!! How different would students’ understanding be of all of the addition and subtraction situations if we continually pointed out that we always, always, always add and subtract like things.
For young children, we must add things like apples to apples, birds to birds, pencils to pencils, etc.. I once worked with a young girl and asked her to make up a word problem for 2 + 5. She looked out the window from where we were sitting and she saw some birds in an apple tree. She began her story, “There were 2 birds in a tree and 5 apples…”.Then she paused, looked at me, and stated: “That doesn’t make sense!” I asked her what would make sense, and she responded, ” I think I should have apples and apples.” She was right. It only makes sense to add things that are the same. Even when we add 12 boys in a class to 14 girls, we end up saying there are 26 children or students. We actually had to find a “same name” or a “like term” for them if we were going to add them.
When it comes to adding multi-digit whole numbers, why is it important to “line up the right side”? Because in writing the problem in that manner we are setting it up so that we can add ones to ones, tens to tens, hundreds to hundreds, etc. This principle is the same for why we “line up the decimal points” when we add decimal numbers. We are ensuring we add ones to ones, tenths to tenths, hundredths to hundredths, etc. In our place-value-based number system, we must add like values, or like terms.
Students are often confused when doing an operation on fractions. They have trouble remembering when they have to change the denominators and when they can go “straight across”. I feel there would be less confusion regarding fractions if they knew the big idea about addition and subtraction: namely, that you always add and subtract like things.
By the time students begin doing algebra, it should be easy understanding about adding “like terms” since that is what they have been doing the whole time in addition, if only we would help them see that!
Mathematically yours,
Carollee
### Combined (“Split”) Grade Math Learning Outcomes: BC/WNCPOctober 20, 2011
Filed under: General Math — Focus on Math @ 11:56 am
Tomorrow morning I am doing a professional development day session about teaching math in “split” or combined grades. It is helpful for teachers of such “split” grades to have side-by-side learning outcomes for their particular grades.
Teachers from the Vancouver IslandNet worked to prepare tables with the learning outcomes for different grades “matched up” in that manner. In other words, for the two (or three) specific grades, all of the outcomes are listed, each beside the outcome(s) that correlate between those grades. If an outcome in one grade does not specifically correlate to the other, then it is suggested that students in the other grade may either explore or review the concept. Such a chart is a very handy tool to have when planning!
This link will take you the the Vancouver IslandNet At-A-Glance Project. The learning outcomes for British Columbia (and any other province or territory using the Western and Northern Canadian Protocol [WNCP] math outcomes) are available here for single grades, combined grades (two), or mulitgrades (three).
Our thanks to the IslandNet group for providing this valuable resource!
Mathematically yours,
Carollee
### The Importance of Speaking Clearly
Filed under: General Math,Ideas from Carollee's Workshops — Focus on Math @ 10:10 am
The toolkit workshop last week was a success overall. We looked (and “played with”) a variety of items that students can use as tools for solving problems.
Along the way, however, I learned a valuable lesson about speaking clearly. One of the tools we used at the workshop was the 100-dot array. Now, as it turned out, the participants had a list of possible toolkit items, and we were exploring a number of items on the list, but I was not introducing the items in any particular order. When I began to speak of the 100-dot array and ask the participants to find the samples provided on the tables, quite a few folks looked at me rather strangely. I asked them again to each get a sample of the 100-dot array to use, and finally several folks stopped me to ask me what, exactly, did I want them to use!
I tend to talk quickly in a workshop (there is always SOOOOO much math to talk about, and I am trying to squeeze in as much as I can to any given time period!) and in speaking quickly I was not speaking clearly. In fact, one participant handed me a note at the end of the session with these words written on it:
• 100 dotter, eh?
• 100 daughter a?
• 100 daughter ray?
• 100 dotteray?
Clearly, in knowing for myself what I was referring to when I said “100-dot array” I assumed that everyone else was hearing what I was saying. They were not!
In the end we sorted it out, I pronounced the tool’s name slowly and clearly, and we went on with the session. But it made me wonder about other workshops, other sessions, other classes of students, other teachers with other classes of students. Do others hear clearly what we are trying to say? And what about the students who assume that if they aren’t “getting it” that it is something wrong with them, and not the speaker? There are so many math terms. It is important that we speak them clearly.
I hope I have learned my lesson from the “!00-dot array”. Maybe the next workshop will provide the evidence.
Mathematically yours,
Carollee
### A Patterning Problem: Finding Sums of Consecutive NumbersOctober 6, 2011
This problem is one of my favourites! I first came across this problem about the sums of consecutive numbers in Marilyn Burns‘ book About Teaching Mathematics: A K-8 Resource (Sausalito, California: Math Solutions) and I thought it was an intriguing question. Burns presents the problem to be solved this way:
“Ask the students, in their groups, to find all the ways to write the numbers from one to twenty-five as the sum of consecutive numbers. (For younger children, finding the sums for the numbers from one to fifteen may be sufficient.) Tell them that some of the numbers are impossible; challenge them to see if they can find the pattern of those numbers. Direct them to search for other patterns as well.” p. 58
I have done this problem over the last few years with quite a few classes in a range of grades (usually somewhere in the grade 3-8 span). For the older grades I have extended the problem, asking students to write sums for numbers up to 35.) I feel that it is not a problem to hurry the students through, and I often take more than one day with the task. What I like most about the problem is that it is full of patterns, and that, in finding the patterns, one “unlocks” the problem. It becomes so much easier to find and predict the various sums when one notices the patterns that are produced.
I also like this problem because we tend to teach patterning to students isolated from problems, and I think that there is something quite powerful about a problem that uses patterns to solve it!
The task of finding sums of consecutive numbers provides a good opportunity for students to develop some problem solving strategies. Burns suggests a list of useful problem-solving strategies, similar to lists proposed by other math authors, naming the major strategies useful to untangle problems:
• look for a pattern
• construct a table (chart)
• make an organized list
• draw a picture
• use objects
• guess and check
• work backward
• write an equation
• solve a simpler (or similar) problem
• make a model
I have found that students doing this problem tend to make use of a number of the above strategies including these: look for a pattern, make an organized list, use objets, guess and check, and work backward. Any time that students are engaged in problem solving It is important to discuss with them both the specific strategies they use to solve the problem and why those strategies were (or were not) effective choices. Additionally, this problem is simple enough on its most basic level that everyone has the chance to delve in and come up with some of the consecutive number sums. At the same time it is quite sophisticated and offers a challenge to the bright students in the class.
Burns offers ideas for extensions for this activity, too. For instance, she suggests students try to predict how many ways thirty-six can be written as the sum of consecutive numbers. Going further, she asks if a prediction can be made for any particular number.
So, I encourage you to give this one a try with your students — and it is OK if YOU do not have it all figured out ahead of time. Let your students know that you are solving the problem along with them.
Mathematically yours,
Carollee
### Upcoming Math ConferenceOctober 3, 2011
The British Columbia Association of Mathematics Teaacher’s annual conference will be taking place October 21 in Burnaby, BC. Here is the schedule for the day (at least as posted presently — some changes/additions will likely yet be made) and the link for registering for the conference. There are a wide range of speakers, both for grade levels and for topics. I know you will find sessions that interest you and I encourage you to go to the conference if you can!
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# How do you find the axis of symmetry, and the maximum or minimum value of the function y = x^2 + 4x + 4?
Feb 10, 2016
The axis of symmetry is x = -2
The minimum of the function is 0
#### Explanation:
The formula for the axis of symmetry is given by $x = - \frac{b}{2 a}$
where the quadratic is $y = a {x}^{2} + b x + c$.
Therefore just plug the values in:
$- \frac{4}{2}$ => $x = - 2$ which is the axis of symmetry
Now that we have the x value, we can plug this in to find the y value, which will always be either a max or a min for the axis of symmetry.
We find that y = 0.
If a is positive, there will be a min
If a is negative, there will be a max
(just think back to the parent function $y = {x}^{2}$ ... it opens upwards with a minimum value)
Therefore 0 is the minimum value.
The axis of symmetry is x = -2
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#### Please Solve R.D.Sharma class 12 Chapter 9 Differentiability Exercise 9.2 Question 2 Maths textbook Solution.
Answer: $f'\left ( 5 \right )=2\: f'\left ( \frac{7}{2} \right )\Rightarrow L.H.S=R.H.S$
Hint: First we will f (x) then put x = 5 and x = $\frac{7}{2}$ in f (x). Then, we will check L.H.S and R.H.S are equal or not.
Given: $f'\left ( x \right )=x^{2}-4x+7$
Solution:
Differentiating f(x) w.r.t x then,
\begin{aligned} &\Rightarrow \frac{d}{d x}\{\mathrm{f}(\mathrm{x})\}=\frac{d}{d x}\left(\mathrm{x}^{2}-4 \mathrm{x}+7\right) \\ &\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=\frac{d}{d x}\left(\mathrm{x}^{2}\right)+\frac{d}{d x}(-4 \mathrm{x})+\frac{d}{d x}(7) \\ &{\left[\because \frac{d}{d x}\left(a x^{2}+b x+c\right)=\frac{d}{d x}\left(a x^{2}\right)+\frac{d}{d x}(b x)+\frac{d}{d x}(c)\right]} \end{aligned}
$\Rightarrow f'\left ( x \right )=2x^{2-1}-4\left ( 1.x^{1-1} \right )+0$ $\left [ \because \frac{d}{dx}\left ( x^{n} \right ) nx^{n-1}\right ]$
$\Rightarrow f'\left ( x \right )=2x-4$ $\left [ \because \frac{d}{dx} \left ( constant \right )=0\right ]$
Now, put x = 5 and x = $\frac{7}{2}$ in f `(x).
$f'\left ( 5 \right )=2\left ( 5 \right ) and x=\frac{7}{2}in f'\left ( x \right )$
$f'\left ( 5 \right )=2\left ( 5 \right )-4$
$=10-4$
$=6$
$f'\left ( \frac{7}{2} \right )=2\frac{7}{2}-4$
$=7-4$
$=3$
L.H.S
$f'(5)=6$
R.H.S
$2\: f'\left ( \frac{7}{2} \right )=2\times 3=6$
$\therefore$ L.H.S=R.H.S
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Ex.1.5 Q2 Number System Solution - NCERT Maths Class 9
Go back to 'Ex.1.5'
Question
Simplify each of the following expressions
(i) $$(3 + \sqrt 3 )\,(2 + \sqrt 2 )$$
(ii) $$(3 + \sqrt 3 )\,(3 - \sqrt 3 )$$
(iii)$${(\sqrt 5 + \sqrt 2 )^2}$$
(iv) $$(5 - \sqrt 2 )\,(5 + \sqrt 2 )$$
Video Solution
Number Systems
Ex 1.5 | Question 2
Text Solution
Steps:
(i) \begin{align}(3 + \sqrt 3 )\,(2 + \sqrt 2 )\end{align}
By Distributive property
\begin{align}=6+3 \sqrt{2}+2 \sqrt{3}+\sqrt{6} \end{align}
(ii) \begin{align}(3 + \sqrt 3 )\,(3 - \sqrt 3 )\end{align}
Using the identity:
\begin{align}(a+b)(a-b) &=a^{2}-b^{2} \3+\sqrt{3}) 3-\sqrt{3} &=9-3 \\ &=6 \end{align} (iii) \(\begin{align}{(\sqrt 5 + \sqrt 2 )^2}\,\end{align}
Steps:
Using the identity: $$(a+b)^{2} =a^{2}+2 a b+b^{2}$$
\begin{align}&=(\sqrt{5})^{2}+2 \times \sqrt{5} \times \sqrt{2}+(\sqrt{2})^{2} \\ &=(5+2 \sqrt{10}+2) \\ &=7+2 \sqrt{10} \end{align}
(iv) \begin{align}(5 - \sqrt 2 )\,(5 + \sqrt 2 )\end{align}
Steps:
Using the identity: $$(a+b)(a-b) =a^{2}-b^{2}$$
\begin{align}&(\sqrt{5})^{2}-(\sqrt{2})^{2} \\ &=5-2 \\ &=3 \end{align}
Video Solution
Number Systems
Ex 1.5 | Question 2
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# Can You Get a Negative out of a Square Root?
Unrivaled math delimiters. Including last one for you.
The basic answer is: yes you can get negative numbers out of square roots. Truth be told, should you wish to locate the square base of any positive genuine numbers, you will get two outcomes: the positive and negative variants of a similar number.
Writing a Square Root Equation for Positive and Negative Results
Think about the accompanying:
16 = 4 * 4 = (- 4) * (- 4)
In the condition above, you can either increase 4 without anyone else or duplicate (- 4) without anyone else to get the aftereffect of 16. Along these lines, the square base of 16 would be:
The √ image is called the radical image, while the number or articulation inside the image—for this situation, 16—is called a radicand.
For what reason would we need a ± image before 4? Indeed, as we have talked about previously, the square underlying foundations of 16 can be either 4 or (- 4).
A great many people would simply write this equation basically as . While it is in fact obvious and there’s nothing amiss with it, it doesn’t recount to the entire story.
Rather, you can likewise compose the condition so that it expressly demonstrates that you need both the positive and negative square root’s outcomes:
Along these lines, others can without much of a stretch tell that the person who composes the condition wishes to have constructive and contrary numbers as the outcome.
Great and Imperfect squares
As you may know, 16 is a perfect square. Immaculate squares are radicands as an integer, or a entire number, that has a square foundation of another whole number. In the model over, 16 is an ideal square since it has the number 4 as its square root.
Positive genuine numbers are not generally flawless squares. There are likewise different numbers, for example, 3, 5, or 13 that are alluded to as imperfect squares. If a radicand is anything but an ideal square, at that point the square foundation of the radicand won’t result is a whole number. Investigate the condition underneath.
5 is certainly not an ideal square. Consequently, its square root won’t be a whole number. Also, the square root isn’t so much as a levelheaded number. Rational numbers are numbers that can be communicated as divisions made out of two whole numbers, e.g. 7/2, 50/4, and 100/3.
The square foundation of 5 is an irrational number since it can’t be communicated as portions. The numbers right of the decimal 2.236067 … would proceed on interminably without any rehashing design. All things considered, both nonsensical and sane numbers are a piece of genuine numbers, which means they have unmistakable qualities and exist on the number line.
Square Roots of Zero and Negative Numbers
We referenced before that any positive genuine numbers have two square roots, the positive one and the negative one. Shouldn’t something be said about negative numbers and zero?
For zero, it just has one square root, which is itself, 0.
Then again, negative numbers don’t have any genuine square roots. Any genuine number—regardless of whether it’s sure or negative—that is increased without anyone else is consistently equivalent to a positive number, aside from 0. Rather, the square base of every single negative number is an imaginary number.
or then again
<img src=”https://lh3.googleusercontent.com/pw/ACtC-3eZFkyDcARtKvAZuSWAZK9MeFq5u0WEOfeRiPkY08R2DKOLaTeX: language structure blunder (Unable to order LaTeX recipe,
{k}_{U}{y}{n}{95}{I}{q}{y}{I}{Q}{t}{U}{8}{X}{H}{2}{I}{c}{J}-{B}{E}{j}times{8}{y}{K}{2}{l}{k}{w}{j}{O}{p}{j}{p}{t}{D}{F}{E}{05}{Z}{S}{p}{j}{50}{y}{C}{i}{3}{n}{w}{C}{a}{B}{X}{W}{G}{B}{Y}{D}{r}{U}{K}{8}{Q}{I}{u}{Z}{55}{g{K}}{K}{q}{w}{J}{E}{T}{s}{t}{6}{n}{p}{h}{A}{Y}{K}{K}{3}{h}{t}{1}{J}{Z}{L}{k}{9}={w}{50}-{h}{19}-{n}{o}?{a}{u}{t}{h}{u}{s}{e}{r}={0} text{ alt=} text{ width=}{50} text{ height=}{19} text{ information widget=}{i}{m}{a}ge text{/>
By definition, the square foundation of (- 1) is i, which is a fanciful unit. As a side note, nonexistent numbers don’t have an unmistakable worth. They are not part of real numbers in the feeling that they can’t be measured on the number line. In any case, they are as yet utilized in math and the investigation of sciences including quantum mechanics, power, and that’s just the beginning.
To show signs of improvement understanding, how about we investigate a model. For example, suppose we need to recognize the square base of (- 9), what might it be?
In this way, is equivalent to 3i. This outcome can likewise be composed as
What about the square foundation of (- 3)?
Thus, is equivalent to }.{O}{r},{i}{t}{c}{a}{n}{a}{l}{s}{o}{b}{e}{w}{r}{i}{mathtt{e}}{n}{a}{s}-{3}={3}{i}{2}frac{.frac{{C}{o}{n}{c}{l}{u}{s}{i}{o}{n}frac{{I}{f{y}}{o}{u}{w}{a}{n}{t}toge{t}{a}neg{a}{t}{i}{v}{e}{o}{u}{t}{o}{f{a}}boxempty{sqrt[,]{j}}{u}{s}{t}{r}{e}{m}{e}{m}{b}{e}{r}to{w}{r}{i}{t}{e}{y}{o}{u}{r}boxempty{sqrt[{e}]{q}}{u}{a}{t}{i}{o}{n}{u}{sin{g{t}}}{h}{e} ±{s}{y}{m}{b}{o}{l}.{T}{h}{i}{s}{w}{i}{l}{l}le{t}{y}{o}{u}{quadtext{and}quad}{o}{t}{h}{e}{r}{s}{k}{n}{o}{w}{t}{hat{y}}{o}{u}{a}{r}{e}{l}infty{k}in{g{{f{quadtext{or}quad}}}}{i}{t}- – {quadtext{and}quad}{i}{t}{i}{s}in{t}{h}{e}{r}{e}!frac{
Check sentence structure.)
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# Making New Functions
Algebra students usually learn at some point in their studies that you can dilate a function like $$\mathtt{f(x)=x^2}$$ by multiplying it by a constant value. Usually the multiplier is written as $$\mathtt{A}$$, so you get $$\mathtt{f(x)=Ax}$$, which can be $$\mathtt{f(x)=2x^2}$$ or $$\mathtt{f(x)=3x^2}$$ or $$\mathtt{f(x)=\frac{1}{2}x^2}$$, and so on.
What we focus on at the beginning is how this change affects the graph of the function—and, importantly, how we can consistently describe that change as it applies to any function.
So, for example, the brightest (orange-brown) curve in the image at the right represents $$\mathtt{f(x)=x^2}$$, or $$\mathtt{f(x)=1x^2}$$. And when we increase the A-value, from 2 to 3, the curve gets narrower. Decreasing the A-value, to $$\mathtt{\frac{1}{2}}$$ for example, causes the curve to get wider. The same kind of “squishing” happens with every function type. (Check out this article by Better Explained for a really nice explanation.)
Making New Functions
We can also make higher-degree functions from lower-degree functions using dilations. The only change we make to the process of dilation shown above is that now we multiply each point of a function by a non-constant value. More specifically, we can multiply the y-coordinate of each point by its x-coordinate to get its new y-coordinate.
At the left, we transform the constant function $$\mathtt{f(x)=5}$$ into a higher-order function in this way. By multiplying the y-coordinate of each original point by its x-coordinate, we change the function from $$\mathtt{f(x)=5}$$ to $$\mathtt{f(x)=(x)(5)}$$, or $$\mathtt{f(x)=5x}$$. Another multiplication of all the points by x would get us $$\mathtt{f(x)=5x^2}$$. In that case, you can see that all the points to the left of the y-axis have to reflect across the x-axis, since each y-coordinate would be a negative times a negative.
Another idea that becomes clearer when working with non-constant dilations in this way is that zeros start to make a little more sense.
Try it with other dilations (say, $$\mathtt{x \cdot (x+3)}$$ or even $$\mathtt{x \cdot (x-1)^2)}$$) and pay attention to what happens to those points that wind up getting multiplied by 0.
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