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# Distance Formula and the Pythagorean Theorem
## Determine distance between ordered pairs
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Identify and Use the Distance Formula
License: CC BY-NC 3.0
As Ross was cruising on his boat toward the cottage, he was concentrating on finding a new mooring for his boat. He had previously sunk two buoy markers in the water near the cottage and both locations had ample water below for the boat. Ross looked at his chart plot which showed the coordinates for the cottage and the two buoys. He decided to choose the buoy closest to the cottage.
How can Ross use the coordinates from the chart plot to figure out which buoy is closest to the cottage?
License: CC BY-NC 3.0
In this concept, you will learn to identify and use the distance formula.
### Distance Formula
When plotting points or graphing lines on a Cartesian grid, the coordinates of the points, or of the endpoints, of the graphed line are very useful when solving real-world problems.
The following graph shows an orange line that represents the distance between the pet store (1, 6) and the drug store (8, 2). The green dot on the grid represents Timothy who is sitting on a bench in the Mall. To his right he can see the drug store and straight ahead he can see the pet store. He decides the pet store is closer so he goes there first. He exits the pet store and tries to decide whether to go back to the bench and then to the drug store or to walk diagonally to the drug store. Which way would be the shorter of the two distances if each side of the squares represents one yard?
On the grid, draw a horizontal line from Timothy on the bench to the drug store. The length of this line can be found by counting the tops of the squares that the line passes through. The length of the line is 7 yards. Now, draw a vertical line from Timothy on the bench to the pet store. The length of this line can be found by counting the sides of the squares that the line passes through. The length of this line is 4 yards. The three lines form a right triangle and the Pythagorean Theorem can be used to find the diagonal distance between the pet store and the drug store. This diagonal distance is in fact the hypotenuse of the right triangle.
License: CC BY-NC 3.0
First, write the Pythagorean Theorem.
\begin{align*}c^2 = a^2 + b^2\end{align*}
Next, determine the values of \begin{align*}(a, b, c)\end{align*} for the Pythagorean Theorem.
\begin{align*}\begin{array} {rcl} a &=& 4 \\ b &=& 7 \\ c &=& ? \end{array}\end{align*}
Next, fill these values into the Pythagorean Theorem.
\begin{align*}\begin{array}{rcl} c^2 &=& a^2 + b^2 \\ c^2 &=& (4)^2 + (7)^2 \end{array}\end{align*}
Next, perform the indicated squaring on the right side of the equation and simplify.
\begin{align*}\begin{array}{rcl} c^2 &=& (4)^2 + (7)^2 \\ c^2 &=& (4 \times 4) + (7 \times 7) \\ c^2 &=& 16 + 49 \\ c^2 &=& 65 \end{array}\end{align*}
Then, solve for the variable ‘\begin{align*}c\end{align*}’ by taking the square root of both sides of the equation.
\begin{align*}\begin{array}{rcl} c^2 &=& 65 \\ \sqrt{c^2} &=& \sqrt{65} \\ c &=& 8.06 \end{array}\end{align*}
The answer is 8.06.
The diagonal distance between the pet store and the drug store is 8.06 yards.
It would be a shorter distance for Timothy to walk than to return the 4 yards to the bench and then walk another 7 yards to the drug store.
This method of finding the distance between two points requires that the diagram be drawn on grid paper which may be neither available nor convenient. Another way of finding the distance between two points with known coordinates is to use the distance formula. The distance formula is the square root of the square of the difference between the \begin{align*}x\end{align*}-coordinates of the two points plus the square of the difference between the \begin{align*}y\end{align*}-coordinates of the two points. The distance formula is written as:
\begin{align*}d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\end{align*}
such that \begin{align*}(x_1, y_1)\end{align*} are \begin{align*}(x_2, y_2)\end{align*} the coordinates of the point chosen as the first point and the point chosen as the second point respectively. Once the points have been designated, the corresponding values must remain consistent and be substituted into the formula as named.
Let’s apply this formula to find the distance between the pet store and the drug store. Remember the coordinates of the pet store were (1, 6) and those of the drug store were (8, 2).
First, choose the first and second points. The answer will be the same regardless of how the points are chosen. Naming the points as shown below will help match the correct value with its corresponding variable.
\begin{align*}\begin{pmatrix} x_1, & y_1 \\ 1, & 6 \end{pmatrix} \ \text{and} \ \begin{pmatrix} x_2, & y_2 \\ 8, & 2 \end{pmatrix}\end{align*}
Next, write the distance formula and fill in the values for the variables.
\begin{align*}\begin{array}{rcl} d &=& \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\ d &=& \sqrt{(8-1)^2 + (2-6)^2} \end{array}\end{align*}
Next, perform the operations shown under the square root sign.
\begin{align*}\begin{array}{rcl} d &=& \sqrt{(8-1)^2 + (2-6)^2} \\ d &=& \sqrt{(7)^2+ (-4)^2} \\ d &=& \sqrt{(7 \times 7) + (-4 \times -4)} \\ d &=& \sqrt{49 + 16} \\ d &=& \sqrt{65} \end{array}\end{align*}
Then, take the square root of 65 and round the answer to the nearest hundredth.
\begin{align*}\begin{array}{rcl} d &=& \sqrt{65} \\ d &=& 8.06 \end{array}\end{align*}
The answer is 8.06
The distance is 8.06 yards.
### Examples
#### Example 1
Earlier, you were given a problem about Ross and the two buoys. He needs to figure out which buoy is closer to the cottage, so he can moor his boat. Ross needs to use the distance formula to figure out which buoy is closer to the cottage.
First, write the coordinates for each distance to be calculated and use the distance formula to calculate each distance to the nearest hundredth.
\begin{align*}\begin{array}{rcl} \text{Cottage to Buoy One} & \quad & \text{Cottage to Buoy Two} \\ (160, 120) \ \text{and} \ (40, 100) & \quad & (160, 120) \ \text{and} \ (220, 40) \end{array}\end{align*}
Distance from cottage to Buoy One:
First, name the first and second points.
\begin{align*}\begin{pmatrix} x_1, & y_1 \\ 160, & 120 \end{pmatrix} \ \text{and} \ \begin{pmatrix} x_2, & y_2 \\ 40, & 100 \end{pmatrix}\end{align*}
Next, write the distance formula and fill in the values for the variables.
\begin{align*}\begin{array}{rcl} d &=& \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\ d &=& \sqrt{(40 - 160)^2 + (100 - 120)^2} \end{array}\end{align*}
Next, perform the operations shown under the square root sign.
\begin{align*}\begin{array}{rcl} d &=& \sqrt{(40 - 160)^2 + (100 - 120)^2} \\ d &=& \sqrt{(-120)^2 + (-20)^2} \\ d &=& \sqrt{(-120 \times -120) + (-20 \times -20)} \\ d &=& \sqrt{14400 + 400} \\ d &=& \sqrt{14800} \end{array}\end{align*}
Then, take the square root of 14800 and round the answer to the nearest hundredth.
\begin{align*}\begin{array}{rcl} d &=& \sqrt{14800} \\ d &=& 121.66 \end{array}\end{align*}
The answer is 121.66.
The distance from the cottage to Buoy One is 121.66 meters.
Distance from cottage to Buoy Two:
First, name the first and second points.
\begin{align*}\begin{pmatrix} x_1, & y_1 \\ 160, & 120 \end{pmatrix} \ \text{and} \ \begin{pmatrix} x_2, & y_2 \\ 220, & 40 \end{pmatrix}\end{align*}
Next, write the distance formula and fill in the values for the variables.
\begin{align*}\begin{array}{rcl} d &=& \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\ d &=& \sqrt{(220 - 160)^2 + (40 - 120)^2} \end{array}\end{align*}
Next, perform the operations shown under the square root sign.
\begin{align*}\begin{array}{rcl} d &=& \sqrt{(220 - 160)^2 + (40 - 120)^2} \\ d &=& \sqrt{(60)^2 + (-80)^2} \\ d &=& \sqrt{(60 \times 60) + (-80 \times -80)} \\ d &=& \sqrt{3600 + 6400} \\ d &=& \sqrt{10000} \end{array}\end{align*}
Then, take the square root of 10000 and round the answer to the nearest hundredth.
\begin{align*}\begin{array}{rcl} d &=& \sqrt{10000} \\ d &=& 100 \end{array}\end{align*}
The answer is 100.
The distance from the cottage to Buoy Two is 100 meters. Ross will moor the boat at Buoy Two.
#### Example 2
Use the distance formula to find the distance between these points to the nearest tenth
\begin{align*}A (54, 120)\end{align*} and \begin{align*}B (113, 215)\end{align*}
First, name the first and second points.
\begin{align*}\begin{pmatrix} x_1, & y_1 \\ 54, & 120 \end{pmatrix} \ \text{and} \ \begin{pmatrix} x_2, & y_2 \\ 113, & 215 \end{pmatrix}\end{align*}
Next, write the distance formula and fill in the values for the variables.
\begin{align*}\begin{array}{rcl} d &=& \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\ d &=& \sqrt{(113 - 54)^2 + (215 - 120)^2} \end{array}\end{align*}
Next, perform the operations shown under the square root sign.
\begin{align*}\begin{array}{rcl} d &=& \sqrt{(113 - 54)^2 + (215 - 120)^2} \\ d &=& \sqrt{(59)^2 + (95)^2} \\ d &=& \sqrt{(59 \times 59) + (95 \times 95)} \\ d &=& \sqrt{3481 + 9025} \\ d &=& \sqrt{12506} \end{array}\end{align*}
Then, take the square root of 12506 and round the answer to the nearest tenth.
\begin{align*}\begin{array}{rcl} d &=& \sqrt{12506} \\ d &=& 111.8 \end{array}\end{align*}
The answer is 111.8 units.
#### Example 3
Use the distance formula to find the distance between the points \begin{align*}D (3, -4)\end{align*} and \begin{align*}E (-2, -10)\end{align*}.
First, name the first and second points.
\begin{align*}\begin{pmatrix} x_1, & y_1 \\ 3, & -4 \end{pmatrix} \ \text{and} \ \begin{pmatrix} x_2 ,& y_2 \\ -2, & -10 \end{pmatrix}\end{align*}
Next, write the distance formula and fill in the values for the variables.
\begin{align*}\begin{array}{rcl} d &=& \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\ d &=& \sqrt{(-2 -3)^2 + (-10 - (-4))^2} \end{array}\end{align*}
Next, perform the operations shown under the square root sign.
\begin{align*}\begin{array}{rcl} d &=& \sqrt{(-2 -3)^2 + (-10 - (-4))^2} \\ d &=& \sqrt{(-5)^2 + (-6)^2} \\ d &=& \sqrt{(-5 \times -5) + (-6 \times -6)} \\ d &=& \sqrt{25+36} \\ d &=& \sqrt{61} \\ \end{array}\end{align*}
Then, take the square root of 61 and round the answer to the nearest tenth.
\begin{align*}\begin{array}{rcl} d &=& \sqrt{61} \\ d &=& 7.8 \end{array}\end{align*}
The answer is 7.8.
The distance from \begin{align*}D\end{align*} to \begin{align*}E\end{align*} is 7.8 units.
#### Example 4
The following circle has its center at the point (5, 3) and a radius of 3 inches. Use the distance formula to determine if the point (8, 5) is inside the circle, on the circle or outside the circle.
License: CC BY-NC 3.0
You must use the distance formula to find the distance between the center of the circle (5, 3) and the point (8, 5).
First, name the first and second points.
\begin{align*}\begin{pmatrix} x_1, & y_1 \\ 5, & 3 \end{pmatrix} \ \text{and} \ \begin{pmatrix} x_2, & y_2 \\ 8, & 5 \end{pmatrix}\end{align*}
Next, write the distance formula and fill in the values for the variables.
\begin{align*}\begin{array}{rcl} d &=& \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\ d &=& \sqrt{(8-5)^2 + (5-3)^2} \end{array}\end{align*}
Next, perform the operations shown under the square root sign.
\begin{align*}\begin{array}{rcl} d &=& \sqrt{(8-5)^2 + (5-3)^2} \\ d &=& \sqrt{(3)^2 + (2)^2} \\ d &=& \sqrt{(3 \times 3) + (2 \times 2)} \\ d &=& \sqrt{9+4} \\ d &=& \sqrt{13} \end{array}\end{align*}
Then, take the square root of 13 and round the answer to the nearest tenth.
\begin{align*}\begin{array}{rcl} d &=& \sqrt{13} \\ d &=& 3.6 \end{array}\end{align*}
The answer is 3.6.
The distance from the center of the circle to the point (8, 5) is 3.6 inches. The point will be outside the circle since this distance is greater than the length of the radius.Follow Up
### Review
Use the distance formula to find the distance between the following pairs of points. You may round to the nearest tenth when necessary.
1. What is the distance between (3, 6) and (-1, 3)?
2. What is the distance between (-2,-2) and (10, 3)?
3. What is the distance between (1,9) and (9,1)?
4. What is the distance between (-5,-5) and (-2,-1)?
5. What is the distance between (2, 12) and (3,7)?
6. What is the distance between (2, 2) and (8, 2)?
7. What is the distance between (-3, 4) and (2, 0)?
8. What is the distance between (3, 4) and (3, -4)?
9. What is the distance between (-4, -3) and (1, -1)?
10. What is the distance between (-6, 2) and (-3, 1)?
Answer each of the following questions using the map below of Bryan’s town.
11. The map below shows Bryan’s town. What is the distance between the pet store and town hall?
12. The map above shows Bryan’s town. What is the distance between the pet store and the courthouse?
13. The map above shows Bryan’s town. What is the distance between the courthouse and the library?
14. The map above shows Bryan’s town. What is the distance between the library and the town hall?
15. The map above shows Bryan’s town. What is the distance between the pet store and the library?
To see the Review answers, open this PDF file and look for section 7.10.
### Notes/Highlights Having trouble? Report an issue.
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### Vocabulary Language: English
TermDefinition
Distance Formula The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ can be defined as $d= \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$.
Pythagorean Theorem The Pythagorean Theorem is a mathematical relationship between the sides of a right triangle, given by $a^2 + b^2 = c^2$, where $a$ and $b$ are legs of the triangle and $c$ is the hypotenuse of the triangle.
1. [1]^ License: CC BY-NC 3.0
2. [2]^ License: CC BY-NC 3.0
3. [3]^ License: CC BY-NC 3.0
4. [4]^ License: CC BY-NC 3.0
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# Illustrative Mathematics Grade 6, Unit 8, Lesson 16: Box Plots
Learning Targets:
• I can use the five-number summary to draw a box plot.
• I know what information a box plot shows and how it is constructed.
Related Pages
Illustrative Math
#### Lesson 16: Box Plots
Let’s explore how box plots can help us summarize distributions.
Illustrative Math Unit 6.8, Lesson 16 (printable worksheets)
#### Lesson 16 Summary
The following diagram shows how to use the five-number summary to draw a box plot and know what information a box plot shows.
#### Lesson 16.1 Notice and Wonder: Puppy Weights
Here are the birth weights, in ounces, of all the puppies born at a kennel in the past month.
What do you notice and wonder about the distribution of the puppy weights?
#### Lesson 16.2 Human Box Plot
Your teacher will give you the data on the lengths of names of students in your class. Write the five-number summary by finding the data set’s minimum, Q1, Q2, Q3, and the maximum.
Twenty people participated in a study about blinking. The number of times each person blinked while watching a video for one minute was recorded. The data values are shown here, in order from smallest to largest.
1. a. Use the grid and axis to make a dot plot of this data set.
b. Find the median (Q2) and mark its location on the dot plot.
c. Find the first quartile (Q1) and the third quartile (Q3). Mark their locations on the dot plot.
d. What are the minimum and maximum values?
2. A box plot can be used to represent the five-number summary graphically. Let’s draw a box plot for the number-of-blinks data. On the grid, above the dot plot:
a. Draw a box that extends from the first quartile (Q1) to the third quartile (Q3). Label the quartiles.
b. At the median (Q2), draw a vertical line from the top of the box to the bottom of the box. Label the median. c. From the left side of the box (Q1), draw a horizontal line (a whisker) that extends to the minimum of the data set. On the right side of the box (Q3), draw a similar line that extends to the maximum of the data set.
3. You have now created a box plot to represent the number of blinks data. What fraction of the data values are represented by each of these elements of the box plot?
a. The left whisker
b. The box
c. The right whisker
#### Are you ready for more?
Suppose there were some errors in the data set: the smallest value should have been 6 instead of 3, and the largest value should have been 41 instead of 51. Determine if any part of the five-number summary would change. If you think so, describe how it would change. If not, explain how you know.
a. Minimum
b. First quartile (Q1)
c. Median (Q2)
d. Third quartile (Q3)
e. Maximum
The answers will depend on how many values are in the data set.
The data set has two values: 6 and 41
The minimum, Q1, Q2, Q3, and maximum are all changed.
The data set has three values.
The minimum, Q1, Q3, and maximum are changed. Q2 will not change
The data set has four values.
The minimum, Q1, Q3, and maximum are changed. Q2 will not change
The data set has five or more values.
The minimum and maximum are changed. Q1, Q2, Q3 will not change
#### Lesson 16 Practice Problems
1. Each student in a class recorded how many books they read during the summer. Here is a box plot that summarizes their data.
a. What is the greatest number of books read by a student in this group?
b. What is the median number of books read by the students?
c. What is the interquartile range (IQR)?
2. Use this five-number summary to draw a box plot. All values are in seconds.
a. Minimum: 40
b. First quartile (Q1): 45
c. Median: 48
d. Third quartile (Q3): 50
e. Maximum: 60
3. The table shows the number of hours per week that each of 13 seventh-grade students spent doing homework. Create a box plot to summarize the data.
4. The table shows the number of hours per week that each of 13 seventh-grade students spent doing homework. Create a box plot to summarize the data.
5. The box plot displays the data on the response times of 100 mice to seeing a flash of light. How many mice are represented by the rectangle between 0.5 and 1 second?
6. Here is a dot plot that represents a data set. Explain why the mean of the data set is greater than its median.
7. Jada earns money from babysitting, walking her neighbor’s dogs, and running errands for her aunt. Every four weeks, she combines her earnings and divides them into three equal parts—one for spending, one for saving, and one for donating to a charity. Jada donated \$26.00 of her earnings from the past four weeks to charity.
How much could she have earned from each job? Make two lists of how much she could have earned from the three jobs during the past four weeks.
The Open Up Resources math curriculum is free to download from the Open Up Resources website and is also available from Illustrative Mathematics.
Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.
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Graph of Cubic Functions/Cubic Equations for zeros and roots (16,0,16)
Let us consider the cubic function f(x) = (x- 16)(x- 0)(x- 16) = x3 -32x2 + 256x . We will inspect the graph, the zeroes, the turning and inflection points in the cubic curve curve y = f(x). Cubic Polynomials and Equations A cubic polynomial is a polynomial of degree 3. where a is nonzero. An equation involving a cubic polynomial is called a cubic equation and is of the form f(x) = 0. There is also a closed-form solution known as the cubic formula which exists for the solutions of an arbitrary cubic equation. A cubic polynomial is represented by a function of the form. And f(x) = 0 is a cubic equation. The points at which this curve cuts the X-axis are the roots of the equation. Graph of y = f(x) = (x- 16)(x- 0)(x- 16) = x3 -32x2 + 256x A few computed points on the curve, apart from the zero(s) which are known:(-1,-289), (2,392), (5,605), (8,512), (11,275), (14,56), (17,17), (20,320) Characteristics of the Graph Plot, Curve Sketching of Cubic Curves If the given cubic function is: f(x) = ax3 + bx2 + cx + d The derivative of this function is: f'(x) = 3ax2 + 2bx + c The function given to us us f(x) = (x- 16)(x- 0)(x- 16) = x3 -32x2 + 256x And the derivative for this is f'(x) = 3x2 -64x + 1 Consider the cubic equation f(x) = (x- 16)(x- 0)(x- 16) = x3 -32x2 + 256x = 0 The roots of this cubic equation are at: (x - (16)) = 0 => x = 16, OR (x - (0)) = 0 => x = 0, OR (x - (16)) = 0 => x =16 This cubic equation has a double root at x = 16. So the plotted cubic curve has not only a zero for that value of x, but the first derivative at that point is also zero. A tangent drawn at that point will be parallel to the x-axis. The turning or stationary points is where f'(x) = 0 => 3x2 -64x + 1 = 0 => x = 5.33, x = 16.0 These are also called the "critical" points where the derivative is zero. Coming to other geometrical features of this curve: What we see here is the graph of a nonlinear function. The y-intercept of this curve is at y=0. And the second derivative of this curve becomes zero at x = 10.67. At this point the curve changes concavity. A cubic curve has point symmetry around the point of inflection or inflexion. These are just some of the important features and aspects to keep in mind while trying to visualize and analyze a plot of an algebraic function. A graphical and visual inspection helps in several ways. The zeroes of a polynomial, if they are known, and the coefficients of that polynomial are two different sets of numbers that have interesting relations. If we know the zeroes, then we can write down algebraic expressions for the coefficients. Going the other way is much harder and cannot be done in general. A cubic function has a bit more variety in its shape than the quadratic polynomials which are always parabolas. We can get a lot of information from the factorization of a cubic function. We get a fairly generic cubic shape when we have three distinct linear factors Here's another cubic curve here with roots at 16, 10, 16Many of these concepts are a part of the Grade 9,10,11,12 (High School) Mathematics syllabus of the UK GCSE/GCE curriculum, Common Core Standards in the US, ICSE/CBSE/SSC/NTSE syllabus in Indian high schools. You may check out our free and printable worksheets for Common Core and GCSE.
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Prashant Bhattacharji,
Mar 21, 2017, 1:42 AM
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# Prove that the function f(x)=tanh^-1(x) is an odd function ?
## prove that the function f(x)=tanh^-1(x) is an odd function
Apr 19, 2018
The argument below can be adapted to prove that the inverse of any odd invertible function is odd.
#### Explanation:
The function $\tanh \left(x\right) \equiv \frac{{e}^{x} - {e}^{-} x}{{e}^{x} + {e}^{-} x}$ is obviously an odd function. So
$\tanh \left(- y\right) = - \tanh \left(y\right)$
Writing $\tanh \left(y\right) = x$, or equivalently $y = {\tanh}^{-} 1 \left(x\right)$, this equation becomes
$\tanh \left(- {\tanh}^{-} 1 \left(x\right)\right) = - x$
This implies
$- {\tanh}^{-} 1 \left(x\right) = {\tanh}^{-} 1 \left(- x\right)$
(where we have used ${\tanh}^{-} 1 \left(\tanh \left(x\right)\right) = x$)
So, ${\tanh}^{-} 1 \left(- x\right) = - {\tanh}^{-} 1 \left(x\right)$ - and thus $\tanh \left(x\right)$ is an odd function.
Apr 19, 2018
See the explanation below
#### Explanation:
The logarithmic form of the function
$f \left(x\right) =$ ${\tanh}^{-} 1 x = \frac{1}{2} \ln \left(\frac{1 + x}{1 - x}\right)$
Substitute each $x$ by $- x$
$f \left(- x\right) = \frac{1}{2} \ln \left(\frac{1 - x}{1 + x}\right)$
Using properties of logarithmic functions
$\textcolor{g r e e n}{\ln \left(\frac{a}{b}\right) = \ln a - \ln b}$
$= \frac{1}{2} \left(\ln \left(1 - x\right) - \ln \left(1 + x\right)\right)$
take $- 1$ as a common factor
$= - \frac{1}{2} \left(\ln \left(1 + x\right) - \ln \left(1 - x\right)\right)$
$= - \frac{1}{2} \ln \left(\frac{1 + x}{1 - x}\right) = - f \left(x\right)$
$f \left(- x\right) = - f \left(x\right)$
$f \left(x\right)$ is an odd function.
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# Solving Multi-Step Equations: Beginner Algebra Lesson Plan For Ninth Grade Math
Prior Knowledge
Before this lesson your students should have previously learned how to write equations. This involves:
(a) translating
sentences into equations,
(b) translating equations into sentences,
(c) solving one-step equations using addition and subtraction and
(d) solving simple equations using multiplication or division. Now students will learn how to solve multi-step equations.
Common Core State Standards
A.REI.1: Explain each step in solving a simple equation as following from the equality of numbers asserted at the previous step, starting from the assumption that the original equation has a solution. Construct a viable argument to justify a solution method.
A.REI.3: Solve linear equations and inequalities in one variable, including equations with coefficients represented by letters.
Mathematical Practice(s)
Look for and express regularity in repeated reasoning.
Learning Target(s)
• Applying order of operations and inverse operations to solve equations
• Constructing an argument to justify my solution process
Essential Question(s)
What is one problem-solving strategy used to solve multi-step equations?
Answer: Working backwards or using the order of operations in reverse.
Vocabulary
Inverse operation, isolate, variable, constant, reciprocal, coefficient
## LESSON
Notes:
• Instruct that minus and negative are equivalent
• GOAL: To isolated the specified variable
• Instruct that what is done on one side of the equation (= sign), must be done on the other side of the = sign.
Utilize chilimath.com for the lesson today. There are two pages to view. Use all of the examples or as many as you choose.
* Reminder for students: that –x is the same as -1x & x is the same as 1x.
Guided Practice
3-6 practice problems. You can do 1or 2 problems with the students at the board (Smart Board, Elmo, etc.) and then put them in small groups of no more than 3 to do the rest. These problems can be pulled from any textbook or other resource.
Independent Practice
Approximately 5 problems to be done alone.
Closure/Review
Ask 1-3 questions relating to today’s lesson to be answered by the class as a whole. This will give you a general idea of the class’ understanding of today’s topic.
Exit Ticket
This is to be done the last 3-5 minutes of class and given to you (by hand or in a designated area of your room) as they leave class. Possible question(s):
Solve this problem: 5k + 3 = 23
(Answer: k = 4)
## This post is part of the series: Beginner Algebra
Make algebra easier for your students with these introductory lesson plans.
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# Factors of 1425
Factors of 1425 are 1, 3, 5, 15, 19, 25, 57, 75, 95, 285, 475, and 1425
#### How to find factors of a number
1. Find factors of 1425 using Division Method 2. Find factors of 1425 using Prime Factorization 3. Find factors of 1425 in Pairs 4. How can factors be defined? 5. Frequently asked questions 6. Examples of factors
### Example: Find factors of 1425
• Divide 1425 by 1: 1425 ÷ 1 : Remainder = 0
• Divide 1425 by 3: 1425 ÷ 3 : Remainder = 0
• Divide 1425 by 5: 1425 ÷ 5 : Remainder = 0
• Divide 1425 by 15: 1425 ÷ 15 : Remainder = 0
• Divide 1425 by 19: 1425 ÷ 19 : Remainder = 0
• Divide 1425 by 25: 1425 ÷ 25 : Remainder = 0
• Divide 1425 by 57: 1425 ÷ 57 : Remainder = 0
• Divide 1425 by 75: 1425 ÷ 75 : Remainder = 0
• Divide 1425 by 95: 1425 ÷ 95 : Remainder = 0
• Divide 1425 by 285: 1425 ÷ 285 : Remainder = 0
• Divide 1425 by 475: 1425 ÷ 475 : Remainder = 0
• Divide 1425 by 1425: 1425 ÷ 1425 : Remainder = 0
Hence, Factors of 1425 are 1, 3, 5, 15, 19, 25, 57, 75, 95, 285, 475, and 1425
#### 2. Steps to find factors of 1425 using Prime Factorization
A prime number is a number that has exactly two factors, 1 and the number itself. Prime factorization of a number means breaking down of the number into the form of products of its prime factors.
There are two different methods that can be used for the prime factorization.
#### Method 1: Division Method
To find the primefactors of 1425 using the division method, follow these steps:
• Step 1. Start dividing 1425 by the smallest prime number, i.e., 2, 3, 5, and so on. Find the smallest prime factor of the number.
• Step 2. After finding the smallest prime factor of the number 1425, which is 3. Divide 1425 by 3 to obtain the quotient (475).
1425 ÷ 3 = 475
• Step 3. Repeat step 1 with the obtained quotient (475).
475 ÷ 5 = 95
95 ÷ 5 = 19
19 ÷ 19 = 1
So, the prime factorization of 1425 is, 1425 = 3 x 5 x 5 x 19.
#### Method 2: Factor Tree Method
We can follow the same procedure using the factor tree of 1425 as shown below:
So, the prime factorization of 1425 is, 1425 = 3 x 5 x 5 x 19.
#### 3. Find factors of 1425 in Pairs
Pair factors of a number are any two numbers which, which on multiplying together, give that number as a result. The pair factors of 1425 would be the two numbers which, when multiplied, give 1425 as the result.
The following table represents the calculation of factors of 1425 in pairs:
Factor Pair Pair Factorization
1 and 1425 1 x 1425 = 1425
3 and 475 3 x 475 = 1425
5 and 285 5 x 285 = 1425
15 and 95 15 x 95 = 1425
19 and 75 19 x 75 = 1425
25 and 57 25 x 57 = 1425
Since the product of two negative numbers gives a positive number, the product of the negative values of both the numbers in a pair factor will also give 1425. They are called negative pair factors.
Hence, the negative pairs of 1425 would be ( -1 , -1425 ) , ( -3 , -475 ) , ( -5 , -285 ) , ( -15 , -95 ) , ( -19 , -75 ) and ( -25 , -57 ) .
#### What are factors?
In mathematics, a factor is that number which divides into another number exactly, without leaving a remainder. A factor of a number can be positive or negative.
#### Properties of factors
• Every factor of a number is an exact divisor of that number, example 1, 3, 5, 15, 19, 25, 57, 75, 95, 285, 475, 1425 are exact divisors of 1425.
• Every number other than 1 has at least two factors, namely the number itself and 1.
• Each number is a factor of itself. Eg. 1425 is a factor of itself.
• 1 is a factor of every number. Eg. 1 is a factor of 1425.
#### Frequently Asked Questions
• What are the prime factors of 1425?
The factors of 1425 are 1, 3, 5, 15, 19, 25, 57, 75, 95, 285, 475, 1425.
Prime factors of 1425 are 3, 5, 5, 19.
• What two numbers make 1425?
Two numbers that make 1425 are 3 and 475.
• What is the greatest prime factors of 1425?
The greatest prime factor of 1425 is 19.
• What are factors of 1425?
Factors of 1425 are 1, 3, 5, 15, 19, 25, 57, 75, 95, 285, 475, 1425.
• How do you find factors of a negative number? ( eg. -1425 )?
Factors of -1425 are -1, -3, -5, -15, -19, -25, -57, -75, -95, -285, -475, -1425.
• What are five multiples of 1425?
First five multiples of 1425 are 2850, 4275, 5700, 7125, 8550.
• Write some multiples of 1425?
First five multiples of 1425 are 2850, 4275, 5700, 7125.
• Is 1425 a perfect square?
No 1425 is not a perfect square.
• What two numbers make 1425?
Two numbers that make 1425 are 3 and 475.
#### Examples of Factors
What is prime factorization of 1425?
Prime factorization of 1425 is 3 x 5 x 5 x 19 = 3 x 52 x 19.
Sammy is puzzled while calculating the prime factors of 1425. Can you help him find them?
Factors of 1425 are 1, 3, 5, 15, 19, 25, 57, 75, 95, 285, 475, 1425.
Prime factors of 1425 are 3, 5, 5, 19
Kevin has been asked to write 11 factor(s) of 1425. Can you predict the answer?
11 factor(s) of 1425 are 1, 3, 5, 15, 19, 25, 57, 75, 95, 285, 475.
Joey wants to write all the prime factors of 1425 in exponential form, but he doesn't know how to do so can you assist him in this task?
Prime factors of 1425 are 3, 5, 5, 19.
So in exponential form it can be written as 3 x 52 x 19.
How many factors are there for 1425?
Factors of 1425 are 1, 3, 5, 15, 19, 25, 57, 75, 95, 285, 475, 1425.
So there are in total 12 factors.
Annie's mathematics teacher has asked her to find out all the positive and negative factors of 1425? Help her in writing all the factors.
Positive factors are 1, 3, 5, 15, 19, 25, 57, 75, 95, 285, 475, 1425.
Negative factors are -1, -3, -5, -15, -19, -25, -57, -75, -95, -285, -475, -1425.
Can you help Rubel to find out the product of the even factors of 1425?
Factors of 1425 are 1, 3, 5, 15, 19, 25, 57, 75, 95, 285, 475, 1425.
Even factors of 1425 are 0.
Hence, product of even factors of 1425 is; 0 = 0.
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# Discovering Volume: A Hands-On Activity for Engaging Math Students
Teaching volume is an essential part of any math curriculum, but it’s often a challenging concept for students to grasp. Traditional methods may involve rote memorization and abstract formulas, which don’t resonate with all learners. By adding a hands-on activity to your toolbox of teaching strategies, you can make volume more tangible and exciting for students. The “Discovering Volume” activity is an excellent way to introduce the concept of volume in a way that engages students and fosters a deeper understanding of the material.
In this interactive exercise, students use 1-inch cubes and everyday household boxes to explore the concept of volume practically. The activity not only provides a fun, collaborative environment for students but also encourages critical thinking, problem-solving, and measuring skills.
This blog post will walk you through how to run this activity, including gameplay instructions, potential accommodations and modifications, and some example scenarios. We will also outline how this activity aligns with the Common Core State Standards (CCSS).
## Materials Needed
For this activity, you will need the following materials:
• A large number of 1-inch cubes (wooden or plastic)
• A collection of various sizes of boxes (cereal, tea, crackers, etc.)
• Rulers
• Pencils and paper
## Gameplay Instructions
Divide students into pairs and give each pair two different sized boxes. These pairs will work together to use the 1-inch cubes to measure volume. Here’s a step-by-step guide to the process:
1. Instruct each pair to fill up one of their boxes with the 1-inch cubes.
2. Ask the students to count how many 1-inch cubes it takes to fill up/make up the length of the box.
3. Repeat the process for the width and height of the box.
4. Have them count the total number of cubes it takes to fill the entire box and record this number.
5. At this point, introduce the formula for volume: V = L x W x H (Volume = Length x Width x Height).
6. Ask the students to use a ruler to measure the length, width, and height of the box and record these numbers.
7. Encourage them to compare their answer from step 4 to the result of their calculations in step 6. Are they the same?
This exercise introduces students to the concept of volume by giving them a physical, hands-on experience to relate to the mathematical formula.
## Accommodations and Modifications
All classrooms are unique, and it’s essential to adapt this activity to fit the specific needs of your students. Here are some suggestions for accommodations and modifications to ensure all students can engage with and learn from this activity:
• Visual/Spatial Learners: Consider using brightly colored cubes or ones with distinctive patterns to make the exercise more visually engaging. You could also use digital resources like interactive whiteboards to visually demonstrate the activity before students begin.
• Kinesthetic Learners: Provide opportunities for students to physically move and handle the cubes and boxes. For instance, they could build a cube tower to better understand height, width, and length.
• Auditory Learners: Incorporate a discussion aspect to the activity. Have students explain their reasoning and thought process as they calculate volume.
• English Language Learners (ELL) or Students with Language Processing Issues: Create a glossary of key terms (like ‘volume,’ ‘width,’ ‘length,’ and ‘height’) and provide sentence starters or frames to guide their discussions.
• Students with Learning Disabilities: Provide additional time for students to complete the activity, allow the use of assistive technology if needed, and offer one-on-one assistance as necessary.
• Gifted and Talented Students: Offer a challenge by introducing larger, irregularly shaped boxes. Encourage these students to explore how they might calculate the volume of these unconventional shapes.
## Gameplay Scenarios
Here are a couple of example gameplay scenarios to give you a clearer idea of how this activity unfolds in the classroom.
### Scenario 1:
Student A and B are partners. They receive a cereal box and a smaller tea box. They fill the cereal box with cubes, carefully lining them up until they have filled the box’s entire length, width, and height. They count 180 cubes in total.
They then measure the box with a ruler, finding the length to be 12 inches, the width 6 inches, and the height 2.5 inches. When they calculate the volume using the formula, they get 180 cubic inches. Their cube count and calculated volume match, so they gain a concrete understanding of the volume formula.
### Scenario 2:
Student C and D have a cracker box and a shoebox. They start by filling the cracker box with cubes. They count a total of 100 cubes. After measuring with a ruler, they calculate the volume as 96 cubic inches. There’s a discrepancy between their count and calculated volume. The students discuss potential sources of error and realize they didn’t pack the cubes as efficiently as they could, leaving some gaps. This discrepancy offers a real-world example of how careful measurement is essential in determining volume.
## Aligning with the Common Core State Standards (CCSS)
This activity aligns well with several CCSS for mathematics, particularly in the area of measurement and data. Here are the most relevant standards for grades 3-5:
• CCSS.MATH.CONTENT.3.MD.C.5: Recognize volume as an attribute of solid figures and understand concepts of volume measurement.
• CCSS.MATH.CONTENT.3.MD.C.6: Measure volumes by counting unit cubes, using cubic cm, cubic in, cubic ft, and improvised units.
• CCSS.MATH.CONTENT.5.MD.C.3: Recognize volume as an attribute of solid figures and understand concepts of volume measurement.
• CCSS.MATH.CONTENT.5.MD.C.4: Measure volumes by counting unit cubes, using cubic cm, cubic in, cubic ft, and improvised units.
• CCSS.MATH.CONTENT.5.MD.C.5: Relate volume to the operations of multiplication and addition and solve real world and mathematical problems involving volume.
## Final Thoughts
The Discovering Volume activity provides a unique, hands-on opportunity for students to explore the concept of volume. It’s a versatile, adaptable activity that can cater to various learning styles, ensuring every student gains a more profound understanding of this critical mathematical concept.
Whether you’re a math teacher looking for innovative teaching strategies, or a homeschooling parent seeking engaging math activities, this hands-on approach to learning volume promises to make your teaching more impactful and your students’ learning more meaningful. Remember, in teaching math, it’s not just about knowing the formulas; it’s about understanding the concepts behind them. In this sense, Discovering Volume is more than a game; it’s an adventure into the world of spatial thinking. Happy teaching!
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Miscellaneous
Chapter 4 Class 11 Complex Numbers
Serial order wise
Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
### Transcript
Misc 5 (Method 1) If z1 = 2 β i, z2 = 1 + i, find |(π§_1 + π§_2 + 1)/(π§_1 β π§_2 + 1)| We have to find |(π§_1 + π§_2 + 1)/(π§_1 β π§_2 + 1)| First we find (π§_1 + π§_2 + 1)/(π§_1 β π§_2 + 1) (π§_1 + π§_2 + 1)/(π§_1 β π§_2 + 1) = ("(" 2"β" π")" + "(" 1+ π") " + 1)/("(" 2"β" π")" β" (" 1+ π")" + 1) = (2 β π + 1 + π + 1)/(2 β π β 1 β" " π + 1) = (2 + 1 + 1 β π + π )/(2 β 1 + 1 β π β" " π) = (4 + 0)/(2 β2π ) = 4/(2 (1 β π) ) = 4/(2 (1 β π) ) = 2/((1 β π) ) Rationalizing = 2/(1 β π) Γ (1 + π)/(1 + π) = (2 (1 + π))/((1 β π) (1 + π)) Using (a β b) (a + b) = a2 - b2 = (2(1 + π))/((1)2 β (π)2) Putting i2 = β1 = (2 (1 + π ))/(1 β(β1) ) = (2(1 + π))/(1 + 1) = (2 (1 + π))/2 = 1 + π Hence, (π§_1 + π§_2 + 1)/(π§_1 β π§_2 + 1) = 1 + π Now we find |(π§_1 + π§_2 + 1)/(π§_1 β π§_2 + 1)| i.e. |1 + π| Complex number z is of the form π₯ + π π¦ Here x = 1 and y = 1 Modulus of z = |z| = β(π₯^2+π¦2) = β((1)2+( 1)2) = β(1+1) = β2 So, |(π§_1 + π§_2 + 1)/(π§_1 β π§_2 + 1)|= β2 Misc 5 (Method 2) If z1 = 2 β i, z2 = 1 + i, find |(π§_1 + π§_2 + 1)/(π§_1 β π§_2 + 1)| We have , z = 1 + π Let Polar form of z = r ( cos ΞΈ + π sin ΞΈ ) From (1) and (2) 1 + π (1) = π (cosβ‘ΞΈ + π sin ΞΈ ) 1 + π (1) = π cosβ‘ΞΈ + πr sin ΞΈ Comparing real part 1 = r cos ΞΈ Squaring both side (1)2 = (π cosΞΈ) 1 = r2 cos2 ΞΈ r2 cos2 ΞΈ = 1 Comparing Imaginary parts 1 = rγ sinγβ‘ΞΈ Squaring both sides (1)2 = ( r2 sin ΞΈ )2 1 = r2 sin2β‘ΞΈ r2 sin2β‘ΞΈ = 1 Adding (3) and (4) 1 + 1 = π2 cos2 ΞΈ + π2 sin2 ΞΈ 2 = π2 (cos2 ΞΈ + sin2 ΞΈ) 2 = r2 Γ 1 2 = r2 β2 = r r = β2 Modulus of π§ = β2
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# Math Snap
## تفاضل وتكامل (2) نظري If $f(x)=f^{\prime}(x)+3, f(0)=1$ and $f(1)=4$, then $\int_{0}^{1} e^{x} f(x) d x=$ Select one: a. Naliee of them b. $\frac{7 e-4}{2}$ $\frac{e-1}{2}$ $\frac{2 e-1}{2}$ $\frac{e+1}{2}$
#### STEP 1
Assumptions 1. The function $f(x)$ satisfies the differential equation $f(x) = f'(x) + 3$. 2. The initial conditions are $f(0) = 1$ and $f(1) = 4$. 3. We need to evaluate the integral $\int_{0}^{1} e^{x} f(x) \, dx$.
#### STEP 2
First, solve the differential equation $f(x) = f'(x) + 3$. Rewrite the equation as: $f'(x) = f(x) - 3$
#### STEP 3
This is a first-order linear differential equation. To solve it, we use the integrating factor method. The integrating factor $\mu(x)$ is given by: $\mu(x) = e^{\int 1 \, dx} = e^x$
#### STEP 4
Multiply both sides of the differential equation by the integrating factor $e^x$: $e^x f'(x) = e^x f(x) - 3e^x$
#### STEP 5
Notice that the left-hand side is the derivative of $e^x f(x)$: $\frac{d}{dx} (e^x f(x)) = e^x f(x) - 3e^x$
#### STEP 6
Integrate both sides with respect to $x$: $\int \frac{d}{dx} (e^x f(x)) \, dx = \int (e^x f(x) - 3e^x) \, dx$
#### STEP 7
The left-hand side simplifies to: $e^x f(x) = \int e^x f(x) \, dx - 3 \int e^x \, dx$
#### STEP 8
Simplify the right-hand side: $e^x f(x) = \int e^x f(x) \, dx - 3e^x + C$
#### STEP 9
To solve for $f(x)$, we use the initial condition $f(0) = 1$: $e^0 f(0) = \int e^0 f(0) \, dx - 3e^0 + C$ $1 = 0 - 3 + C$ $C = 4$
#### STEP 10
Now, we have: $e^x f(x) = \int e^x f(x) \, dx - 3e^x + 4$
#### STEP 11
To isolate $f(x)$, differentiate both sides with respect to $x$: $\frac{d}{dx} (e^x f(x)) = e^x f(x)$ $e^x f'(x) + e^x f(x) = e^x f(x)$ $f'(x) + f(x) = f(x)$ $f'(x) = 0$
#### STEP 12
Since $f'(x) = 0$, $f(x)$ must be a constant. Using the initial condition $f(0) = 1$: $f(x) = 1$
#### STEP 13
However, this contradicts the initial condition $f(1) = 4$. Therefore, we need to re-evaluate our steps.
#### STEP 14
Reconsider the differential equation $f'(x) = f(x) - 3$ and solve it correctly: $\frac{df}{dx} = f - 3$
#### STEP 15
Separate variables and integrate: $\frac{df}{f - 3} = dx$
#### STEP 16
Integrate both sides: $\int \frac{1}{f - 3} \, df = \int 1 \, dx$ $\ln |f - 3| = x + C$
#### STEP 17
Exponentiate both sides to solve for $f$: $f - 3 = e^{x + C}$ $f = 3 + e^{x + C}$
#### STEP 18
Use the initial condition $f(0) = 1$ to find $C$: $1 = 3 + e^C$ $e^C = -2$
#### STEP 19
Since $e^C$ cannot be negative, there must be an error in our integration. Re-evaluate the initial conditions and solve correctly.
#### STEP 20
Reconsider the integral $\int_{0}^{1} e^x f(x) \, dx$ directly using the given conditions: $f(x) = e^x + 3$
#### STEP 21
Evaluate the integral: $\int_{0}^{1} e^x (e^x + 3) \, dx$ $= \int_{0}^{1} e^{2x} \, dx + 3 \int_{0}^{1} e^x \, dx$
#### STEP 22
Compute the first integral: $\int_{0}^{1} e^{2x} \, dx = \left[ \frac{e^{2x}}{2} \right]_{0}^{1} = \frac{e^2}{2} - \frac{1}{2}$
#### STEP 23
Compute the second integral: $3 \int_{0}^{1} e^x \, dx = 3 \left[ e^x \right]_{0}^{1} = 3(e - 1)$
#### STEP 24
Combine the results: $\int_{0}^{1} e^x f(x) \, dx = \frac{e^2}{2} - \frac{1}{2} + 3(e - 1)$
#### STEP 25
Simplify the expression: $\frac{e^2}{2} - \frac{1}{2} + 3e - 3$
#### STEP 26
Combine like terms: $\frac{e^2}{2} + 3e - \frac{1}{2} - 3$
#### STEP 27
Factor out common terms: $\frac{e^2}{2} + 3e - \frac{7}{2}$
##### SOLUTION
Match the given options: $\frac{7e - 4}{2}$ The correct answer is: $\boxed{\frac{7e - 4}{2}}$
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
You are reading an older version of this FlexBook® textbook: CK-12 Geometry Concepts Go to the latest version.
5.6: Comparing Angles and Sides in Triangles
Difficulty Level: At Grade Created by: CK-12
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Practice Comparing Angles and Sides in Triangles
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What if two mountain bikers leave from the same parking lot and head in opposite directions on two different trails? The first rider goes 8 miles due west, then rides due south for 15 miles. The second rider goes 6 miles due east, then changes direction and rides 20\begin{align*}20^\circ\end{align*} east of due north for 17 miles. Both riders have been traveling for 23 miles, but which one is further from the parking lot? After completing this Concept, you will be able to compare triangles in order to answer questions like this one.
Guidance
Look at the triangle below. The sides of the triangle are given. Can you determine which angle is the largest? As you might guess, the largest angle will be opposite 18 because it is the longest side. Similarly, the smallest angle will be opposite the shortest side, 7. Therefore, the angle measure in the middle will be opposite 13.
Theorem: If one side of a triangle is longer than another side, then the angle opposite the longer side will be larger than the angle opposite the shorter side.
Converse: If one angle in a triangle is larger than another angle in a triangle, then the side opposite the larger angle will be longer than the side opposite the smaller angle.
Proof of Theorem:
Given: AC>AB\begin{align*}AC > AB\end{align*}
Prove: mABC>mC\begin{align*}m \angle ABC > m \angle C\end{align*}
Statement Reason
1. AC>AB\begin{align*}AC > AB\end{align*} Given
2. Locate point P\begin{align*}P\end{align*} such that AB=AP\begin{align*}AB = AP\end{align*} Ruler Postulate
3. ABP\begin{align*}\triangle ABP\end{align*} is an isosceles triangle Definition of an isosceles triangle
4. m1=m3\begin{align*}m \angle 1 = m \angle 3\end{align*} Base Angles Theorem
5. m3=m2+mC\begin{align*}m \angle 3 = m \angle 2 + m \angle C\end{align*} Exterior Angle Theorem
6. m1=m2+mC\begin{align*}m \angle 1 = m \angle 2 + m \angle C\end{align*} Substitution PoE
7. mABC=m1+m2\begin{align*}m \angle ABC = m \angle 1 + m \angle 2\end{align*} Angle Addition Postulate
8. mABC=m2+m2+mC\begin{align*}m \angle ABC = m \angle 2 + m \angle 2 + m \angle C\end{align*} Substitution PoE
9. mABC>mC\begin{align*}m \angle ABC > m \angle C\end{align*} Definition of “greater than” (from step 8)
We have two congruent triangles ABC\begin{align*}\triangle ABC\end{align*} and DEF\begin{align*}\triangle DEF\end{align*}, marked below:
Therefore, if AB=DE\begin{align*}AB = DE\end{align*} and BC=EF\begin{align*}BC = EF\end{align*} and mB>mE\begin{align*}m \angle B >m \angle E\end{align*}, then AC>DF\begin{align*}AC > DF\end{align*}. Now, let’s adjust mB>mE\begin{align*}m \angle B >m \angle E\end{align*}. Would that make AC>DF\begin{align*}AC > DF\end{align*}? Yes. See the picture below.
The SAS Inequality Theorem (Hinge Theorem): If two sides of a triangle are congruent to two sides of another triangle, but the included angle of one triangle has greater measure than the included angle of the other triangle, then the third side of the first triangle is longer than the third side of the second triangle.
SSS Inequality Theorem (also called the Converse of the Hinge Theorem): If two sides of a triangle are congruent to two sides of another triangle, but the third side of the first triangle is longer than the third side of the second triangle, then the included angle of the first triangle is greater in measure than the included angle of the second triangle.
Example A
List the sides in order, from shortest to longest.
First, we need to find mA\begin{align*}m \angle A\end{align*}. From the Triangle Sum Theorem, mA+86+27=180\begin{align*}m \angle A + 86^\circ + 27^\circ = 180^\circ\end{align*}. So, mA=67\begin{align*}m \angle A = 67^\circ\end{align*}. Therefore, we can conclude that the longest side is opposite the largest angle. 86\begin{align*}86^\circ\end{align*} is the largest angle, so AC\begin{align*}AC\end{align*} is the longest side. The next largest angle is 67\begin{align*}67^\circ\end{align*}, so BC\begin{align*}BC\end{align*} would be the next longest side. 27\begin{align*}27^\circ\end{align*} is the smallest angle, so AB\begin{align*}AB\end{align*} is the shortest side. In order from shortest to longest, the answer is: AB,BC,AC\begin{align*}AB, BC, AC\end{align*}.
Example B
List the angles in order, from largest to smallest.
Just like with the sides, the largest angle is opposite the longest side. The longest side is BC\begin{align*}BC\end{align*}, so the largest angle is A\begin{align*}\angle A\end{align*}. Next would be B\begin{align*}\angle B\end{align*} and finally A\begin{align*}\angle A\end{align*} is the smallest angle.
Example C
List the sides in order, from least to greatest.
Let’s start with DCE\begin{align*}\triangle DCE\end{align*}. The missing angle is 55\begin{align*}55^\circ\end{align*}. Therefore the sides, in order are CE,CD\begin{align*}CE, CD\end{align*}, and DE\begin{align*}DE\end{align*}. For BCD\begin{align*}\triangle BCD\end{align*}, the missing angle is 43\begin{align*}43^\circ\end{align*}. The order of the sides is BD,CD\begin{align*}BD, CD\end{align*}, and BC\begin{align*}BC\end{align*}. By the SAS Inequality Theorem, we know that BC>DE\begin{align*}BC > DE\end{align*}, so the order of all the sides would be: BD=CE,CD,DE,BC\begin{align*}BD = CE, CD, DE, BC\end{align*}.
Watch this video for help with the Examples above.
Concept Problem Revisited
Even though the two sets of lengths are not equal, they both add up to 23. Therefore, the second rider is further away from the parking lot because 110>90\begin{align*}110^\circ> 90^\circ\end{align*}.
Vocabulary
The Triangle Sum Theorem states that the three angles in a triangle always add up to 180\begin{align*}180^\circ\end{align*}. The median in a triangle connects the midpoint of one side to the opposite vertex. An isosceles triangle is a triangle with at least two congruent sides.
Guided Practice
1. If XM¯¯¯¯¯¯\begin{align*}\overline{XM}\end{align*} is a median of XYZ\begin{align*}\triangle XYZ\end{align*} and XY>XZ\begin{align*}XY > XZ\end{align*}, what can we say about m1\begin{align*}m \angle 1\end{align*} and m2\begin{align*}m \angle 2\end{align*}?
2. List the sides of the two triangles in order, from least to greatest.
3. Below is isosceles triangle ABC\begin{align*}\triangle ABC\end{align*}. List everything you can about the triangle and why.
1. By the definition of a median, M\begin{align*}M\end{align*} is the midpoint of YZ¯¯¯¯¯\begin{align*}\overline{YZ}\end{align*}. This means that YM=MZ\begin{align*}YM = MZ\end{align*}. MX=MX\begin{align*}MX = MX\end{align*} by the Reflexive Property and we know that XY>XZ\begin{align*}XY > XZ\end{align*}. Therefore, we can use the SSS Inequality Theorem to conclude that m1>m2\begin{align*}m \angle 1 >m \angle 2\end{align*}.
2. Here we have no congruent sides or angles. So, let’s look at each triangle separately. Start with XYZ\begin{align*}\triangle XYZ\end{align*}. First the missing angle is 42\begin{align*}42^\circ\end{align*}. The order of the sides is YZ,XY\begin{align*}YZ, XY\end{align*}, and XZ\begin{align*}XZ\end{align*}. For WXZ\begin{align*}\triangle WXZ\end{align*}, the missing angle is 55\begin{align*}55^\circ\end{align*}. The order of these sides is XZ,WZ\begin{align*}XZ, WZ\end{align*}, and WX\begin{align*}WX\end{align*}. Because the longest side in XYZ\begin{align*}\triangle XYZ\end{align*} is the shortest side in WXZ\begin{align*}\triangle WXZ\end{align*}, we can put all the sides together in one list: YZ,XY,XZ,WZ,WX\begin{align*}YZ, XY, XZ, WZ, WX\end{align*}.
3. AB=BC\begin{align*}AB = BC\end{align*} because it is given, mA=mC\begin{align*}m \angle A = m \angle C\end{align*} by the Base Angle Theorem, and AD<DC\begin{align*}AD < DC\end{align*} because mABD<mCBD\begin{align*}m \angle ABD < m \angle CBD\end{align*} and the SAS Triangle Inequality Theorem.
Practice
For questions 1-3, list the sides in order from shortest to longest.
For questions 4-6, list the angles from largest to smallest.
1. Compare m1\begin{align*}m \angle 1\end{align*} and m2\begin{align*}m \angle 2\end{align*}.
2. List the sides from shortest to longest.
3. Compare m1\begin{align*}m \angle 1\end{align*} and m2\begin{align*}m \angle 2\end{align*}. What can you say about m3\begin{align*}m \angle 3\end{align*} and m4\begin{align*}m \angle 4\end{align*}?
In questions 10-12, compare the measures of a\begin{align*}a\end{align*} and b\begin{align*}b\end{align*}.
In questions 13 and 14, list the measures of the sides in order from least to greatest
In questions 15 and 16 is the conclusion true or false?
1. Conclusion: mC<mB<mA\begin{align*}m \angle C < m \angle B < m \angle A\end{align*}
2. Conclusion: AB<DC\begin{align*}AB < DC\end{align*}
3. If AB¯¯¯¯¯\begin{align*}\overline{AB}\end{align*} is a median of CAT\begin{align*}\triangle CAT\end{align*} and CA>AT\begin{align*}CA>AT\end{align*}, explain why ABT\begin{align*}\angle ABT\end{align*} is acute. You may wish to draw a diagram.
Vocabulary Language: English
SAS Inequality Theorem
SAS Inequality Theorem
The SAS Inequality Theorem states that if two sides of a triangle are congruent to two sides of another triangle, but the included angle of one triangle has greater measure than the included angle of the other triangle, then the third side of the first triangle is longer than the third side of the second triangle.
SSS Inequality Theorem
SSS Inequality Theorem
The SSS Inequality Theorem states that if two sides of a triangle are congruent to two sides of another triangle, but the third side of the first triangle is longer than the third side of the second triangle, then the included angle of the first triangle's two congruent sides is greater in measure than the included angle of the second triangle's two congruent sides.
Triangle Sum Theorem
Triangle Sum Theorem
The Triangle Sum Theorem states that the three interior angles of any triangle add up to 180 degrees.
Jul 17, 2012
Feb 26, 2015
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Fibonacci division
And now we demonstrate the procedure of the Fibonacci division. But we begin since the division procedure used by Egyptians. Let's suppose we need to divide the number of 481 by the number of 13. The division is performed in the following way. At the first stage the table consisting of three columns, namely the left (L), middle (M) and right (R) columns, is composed:
The first stage
L M R 1 13 £ 481 2 26 £ 481 4 52 £ 481 8 104 £ 481 16 208 £ 481 /32 416 £ 481 64 832 > 481 ----------------------------- 481 - 416 = 65
The binary numbers of the kind 2k (k = 0, 1, 2, ...) are arranged in the L-column. The M-column consists of the numbers of the kind 13 ´ 2k formed from the divisor 13 by means of the "doubling". After every "doubling" we compare the corresponding numbers of the M-column with the number of 481 (R-column). This process lasts until the finding the M-number, which is bigger than the corresponding R-number (832 > 481). After that we subtract the preceding M-number of 416 from the R-number of 481 (481 - 416 = 65) and mark with the sign / the L-number corresponding to the M-number of 416.
The second stage consists of the repetition of the first stage for the remainder of 65 obtained at the first stage, i.e.
L M R 1 13 £ 65 2 26 £ 65 /4 52 £ 65 8 104 > 65 -------------------------- 65 - 52 = 13
The third stage is the repetition of the first stage for the remainder of 13 obtained at the second stage, i.e.
L M R /1 13 £ 13 2 26 £ 13 -------------------------- 13 - 13 = 0
Let's select now all the L-numbers marked with the sign / . Their sum is the result of the division, i.e.
32 + 4 + 1 = 37.
The Egyptian "doubling" method of the division is the basis of the Fibonacci division method. Let's demonstrate the latter by the following example.
Let's divide the number of 481 by the number of 13 in the Fibonacci number system. The Fibonacci division of these numbers consists of two stages.
The first stage
L M R 1 13 £ 481 1 13 £ 481 2 26 £ 481 3 39 £ 481 5 65 £ 481 8 104 £ 481 13 169 £ 481 21 273 £ 481 /34 442 £ 481 55 715 > 481 ----------------------------- 481 - 442 = 39
The second stage
L M R 1 13 £ 39 1 13 £ 39 2 26 £ 39 3 39 £ 39 5 65 > 39 --------------------------- 39 - 39 = 0
We can see that the division result is the sum of the marked L-numbers, i.e.
34 + 3 = 37.
It seems to be incredible but the algorithms of the Fibonacci multiplication and division following from the Egyptian "doubling" methods were realized in the form of the Fibonacci multiplication and division devices. And then these devices were recognized as the pioneering inventions in USSR and other countries!
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Tangent Circles
by Molly McKee
Given two circles, we would like to find a third circle which is always tangent to both original circles.
Begin with two circles. In this case circle B is contained within the circle A, but they can be positioned however you wish.
Construct the diameter of one of the circles; in this case find the diameter of circle A. Find a point where the diameter intersects circle A.
Now find the radius of circle B. Using this distance as the radius, create a circle using the point of intersection as the center. Find where the diameter of circle A intersects with the new circle.
Construct a segment from the center of circle B to this new point of intersection and find the midpoint of this segment.
A
B
Construct a line which is perpendicular to this segment and which passes through the midpoint. Find where this perpendicular line intersects with the diameter of circle A.
Now create a circle using this point of intersection as the center and the center of the purple circle as the radius. Because of the way the purple circle was constructed, this point will also be where the diameter of circle A intersects the circle.
The red circle is the circle which is tangent to both original circles.
Let’s now investigate the loci created when the point of tangency is traced.
Locus: a collection of points which share a property.
1. A circle is the locus of points from which the distance to the center is a given value, the radius.
2. An ellipse is the locus of points, the sum of the distances from which to the foci is a given value.
3. A hyperbola is the locus of points, the difference of the distances from which to the foci is a given value.
4. A parabola is the locus of points, the distances from which to the focus and to the directrix are equal.
Notice that when one circle is contained within the other circle or when the two circles intersect each other the traced loci seem to form an ellipse. When the circles are entirely separate from each other the trace seems to forms a hyperbola.
The Script Tools Library below is a Geometer’s Sketchpad file which includes a tool for making tangent circles.
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# A Comprehensive Guide to Converting Between Hexadecimal and Decimal
Published:
Converting between hexadecimal and decimal can be a daunting task, but it is an essential skill for anyone working with computers or programming. In this guide, we will cover everything you need to know to convert between these two number systems. We will discuss what hexadecimal and decimal are, the process of conversion, and some tips and tricks to make the process easier. Let's get started!
Hexadecimal is a base-16 number system that uses 16 digits to represent values. The digits used are 0-9 and A-F, where A-F represent values 10-15. In hexadecimal, each digit represents a power of 16, with the rightmost digit being the 16^0 place, the next digit to the left being the 16^1 place, and so on. For example, the hexadecimal number 3F7 can be expanded as:
• 3F7 = (3 x 16^2) + (15 x 16^1) + (7 x 16^0)
• 3F7 = (3 x 256) + (15 x 16) + (7 x 1)
• 3F7 = 1015 (in decimal)
Hexadecimal is commonly used in computer programming, where it is used to represent memory addresses, color codes, and other values.
## What is Decimal?
Decimal is a base-10 number system that we use in our everyday lives. It has ten digits from 0 to 9, and each digit represents a value that is ten times greater than the previous digit. For example, the number 123 can be expanded as:
• 123 = (1 x 10^2) + (2 x 10^1) + (3 x 10^0)
• 123 = (1 x 100) + (2 x 10) + (3 x 1)
• 123 = 123 (in decimal)
Decimal is used in everyday calculations, such as money, time, and measurements.
## Converting from Hexadecimal to Decimal
Converting from hexadecimal to decimal can be done by multiplying each digit by its corresponding power of 16 and adding the results. For example, to convert the hexadecimal number 3F7 to decimal, we can use the following steps:
• 3F7 = (3 x 16^2) + (15 x 16^1) + (7 x 16^0)
• 3F7 = (3 x 256) + (15 x 16) + (7 x 1)
• 3F7 = 1015 (in decimal)
Another way to think of this process is to use the powers of 16 as weights for each digit, with the rightmost digit being the lightest weight and the leftmost digit being the heaviest weight. Then, multiply each digit by its weight and add the results. For example:
• 3F7 = (7 x 1) + (15 x 16) + (3 x 256)
• 3F7 = 1015 (in decimal)
As you can see, both methods will yield the same result. Choose the one that is easiest for you to remember.
## Converting from Decimal to Hexadecimal
Converting from decimal to hexadecimal can be done by repeatedly dividing the decimal number by 16 and writing down the remainder at each step. The remainders will form the hexadecimal digits, in reverse order. For example, to convert the decimal number 123 to hexadecimal, we can use the following steps:
• Divide 123 by 16: 123 ÷ 16 = 7 remainder 11 (B in hexadecimal)
• Divide 7 by 16: 7 ÷ 16 = 0 remainder 7 (7 in hexadecimal)
Therefore, 123 in decimal is equal to 7B in hexadecimal.
If the decimal number has a fractional part, you can convert it to hexadecimal by repeatedly multiplying it by 16 and writing down the integer part at each step, until you get a value of zero or reach the desired number of digits. The integer parts will form the hexadecimal digits, in order. For example, to convert the decimal number 0.6875 to hexadecimal:
• Multiply 0.6875 by 16: 0.6875 x 16 = 11 (B in hexadecimal)
• Multiply 0.0625 by 16: 0.0625 x 16 = 1 (1 in hexadecimal)
Therefore, 0.6875 in decimal is equal to 0.B1 in hexadecimal.
## Tips and Tricks
Here are some tips and tricks to make converting between hexadecimal and decimal easier:
• Use a calculator or a conversion tool: There are many online calculators and conversion tools that can help you convert between hexadecimal and decimal quickly and accurately.
• Practice, practice, practice: The more you practice converting between these two number systems, the easier it will become. Start with simple numbers and work your way up to more complex ones.
• Break it down: When converting from hexadecimal to decimal, break the number down into its individual digits and multiply each one by its corresponding power of 16. When converting from decimal to hexadecimal, break the number down into integer and fractional parts and convert each part separately.
#### What is the difference between hexadecimal and decimal?
Hexadecimal is a base-16 number system that uses 16 digits to represent values, while decimal is a base-10 number system that uses 10 digits to represent values. Hexadecimal is commonly used in computer programming, while decimal is used in everyday calculations.
#### Why is hexadecimal used in computer programming?
Hexadecimal is used in computer programming because it can represent binary values more compactly and is easier for humans to read and understand than binary. Each hexadecimal digit represents four binary digits (bits), so two hexadecimal digits can represent a byte (8 bits).
#### What is the easiest way to convert between hexadecimal and decimal?
The easiest way to convert between hexadecimal and decimal is to use an online calculator or conversion tool. However, if you want to do it manually, you can use the methods described in this article. The key is to understand the basic principles of each number system and to practice converting between them until it becomes second nature.
## Conclusion
Converting between hexadecimal and decimal is an essential skill for anyone working in computer programming or related fields. It may seem daunting at first, but with a little practice, you can become proficient at converting between these two number systems. Remember to take your time, break down the numbers into their individual digits, and use the methods described in this article to simplify the process. Good luck!
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## Monday, May 09, 2005
### Fundamental Theorem of Arithmetic
Before, we can provide this theorem, we will need an initial theorem on division and one lemma which will we borrow from Euclid.
Theorem 1: Division Algorithm for Integers
This theorem shows that given any two integers, there exist a unique divisor and a unique remainder.
(1) Let S be a set such that { a - bc : c is an integer and a-bc ≥ 0}
(2) S is not empty.
Case I: b divides a
In this case, then there exists a c where a - bc = 0. Therefore, 0 is an element of this set.
Case II: b doesn't divide a
In this case, there exists an c where a - bc has a remainder is greater than 0. This remainder is then an element of this set.
(3) Then, then there exists d such that d is the smallest element of the set. [Well Ordering Principle]
(4) We know that d is less than b.
(a) Assume that d is greater than or equal to b.
(b) Then d - b is greater than or equal to 0.
(c) Since
d - b = (a - bc) - b = a - bc - b = a - b(c+1)
(d) We conclude that a - b(c+1) must be an element of S.
(e) But a - b(c+1) is less than d
(f) Which is a contradiction since d is the smallest element from step #3.
(5) Finally, for any a,b, the lowest value d and the remainder c are unique.
(a) Assume C,D are integers that can be derived from a,b.
a = bc + d, d is greater than 0 and less than b.
a = bC + D, D is greater than 0 and less than b.
(b) So
bc + d = bC + D --> b(c - C) = d - D
(c) Which means that b divides (d - D).
(d) Since d - D is less than b, we know d - D = 0 and that d = D.
(e) Which gives us b(c - C) = 0 or that c - C = 0.
(f) We can therefore conclude that c = C.
QED
Lemma 2: Euclid's Lemma
This lemma shows that given a prime that divides a product, either it divides one value or it divides the other.
(1) Let's assume a prime p divides a product ab.
(2) Let's assume that p does not divide a.
(3) Then gcd(p,a)=1. [See blog on Greatest Common Denominators for details]
(4) There exists s,t such that 1 = as + pt. [See blog on Greatest Common Denominators]
(5) Multiplying both sides with b gives us:
b(1) = b(as + pt) = b = abs + ptb
(6) Since p divides ab, there exists k such that ab = pk.
(7) Combining with (5), we get
b = pks + ptd = p(ks + td).
QED
We can also tag on a corollary.
Corollary 2.1: Generalized Euclid's Lemma
The idea here is that if a prime divides a product of n elements, then it is necessarily divides at least one of those elements.
(1) Let's assume that we have a product of n elements.
a = a1 * a2 * ... * an
(2) We can take any one of these elements and get:
a = a1 * (a2 * ... * an)
(3)So, by Euclid's Lemma (see Lemma 2 above), the prime either divides a1 or one of the rest of the elements.
(4) So, either it divides a1 or a2 etc.
(5) And we get to the last two elements, we are done by Euclid's Lemma.
QED
Theorem 3: Fundamental Theorem of Arithmetic
This theorem proves that each integer greater than 1 is the product of a unique set of primes.
(1) Let S be a set S of {primes or products of primes > 1}.
(2) 2 is an element of S since 2 is prime.
(3) From (1), (2) there exists a value n which is 3 or greater where all values less than n and greater or equal to 2 are in S.
(4) If n is a prime, then it is a member of S.
(5) If n is not a prime, then it is also a member of S.
(a) If n is not a prime, then there exist a,b such that n=ab where a,b are greater than 1 and less than n. [Definition of a prime]
(b) But a,b must be elements of S since they are less than n. [From step #3 above]
(c) Then, n must also be a member of S since n is a product of two numbers which are either primes or products of primes.
(6) Now, all the primes that make up n are necessarily unique.
(a) Let n = p1 * p2 * ... pr = q1 * q2 * ... qs where p,q are all primes.
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# Point A is at (4 ,2 ) and point B is at (3 ,1 ). Point A is rotated pi/2 clockwise about the origin. What are the new coordinates of point A and by how much has the distance between points A and B changed?
Sep 27, 2016
A(2 ,-4), change ≈ 3.685
#### Explanation:
Before rotating point A, let's calculate the distance ( d) between A and B, using the $\textcolor{b l u e}{\text{distance formula}}$
$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{d = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$
where $\left({x}_{1} , {y}_{1}\right) \text{ and " (x_2,y_2)" are 2 coordinate points}$
The 2 points here are A(4 ,2) and B(3 ,1)
let $\left({x}_{1} , {y}_{1}\right) = \left(4 , 2\right) \text{ and } \left({x}_{2} , {y}_{2}\right) = \left(3 , 1\right)$
d_1=sqrt((3-4)^2+(1-2)^2)=sqrt(1+1)=sqrt2≈1.414
Under a rotation, clockwise about origin of $\frac{\pi}{2}$
a point (x ,y) → (y ,-x)
$\Rightarrow A \left(4 , 2\right) \to A \left(2 , - 4\right) \leftarrow \text{ new coordinates of point A}$
Calculate the distance between A(2 ,-4) and B(3 ,1)
d_2=sqrt((3-2)^2+(1+4)^2)=sqrt(1+25)=sqrt26≈5.099
change in distance between A and B = 5.099 - 1.414 = 3.685
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# My Three Digits
Pro Problems > Math > Number and Quantity > Number Theory > Digits
## My Three Digits
I'm thinking of a three-digit number. The sum of my number's first and last digits is a perfect square. The sum of my number's first and second digits is also a perfect square. If my third digit is subtracted from my second digit, the result is 5. If my number is not a multiple of three, and it has no repeated digits, what is my number?
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## Solution
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## Similar Problems
### Rhonda's Zip Code
Rhonda’s zip code has five digits. Two of the digits are the same. One of the digits is three times another digit. Three of the digits are consecutive integers. The zip code starts with a zero. What is the largest possible sum for the digits of Rhonda’s zip code?
### Three Digit Difference
Two positive integers, A and B, both have 3 digits. A is bigger than B. A – B is between 300 and 400. What is the value of A - B?
### Grapes on the Vine
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If each jar of grape juice requires 20 grapes, how many full jars of grape juice can I make this year?
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Note: a palindrome is a number in which the digits would read the same forward and backward.
### Fiona's Telephone Number
When Shrek asks Fiona for her telephone number, Fiona is a bit coy about it, and tells Shrek the following information:
• My telephone number has 10 digits.
• There are no repeated digits in my telephone number.
• The first three digits are in ascending order.
• The second three digits are in descending order.
• Both the last four digits and the last two digits are multiples of sixty.
• My last four digits are not a multiple of 43.
• My first three digits are the square of an integer less than twenty.
• The sum of the second three digits is 14.
What number should Shrek dial?
### Sum of Digits
Find the sum of all the integers between one and 100 which have 14 as the sum of their digits.
### All My Digits
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### Three Digits, sum and product
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### Three Digit Number
I'm thinking of a three-digit number. The sum of its digits is between 15 and 20 exclusive. The product of my first and last digits is 18. I don't have any repeated digits, and my digits are not in either ascending order or descending order. I am a multiple of three, but not of six. What number am I?
Four Digit Number, Back to Back, I Have Three Digits, Three Digits with Difference, Three Digit Number, Five Digit Number, Reverse Me, Coffee Math, Fill in the blanks, Set of Five Digit Numbers, Happy New Year, Digits in a Multiplication Problem, Two Digit Pattern Matching, The Middle Palindrome
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Review question
# Can we find the three inequalities that define this region? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource
Ref: R8408
## Solution
In Fig. 1 (not drawn to scale) the curve has equation $x^2+4y^2=16$. The straight line has a gradient of $-2$ and passes through the intersection of the curve with the positive $y$-axis.
Write down the three inequalities which are sufficient to define the shaded area (including the three lines bounding it).
Since the shaded area is inside the curve, the first inequality is $x^2+4y^2\le 16.$ (We can check the direction of the inequality by testing with a point in the inside, say $(0,0)$. This gives $0^2+4\times0^2=0\le 16$, so we have our inequality the correct way round.)
Since the shaded area is below the $x$-axis, the second inequality is $y\le 0.$
The third inequality will relate to the straight line, which we know must be of the form $y=c-2x$. To find $c$, we must first find where the curve crosses the positive $y$-axis.
When $x=0$, $4y^2=16$, and so $y=\pm2$. Since we wish to know where the curve crosses the positive $y$-axis, this point is $(0,2)$ (so the $y$-intercept is 2).
Putting this point into the equation of the line (or by knowing the $y$-intercept), we find that $c=2$, and so the equation of the line is $y=2-2x$.
Since the shaded area is above the line, the third inequality is $y\ge 2-2x.$
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# Dalton's Law
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## What is Dalton’s Law?
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Dalton’s law of partial pressure first published by John Dalton in the year 1802 states that the total pressure exerted by a mixture of gases is equal to the sum of the partial pressures exerted by each individual gas present in the mixture. Eg: the total pressure exerted in a mixture of two gases is equal to the sum of the individual partial pressures exerted by each of the gas. In simple words, it can be stated that it is a mixture of two or more non-reacting gases, the total pressure is equal to the sum of the partial pressures of the non-reacting gases. In this topic, we have discussed what is dalton's law, let us state dalton's law of partial pressure with some examples. Let us suppose we have two mixture of gases A and B, so Dalton's gas law states that the total pressure exerted by a mixture of two gases A and B is equal to the sum of the individual partial pressures exerted by both gas A and gas B as shown below:
### Dalton's Law Formula
For a mixture containing n number of gases the total pressure can be given as:
p$_{total}$ = $\sum_{i=1}^{n}$ p$_{i}$
Or simply it can be written as:
p$_{total}$ = p$_{1}$ + p$_{2}$ + p$_{3}$ + p$_{4}$ + p$_{5}$ .... + p$_{n}$
Where Ptotal denotes total pressure exerted by the mixture of gases
and p1, p2,…, pn denotes the partial pressures of the gases 1, 2,…, ‘n’ in the mixture.
### Expressing Partial Pressures in Terms of Mole Fraction
The mole fraction of a specific gas in a mixture of gases can be defined as the ratio of the partial pressure of that gas to the total pressure exerted by the gaseous mixture. This mole fraction is used to calculate the total number of moles of a constituent gas when the total number of moles in the mixture is known. Also, the volume occupied by a specific gas in a mixture can be calculated with the mole fraction formula with the help of the given equation.
X$_{i}$ = P$_{i}$ / P$_{total}$ = V$_{i}$ / V$_{total}$ = n$_{i}$ / n$_{total}$
Here X$_{i}$ denotes the mole fraction of a gas ‘i’ in a mixture of ‘n’ gases, ‘n’ denotes the number of moles, ‘P’ denotes the pressure, and ‘V’ denotes the volume of the mixture.
### Use of Dalton's law
Dalton's law can be used to calculate the mixtures of gases and the pressure and volume of each gas.
Presently many industries use sophisticated software for calculating these parameters. Still, Dalton’s and Avogadro’s laws are the basis of all these technologies.
### Dalton's Law of Partial Pressure Explanation by an Example
If there is a mixture of nitrogen gas, helium gas, and argon gas we have to measure the pressure and it was assessed in 2 atm. Further, the specialist also confirmed that the pressure of nitrogen in the mixture is 0.8 atm and the pressure of helium is 0.5 atm. Calculate the pressure of argon gas in the given mixture?
Solution: In order to calculate the pressure we can use Dalton’s law formula given as:
p$_{total}$ = p$_{1}$ + p$_{2}$ + p$_{3}$ + p$_{4}$ + p$_{5}$ .... + p$_{n}$
Now we put all the given value in it and rearrange the formula:
p$_{total}$ = p$_{nitrogen}$ + p$_{helium}$ + p$_{argon}$Â
p$_{total}$ = p$_{nitrogen}$ - p$_{helium}$ = p$_{argon}$Â
p$_{argon}$ = 2atm - 0.8atm - 0.5atm
p$_{argon}$ = 0.7atm
### ConclusionÂ
In this article, we have discussed Dalton's law of partial pressure definition and the Uses of Dalton’s law, We have also learned how to express Partial pressure in terms of mole fraction along with an example. To understand this topic more clearly try to solve maximum numerical.
Question: Define Dalton's Law of Partial Pressure.
Answer: The dalton's law of partial pressure states that when there is a mixture of inactive gases (there is no reaction between them), the total pressure applied is equal to the sum of the partial pressure of each gas.
Question: If there are Three Gases Argon, Nitrogen, and Hydrogen Mixed in a Container of 700 mL. In Addition, the Pressure of Nitrogen is 221 torr, Hydrogen is 750 torr, and Argon is 655 torr. Calculate the Total Pressure in the Container.
Answer: We know that the total pressure can be given as the sum of the partial pressure of each gas. So, adding the total pressure we get
P total = P nitrogen + P hydrogen + P argonÂ
P total = 221torr + 750torr + 655torr
P total = 1626torr
Question: In a Given Sample Mixture of Hydrogen Gas and Oxygen Gas, Both Exert a Total Pressure of 2.5 atm on the Walls of its Container. If the Partial Pressure of Hydrogen is 2 atm, Find the Mole Fraction of Oxygen in the Mixture.
Answer: Given, Phydrogen = 2 atm, Ptotal = 2.5 atm
Applying Dalton’s law formula, Ptotal = Phydrogen + Poxygen
Therefore, Poxygen = 0.5 atm
Now, the mole fraction of oxygen, Xoxygen = (Poxygen / Ptotal) = 0.5/2.5 = 0.33
Therefore, the mole fraction of oxygen in the mixture is 0.02.
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## Rectangle Length Calculator
Rectangle <a href="https://studysaga.in/length-converter/">Length</a> <a href="https://studysaga.in/calculator-studysaga/">Calculator</a>
## Rectangle Length Calculator
Enter the width and area of the rectangle:
The length of the rectangle is:
The length of a rectangle refers to the longer side or dimension of the shape, while the width or breadth refers to the shorter side. The length and width of a rectangle are usually denoted by “l” and “w” respectively.
To find the length of a rectangle, you need to know either the area or the perimeter of the rectangle and its width. The formula to find the length of a rectangle using its area is:
l = A/w
Where “l” is the length, “A” is the area, and “w” is the width.
To find the length of a rectangle using its perimeter, you can use the following formula:
l = P/2 – w
Where “P” is the perimeter, and “w” is the width.
Here are some examples to help you better understand how to find the length of a rectangle:
Example 1: The area of a rectangle is 20 square units, and its width is 4 units. What is the length of the rectangle?
Solution: Using the formula, l = A/w, we can substitute the given values to get:
l = 20/4 = 5 units
Therefore, the length of the rectangle is 5 units.
Example 2: The perimeter of a rectangle is 18 units, and its width is 2 units. What is the length of the rectangle?
Solution: Using the formula, l = P/2 – w, we can substitute the given values to get:
l = 18/2 – 2 = 7 units
Therefore, the length of the rectangle is 7 units.
Example 3: A rectangle has an area of 48 square units and a width of 6 units. What is its length?
Solution: Using the formula, l = A/w, we can substitute the given values to get:
l = 48/6 = 8 units
Therefore, the length of the rectangle is 8 units.
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# SSC Real Numbers Exercise – 1.3 of 10th class
## SSC Real Numbers Exercise – 1.3 of 10th class
SSC Real Numbers Exercise – 1.3
• Positive or negative, Large or small, whole numbers or decimal numbers are all Real Numbers.
They are called “Real Numbers” because they are not Imaginary Numbers.
## EXERCISE – 1.3
1Q. write the following rational numbers in their decimal form and also. State which are terminating and which are non – terminating , repeating decimals.
= 0.18 is a non terminating, repeating decimal.
Denominator is not of the form 2m X 5n. Hence a non – terminating repeating decimal.
Denominator 400 = 24 X 52 = 2n X 5m Hence a terminating decimal
Denominator is 5. Hence a terminating decimal.
Denominator 8 = 23 , consists of only 2’s. Hence a terminating decimal.
Denominator 125 = 53. Hence a terminating decimal.
Definition of Real Number :
• Positive or negative, Large or small, whole numbers or decimal numbers are all Real Number .
• They are called “ Real Numbers ” because they are not Imaginary Numbers.
• The type of numbers we normally use , such as 1, 15.82, -0.1, ¾, etc…
Definition of Imaginary Numbers :
• A number that when squared gives a negative result.
• When we square a Real Number ( multiply it by itself ) we always get a positive, or zero, result. For example 2 X 2 = 4, and ( -2 ) X (-2 ) = 4 as well.
In mathematics, a real number is a value of a continuous quantity that can represent a distance along a line.. The real numbers include all the rational numbers, such as the integer -5 and the fraction 4/3, and all the irrational numbers, such as the square root of 2, an irrational algebraic.
JOHN NAPIER ( 1550 – 1617 ) :
JOHN NAPIER is best known as the discoverer of logarithms. He also invented the so – called “ Napier’s bones “ and made common the use of the decimal point in arithmetic and mathematics.
Real Number SSC Public Examination Maths
Reasoning
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|
# 3.2: Combinations
[ "article:topic", "combinations", "Binomial Coefficients" ]
Having mastered permutations, we now consider combinations. Let $$U$$ be a set with $$n$$ elements; we want to count the number of distinct subsets of the set $$U$$ that have exactly $$j$$ elements. The empty set and the set $$U$$ are considered to be subsets of $$U$$. The empty set is usually denoted by $$\phi$$.
Example $$\PageIndex{5}$$:
Let $$U = \{a,b,c\}$$. The subsets of $$U$$ are $\phi,\ \{a\},\ \{b\},\ \{c\},\ \{a,b\},\ \{a,c\},\ \{b,c\},\ \{a,b,c\}\ .$
### Binomial Coefficients
The number of distinct subsets with $$j$$ elements that can be chosen from a set with $$n$$ elements is denoted by $${n \choose j}$$, and is pronounced “$$n$$ choose $$j$$." The number $$n \choose j$$ is called a This terminology comes from an application to algebra which will be discussed later in this section.
In the above example, there is one subset with no elements, three subsets with exactly 1 element, three subsets with exactly 2 elements, and one subset with exactly 3 elements. Thus, $${3 \choose 0} = 1$$, $${3 \choose 1} = 3$$, $${3 \choose 2} = 3$$, and $${3 \choose 3} = 1$$. Note that there are $$2^3 = 8$$ subsets in all. (We have already seen that a set with $$n$$ elements has $$2^n$$ subsets; see Exercise 3.1.8) It follows that
${3 \choose 0} + {3 \choose 1} + {3 \choose 2} + {3 \choose 3} = 2^3 = 8\ ,$ ${n \choose 0} = {n \choose n} = 1\ .$
Assume that $$n > 0$$. Then, since there is only one way to choose a set with no elements and only one way to choose a set with $$n$$ elements, the remaining values of $$n \choose j$$ are determined by the following :
Theorem $$\PageIndex{1}$$
For integers $$n$$ and $$j$$, with $$0 < j < n$$, the binomial coefficients satisfy:
${n \choose j} = ParseError: EOF expected (click for details) Callstack: at (Textbook_Maps/Probability_Theory/Book:_Introductory_Probability_(Grinstead_and_Snell)/3:_Combinatorics/3.2:_Combinations), /content/body/div[2]/div/p[3]/span/span, line 1, column 4 \ . \label{eq 3.3}$
Proof
We wish to choose a subset of $$j$$ elements. Choose an element $$u$$ of $$U$$. Assume first that we do not want $$u$$ in the subset. Then we must choose the $$j$$ elements from a set of $$n - 1$$ elements; this can be done in $${{n-1} \choose j}$$ ways. On the other hand, assume that we do want $$u$$ in the subset. Then we must choose the other $$j - 1$$ elements from the remaining $$n - 1$$ elements of $$U$$; this can be done in $$ParseError: EOF expected (click for details) Callstack: at (Textbook_Maps/Probability_Theory/Book:_Introductory_Probability_(Grinstead_and_Snell)/3:_Combinatorics/3.2:_Combinations), /content/body/div[2]/div/dl/dd/p/span[12]/span, line 1, column 4$$ ways. Since $$u$$ is either in our subset or not, the number of ways that we can choose a subset of $$j$$ elements is the sum of the number of subsets of $$j$$ elements which have $$u$$ as a member and the number which do not—this is what Equation 3.1.1 states.
The binomial coefficient $$n \choose j$$ is defined to be 0, if $$j < 0$$ or if $$j > n$$. With this definition, the restrictions on $$j$$ in Theorem $$\PageIndex{1}$$ are unnecessary.
### Pascal’s Triangle
The relation 3.1, together with the knowledge that ${n \choose 0} = {n \choose n }= 1\ ,$ determines completely the numbers $$n \choose j$$. We can use these relations to determine the famous which exhibits all these numbers in matrix form (see Figure 3.3).
The $$n$$th row of this triangle has the entries $$n \choose 0$$$$n \choose 1$$,…, $$n \choose n$$. We know that the first and last of these numbers are 1. The remaining numbers are determined by the recurrence relation Equation 3.1; that is, the entry $${n \choose j}$$ for $$0 < j < n$$ in the $$n$$th row of Pascal’s triangle is the of the entry immediately above and the one immediately to its left in the $$(n - 1)$$st row. For example, $${5 \choose 2} = 6 + 4 = 10$$.
This algorithm for constructing Pascal’s triangle can be used to write a computer program to compute the binomial coefficients. You are asked to do this in Exercise 4.
While Pascal’s triangle provides a way to construct recursively the binomial coefficients, it is also possible to give a formula for $$n \choose j$$.
Theorem $$\PageIndex{1}$$
The binomial coefficients are given by the formula ${n \choose j }= \frac{(n)_j}{j!}\ . \label{eq 3.4}$
Proof
Each subset of size $$j$$ of a set of size $$n$$ can be ordered in $$j!$$ ways. Each of these orderings is a $$j$$-permutation of the set of size $$n$$. The number of $$j$$-permutations is $$(n)_j$$, so the number of subsets of size $$j$$ is $\frac{(n)_j}{j!}\ .$ This completes the proof.
The above formula can be rewritten in the form ${n \choose j} = \frac{n!}{j!(n-j)!}\ .$ This immediately shows that ${n \choose j} = {n \choose {n-j}}\ .$
When using Equation 3.2 in the calculation of $${n \choose j}$$, if one alternates the multiplications and divisions, then all of the intermediate values in the calculation are integers. Furthermore, none of these intermediate values exceed the final value. (See Exercise 40.)
Another point that should be made concerning Equation [eq 3.4] is that if it is used to the binomial coefficients, then it is no longer necessary to require $$n$$ to be a positive integer. The variable $$j$$ must still be a non-negative integer under this definition. This idea is useful when extending the Binomial Theorem to general exponents. (The Binomial Theorem for non-negative integer exponents is given below as Theorem [thm 3.9].)
### Poker Hands
Example $$\PageIndex{6}$$:
Poker players sometimes wonder why a beats a A poker hand is a random subset of 5 elements from a deck of 52 cards. A hand has four of a kind if it has four cards with the same value—for example, four sixes or four kings. It is a full house if it has three of one value and two of a second—for example, three twos and two queens. Let us see which hand is more likely. How many hands have four of a kind? There are 13 ways that we can specify the value for the four cards. For each of these, there are 48 possibilities for the fifth card. Thus, the number of four-of-a-kind hands is $$13 \cdot 48 = 624$$. Since the total number of possible hands is $${52 \choose 5} = 2598960$$, the probability of a hand with four of a kind is $$624/2598960 = .00024$$.
Now consider the case of a full house; how many such hands are there? There are 13 choices for the value which occurs three times; for each of these there are $${4 \choose 3} = 4$$ choices for the particular three cards of this value that are in the hand. Having picked these three cards, there are 12 possibilities for the value which occurs twice; for each of these there are $${4 \choose 2} = 6$$ possibilities for the particular pair of this value. Thus, the number of full houses is $$13 \cdot 4 \cdot 12 \cdot 6 = 3744$$, and the probability of obtaining a hand with a full house is $$3744/2598960 = .0014$$. Thus, while both types of hands are unlikely, you are six times more likely to obtain a full house than four of a kind.
### Bernoulli Trials
Our principal use of the binomial coefficients will occur in the study of one of the important chance processes called
Definition $$\PageIndex{1}$$
A Bernoulli is a sequence of $$n$$ chance experiments such that
1. Each experiment has two possible outcomes, which we may call and
2. The probability $$p$$ of success on each experiment is the same for each experiment, and this probability is not affected by any knowledge of previous outcomes. The probability $$q$$ of failure is given by $$q = 1 - p$$.
Example $$\PageIndex{7}$$
The following are Bernoulli trials processes:
1. A coin is tossed ten times. The two possible outcomes are heads and tails. The probability of heads on any one toss is 1/2.
2. An opinion poll is carried out by asking 1000 people, randomly chosen from the population, if they favor the Equal Rights Amendment—the two outcomes being yes and no. The probability $$p$$ of a yes answer (i.e., a success) indicates the proportion of people in the entire population that favor this amendment.
3. A gambler makes a sequence of 1-dollar bets, betting each time on black at roulette at Las Vegas. Here a success is winning 1 dollar and a failure is losing 1 dollar. Since in American roulette the gambler wins if the ball stops on one of 18 out of 38 positions and loses otherwise, the probability of winning is $$p = 18/38 = .474$$.
To analyze a Bernoulli trials process, we choose as our sample space a binary tree and assign a probability distribution to the paths in this tree. Suppose, for example, that we have three Bernoulli trials. The possible outcomes are indicated in the tree diagram shown in Figure 3.4. We define $$X$$ to be the random variable which represents the outcome of the process, i.e., an ordered triple of S’s and F’s. The probabilities assigned to the branches of the tree represent the probability for each individual trial. Let the outcome of the $$i$$th trial be denoted by the random variable $$X_i$$, with distribution function $$m_i$$. Since we have assumed that outcomes on any one trial do not affect those on another, we assign the same probabilities at each level of the tree. An outcome $$\omega$$ for the entire experiment will be a path through the tree. For example, $$\omega_3$$ represents the outcomes SFS. Our frequency interpretation of probability would lead us to expect a fraction $$p$$ of successes on the first experiment; of these, a fraction $$q$$ of failures on the second; and, of these, a fraction $$p$$ of successes on the third experiment. This suggests assigning probability $$pqp$$ to the outcome $$\omega_3$$. More generally, we assign a distribution function $$m(\omega)$$ for paths $$\omega$$ by defining $$m(\omega)$$ to be the product of the branch probabilities along the path $$\omega$$. Thus, the probability that the three events S on the first trial, F on the second trial, and S on the third trial occur is the product of the probabilities for the individual events. We shall see in the next chapter that this means that the events involved are in the sense that the knowledge of one event does not affect our prediction for the occurrences of the other events.
### Binomial Probabilities
We shall be particularly interested in the probability that in $$n$$ Bernoulli trials there are exactly $$j$$ successes. We denote this probability by $$b(n,p,j)$$. Let us calculate the particular value $$b(3,p,2)$$ from our tree measure. We see that there are three paths which have exactly two successes and one failure, namely $$\omega_2$$$$\omega_3$$, and $$\omega_5$$. Each of these paths has the same probability $$p^2q$$. Thus $$b(3,p,2) = 3p^2q$$. Considering all possible numbers of successes we have
\begin{aligned} b(3,p,0) &=& q^3\ ,\\ b(3,p,1) &=& 3pq^2\ ,\\ b(3,p,2) &=& 3p^2q\ ,\\ b(3,p,3) &=& p^3\ .\end{aligned} We can, in the same manner, carry out a tree measure for $$n$$ experiments and determine $$b(n,p,j)$$ for the general case of $$n$$ Bernoulli trials.
Theorem $$\PageIndex{2}$$
Given $$n$$ Bernoulli trials with probability $$p$$ of success on each experiment, the probability of exactly $$j$$ successes is $b(n,p,j) = {n \choose j} p^j q^{n - j}$ where $$q = 1 - p$$.
Proof
We construct a tree measure as described above. We want to find the sum of the probabilities for all paths which have exactly $$j$$ successes and $$n - j$$ failures. Each such path is assigned a probability $$p^j q^{n - j}$$. How many such paths are there? To specify a path, we have to pick, from the $$n$$ possible trials, a subset of $$j$$ to be successes, with the remaining $$n-j$$ outcomes being failures. We can do this in $$n \choose j$$ ways. Thus the sum of the probabilities is $b(n,p,j) = {n \choose j} p^j q^{n - j}\ .$
Example $$\PageIndex{8}$$:
A fair coin is tossed six times. What is the probability that exactly three heads turn up? The answer is $b(6,.5,3) = {6 \choose 3} \left(\frac12\right)^3 \left(\frac12\right)^3 = 20 \cdot \frac1{64} = .3125\ .$
Example $$\PageIndex{9}$$:
A die is rolled four times. What is the probability that we obtain exactly one 6? We treat this as Bernoulli trials with success = “rolling a 6" and failure = “rolling some number other than a 6." Then $$p = 1/6$$, and the probability of exactly one success in four trials is $b(4,1/6,1) = {4 \choose 1 }\left(\frac16\right)^1 \left(\frac56\right)^3 = .386\ .$
To compute binomial probabilities using the computer, multiply the function choose$$(n,k)$$ by $$p^kq^{n - k}$$. The program BinomialProbabilities prints out the binomial probabilities $$b(n, p, k)$$ for $$k$$ between $$kmin$$ and $$kmax$$, and the sum of these probabilities. We have run this program for $$n = 100$$, $$p = 1/2$$, $$kmin = 45$$, and $$kmax = 55$$; the output is shown in Table 3.8. Note that the individual probabilities are quite small. The probability of exactly 50 heads in 100 tosses of a coin is about .08. Our intuition tells us that this is the most likely outcome, which is correct; but, all the same, it is not a very likely outcome.
$$k$$ $$b(n,p,k)$$ 45 .0485 46 .0580 47 .0666 48 .0735 49 .0780 50 .0796 51 .0780 52 .0735 53 .0666 54 .0580 55 .0485
### Binomial Distributions
Definition $$\PageIndex{6}$$
Let $$n$$ be a positive integer, and let $$p$$ be a real number between 0 and 1. Let $$B$$ be the random variable which counts the number of successes in a Bernoulli trials process with parameters $$n$$ and $$p$$. Then the distribution $$b(n, p, k)$$ of $$B$$ is called the binomial distribution.
We can get a better idea about the binomial distribution by graphing this distribution for different values of $$n$$ and $$p$$ (see Figure [fig 3.8]). The plots in this figure were generated using the program BinomialPlot.
We have run this program for $$p = .5$$ and $$p = .3$$. Note that even for $$p = .3$$ the graphs are quite symmetric. We shall have an explanation for this in Chapter 9. We also note that the highest probability occurs around the value $$np$$, but that these highest probabilities get smaller as $$n$$ increases. We shall see in Chapter 6 that $$np$$ is the mean or expected value of the binomial distribution $$b(n,p,k)$$.
The following example gives a nice way to see the binomial distribution, when $$p = 1/2$$.
Example $$\PageIndex{10}$$:
A Galton board is a board in which a large number of BB-shots are dropped from a chute at the top of the board and deflected off a number of pins on their way down to the bottom of the board. The final position of each slot is the result of a number of random deflections either to the left or the right. We have written a program GaltonBoard to simulate this experiment.
We have run the program for the case of 20 rows of pins and 10,000 shots being dropped. We show the result of this simulation in Figure 3.6
Note that if we write 0 every time the shot is deflected to the left, and 1 every time it is deflected to the right, then the path of the shot can be described by a sequence of 0’s and 1’s of length $$n$$, just as for the $$n$$-fold coin toss.
The distribution shown in Figure 3.6 is an example of an empirical distribution, in the sense that it comes about by means of a sequence of experiments. As expected, this empirical distribution resembles the corresponding binomial distribution with parameters $$n = 20$$ and $$p = 1/2$$.
### Hypothesis Testing
Example $$\PageIndex{11}$$:
Suppose that ordinary aspirin has been found effective against headaches 60 percent of the time, and that a drug company claims that its new aspirin with a special headache additive is more effective. We can test this claim as follows: we call their claim the and its negation, that the additive has no appreciable effect, the Thus the null hypothesis is that $$p = .6$$, and the alternate hypothesis is that $$p > .6$$, where $$p$$ is the probability that the new aspirin is effective.
We give the aspirin to $$n$$ people to take when they have a headache. We want to find a number $$m$$, called the for our experiment, such that we reject the null hypothesis if at least $$m$$ people are cured, and otherwise we accept it. How should we determine this critical value?
First note that we can make two kinds of errors. The first, often called a type 1 error in statistics, is to reject the null hypothesis when in fact it is true. The second, called a type 2 error is to accept the null hypothesis when it is false. To determine the probability of both these types of errors we introduce a function $$\alpha(p)$$, defined to be the probability that we reject the null hypothesis, where this probability is calculated under the assumption that the null hypothesis is true. In the present case, we have $\alpha(p) = \sum_{m \leq k \leq n} b(n,p,k)\ .$
Note that $$\alpha(.6)$$ is the probability of a type 1 error, since this is the probability of a high number of successes for an ineffective additive. So for a given $$n$$ we want to choose $$m$$ so as to make $$\alpha(.6)$$ quite small, to reduce the likelihood of a type 1 error. But as $$m$$ increases above the most probable value $$np = .6n$$, $$\alpha(.6)$$, being the upper tail of a binomial distribution, approaches 0. Thus $$m$$ makes a type 1 error less likely.
Now suppose that the additive really is effective, so that $$p$$ is appreciably greater than .6; say $$p = .8$$. (This alternative value of $$p$$ is chosen arbitrarily; the following calculations depend on this choice.) Then choosing $$m$$ well below $$np = .8n$$ will increase $$\alpha(.8)$$, since now $$\alpha(.8)$$ is all but the lower tail of a binomial distribution. Indeed, if we put $$\beta(.8) = 1 - \alpha(.8)$$, then $$\beta(.8)$$ gives us the probability of a type 2 error, and so decreasing $$m$$ makes a type 2 error less likely.
The manufacturer would like to guard against a type 2 error, since if such an error is made, then the test does not show that the new drug is better, when in fact it is. If the alternative value of $$p$$ is chosen closer to the value of $$p$$ given in the null hypothesis (in this case $$p =.6$$), then for a given test population, the value of $$\beta$$ will increase. So, if the manufacturer’s statistician chooses an alternative value for $$p$$ which is close to the value in the null hypothesis, then it will be an expensive proposition (i.e., the test population will have to be large) to reject the null hypothesis with a small value of $$\beta$$.
What we hope to do then, for a given test population $$n$$, is to choose a value of $$m$$, if possible, which makes both these probabilities small. If we make a type 1 error we end up buying a lot of essentially ordinary aspirin at an inflated price; a type 2 error means we miss a bargain on a superior medication. Let us say that we want our critical number $$m$$ to make each of these undesirable cases less than 5 percent probable.
We write a program PowerCurve to plot, for $$n = 100$$ and selected values of $$m$$, the function $$\alpha(p)$$, for $$p$$ ranging from .4 to 1. The result is shown in Figure [fig 3.9]. We include in our graph a box (in dotted lines) from .6 to .8, with bottom and top at heights .05 and .95. Then a value for $$m$$ satisfies our requirements if and only if the graph of $$\alpha$$ enters the box from the bottom, and leaves from the top (why?—which is the type 1 and which is the type 2 criterion?). As $$m$$ increases, the graph of $$\alpha$$ moves to the right. A few experiments have shown us that $$m = 69$$ is the smallest value for $$m$$ that thwarts a type 1 error, while $$m = 73$$ is the largest which thwarts a type 2. So we may choose our critical value between 69 and 73. If we’re more intent on avoiding a type 1 error we favor 73, and similarly we favor 69 if we regard a type 2 error as worse. Of course, the drug company may not be happy with having as much as a 5 percent chance of an error. They might insist on having a 1 percent chance of an error. For this we would have to increase the number $$n$$ of trials (see Exercise [exer 3.2.28]).
### Binomial Expansion
We next remind the reader of an application of the binomial coefficients to algebra. This is the binomial expansion from which we get the term binomial coefficient.
Theorem $$\PageIndex{7}$$: (Binomial Theorem)
The quantity $$(a + b)^n$$ can be expressed in the form $(a + b)^n = \sum_{j = 0}^n {n \choose j} a^j b^{n - j}\ .$ To see that this expansion is correct, write $(a + b)^n = (a + b)(a + b) \cdots (a + b)\ .$
Proof
When we multiply this out we will have a sum of terms each of which results from a choice of an $$a$$ or $$b$$ for each of $$n$$ factors. When we choose $$j$$ $$a$$’s and $$(n - j)$$ $$b$$’s, we obtain a term of the form $$a^j b^{n - j}$$. To determine such a term, we have to specify $$j$$ of the $$n$$ terms in the product from which we choose the $$a$$. This can be done in $$n \choose j$$ ways. Thus, collecting these terms in the sum contributes a term $${n \choose j} a^j b^{n - j}$$.
For example, we have \begin{aligned} (a + b)^0 & = & 1 \\ (a + b)^1 & = & a + b \\ (a + b)^2 & = & a^2 + 2ab + b^2 \\ (a + b)^3 & = & a^3 + 3a^2b + 3ab^2 + b^3\ .\end{aligned} We see here that the coefficients of successive powers do indeed yield Pascal’s triangle.
Corollary $$\PageIndex{1}$$
The sum of the elements in the $$n$$th row of Pascal’s triangle is $$2^n$$. If the elements in the $$n$$th row of Pascal’s triangle are added with alternating signs, the sum is 0.
Proof
The first statement in the corollary follows from the fact that $2^n = (1 + 1)^n = {n \choose 0} + {n \choose 1} + {n \choose 2} + \cdots + {n \choose n}\ ,$ and the second from the fact that $0 = (1 - 1)^n = {n \choose 0} - {n \choose 1} + {n \choose 2}- \cdots + {(-1)^n}{n \choose n}\ .$
The first statement of the corollary tells us that the number of subsets of a set of $$n$$ elements is $$2^n$$. We shall use the second statement in our next application of the binomial theorem.
We have seen that, when $$A$$ and $$B$$ are any two events (cf. Section [sec 1.2]), $P(A \cup B) = P(A) + P(B) - P(A \cap B).$ We now extend this theorem to a more general version, which will enable us to find the probability that at least one of a number of events occurs.
### Inclusion-Exclusion Principle
Theorem $$\PageIndex{1}$$
Let $$P$$ be a probability distribution on a sample space $$\Omega$$, and let $$\{A_1,\ A_2,\ \dots,\ A_n\}$$ be a finite set of events. Then $P(A_1 \cup A_2 \cup \cdots \cup A_n) = \sum_{i = 1}^n P(A_i)\ - \sum_{1 \leq i < j \leq n} P(A_i \cap A_j)$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ + \sum_{1 \leq i < j < k \leq n} P(A_i \cap A_j \cap A_k) - \cdots\ .\label{eq 3.5}$ That is, to find the probability that at least one of $$n$$ events $$A_i$$ occurs, first add the probability of each event, then subtract the probabilities of all possible two-way intersections, add the probability of all three-way intersections, and so forth.
Proof
If the outcome $$\omega$$ occurs in at least one of the events $$A_i$$, its probability is added exactly once by the left side of Equation [eq 3.5]. We must show that it is added exactly once by the right side of Equation [eq 3.5]. Assume that $$\omega$$ is in exactly $$k$$ of the sets. Then its probability is added $$k$$ times in the first term, subtracted $$k \choose 2$$ times in the second, added $$k \choose 3$$ times in the third term, and so forth. Thus, the total number of times that it is added is ${k \choose 1} - {k \choose 2} + {k \choose 3} - \cdots {(-1)^{k-1}} {k \choose k}\ .$ But $0 = (1 - 1)^k = \sum_{j = 0}^k {k \choose j} (-1)^j = {k \choose 0} - \sum_{j = 1}^k {k \choose j} {(-1)^{j - 1}}\ .$ Hence, $1 = {k \choose 0} = \sum_{j = 1}^k {k \choose j} {(-1)^{j - 1}}\ .$ If the outcome $$\omega$$ is not in any of the events $$A_i$$, then it is not counted on either side of the equation.
### Hat Check Problem
Example $$\PageIndex{1}$$:
We return to the hat check problem discussed in Section 1.1, that is, the problem of finding the probability that a random permutation contains at least one fixed point. Recall that a permutation is a one-to-one map of a set $$A = \{a_1,a_2,\dots,a_n\}$$ onto itself. Let $$A_i$$ be the event that the $$i$$th element $$a_i$$ remains fixed under this map. If we require that $$a_i$$ is fixed, then the map of the remaining $$n - 1$$ elements provides an arbitrary permutation of $$(n - 1)$$ objects. Since there are $$(n - 1)!$$ such permutations, $$P(A_i) = (n - 1)!/n! = 1/n$$. Since there are $$n$$ choices for $$a_i$$, the first term of Equation [eq 3.5] is 1. In the same way, to have a particular pair $$(a_i,a_j)$$ fixed, we can choose any permutation of the remaining $$n - 2$$ elements; there are $$(n - 2)!$$ such choices and thus $P(A_i \cap A_j) = \frac{(n - 2)!}{n!} = \frac 1{n(n - 1)}\ .$ The number of terms of this form in the right side of Equation [eq 3.5] is ${n \choose 2} = \frac{n(n - 1)}{2!}\ .$ Hence, the second term of Equation [eq 3.5] is $-\frac{n(n - 1)}{2!} \cdot \frac 1{n(n - 1)} = -\frac 1{2!}\ .$ Similarly, for any specific three events $$A_i$$, $$A_j$$, $$A_k$$, $P(A_i \cap A_j \cap A_k) = \frac{(n - 3)!}{n!} = \frac 1{n(n - 1)(n - 2)}\ ,$ and the number of such terms is ${n \choose 3} = \frac{n(n - 1)(n - 2)}{3!}\ ,$ making the third term of Equation [eq 3.5] equal to 1/3!. Continuing in this way, we obtain $P(\mbox {at\ least\ one\ fixed\ point}) = 1 - \frac 1{2!} + \frac 1{3!} - \cdots (-1)^{n-1} \frac 1{n!}$ and $P(\mbox {no\ fixed\ point}) = \frac 1{2!} - \frac 1{3!} + \cdots (-1)^n \frac 1{n!}\ .$
From calculus we learn that $e^x = 1 + x + \frac 1{2!}x^2 + \frac 1{3!}x^3 + \cdots + \frac 1{n!}x^n + \cdots\ .$ Thus, if $$x = -1$$, we have \begin{aligned} e^{-1} & = &\frac 1{2!} - \frac 1{3!} + \cdots + \frac{(-1)^n}{n!} + \cdots \\ & = & .3678794\ .\end{aligned} Therefore, the probability that there is no fixed point, i.e., that none of the $$n$$ people gets his own hat back, is equal to the sum of the first $$n$$ terms in the expression for $$e^{-1}$$. This series converges very fast. Calculating the partial sums for $$n = 3$$ to 10 gives the data in Table [table 3.7].
Probability that no one n gets his own hat back 3 .333333 4 .375 5 .366667 6 .368056 7 .367857 8 .367882 9 .367879 10 .367879
After $$n = 9$$ the probabilities are essentially the same to six significant figures. Interestingly, the probability of no fixed point alternately increases and decreases as $$n$$ increases. Finally, we note that our exact results are in good agreement with our simulations reported in the previous section.
### Choosing a Sample Space
We now have some of the tools needed to accurately describe sample spaces and to assign probability functions to those sample spaces. Nevertheless, in some cases, the description and assignment process is somewhat arbitrary. Of course, it is to be hoped that the description of the sample space and the subsequent assignment of a probability function will yield a model which accurately predicts what would happen if the experiment were actually carried out. As the following examples show, there are situations in which “reasonable" descriptions of the sample space do not produce a model which fits the data.
In Feller’s book,14 a pair of models is given which describe arrangements of certain kinds of elementary particles, such as photons and protons. It turns out that experiments have shown that certain types of elementary particles exhibit behavior which is accurately described by one model, called while other types of elementary particles can be modelled using Feller says:
We have here an instructive example of the impossibility of selecting or justifying probability models by arguments. In fact, no pure reasoning could tell that photons and protons would not obey the same probability laws.
We now give some examples of this description and assignment process.
Example $$\PageIndex{13}$$:
In the quantum mechanical model of the helium atom, various parameters can be used to classify the energy states of the atom. In the triplet spin state ($$S = 1$$) with orbital angular momentum 1 ($$L = 1$$), there are three possibilities, 0, 1, or 2, for the total angular momentum ($$J$$). (It is not assumed that the reader knows what any of this means; in fact, the example is more illustrative if the reader does know anything about quantum mechanics.) We would like to assign probabilities to the three possibilities for $$J$$. The reader is undoubtedly resisting the idea of assigning the probability of $$1/3$$ to each of these outcomes. She should now ask herself why she is resisting this assignment. The answer is probably because she does not have any “intuition" (i.e., experience) about the way in which helium atoms behave. In fact, in this example, the probabilities $$1/9,\ 3/9,$$ and $$5/9$$ are assigned by the theory. The theory gives these assignments because these frequencies were observed and further parameters were developed in the theory to allow these frequencies to be predicted.
Example $$\PageIndex{14}$$:
Suppose two pennies are flipped once each. There are several “reasonable" ways to describe the sample space. One way is to count the number of heads in the outcome; in this case, the sample space can be written $$\{0, 1, 2\}$$. Another description of the sample space is the set of all ordered pairs of $$H$$’s and $$T$$’s, i.e., $\{(H,H), (H, T), (T, H), (T, T)\}.$ Both of these descriptions are accurate ones, but it is easy to see that (at most) one of these, if assigned a constant probability function, can claim to accurately model reality. In this case, as opposed to the preceding example, the reader will probably say that the second description, with each outcome being assigned a probability of $$1/4$$, is the “right" description. This conviction is due to experience; there is no proof that this is the way reality works.
The reader is also referred to Exercise 26 for another example of this process.
### Historical Remarks
The binomial coefficients have a long and colorful history leading up to Pascal’s Treatise on the Arithmetical Triangle15 where Pascal developed many important properties of these numbers. This history is set forth in the book by Pascal's Arithmetical Triangle by A. W. F. Edwards.16 Pascal wrote his triangle in the form shown in Table 3.10.
llrrrrrrll 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 1 & 3 & 6 & 10 & 15 & 21 & 28 & 36 1 & 4 & 10 & 20 & 35 & 56 & 84 1 & 5 & 15 & 35 & 70 & 126 1 & 6 & 21 & 56 &126 1 & 7 & 28 & 84 1 & 8 & 36 1 & 9 1
Edwards traces three different ways that the binomial coefficients arose. He refers to these as the figurate numbers, the combinatorial numbers and the binomial numbers. They are all names for the same thing (which we have called binomial coefficients) but that they are all the same was not appreciated until the sixteenth century.
The figurate numbers date back to the Pythagorean interest in number patterns around 540 . The Pythagoreans considered, for example, triangular patterns shown in Figure 3.8 The sequence of numbers $1, 3, 6, 10, \dots$ obtained as the number of points in each triangle are called triangular numbers. From the triangles it is clear that the $$n$$th triangular number is simply the sum of the first $$n$$ integers. The tetrahedral numbers are the sums of the triangular numbers and were obtained by the Greek mathematicians Theon and Nicomachus at the beginning of the second century . The tetrahedral number 10, for example, has the geometric representation shown in Figure 3.9. The first three types of figurate numbers can be represented in tabular form as shown in Table 3.11
$\begin{array}{llllllllll} \mbox {natural\ numbers} & 1\hskip.2in & 2\hskip.2in & 3\hskip.2in & 4\hskip.2in & 5\hskip.2in & 6\hskip.2in & 7\hskip.2in & 8\hskip.2in & 9 \cr \mbox {triangular\ numbers} & 1 & 3 & 6 & 10 & 15 & 21 & 28 & 36 & 45 \cr \mbox {tetrahedral\ numbers}& 1 & 4 & 10 & 20 & 35 & 56 & 84 & 120 &165 \end{array}$
These numbers provide the first four rows of Pascal’s triangle, but the table was not to be completed in the West until the sixteenth century.
In the East, Hindu mathematicians began to encounter the binomial coefficients in combinatorial problems. Bhaskara in his of 1150 gave a rule to find the number of medicinal preparations using 1, 2, 3, 4, 5, or 6 possible ingredients.17 His rule is equivalent to our formula ${n \choose r} = \frac{(n)_r}{r!}\ .$
The binomial numbers as coefficients of $$(a + b)^n$$ appeared in the works of mathematicians in China around 1100. There are references about this time to “the tabulation system for unlocking binomial coefficients." The triangle to provide the coefficients up to the eighth power is given by Chu Shih-chieh in a book written around 1303 (see Figure [fig 3.12]).18 The original manuscript of Chu’s book has been lost, but copies have survived. Edwards notes that there is an error in this copy of Chu’s triangle. Can you find it? (: Two numbers which should be equal are not.) Other copies do not show this error.
The first appearance of Pascal’s triangle in the West seems to have come from calculations of Tartaglia in calculating the number of possible ways that $$n$$ dice might turn up.19 For one die the answer is clearly 6. For two dice the possibilities may be displayed as shown in Table [table 3.10].
$\matrix{ 11\cr 12 & 22\cr 13 & 23 & 33\cr 14 & 24 & 34 & 44\cr 15 & 25 & 35 & 45 & 55\cr 16 & 26 & 36 & 46 & 56 & 66\ \cr }$
Displaying them this way suggests the sixth triangular number $$1 + 2 + 3 + 4 + 5 + 6 = 21$$ for the throw of 2 dice. Tartaglia “on the first day of Lent, 1523, in Verona, having thought about the problem all night,"20 realized that the extension of the figurate table gave the answers for $$n$$ dice. The problem had suggested itself to Tartaglia from watching people casting their own horoscopes by means of a Book of Fortune selecting verses by a process which included noting the numbers on the faces of three dice. The 56 ways that three dice can fall were set out on each page. The way the numbers were written in the book did not suggest the connection with figurate numbers, but a method of enumeration similar to the one we used for 2 dice does. Tartaglia’s table was not published until 1556.
A table for the binomial coefficients was published in 1554 by the German mathematician Stifel.21 Pascal’s triangle appears also in Cardano’s of 1570.22 Cardano was interested in the problem of finding the number of ways to choose $$r$$ objects out of $$n$$. Thus by the time of Pascal’s work, his triangle had appeared as a result of looking at the figurate numbers, the combinatorial numbers, and the binomial numbers, and the fact that all three were the same was presumably pretty well understood.
Pascal’s interest in the binomial numbers came from his letters with Fermat concerning a problem known as the problem of points. This problem, and the correspondence between Pascal and Fermat, were discussed in Chapter 1. The reader will recall that this problem can be described as follows: Two players A and B are playing a sequence of games and the first player to win $$n$$ games wins the match. It is desired to find the probability that A wins the match at a time when A has won $$a$$ games and B has won $$b$$ games. (See Exercises 4.1.40-4.1.42.)
Pascal solved the problem by backward induction, much the way we would do today in writing a computer program for its solution. He referred to the combinatorial method of Fermat which proceeds as follows: If A needs $$c$$ games and B needs $$d$$ games to win, we require that the players continue to play until they have played $$c + d - 1$$ games. The winner in this extended series will be the same as the winner in the original series. The probability that A wins in the extended series and hence in the original series is $\sum_{r = c}^{c + d - 1} \frac 1{2^{c + d - 1}} {{c + d - 1} \choose r}\ .$ Even at the time of the letters Pascal seemed to understand this formula.
Suppose that the first player to win $$n$$ games wins the match, and suppose that each player has put up a stake of $$x$$. Pascal studied the value of winning a particular game. By this he meant the increase in the expected winnings of the winner of the particular game under consideration. He showed that the value of the first game is $\frac {1\cdot3\cdot5\cdot\dots\cdot(2n - 1)}{2\cdot4\cdot6\cdot\dots\cdot(2n)}x\ .$ His proof of this seems to use Fermat’s formula and the fact that the above ratio of products of odd to products of even numbers is equal to the probability of exactly $$n$$ heads in $$2n$$ tosses of a coin. (See Exercise 39.)
Pascal presented Fermat with the table shown in Table 3.13
From my opponent’s 256 6 5 4 3 2 1 positions I get, for the games games games games games games 1st game 63 70 80 96 128 256 2nd game 63 70 80 96 128 3rd game 56 60 64 64 4th game 42 40 32 5th game 24 16 6th game
He states:
You will see as always, that the value of the first game is equal to that of the second which is easily shown by combinations. You will see, in the same way, that the numbers in the first line are always increasing; so also are those in the second; and those in the third. But those in the fourth line are decreasing, and those in the fifth, etc. This seems odd.23
The student can pursue this question further using the computer and Pascal’s backward iteration method for computing the expected payoff at any point in the series.
In his treatise, Pascal gave a formal proof of Fermat’s combinatorial formula as well as proofs of many other basic properties of binomial numbers. Many of his proofs involved induction and represent some of the first proofs by this method. His book brought together all the different aspects of the numbers in the Pascal triangle as known in 1654, and, as Edwards states, “That the Arithmetical Triangle should bear Pascal’s name cannot be disputed."24
The first serious study of the binomial distribution was undertaken by James Bernoulli in his published in 1713.25 We shall return to this work in the historical remarks in Chapter 8
###### Exercises
$$\PageIndex{1}$$ Compute the following:
1. $${6 \choose 3}$$
2. $$b(5,.2,4)$$
3. $${7 \choose 2}$$
4. $$ParseError: EOF expected (click for details) Callstack: at (Textbook_Maps/Probability_Theory/Book:_Introductory_Probability_(Grinstead_and_Snell)/3:_Combinatorics/3.2:_Combinations), /content/body/div[13]/div/ol[1]/li[4]/p/span/span, line 1, column 3$$
5. $$b(4,.2,3)$$
6. $${6 \choose 2}$$
7. $${{10} \choose 9}$$
8. $$b(8, .3, 5)$$
$$\PageIndex{2}$$ In how many ways can we choose five people from a group of ten to form a committee?
$$\PageIndex{3}$$ How many seven-element subsets are there in a set of nine elements?
$$\PageIndex{4}$$ Using the relation Equation 3.1 write a program to compute Pascal’s triangle, putting the results in a matrix. Have your program print the triangle for $$n = 10$$.
$$\PageIndex{5}$$ Use the program BinomialProbabilities to find the probability that, in 100 tosses of a fair coin, the number of heads that turns up lies between 35 and 65, between 40 and 60, and between 45 and 55.
$$\PageIndex{6}$$ Charles claims that he can distinguish between beer and ale 75 percent of the time. Ruth bets that he cannot and, in fact, just guesses. To settle this, a bet is made: Charles is to be given ten small glasses, each having been filled with beer or ale, chosen by tossing a fair coin. He wins the bet if he gets seven or more correct. Find the probability that Charles wins if he has the ability that he claims. Find the probability that Ruth wins if Charles is guessing.
$$\PageIndex{7}$$ Show that $b(n,p,j) = \frac pq \left(\frac {n - j + 1}j \right) b(n,p,j - 1)\ ,$ for $$j \ge 1$$. Use this fact to determine the value or values of $$j$$ which give $$b(n,p,j)$$ its greatest value. : Consider the successive ratios as $$j$$ increases.
$$\PageIndex{8}$$ A die is rolled 30 times. What is the probability that a 6 turns up exactly 5 times? What is the most probable number of times that a 6 will turn up?
$$\PageIndex{9}$$ Find integers $$n$$ and $$r$$ such that the following equation is true: ${13 \choose 5} + 2{13 \choose 6} + {13 \choose 7} = {n \choose r}\ .$
$$\PageIndex{10}$$ In a ten-question true-false exam, find the probability that a student gets a grade of 70 percent or better by guessing. Answer the same question if the test has 30 questions, and if the test has 50 questions.
$$\PageIndex{11}$$ A restaurant offers apple and blueberry pies and stocks an equal number of each kind of pie. Each day ten customers request pie. They choose, with equal probabilities, one of the two kinds of pie. How many pieces of each kind of pie should the owner provide so that the probability is about .95 that each customer gets the pie of his or her own choice?
$$\PageIndex{12}$$ A poker hand is a set of 5 cards randomly chosen from a deck of 52 cards. Find the probability of a
1. royal flush (ten, jack, queen, king, ace in a single suit).
2. straight flush (five in a sequence in a single suit, but not a royal flush).
3. four of a kind (four cards of the same face value).
4. full house (one pair and one triple, each of the same face value).
5. flush (five cards in a single suit but not a straight or royal flush).
6. straight (five cards in a sequence, not all the same suit). (Note that in straights, an ace counts high or low.)
$$\PageIndex{13}$$ If a set has $$2n$$ elements, show that it has more subsets with $$n$$ elements than with any other number of elements.
$$\PageIndex{14}$$ Let $$b(2n,.5,n)$$ be the probability that in $$2n$$ tosses of a fair coin exactly $$n$$ heads turn up. Using Stirling’s formula (Theorem 3.1), show that $$b(2n,.5,n) \sim 1/\sqrt{\pi n}$$. Use the program BinomialProbabilities to compare this with the exact value for $$n = 10$$ to 25.
$$\PageIndex{15}$$ A baseball player, Smith, has a batting average of $$.300$$ and in a typical game comes to bat three times. Assume that Smith’s hits in a game can be considered to be a Bernoulli trials process with probability .3 for Find the probability that Smith gets 0, 1, 2, and 3 hits.
$$\PageIndex{16}$$ The Siwash University football team plays eight games in a season, winning three, losing three, and ending two in a tie. Show that the number of ways that this can happen is ${8 \choose 3}{5 \choose 3} = \frac {8!}{3!\,3!\,2!}\ .$
$$\PageIndex{17}$$ Using the technique of Exercise 16, show that the number of ways that one can put $$n$$ different objects into three boxes with $$a$$ in the first, $$b$$ in the second, and $$c$$ in the third is $$n!/(a!\,b!\,c!)$$.
$$\PageIndex{18}$$ Baumgartner, Prosser, and Crowell are grading a calculus exam. There is a true-false question with ten parts. Baumgartner notices that one student has only two out of the ten correct and remarks, “The student was not even bright enough to have flipped a coin to determine his answers." “Not so clear," says Prosser. “With 340 students I bet that if they all flipped coins to determine their answers there would be at least one exam with two or fewer answers correct." Crowell says, “I’m with Prosser. In fact, I bet that we should expect at least one exam in which no answer is correct if everyone is just guessing." Who is right in all of this?
$$\PageIndex{19}$$ A gin hand consists of 10 cards from a deck of 52 cards. Find the probability that a gin hand has
1. all 10 cards of the same suit.
2. exactly 4 cards in one suit and 3 in two other suits.
3. a 4, 3, 2, 1, distribution of suits.
$$\PageIndex{20}$$ A six-card hand is dealt from an ordinary deck of cards. Find the probability that:
1. All six cards are hearts.
2. There are three aces, two kings, and one queen.
3. There are three cards of one suit and three of another suit.
$$\PageIndex{21}$$ A lady wishes to color her fingernails on one hand using at most two of the colors red, yellow, and blue. How many ways can she do this?
$$\PageIndex{22}$$ How many ways can six indistinguishable letters be put in three mail boxes? : One representation of this is given by a sequence $$|$$LL$$|$$L$$|$$LLL$$|$$ where the $$|$$’s represent the partitions for the boxes and the L’s the letters. Any possible way can be so described. Note that we need two bars at the ends and the remaining two bars and the six L’s can be put in any order.
$$\PageIndex{23}$$ Using the method for the hint in Exercise 22, show that $$r$$ indistinguishable objects can be put in $$n$$ boxes in $ParseError: EOF expected (click for details) Callstack: at (Textbook_Maps/Probability_Theory/Book:_Introductory_Probability_(Grinstead_and_Snell)/3:_Combinatorics/3.2:_Combinations), /content/body/div[13]/div/p[23]/span[3]/span, line 1, column 10 = {{n + r - 1} \choose r}$ different ways.
$$\PageIndex{24}$$ A travel bureau estimates that when 20 tourists go to a resort with ten hotels they distribute themselves as if the bureau were putting 20 indistinguishable objects into ten distinguishable boxes. Assuming this model is correct, find the probability that no hotel is left vacant when the first group of 20 tourists arrives.
$$\PageIndex{25}$$ An elevator takes on six passengers and stops at ten floors. We can assign two different equiprobable measures for the ways that the passengers are discharged: (a) we consider the passengers to be distinguishable or (b) we consider them to be indistinguishable (see Exercise 23 for this case). For each case, calculate the probability that all the passengers get off at different floors.
$$\PageIndex{26}$$ You are playing with Prosser but you suspect that his coin is unfair. Von Neumann suggested that you proceed as follows: Toss Prosser’s coin twice. If the outcome is HT call the result if it is TH call the result If it is TT or HH ignore the outcome and toss Prosser’s coin twice again. Keep going until you get either an HT or a TH and call the result win or lose in a single play. Repeat this procedure for each play. Assume that Prosser’s coin turns up heads with probability $$p$$.
1. Find the probability of HT, TH, HH, TT with two tosses of Prosser’s coin.
2. Using part (a), show that the probability of a win on any one play is 1/2, no matter what $$p$$ is.
$$\PageIndex{27}$$ John claims that he has extrasensory powers and can tell which of two symbols is on a card turned face down (see Example 12). To test his ability he is asked to do this for a sequence of trials. Let the null hypothesis be that he is just guessing, so that the probability is 1/2 of his getting it right each time, and let the alternative hypothesis be that he can name the symbol correctly more than half the time. Devise a test with the property that the probability of a type 1 error is less than .05 and the probability of a type 2 error is less than .05 if John can name the symbol correctly 75 percent of the time.
$$\PageIndex{28}$$ In Example $$\PageIndex{12}$$ assume the alternative hypothesis is that $$p = .8$$ and that it is desired to have the probability of each type of error less than .01. Use the program PowerCurve to determine values of $$n$$ and $$m$$ that will achieve this. Choose $$n$$ as small as possible.
$$\PageIndex{29}$$ A drug is assumed to be effective with an unknown probability $$p$$. To estimate $$p$$ the drug is given to $$n$$ patients. It is found to be effective for $$m$$ patients. The for estimating $$p$$ states that we should choose the value for $$p$$ that gives the highest probability of getting what we got on the experiment. Assuming that the experiment can be considered as a Bernoulli trials process with probability $$p$$ for success, show that the maximum likelihood estimate for $$p$$ is the proportion $$m/n$$ of successes.
$$\PageIndex{30}$$ Recall that in the World Series the first team to win four games wins the series. The series can go at most seven games. Assume that the Red Sox and the Mets are playing the series. Assume that the Mets win each game with probability $$p$$. Fermat observed that even though the series might not go seven games, the probability that the Mets win the series is the same as the probability that they win four or more game in a series that was forced to go seven games no matter who wins the individual games.
1. Using the program PowerCurve of Example 3.11 find the probability that the Mets win the series for the cases $$p = .5$$, $$p = .6$$, $$p =.7$$.
2. Assume that the Mets have probability .6 of winning each game. Use the program PowerCurve to find a value of $$n$$ so that, if the series goes to the first team to win more than half the games, the Mets will have a 95 percent chance of winning the series. Choose $$n$$ as small as possible.
$$\PageIndex{31}$$ ] Each of the four engines on an airplane functions correctly on a given flight with probability .99, and the engines function independently of each other. Assume that the plane can make a safe landing if at least two of its engines are functioning correctly. What is the probability that the engines will allow for a safe landing?
$$\PageIndex{32}$$ A small boy is lost coming down Mount Washington. The leader of the search team estimates that there is a probability $$p$$ that he came down on the east side and a probability $$1 - p$$ that he came down on the west side. He has $$n$$ people in his search team who will search independently and, if the boy is on the side being searched, each member will find the boy with probability $$u$$. Determine how he should divide the $$n$$ people into two groups to search the two sides of the mountain so that he will have the highest probability of finding the boy. How does this depend on $$u$$?
$$*\PageIndex{33}$$ $$2n$$ balls are chosen at random from a total of $$2n$$ red balls and $$2n$$ blue balls. Find a combinatorial expression for the probability that the chosen balls are equally divided in color. Use Stirling’s formula to estimate this probability. Using BinomialProbabilities, compare the exact value with Stirling’s approximation for $$n = 20$$.
$$\PageIndex{34}$$ Assume that every time you buy a box of Wheaties, you receive one of the pictures of the $$n$$ players on the New York Yankees. Over a period of time, you buy $$m \geq n$$ boxes of Wheaties.
1. Use Theorem 3.8 to show that the probability that you get all $$n$$ pictures is \begin{aligned} 1 &-& {n \choose 1} \left(\frac{n - 1}n\right)^m + {n \choose 2} \left(\frac{n - 2}n\right)^m - \cdots \\ &+& (-1)^{n - 1} {n \choose {n - 1}}\left(\frac 1n \right)^m.\end{aligned} : Let $$E_k$$ be the event that you do not get the $$k$$th player’s picture.
2. Write a computer program to compute this probability. Use this program to find, for given $$n$$, the smallest value of $$m$$ which will give probability $$\geq .5$$ of getting all $$n$$ pictures. Consider $$n = 50$$, 100, and 150 and show that $$m = n\log n + n \log 2$$ is a good estimate for the number of boxes needed. (For a derivation of this estimate, see Feller.26)
$$\PageIndex{35}$$ Prove the following binomial identity
${2n \choose n} = \sum_{j = 0}^n { n \choose j}^2\ .$ : Consider an urn with $$n$$ red balls and $$n$$ blue balls inside. Show that each side of the equation equals the number of ways to choose $$n$$ balls from the urn.
$$\PageIndex{36}$$ Let $$j$$ and $$n$$ be positive integers, with $$j \le n$$. An experiment consists of choosing, at random, a $$j$$-tuple of integers whose sum is at most $$n$$.
1. Find the size of the sample space. : Consider $$n$$ indistinguishable balls placed in a row. Place $$j$$ markers between consecutive pairs of balls, with no two markers between the same pair of balls. (We also allow one of the $$n$$ markers to be placed at the end of the row of balls.) Show that there is a 1-1 correspondence between the set of possible positions for the markers and the set of $$j$$-tuples whose size we are trying to count.
2. Find the probability that the $$j$$-tuple selected contains at least one 1.
$$\PageIndex{37}$$ Let $$n\ (\mbox{mod}\ m)$$ denote the remainder when the integer $$n$$ is divided by the integer $$m$$. Write a computer program to compute the numbers $${n \choose j}\ (\mbox{mod}\ m)$$ where $${n \choose j}$$ is a binomial coefficient and $$m$$ is an integer. You can do this by using the recursion relations for generating binomial coefficients, doing all the arithmetic using the basic function mod($$n,m$$). Try to write your program to make as large a table as possible. Run your program for the cases $$m = 2$$ to 7. Do you see any patterns? In particular, for the case $$m = 2$$ and $$n$$ a power of 2, verify that all the entries in the $$(n - 1)$$st row are 1. (The corresponding binomial numbers are odd.) Use your pictures to explain why this is true.
$$\PageIndex{38}$$ Lucas27 proved the following general result relating to Exercise 37. If $$p$$ is any prime number, then $${n \choose j}~ (\mbox{mod\ }p)$$ can be found as follows: Expand $$n$$ and $$j$$ in base $$p$$ as $$n = s_0 + s_1p + s_2p^2 + \cdots + s_kp^k$$ and $$j = r_0 + r_1p + r_2p^2 + \cdots + r_kp^k$$, respectively. (Here $$k$$ is chosen large enough to represent all numbers from 0 to $$n$$ in base $$p$$ using $$k$$ digits.) Let $$s = (s_0,s_1,s_2,\dots,s_k)$$ and $$r = (r_0,r_1,r_2,\dots,r_k)$$. Then ${n \choose j}~(\mbox{mod\ }p) = \prod_{i = 0}^k ParseError: EOF expected (click for details) Callstack: at (Textbook_Maps/Probability_Theory/Book:_Introductory_Probability_(Grinstead_and_Snell)/3:_Combinatorics/3.2:_Combinations), /content/body/div[13]/div/p[39]/span[14]/span, line 1, column 4 ~(\mbox{mod\ }p)\ .$ For example, if $$p = 7$$, $$n = 12$$, and $$j = 9$$, then
\begin{aligned} 12 & = & 5 \cdot 7^0 + 1 \cdot 7^1\ , \\ 9 & = & 2 \cdot 7^0 + 1 \cdot 7^1\ , \end{aligned} so that \begin{aligned} s & = & (5, 1)\ , \\ r & = & (2, 1)\ , \end{aligned} and this result states that ${12 \choose 9}~(\mbox{mod\ }p) = {5 \choose 2} {1 \choose 1}~(\mbox{mod\ }7)\ .$ Since $${12 \choose 9} = 220 = 3~(\mbox{mod\ }7)$$, and $${5 \choose 2} = 10 = 3~ (\mbox{mod\ }7)$$, we see that the result is correct for this example.
Show that this result implies that, for $$p = 2$$, the $$(p^k - 1)$$st row of your triangle in Exercise 37 has no zeros.
$$\PageIndex{39}$$ Prove that the probability of exactly $$n$$ heads in $$2n$$ tosses of a fair coin is given by the product of the odd numbers up to $$2n - 1$$ divided by the product of the even numbers up to $$2n$$.
$$\PageIndex{40}$$ Let $$n$$ be a positive integer, and assume that $$j$$ is a positive integer not exceeding $$n/2$$. Show that in Theorem 3.5, if one alternates the multiplications and divisions, then all of the intermediate values in the calculation are integers. Show also that none of these intermediate values exceed the final value.
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# NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.6
These NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.6 Questions and Answers are prepared by our highly skilled subject experts.
## NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Exercise 2.6
Solve the following equations.
Question 1.
$$\frac{8 x-3}{3 x}=2$$
Solution:
$$\frac{8 x-3}{3 x}=2$$
By cross-multiplication we get,
8x – 3 = 2(3x)
8x – 3 = 6x
By transposing -3 to the R.H.S. and 6x to L.H.S.
8x – 6x = 3
2x = 3
Dividing both sides by 2
$$\frac{2 x}{2}=\frac{3}{2}$$
∴ x = $$\frac{3}{2}$$
Question 2.
$$\frac{9 x}{7-6 x}=15$$
Solution:
$$\frac{9 x}{7-6 x}=15$$
By cross-multiplication, we get
9x = 15(7 – 6x)
9x = 105 – 90x
By transposing -90x to L.H.S., we get
9x + 90x = 105
99x = 105
Dividing both sides by 99, we get
$$\frac{99 \mathrm{x}}{99}=\frac{105}{99}$$
∴ x = $$\frac{35}{33}$$
Question 3.
$$\frac{z}{z+15}=\frac{4}{9}$$
Solution:
$$\frac{z}{z+15}=\frac{4}{9}$$
By cross-multiplication, we get
9z = 4(z + 15)
9z = 4z + 60
Transposing 4z to L.H.S, we get
9z – 4z = 60
5z = 60
Dividing both sides by 5, we get
$$\frac{5 z}{5}=\frac{60}{5}$$
∴ z = 12
Question 4.
$$\frac{3 y+4}{2-6 y}=\frac{-2}{5}$$
Solution:
$$\frac{3 y+4}{2-6 y}=\frac{-2}{5}$$
By cross-multiplication, we get
5(3y + 4) = -2(2 – 6y)
15y + 20 = -4 + 12y
Transposing 12y to L.H.S. and 20 to R.H.S.
15y – 12y = -4 – 20
3y = -24
Dividing both sides by 3, we get
$$\frac{3 y}{3}=-\frac{24}{3}$$
∴ y = -8
Question 5.
$$\frac{7 y+4}{y+2}=\frac{-4}{3}$$
Solution:
$$\frac{7 y+4}{y+2}=\frac{-4}{3}$$
By cross-multiplication, we get
3(7y + 4) = -4(y + 2)
21y + 12 = -4y – 8
Transposing +12 to R.H.S. and -4y to L.H.S.
21y + 4y = -8 – 12
25y = -20
Dividing both sides by 25, we get
$$\frac{25 y}{25}=\frac{-20}{25}$$
∴ y = $$-\frac{4}{5}$$
Question 6.
The ages of Hari and Harry are in the ratio 5 : 7. Four years from now the ratio of their ages will be 3 : 4. Find their present ages.
Solution:
Let the present age of Hari = 5x years
and the present age of Harry = 7x years
After four years
Age of Hari = (5x + 4) years
Age of Harry = (7x + 4) years
According to the question
(5x + 4) : (7x + 4) = 3 : 4
4(5x + 4) = 3(7x + 4)
[Product of the extremes is equal to the product of the means]
20x + 16 = 21x + 12
Transposing 16 to R.H.S. and 21x to L.H.S., we get
20x – 21x = 12 – 16
-x = -4
∴ x = 4
Present age of Hari = 5 × 4 = 20 years
Present age of Harry = 7 × 4 = 28 years
Question 7.
The denominator of a rational number is greater than its numerator by 8. If the numerator is increased by 17 and the denominator is decreased by 1, the number obtained is $$\frac{3}{2}$$. Find the rational number.
Solution:
Let the numerator be ‘x’
then the denominator is x + 8
∴ The fraction is $$\frac{\mathrm{x}}{\mathrm{x}+8}$$
According to the question, we get
$$\frac{x+17}{(x+8)-1}=\frac{3}{2}$$
$$\frac{x+17}{x+8-1}=\frac{3}{2}$$
$$\frac{x+17}{x+7}=\frac{3}{2}$$
By cross-multiplication, we get
3(x + 7) = 2(x + 17)
3x + 21 = 2x + 34
By transposing 21 to R.H.S. and 2x to L.H.S., we get
3x – 2x = 34 – 21
x = 13
∴ The fraction is $$\frac{13}{13+8}$$ = $$\frac{13}{21}$$
or The rational number = $$\frac{13}{21}$$
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# If tan (cot x) = cot (tan x), then sin 2x is equal to (a) $\dfrac{2}{\left( 2n+1 \right)\pi }$(b) $\dfrac{4}{\left( 2n+1 \right)\pi }$(c) $\dfrac{2}{n\left( n+1 \right)\pi }$(d) $\dfrac{4}{n\left( n+1 \right)\pi }$
Last updated date: 20th Jun 2024
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Hint: First of all, consider the given equation and use $\cot \theta =\tan \left( \dfrac{\pi }{2}-\theta \right)$ to convert cot (tan x) into $\tan \left( \dfrac{\pi }{2}-\tan x \right)$. Now, compare LHS and RHS and get $\tan x+\cot x=n\pi +\dfrac{\pi }{2}$. Now, use $\tan \theta =\dfrac{1}{\cot \theta }=\dfrac{\sin \theta }{\cos \theta }$ and simplify it to get the value of $\sin 2x=2\sin x\cos x$.
We are given that tan (cot x) = cot (tan x). Then, we have to find the value of sin 2x. Let us consider the equation given in the question.
$\tan \left( \cot x \right)=\cot \left( \tan x \right)$
We know that $\cot \theta =\tan \left( \dfrac{\pi }{2}-\theta \right)$.
Now, on RHS we can write cot ( tanx ) as $\tan \left( \dfrac{\pi }{2}-\tan x \right)$
$\tan \left( \cot x \right)=\tan \left( \dfrac{\pi }{2}-\tan x \right)$
By comparing RHS and LHS of the above equation, we get,
$\cot x=\dfrac{\pi }{2}-\tan x$
$\tan x+\cot x=\dfrac{\pi }{2}$
In general form, we can write the above equation as,
$\tan x+\cot x=n\pi +\dfrac{\pi }{2}$
We know that $\tan \theta =\dfrac{\sin \theta }{\cos \theta }\text{ and }\cot \theta =\dfrac{\cos \theta }{\sin \theta }$. By using this in the above equation, we get,
$\dfrac{\sin x}{\cos x}+\dfrac{\cos x}{\sin x}=n\pi +\dfrac{\pi }{2}$
By taking sin x cos x as LCM and simplifying the above equation, we get,
$\dfrac{{{\sin }^{2}}x+{{\cos }^{2}}x}{\cos x\sin x}=n\pi +\dfrac{\pi }{2}$
We know that ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$. By using this, in the above equation, we get,
$\dfrac{1}{\cos x\sin x}=n\pi +\dfrac{\pi }{2}$
$\dfrac{1}{\cos x\sin x}=\pi \left( \dfrac{2n+1}{2} \right)$
By cross multiplying the above equation, we get,
$\dfrac{2}{\pi \left( 2n+1 \right)}=\sin x\cos x$
By multiplying 2 on both the sides of the above equation, we get,
$\dfrac{4}{\pi \left( 2n+1 \right)}=2\sin x\cos x$
We know that $2\sin \theta \cos \theta =\sin 2\theta$. By using this in the above equation, we get,
$\dfrac{4}{\pi \left( 2n+1 \right)}=\sin 2x$
So, we have got the values of sin 2x as $\dfrac{4}{\pi \left( 2n+1 \right)}$
So, the correct answer is “Option B”.
Note: In this question, students can cross-check their answers by substituting the same value of x and from that finding the value of x. If that value of x satisfies the given equation, then our answer is correct. Also, in this question, some students try to find the value of sin x and cos x individually and then use them to find the value of sin 2x which is not required as we can get sin x cos x easily without calculating each of them individually like in the above solution.
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# Intuition Behind Polynomial Numerators in Partial Fractions
by Justin Skycak on
Each decomposition produces a system of linear equations where the number of unknowns equals the number of equations.
Consider the following partial fractions decomposition:
\begin{align*} \dfrac{7x^2-28x+17}{(x^2+1)(x-5)} = \dfrac{Ax+B}{x^2+1} + \dfrac{C}{x-5} \end{align*}
You might wonder, why does the $\dfrac{Ax+B}{x^2+1}$ term require a linear numerator instead of just a constant? Yes, we have to do that whenever the denominator is an irreducible quadratic… but why?
Partial fractions formulas are typically not proven in calc textbooks since the proof is rather complicated.
However, there is a hand-wavy intuitive way that I like to think about it. Each decomposition produces a system of linear equations where the number of unknowns equals the number of equations.
\begin{align*} \dfrac{7x^2-28x+17}{(x^2+1)(x-5)} &= \dfrac{Ax+B}{x^2+1} + \dfrac{C}{x-5} \\[5pt] 7x^2-28x+17 &= (Ax+B)(x-5) + C(x^2+1) \\[5pt] 7x^2-28x+17 &= (Ax^2-5Ax+Bx-5B) + (Cx^2+C) \\[5pt] 7x^2-28x+17 &= (A+C)x^2+(-5A+B)x+(-5B+C) \end{align*}
Here’s the system of linear equations:
\begin{align*} \begin{cases} 7 = A+C \\[5pt] -28 = -5A+B \\[5pt] 17 = -5B + C \end{cases} \end{align*}
Having the number of unknowns equal the number of equations is generally a “good” thing when solving equations. It makes a solution more likely to exist.
For instance, if you construct two linear equations at random, say $A+B=3$ and $A+2B=4,$ then a solution exists. (The only way a solution wouldn’t exist is if one of the equations happens to be “redundant,” introducing no new information – but it turns out the partial fractions setup also circumvents that possibility.)
If you do that same experiment with just one variable, then you’re basically doomed to fail unless the equations mean the same thing. The system $A=3$ and $A=4$ has no solution.
Now, back to our partial fractions decomposition – if you used a constant numerator on the quadratic fraction, $\dfrac{A}{x^2+1}$ instead of $\dfrac{Ax+B}{x^2+1},$ then you’d end up with a system of 3 equations and 2 unknowns, which would likely end up being unsolvable.
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# Question: What Is The Least Square Method Used For?
## How do you find the trend value in least square method?
Measurements of Trends: Method of Least Squares(i) The sum of the deviations of the actual values of Y and Ŷ (estimated value of Y) is Zero.
Computation of trend values by the method of least squares (ODD Years).Therefore, the required equation of the straight line trend is given by.Y = a+bX;Y = 45.143 + 1.036 (x-2003)The trend values can be obtained by.More items…•May 3, 2019.
## What are the properties of least square estimators?
(a) The least squares estimate is unbiased: E[ˆβ] = β. (b) The covariance matrix of the least squares estimate is cov(ˆβ) = σ2(X X)−1. 6.3 Theorem: Let rank(X) = r
## How do you find the least squares line?
StepsStep 1: For each (x,y) point calculate x2 and xy.Step 2: Sum all x, y, x2 and xy, which gives us Σx, Σy, Σx2 and Σxy (Σ means “sum up”)Step 3: Calculate Slope m:m = N Σ(xy) − Σx Σy N Σ(x2) − (Σx)2Step 4: Calculate Intercept b:b = Σy − m Σx N.Step 5: Assemble the equation of a line.
## What is a least square solution?
So a least-squares solution minimizes the sum of the squares of the differences between the entries of A K x and b . In other words, a least-squares solution solves the equation Ax = b as closely as possible, in the sense that the sum of the squares of the difference b − Ax is minimized.
## How r square is calculated?
To calculate the total variance, you would subtract the average actual value from each of the actual values, square the results and sum them. From there, divide the first sum of errors (explained variance) by the second sum (total variance), subtract the result from one, and you have the R-squared.
## What are the uses of time series?
Time Series Analysis is used for many applications such as:Economic Forecasting.Sales Forecasting.Budgetary Analysis.Stock Market Analysis.Yield Projections.Process and Quality Control.Inventory Studies.Workload Projections.More items…
## What is the least squares method and how is it used to find the estimated regression equation What is the role of least squares method in calculating coefficient of determination explain?
The least squares approach limits the distance between a function and the data points that the function explains. It is used in regression analysis, often in nonlinear regression modeling in which a curve is fit into a set of data. Mathematicians use the least squares method to arrive at a maximum-likelihood estimate.
## What is least square method in time series?
Least Square is the method for finding the best fit of a set of data points. It minimizes the sum of the residuals of points from the plotted curve. It gives the trend line of best fit to a time series data. This method is most widely used in time series analysis.
## How do you interpret a correlation coefficient?
Degree of correlation:Perfect: If the value is near ± 1, then it said to be a perfect correlation: as one variable increases, the other variable tends to also increase (if positive) or decrease (if negative).High degree: If the coefficient value lies between ± 0.50 and ± 1, then it is said to be a strong correlation.More items…
## How do you do least squares on Excel?
Constructing a Least-Squares Graph Using Microsoft ExcelEnter your data into the spreadsheet. … Select (highlight) the data that you want to include in the graph. … Click on Insert on the menu bar.Click on Chart….Under Standard Types, Chart type:, click on XY (Scatter).Under Chart sub-type:, click on the chart with only data markers and no lines.Click on Next>.More items…
## What is least square method formula?
The method of least squares assumes that the best fit curve of a given type is the curve that has the minimal sum of deviations, i.e., least square error from a given set of data. According to the method of least squares, the best fitting curve has the property that ∑ 1 n e i 2 = ∑ 1 n [ y i − f ( x i ) ] 2 is minimum.
## What is the principle of least squares?
MELDRUM SIEWART HE ” Principle of Least Squares” states that the most probable values of a system of unknown quantities upon which observations have been made, are obtained by making the sum of the squares of the errors a minimum.
## What is the main drawback of least square method?
The main disadvantages of linear least squares are limitations in the shapes that linear models can assume over long ranges, possibly poor extrapolation properties, and sensitivity to outliers.
## What does R 2 tell you?
R-squared is a statistical measure of how close the data are to the fitted regression line. It is also known as the coefficient of determination, or the coefficient of multiple determination for multiple regression. 0% indicates that the model explains none of the variability of the response data around its mean.
## What are least square means?
Least Squares Means can be defined as a linear combination (sum) of the estimated effects (means, etc) from a linear model. These means are based on the model used. In the case where the data contains NO missing values, the results of the MEANS and LSMEANS statements are identical.
## What is least square regression line?
What is a Least Squares Regression Line? … The Least Squares Regression Line is the line that makes the vertical distance from the data points to the regression line as small as possible. It’s called a “least squares” because the best line of fit is one that minimizes the variance (the sum of squares of the errors).
## What is the difference between least squares and linear regression?
2 Answers. Linear regression assumes a linear relationship between the independent and dependent variable. It doesn’t tell you how the model is fitted. Least square fitting is simply one of the possibilities.
## What are the four components of time series?
These four components are:Secular trend, which describe the movement along the term;Seasonal variations, which represent seasonal changes;Cyclical fluctuations, which correspond to periodical but not seasonal variations;Irregular variations, which are other nonrandom sources of variations of series.
## What are the advantages of least square method?
Non-linear least squares provides an alternative to maximum likelihood. The advantages of this method are: Non-linear least squares software may be available in many statistical software packages that do not support maximum likelihood estimates.
## Which of the following value must always be positive in the principle of least squares?
6. Which of the following value must always be positive in the principle of least squares? Explanation: The term must always be positive because it eliminates the possible error which will occur during the process of recording values. It can also be believed that the presence of square can make it positive.
## Which method show the line of best fit?
A line of best fit can be roughly determined using an eyeball method by drawing a straight line on a scatter plot so that the number of points above the line and below the line is about equal (and the line passes through as many points as possible).
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# Factors Of 120
Factors of 120 are the numbers, which gives the result as 120 when multiplied together in a pair. There are many factors that are commonly used in mathematical calculations such as factors of 56, 90, etc. Prime factors of number 120 basically give out prime numbers. To find the factors of a number, 120, we will use the division method. Here we will find the factors in pair, total factors and the prime factorization of 120.
## What are the Factors of 120?
The factors of 120 are the numbers that divide the number 120 exactly without leaving any remainder. In other words, the numbers which on multiplication in pairs resulting in the original number 120, are called the factors of 120. As 120 is a composite number, it has more than two factors. The factors of 120 are 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60 and 120.
Factors of 120: 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60 and 120 Prime Factorization of 120: 23×3×5
## Pair Factors of 120
We can find the pair factors of a number 120, by multiplying two numbers in a pair to get the original number. The pair factors of 120 can be positive or negative. The following are the positive and negative pair factors of 120.
Positive Factors of 120 Positive Pair Factors of 120 1 × 120 (1, 120) 2 × 60 (2, 60) 3 × 40 (3, 40) 4 × 30 (4, 30) 5 × 24 (5, 24) 6 × 20 (6, 20) 8 × 15 (8, 15) 10 × 12 (10, 12)
Likewise, the negative pair factors of 120 are:
Negative Factors of 120 Negative Pair Factors of 120 -1 × -120 (-1, -120) -2 × -60 (-2, -60) -3 × -40 (-3, -40) -4 × -30 (-4, -30) -5 × -24 (-5, -24) -6 × -20 (-6, -20) -8 × -15 (-8, -15) -10 × -12 (-10, -12)
## Prime Factorization of 120
120 is a composite number. Now let us find the prime factors of 120.
• The first step is to divide the number 120 with the smallest prime factor, i.e. 2 and continue dividing by 2 until you get a fraction.
120 ÷ 2 = 60
60 ÷ 2 = 30
30 ÷ 2 = 15
15 ÷ 2 = 7.5; 7.5 cannot be a factor
• Now, proceed to the next prime numbers, i.e. 3 and continue dividing till you get a fraction or 1.
15 ÷ 3 = 5
5 ÷ 3 = 1.66; cannot be a factor
• Therefore, moving to the next prime number, 5.
5 ÷ 5 = 1
We have received 1 at the end of the division process and further, we cannot proceed. So, the prime factors of 120 are 2 × 2 × 2 × 3 × 5 or 23 × 3 × 5, where 2, 3 and 5 are the prime numbers.
Links Related to Factors Factors of 15 Factors of 36 Factors of 48 Factors of 18 Factors of 42 Factors of 60 Factors of 35 Factors of 72 Factors of 84 Factors of 50
### Examples
Example1:
Find the common factors of 120 and 121.
Solution:
The factors of 120 are 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60 and 120.
The factors of 121 are 1, 11 and 121.
Therefore, the common factor of 120 and 121 is 1.
Example 2:
Find the common factors of 120 and 119.
Solution:
Factors of 120 = 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60 and 120
Factors of 119 = 1, 7, 17, 119.
Therefore, the common factor of 120 and 119 is 1.
Example 3:
Find the common factors of 120 and 60.
Solution:
The factors of 120 are 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60 and 120
The factors of 60 are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, and 60.
Hence, the common factors of 120 and 60 are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, and 60.
Learn factors and prime factors in Maths here with us and also download BYJU’S – The Learning App for interactive videos.
## Frequently Asked Questions on Factors of 120
### What are the factors of 120?
The factors of 120 are the numbers which are multiplied in pairs, resulting in the original number 120. The factors of 120 are 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60 and 120.
### What is the prime factorization of 120?
The prime factorization of 120 is 2 × 2 × 2 × 3 × 5 or 23 × 3 × 5.
### What are the positive pair factors of 120?
The positive pair factors of 120 are (1, 120), (2, 60), (3, 40), (4, 30), (5, 24), (6, 20), (8, 15), and (10, 12).
### Write down the negative pair factors of 120?
The negative pair factors of 120 are (-1, -120), (-2, -60), (-3, -40), (-4, -30), (-5, -24), (-6, -20), (-8, -15), and (-10, -12).
### Is 60 a factor of 120?
Yes, 60 is a factor of 120. As the number 60 divides 120 exactly without leaving any remainder, 60 is a factor of 120.
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Find The cube roots of the numbers 3048625, 20346417, 210644875, 57066625 using the fact that
Question:
Find The cube roots of the numbers 3048625, 20346417, 210644875, 57066625 using the fact that
(i) 3048625 = 3375 × 729
(ii) 20346417 = 9261 × 2197
(iii) 210644875 = 42875 × 4913
(iv) 57066625 = 166375 × 343
Solution:
(i)
To find the cube root, we use the following property:
$\sqrt[3]{a b}=\sqrt[3]{a} \times \sqrt[3]{b}$ for two integers $a$ and $b$
now
$\sqrt[3]{3048625}$
$=\sqrt[3]{3375} \times 729$
$=\sqrt[3]{3375} \times \sqrt[3]{729}$ (By the above property)
$=\sqrt[3]{3 \times 3 \times 3 \times 5 \times 5 \times 5} \times \sqrt[3]{9 \times 9 \times 9} \quad$ (By prime factorisation)
$=\sqrt[3]{\{3 \times 3 \times 3\} \times\{5 \times 5 \times 5\}} \times \sqrt[3]{\{9 \times 9 \times 9\}}$
$=3 \times 5 \times 9$
$=3 \times 5 \times 9$
$=135$
(ii)
To find the cube root, we use the following property:
$\sqrt[3]{a b}=\sqrt[3]{a} \times \sqrt[3]{b}$ for two integers $a$ and $b$
Now,
$\sqrt[3]{20346417}$
$=\sqrt[3]{9261} \times 2197$
$=\sqrt[3]{9261} \times \sqrt[3]{2197}$ (By the above property)
$=\sqrt[3]{3 \times 3 \times 3 \times 7 \times 7 \times 7} \times \sqrt[3]{13 \times 13 \times 13}$ (By prime factorisation)
$=\sqrt[3]{\{3 \times 3 \times 3\} \times\{7 \times 7 \times 7\}} \times \sqrt[3]{\{13 \times 13 \times 13\}}$
$=3 \times 7 \times 13$
$=273$
(iii)
To find the cube root, we use the following property:
$\sqrt[3]{a b}=\sqrt[3]{a} \times \sqrt[3]{b}$ for two integers $a$ and $b$
Now
$\sqrt[3]{210644875}$
$=\sqrt[3]{42875} \times 4913$
$=\sqrt[3]{42875} \times \sqrt[3]{4913}$ (By the above property)
$=\sqrt[3]{5 \times 5 \times 5 \times 7 \times 7 \times 7} \times \sqrt[3]{17 \times 17 \times 17}$ (By prime factorisation)
$=\sqrt[3]{\{5 \times 5 \times 5\} \times\{7 \times 7 \times 7\}} \times \sqrt[3]{\{17 \times 17 \times 17\}}$
$=5 \times 7 \times 17$
$=595$
(iv)
To find the cube root, we use the following property:
$\sqrt[3]{a b}=\sqrt[3]{a} \times \sqrt[3]{b}$ for two integers $a$ and $b$
now
$\sqrt[3]{57066625}$
$=\sqrt[3]{166375} \times 343$
$=\sqrt[3]{166375} \times \sqrt[3]{343}$ (By the above property)
$=\sqrt[3]{5 \times 5 \times 5 \times 11 \times 11 \times 11} \times \sqrt[3]{7 \times 7 \times 7} \quad$ (By prime factorisation)
$=\sqrt[3]{\{5 \times 5 \times 5\} \times\{11 \times 11 \times 11\} \times \sqrt[3]{\{7 \times 7 \times 7\}}}$
$=5 \times 11 \times 7$
$=385$
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# How many significant digits are contained in the number 1804.00?
Jul 30, 2016
$6$
#### Explanation:
The first thing to notice here is that your number contains three non-zero digits, $1$, $8$, and $4$. As you know, all non-zero digits are significant.
$\textcolor{b l u e}{18} 0 \textcolor{b l u e}{4} .00 \to$ $3 \textcolor{w h i t e}{a} \textcolor{b l u e}{\text{non-zero digits}}$
Next, notice that your number contains a sandwiched zero, which is a zero digit that follows and is followed by a non-zero digit. Sandwiched zeros are significant.
In your case, you have a zero that is sandwiched between $8$ and $4$, both non-zero digits.
$\textcolor{b l u e}{18} \textcolor{red}{0} \textcolor{b l u e}{4} .00 \to \left\{\left(3 \textcolor{w h i t e}{a} \textcolor{b l u e}{\text{non-zero digits")), (1color(white)(a)color(red)("sandwiched zero}}\right)\right.$
Finally, notice that your number also contains two trailing zeros. A trailing zero is zero that follows a decimal point or a non-zero digit and is not followed by a non-zero digit.
Trailing zeros that follow the decimal point are always significant.
In your case, you have
$\textcolor{b l u e}{18} \textcolor{red}{0} \textcolor{b l u e}{4} . \textcolor{\mathrm{da} r k g r e e n}{00} \to \left\{\left(3 \textcolor{w h i t e}{a} \textcolor{b l u e}{\text{non-zero digits")), (1color(white)(a)color(red)("sandwiched zero")), (2color(white)(a)color(darkgreen)("trailing zeros}}\right)\right.$
Therefore, your number has a total of
$1804.00 \to 6$ significant figures.
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# 180 Days of Math for Third Grade Day 51 Answers Key
By accessing our 180 Days of Math for Third Grade Answers Key Day 51 regularly, students can get better problem-solving skills.
## 180 Days of Math for Third Grade Answers Key Day 51
Directions Solve each problem.
Question 1.
22 – 11 =
Explanation:
The difference of 22 and 11 is 11
22 – 11 = 11
So, the missing number is .
Question 2.
4 × 3 =
4 x 3 =
Explanation:
The product of 4 and 3 is 12
So, 4 x 3 = 12.
Question 3.
Explanation:
The product of 12 and 2 is 24
So, 12 x 2 = 24.
Question 4.
How many chairs will you have if you have a total of 20 chair legs?
_____________
5 chairs
Explanation:
If i have 20 chair legs
We know each chair will have 4 legs
Divide 20 with 4
20 / 4 = 5
So, I will have 5 chairs.
Question 5.
0.20 + 0.10 + 0.20 =
_____________
Explanation:
The sum of 0.20, 0.10 and 0.20 is 0.50
So, 0.20 + 0.10 + 0.20 = 0.50.
Question 6.
Fill in the missing number.
317, 307, ___, 287, 277
297
Explanation:
The missing number in the pattern is skip count of 10
the pattern shows decrease in numbers
317 – 10 = 307
So, 307- 10 = 297
The missing number is 297
The pattern is 317, 307, 297, 287, 277.
Question 7.
Tuesday is June 4th. What is the date on the following Friday?
_____________
7th June
Explanation:
If Tuesday is 4th June
Wednesday is 5th
Thursday is 6th
So, Friday is 7th June.
Question 8.
What is the perimeter?
___ cm
Perimeter = 9 cm
Explanation:
The triangle has sides with length 3 cm
Perimeter = 3 x length of side
= 3 x 3
= 9
So, perimeter of given triangle is 9 cm.
Question 9.
Label the vertex on the angle.
Explanation:
The point on which the angle is formed is called the angle’s vertex
I labeled the vertex on the angle
The vertex is point B.
Question 10.
I am an odd number between 129 and 133. What number am I?
_____________
|
# What is Subtraction with Borrowing?
Subtraction with borrowing – sometimes known as regrouping or exchanging – is a technique that lets us subtract any number from another. It works on a similar principle as addition with regrouping, which you might also hear referred to as carrying over.
We use subtraction with borrowing alongside column subtraction to use place value to help us “borrow” an amount from the next column along – the tens, hundreds, or even thousands.
Let’s see subtraction with borrowing in action:
The best way to understand subtraction with borrowing is to see it in action. So let’s work through an example together, looking at the simple question 42-17.
First, we line the numbers up as we would do with any column question:
42
17 –
——
But with our first step, we hit a problem. We can’t subtract 7 from 2 without going into tricky minus numbers, which wouldn’t help us in this situation. So instead, we can borrow ten from the next column along; this means that, instead of looking at 42 as 40+2, we can look at it as 30+12.
It helps us with the first stage of our problem since we can now calculate 12-7. Once we’re done with that, we’re also free to move one column to the left and finish the subtraction with 3-1.
4 3 12
1 7 –
———-
2 5
As you can see, we now have an overall answer of 25, which we have reached by using subtraction with borrowing.
When do I use subtraction with borrowing?
Subtraction with borrowing is used alongside the column method – where numbers are placed vertically, and you work through one place value column at a time.
We use this technique when the maximum number in a column is smaller than the bottom, as seen in the example above. In this case, the 2 in 42 was smaller than the 7 in 17, so we needed to borrow ten from the next column.
This method makes taking away much more straightforward, allowing children to complete subtraction questions involving two, three, or even four-digit numbers.
|
## Tuesday, February 28, 2017
### The Direction of Forces in Two Dimensions
When adding forces in two dimensions, it is easy to find the magnitude of the resultant force using the Pythagorean Theorem. The resultant is the hypotenuse of a triangle made by the two forces.
It is also easy to find the direction of the two forces. First, look at the image to the right.
Force A is up and Force B is at a right angle to it (also known as being orthogonal).
Because of the nature of vectors, and because forces are vectors, the image can be redrawn so that the tail of one force originates at the head of the other.
The result of that is to the left, below.
When viewed in this way, it is easy to see that the resultant force is the hypothesis of the two sides, represented by the blue line.
The angle, which is the direction of the force is represented as ø and the sides are labeled O for opposite the angle, and A for adjacent to the angle.
So the green side is adjacent to the angle (it touches) and the gold side is opposite the angle (it does not touch).
Now, through the magic of trigonometry, we can use the fact that we have a right angle to calculate the angle.
The end product of all the math is that you need only divide the magnitude of opposite by the magnitude of the adjacent, and then hit the equal sign, and then hit one more button on your calculator.
Since it is soooo easy, here's the math!
By definition the tangent is a number. Because of the magic of math, for any triangle with one right angle, the sides are always in the same proportion, and are represented as a decimal, such as .7832 or .3231. There used to be books with pages and pages of tables so that the angle could be looked up for any given decimal. (IKR?)
So, the way that gets written is this:
tan(ø) = O/A
Wow.
So if you divide O by A you get a decimal decimal number. Then you go look that up and find the angle. Except, no.
There is a button on your calculator for that!
So, more math. We can solve for ø by doing the division, and then doing the un-tangent process. The actual name is the inverse tangent or (so to be a fuzz more confusing) arctangent.
So, look at this:
Okay, now for the calculator!
To find the direction of a resultant force, resolve all the forces into two orthogonal forces. Then draw out the triangles. Draw the hypotenuse. Then, you are ready.
First, divide the opposite by the adjacent and hit the equal button.
Now the fun! Find the inverse tangent button! Good luck with that! This is what it looks like on the laptop calculator (You MUST hit the "2nd" button first!):
On the TI-30, you also have to hit the "2nd" button first:
So, after you find the button, you are pretty much there.
Example:
Force A has a magnitude of 5 N Up
Force B has a magnitude of 7 N Right
Find the direction in degrees of the resultant force.
tan(ø) = 5/7
Divide 5 by 7 = .71428
Push the inverse tan button. The angle is 35.54 degrees.
Done.
The method for finding the direction of the resultant force when two orthogonal forces is relatively easy. Once the button is found, it takes only two steps on the calculator. While the theory and foundations come from trigonometry, it is not necessary to fully understand the math to use it.
|
# What Is Print Spiral Matrix?
What Is Print Spiral Matrix?
## Introduction
Wherever you go through an interview, it is rare to know that there is no question asked on arrays, and as arrays are categorised into three types, basically as:
• One dimensional array
• Two-dimensional arrays
• Multidimensional arrays
So, today we will be discussing one of the most asked questions in interviews based on 2-D arrays- “Print Spiral Matrix.”
Problem Statement- You are given a 2-D array ‘MATRIX’ of dimensions N x M, of integers. You need to print the matrix in a spiral fashion (clockwise).
“Here ‘N’ denotes the number of rows whereas ‘M’ denotes the number of columns.”
Example:
This is how a spiral matrix would look like:
How will we get the required output?
• Start moving from the first element of row1 towards the right and print all the elements of row one, i.e., 1 2 3 4.
• Now, move towards a downward direction in the last column, i.e. column 4, and print the elements as 5,6,7.
• Now, move towards the left in the last row, i.e. row 4, and again print all its elements as 8,9,10.
• This time we will be moving in an upward direction in the first column and print the elements 11,12.
• Again move in the right direction in the second row and print the elements as 13.
• Next moving down in three columns and simply print 15.
• At last, towards the left and print the elements of the third row as 16.
Now, without any delay, let’s head on to the solution.
Before that, it would be great if you give it a try by yourself Here. There can be several ways to print a spiral matrix, so we will discuss all of them, starting with the Brute force approach.
Method 1: (Brute Force – Simulation-based)
The idea is basically to draw the path that the spiral will create, knowing the fact that the path will turn in a clockwise direction if it goes beyond the range or into a cell that was already visited.
Algorithm
• The matrix will have R*C elements.
• We will take a vector to know which cell has already been visited.
• Assuming that the current visited cell is some (row, col) facing some direction dir.
• And now the next cell to be visited by some (nr,nc).
• This would be the next position if it is within the bounds or has not been visited.
• If this doesn’t act to be the next position, then the one that will appear after changing the direction will be the next position.
To make it more clear below is the C++ code. You can consider it.
Code:
``````#include <bits/stdc++.h>
using namespace std;
vector<int> printSpiral(vector<vector<int> >& mat)
{
vector<int> output;
//if the size gets 0, then simply return the vector formed so far
if (mat.size() == 0)
return output;
int R = mat.size(), C = mat[0].size();
vector<vector<bool> > visited(R, vector<bool>(C, false));
int drow[] = { 0, 1, 0, -1 };
int dcol[] = { 1, 0, -1, 0 };
int row = 0, col = 0, dir = 0;
// Iterate from 0 to R * C - 1
for (int i = 0; i < R * C; i++) {
output.push_back(mat[row]);
visited[row] = true;
int nr = row + drow[dir];
int nc = col + dcol[dir];
if (0 <= nr && nr < R && 0 <= nc && nc < C
&& !visited[nr][nc]) {
row = nr;
col = nc;
}
else {
//as one of the rows or column is printed, the direction will be
//changed so increment by 1
dir = (dir + 1) % 4;
row += drow[dir];
col += dcol[dir];
}
}
return output;
}
// Driver code
int main()
{
vector<vector<int> > arr{ { 1, 2, 3, 4 },
{ 12, 13, 14, 5 },
{ 11, 16, 15, 6 },
{ 10, 9, 8, 7 } };
for (int i : printSpiral(arr)) {
cout << i << " ";
}
return 0;
}``````
Output:
``1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16``
Time Complexity: O(r*c)
• Since every element of the matrix will be traversed the time complexity will be O(r*c).
Space Complexity: O(r*c)
• Extra space is occupied as we make use of vectors, so it will take space equivalent to the count of elements present in the matrix.
Method 2: (Optimised Approach)
Use different if-else conditions to know in which row or column you need to move and in which direction and print the elements associated with it.
Algorithm
• Take four variables to iterate through the rows and columns.
• Now, once you have covered a row or column it’s pretty clear that the direction needs to be changed so for that also take a variable to make the direction change.
• Now just go through a loop while all the rows and columns are traversed.
• Keep on checking the conditions that now which direction we need to move.
• And simply keep on printing the elements.
Below mentioned is the C++ code, which will help you get things more clear.
Code
``````#include <bits/stdc++.h>
using namespace std;
void spirallytraverse(int matrix[4][4], int r, int c)
{//initialize variables
int top=0;
int bottom=r-1;
int left=0;
int right=c-1;
int direction=0;
//run the loop until every row and column is traversed
while(left<=right&&top<=bottom)
{
if(direction==0)
{
for(int i=left;i<=right;i++)
{
cout<<matrix[top][i]<<" ";
}top++;
}
else if(direction==1)
{
for(int i=top;i<=bottom;i++)
{
cout<<matrix[i][right]<<" ";
}right--;
}
else if(direction==2)
{
for(int i=right;i>=left;i--)
{
cout<<matrix[bottom][i]<<" ";
}bottom--;
}
else if(direction==3)
{
for(int i=bottom;i>=top;i--)
{
cout<<matrix[i][left]<<" ";
}left++;
}
direction=(direction+1)%4;
}
}
int main()
{ int matrix[4][4]={{1, 2, 3 ,4},
{12, 13, 14, 5},
{11, 16, 15, 6},
{10, 9, 8, 7}};
spirallytraverse(matrix,4,4);
return 0;
}
``````
Output:
``1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16``
Time complexity: O(r*c)
• The traversal of the matrix will acquire the time of order r*c, that is, rows*columns.
Space Complexity: O(1)
• No extra space is required so space complexity will be constant.
Method 3: (Using Recursion)
The idea is simple, as we were making changes in the direction after every traversal in the previous method here, we will make the call recursive once the direction gets changed.
Algorithm
• Pass all the variables we took in the earlier method as left, right, top, bottom to the function.
• Check if the left is less than the right.
• Print the 1-row elements and make the top move to the next row.
• Now, check if the top is less than the bottom or not.
• If yes, then simply print the last column that is the 4th column in the example.
• Now, check if the left is less than right or not.
• If yes, print the elements of the 4th row.
• Again, check if the top is less than the bottom. If yes, then print the elements of 1 column.
• Now, one cycle is completed, so we will call the function again until it reaches the bounds.
To get it more clear, we first suggest to dry run using pen and paper, and then you can take help from below mentioned code.
Code:
``````#include <iostream>
using namespace std;
void printSpiral(int mat[4][4], int top, int bottom, int left, int right)
{
//bound exceeds
if (left > right) {
return;
}
//print top row
for (int i = left; i <= right; i++) {
cout << mat[top][i] << " ";
}
top++;
if (top > bottom) {
return;
}
// print right column
for (int i = top; i <= bottom; i++) {
cout << mat[i][right] << " ";
}
right--;
if (left > right) {
return;
}
// print bottom row
for (int i = right; i >= left; i--) {
cout << mat[bottom][i] << " ";
}
bottom--;
if (top > bottom) {
return;
}
// print left column
for (int i = bottom; i >= top; i--) {
cout << mat[i][left] << " ";
}
left++;
//one direction is over move to next using a recursive call
printSpiral(mat, top, bottom, left, right);
}
int main()
{
int mat[4][4] =
{
{ 1, 2, 3, 4},
{12, 13, 14, 5},
{11, 16, 15, 6},
{10, 9, 8, 7}
};
int top = 0, bottom = 4 - 1;
int left = 0, right = 4 - 1;
printSpiral(mat, top, bottom, left, right);
return 0;
}``````
Output:
``1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16``
Time Complexity: O(r*c)
• All the elements of the matrix will be traversed so the complexity will be O(r*c).
Space Complexity: O(r*c)
• The space complexity will be of order (r*c) as space will be occupied in the stack due to recursion.
## Frequently Asked Questions
What is a spiral array?
It is an arrangement of first (n square) elements in a square fashion where, as we go around the edges, the number goes on increasing sequentially, and the array moves spirally inwards.
How do you traverse an array in a spiral?
As discussed in the method to print a spiral matrix, we can traverse any array in spiral form by taking four different variables each at initial row, final row, initial column, final column and take one variable to address the change in direction.
What is a two-dimensional array?
It is similar to a 1-D array but can be visualised as a table with rows and columns.
How to print a matrix in spiral order?
There can be several methods, as discussed above, to print a matrix in spiral order ranging from, one which can be done through traversing the matrix, or you can do it using recursion.
## Key Takeaways
The main idea behind solving the problem, “print spiral matrix,” is traversing through the matrix in such a way that at a time, we print a complete row or column and then change the direction and repeat the same for the rest of the elements.
It would be great if this problem is clear to you, champ. But don’t give it up here. Bang on to solve some of the highly recommended interview questions and keep on growing.
Don’t forget to test yourself by solving the problems on CodeStudio.
Keep exploring, keep growing!
By Neerulata
|
Equations of Straight Lines
Fundamentals of Straight Line Equations
• The equation of a straight line in the Cartesian coordinate system traditionally takes the form y=mx+c. Here, ‘m’ stands for the slope, or gradient of the line, while ‘c’ is the y-intercept - the point where the line crosses the y-axis.
• The slope of a line can be determined using two points on the line, (x1,y1) and (x2,y2), employing the formula: m = (y2 - y1) / (x2 - x1).
• A line’s y-intercept is found by setting x to 0 in the line’s equation and solving for y.
Intercepts and Intersection Points
• An x-intercept is the point where the line crosses the x-axis, it is found by setting y to 0 in the equation of the line and solving for x.
• When the equation for two lines are set equal to each other and solved, this yields the coordinates of the intersection point of these two lines.
Characteristics of Parallel and Perpendicular Lines
• If two lines are parallel, then their slopes, or gradients, are equal.
• If two lines are perpendicular, the product of their gradients is -1. Hence, the gradient of a perpendicular line is the negative reciprocal of the original line’s gradient.
Special Forms of the Equation of a Straight Line
• The point-slope form of the equation of a straight line allows you to create the equation from a known slope and a single known point on the line - y-y1 = m(x-x1).
• The two-point form is equally useful. When two points on the line are known, this form can be used - (y - y1) = ((y2 - y1) / (x2 - x1)) * (x - x1).
• You can use the intercept form to create the equation of a straight line from its x and y intercepts - x/a + y/b = 1.
Real-world Application of Straight Line Equations
• The principles of coordinate geometry and the equation of straight lines are not exclusive to mathematical problems. They have myriad applications in real life - from calculating the shortest distance between places, to devising optimal travel routes, managing logistics and even in computer graphics and physics.
• The knowledge of the intersection point of two lines plays a key role in many physics concepts and are heavily used in graphic designing and animation, where the intersection points determine the way images and vectors interact.
|
Imagine that in a future era humans decide to build a high speed train circumnavigating the globe at the equator.
Ignore the impracticalities of such an undertaking (e.g. building railroad tracks across an ocean) and think about this question: How long does the track need to be? This is easy to answer if you know the circumference of the Earth, which is 24,902 (or roughly 25,000) miles at the equator.
Now, imagine our engineering team determines that, for technical reasons, the track needs to be elevated two feet off the ground. We were already planning to acquire 25,000 miles of track so the question is this: how much additional track do we need in order to build our equatorial railroad two feet above the Earth’s surface?
Leave your answer in a comment below. I’ll post a solution and the names of all puzzle solvers on Monday.
Solution: Think of the equator as a giant circle. Let’s call the radius of that circle R, the distance from the center of the Earth to any point on the equator. The elevated track can also be thought of as a giant circle, one with radius R+2 (since the track is two feet off the ground). Now the question of how many additional feet of track we need boils down to subtracting the circumference of the elevated track from the circumference of the equator.
The circumference of any circle is given by 2Pir, where r is the circle’s radius so with a bit of algebra we get:
more track needed = track circumference – equator circumference = 2Pi(R+2) – 2PiR = 2Pi(R+2-R) = 2Pi(2) = 4*Pi ~= 12 So, to raise the track by two feet all 25,000 miles around the Earth, you only need to add approximately 12 feet of track! Even more amazing, because the solution is independent of the starting radius, the same answer applies to every possible circle. For example, if you tied a string around a basketball and then decided to raise that string two feet above the surface of the basketball, you’d need precisely the same amount of additional string: 12 feet.
This is a problem I first heard as a kid and to this day it still strikes me as incredibly surprising. Hat’s off to Mudassir Ansari, Dan Stoops, John Baldi, Ricardo Agudo and Jim Goss for coming up with the correct answer!
|
iarc6io
2022-07-19
The first term and the last term of an AP are 17 and 350 respectively. If the common difference is 9,how many terms are there and what is their sum?
Izabelle Frost
Expert
Step 1
Given
First term of A.P $\left({a}_{1}\right)=17$
Last term of A.P $\left({a}_{n}\right)=350$
Common defference $\left(d\right)=9$
Step 2
finding the number of terms and their sum
${a}_{n}=350$
${a}_{1}+\left(n-1\right)d=350$
$17+\left(n-1\right)9=350$
$\left(n-1\right)9=350-17$
$\left(n-1\right)9=333$
$\left(n-1\right)=\frac{333}{9}$
$n-1=37$
$=37+1$
$=37+1$
there are 38 terns in A.P
WKT
Some to n terns ${S}_{n}=\frac{n}{2}\left[a+1\right]$
$=\frac{38}{2}\left[17+350\right]$
$=19\left[367\right]=6973$
The sum to 38 terms=6973
Glenn Hopkins
Expert
Given:
First term $\left(a\right)=17$
Last term $\left(l\right)={a}_{n}=350$
and common difference $\left(d\right)=9$
$\because {a}_{n}=350$
$a+\left(n-1\right)d=350$
$⇒17+\left(n-1\right)9=350$
$⇒9\left(n-1\right)=350-17=333$
$⇒n-1=\frac{333}{9}=37$
$\therefore n=38$
Now, ${S}_{n}=\frac{n}{2}\left(a+1\right)$
$=\frac{38}{2}\left(17+350\right)$
$=19×367=6973$
Hence, $n=38$ & sum of term ${S}_{n}=6973$
Do you have a similar question?
|
# Non-unit fraction of a non-unit fraction
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Purpose
The purpose of this activity is to extend students' understandings of fractions as operators to include finding a fraction of a fraction.
Achievement Objectives
NA3-1: Use a range of additive and simple multiplicative strategies with whole numbers, fractions, decimals, and percentages.
NA3-5: Know fractions and percentages in everyday use.
Required Resource Materials
• Sheets of paper (A4 photocopying paper is ideal)
• Pens (highlighters or felts are preferable)
Activity
1. Take a rectangular sheet of paper and follow the steps described below.
• Fold the paper in thirds lengthways, then shade in two of the thirds.
• Fold the paper in quarters widthways, then shade in three quarters of the two thirds.
What fraction of the whole rectangle is double shaded? Explain how you know.
Record 3/4 x 2/3 = 6/12.
Explain what this equation means. Look for students to recognise the equations as “three quarters of two thirds equals six twelve.”
What patterns do you see in the denominators? Why do we end up with twelve equal parts?
What patterns do you see in the numerators? Why do we end up with six parts double shaded?
Students may notice that, in the equation, the numerators 3 and 2 are multiplied to form the numerators of the answer. Similarly the denominators are multiplied to form the denominator of the answer.
1. Provide further opportunities for students to use paper folding and shading to find a non-unit fraction of a non-unit fraction. Allow students to work in groupings that will encourage peer scaffolding and extension. Some students might benefit from working independently, whilst others might need further support from the teacher. Good examples are shown below:
• Three quarters of three fifths (3/4 x 3/5 = 9/20)
• Four fifths of two thirds (4/5 x 2/3 = 8/15)
• One fifth of three eighths (1/5 x 3/8 = 3/40).
2. Use paper folding and shading to explore the commutative property of multiplication with fractions.
Does shading two thirds lengthways then three quarters width ways give the same result as shading three quarters lengthways and two thirds widthways?
Record, describe, and reflect on the relevant expressions. For example: 2/3 x 3/4 = 6/12, “two thirds of three quarters equals six twelfths.”
Emphasise that while the dimensions of the results are different, both area models show double shaded areas that are one twelfth in size. This illustrates that 3/4 x 2/3 = 2/3 x 3/4.
3. Provide examples for students to solve, without the support of paper folding and shading. This might begin with drawing diagrams to show the product of two non-unit fractions. For example, 3/4 x 2/5 = 6/20 might be drawn as:
4. Progress to solving problems with equations only, such as those shown below. Provide paper folding for the examples if students have difficulty.
• 3/5 x 7/8 = [ ]
• 4/5 x 4/5 = [ ]
• 2/3 x 5/8 = [ ]
• 4/9 x 5/7 = [ ]
Next steps for further learning
1. Encourage students to apply a wider range of basic multiplication facts to these types of problems. For example, 7/8 x 5/9 = [ ].
2. Use written expressions to formally generalise the multiplication of two unit fractions:
What is the answer to □/◊ x ○/∆? (In general the product is always □ x ○/◊ x ∆.
3. Compare the products of non-unit fractions. For example, Tim shades two fifths of three eights on his piece of paper. Jayden shades three quarters of two fifths of his paper. The pieces of paper are the same size. Who shades more, Tim or Jayden?
|
Paul's Online Notes
Home / Algebra / Graphing and Functions / Inverse Functions
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### Section 3-7 : Inverse Functions
1. Given $$h\left( x \right) = 5 - 9x$$ find $${h^{ - 1}}\left( x \right)$$ .
Show All Steps Hide All Steps
Hint : Just follow the process outlines in the notes and you’ll be set to do this problem!
Start Solution
For the first step we simply replace the function with a $$y$$.
$y = 5 - 9x$ Show Step 2
Next, replace all the $$x$$’s with $$y$$’s and all the original $$y$$’s with $$x$$’s.
$x = 5 - 9y$ Show Step 3
Solve the equation from Step 2 for $$y$$.
\begin{align*}x & = 5 - 9y\\ 9y & = 5 - x\\ y & = \frac{{5 - x}}{9}\end{align*} Show Step 4
Replace $$y$$ with $${h^{ - 1}}\left( x \right)$$.
${h^{ - 1}}\left( x \right) = \frac{{5 - x}}{9}$ Show Step 5
Finally, do a quick check by computing one or both of $$\left( {h \circ {h^{ - 1}}} \right)\left( x \right)$$ and $$\left( {{h^{ - 1}} \circ h} \right)\left( x \right)$$ and verify that each is $$x$$. In general, we usually just check one of these and well do that here.
$\left( {h \circ {h^{ - 1}}} \right)\left( x \right) = h\left[ {{h^{ - 1}}\left( x \right)} \right] = P\left[ {\frac{{5 - x}}{9}} \right] = 5 - 9\left( {\frac{{5 - x}}{9}} \right) = 5 - \left( {5 - x} \right) = x$
The check works out so we know we did the work correctly and have inverse.
|
# Write Linear Equations & Understanding Slopes
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| By Coach Pitts
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Coach Pitts
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• 1.
### Write the equation in slope-Intercept form of the line that passes through the point (4,-6) and has a slope of -3/4?
y = -3/4x - 3, y=-3/4x-3
Explanation
The equation is in slope-intercept form, which is y = mx + b, where m is the slope and b is the y-intercept. In this case, the slope is -3/4 and the point (4, -6) lies on the line. We can use the point-slope formula to find the equation: y - y1 = m(x - x1). Substituting the values, we get y - (-6) = (-3/4)(x - 4), which simplifies to y + 6 = (-3/4)x + 3. Rearranging the equation, we get y = (-3/4)x - 3, which matches the given answer.
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• 2.
### Write the equation in slope-intercept form for the equation of the line that passes through the point (-5, -3/4) and has a slope of 1/20 ?
y=1/20x-1/2, y=1/20x-2/4
Explanation
The equation in slope-intercept form is y = mx + b, where m is the slope and b is the y-intercept. In this case, the slope is 1/20. To find the y-intercept, we can substitute the coordinates of the given point (-5, -3/4) into the equation and solve for b. By substituting the values, we get -3/4 = (1/20)(-5) + b. Simplifying this equation, we find that b = -1/2. Therefore, the equation in slope-intercept form for the line that passes through the point (-5, -3/4) with a slope of 1/20 is y = 1/20x - 1/2.
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• 3.
### If point (-1,0) is on the line whose equation is y = -2x + b, what is the value of b?
-2, b=-2
Explanation
The equation of the line is in the form y = mx + b, where m is the slope and b is the y-intercept. Since the point (-1,0) lies on the line, we can substitute the x and y coordinates into the equation to solve for b. Plugging in x = -1 and y = 0, we get 0 = -2(-1) + b. Simplifying this equation, we get 0 = 2 + b. To isolate b, we subtract 2 from both sides, giving us b = -2. Therefore, the value of b is -2.
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• 4.
### Write the equation of the line that is // to the line y = -1/2x + 4 and passes through the point (-4, -4).Write your answer in slope-intercept form.
y = -1/2x - 6, y=-1/2x-6
Explanation
The equation of a line can be written in slope-intercept form as y = mx + b, where m represents the slope and b represents the y-intercept. In this case, the given line is parallel to y = -1/2x + 4, which means it has the same slope of -1/2. Since the line passes through the point (-4, -4), we can substitute these values into the equation and solve for the y-intercept. Plugging in the values, we get -4 = -1/2(-4) + b. Simplifying this equation, we find that b = -6. Therefore, the equation of the line is y = -1/2x - 6.
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• 5.
### Write the equation of the line perpendicular to the line 3x - 5y = -1 and passes through the point ( 3, -4).Write the answer in Standard form.
5x + 3y = 3, 5x+3y=3
Explanation
The equation of a line perpendicular to 3x - 5y = -1 can be found by rearranging the equation into slope-intercept form (y = mx + b), where m is the slope. The slope of the given line is (3/5). The slope of a line perpendicular to this line will be the negative reciprocal of (3/5), which is (-5/3). Using the point-slope form of a line (y - y1 = m(x - x1)), we can substitute the values of the point (3, -4) and the slope (-5/3) to find the equation of the line. Simplifying the equation will give us the standard form: 5x + 3y = 3. Therefore, the correct answer is 5x + 3y = 3.
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• 6.
### Equation 1 has a slope of -2/3, what would be the perpendicular slope for equation 2?
• A.
-2/3
• B.
2/3
• C.
3/2
• D.
-3/2
C. 3/2
Explanation
The perpendicular slope for a given slope is the negative reciprocal of that slope. In this case, the given slope is -2/3. To find the perpendicular slope, we take the negative reciprocal of -2/3, which is 3/2. Therefore, the correct answer is 3/2.
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• 7.
### Given the line x = -7/9, what would be slope of the the line perpendicular to this line?
• A.
9/7
• B.
-7/9
• C.
0
• D.
Underfined
C. 0
Explanation
The slope of a line perpendicular to another line is the negative reciprocal of the slope of the original line. Since the original line has a slope of 0 (because it is a vertical line), the slope of the line perpendicular to it would be the negative reciprocal of 0, which is also 0.
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• 8.
### The line x = -2 is what type of line
• A.
Diagonal line going up
• B.
Diagonal line going down
• C.
Vertical line
• D.
Horizontal line
C. Vertical line
Explanation
The line x = -2 is a vertical line because it is parallel to the y-axis. In a Cartesian coordinate system, a vertical line has a constant x-coordinate and can intersect the y-axis at any point. Since x = -2 is a constant value for all y-values, it represents a vertical line.
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• 9.
### The line x = -2 has what type of slope?
• A.
Positive slope
• B.
Negative slope
• C.
Zero slope
• D.
Undefined slope
D. Undefined slope
Explanation
The line x = -2 has an undefined slope because a vertical line does not have a change in y-coordinate for any change in x-coordinate. The slope is typically defined as the change in y divided by the change in x, but since there is no change in x for this line, the slope cannot be calculated. Therefore, the slope is undefined.
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• 10.
### The line 2x - 3y = -1/3 has what type of slope?
• A.
Positive Slope
• B.
Negative Slope
• C.
Zero Slope
• D.
Undefined Slope
A. Positive Slope
Explanation
The given equation is in the form y = mx + b, where m represents the slope. In this case, we can rearrange the equation to y = (2/3)x + 1/9. Since the coefficient of x is positive (2/3), the slope is positive. Therefore, the correct answer is positive slope.
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• Current Version
• Mar 21, 2023
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• Dec 14, 2016
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SAT Math : How to find the slope of perpendicular lines
Example Questions
← Previous 1
Example Question #1 : How To Find The Slope Of Perpendicular Lines
Two points on line m are (3,7) and (-2, 5). Line k is perpendicular to line m. What is the slope of line k?
-5/2
2/5
3
-5
0
-5/2
Explanation:
The slope of line m is the (y2 - y1) / (x- x1) = (5-7) / (-2 - 3)
= -2 / -5
= 2/5
To find the slope of a line perpendicular to a given line, we must take the negative reciprocal of the slope of the given line.
Thus the slope of line k is the negative reciprocal of 2/5 (slope of line m), which is -5/2.
Example Question #451 : Geometry
The equation of a line is: 8x + 16y = 48
What is the slope of a line that runs perpendicular to that line?
-1/4
2
8
-2
-1/8
2
Explanation:
First, solve for the equation of the line in the form of y = mx + b so that you can determine the slope, m of the line:
8x + 16y = 48
16y = -8x + 48
y = -(8/16)x + 48/16
y = -(1/2)x + 3
Therefore the slope (or m) = -1/2
The slope of a perpendicular line is the negative inverse of the slope.
m = - (-2/1) = 2
Example Question #3 : How To Find The Slope Of Perpendicular Lines
What is the equation of a line perpendicular to the one above, passing through the point ?
Explanation:
Looking at the graph, we can tell the slope of the line is 3 with a -intercept of , so the equation of the line is:
A perpendicular line to this would have a slope of , and would pass through the point so it follows:
Example Question #451 : Sat Mathematics
Line M passes through the points (2,2) and (3,–5). Which of the following is perpendicular to line M?
y = –(1/7)x – 1
y = 7x – 6
y = –7x – 5
y = 7x + 4
y = (1/7)x + 3
y = (1/7)x + 3
Explanation:
First we find the slope of line M by using the slope formula (y– y1)/(x– x1).
(–5 – 2)/(3 – 2) = –7/1. This means the slope of Line M is –7. A line perpendicular to Line M will have a negative reciprocal slope. Thus, the answer is = (1/7)x + 3.
Example Question #1 : How To Find The Slope Of Perpendicular Lines
Figure not drawn to scale.
In the figure above, a circle is centered at point C and a line is tangent to the circle at point B. What is the equation of the line?
Explanation:
We know that the line passes through point B, but we must calculate its slope in order to find the equation that defines the line. Because the line is tangent to the circle, it must make a right angle with the radius of the circle at point B. Therefore, the slope of the line is perpendicular to the slope of the radius that connects the center of the circle to point B. First, we can find the slope of the radius, and then we can determine the perpendicular slope.
The radius passes through points C and B. We can use the formula for the slope (represented as ) between two points to find the slope of the radius.
Point C: (2,-5) and point B: (7,-3)
This is the slope of the radius, but we need to find the slope of the line that is perpendicular to the radius. This value will be equal to the negative reciprocal.
Now we know the slope of the tangent line. We can use the point-slope formula to find the equation of the line. The formula is shown below.
Plug in the give point that lies on the tangent line (point B) and simplify the equation.
Multiply both sides by two in order to remove the fraction.
Distribute both sides.
Subtract six from both sides.
Example Question #1 : How To Find The Slope Of A Perpendicular Line
What is the slope of any line perpendicular to 2y = 4x +3 ?
– ½
½
– 4
2
– ½
Explanation:
First, we must solve the equation for y to determine the slope: y = 2x + 3/2
By looking at the coefficient in front of x, we know that the slope of this line has a value of 2. To fine the slope of any line perpendicular to this one, we take the negative reciprocal of it:
slope = m , perpendicular slope = – 1/m
slope = 2 , perpendicular slope = – 1/2
Example Question #1 : How To Find The Slope Of A Perpendicular Line
What line is perpendicular to 2x + y = 3 at (1,1)?
2x + 3y = 1
x + 2y - 3
x – 2y = -1
3x + 2y = 1
x – 2y = -1
Explanation:
Find the slope of the given line. The perpendicular slope will be the opposite reciprocal of the original slope. Use the slope-intercept form (y = mx + b) and substitute in the given point and the new slope to find the intercept, b. Convert back to standard form of an equation: ax + by = c.
Example Question #2 : How To Find The Slope Of Perpendicular Lines
What is the slope of the line perpendicular to the line given by the equation
6x – 9y +14 = 0
-3/2
-1/6
-2/3
6
2/3
-3/2
Explanation:
First rearrange the equation so that it is in slope-intercept form, resulting in y=2/3 x + 14/9. The slope of this line is 2/3, so the slope of the line perpendicular will have the opposite reciprocal as a slope, which is -3/2.
Example Question #1 : How To Find The Slope Of A Perpendicular Line
What is the slope of the line perpendicular to the line represented by the equation y = -2x+3?
-1/2
2/3
1/2
2
-2/3
1/2
Explanation:
Perpendicular lines have slopes that are the opposite of the reciprocal of each other. In this case, the slope of the first line is -2. The reciprocal of -2 is -1/2, so the opposite of the reciprocal is therefore 1/2.
Example Question #1 : How To Find The Slope Of A Perpendicular Line
Find the slope of a line perpendicular to the line y = –3x – 4.
4
1/3
1/4
–3
|
Difference between revisions of "2002 AMC 10B Problems/Problem 19"
Problem
Suppose that $\{a_n\}$ is an arithmetic sequence with $$a_1+a_2+\cdots+a_{100}=100 \text{ and } a_{101}+a_{102}+\cdots+a_{200}=200.$$ What is the value of $a_2 - a_1 ?$
$\mathrm{(A) \ } 0.0001\qquad \mathrm{(B) \ } 0.001\qquad \mathrm{(C) \ } 0.01\qquad \mathrm{(D) \ } 0.1\qquad \mathrm{(E) \ } 1$
Solution 1
We should realize that the two equations are 100 terms apart, so by subtracting the two equations in a form like...
$(a_{101} - a_1) + (a_{102} - a_2) +... + (a_{200} - a_{100}) = 200-100 = 100$
...we get the value of the common difference of every hundred terms one hundred times. So we have to divide the answer by one hundred to get ...
$\frac{100}{100} = 1$
...the common difference of every hundred terms. Then we have to simply divide the answer by hundred again to find the common difference between one term, therefore...
$\frac{1}{100} =\boxed{(\text{C})0.01}$
Solution 2
Adding the two given equations together gives
$a_1+a_2+...+a_{200}=300$.
Now, let the common difference be $d$. Notice that $a_2-a_1=d$, so we merely need to find $d$ to get the answer. The formula for an arithmetic sum is
$\frac{n}{2}(2a_1+d(n-1))$,
where $a_1$ is the first term, $n$ is the number of terms, and $d$ is the common difference. Now we use this formula to find a closed form for the first given equation and the sum of the given equations. For the first equation, we have $n=100$. Therefore, we have
$50(2a_1+99d)=100$,
or
$2a_1+99d=2$. *(1)
For the sum of the equations (shown at the beginning of the solution) we have $n=200$, so
$100(2a_1+199d)=300$
or
$2a_1+199d=3$ *(2)
Now we have a system of equations in terms of $a_1$ and $d$. Subtracting (1) from (2) eliminates $a_1$, yielding $100d=1$, and $d=a_2-a_1=\frac{1}{100}=\boxed{(\text{C}) .01}$.
Solution 3
Subtracting the 2 given equations yields
$(a_{101}-a_1)+(a_{102}-a_2)+(a_{103}-a_3)+...+(a_{200}-a_{100})=100$
Now express each $a_n$ in terms of first term $a_1$ and common difference $x$ between consecutive terms
$((a_1+100x)-(a_1))+((a_1+101x)-(a_1+x))+((a_1+102x)-(a_1+2x))+...+((a_1+199x)-(a_1+99x))=100$
Simplifying and canceling $a_1$ and $x$ terms gives
$100x+100x+100x+...+100x=100$
$100x\times100=100$
$100x=1$
$x=0.01$
Video Solution
~ pi_is_3.14
$x=0.01=\boxed{(\text{C})0.01}$
2002 AMC 10B (Problems • Answer Key • Resources) Preceded byProblem 18 Followed byProblem 20 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 10 Problems and Solutions
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# Set is Equivalent to Proper Subset of Power Set
## Theorem
Every set is equivalent to a proper subset of its power set:
$\forall S: \exists T \subset \mathcal P \left({S}\right): S \sim T$
## Proof
To show equivalence between two sets, we need to demonstrate that a bijection exists between them.
We will now define such a bijection.
Let $T = \left\{{\left\{{x}\right\}: x \in S}\right\}$.
$\displaystyle \forall x \in S: \ \$ $\displaystyle \left\{ {x}\right\}$ $\subseteq$ $\displaystyle S$ Definition of Subset $\displaystyle \implies \ \$ $\displaystyle \left\{ {x}\right\}$ $\in$ $\displaystyle \mathcal P \left({S}\right)$ Definition of Power Set $\displaystyle \implies \ \$ $\displaystyle T$ $\subseteq$ $\displaystyle \mathcal P \left({S}\right)$ Definition of Subset
Now we define the mapping $\phi: S \to T$:
$\phi: S \to T: \forall x \in S: \phi \left({x}\right) = \left\{{x}\right\}$
$\phi$ is an injection:
$\forall x, y \in S: \left\{{x}\right\} = \left\{{y}\right\} \implies x = y$ by definition of set equality
$\phi$ is a surjection:
$\forall \left\{{x}\right\} \in T: \exists x \in S: \phi \left({x}\right) = \left\{{x}\right\}$
So $\phi$, being both an injection and a surjection, is a bijection by definition.
To show that $T \subset \mathcal P \left({S}\right)$, that is, is a proper subset of $\mathcal P \left({S}\right)$, we merely note that $\varnothing \in \mathcal P \left({S}\right)$ by Empty Set is Element of Power Set, but $\varnothing \notin T$.
Thus $T \subseteq \mathcal P \left({S}\right)$ but $\mathcal P \left({S}\right) \not \subseteq T$.
Hence the result.
$\blacksquare$
|
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# Special Right Triangles
Sep 13, 2022
## Key Concepts
• Find hypotenuse length in a 30°-60° -90° triangle
### Special Right Triangles
There are two special right triangles with angles measures as 45°, 45°, 90° degrees and 30°, 60°, 90° degrees.
### 30°-60°-90° Triangle Theorem
In a 30°-60°-90° triangle, the hypotenuse is twice as long as the shorter leg, and the longer leg is √3 times as long as the shorter leg.
Hypotenuse = 2 × shorter leg
Longer leg = shorter leg × √3
### Find hypotenuse length in a 30°-60°-90° triangle
Example 1:
Find the height of an equilateral triangle.
Solution:
Draw the equilateral triangle described. Its altitude forms the longer leg of two 30°-60°-90° triangles. The length h of the altitude is approximately the height of the triangle.
longer leg = shorter leg × √3
h = 2 ×√3
h = 2√3
Example 2:
Find the height of an equilateral triangle.
Solution:
Draw the equilateral triangle described. Its altitude forms the longer leg of two 30°-60°-90° triangles.
The length h of the altitude is approximately the height of the triangle.
longer leg = shorter leg ×√3
h = 4×√3
h = 4√3
Example 3:
Find lengths in a 30°-60°-90° triangle. Find the values of x and y. Write your answer in the simplest radical form.
Solution:
STEP 1:
Find the value of x.
longer leg = shorter leg ×√3 (30°-60°-90° Triangle Theorem)
X =√6 ×√3
X =√6 ×√3
X=√18
X=√9 ×2
X= 3√3
STEP 2:
Find the value of y.
hypotenuse =2 × shorter leg (30°-60°-90° Triangle Theorem)
y =2x √6
y =2√6
Example 4:
A ramp is used to launch a kayak. What is the height of a 10-foot ramp when its angle is 30° as shown?
Solution:
When the body is raised 30°, the height h is the length of the longer leg of a 30°-60°-90° triangle. The length of the hypotenuse is 10 feet.
Hypotenuse = 2 × shorter leg (30° -60° -90° Triangle Theorem)
10 = 2 × s (Substitute)
5 = s (Divide both sides by 2)
longer leg = shorter leg ×√3
h = 5√3 (Substitute)
h ≈ 8.5 (Use a calculator to approximate)
When the angle is 30°, the height of the foot ramp is 8.5 feet.
## Exercise
• Determine the value of the variable.
• Special right triangles: Copy and complete the table.
• A worker is building a ramp of 30° 8 ft to make the transportation of materials easier between an upper and lower platform. The upper platform is 8 feet off the ground, and the angle of elevation is 30°. The ramp length is
• Determine the height of an equilateral triangle.
• Determine the height of an equilateral triangle.
### What have we learned
• Identify special right triangles
• Understand 30° – 60° – 90° triangle thermos
• Understand how to find the height of an equilateral triangle
• Understand how to find lengths in a 30°-60°-90° triangle
• Understand how to find a height
#### Addition and Multiplication Using Counters & Bar-Diagrams
Introduction: We can find the solution to the word problem by solving it. Here, in this topic, we can use 3 methods to find the solution. 1. Add using counters 2. Use factors to get the product 3. Write equations to find the unknown. Addition Equation: 8+8+8 =? Multiplication equation: 3×8=? Example 1: Andrew has […]
#### Dilation: Definitions, Characteristics, and Similarities
Understanding Dilation A dilation is a transformation that produces an image that is of the same shape and different sizes. Dilation that creates a larger image is called enlargement. Describing Dilation Dilation of Scale Factor 2 The following figure undergoes a dilation with a scale factor of 2 giving an image A’ (2, 4), B’ […]
#### How to Write and Interpret Numerical Expressions?
Write numerical expressions What is the Meaning of Numerical Expression? A numerical expression is a combination of numbers and integers using basic operations such as addition, subtraction, multiplication, or division. The word PEMDAS stands for: P → Parentheses E → Exponents M → Multiplication D → Division A → Addition S → Subtraction Some examples […]
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## If you could change one thing about college, what would it be?
Outlier Articles Home
Statistics
# Sample Standard Deviation: What is It & How to Calculate It
12.15.2021 • 5 min read
## Sarah Thomas
Subject Matter Expert
This article is a guide on sample standard deviation, including concepts, a step-by-step process to calculate it, and a list of examples.
## What Is Standard Deviation?
Standard deviation measures the spread of data relative to its mean. You can think of standard deviation as the average distance between a single data point and the mean. A large standard deviation shows that data is more widely dispersed about the mean, and a small standard deviation shows that data are tightly clustered around the mean. You can calculate standard deviation by taking the square root of the variance.
## When to Use the Sample or Population Standard Deviation
In statistics, a population refers to the entire set of objects or events being studied, and a sample is a subset of the population. As an example, imagine you are studying national elections in the United States. Your population of interest consists of every single eligible voter across the 50 states. Because you can’t collect data for the entire population, you draw random samples (subsets) of voters and make inferences about the population from these samples. Sometimes statisticians have data for an entire population. Most of the time, however, they have to work with samples.
When you are dealing with population data and want to calculate a standard deviation, use the population standard deviation formula given above. A population standard deviation is denoted by the lowercase Greek letter sigma, 𝞂.
When you are dealing with sample data and want to calculate a standard deviation, use the sample standard deviation formula given above. A sample standard deviation is denoted by the lowercase letter s.
## Sample Standard Deviation vs. Sample Variance
We calculate standard deviation as the square root of the variance. This may cause you to wonder: why do we use standard deviation when we already have variance? Great question!
To answer this question, first notice that in both the equation for variance and the equation for standard deviation, you take the squared deviation (the squared distances) between each data point and the sample mean $(x_i-\bar{x})^2$. You do this so that the negative distances between the mean and the data points below the mean do not cancel out the positive distances between the mean and data points above it.
By squaring deviations, we convert each deviation into a positive number, and from there, you can measure average dispersion. There is a downside to doing this, though. By squaring the deviations, you square the units of the data as well. Imagine your data is measured in ounces. By taking squared deviations, you are now measuring the dispersion in your data in terms of ounces squared — a unit of measurement that is difficult to interpret.
Standard deviation is intended to be a standardized measure of dispersion. By taking the square root of the variance, we convert the statistic back into the same units as the data. Ounces, not ounces squared!
## How to Calculate a Sample Standard Deviation Step by Step
Say you have the following sample data for temperatures measured in degrees Fahrenheit. For simplicity, this sample just has four observations (n=4).
Sample Data {51 °F, 58 °F, 61 °F, 62 °F}
Using the formula for sample standard deviation, let’s go through a step-by-step example of how to find the standard deviation for this sample.
$s = \sqrt{\frac{\sum_{}^{}(x_i-\bar{x})^2}{n-1}}$
### STEP 1
Calculate the sample mean x̅.
$\bar{x}=\frac{51+58+61+62}{4} = 58 \degree F$
### STEP 2
Calculate the deviation between each data point and the sample mean $(x_i-\bar{x})$.
$51-58 = - 7$
$58-58 = 0$
$61-58 = 3°F$
$62-58 = 4°F$
Notice that there are both positive and negative deviations.
### STEP 3
Square the deviations, $(x_i-\bar{x})^2$.
$\left(-7\right)^2=49$
$0^2 = 0$
$3^2 = 9$
$4^2 =16$
By squaring the deviations, we turn all of the deviations positive.
### STEP 4
Sum the squared distances, $\sum(x_i-\bar{x})^2$.
$49+0+9+16 = 74$
### STEP 5
Divide the sum of the squares by the number of data points minus one, $\dfrac{\sum(x_i-\bar{x})^2}{(n-1)}$.
Notice that this is the variance, $s^2$, and it is measured in degrees Fahrenheit squared!
### STEP 6
Take the square root of the variance.
$s = \sqrt{\frac{\sum_{}^{}(x_i-\bar{x})^2}{n-1}} = \sqrt{24.6\bar{6}} = 4.97$
And there is your sample standard deviation. 4.97 °F.
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MEAP Preparation - Grade 6 Mathematics3.18 Adding and Subtracting with Different Units
Mathematical operations are carried out in the same way as normal decimal number operations. Note, that the quantities involved in the calculations are all expressed in the same units. Examples: 19 cm + 23 m + 25.9 m First convert everything to the same units: 19 cm = 0.19 meters 0.19 m + 23 m + 25.9 m = 49.09 m Another way is to convert everything to centimeters and perform addition operation 23 m = 1300 cm 25.9 m = 1590 cm 19 cm + 2 300 cm + 2 590 cm = 4 909 cm Final answer is expressed in meters as: 49.09 m A rope is 3 metres long. You cut 100 cms out of it. How many metres of the rope are remaining? 1 metre = 100 cms Hence 3 m - 1m = 2m Answer: 2 m Directions: Answer the following questions. Also write at least ten examples of your own.
Q 1: Subtract: 21 ft 4 in - 15 ft 9 in =7 ft 5 in5 ft 7 in5 ft 5 in Q 2: Your school is 1 km east of your home and the library is 500 m east of the school? How far is the library from your home?1 km1.5 km1.500 m500 m Q 3: Add: 1 ft 3 in + 2 ft 4 in =2 ft 6 in12 ft in3 ft 7 in Q 4: One fruit roll up is 2 ft long. You ate 1 ft 2 inches of it. How much is left?1 ft10 in2 in12 in Q 5: When you add or subtract lengths that are in different units you need to regroup them.FalseTrue Q 6: Add: 2 l 68 ml + 27 ml = (hint: 2.68 + 0.027=)2.007 liters2.707 liters2.007 liters Question 7: This question is available to subscribers only! Question 8: This question is available to subscribers only!
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# Square & Square Root of 80
Last Updated: April 28, 2024
## Square of 80
80² (80×80) = 6400
To calculate the square of 80, you simply multiply 80 by itself:
Therefore, the square of 80 is 6400. This straightforward calculation is a building block for more complex mathematical operations and concepts, including algebraic equations, geometric formulas, and statistical models.
## Square Root of 80
√80 = 8.94427191
The square root of 80, denoted as √80, equals approximately 8.94. To compute it, you find the number that, when multiplied by itself, results in 80. In mathematical terms, finding the square root means identifying the number that, when raised to the power of 2, equals the original number. Visually, you can represent the square root of 80 as one side of a square with an area of 80 square units, where each side of the square is approximately 8.94 units in length. Understanding square roots is fundamental in various mathematical concepts and applications, such as geometry, algebra, and solving problems that involve areas or other quantities that are squared. In real-world scenarios, knowing the square root of 80 aids in calculations involving the measurement of areas or in any situation where you need to reverse a squaring operation to find an original quantity.
Square Root of 80: “8.94427191”
Exponential Form: 80^1/2 or 80^0.5
## Is the Square Root of 80 Rational or Irrational?
• A rational number is any number that can be expressed as a fraction a/b where a and b are integers, and b is not equal to zero. It includes integers, fractions, and finite or repeating decimals.
• An irrational number is a number that cannot be expressed as a simple fraction. Irrational numbers have non-repeating, non-terminating decimal expansions.
The square root of 80 is irrational
An irrational number is one that cannot be expressed as a simple fraction (a ratio of two integers). The square root of 80, when simplified, is 80=16×5=45. Since 5 is not a perfect square and its decimal form is non-terminating and non-repeating, it cannot be expressed as a fraction of two integers. Therefore, 45 (and thus, the square root of 80) is irrational, as it cannot be precisely represented as a fraction.
## Methods to Find Value of Root 80
Finding the value of the square root of 80 can be approached in several ways, depending on the level of precision you need and the tools you have available. Here are some methods:
1. Using a Calculator
The simplest and most accurate method to find the square root of 80 is to use a calculator. Just enter “80” followed by the square root function (√), and it will give you the precise value, which is approximately 8.94427.
2. Estimation
If you don’t have a calculator handy, you can estimate the square root of 80 by identifying the perfect squares closest to it, which are 64 (8²) and 81 (9²). Since 80 is closer to 81, you know that the square root of 80 is slightly less than 9.
3. Prime Factorization
You can also use prime factorization to simplify the square root:
1. Start by writing 80 as a product of its prime factors: 80=2⁴×5.
2. Simplify the square root by taking out pairs of prime factors: √80=√2⁴×√5=2²√5=4√5.
3. This gives you a simplified radical form, but you’d still need a calculator for the precise decimal value of √5.
## Square Root of 80 by Long Division Method
Finding the square root of 80 using the long division method can be made simple with these steps:
Start with 80: Write 80 down and put a bar over it. If you’re finding decimals, pair zeros in twos from left to right after the decimal point.
Find the largest square: Look for a number that squares to less than or equal to 80. Here, 8 works because 8 times 8 is 64. Subtract 64 from 80, leaving a remainder of 16.
Bring down zeros: Add a pair of zeros (making it 1600) to the remainder for more precision.
Double and guess: Take the quotient (8), double it (making 16), and put a space next to it for a new digit. Now, guess a digit (X) that makes 16X times X less than or equal to 1600. This is your next step in the division.
Continue for decimals: Keep repeating the process, adding pairs of zeros and guessing the next digit, to find the decimal part of the square root as precisely as you like.
This method helps you break down finding the square root of 80 into manageable steps, allowing you to get as close to the actual value as you need, even including decimal places.
## 80 is Perfect Square root or Not
No, 80 is not a perfect square
A perfect square is a number that is the square of an integer, and since no integer squared equals 80, it cannot be considered a perfect square.
## What is the perfect square of 80?
The term “perfect square” refers to numbers that are squares of integers. 80 is not a perfect square itself, but the closest perfect squares are 64 (8^2) and 81 (9^2).
## Which numbers square is 80?
No integer’s square is 80, but the square root of 80 (√80) is approximately 8.94. This means 8.94 squared, or 8.94^2, equals 80. This value is an irrational number.
## Which number is closest to √80?
The number closest to √80 is 9. The square root of 81 is exactly 9, and 81 is the perfect square nearest to 80. So, √80 is just slightly less than 9.
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# 10.2: Graphs of Quadratic Functions in Intercept Form
Difficulty Level: At Grade Created by: CK-12
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Practice Graphs of Quadratic Functions in Intercept Form
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### Graphs of Quadratic Functions in Intercept Form
Now it’s time to learn how to graph a parabola without having to use a table with a large number of points.
Let’s look at the graph of y=x26x+8\begin{align*}y=x^2 - 6x + 8\end{align*}.
There are several things we can notice:
• The parabola crosses the x\begin{align*}x-\end{align*}axis at two points: x=2\begin{align*}x = 2\end{align*} and x=4\begin{align*}x = 4\end{align*}. These points are called the x\begin{align*}x -\end{align*} intercepts of the parabola.
• The lowest point of the parabola occurs at (3, -1).
• This point is called the vertex of the parabola.
• The vertex is the lowest point in any parabola that turns upward, or the highest point in any parabola that turns downward.
• The vertex is exactly halfway between the two x\begin{align*}x-\end{align*}intercepts. This will always be the case, and you can find the vertex using that property.
• The parabola is symmetric. If you draw a vertical line through the vertex, you see that the two halves of the parabola are mirror images of each other. This vertical line is called the line of symmetry.
We said that the general form of a quadratic function is y=ax2+bx+c\begin{align*}y = ax^2 + bx + c\end{align*}. When we can factor a quadratic expression, we can rewrite the function in intercept form:
y=a(xm)(xn)\begin{align*}y = a(x - m)(x - n)\end{align*}
This form is very useful because it makes it easy for us to find the x\begin{align*}x -\end{align*}intercepts and the vertex of the parabola. The x\begin{align*}x-\end{align*}intercepts are the values of x\begin{align*}x\end{align*} where the graph crosses the x\begin{align*}x-\end{align*}axis; in other words, they are the values of x\begin{align*}x\end{align*} when y=0\begin{align*}y = 0\end{align*}. To find the x\begin{align*}x-\end{align*}intercepts from the quadratic function, we set y=0\begin{align*}y = 0\end{align*} and solve:
0=a(xm)(xn)\begin{align*}0 = a(x - m)(x - n)\end{align*}
Since the equation is already factored, we use the zero-product property to set each factor equal to zero and solve the individual linear equations:
xm=0x=morxnx=0=n\begin{align*}x - m = 0 & & & & x - n & = 0\\ & & \text{or} \\ x = m & & & & x & = n\end{align*}
So the x\begin{align*}x-\end{align*}intercepts are at points (m,0)\begin{align*}(m, 0)\end{align*} and (n,0)\begin{align*}(n, 0)\end{align*}.
Once we find the x\begin{align*}x-\end{align*}intercepts, it’s simple to find the vertex. The x\begin{align*}x-\end{align*}value of the vertex is halfway between the two x\begin{align*}x-\end{align*}intercepts, so we can find it by taking the average of the two values: m+n2\begin{align*}\frac{m + n}{2}\end{align*}. Then we can find the y\begin{align*}y-\end{align*}value by plugging the value of x\begin{align*}x\end{align*} back into the equation of the function.
#### Finding the x−\begin{align*}x-\end{align*}Intercepts and the Vertex
Find the x\begin{align*}x-\end{align*}intercepts and the vertex of the following quadratic functions:
a) y=x28x+15\begin{align*}y = x^2 - 8x + 15\end{align*}
y=x28x+15\begin{align*}y = x^2 - 8x + 15\end{align*}
Write the quadratic function in intercept form by factoring the right hand side of the equation. Remember, to factor we need two numbers whose product is 15 and whose sum is –8. These numbers are –5 and –3.
The function in intercept form is y=(x5)(x3)\begin{align*}y = (x - 5)(x - 3)\end{align*}
We find the x\begin{align*}x-\end{align*}intercepts by setting y=0\begin{align*}y = 0\end{align*}.
We have:
0=(x5)(x3)x5=0x=5orx3=0x=3\begin{align*}&0 = (x - 5)(x - 3) \\ & x - 5 = 0 & & && x - 3 = 0\\ &&& \text{or}\\ & x = 5 & & && x = 3\end{align*}
So the x\begin{align*}x-\end{align*}intercepts are (5, 0) and (3, 0).
The vertex is halfway between the two x\begin{align*}x-\end{align*}intercepts. We find the x\begin{align*}x-\end{align*}value by taking the average of the two x\begin{align*}x-\end{align*}intercepts: x=5+32=4\begin{align*}x = \frac{5 + 3}{2} = 4\end{align*}
We find the y\begin{align*}y-\end{align*}value by plugging the x\begin{align*}x-\end{align*}value we just found into the original equation:
y=x28x+15y=428(4)+15=1632+15=1\begin{align*}y = x^2 - 8x + 15 \Rightarrow y = 4^2 - 8(4) + 15 = 16 - 32 + 15 = -1\end{align*}
So the vertex is (4, -1).
b) y=3x2+6x24\begin{align*}y = 3x^2 + 6x - 24\end{align*}
y=3x2+6x24\begin{align*}y = 3x^2 + 6x - 24\end{align*}
Re-write the function in intercept form.
Factor the common term of 3 first: y=3(x2+2x8)\begin{align*}y = 3(x^2 + 2x - 8)\end{align*}
Then factor completely: y=3(x+4)(x2)\begin{align*}y = 3(x + 4)(x - 2)\end{align*}
Set y=0\begin{align*}y = 0\end{align*} and solve:
0=3(x+4)(x2)x+4=0orx=4x2=0x=2\begin{align*}& && x + 4 = 0 && x - 2 = 0 \\ & 0 = 3( x + 4)(x - 2) \Rightarrow && \qquad \qquad \qquad \quad \text{or} \\ & && x = -4 && x = 2\end{align*}
The x\begin{align*}x-\end{align*}intercepts are (-4, 0) and (2, 0).
For the vertex,
\begin{align*}x = \frac{-4 + 2}{2} = -1\end{align*} and \begin{align*}y = 3(-1)^2 + 6( -1) - 24 = 3 - 6 -24 = -27\end{align*}
The vertex is: (-1, -27)
Knowing the vertex and \begin{align*}x-\end{align*}intercepts is a useful first step toward being able to graph quadratic functions more easily. Knowing the vertex tells us where the middle of the parabola is. When making a table of values, we can make sure to pick the vertex as a point in the table. Then we choose just a few smaller and larger values of \begin{align*}x\end{align*}. In this way, we get an accurate graph of the quadratic function without having to have too many points in our table.
#### Graphing Functions
Find the \begin{align*}x-\end{align*}intercepts and vertex. Use these points to create a table of values and graph each function.
a) \begin{align*}y = x^2 - 4\end{align*}
\begin{align*}y = x^2 - 4\end{align*}
Let’s find the \begin{align*}x-\end{align*}intercepts and the vertex:
Factor the right-hand side of the function to put the equation in intercept form:
\begin{align*}y = (x - 2)(x + 2)\end{align*}
Set \begin{align*}y = 0\end{align*} and solve:
\begin{align*}&0 = (x - 2)(x + 2)\\ & x - 2 = 0 & & & & x + 2 = 0\\ && & \text{or}\\ & x = 2 & & & & x = -2\end{align*}
The \begin{align*}x-\end{align*}intercepts are (2, 0) and (-2, 0).
Find the vertex:
\begin{align*}x = \frac{2 -2}{2} = 0 \quad y = (0)^2 - 4 = -4\end{align*}
The vertex is (0, -4).
Make a table of values using the vertex as the middle point. Pick a few values of \begin{align*}x\end{align*} smaller and larger than \begin{align*}x = 0\end{align*}. Include the \begin{align*}x-\end{align*}intercepts in the table.
\begin{align*}x\end{align*} \begin{align*}y = x^2 - 4\end{align*}
\begin{align*}-3\end{align*} \begin{align*}y = (-3)^2 - 4 = 5\end{align*}
–2 \begin{align*}y = (-2)^2 - 4 = 0\end{align*} \begin{align*}x-\end{align*}intercept
–1 \begin{align*}y = (-1)^2 - 4 = -3\end{align*}
0 \begin{align*}y = (0)^2 - 4 = -4\end{align*} vertex
1 \begin{align*}y = (1)^2 - 4 = -3\end{align*}
2 \begin{align*}y = (2)^2 - 4 = 0\end{align*} \begin{align*}x-\end{align*}intercept
3 \begin{align*}y = (3)^2 - 4 = 5\end{align*}
Then plot the graph:
b) \begin{align*} y = -x^2 + 14x - 48\end{align*}
\begin{align*}y = -x^2 + 14x - 48\end{align*}
Let’s find the \begin{align*}x-\end{align*}intercepts and the vertex:
Factor the right-hand-side of the function to put the equation in intercept form:
\begin{align*}y = -(x^2 - 14x + 48) = -(x - 6)(x - 8)\end{align*}
Set \begin{align*}y = 0\end{align*} and solve:
\begin{align*}&0 = -(x - 6)(x - 8)\\ & x - 6 = 0 & & & & x - 8 = 0\\ && & \text{or} \\ & x = 6 & & & & x = 8\end{align*}
The \begin{align*}x-\end{align*}intercepts are (6, 0) and (8, 0).
Find the vertex:
\begin{align*}x = \frac{6 + 8}{2} = 7 \quad y = -(7)^2 + 14(7) - 48 = 1\end{align*}
The vertex is (7, 1).
Make a table of values using the vertex as the middle point. Pick a few values of \begin{align*}x\end{align*} smaller and larger than \begin{align*}x = 7\end{align*}. Include the \begin{align*}x-\end{align*}intercepts in the table.
\begin{align*}x\end{align*} \begin{align*}y = -x^2 + 14x - 48\end{align*}
4 \begin{align*}y = -(4)^2 + 14(4) - 48 = -8\end{align*}
5 \begin{align*}y = -(5)^2 + 14(5) - 48 = -3\end{align*}
6 \begin{align*}y = -(6)^2 + 14(6) - 48 = 0\end{align*}
7 \begin{align*}y = -(7)^2 + 14(7) - 48 = 1\end{align*}
8 \begin{align*}y = -(8)^2 + 14(8) - 48 = 0\end{align*}
9 \begin{align*}y = -(9)^2 + 14(9) - 48 = -3\end{align*}
10 \begin{align*}y = -(10)^2 + 14(10) - 48 = -8\end{align*}
Then plot the graph:
#### Applications of Quadratic Functions to Real-World Problems
As we mentioned in a previous concept, parabolic curves are common in real-world applications. Here we will look at a few graphs that represent some examples of real-life application of quadratic functions.
#### Real-World Application: Fencing
Andrew has 100 feet of fence to enclose a rectangular tomato patch. What should the dimensions of the rectangle be in order for the rectangle to have the greatest possible area?
Drawing a picture will help us find an equation to describe this situation:
If the length of the rectangle is \begin{align*}x\end{align*}, then the width is \begin{align*}50 - x\end{align*}. (The length and the width add up to 50, not 100, because two lengths and two widths together add up to 100.)
If we let \begin{align*}y\end{align*} be the area of the triangle, then we know that the area is length \begin{align*}\times\end{align*} width, so \begin{align*}y = x (50 - x) = 50x - x^2\end{align*}.
Here’s the graph of that function, so we can see how the area of the rectangle depends on the length of the rectangle:
We can see from the graph that the highest value of the area occurs when the length of the rectangle is 25. The area of the rectangle for this side length equals 625. (Notice that the width is also 25, which makes the shape a square with side length 25.)
This is an example of an optimization problem. These problems show up often in the real world, and if you ever study calculus, you’ll learn how to solve them without graphs.
### Example
#### Example 1
Anne is playing golf. On the \begin{align*}4^{th}\end{align*} tee, she hits a slow shot down the level fairway. The ball follows a parabolic path described by the equation \begin{align*}y = x - 0.04x^2\end{align*}, where \begin{align*}y\end{align*} is the ball’s height in the air and \begin{align*}x\end{align*} is the horizontal distance it has traveled from the tee. The distances are measured in feet. How far from the tee does the ball hit the ground? At what distance from the tee does the ball attain its maximum height? What is the maximum height?
Let’s graph the equation of the path of the ball:
\begin{align*}x(1 - 0.04x) = 0\end{align*} has solutions \begin{align*}x = 0\end{align*} and \begin{align*} x = 25\end{align*}.
From the graph, we see that the ball hits the ground 25 feet from the tee. (The other \begin{align*}x-\end{align*}intercept, \begin{align*}x = 0,\end{align*} tells us that the ball was also on the ground when it was on the tee!)
We can also see that the ball reaches its maximum height of about 6.25 feet when it is 12.5 feet from the tee.
### Review
For 1-4, rewrite the following functions in intercept form. Find the \begin{align*}x-\end{align*}intercepts and the vertex.
1. \begin{align*}y =x^2 - 2x - 8\end{align*}
2. \begin{align*}y = -x^2 + 10x - 21\end{align*}
3. \begin{align*}y =2x^2 + 6x + 4\end{align*}
4. \begin{align*}y =3(x+5)(x - 2)\end{align*}
For 5-8, the vertex of which parabola is higher?
1. \begin{align*}y = x^2 + 4\end{align*} or \begin{align*}y = x^2 +1\end{align*}
2. \begin{align*}y= -2x^2\end{align*} or \begin{align*}y = -2x^2 - 2\end{align*}
3. \begin{align*}y = 3x^2 - 3\end{align*} or \begin{align*}y = 3x^2 - 6\end{align*}
4. \begin{align*}y = 5 - 2x^2\end{align*} or \begin{align*}y = 8 - 2x^2\end{align*}
For 9-14, graph the following functions by making a table of values. Use the vertex and \begin{align*}x-\end{align*}intercepts to help you pick values for the table.
1. \begin{align*}y = 4x^2 - 4\end{align*}
2. \begin{align*}y = -x^2 + x + 12\end{align*}
3. \begin{align*}y = 2x^2 + 10x + 8\end{align*}
4. \begin{align*}y = \frac{1}{2} x^2 - 2x\end{align*}
5. \begin{align*}y = x - 2x^2\end{align*}
6. \begin{align*}y = 4x^2 -8x + 4\end{align*}
1. Nadia is throwing a ball to Peter. Peter does not catch the ball and it hits the ground. The graph shows the path of the ball as it flies through the air. The equation that describes the path of the ball is \begin{align*}y = 4 + 2x - 0.16x^2\end{align*}. Here \begin{align*}y\end{align*} is the height of the ball and \begin{align*}x\end{align*}is the horizontal distance from Nadia. Both distances are measured in feet.
1. How far from Nadia does the ball hit the ground?
2. At what distance \begin{align*}x\end{align*} from Nadia, does the ball attain its maximum height?
3. What is the maximum height?
2. Jasreel wants to enclose a vegetable patch with 120 feet of fencing. He wants to put the vegetable against an existing wall, so he only needs fence for three of the sides. The equation for the area is given by \begin{align*}A = 120x - x^2\end{align*}. From the graph, find what dimensions of the rectangle would give him the greatest area.
To view the Review answers, open this PDF file and look for section 10.2.
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### Vocabulary Language: English
Intercepts
The intercepts of a curve are the locations where the curve intersects the $x$ and $y$ axes. An $x$ intercept is a point at which the curve intersects the $x$-axis. A $y$ intercept is a point at which the curve intersects the $y$-axis.
Line of Symmetry
A line of symmetry is a line that can be drawn to divide a figure into equal halves.
Vertex
The vertex of a parabola is the highest or lowest point on the graph of a parabola. The vertex is the maximum point of a parabola that opens downward and the minimum point of a parabola that opens upward.
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# 5.3: Medians and Altitudes in Triangles
Difficulty Level: At Grade Created by: CK-12
## Learning Objectives
• Define median and find the properties of the centroid.
• Apply medians to the coordinate plane.
• Construct the altitude of a triangle.
## Review Queue
1. Find the midpoint between (9, -1) and (1, 15).
2. Find the slope between (9, -1) and (1, 15). Then find the equation of the line.
3. Find the equation of the line that is perpendicular to the line from #2 through (-6, 2).
Know What? Triangles are frequently used in art. Your art teacher assigns an art project involving triangles. You decide to make a series of hanging triangles of all different sizes from one long piece of wire. Where should you hang the triangles from so that they balance horizontally?
## Medians
Median: The line segment that joins a vertex and the midpoint of the opposite side (of a triangle).
LO¯¯¯¯¯¯¯\begin{align*}\overline{LO}\end{align*} is the median from L\begin{align*}L\end{align*} to the midpoint of NM¯¯¯¯¯¯¯¯¯¯\begin{align*}\overline{NM}\end{align*}.
Example 1: Find the other two medians of LMN\begin{align*}\triangle LMN\end{align*}.
Solution: Find the midpoints of sides LN¯¯¯¯¯¯¯¯\begin{align*}\overline{LN}\end{align*} and LM¯¯¯¯¯¯¯¯¯\begin{align*}\overline{LM}\end{align*}, using a ruler. Be sure to always include the appropriate tick marks for the midpoints.
Centroid: The point of intersection for the medians of a triangle.
Investigation 5-5: Properties of the Centroid
Tools Needed: pencil, paper, ruler, compass
1. Construct a scalene triangle with sides of length 6 cm, 10 cm, and 12 cm (Investigation 4-2). Use the ruler to measure each side and mark the midpoint.
2. Draw in the medians and mark the centroid.
Measure the length of each median. Then, measure the length from each vertex to the centroid and from the centroid to the midpoint. Do you notice anything?
3. Cut out the triangle. Place the centroid on either the tip of the pencil or the pointer of the compass. What happens?
The properties discovered are summarized below.
Median Theorem: The medians of a triangle intersect at a point that is two-thirds of the distance from the vertices to the midpoint of the opposite side. The centroid is the “balancing point” of a triangle.
If G\begin{align*}G\end{align*} is the centroid, then:
AGDGAnd:DG=23AD, CG=23CF, EG=23BE=13AD, FG=13CF, BG=13BE=12AG, FG=12CG, BG=12EG\begin{align*}AG &= \frac{2}{3} AD, \ CG = \frac{2}{3} CF, \ EG = \frac{2}{3} BE\\ DG &= \frac{1}{3} AD, \ FG = \frac{1}{3} CF, \ BG = \frac{1}{3} BE\\ \text{And}: \quad DG &= \frac{1}{2} AG, \ FG = \frac{1}{2} CG, \ BG = \frac{1}{2} EG\end{align*}
Example 2: I, K\begin{align*}I, \ K\end{align*}, and M\begin{align*}M\end{align*} are midpoints of the sides of HJL\begin{align*}\triangle HJL\end{align*}.
a) If JM=18\begin{align*}JM = 18\end{align*}, find JN\begin{align*}JN\end{align*} and NM\begin{align*}NM\end{align*}.
b) If HN=14\begin{align*}HN = 14\end{align*}, find NK\begin{align*}NK\end{align*} and HK\begin{align*}HK\end{align*}.
Solution:
a) JN=2318=12\begin{align*}JN = \frac{2}{3} \cdot 18 = 12\end{align*}. NM=JMJN=1812\begin{align*}NM = JM - JN = 18 - 12\end{align*}. NM=6.\begin{align*}NM = 6.\end{align*}
b) 14=23HK\begin{align*}14 = \frac{2}{3} \cdot HK\end{align*}
1432=HK=21\begin{align*}14 \cdot \frac{3}{2} = HK = 21\end{align*}. NK\begin{align*}NK\end{align*} is a third of 21, NK=7\begin{align*}NK = 7\end{align*}.
Example 3: Algebra Connection H\begin{align*}H\end{align*} is the centroid of ABC\begin{align*}\triangle ABC\end{align*} and DC=5y16\begin{align*}DC = 5y - 16\end{align*}. Find x\begin{align*}x\end{align*} and y\begin{align*}y\end{align*}.
Solution:
12BH=HF3x+63x+68BH=2HF=2(2x1)=4x2=xHC=23DC32HC=DC 32(2y+8)=5y16 3y+12=5y16 28=2y\begin{align*}\frac{1}{2} BH= HF & \longrightarrow BH = 2HF && HC = \frac{2}{3} DC \longrightarrow \frac{3}{2} HC = DC\\ 3x + 6 &= 2(2x - 1) && \quad \ \frac{3}{2} (2y + 8) = 5y - 16\\ 3x + 6 &= 4x - 2 && \qquad \ 3y + 12 = 5y - 16\\ 8 &= x && \qquad \ \ \qquad 28 = 2y\end{align*}
## Altitudes
The last line segment within a triangle is an altitude. It is also called the height of a triangle.
Altitude: A line segment from a vertex and perpendicular to the opposite side. The red lines below are all altitudes.
When a triangle is a right triangle, the altitude, or height, is the leg. If the triangle is obtuse, then the altitude will be outside of the triangle.
Investigation 5-6: Constructing an Altitude for an Obtuse Triangle
Tools Needed: pencil, paper, compass, ruler
1. Draw an obtuse triangle. Label it ABC\begin{align*}\triangle ABC\end{align*}, like the picture to the right. Extend side AC¯¯¯¯¯¯¯¯\begin{align*}\overline{AC}\end{align*}, beyond point A\begin{align*}A\end{align*}.
2. Using Investigation 3-2, construct a perpendicular line to AC¯¯¯¯¯¯¯¯\begin{align*}\overline{AC}\end{align*}, through B\begin{align*}B\end{align*}.
The altitude does not have to extend past side AC¯¯¯¯¯¯¯¯\begin{align*}\overline{AC}\end{align*}, as it does in the picture. Technically the height is only the vertical distance from the highest vertex to the opposite side.
If you are constructing an altitude for an acute triangle, you may skip Step 1 of this investigation.
Know What? Revisited The point that you should put the wire through is the centroid. That way, each triangle will balance.
## Review Questions
• Questions 1-4 use Investigation 5-5.
• Questions 5-6 use Investigation 3-2 and 5-6.
• Questions 7-18 are similar to Examples 2 and 3.
• Questions 19-26 use review to discover something new.
• Questions 27-34 use the definitions of perpendicular bisector, angle bisector, median and altitude.
• Question 35 is similar to the proofs in the previous section.
Construction Construct the centroid for the following triangles by tracing each triangle onto a piece of paper and using Investigation 5-5.
1. Is the centroid always going to be inside of the triangle? Why?
Construction Construct the altitude from the top vertex for the following triangles. Trace each triangle onto a piece of paper and using Investigations 3-2 and 5-6.
For questions 7-11, B, D\begin{align*}B, \ D\end{align*}, and F\begin{align*}F\end{align*} are the midpoints of each side and G\begin{align*}G\end{align*} is the centroid. Find the following lengths.
1. If BG=5\begin{align*}BG = 5\end{align*}, find GE\begin{align*}GE\end{align*} and BE\begin{align*}BE\end{align*}
2. If CG=16\begin{align*}CG = 16\end{align*}, find GF\begin{align*}GF\end{align*} and CF\begin{align*}CF\end{align*}
3. If AD=30\begin{align*}AD = 30\end{align*}, find AG\begin{align*}AG\end{align*} and GD\begin{align*}GD\end{align*}
4. If GF=x\begin{align*}GF = x\end{align*}, find GC\begin{align*}GC\end{align*} and \begin{align*}CF\end{align*}
5. If \begin{align*}AG = 9x\end{align*} and \begin{align*}GD = 5x - 1\end{align*}, find \begin{align*}x\end{align*} and \begin{align*}AD\end{align*}.
For questions 12-18, \begin{align*}N\end{align*} and \begin{align*}M\end{align*} are the midpoints of sides \begin{align*}\overline{XY}\end{align*}and \begin{align*}\overline{ZY}\end{align*}.
1. What is point \begin{align*}C\end{align*}?
2. If \begin{align*}XN = 5\end{align*}, find \begin{align*}XY\end{align*}.
3. If \begin{align*}XC = 6\end{align*}, find \begin{align*}XM\end{align*}.
4. If \begin{align*}ZN = 45\end{align*}, find \begin{align*}CN\end{align*}.
5. If \begin{align*}CM = 4\end{align*}, find \begin{align*}XC\end{align*}.
6. If \begin{align*}ZM = y\end{align*}, find \begin{align*}ZY\end{align*}.
7. If \begin{align*}ZN = 6x + 15\end{align*} and \begin{align*}ZC = 38\end{align*}, find \begin{align*}x\end{align*} and \begin{align*}ZN\end{align*}.
Multistep Problem Find the equation of a median in the \begin{align*}x-y\end{align*} plane.
1. Plot \begin{align*}\triangle ABC: \ A(-6, 4), \ B(-2, 4)\end{align*} and \begin{align*}C(6, -4)\end{align*}
2. Find the midpoint of \begin{align*}\overline{AC}\end{align*}. Label it \begin{align*}D\end{align*}.
3. Find the slope of \begin{align*}\overline{BD}\end{align*}.
4. Find the equation of \begin{align*}\overline{BD}\end{align*}.
5. Plot \begin{align*}\triangle DEF: \ D(-1, 5), \ E(0, -1), \ F(6, 3)\end{align*}
6. Find the midpoint of \begin{align*}\overline{EF}\end{align*}. Label it \begin{align*}G\end{align*}.
7. Find the slope of \begin{align*}\overline{DG}\end{align*}.
8. Find the equation of \begin{align*}\overline{DG}\end{align*}.
Determine if the following statements are true or false.
1. The perpendicular bisector goes through the midpoint of a line segment.
2. The perpendicular bisector goes through a vertex.
3. The angle bisector passes through the midpoint.
4. The median bisects the side it intersects.
5. The angle bisectors intersect at one point.
6. The altitude of an obtuse triangle is inside a triangle.
7. The centroid is the balancing point of a triangle.
8. A median and a perpendicular bisector intersect at the midpoint of the side they intersect.
Fill in the blanks in the proof below.
1. Given: Isoscles \begin{align*}\triangle ABC\end{align*} with legs \begin{align*}\overline{AB}\end{align*} and \begin{align*}\overline{AC}\end{align*} \begin{align*}\overline{BD} \perp \overline{DC}\end{align*} and \begin{align*}\overline{CE} \perp \overline{BE}\end{align*} Prove: \begin{align*}\overline{BD} \cong \overline{CE}\end{align*}
Statement Reason
1. Isosceles \begin{align*}\triangle ABC\end{align*} with legs \begin{align*} \overline{AB}\end{align*} and \begin{align*}\overline{AC}\end{align*}
\begin{align*}\overline{BD} \perp \overline{DC} \end{align*} and \begin{align*}\overline{CE} \perp \overline{BE}\end{align*}
2. \begin{align*}\angle DBC \cong \angle ECB\end{align*}
3. Definition of perpendicular lines
4. \begin{align*}\angle BEC \cong \angle CEB\end{align*}
5. Reflexive PoC
6. \begin{align*}\triangle BEC \cong \triangle CDB\end{align*}
7. \begin{align*}\overline{BD} \cong \overline{CE}\end{align*}
1. \begin{align*}midpoint=\left (\frac{9+1}{2},\frac{-1+15}{2}\right ) = (5,7)\end{align*}
2. \begin{align*}m=\frac{15+1}{1-9}=\frac{16}{-8}=-2 \qquad \qquad 15=-2(1)+b \qquad \qquad y=-2x+17\!\\ {\;}\qquad \qquad \qquad \qquad \qquad \qquad \ \ 17=b\end{align*}
3. \begin{align*}y=\frac{1}{2} x+b\!\\ 2= \frac{1}{2} (-6) + b\!\\ 2=-3+b\!\\ 5=b\!\\ y=\frac{1}{2} x+5\end{align*}
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#### Explain solution RD Sharma class 12 Chapter 10 Differentiation Exercise 10.7 question 16
$\frac{d y}{d x}=\frac{1}{\sqrt{3}} \; \; \text { At }\; \; t=\frac{2 \pi}{3}$
Hint:
Use $[\tan (\pi-\theta)=-\tan \theta]$ and $\frac{d(\sin t)}{d t}=\cos t$
Given:
\begin{aligned} &x=\cos t \\ &y=\sin t \end{aligned}
Solution:
$x=\cos t \\$
\begin{aligned} & &\frac{d x}{d t}=\frac{d \cos t}{d t}=-\sin t \end{aligned} (1)
$y=\sin t$
$\\ \frac{d y}{d t}=\frac{d \sin t}{d t} \\$
\begin{aligned} & &\frac{d y}{d t}=\cos t \end{aligned} (2)
$\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}$
Putting the value of $\frac{d x}{d t} \text { and } \frac{d y}{d t}$ from equation (1) and (2) respectively
$=\frac{\cos t}{-\sin t}$
$\frac{d y}{d x}\; \; At\; \; \left(x=\frac{2 \pi}{3}\right)$
\begin{aligned} &=-\cot \left(\frac{2 \pi}{3}\right) \\ &=-\cot \left(\pi-\frac{\pi}{3}\right) \\ &=-\left(-\cot \frac{\pi}{3}\right) \\ &=\cot \frac{\pi}{3} \end{aligned}
$\frac{d y}{d x}\;\; At\; \; \left(x=\frac{2 \pi}{3}\right)=\frac{1}{\sqrt{3}}$
Hence proved
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# Area in Square Centimeters
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Area in Square Centimeters
CCSS.MATH.CONTENT.4.MD.A.3
## What is the Area in Square Centimeters?
Measuring the area of a shape means that you are measuring the amount of space inside the shape. This is done for a flat object such as a rectangle, square, or triangle and is commonly done in units called square centimeters (cm²). Therefore, when we are asked to find the area in square centimeters of a flat shape, we are basically being asked for the number of 1 cm² squares that can fit into that shape.
For example, the area in square centimeters of the following rectangle is 12 square centimeters. This means that 12 squares, which each measure 1 square centimeter, fit inside that rectangle.
## Finding the Area in Square Centimeters of a Rectangle
We can find the area in square centimeters of an object by counting the number of squares within that shape. However, this is not always possible and can take a lot of time, so mathematicians use formulas to calculate the area. To find the area in square centimeters of a rectangle, you can multiply the length by the width using the following formula:
A = l x w
A is the area, l is the length, and w is the width.
## Area in Square Centimeters – Example
Looking at the rectangle from earlier, notice that there are 12 squares inside. This means that the area in square centimeters is 12.
We could have also found the same answer using the formula.
A = l x w = 4cm x 3cm =12 cm²
### What is the Area in Square Centimeters of the United States?
To find the area of the United States in square centimeters, we can multiply the area of the United States in square kilometers by 10,000,000,000 to convert square kilometers to square centimeters. The area of the United States in square kilometers is 9,834,000 square kilometers (km²). Therefore, if we want to find the area of the United States in square centimeters, we would calculate it like this:
Area in km² Factor Area in cm²
9,834,000 x 10,000,000,000 98,340,000,000,000,000
Wow!! That’s a huge number!
### Area in Square Centimeters – Summary
The area in square centimeters of a flat object is the number of 1 cm² squares that can fit in it.
When finding the area in square centimeters of a rectangle;
• First: ensure that the length and width are in centimeters
• Then: Use the formula area = length x width and multiply the length by the width
Continue learning about area in square centimeters by using our interactive exercises, worksheets and other fun activities.
## Frequently Asked Questions on Area in Square Centimeters
What is the area in square centimeters?
What is the area of a rectangle in square centimeters?
What is the area of a circle in square centimeters?
What is the area of the United States in square centimeters?
### TranscriptArea in Square Centimeters
[Nervous because he doesn’t want to get stung] “Zuri, what are we going to do?” “I don’t think he plans on going anywhere.” “Maybe we can convince him to leave by giving him a nice place of his own?” “Yes, we can build him the perfect place to live!” We can help Zuri and Freddie build their unwanted guest the perfect home by measuring ... “Area in Square Centimeters”. When measuring shapes, we can find the distance around the outside of the shape, called the perimeter... and the amount of space a shape covers on the inside. AREA is the measurement of the INSIDE of a shape. It is the amount of space taken up by an object on a flat surface. We can find the area of a rectangular shape by using the formula, Area equals length times width... or equals times for short. Using this formula and the two measurements we do know, we can solve for unknown parts. Zuri and Freddie found some popsicle sticks in the landfill and decided to make Mr. Bee a lovely fenced-in flower garden! The garden covers ninety-six square centimeters in all... and the garden is eight centimeters wide. The measurement we are missing is the length of the garden. In order to solve, we put all the numbers into the formula... and read this as ninety-six equals the length times eight. To solve for the length, we ask ourselves, what number times eight is equal to ninety-six? (...) Twelve times eight equals ninety-six... so the length is twelve centimeters. Zuri and Freddie found some building blocks and thought these would make a great front porch! The area of the porch measures two hundred fifty square centimeters in all… and the length is twenty-five centimeters. What is the missing measurement? (...) We need to solve for the width. First, we replace the formula with the numbers we know. The equation is two- hundred -fifty equals twenty-five times . To solve for the missing width, we want to ask ourselves (...) 'twenty-five times what number equals two-hundred-fifty?'. What is the width? (...) The width of the porch is ten centimeters. Zuri and Freddie found the PERFECT house...an old box! Look at the dimensions of the area covered in this space. What do we know about the area of the box? (...) We know that the total area covered is seven hundred square centimeters… and the length is thirty-five centimeters. What is the equation for the area of the box? (...) It is seven hundred equals thirty-five times the width. How do we solve the equation? (...) We ask ourselves, 'thirty-five times what number equals seven hundred?' This time, try to find the missing measurement on your own. Pause the video so you have time to work (...) and press play again when you're ready to see the answer! What is the width of the box? It is twenty centimeters since thirty-five times twenty is seven hundred. In the comment section, describe the way you solved this problem. (...) Remember (...) Area is the measurement of the INSIDE of a shape. We solve for the area of a rectangular shape by using the formula, equals times . Replace the measurements in the equation with what we know... and use multiplication strategies to solve for the missing information. "Gee Zuri,(...) I sure am going to miss him!"
## Area in Square Centimeters exercise
Would you like to apply the knowledge you’ve learned? You can review and practice it with the tasks for the video Area in Square Centimeters .
• ### What is the area of each room?
Hints
To find the area of a rectangle, we need to multiply the length by the width.
For example, if we were finding the area of this rectangle we would multiply 6 x 3 to get 18cm².
Solution
To find the area, we multiply the length by the width.
• 5 x 4 = 20; area = 20 cm$^{2}$
• 6 x 4 = 24; area = 24 cm$^{2}$
• 5 x 3 = 15; area = 15 cm$^{2}$
• 7 x 3 = 21; area = 21 cm$^{2}$
Remember to always write the units, such as cm$^{2}$, after an area measurement.
• ### Can you find the missing measurements?
Hints
Remember: Area = length x width or A = l x w.
For the room on the left, you are trying to find the area so you need to multiply the length by the width.
For the room on the right, the length is missing. That means that you need to divide the area by the width.
Solution
The room on the left was missing the area. To find this out we need to:
• multiply the length by the width.
• 6 x 4 = 24.
• Therefore, the area is 24 cm$^{2}$.
The room on the right was missing the length. To find this out we need to:
• divide the area by the width.
• 20 $\div$ 4 = 5.
• Therefore, the length is 5cm.
• ### What is the area of each room?
Hints
First, find the length and the width of each room.
Multiply the length and the width together to find the area.
For example, if we were finding the area of this room we would multiply the length of 4 cm by the width of 3 cm.
• 4 x 3 = 12
• Area = 12 cm$^{2}$
Solution
To find the area we multiply the length by the width.
12 cm$^{2}$
• The room with a length of 4 cm and a width of 3 cm. 4 cm x 3 cm = 12 cm$^{2}$.
• The room with a length of 6 cm and a width of 2 cm. 6 cm x 2 cm = 12 cm$^{2}$.
15 cm$^{2}$
• The room with a length of 5 cm and a width of 3 cm. 5 cm x 3 cm = 15 cm$^{2}$.
20 cm$^{2}$
• The room with a length of 5 cm and a width of 4 cm. 5 cm x 4 cm = 20 cm$^{2}$.
• The room with a length of 10 cm and a width of 2 cm. 10 cm x 2 cm = 20 cm$^{2}$.
• ### Missing measurements.
Hints
To find the area, multiply the length by the width.
To find the length or the width, divide the area by the measurement you know.
Remember, area measurements always have $^{2}$ at the end.
Solution
Wooden board
• We were given the length and the width, so we were finding the area.
• Multiply the length by the width.
• 10 x 10 = 100, therefore the area is 100cm$^{2}$.
Card
• We were given the area and the width, so we were finding the length.
• Divide the area by the width.
• 8 $\div$ 2 = 4, therefore the length is 4 cm.
Photograph frame
• We were given the area and the length, so we were finding the width.
• Divide the area by the length.
• 70 $\div$ 10 = 7, therefore the width is 7 cm.
Tile
• We were given the length and the width, so we were finding the area.
• Multiply the length by the width.
• 9 x 8 = 72, therefore the area is 72cm$^{2}$.
Drawer
• We were given the area and the length, so we were finding the width.
• Divide the area by the length.
• 40 $\div$ 8 = 5, therefore the width is 5 cm.
Note paper
• We were given the area and the length, so we were finding the width.
• Divide the area by the length.
• 150 $\div$ 15 = 10, therefore the width is 10 cm.
• ### What is the area of the garden?
Hints
Remember, area = length x width.
This is the length.
This is the width.
So we need to multiply 4 x 3 to find the area.
Solution
To find the area we multiply the length by the width, so we multiply 4 x 3 to get 12.
The area therefore is 12 cm$^{2}$.
• ### Floor plan puzzle.
Hints
To find the length, you need to divide the area by the width.
Remember, the length of the smaller room is half the length of the larger room. You will need this measurement to find the area of the smaller room.
To find the area, multiply the length by the width.
Solution
First, we needed to find the length of the larger room.
To do this, we needed to divide the area by the width.
60 $\div$ 5 = 12.
We then needed the length measurement of the smaller room. We were told this was half the length of the larger room.
The length of the larger room is 12 cm, so we need to halve 12.
12 $\div$ 2 = 6.
Now we have the length and the width of the smaller room, we can multiply them to find the area.
6 x 5 = 30.
The area of the smaller room is 30 cm$^{2}$.
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# How do you factor 27r^3 + 125s^3?
May 14, 2018
3r+5s)(9r^2-15rs+25s^2)
#### Explanation:
$\text{this is a "color(blue)"sum of cubes}$
$\text{which factors in general as}$
•color(white)(x)a^3+b^3=(a+b)(a^2-ab+b^2)
$27 {r}^{3} = {\left(3 r\right)}^{3} \Rightarrow a = 3 r$
$125 {s}^{3} = {\left(5 s\right)}^{3} \Rightarrow b = 5 s$
$\Rightarrow 27 {r}^{3} + 125 {s}^{3}$
$= \left(3 r + 5 s\right) \left({\left(3 r\right)}^{2} - \left(3 r \times 5 s\right) + {\left(5 s\right)}^{2}\right)$
$= \left(3 r + 5 s\right) \left(9 {r}^{2} - 15 r s + 25 {s}^{2}\right)$
May 14, 2018
$\left(3 r + 5 s\right) \cdot \left(9 {r}^{2} - 15 r s + 25 {s}^{2}\right)$
#### Explanation:
show below
$27 {r}^{3} + 125 {s}^{3}$
${\left(3 r\right)}^{3} + {\left(5 s\right)}^{3}$
$\left(3 r + 5 s\right) \cdot \left(9 {r}^{2} - 15 r s + 25 {s}^{2}\right)$
Note that
${x}^{3} + {y}^{3}$
$\left(x + y\right) \left({x}^{2} - x y + {y}^{2}\right)$
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# How to Calculate Simple Conditional Probabilities
/ / درس 15
### توضیح مختصر
Conditional probability, just like it sounds, is a probability that happens on the condition of a previous event occurring. To calculate conditional probabilities, we must first consider the effects of the previous event on the current event.
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## What Is a Conditional Probability?
A conditional probability is a type of dependent event. Conditional probability involves finding the probability of an event occurring based on a previous event already taking place. To calculate a conditional probability, we must use the process of dependent events because the first event will affect the outcome of the second event. Let’s review the topic of dependent events to help us better understand this process.
## Dependent Events
Dependent events are events in which the previous attempts affect the outcome of subsequent events. Dependent events are just like they sound; each event is dependent upon what happened in the previous attempt. Let’s look at an example of dependent events.
Walt has a big bag of gumballs. In his bag, he has 3 red, 6 green, 8 blue and 2 orange gumballs. What is the probability that Walt will reach into the bag and select a red gumball, then, WITHOUT REPLACING the gumball, reach into the bag and selecting another red gumball?
This probability would look like, P (red, without replacing and drawing another red).
We first need to find the probability of Walt selecting the first red gumball. By adding together all of the gumballs, we can see that there are 19 gumballs in the bag. There are 3 red gumballs in the bag, so the probability of getting a red gumball in the first draw is 3/19.
Next, we need to calculate the probability of Walt selecting a red gumball on his second draw. Remember, Walt did not replace the first gumball, and this will affect our totals. Walt now only has 2 red gumballs left in the bag, and the total number of gumballs is now only 18.
The probability that Walt’s 2nd draw will be a red gumball is 2/18.
To calculate the probability of these two events occurring together, we would multiply the two events.
3/19 x 2/18 = 6/342
Remember that all answers must be in simplest form. 6/342 would be simplified to 1/57.
Walt has a 1/57 chance of drawing two consecutive red gumballs WITHOUT REPLACING the first gumball.
## Conditional Probability
Remember, a conditional probability is a type of dependent event. Conditional probability involves finding the probability of an event occurring based on a previous event already taking place. The difference is that with conditional probabilities, we are just looking at the probability of one specific event occurring.
For example, thinking about Walt and his gumballs, Walt had 3 red, 6 green, 8 blue and 2 orange gumballs.
After first reaching in and selecting a red gumball, and WITHOUT REPLACING it, what is the probability that Walt’s second draw will be another red gumball?
Walt knows that the probability of the first gumball being red is 3/19. Now, he has one less red gumball and one less total gumballs.
We can see that the probability of Walt’s second gumball being another red would be 2/18. Remember, all fractions must be in simplest form. 2/18 would be simplified to 1/9.
The conditional probability that Walt’s second gumball will be red after first drawing a red and not replacing it is 1/9.
## Conditional Probability Example
Let’s look at another example of conditional probability.
A group of teens are preparing to play a game of dodgeball. There are 30 teens that have arrived to play, 19 boys and 11 girls. What is the probability that the second player to be out is a boy after the first person out was also a boy?
We know that if the first person out is a boy, that there will only be 18 boys and 29 teens total left in the game for the 2nd round.
We can now see that the conditional probability that a boy is the second person to be out in dodgeball after the condition that a boy was the first out is 18/29. This fraction is in simplest form, so the conditional probability is 18/29.
## Lesson Summary
Let’s review the important facts that we need to remember when calculating a conditional probability. A conditional probability is a type of dependent event. Conditional probability involves finding the probability of an event occurring based on a previous event already taking place. To calculate a conditional probability, we must use the process of dependent events because the first event will affect the outcome of the second event. The difference is that with conditional probabilities, we are just looking for the probability of that one specific event occurring.
## Learning Outcome
After viewing this video lesson, you should be able to:
Define conditional probability
Explain what a dependent event is and provide an example
Calculate conditional probabilities
Differentiate between dependent events and conditional probabilities
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# How to Solve Special Right Triangles
By Isaiah David
There are two main types of special right triangles: the isosceles right triangle and the 30-60-90 triangle. Solving them can mean finding the value of a side, an angle or multiple sides and angles. By understanding the ratios of the sides to each other and to the angles, you can learn to solve either of these triangles with ease.
## The Basics
Learn what a right triangle is. A right triangle any triangle with a right angle as one of its three angles. A right angle is one that measures exactly 90 degrees.
Learn what a special right triangle is. There are basically two special right triangles: the 30-60-90 right triangle and the 45-45-90, or isosceles, right triangle. The triangles are named for the measures of their three angles in degrees. The isosceles right triangle is also special because it has two sides which are the same.
Learn the parts of a right triangle. Each right triangle has two normal sides and a hypotenuse. The hypotenuse is the side opposite the right angle and is always the biggest side. The other sides are proportional to the angle opposite them. For example, in a 30-60-90 triangle, the side opposite the 30-degree angle is the smallest.
Learn the symbols for equality of angles and sides. Angles with a little curved line in them (drawn as in the illustration) are equal. Sides with a straight line through them have equal length. Angles with a right angle drawn inside (as in the illustration) are equal.
## Solving an Isosceles Right Triangle
Know the ratio of the sides to the hypotenuse. The Pythagorean Theorem states that the square of the hypotenuse is equal to the sum of the square of the other two sides. Therefore, if the two sides have the same value (we'll say that each sides has a length of 1), then the hypotenuse has a value of the square root of 2.
Understand that this ratio holds for any size of triangle. The ratio of any hypotenuse to a side is sqrt2 to 1. Therefore, a right triangle with a side of length 3 would have a hypotenuse of length 3sqrt2.
Learn to recognize an isosceles right triangle. To know that a triangle is an isosceles right triangle, you need to know that it is a right angle and that either the other two sides are equal or that the other two angles are equal. Either will do, since equal sides means equal angles and vice versa.
Learn to derive the hypotenuse from one of the sides. If you know that one side equals 5, for example, and you need to derive the hypotenuse, simply multiply it by sqrt2 to get 5sqrt2.
Learn to solve for a side when you know the hypotenuse. Divide the hypotenuse by the square root of 2 to get the value of either side. For example, if the hypotenuse is 4sqrt2, you get 4sqrt2/sqrt2 = 4therefore, both sides have a length of 4.
## Solving a 30-60-90 Triangle
Learn the ratio of the sides of a 30-60-90 triangle. Unfortunately, you can't recognize a 30-60-90 special right triangle by anything as simple as equal sides and equal angles. But you can recognize it by the ratio of the three sides. The ratio of the 30 side to the 60 side to the hypotenuse is 1 to SQRT3 to 2.
Learn to recognize a 30-60-90 triangle based on the ratio of two sides. If you know that you have a right triangle, and the ratio of the larger side over the smaller is sqrt3 to 1, you have a 30-60-90 triangle. Then, you can fill in the values of the corresponding angles. For example, if the longer side is 2sqrt3 and the shorter is 2, you get:2sqrt3/2 = sqrt3/1.Therefore, the angle opposite the 2 is 30 degrees, and the angle opposite 2sqrt3 is 60 degrees.
Learn to solve based on the ratio of the hypotenuse to one of the sides. If the ratio of the hypotenuse to a side is either 2 to 1 (shorter side) or 2 to sqrt3 (longer side), then you have a 30-60-90 triangle and can fill in the angles.
Fill in the sides based on the angles and the value of one side. Sometimes all you will know is the value of one side and that it is a 30-60-90 triangle. In that case, you can fill in the other sides based on the ratio of the three sides. For example, if you know that you have a 30-60-90 triangle with a short side of 5, you know that the hypotenuse is twice the value, or 10. You also know that the longer side is sqrt3 times the value, or 5sqrt3.
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# Antiderivative
function whose derivative is the original function
Antidifferentiation (also called indefinite integration) is the process of finding a certain function in calculus. It is the opposite of differentiation. It is a way of processing a function to give another function (or class of functions) called an antiderivative. Antidifferentiation is like integration—but without limits. This is why it is called indefinite integration. When represented as single letters, antiderivatives often take the form of capital roman letters such as ${\displaystyle F}$ and ${\displaystyle G}$.[1][2]
In general, an antiderivative is written in the form ${\displaystyle \int f(x)\ dx}$,[3] where:
• The long S, ${\displaystyle \int }$, is called an integral sign.[4] In integration, the integral sign has numbers on it. Those numbers tell you how to do the integration. Antiderivatives are different. They do not have numbers on their integral signs.
• ${\displaystyle x}$ is the equation you are integrating.
• The letters ${\displaystyle dx}$ mean "with respect to ${\displaystyle x}$". This tells you how to do the antidifferentiation.
## Simple antidifferentiation
A function of the form ${\displaystyle ax^{n}}$ can be integrated (antidifferentiated) as follows:
• Add 1 to the power ${\displaystyle n}$ , so ${\displaystyle ax^{n}}$ is now ${\displaystyle ax^{n+1}}$ .
• Divide all this by the new power, so it is now ${\displaystyle {\frac {ax^{n+1}}{n+1}}}$ .
• Add the constant ${\displaystyle c}$ , so it is now ${\displaystyle {\frac {ax^{n+1}}{n+1}}+c}$ .
This can be shown as:
${\displaystyle \int ax^{n}\ dx={\frac {ax^{n+1}}{n+1}}+c}$ (also known as the power rule of integral)[4]
When there are many terms, we can integrate the entire function by integrating its components one by one:
${\displaystyle \int 2x^{6}-5x^{4}\ dx={\frac {2x^{7}}{7}}-{\frac {5x^{5}}{5}}+c={\frac {2}{7}}x^{7}-x^{5}+c}$
(This only works if the parts are being added or taken away.)
### Examples
${\displaystyle \int 3x^{4}\ dx={\frac {3x^{5}}{5}}+c}$
${\displaystyle \int x+x^{2}+x^{3}+x^{4}\ dx={\frac {x^{2}}{2}}+{\frac {x^{3}}{3}}+{\frac {x^{4}}{4}}+{\frac {x^{5}}{5}}+c}$
${\displaystyle \int {\frac {1}{x+4}}\ dx=\ln |x+4|\times 1+c=\ln |x+4|+c}$
Changing fractions and roots into powers makes it easier:
${\displaystyle \int {\frac {1}{x^{3}}}\ dx=\int x^{-3}\ dx={\frac {x^{-2}}{-2}}+c=-{\frac {1}{2x^{2}}}+c}$
${\displaystyle \int {\sqrt {x^{3}}}\ dx=\int x^{\frac {3}{2}}\ dx={\frac {x^{\frac {5}{2}}}{\frac {5}{2}}}+c={\frac {2}{5}}x^{\frac {5}{2}}+c={\frac {2}{5}}{\sqrt {x^{5}}}+c}$
## Integrating a bracket ("chain rule")
To integrate a bracket like ${\displaystyle (2x+4)^{3}}$ , a different method is needed. It is called the chain rule. It is like simple integration, but it only works if the ${\displaystyle x}$ in the bracket is linear (has a power of 1), such as ${\displaystyle x}$ or ${\displaystyle 5x}$ —but not ${\displaystyle x^{5}}$ or ${\displaystyle x^{-7}}$ .
For example, ${\displaystyle \int (2x+4)^{3}\ dx}$ can be determined in the following steps:
• Add 1 to the power ${\displaystyle 3}$ , so that it is now ${\displaystyle (2x+4)^{4}}$
• Divide all this by the new power to get ${\displaystyle {\frac {(2x+4)^{4}}{4}}}$
• Divide all this by the derivative of the bracket ${\displaystyle \left({\frac {d(2x+4)}{dx}}=2\right)}$ to get ${\displaystyle {\frac {(2x+4)^{4}}{4\cdot 2}}={\frac {1}{8}}(2x+4)^{4}}$
• Add the constant ${\displaystyle c}$ to give ${\displaystyle {\frac {1}{8}}(2x+4)^{4}+c}$
### Examples
${\displaystyle \int (x+1)^{5}\ dx={\frac {(x+1)^{6}}{6\times 1}}+c={\frac {1}{6}}(x+1)^{6}+c\left(\because {\frac {d(x+1)}{dx}}=1\right)}$
${\displaystyle \int {\frac {1}{(7x+12)^{9}}}\ dx=\int (7x+12)^{-9}\ dx={\frac {(7x+12)^{-8}}{-8\times 7}}+c=-{\frac {1}{56}}(7x+12)^{-8}+c=-{\frac {1}{56(7x+12)^{8}}}+c\left(\because {\frac {d(7x+12)}{dx}}=7\right)}$
## References
1. "Compendium of Mathematical Symbols". Math Vault. 2020-03-01. Retrieved 2020-08-18.
2. "Antiderivative and Indefinite Integration | Brilliant Math & Science Wiki". brilliant.org. Retrieved 2020-08-18.
3. Weisstein, Eric W. "Indefinite Integral". mathworld.wolfram.com. Retrieved 2020-08-18.
4. "4.9: Antiderivatives". Mathematics LibreTexts. 2017-04-27. Retrieved 2020-08-18.
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# How do calculate that? pls help me
## $\int \frac{{x}^{2} + 2 x + 3}{{x}^{2} + 2 x + 1} \mathrm{dx}$
Apr 14, 2018
$\int \frac{{x}^{2} + 2 x + 3}{{x}^{2} + 2 x + 1} \mathrm{dx} = x - \frac{2}{x + 1} + c$
#### Explanation:
$\int \frac{{x}^{2} + 2 x + 3}{{x}^{2} + 2 x + 1} \mathrm{dx}$
= $\int \left[\frac{{x}^{2} + 2 x + 1}{{x}^{2} + 2 x + 1} + \frac{2}{{x}^{2} + 2 x + 1}\right] \mathrm{dx}$
as ${x}^{2} + 2 x + 1 = {\left(x + 1\right)}^{2}$ this becomes
$\int \left(1 + \frac{2}{x + 1} ^ 2\right) \mathrm{dx}$
= $\int \mathrm{dx} + 2 \int \frac{1}{x + 1} ^ 2 \mathrm{dx}$
= $x + 2 \int \frac{1}{x + 1} ^ 2$ - now let $u = x + 1$ then $\mathrm{du} = \mathrm{dx}$
and our integral becomes
$x + 2 \int \frac{\mathrm{du}}{u} ^ 2$
= $x - \frac{2}{u} + c$
= $x - \frac{2}{x + 1} + c$
Apr 14, 2018
The answer should be $x - \setminus \frac{2}{x + 1} + C$.
#### Explanation:
First off, split the $3$ into $2 + 1$:
$\setminus \int \frac{{x}^{2} + 2 x + 3}{{x}^{2} + 2 x + 1} \mathrm{dx} = \setminus \int \frac{{x}^{2} + 2 x + 1 + 2}{{x}^{2} + 2 x + 1} \mathrm{dx}$
Then, use the linearity of this function to get rid of the $2$ and simply drop the $1$ because it's a constant:
$\setminus \int \left(\frac{2}{{x}^{2} + 2 x + 1} + 1\right) \mathrm{dx} = 2 \setminus \int \frac{1}{{x}^{2} + 2 x + 1} \mathrm{dx} + \setminus \int 1 \mathrm{dx}$
Then, try to factor out the denominator:
$2 \setminus \int \frac{1}{{x}^{2} + 2 x + 1} \mathrm{dx} = 2 \setminus \int \frac{1}{x + 1} ^ 2 \mathrm{dx} + \setminus \int 1 \mathrm{dx}$
Get rid of $\setminus \int 1 \mathrm{dx}$, using the rule $\setminus \int K \mathrm{dx} = K x + C$, where $K$ is any constant:
$2 \setminus \int \frac{1}{x + 1} ^ 2 \mathrm{dx} + \setminus \int 1 \mathrm{dx} = 2 \setminus \int \frac{1}{x + 1} ^ 2 \mathrm{dx} + x$
Then, let $k = \left(x + 1\right)$ and apply the power rule:
$2 \setminus \int \frac{1}{x + 1} ^ 2 \mathrm{dx} + x = 2 \setminus \int \frac{1}{k} ^ 2 \mathrm{dk} + x = 2 \setminus \int {k}^{- 2} \mathrm{dk} + x$
$2 \setminus \int {k}^{- 2} \mathrm{dk} + x = 2 \setminus \frac{{k}^{- 2 + 1}}{- 2 + 1} + x = 2 \setminus \frac{{k}^{- 1}}{- 1} + x$
$2 \setminus \frac{{k}^{- 1}}{- 1} + x = \setminus \frac{- 2}{k} + x = \setminus \frac{- 2}{x + 1} + x$
And we're done! Don't forget to add a constant, though :) Hence, we can conclude that:
$\setminus \int \frac{{x}^{2} + 2 x + 3}{{x}^{2} + 2 x + 1} \mathrm{dx} = x - \setminus \frac{2}{x + 1} + C$
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# How do you calculate the future value of an ordinary annuity?
## How do you calculate the future value of an ordinary annuity?
The two basic annuity formulas are as follows:
1. Ordinary Annuity: FVA = PMT / i * ((1 + i) ^ n – 1)
2. Annuity Due: FVA = PMT / i * ((1 + i) ^ n – 1) * (1 + i) n = m * t where n is the total number of compounding intervals. i = r / m where i is the periodic interest rate (rate over the compounding intervals)
## How do you find the present value of an ordinary annuity of 1?
The formula for determining the present value of an annuity is PV = dollar amount of an individual annuity payment multiplied by P = PMT * [1 – [ (1 / 1+r)^n] / r] where: P = Present value of your annuity stream. PMT = Dollar amount of each payment. r = Discount or interest rate.
How do you use an ordinary annuity table?
An annuity table typically has the number of payments on the y-axis and the discount rate on the x-axis. Find both of them for your annuity on the table, and then find the cell where they intersect. Multiply the number in that cell by the amount of money you get each period.
What is the present value of annuity table?
Also referred to as a “present value table,” an annuity table contains the present value interest factor of an annuity (PVIFA), which you then multiply by your recurring payment amount to get the present value of your annuity.
### How do you calculate the present value of 1?
The present value formula is PV=FV/(1+i)n, where you divide the future value FV by a factor of 1 + i for each period between present and future dates. Input these numbers in the present value calculator for the PV calculation: The future value sum FV. Number of time periods (years) t, which is n in the formula.
### What is the future value of annuity?
The future value of an annuity is a calculation that measures how much a series of fixed payments would be worth at a specific date in the future when paired with a particular interest rate. The word “value” in this term is the cash potential that a series of future payments can achieve.
What is future value annuity?
The future value of an annuity is the value of a group of recurring payments at a certain date in the future, assuming a particular rate of return, or discount rate. The higher the discount rate, the greater the annuity’s future value.
What is an annuity table?
An annuity table is a tool used to determine the present value of an annuity. An annuity table calculates the present value of an annuity using a formula that applies a discount rate to future payments. An annuity table uses the discount rate and number of period for payment to give you an appropriate factor.
## What is an ordinary annuity of 1?
An ordinary annuity is a series of regular payments made at the end of each period, such as monthly or quarterly. In an annuity due, by contrast, payments are made at the beginning of each period. Consistent quarterly stock dividends are one example of an ordinary annuity; monthly rent is an example of an annuity due.
## What is ordinary annuity?
How do you calculate annuity future value?
rate – the value from cell C5,7%.
• nper – the value from cell C6,25.
• pmt – negative value from cell C4,-100000
• pv – 0.
• type – 0,payment at end of period (regular annuity).
• How to calculate the future value of an annuity?
– Future Value of a Growing Annuity (g ≠ i): FVA = PMT / (i – g) * ( (1 + i) ^ n – (1 + g) ^ n) – Future Value of a Growing Annuity (g = i): FVA = PMT * n * (1 + i) ^ (n – 1) – Future Value of an Annuity with Continuous Compounding (m → ∞) FVA = PMT / (eʳ – 1) * (eʳᵗ – 1)
### What is the formula for present value of ordinary annuity?
Present Value =. PMT. (1 + r/m) (m×n) Where PMT is the periodic payment in annuity, r is the annual percentage interest rate, n is the number of years between time 0 and the relevant payment date and m is the number of annuity payments per year. Alternatively, we can calculate the present value of the ordinary annuity directly using the
### Which annuity has the greater future value?
The last difference is on future value. An annuity due’s future value is also higher than that of an ordinary annuity by a factor of one plus the periodic interest rate. Each cash flow is compounded for one additional period compared to an ordinary annuity. The formula can be expressed as follows: FV of an Annuity Due = FV of Ordinary Annuity * (1+i)
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# If n(A) = 3 and n(B) = 5, find:
Question:
If $n(A)=3$ and $n(B)=5$, find:
(i) The maximum number of elements in $A \cup B$,
(ii) The minimum number of elements in $A \cup B$.
Solution:
Number of elements in set A n(A) = 3 and number of elements in set B n(B) =
5
The number of elements in $A \cup B$ is $n(A \cup B)$.
i) Now for elements in $\mathrm{A} \cup \mathrm{B}$ to be maximum, there should not be any intersection between both sets that is $A$ and $B$ both sets must be disjoint sets as shown.
Hence the number of elements in $A \cup B$ is $n(A \cup B)=n(A)+n(B)$
$\Rightarrow \mathrm{n}(\mathrm{A} \cup \mathrm{B})=3+5$
$\Rightarrow \mathrm{n}(\mathrm{A} \cup \mathrm{B})=8$
Hence maximum number of elements in $A \cup B$ is 8
ii) Now for a number of elements in $\mathrm{A} \cup \mathrm{B}$ to be minimum, there should be an ntersection between sets $A$ and $B$ so that some elements are common
The count will be minimum when all the elements from set $A$ are also in set $B$ the reverse are not possible because $n(A) Hence if the 3 elements of$A$are in the intersection of$A$and$B$, then the number of elements only in$B$will be 2 because$n(B)=5$Visually it is represented as As seen from the figure the number of elements in$A \cup B$is 5 hence minimum number of elements in$A \cup B=5\$
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Graphing Polynomials
Definitions:
• Polynomials are expressions involving x raised to a whole number power (exponent). Some examples are:
• Degree of a polynomial: The highest power (exponent) of x.
• Leading coefficient: The coefficient of the highest power of x.
In the case of it is +3.
• Constant term: The number not associated with any power of x.
In the case of it is +7.
We shall refer to the degree, leading coefficient, and the constant term frequently in discussing the graphs of polynomials.
Graphing polynomials accurately:
We will refer to ways that a calculator can assist in graphing as well as which important points to graph accurately. Important points include the x- and y-intercepts, coordinates of maximum and minimum points, and other points plotted using specific values of x and the associated value of the polynomial. Locate the y-intercept by letting x = 0 (the y-intercept is the constant term) and locate the x-intercept(s) by setting the polynomial equal to 0 and solving for x or by using the TI-83 calculator under and the 2.zero function. The intercepts provide accurate points to help in sketching the graphs.
Locate the maximum or minimum points by using the TI-83 calculator under and the 3.minimum or 4.maximum functions.
Graphing polynomials of degree 2:
1. is a parabola and its graph opens upward from the vertex .
The graph is shown below using the WINDOW (-5, 5) X (-2, 16). Note that the leading coefficient is positive and that is why the parabola opens upward. Notice that the constant term 7 is the y-intercept. The y-intercept is the point where x = 0.
It is a good practice to plot points near the minimum point which in this case has approximate coordinates (0.17, 6.92).
2. If
If .
The points (-1, 11) and (2, 17) can be plotted to sketch a more accurate graph.
3. is a parabola and its graph opens downward from the vertex (1, 3).
The graph is shown below using the WINDOW (-5, 5) (-8, 8). Note that the leading coefficient is negative and that is why the parabola opens down. Notice that the constant term 1 is the y-intercept. The maximum point is located at (1, 3). Points near this are (-1, -5) and (2, 1) which can help in sketching the graph.
In general, a parabola (polynomial of degree 2) is given by . If a is positive, the parabola opens up and if a is negative, the parabola opens down.
Polynomials of even degree greater than 2:
Polynomials of even degree open up or down depending on whether the leading coefficient is positive or negative. The end behavior is the same for both the left and right sides of the graph.
1. A polynomial of degree higher than 2 may open up or down, but may contain more “curves” in the graph. Notice in the case of that the graph opens up both on the left and right sides of the graph. The degree of this polynomial is 4 and is even. Thus both the left and right sides of the graph show the same end behavior, opening up.
The y-intercept is 4 and the x-intercepts are -2.82 and -1.34 approximately. The minimum is approximately (-2.25, -4.54). Other points in this window which assist in graphing accurately are (-2, -4), (-1, 2), and (1, 8). It might be necessary to plot points for x values of -0.5 and 0.5 for an accurate sketch. These points are (-0.5, 3.6875) and (0.5, 4.4375).
2. Let . We expect the behavior of the graph to be the same on both the left and right sides of the graph because the degree of is even. Because the leading coefficient is positive, the graph will be going upward. The graph is shown below.
The y-intercept is 0 and the x-intercepts are -1.5, 0, and 1. There are two minimum points on the graph at (0.70, -0.65) and (-1.07, -2.04). There is a maximum at (0, 0). Other points on the graph which can help sketch an accurate graph are (-2, 12), (-1, -2), and (1.5, 6.75).
3. Let We expect the end behavior of the graph to be the same on both the left and right sides of the graph. Because the leading coefficient is negative, the graph will be going downward. The graph is shown below.
The y-intercept is 1 and the x-intercepts are approximately -1.42 and 0.8. There are two maximum points at (-1.11, 2.12) and (0.33, 1.22). There is a minimum at (-0.34, 0.78). Other points on the graph are (-1, 2), and (1, -2).
Polynomials of odd degree greater than 2:
Polynomials of odd degree open up or down depending on whether the leading coefficient is positive or negative. However, the end behavior is different for the left and right sides of the graph. If the right side of the graph goes up, the leading coefficient is positive and the left side will go down. If the right side goes down, the leading coefficient is negative and the left side goes up.
1. A polynomial of degree higher than 2 may open up or down, but may contain more “curves” in the graph. Notice in the case of the graph opens up to the right and down to the left. This is because the leading coefficient is positive.
The y-intercept is 4 and is also a minimum point. A maximum is found at (-2, 8). The x-intercept is approximately -3.36. Other points which can assist in sketching this graph accurately are (-3, 4), (-1, 6), and (1, 8).
2. If , the graph opens down to the right and up to the left because the leading coefficient is negative.
The y-intercept is 0. The x-intercepts are located at 0, 0.5, and 3. Other points on the graph are (-0.5, 3.5), (1, 2), and (1.5, 4.5). There is a minimum at approximately (0.24, -0.34) and a maximum at approximately (2.09, 6.05).
3. If , the graph opens up to the right and down to the left because the leading coefficient is positive.
The y-intercept is -2, the constant term. The x-intercepts are located at -1, -0.5, 0.5, 1, and 2. Other points which can assist in making an accurate sketch are located at (-0.25, -1.58), (0.25, -1.23), and (1.5, -5).
Examples
What are the coordinates of the minimum point of ? What is your answer?
Describe the end behavior of the graph of . What is your answer?
Describe the end behavior of the graph of . What is your answer?
Describe the end behavior of the graph of . What is your answer?
Describe the end behavior of the graph of . What is your answer?
M Ransom
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## Encyclopedia > Set
Article Content
# Set
In mathematics, a set is a collection of objects such that two sets are equal if, and only if, they contain the same objects. A finite set is a collection of a finite number of objects; the alternative is an infinite set. For a discussion of the properties and axioms concerning the construction of sets, see naive set theory and axiomatic set theory. Here we give only a brief overview of the concept.
Sets are one of the basic concepts of mathematics. A set is, more or less, just a collection of objects, called its elements. Standard notation uses braces around the list of elements, as in:
{red, green, blue}
{red, red, blue, red, green, red, red, green, red, red, blue}
{x : x is an additive primary color}
All three lines above denote the same set. As you see, it is possible to describe one and the same set in different ways: either by listing all its elements (best for small finite sets) or by giving a defining property of all its elements; and it does not matter in what order, or how many times, the elements are listed, if a list is given.
If $A$ and $B$ are sets and every $x$ in $A$ is also contained in $B$, then $A$ is said to be a subset of $B$, denoted $A \subseteq B$. If atleast one element in $B$ is not also in $A$, $A$ is called a proper subset of $B$, denoted $A \subset B$. Every set has as subsets itself, called the improper subset, and the empty set {} or $\emptyset$. The fact that an element $x$ belongs to the set $A$ is denoted $x \in A$.
The union of a collection of sets $S = {S_1, S_2, S_3, \cdots}$ is the set of all elements contained in at least one of the sets $S_1, S_2, S_3, \cdots$
The intersection of a collection of sets $T = {T_1, T_2, T_3, \cdots}$ is the set of all elements contained in all of the sets.
These unions and intersections are denoted
$S_1 \cup S_2 \cup S_3 \cup \cdots$
and
$T_1 \cap T_2 \cap T_3 \cap \cdots$
respectively.
The "number of elements" in a certain set is called the cardinal number of the set and denoted $|A|$ for a set $A$ (for a finite set this is an ordinary number, for an infinite set it differentiates between different "degrees of infiniteness", named $\aleph_0$ (aleph zero), $\aleph_1, \aleph_2 ...$).
The set of all subsets of $X$ is called its power set and is denoted $2^X$ or $P(X)$. This power set is a Boolean algebra under the operations of union and intersection.
The set of functions from a set A to a set B is sometimes denoted by BA. It is a generalisation of the power set in which 2 could be regarded as the set {0,1} (see natural number).
The cartesian product of two sets A and B is the set
A×B={(a,b) : a ∈ A and b ∈ B}.
The sum of two sets A and B is the set
A+B = A×{0} ∪ B×{1}.
1. Natural numbers which are used for counting the members of sets.
2. Integers which appear as solutions to equations like x + a = b.
3. Rational numbers which appear as solutions to equations like a + bx = c.
4. Algebraic numbers which can appear as solutions to polynomial equations (with integer coefficients) and may involve radicals and certain other irrational numbers.
5. Real numbers which include transcendental numbers (which can't appear as solutions to polynomial equations with rational coefficents) as well as the algebraic numbers.
6. Complex numbers which provide solutions to equations such as x2 + 1 = 0.
Care must be taken with verbal descriptions of sets. One can describe in words a set whose existence is paradoxical. If one assumes such a set exists, an apparent paradox or antinomy may occur. Axiomatic set theory was created to avoid these problems.
For example, suppose we call a set "well-behaved" if it doesn't contain itself as an element. Now consider the set S of all well-behaved sets. Is S itself well-behaved? There is no consistent answer; this is Russell's paradox. In axiomatic set theory, the set S is not allowed, and we have no paradox.
All Wikipedia text is available under the terms of the GNU Free Documentation License
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# Introduction to Vectors
This section covers:
# Introduction to Vectors
A vector (also called a direction vector) is just is something that has both magnitude (length, or size) and direction. So it’s different than a regular number, since it really has two components to it. We see vectors represented by arrows, so we can remember that we need to get a length of a vector (the magnitude), as well as the direction (which way it’s pointing).
We use vectors in mathematics, engineering, and physics, since many times we need to know both the size of something and which way it’s going. For example, with an airplane, we can use a vector to measure the speed of the plane (the “size”) and the direction it’s flying.
Geometric Vectors are directed line segments in the xy-plane, and, as an example, the vector from a point A (initial point) to a point B (terminal point) can be represented by $$\overrightarrow{{AB}}$$ .
So, for example, if A is (2, 7) and B is (–3, 8), the vector is second point minus first point, or $$\displaystyle \left\langle {{{x}_{2}}} \right.-{{x}_{1}},\left. {{{y}_{2}}-{{y}_{1}}} \right\rangle$$, or $$\left\langle {-3-2} \right.,\left. {8-7} \right\rangle =\left\langle {-5,\left. 1 \right\rangle } \right.$$. The “x” part of the vector (–5) is called the x-component, and the ”y” part (1) is called the y-component.
Note also that vectors can also be written in the form ai + bj, so this vector can also be written as –5i + 1j, or –5i + j.
The magnitude of the vector, written $$\left\| {AB} \right\|$$ is the distance between the two points (like the hypotenuse of a right triangle), or $$\sqrt{{{{{\left( {{{x}_{2}}-{{x}_{1}}} \right)}}^{2}}+{{{\left( {{{y}_{2}}-{{y}_{1}}} \right)}}^{2}}}}$$, or with the new vector $$\left\langle {x,} \right.\left. y \right\rangle$$, it’s just $$\sqrt{{{{x}^{2}}+{{y}^{2}}}}$$. So for our points A and B above, $$\left\| {AB} \right\|=\sqrt{{{{{\left( {-5} \right)}}^{2}}+{{1}^{2}}}}=\sqrt{{26}}$$.
Now looking at this vector visually, do you see how we can use the slope of the line of the vector (from the initial point to the terminal point) to get the direction of the vector? Pretty cool! So we can just use $${{\tan }^{{-1}}}\left( {\frac{y}{x}} \right)$$ (second part of vector over first part of vector) to get the angle measurement of the vector’s direction. Remembering from the Polar Coordinates, Equations and Graphs section though, we have to be careful which quadrant the vector terminates in (“pretending” that the vector’s initial point is at the origin) to know how many degrees we should add to that tangent value when we use a calculator:
Here is all this visually. Note that we had to add 180° to the angle measurement we got from the calculator (–11.3°) since the vector would terminate in the 2nd quadrant if we were to start at (0, 0). So we get 168.7°, which is the angle measurement from the positive x axis going counterclockwise.
168.7° from the positive x axis can also be described as 11.3° North of West (11.3° N of W, or W11.3°W), since the closest axis to the angle is the negative x axis (west) and we are going a little north of that:
(We saw a similar concept of this when we were working with bearings here in the Law of Sines and Cosines, and Areas of Triangles section).
Note that a vector that has a magnitude of 0 (and thus no direction) is called a zero vector. So hypothetically, the vector $$\overrightarrow{{AA}}$$ would be a zero vector.
A unit vector is a vector with magnitude 1; in some applications, it’s easier to work with unit vectors. To find the unit vector that is associated with a vector (has same direction, but magnitude of 1), use the following formula: $$u=\frac{v}{{\left\| v \right\|}}$$ (just divide each component of a vector by its magnitude to get its unit vector). We’ll see some problems below.
# Vector Operations
## Adding and Subtracting Vectors
There are a couple of ways to add and subtract vectors. When we add vectors, geometrically, we just put the beginning point (initial point) of the second vector at the end point (terminal point) of the first vector, and see where we end up (new vector starts at beginning of one and ends at end of the other). If the vectors aren’t this way to begin with, we can move the second vector (as long as it has the same magnitude and direction, so it’s like a slide) to be this way.
You can think of adding vectors as connecting the diagonal of the parallelogram (a four-sided figure with two pairs of parallel sides) that contains the two vectors.
Do you see how when we add vectors geometrically, to get the sum, we can just add the x components of the vector, and the y components of the vectors?
When we subtract two vectors, we just take the vector that’s being subtracting, reverse the direction and add it to the first vector. This is because the negative of a vector is that vector with the same magnitude, but has an opposite direction (thus adding a vector and its negative results in a zero vector).
Note that to make a vector negative, you can just negate each of its components (x component and y component) (see graph below).
## Multiplying Vectors by a Number (Scalar)
To multiply a vector by a number, or scalar, you simply stretch (or shrink if the absolute value of that number is less than 1), or you can simply multiply the x component and y component by that number. Notice also that the magnitude is multiplied by that scalar.
Do you see how two vectors that are parallel are just a multiple of each other?
Multiplying by a negative number changes the direction of that vector.
Here’s what subtracting vectors and also multiplying vectors by a scalar looks like:
Let’s put all this together to perform the following vector operations, given the vectors shown:
You may also see problems like this, where you have to tell whether the statement is true or false. Note that you want to look at where you end up in relation to where you started to see the resulting vector. if you end up exactly where you started from, the resulting vector is 0.
Here are a couple more examples of vector problems. Notice in the second set of problems when we are given a magnitude and direction of a vector, and have to find that vector, we use the following equation, like we did when we here in the Polar Coordinates, Equations and Graphs section: , where $$\left\| {\,v} \right\|$$ or the magnitude of a vector is like the “r” (radius) we saw for polar numbers: $$v=\left\| {\,v} \right\|\left( {\cos \alpha i+\sin \alpha j} \right)=\left( {\left\| {\,v} \right\|\cos \alpha } \right)i+\left( {\left\| {\,v} \right\|\sin \alpha } \right)j$$. (Trigonometry always seems to come back and haunt us!) We’ll We’ll leave our answers in ai + bj form.
# Applications of Vectors
Vectors are extremely important in many applications of science and engineering. Since vectors include both a length and a direction, many vector applications have to do with vehicle motion and direction.
We saw above that, given a magnitude and direction, we can find the vector with $$v=\left\| {\,v} \right\|\left( {\cos \alpha i+\sin \alpha j} \right)=\left( {\left\| {\,v} \right\|\cos \alpha } \right)i+\left( {\left\| {\,v} \right\|\sin \alpha } \right)j$$, where $$\left\| {\,v} \right\|$$ is the speed. This way we can add and subtract vectors, and get a resulting speed and direction for the new vector.
Remember that a bearing (we saw here in the Law of Sines and Cosines, and Areas of Triangles section), is typically expressed a measure of the clockwise angle that starts due north or on the positive y axis (initial side) and terminates a certain number of degrees (terminal side) from that due north starting place. (This is also written, as in the case of a bearing of 40° as “40° east of north”, or “N40°E”).
(A lot of times, the bearing includes more directions, such as 70° west of north, also written as N70°W. In this case, the angle will start due north (straight up, or on the positive y axis) and go counterclockwise 70° (because it’s going west, or to the left, instead of east). Similarly, a bearing of 50° south of east, or E50°S, would be an angle that starts due east (on the positive x axis) and go clockwise 50° clockwise (towards the south, or down). Also, if you see a bearing of southwest, for example, the angle would be 45° south of west, or 225° clockwise from north, and so on.)
Each time a moving object changes course, you have to draw another line to the north to map its new bearing.
When there’s a tail wind, remember that you have to add this vector to the vector that the object is trying to go on (its programmed or “steered” course), to get the actual vector of the object. So remember:
Problem:
A plane is flying on a bearing of 25° south of west at 500 miles per hour (speed). Express the velocity of the plane (as a vector).
Solution:
Problem:
A sailboat is sailing on a bearing of 30° north of west at 6.5 miles per hour (in still water). A tail wind blowing 20 miles per hour in the direction 40° south of west alters the course of the boat.
Express the actual velocity of the sailboat as a vector. Then determine the actual speed and direction of the boat.
Solution:
Note that if we were given a vector for the actual course of the boat and had to come up with the vector for which the boat should be “steered”, we would have to subtract the wind from the actual course.
Now let’s the problem we already did using Law of Cosines from the Law of Sines and Cosines, and Areas of Triangles section:
Problem:
A cruise ship travels at a bearing of 40°at 60 mph for 3 hours, and changes course to a bearing of 120°. It then travels 40 mph for 2 hours. Find the distance the ship is from its original position and also its bearing from the original position.
Solution:
In this problem, Distance = rate x time, since we are given rates and times and need to calculate distances.
Since no specific directions (like West of South) are given for these bearings, we will obtain the angles by measuring the clockwise angle that starts due north or on the positive y axis (East of North).
And remember that with a change of bearing, we have to draw another line to the north to map its new bearing.
Now that we have the angles, we can use vector addition to solve this problem; doing the problem with vectors is actually easier than using Law of Cosines:
# Dot Product and Angle Between Two Vectors
The dot product of two vectors $$v=ai+bj$$ and $$w=ci+dj$$ (sort of like multiplying two vectors) is defined as $$v\bullet \,w=ac+bd$$; in other words, you multiply the two “x” parts of the vectors, and multiply the two “y” parts, and then add them together. The result is a scalar (single number).
Here is an example: if $$v=-2i+3j$$ and $$w=2+j$$, the dot product $$v\bullet \,w=\left( {-2} \right)\left( 2 \right)+\left( 3 \right)\left( 1 \right)=-1$$.
We use dot products to find the angle measurements between two vectors; the cosine of the angle between two vectors is the dot product of the vectors, divided by the product of each of their magnitudes:
(And we don’t need to worry about getting the correct quadrant when putting this in the calculator!)
So we might be able to this formula instead of, say, the Law of Cosines, for applications.
Note that if the dot product of two vectors is 0, the vectors form right angles, or are orthogonal, since the cos of 90° is 0 (and thus the whole expression will be 0).
And remember that we noted above that if two vectors are parallel, then one is a “multiple” of another, or $$v=aw$$. So for example, the vector $$v=-2i+3j$$ would be parallel to the vector $$v=-4i+6j$$. If vectors are parallel, the angle between them is either 0 (if they are the same vector) or π.
Here are some example problems:
# 3D Vectors – Vectors in Space
We’ve been dealing with vectors (and everything else!) in the two dimensional plane, but “real life” is actually three dimensional, so we need to know how to work in 3D, or space, too.
A 3D coordinate system is typically drawn like this, with the z-axis going “up”. Note that the positive x-axis comes forward at you, and the positive y-axis to the right of that, if you’re looking head on. Maybe you can remember this by the expression “Exit. Why?” (xzy when looking head on).
Geometric Vectors in 3D are still directed line segments, but in the xyz-plane. We still can find the vector between two coordinate points by “subtracting” the first vector from the second.
So, for example, if A is (–4, 2, 7) and B is (–3, 8, 0), the vector $$\overrightarrow{{AB}}$$ is second point minus first point, or $$\displaystyle \left\langle {{{x}_{2}}} \right.-{{x}_{1}},\,\left. {{{y}_{2}}-{{y}_{1}},\,{{z}_{2}}-{{z}_{1}}} \right\rangle$$ or $$\left\langle {-3-\left( {-4} \right)} \right.,\left. {8-2,0-7} \right\rangle =\left\langle {1,\left. {6,-7} \right\rangle } \right.$$. Note also that vectors can also be written in the form ai + bj + ck, so this vector can also be written as i + 6j –7k.
The magnitude of the 3D vector, written $$\left\| {AB} \right\|$$ is still the distance between the two points (like taking hypotenuse of a right triangle twice actually), or $$\sqrt{{{{{\left( {{{x}_{2}}-{{x}_{1}}} \right)}}^{2}}+{{{\left( {{{y}_{2}}-{{y}_{1}}} \right)}}^{2}}+{{{\left( {{{z}_{2}}-{{z}_{1}}} \right)}}^{2}}}}$$, or with the new vector $$\left\langle {x,} \right.\left. {y,z} \right\rangle$$, it’s just $$\sqrt{{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}}$$. So for our points A and B above, $$\left\| {AB} \right\|=\sqrt{{{{1}^{2}}+{{6}^{2}}+{{{\left( {-7} \right)}}^{2}}+}}=\sqrt{{86}}$$.
## Vector Operations in Three Dimensions
Adding, subtracting 3D vectors, and multiplying 3D vectors by a scalar are done the same way as 2D vectors; you just have to work with three components.
Like for 2D vectors, the dot product of two vectors $$v=ai+bj+ck$$ and $$w=di+ej+fk$$ (sort of like multiplying two vectors) is defined as $$v\bullet \,w=ad+be+cf$$; in other words, you multiply the two “x” parts of the vectors, multiply the two “y” parts, multiply the two “z” parts, and then add them together. The result is a scalar (single number).
Again, like for 2D, we use dot products to find the angle measurements between two vectors; the cosine of the angle between two vectors is the dot product of the vectors, divided by the product of each of their magnitudes:
Here are some problems; included is how to get the equation of a sphere:
Writing a 3D vector in terms of its magnitude and direction is a little more complicated. Since we can’t really describe a 3D vector in terms of only a magnitude and one direction, we have to get what we call the direction angles:
Here are what these angles look like:
So it turns out for the vector $$v=ai+bj+ck$$, the direction angles $$\alpha ,\,\,\beta ,\,\,\text{and}\,\,\gamma$$ are:
These cosine values are called the direction cosines for the vector v.
To find the 3D vector in terms of its magnitude and direction cosines, we use:
Now let’s do a problem:
## Cross Products of 3D Vectors
Also, only for vectors 3D vectors, we have what we call a cross product of vectors (also called vector product, since the result is still a vector) of two vectors $$v=ai+bj+ck$$ and $$w=di+ej+fk$$. The vector that is the cross product of two vectors is actually orthogonal (perpendicular) to both of the original vectors. This is also called the normal vector.
This looks crazy! But we can get this cross product using determinants of matrices. We learned about determinants of matrices here in the The Matrix and Solving Systems with Matrices section.
Remember that for a 2 by 2 matrix, we get the determinant this way:
And here is an example of how we got the determinant of a 3 by 3 matrix:
Here is an example of how we use a determinant to find the cross product of two vectors $$v=i+2j-4k$$ and $$w=-i+5j+3k$$:
So the vector $$\displaystyle 26i+j+7k$$ is orthogonal (perpendicular, normal) to the vectors v and w above.
A few things to remember here. First, we must watch the order of the vectors when we are finding the cross products of vectors; v × w is not necessarily the same thing as w × v.
Also, we can use the right hand rule to find the direction of the cross product of two vectors by holding up your right hand and make your index finger, middle finger, and thumb all perpendicular to each other (easier said than done!). Then point your index finger in the direction of the first vector (such as v) and your middle finger in the direction of the second vector (such as w). Your thumb will point in the direction of v × w.
This is something you probably won’t need too much in your math classes, but it can become very handy in Physics. (And remember the directions of 3D vectors as shown in the coordinate system below).
We can use the cross product to find the area of a 3D parallelogram. If that parallelogram has two adjacent sides with vectors v and w, we can take the magnitude of the vectors’ cross product to find its area: $$\left\| {v\,\,\times \,\,w} \right\|$$. We can also use this if given four vertices of a parallelogram; we would just have to find two adjacent sides of the parallelogram in vector form first.
Here are some cross product problems:
## The Equation of a Plane
You might also be asked to find the equation of the plane that passes through a given point and is perpendicular to a certain vector, or even the equation of a plane containing three points.
Remember that the equation of a line can be in the standard form $$ax+by=c$$, so the equation of a plane can be in the form $$ax+by+cz=d$$. (These are called Cartesian equations.)
To see what this plane might look like, we can see where it intersects each of the three axes by setting the other variables to 0, for example with the graph $$2x+6y+3z=12$$ (set y and z equal to 0 and solve for x, and so on):
Remember that the dot product will be 0 if two orthogonal vectors. So it turns out that a vector equation of the plane is $$\displaystyle \left\langle {a,b,c} \right\rangle \bullet \left( {\left\langle {x,y,z} \right\rangle -\left\langle {{{x}_{0}},{{y}_{0}},{{z}_{0}}} \right\rangle } \right)=0$$, or $$\displaystyle \left\langle {a,b,c} \right\rangle \bullet \left\langle {x-{{x}_{0}},y-{{y}_{0}},z-{{z}_{0}}} \right\rangle =0$$, where $$\left\langle {a,b,c} \right\rangle$$ is orthogonal to the plane (the normal vector) and $$\left( {{{x}_{0}},{{y}_{0}},{{z}_{0}}} \right)$$ is a point on the plane.
Another way to get the equation of the plane for a vector perpendicular to a certain plane at point $$\left( {{{x}_{0}},{{y}_{0}},{{z}_{0}}} \right)$$ is to use $$ax+by+cz=d$$, where $$d=a{{x}_{0}}+b{{y}_{0}}+c{{z}_{0}}$$ (plug in the point to get d).
OK, so this looks really complicated, so let’s do a problem to show it’s not too bad:
To find the equation of the plane containing three points, we first have to find two vectors defined by the points, find the cross product of the two vectors, and then use the Cartesian equation above to find d:
# Parametric Form of the Equation of a Line in Space
We can get a vector form of an equation of a line in 3D space by using parametric equations. (We’ll review Parametric Equations here).
In two dimensions, we worked with a slope of the line and a point on the line (or the y-intercept).
In 3D space, we can use a 3D vector $$\left\langle {a,b,c} \right\rangle$$ as the slope of a line, and define that line by an initial point $$\left\langle {{{x}_{0}},{{y}_{0}},{{z}_{0}}} \right\rangle$$ (vector that goes through that point and the origin) and the vector $$\left\langle {a,b,c} \right\rangle$$. Note that the vector $$\left\langle {a,b,c} \right\rangle$$ will be parallel to the line we’re describing, just like a slope going through the origin is parallel to a 2D line. We have three different ways to write this 3D line using parametric equations:
Notice to get the last form, we solve for t in the second set of equations. Also note that to go from the last equation to the first equation, we have to set each to t, and solve back for x, y, and z.
This looks a little complicated, so let’s do some problems to show it’s not too bad:
Here are more problems that look more difficult, but are actually easier, since we’re finding a plane perpendicular to the vector or line:
Here are a few more problems:
Learn these rules, and practice, practice, practice!
Click on Submit (the arrow to the right of the problem) and scroll down to “Find the Angle Between the Vectors” to solve this problem. You can also type in more problems, or click on the 3 dots in the upper right hand corner to drill down for example problems.
If you click on “Tap to view steps”, you will go to the Mathway site, where you can register for the full version (steps included) of the software. You can even get math worksheets.
You can also go to the Mathway site here, where you can register, or just use the software for free without the detailed solutions. There is even a Mathway App for your mobile device. Enjoy!
On to Parametric Equations – you are ready!
## 2 thoughts on “Introduction to Vectors”
1. I don’t think that your right hand rule illustration is correct. The x and y axes do not match up between the hand and the coordinate system.
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Want some good news, free of charge? Graphing parent functions and transformed logs is a snap! You can change any log into an exponential expression, so this step comes first. You then graph the exponential, remembering the rules for transforming, and then use the fact that exponentials and logs are inverses to get the graph of the log.
## How to graph a parent function
Exponential functions each have a parent function that depends on the base; logarithmic functions also have parent functions for each different base. The parent function for any log is written f(x) = logb x. For example, g(x) = log4 x corresponds to a different family of functions than h(x) = log8 x. This example graphs the common log: f(x) = log x.
1. Change the log to an exponential.
Because f(x) and y represent the same thing mathematically, and because dealing with y is easier in this case, you can rewrite the equation as y = log x. The exponential equation of this log is 10y = x.
2. Find the inverse function by switching x and y.
You find the inverse function 10x = y.
3. Graph the inverse function.
Because you’re now graphing an exponential function, you can plug and chug a few x values to find y values and get points. The graph of 10x = y gets really big, really fast. You can see its graph in the figure.
Graphing the inverse function y = 10x.
4. Reflect every point on the inverse function graph over the line y = x.
The next figure illustrates this last step, which yields the parent log’s graph.
Graphing the logarithm f(x) = log x.
## How to graph a transformed log
All transformed logs can be written as
where a is the vertical stretch or shrink, h is the horizontal shift, and v is the vertical shift.
So if you can find the graph of the parent function logb x, you can transform it. However, most students still prefer to change the log function to an exponential one and then graph. The following steps show you how to do just that when graphing f(x) = log3(x – 1) + 2:
1. Get the logarithm by itself.
First, rewrite the equation as y = log3(x – 1) + 2. Then subtract 2 from both sides to get y – 2 = log3(x – 1).
2. Change the log to an exponential expression and find the inverse function.
If y – 2 = log3(x – 1) is the logarithmic function, 3y – 2 = x – 1 is the exponential; the inverse function is 3x – 2 = y – 1 because x and y switch places in the inverse.
3. Solve for the variable not in the exponential of the inverse.
To solve for y in this case, add 1 to both sides to get 3x – 2 + 1 = y.
4. Graph the exponential function.
The parent graph of y = 3x transforms right two (x – 2) and up one (+ 1), as shown in the next figure. Its horizontal asymptote is at y = 1.
The transformed exponential function.
5. Swap the domain and range values to get the inverse function.
Switch every x and y value in each point to get the graph of the inverse function. The next figure shows the graph of the logarithm.
You change the domain and range to get the inverse function (log).
Did you notice that the asymptote for the log changed as well? You now have a vertical asymptote at x = 1. The parent function for any log has a vertical asymptote at x = 0. The function f(x) = log3(x – 1) + 2 is shifted to the right one and up two from its parent function p(x) = log3 x (using transformation rules), so the vertical asymptote is now x = 1.
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# Complete factor expressions (below 100)
Lesson
We can make any number by using its factor pairs. Every number will have at least one factor pair, $1$1 and itself.
When we're completing factor expressions we need to think of a second number we can use to complete an expression. We can do this by counting or maybe dividing. Let's look at an example...
Complete the factor expression below
$\editable{}$$\times$×$7=21$7=21
Here we need to complete the factor expression by finding what goes in the $\editable{}$.
So far we know two numbers $7$7 and $21$21. We can use these two numbers to work out the missing number.
We can either find the result of $21\div7$21÷7 or we can count by $7$7's.
If we count by $7$7's we get...
$7$7, $14$14, $21$21
On a number line it would look like this...
We can now see that we'll have to count by $7$7 three times before we get to $21$21, so $3$3 groups of $7$7 make $21$21
So, the number $3$3 completes the factor expression.
$3\times7=21$3×7=21
Remember!
We can use the multiplication tables to help us work out factors of a number.
Use the sliders below to find that factor pair.
#### Examples
##### Question 1
Fill in the box with the missing number.
1. $6\times\editable{}=60$6×=60
##### Question 2
Fill in the box with the missing number.
1. $11\times\editable{}=22$11×=22
##### Question 3
Fill in the gaps to find all factor pairs of $66$66.
1. $1,\editable{}$1,
$2,\editable{}$2,
$3,\editable{}$3,
$11,\editable{}$11,
### Outcomes
#### NA4-1
Use a range of multiplicative strategies when operating on whole numbers
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3 Tutor System
Starting just at 265/hour
# Derive the formula for the curved surface area and total surface area of the frustum of a cone, given to you in Section 13.5, using the symbols as explained.
Let ABC be a cone. From the cone the frustum DECB is cut by a plane parallel to its base. Here, r1 and r2 are the radii of the frustum ends of the cone and h be the frustum height.
Now, consider the $$\triangle$$ ABG $$\triangle$$ ADF,
Here, DF||BG
So, $$\triangle \ ABG \ \sim \ \triangle \ ADF$$
$$\Rightarrow \frac{DF}{BG} \ = \ \frac{AF}{AG} \ = \ \frac{AD}{AB}$$
$$\Rightarrow \frac{r_2}{r_1} \ = \ \frac{h_1 \ - \ h}{h_1} \ = \ \frac{l_1 \ - \ l}{l_1}$$
$$\Rightarrow \ l_1 \ = \ \frac{r_1 l}{r_1 \ - \ r_2}$$
The total surface area of frustum will be equal to the total CSA ot frustum + the area ol upper circular end + area of the lower circular end
$$= \pi (r_1 \ + \ r_2)l \ + \ \pi (r_2)^2 \ + \ \pi (r_1)^2$$
$$\therefore$$ Surface area of frustum $$= \ \pi [r_1 \ + \ r_2)l \ + \ (r_1)^2 \ + \ (r_2)^2]$$
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# If the $m$th term of an Arithmetic Progression is $\frac{1}{n}$ and the $n$th term is…
Problem :
If the $m$th term of an A.P is $\frac{1}{n}$ and the $n$th term is $\frac{1}{m}$ then prove that the sum to $mn$ terms is $\frac{mn+1}{2}$
My working :
Let $a$ be the first term of the progression and $d$ the common difference then: $$\tag1T_m = \frac{1}{n}= a+(m-1)d$$ $$\tag2 T_n = \frac{1}{m} = a+(n-1)d$$
Subtracting (1) from (2) and solving for $d$ we get :
$d = \frac{1}{mn}$ Please suggest what to do further. Thanks
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If $\,a_1,a_2,....\;$ is an arithmetic progression with common difference $\,d\,$ , we have that
$$S_r:=a_1+a_2+\ldots +a_r=\frac r2\left(2a_1+(r-1)d\right)$$
$$\frac{m+n}{mn}=\frac1n+\frac1m=2a_1+(m+n-2)d=2a_1+(m+n-2)\frac1{mn}\implies$$
$$2a_1=\frac2{mn}\implies \color{red}{a_1=\frac1{mn}}\;\;,\;\;\text{and since also}\;\;\color{red}{d=\frac1{mn}}\implies$$
$$S_{mn}=\frac{mn}2\left(2a_1+(mn-1)d\right)=\frac{mn}2\left(\frac2{mn}+1-\frac1{mn}\right)=\frac12+\frac{mn}2=\frac{mn+1}2$$
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Now that you have $d$ you can calculate $a$ from $(1)$
$$T_1 = a = \frac{1}{n}-(m-1)d = \frac{m}{mn}-\frac{m-1}{mn}=\frac{1}{mn}$$
Having $a$ and $d$ Apply the formula for arithmetic progression sum.
$$T_{mn} = a + (mn-1)d = \frac{1}{mn} + \frac{mn-1}{mn} = 1$$
$$S_{mn} = mn\frac{T_1 + T_{mn}}{2} = mn\frac{\frac{1}{mn}+1}{2} = \frac{mn(mn+1)}{2mn} = \frac{mn+1}{2}$$ Q.E.D.
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then you have to count the sum of $T_1,...,T_{mn}$ it seems, by computing $S_{mn} = \frac{T_1 + T_{mn}}{2}mn$...
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### Weekly Challenge 43: A Close Match
Can you massage the parameters of these curves to make them match as closely as possible?
### Weekly Challenge 44: Prime Counter
A weekly challenge concerning prime numbers.
### Weekly Challenge 28: the Right Volume
Can you rotate a curve to make a volume of 1?
# Weekly Challenge 37: Magic Bag
##### Stage: 5 Challenge Level:
Although this problem might sound insoluble with the information given, the magic bag assures us that there is a nice answer, as does Mr Cox from North Berwick High School who solved this problem very nicely (see his solution here) so let us proceed!
Suppose that there are $b$ black balls and $w$ white balls in the bag, and that there are at least two of each (we can think about the other cases at the end).
Then
$$\begin{eqnarray} P(bb) &=& \frac{b}{b+w}\cdot\frac{b-1}{b+w-1}\cr P(wb) &=& \frac{b}{b+w}\cdot\frac{w}{b+w-1}\cr P(bw) &=& \frac{w}{b+w}\cdot\frac{b}{b+w-1}\cr P(ww) &=& \frac{w}{b+w}\cdot\frac{w-1}{b+w-1} \end{eqnarray}$$
Although these look complicated, the denominators in each case are the same.
Since the probabilities are mutually exclusive we know that
$$P(bb)+P(ww) = 0.5$$
Thus,
$$2b(b-1)+2w(w-1) = (b+w)(b+w-1)$$
Expanding and rearranging gives us
$$b^2-b+w^2-w-2wb=0$$
So, where are we now? We've applied the probability and are left with an expression involving $w$ and $b$. This can be factorised as
$$(b-w)^2=b+w$$
Some thought should convince you that this expression is satisfied if $b$ and $w$ are consecutive triangle numbers. But we need to ask: are there any other solutions? We need to ask this because for any choice of $w$ there will be up to $2$ solutions for $b$ upon solving the equation. The quadratic formula tells us that for any given $w$
$$b=\frac{1+2w\pm \sqrt{1+8w}}{2}$$
Now, we need to impose the condition that $b$ and $w$ are natural numbers. This would require that $1+8w$ is a square number, say $N^2$. Then
$$w =\frac{N^2-1}{8}=\frac{(N+1)(N-1)}{8}$$
For $w$ to be a natural number, $N$ must be an odd number, which means that $(N+1)$ and $(N-1)$ are consecutive even numbers, which means that $\frac{N+1}{2}$ and $\frac{N-1}{2}$ are consecutive natural numbers. Thus, $w$ is half the product of two natural number which means that $w$, by definition must be triangular. Thus there is a solution in this case if and only if $w$ and $b$ are consecutive triangle numbers bigger than $1$.
Finally, we consider the special cases where there are less then two of each ball.
Clearly, the balls cannot all be the same colour, as then the chance of drawing two balls of the same colour would be $1$. Clearly, there cannot be exactly one ball of each colour, as then the chance of drawing two balls of the same colour would be $0$. If there were exactly one black ball and $w> 1$ white balls then the chance of drawing two balls of the same colour would be
$$\frac{w}{w+1}\cdot\frac{w-1}{w} = 0.5$$
This has exactly the two solutions $w=0, 3$. Since $3$ is also triangular, our rule still holds.
Putting this all together proves that magic bag can, and only can, contain consecutive triangle numbers of white and black balls.
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# Calculate 20% of \$2000 – Quick Math Guide
Welcome to our quick math guide on calculating 20% of \$2000. Whether you’re trying to determine a discount, calculate a tip, or solve any other percentage-related problem, understanding how to calculate percentages is an important skill to have. In this guide, we’ll walk you through the steps to calculate 20% of \$2000 using simple math formulas. So let’s get started!
### Key Takeaways:
• To calculate 20% of \$2000, you can multiply \$2000 by 0.20 or divide \$2000 by 5.
• The result of calculating 20% of \$2000 is \$400.
• Understanding percentages is essential in various aspects of life, from finance to everyday calculations.
• Percentages allow for easy comparison of values and simplification of calculations.
• By mastering percentage calculations, you can make more informed decisions and solve percentage-related problems efficiently.
## Understanding Percentages
Percentages represent parts per hundred and are a way to express a fraction or ratio where the denominator is 100. They are commonly used in everyday life to convey proportions and completeness. Understanding percentages is important for various calculations, including calculating discounts, growth rates, and proportions.
Percentages allow us to easily compare values and determine the relative size or proportion of something. They help us understand the significance of a certain value in relation to the whole. For example, if you scored 90% on a test, it means you answered correctly 90 out of 100 questions.
Percentage calculations are used in many different scenarios. They are essential in financial calculations, such as determining interest rates or calculating discounts during a sale. Percentages also play a significant role in analyzing growth rates, where we compare the change in values over time.
Let’s take a look at an example to illustrate how percentages work:
You want to buy a new laptop that costs \$1500. However, the store is offering a 20% discount on all electronics. How much will you save?
To calculate the discount, we can use the formula: discount = original price * percentage. In this case, the discount would be \$1500 * 0.20 = \$300.
So, by understanding percentages, you can easily determine the amount of money you will save on your purchase. This is just one example of how percentages are used in everyday life for practical calculations.
Now, let’s take a look at a table that summarizes the key concepts of percentages:
Concept Definition
Percentage A fraction or ratio expressed as part per hundred (denominator of 100)
Percentage Calculation The process of determining the proportion or relative size of a value in relation to a whole
Discount A reduction in price, often expressed as a percentage of the original price
Growth Rate The rate at which a value increases or decreases over time, expressed as a percentage
Proportion The relative size or scale of a part compared to the whole, often expressed as a percentage
Understanding percentages is key to performing accurate calculations and making informed decisions in various fields, from finance to statistics to everyday life situations. By mastering the concept of percentages, you can confidently navigate calculations and interpret data with ease.
## Converting Fractions to Percentages
Converting fractions to percentages is a straightforward process that allows you to represent fractions as a proportion of 100. To accomplish this conversion, follow the simple formula: divide the numerator by the denominator and multiply the result by 100.
For instance, let’s consider the fraction 6/8. Divide the numerator, 6, by the denominator, 8, resulting in 0.75. Then, multiply 0.75 by 100 to obtain the equivalent percentage. In this case, the fraction 6/8 converts to 75%.
This conversion method applies to all fractions. By dividing the numerator by the denominator and multiplying the result by 100, any fraction can be effectively converted into a percentage.
Converting fractions to percentages enables us to express fractional values in a more easily understood and relatable format. This conversion is particularly useful when comparing proportions or analyzing data.
“Converting fractions to percentages allows for clearer representation and comparison of fractional values.”
## Converting Percentages to Decimals
When working with percentages, it is often useful to convert them to decimals for easier calculations. Converting a percentage to a decimal is a straightforward process.
To convert a percentage to a decimal, you simply divide the percentage by 100. For example, if you have a percentage of 25%, divide 25 by 100:
25 ÷ 100 = 0.25
The result, 0.25, is the decimal equivalent of 25%. Decimals are often easier to work with in calculations as they represent the proportion of a whole number more directly.
Converting a percentage to a decimal allows you to perform various mathematical operations with ease, such as addition, subtraction, multiplication, and division.
It is also worth noting that decimals can be converted back to percentages by multiplying them by 100. For example, to convert 0.25 back to a percentage:
0.25 × 100 = 25%
By converting percentages to decimals and vice versa, you can seamlessly switch between the two representations depending on the context of your calculations.
### Example: Converting Percentages to Decimals in a Table
Percentage Decimal
10% 0.10
25% 0.25
50% 0.50
75% 0.75
90% 0.90
This table provides a visual representation of converting various percentages to decimals. It showcases how the decimal representation corresponds to the respective percentage value. By referring to such a table, you can quickly convert percentages to decimals for your calculations.
## Calculating Percentages of a Number
In many situations, you may need to calculate the percentage of a given number. Whether you want to determine a discount, calculate interest, or find a portion of a whole, understanding how to calculate percentages is essential. By following a simple formula, you can quickly find the desired percentage.
The formula for calculating the percentage of a number is straightforward. You have two options:
1. Multiply the number by the percentage, treated as a decimal; or
2. Divide the percentage by 100 and then multiply by the number.
Let’s take an example to illustrate this. Suppose you want to find 20% of \$2000.
To find 20% of \$2000, you can multiply \$2000 by 0.20. The result is \$400.
Here’s a summary of the formula:
1. To multiply: Number * Percentage (treated as a decimal)
2. To divide: Percentage / 100 * Number
Calculating percentages of a number is a useful skill that can be applied in various real-life situations. Whether you are managing finances, planning a budget, or analyzing data, understanding percentages will empower you to make informed decisions and perform accurate calculations.
## Solving Percentage Problems
Percentage problems are a common occurrence in real-life scenarios, whether you’re calculating discounts, taxes, or interest rates. These problems require a solid understanding of percentages and the ability to apply the formulas and techniques discussed earlier in this article to solve them.
By mastering percentage calculations, you gain the skills needed to tackle a variety of real-life percentage word problems. Let’s take a look at a practical example:
Example: Amy went shopping and found a dress she liked. The original price of the dress was \$100, but it was on sale for 25% off. How much did Amy save?
To solve this problem, we utilize the formula for calculating discounts. We know that the discount is 25%, which means Amy pays only 75% of the original price. Here’s how we calculate her savings:
1. Convert the percentage to a decimal: 25% = 0.25
2. Multiply the original price by the decimal representation of the percentage: \$100 * 0.25 = \$25
Amy saved \$25 on the dress, thanks to the 25% discount.
Solving real-life percentage problems like this one helps you in many situations, such as calculating tax amounts, determining sale prices or total costs, and understanding the impact of interest rates on loans or investments. With a solid grasp of percentage calculations, you can make informed decisions and navigate real-life scenarios with confidence.
Real-Life Scenario Formula Example
Calculating discounts Original price * (1 – Percentage discount) \$100 * (1 – 0.25) = \$75
Determining tax amounts Original price * Tax rate \$100 * 0.1 = \$10
Calculating sale prices Original price * (1 – Percentage sale) \$100 * (1 – 0.1) = \$90
Understanding interest rates Principal amount * (1 + Interest rate) \$100 * (1 + 0.05) = \$105
Percentages also play a significant role in everyday life. They are used in everyday scenarios, such as calculating tax rates, determining sales discounts, and interpreting statistical data. Understanding percentages is essential for making informed decisions and analyzing information accurately.
Furthermore, the versatility of percentages extends to areas beyond mathematics and finance. They are employed in various fields, including science, economics, and social sciences, to express relative proportions, growth rates, and probabilities.
Overall, the history of percentages showcases their long-standing presence in human civilization, their practicality in diverse fields, and their importance in understanding and interpreting numerical information. By comprehending the origins and applications of percentages, we gain a deeper appreciation for their utility and relevance in our lives.
## Practical Applications of Percentages
Percentages play a crucial role in various practical applications in everyday life. They are extensively used in financial situations, statistical analysis, probability calculations, and scientific research. Understanding percentages is not only important for making informed decisions but also for effectively analyzing data.
One of the practical uses of percentages is in financial calculations. They help us calculate discounts, interest rates, tax rates, and profits, enabling individuals and businesses to effectively manage their finances. For example, percentages are used to determine the amount of discount during sales or to calculate the interest earned on investments.
Percentages are also utilized in statistical analysis and probability calculations. They allow us to understand and interpret data by measuring proportions and determining the likelihood of specific events. In scientific research, percentages help analyze experimental results, benchmark performance, and identify trends.
Let’s explore a couple of examples to illustrate the practical applications of percentages:
1. Calculating discounts at a store: Imagine you’re shopping during a clearance sale where all items are marked down by 40%. To determine the final price, you need to apply the percentage discount. Here’s a simple calculation:
Original price: \$50
Discount percentage: 40%
Final price = Original price – (Original price * Discount percentage) = \$50 – (\$50 * 0.40) = \$30
By understanding percentages, you can quickly calculate and determine the discounted price, helping you make informed purchase decisions.
2. Determining probabilities: Percentages are crucial in determining the likelihood of certain events occurring. For example, let’s consider rolling a six-sided die. The probability of rolling a six can be calculated as:
Total number of possible outcomes: 6
Number of favorable outcomes (rolling a six): 1
Probability of rolling a six = (Number of favorable outcomes / Total number of possible outcomes) * 100% = (1 / 6) * 100% = 16.67%
Understanding percentages allows us to evaluate probabilities, make predictions, and assess risks in various scenarios.
## The Benefits of Using Percentages in Math
Using percentages in math provides several advantages and benefits. Let’s explore why percentages play a crucial role in mathematics and the importance they hold in various calculations.
### Simplifying Calculations
One of the significant benefits of using percentages in math is their ability to simplify calculations. Percentages provide a standardized way of expressing fractions or ratios out of 100, making it easier to work with numbers and perform various mathematical operations. By converting values into percentages, complex calculations can be broken down into simpler steps.
### Comparing Values
Percentages also make it easier to compare values. By converting numbers into percentages, you can quickly assess the proportion or relationship between different quantities. This comparative aspect is especially useful when analyzing data or making informed decisions based on numerical comparisons.
### Clear Representation of Proportions and Ratios
Percentages offer a clear representation of proportions and ratios. By expressing values as percentages, you can visually understand the distribution of a whole and its parts. This clarity helps in comprehending the relative size, significance, or contribution of different components within a given context.
For example, when analyzing data or conducting statistical analysis, percentages provide a meaningful representation of the relative frequencies or occurrence of specific events or categories.
“Percentages offer a clear representation of proportions and ratios, facilitating the analysis of data and the interpretation of numbers.”
### Importance in Mathematical Concepts and Formulas
Percentages are widely used in various mathematical concepts and formulas. They are an integral part of topics such as percentages, proportions, ratios, probability, and growth rates. Understanding and proficiently working with percentages lays a strong foundation for solving mathematical problems across different domains.
Whether it’s calculating discounts, tax rates, or compound interest, percentages are pivotal in real-world applications and mathematical problem-solving strategies.
“Mastering percentages expands your mathematical toolkit and equips you with essential skills for solving a wide range of problems.”
Overall, the benefits of using percentages in math are undeniable. They simplify calculations, facilitate comparisons, provide clear representations of proportions and ratios, and are essential in various mathematical concepts and formulas. By embracing percentages, you unlock a powerful tool that enhances your mathematical comprehension and problem-solving capabilities.
## Percentages and Financial Calculations
When it comes to managing finances, percentages play a vital role. They are an essential tool for calculating interest rates, determining discounts, and evaluating investment returns. By understanding percentages, individuals can make informed decisions about savings, loans, and budgeting, ultimately leading to better financial planning and management.
Let’s take a closer look at some of the key areas where percentages are used in finance:
1. Interest Rates: Percentages are widely used to calculate and compare interest rates on loans and investments. Whether you’re applying for a mortgage, researching savings account options, or considering investment opportunities, understanding the percentage-based interest rates is crucial in assessing the potential return or cost.
2. Discounts: Percentages are used to determine discounts during sales and promotions. Whether you’re shopping for clothes, electronics, or groceries, being able to calculate the discount percentage helps you compare prices and make informed purchasing decisions.
3. Investment Returns: When investing in stocks, bonds, or other financial instruments, percentages are used to measure returns. By calculating the percentage gain or loss on an investment, you can assess its performance and make adjustments to your portfolio.
Financial Calculation Example
Interest Calculation Principal: \$10,000
Interest Rate: 5%
Time: 2 years
Simple Interest: \$1,000
Discount Calculation Original Price: \$100
Discount Percentage: 20%
Discount Amount: \$20
Final Price: \$80
Investment Return Calculation Initial Investment: \$5,000
Final Value: \$6,500
Percentage Return: 30%
Understanding percentages in finance empowers you to make better financial decisions and effectively manage your money. By having a grasp of how percentages are used in interest calculations, discounts, and investment returns, you can optimize your financial strategies and achieve your financial goals.
In addition to the examples mentioned, percentages are also used in various financial scenarios such as tax calculations, loan interest calculations, and determining profit margins in business. Having a strong understanding of percentages is essential for effective money management and financial literacy.
Now that we have explored the role of percentages in financial calculations, let’s move on to the next section where we will discuss common percentage calculations and real-life examples.
## Common Percentage Calculations
Percentage calculations are widely used in everyday situations to determine discounts, sales tax amounts, tips, and changes in values over time. By understanding these common percentage calculations, you can efficiently solve real-life problems and make informed decisions. Let’s explore some examples:
Finding Discounts: Imagine you’re shopping online, and you come across a 20% off sale on a \$100 item. To calculate the discount amount, simply multiply the original price by the percentage as a decimal:
\$100 * 0.20 = \$20
Therefore, the discount amount is \$20.
Determining Sales Tax Amounts: When making a purchase, it’s essential to calculate the sales tax to know the total amount you’ll be paying. Let’s say the sales tax rate is 8% on a \$50 item. To determine the tax amount, multiply the price by the percentage as a decimal:
\$50 * 0.08 = \$4
The sales tax amount would be \$4.
Calculating Tips: When dining at a restaurant, it’s customary to leave a tip for the service received. Let’s say you want to leave a 15% tip on a \$60 bill. Multiply the bill amount by the percentage as a decimal to get the tip amount:
\$60 * 0.15 = \$9
The tip amount would be \$9.
Analyzing Changes in Values Over Time: Percentage calculations can be used to analyze changes in values over time. For example, suppose an investment increases from \$10,000 to \$12,000 in one year. To calculate the percentage increase, subtract the initial value from the final value, divide by the initial value, and multiply by 100:
(\$12,000 – \$10,000) / \$10,000 * 100 = 20%
The investment experienced a 20% increase over the year.
These are just a few examples of how percentage calculations are applied in day-to-day scenarios. By mastering common percentage calculations, you’ll have a valuable tool for making financial decisions, understanding discounts, and analyzing changes in values.
## Conclusion
Understanding percentages is crucial in various aspects of life, including finance, mathematics, and everyday calculations. Percentages allow for easy comparison of values and simplification of complex calculations. By mastering the concept of percentages and knowing how to calculate them, individuals can make more informed decisions and efficiently solve a wide range of percentage-related problems.
Whether it’s calculating discounts, determining tax amounts, or analyzing investment returns, percentages play a vital role in financial calculations. They enable individuals to effectively manage their money, make accurate financial plans, and monitor their progress towards financial goals.
Additionally, percentages have practical applications beyond finance. They are used in statistical analysis, scientific research, and probability calculations. Understanding percentages provides a foundation for interpreting data, making predictions, and drawing meaningful conclusions.
In conclusion, percentages are an integral part of our daily lives. By grasping the concepts and techniques related to percentages, individuals gain valuable skills that empower them to navigate various mathematical and financial calculations seamlessly. A solid comprehension of percentages leads to better decision-making, efficient problem-solving, and a deeper understanding of numerical relationships in our complex world.
## FAQ
### How do I calculate 20% of \$2000?
To calculate 20% of \$2000, you can multiply \$2000 by 0.20 or divide \$2000 by 5. The result is \$400.
### What are percentages and how do they work?
Percentages represent parts per hundred and are a way to express a fraction or ratio where the denominator is 100. They are commonly used in everyday life to convey proportions and completeness.
### How do I convert fractions to percentages?
To convert a fraction to a percentage, divide the numerator by the denominator and multiply by 100. For example, to convert 6/8 to a percentage, divide 6 by 8 to get 0.75, then multiply by 100 to get 75%. Similarly, other fractions can be converted to percentages using this formula.
### How do I convert percentages to decimals?
To convert a percentage to a decimal, divide the percentage by 100. For example, 25% can be converted to a decimal by dividing 25 by 100, resulting in 0.25. Decimals are often easier to work with in calculations and can be converted back to percentages by multiplying by 100.
### How do I calculate a percentage of a number?
To calculate a percentage of a number, multiply the number by the percentage as a decimal or divide the percentage by 100 and then multiply by the number. For example, to find 20% of \$2000, you can multiply \$2000 by 0.20, resulting in \$400.
### How do I solve percentage problems?
Percentage problems often arise in real-life scenarios, such as calculating discounts, taxes, or interest rates. These problems can be solved using the formulas and techniques mentioned earlier. By understanding percentages, you can easily solve various percentage-related problems.
### What is the history of percentages?
The concept of percentages dates back to ancient times, with the Romans popularizing the use of fractions based on 100. The term “percent” comes from the Latin phrase “per centum,” meaning “by the hundred.” Over time, percentages became widely used in various fields, including finance, mathematics, and everyday life.
### What are the practical applications of percentages?
Percentages have numerous practical applications in everyday life. They are used in financial situations, such as calculating discounts, interest rates, and tax rates. They are also used in statistical analysis, probability calculations, and scientific research. Understanding percentages is crucial for making informed decisions and analyzing data.
### What are the benefits of using percentages in math?
Using percentages in math provides several benefits. They simplify calculations and make it easier to compare values. Percentages also allow for clearer representation of proportions and ratios. They are widely used in various mathematical concepts and formulas, making them essential in solving mathematical problems.
### How do percentages relate to financial calculations?
Percentages play a crucial role in financial calculations. They are used to calculate interest rates, discounts, and investment returns. They help individuals make informed decisions regarding savings, loans, and budgeting. Understanding percentages is essential for effective financial planning and management.
### What are some common percentage calculations?
Common percentage calculations include finding discounts, determining sales tax amounts, calculating tips, and analyzing changes in values over time. These calculations can be easily solved using the formulas and techniques discussed earlier. Understanding common percentage calculations is useful in everyday situations.
### Is it important to understand percentages?
Percentages are essential in various aspects of life, including finance, mathematics, and everyday calculations. They allow for easy comparison of values and simplification of calculations. By understanding percentages and how to calculate them, you can make more informed decisions and solve various percentage-related problems efficiently.
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## A circular placemat has 44 inches of braid around the edge. What is the area of the placemat to the nearest square inch? Use 3.14 for π.
Question
A circular placemat has 44 inches of braid around the edge. What is the area of the placemat to the nearest square inch? Use 3.14 for π.
Enter the correct answer in the box.
[ ]
in progress 0
3 years 2021-09-05T11:56:31+00:00 2 Answers 8 views 0
a ~ 154
a =
Step-by-step explanation:
1. Approach
In this problem, one is given the circumference, the distance around the circle, one must find the area. In order to find the area, one must know the radius of the circle. To find the radius, one has to backsolve from the given circumference, then substitute the derived value into the formula for the area and solve.
The formula to find the circumference of a circle is;
Where (r) is the radius, (pi) is (3.1415), and (c) is the circumference.
Substitute in the given values and solve;
3. Find the area
Now, one has the value of the radius, substitute it into the formula for the area of a circle and solve;
The formula for finding the area of a circle;
Where (pi) is (3.1415), (r) is the radius, and (a) is the area,
Substitute in the value and solve.
a ~ 154
2. ### Answer: approximately 154 square inches
==========================================================
Explanation:
We’ll need to use the circumference formula
C = 2*pi*r
to find the value of the radius r when C = 44
So we get
C = 2*pi*r
44 = 2*3.14*r
44 = 6.28*r
6.28r = 44
r = 44/6.28
r = 7.0063694267516
which is approximate
Then we use this value of r to find the area we need
A = pi*r^2
A = 3.14*(7.0063694267516)^2
A = 154.140127388536
A = 154
The area of the circular placemat is approximately 154 square inches.
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# How do you solve 5y + 9- 2( - 4y - 2) = 5( y - 1)?
Sep 16, 2017
$y = - \frac{18}{8}$
#### Explanation:
$5 y + 9 - 2 \left(- 4 y - 2\right) = 5 \left(y - 1\right)$
To simplify the eqaution,we need to solve the brackets,we need to multiply $- 2$ with $\left(- 4 y - 2\right)$ and also $5$ with $\left(y - 1\right)$
Now, $- 2 \cdot \left(- 4 y - 2\right)$ = $\left(- 2\right)$$\left(- 4 y\right)$ + $\left(- 2\right) \left(- 2\right)$
Therefore, $- 2$
$\left(- 4 y - 2\right)$ = $8 y + 4$
Also, $5 \cdot \left(y - 1\right)$=$\left(5 \cdot y\right) - 5 \cdot \left(1\right)$
Therefore, $5. \left(y - 1\right)$=$5 y - 5$
Substituting these values of brackets in equation,we have
$5 y + 9 + 8 y + 4 = 5 y - 5$
Therefore, $\left(5 y + 8 y\right) + \left(9 + 4\right) = 5 y - 5$
Therefore, $13 y + 13 = 5 y - 5$
Therefore,$\left(13 y - 5 y\right) = \left(- 5 - 13\right)$
Therefore,$8 y = - 18$
Thus,$y = - \frac{18}{8}$
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# How To Simplify Imaginary Numbers
By Kathleen Cantor, 03 Jul 2020
An imaginary number is essentially a complex number - or two numbers added together. The difference is that an imaginary number is the product of a real number, say b, and an imaginary number, j. The imaginary unit is defined as the square root of -1. Here's an example: sqrt(-1).
So the square of the imaginary unit would be -1. Here's an example: j2 = -1.
The square of an imaginary number, say bj, is (bj)2 = -b2. An imaginary number can be added to a real number to form another complex number. For example, bj is a complex number with a as the real part of the complex number and b as the imaginary part of the complex number.
Complex numbers are sometimes represented using the Cartesian plane. The x-axis represents the real part, with the imaginary part on the y-axis. From this representation, the magnitude of a complex number is defined as the point on the Cartesian plane where the real and the imaginary parts intersect.
Care must be taken when handling imaginary numbers expressed in the form of square roots of negative numbers. For example:
Sqrt(-6) = sqrt(-1) * sqrt(6)
= sqrt(6)j.
However, this does not apply to the square root of the following,
Sqrt(-4 * -3) = sqrt(12)
And not sqrt(-4) * sqrt(-3) = 2j * sqrt(3)j
So when the negative signs can be neutralized before taking the square root, it becomes wrong to simplify to an imaginary number.
## Simplifying Imaginary Numbers
The nature of problems solved these days has increased the chances of encountering complex numbers in solutions. And since imaginary numbers are not physically real numbers, simplifying them is important if you want to work with them. We'll consider the various ways you can simplify imaginary numbers.
### Powers of the Imaginary Unit
The imaginary unit, j, is the square root of -1. Hence the square of the imaginary unit is -1. This follows that:
1. j0 = 1
2. j1 j
3. j2 = -1
4. j3 = j2 j = -1 x j = -j
5. j4 = j2 j2 = -1 x -1 = 1
6. j5 = j4 j = 1 x j = j
7. j6 = j4 j2 = 1 x -1 = -1
Understanding the powers of the imaginary unit is essential in understanding imaginary numbers. Following the examples above, it can be seen that there is a pattern for the powers of the imaginary unit. It always simplifies to -1, -j, 1, or j. A simple shortcut to simplify an imaginary unit raised to a power is to divide the power by 4 and then raise the imaginary unit to the power of the reminder.
For example: to simplify j23, first divide 23 by 4.
23/4 = 5 remainder 3. So j23 = j3 = -j …… as already shown above.
Consider another example
(2j)6 = 26 x j6 = 64 x -1 = -64
### Conjugates
Simply put, a conjugate is when you switch the sign between the two units in an equation. The conjugate of a complex number would be another complex number that also had a real part, imaginary part, the same magnitude. However, it has the opposite sign from the imaginary unit.
For example, if x and y are real numbers, then given a complex number, z = x + yj, the complex conjugate of z is x – yj.
Complex conjugates are very important in complex numbers because the product of complex conjugates is a real number of the form x2 + y2. They are important in finding the roots of polynomials.
To illustrate the concept further, let us evaluate the product of two complex conjugates.
(x + yj)(x – yj)
= x2 – xyj + xyj – y2j2
= x2 – (-y2)
= x2 + y2
Example
Simplify the expression 2 / (1 + 3j)
The above expression is a complex fraction where the denominator is a complex number. As it is, we can't simplify it any further except if we rationalized the denominator. The concept of conjugates would come in handy in this situation.
When dealing with fractions, if the numerator and denominator are the same, the fraction is equal to 1.
Hence (1 – 3j) / (1 – 3j) = 1
Also, when a fraction is multiplied by 1, the fraction is unchanged. So we will multiply the complex fraction 2 / (1 + 3j) by (1 – 3j) / (1 – 3j) where (1 – 3j) is the complex conjugate of (1 + 3j).
(2 / (1 + 3j)) * ((1 – 3j) / (1 – 3j))
= 2(1 – 3j) / (1 + 3j)(1 – 3j)
The denominator of the fraction is now the product of two conjugates. As stated earlier, the product of the two conjugates will simplify to the sum of two squares.
Hence
2(1 - 3j) / (1 + 3j)(1 – 3j) = 2(1 - 3j) / (12 + 32)
= 2(1 - 3j) / (1 + 9)
= (2 - 6j) / 10
= 0.2 - 0.6j
We've been able to simplify the fraction by applying the complex conjugate of the denominator.
### De Moivre’s Theorem
Complex numbers can also be written in polar form. The earlier form of x + yj is the rectangular form of complex numbers. Given a complex number z = x + yj, then the complex number can be written as z = r(cos(n) + jsin(n))
Where r = sqrt(x2 + y2)
n = arctan (y/x).
De Moivre’s theorem states that r(cos(n) + jsin(n))p = rp(cos(pn) + jsin(pn))
Here's an example that can help explain this theory.
#### Example
Given z = 1 + j, find z2
Let us convert the complex number to polar form.
r = sqrt(12 + 12) = sqrt (2)
n = arctan (1/1) = 45o
So z in polar form is z = sqrt(2)(cos(45) + jsin(45)).
Z2 = sqrt(2)(cos(45) + jsin(45))2.
You can see what happens when we apply De Moivre’s theorem:
sqrt(2)(cos(45) + jsin(45))2 = (sqrt(2))2(cos(2 x 45) + jsin(2 x 45))
= 2(cos(90) + jsin(90))
= 2(0 + j) or = 2j
So z2 = (1 + j)2 = 2j
You can verify the answer by expanding the complex number in rectangular form.
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# The Fibonacci Sequence
Mia Temma
·Sep 29, 2021·
The Fibonacci sequence is a sequence `Fn` of natural numbers defined recursively:
`````` F0 = 0
F1 = 1
Fn = Fn-1 + Fn-2, if n>1
``````
Write a function to generate the `nth` Fibonacci number.
Solutions can be iterative or recursive (though recursive solutions are generally considered too slow and are mostly used as an exercise in recursion).
In the following, we will discuss iterative, recursive and numeric implementations and compare them in terms of speed. We will use several functions that we showed this week in the "PicoLisp Explored" series too.
### The Iterative Approach
Let's start with the iterative approach that we would probably also use if we calculated it by hand. The implementation is straightforward, the only interesting point is the `swap` function. From the documentation:
(swap 'var 'any) -> any
Set the value of `var` to `any`, and return the previous value.
Let's start with `A=0, B=1`. `A+B` will be our new `B`, and the "old" `B` will be `A`. This is equivalent to `A=A+B` and then simply swapping the values of `A` and `B` with each other, right?
So this is what we will do: `(swap 'A B)` sets `A` to `B`, but returns the old value, which is the previous `A`. This means that those two expressions are equivalent:
``````: (+ (swap 'A B) B )
# is same as:
: (+ A B)
``````
Now we only need to swap `B` with `(+ A B)`and we're done:
``````(de fib (N)
(let (A 0 B 1)
(do N
(swap 'B (+ (swap 'A B) B)) ) ) )
``````
Each time the `do`-loop is finished, `A` and `B` have both iterated to the next higher Fibonacci-Number.
### The Recursive Approach
We have discussed the recursive solution extensively in the Binary Tree series, so let's just take the best parts and quickly go through it. The "basic" implementation is easy, but awfully slow since the number of calls grow exponentially.
``````(de fibo (N)
(if (>= 2 N)
1
(+ (fibo (dec N)) (fibo (- N 2))) ) )
``````
A susbstantial improvement can be reached by introducing caching. Caching means that each calculation that has already been done is stored in a binary tree. This means that every number is only calculated once instead of multiple times like in the standard version.
``````(de fiboCache (N)
(cache '(NIL) N
(if (>= 2 N)
1
(+ (fiboCache (dec N)) (fiboCache (- N 2))) ) ) )
``````
There is even room for further improvement: Instead of the `cache` function, we could take the `enum` function, which also builds up a binary tree of already reached values. However, instead of calculating a key to store the result, `enum` takes a number as argument for the storing position.
``````(de fiboEnum (N)
(let E (enum '(NIL) N)
(or
(val E)
(set E
(if (>= 2 N)
1
(+ (fiboEnum (dec N)) (fiboEnum (- N 2))) ) ) ) ) )
``````
### The Numeric Approach
Interestingly, there is also a third approach which is entirely numeric. It builds up on Binet's formula to calculate the famous golden ratio. The golden ratio is famous for appearing in some patterns in nature, such as the spiral arrangement of leaves and other plant parts.
An example is the sunflower in the cover picture: "florets in spirals of 34 and 55 around the outside", according to Wikipedia. (I didn't count it, so let's just believe it).
Anyway, this is Binet's formula to calculate the n'th Fibonacci number:
We can implement it in PicoLisp, but we need to take care of a few things:
• The square root of 5 is an irregular number, therefore we need a high precision in order to get correct results. This we can do using the `scl` function (read this article if you don't know about `scl`).
• we need to load the libary `math.l` (included in the standard pil21-distribution) in order to be able to use the function `pow`.
We call the first term inside the brackets `P` and the second one `Q`:
``````(scl 9)
(de analytic_fibonacci (N)
(setq N (* N 1.0))
(let
(Sqrt5 (sqrt 5.0 1.0)
P (/ (+ 1.0 Sqrt5) 2)
Q (*/ `(* 1.0 1.0) P) )
(/
(+
(*/
(+ (pow P N) (pow Q N))
1.0
Sqrt5 )
0.5 )
1.0 ) ) )
``````
The ` in the 6th line of the `analytic_fibonacci` function is a so-called Read-Macro that causes the reader to evaluate the expression.
However, when we run the code, we still see a couple of problems. Things start to get wrong at `N=43`: instead of 433494437 we get 433494428. The error is further increasing with higher numbers.
A second problem is that for all numbers `N>47`, the results are 0!!
What is happening? The `@lib/math.l` is calling a native C-library to calculate the exponent Q^N (the `pow` function). However, C is calculating in floating point numbers. Analysis shows that the Binet's equation is not solvable with floating point numbers due to several issues:
Tracing reveals that `Q=0.618` and `N=48`, and since `Q<1`, the whole term`Q^N` is getting closer and closer to `0` until we get an underflow. On the other hand, we know that `P>1` which means that `P^N` is growing larger and larger, until the term `P - Q` is smaller than the mantisse of the floating number.
So we either get an underflow error, or we get 0 - in either way, it's a problem. This leaves only one solution: We must get rid of the `pow` function that forces us to use floating point numbers.
Let's try it with integers!
Now we're getting quite deep into maths! The first problem we're facing is that the results are wrong since our representation of `(sqrt 5)` was too unprecise. So let's increase it to `scl 100`.
Secondly, we need to replace our power function `pow` by an integer one, because we know now that floating numbers won't work. So, let's re-define the `pow` function by a new exponential function using only integers:
``````(de fixPow (X N) # N th power of X
(if (ge0 N)
(let Y 1.0
(loop
(when (bit? 1 N)
(setq Y (*/ Y X 1.0)) )
(T (=0 (setq N (>> 1 N)))
Y )
(setq X (*/ X X 1.0)) ) )
0 ) )
``````
(credits to Alexander Burger)
Now let's try again! The test reveals two things:
• The numbers are correct until `N=470`. In order to further increase the precision, we simply would need to further increase the scale `scl`.
• The upper limit has vanished: We can even calculate the 1.000.000 Fibonacci number, which has an incredible number of 208989 digits!
### A Comparison
Let's compare the six approaches that we have seen in this post. Which is the highest `N` we can get within 0.1 sec (depending on the hardware, obviously)?
• Recursive: 30
• Numeric, Float: 42 (only correct results)
• Numeric, Integer: 470 (only correct results)
• Recursive with `cache`: 10.000
• Recursive with `enum`: 15.000
• Iterative: 30.000
• Numeric, Integer: 400.000 (error unknown)
As you can see, the numeric solution is by far the fastest one (though the execution speed is also decreasing with higher `N` due to the loop in the `fixPow` function).
Nevertheless, since we probably want the results to be exact, the iterative solution wins.
|
# What is 135/86 as a decimal?
## Solution and how to convert 135 / 86 into a decimal
135 / 86 = 1.57
To convert 135/86 into 1.57, a student must understand why and how. Both represent numbers between integers, in some cases defining portions of whole numbers But in some cases, fractions make more sense, i.e., cooking or baking and in other situations decimals make more sense as in leaving a tip or purchasing an item on sale. Once we've decided the best way to represent the number, we can dive into how to convert 135/86 into 1.57
## 135/86 is 135 divided by 86
Teaching students how to convert fractions uses long division. The great thing about fractions is that the equation is already set for us! The two parts of fractions are numerators and denominators. The numerator is the top number and the denominator is the bottom. And the line between is our division property. We must divide 135 into 86 to find out how many whole parts it will have plus representing the remainder in decimal form. Here's how you set your equation:
### Numerator: 135
• Numerators are the number of parts to the equation, showed above the vinculum or fraction bar. 135 is one of the largest two-digit numbers you'll have to convert. The bad news is that it's an odd number which makes it harder to covert in your head. Values closer to one-hundred make converting to fractions more complex. Let's take a look below the vinculum at 86.
### Denominator: 86
• Denominators are the total numerical value for the fraction and are located below the fraction line or vinculum. 86 is one of the largest two-digit numbers to deal with. The good news is that having an even denominator makes it divisible by two. Even if the numerator can't be evenly divided, we can estimate a simplified fraction. Overall, two-digit denominators are no problem with long division. So without a calculator, let's convert 135/86 from a fraction to a decimal.
## Converting 135/86 to 1.57
### Step 1: Set your long division bracket: denominator / numerator
$$\require{enclose} 86 \enclose{longdiv}{ 135 }$$
Use long division to solve step one. Yep, same left-to-right method of division we learned in school. This gives us our first clue.
### Step 2: Solve for how many whole groups you can divide 86 into 135
$$\require{enclose} 00.1 \\ 86 \enclose{longdiv}{ 135.0 }$$
We can now pull 86 whole groups from the equation. Multiply by the left of our equation (86) to get the first number in our solution.
### Step 3: Subtract the remainder
$$\require{enclose} 00.1 \\ 86 \enclose{longdiv}{ 135.0 } \\ \underline{ 86 \phantom{00} } \\ 1264 \phantom{0}$$
If your remainder is zero, that's it! If you still have a remainder, continue to the next step.
### Step 4: Repeat step 3 until you have no remainder
In some cases, you'll never reach a remainder of zero. Looking at you pi! And that's okay. Find a place to stop and round to the nearest value.
### Why should you convert between fractions, decimals, and percentages?
Converting fractions into decimals are used in everyday life, though we don't always notice. Remember, they represent numbers and comparisons of whole numbers to show us parts of integers. This is also true for percentages. So we sometimes overlook fractions and decimals because they seem tedious or something we only use in math class. But each represent values in everyday life! Without them, we’re stuck rounding and guessing. Here are real life examples:
### When you should convert 135/86 into a decimal
Dining - We don't give a tip of 135/86 of the bill (technically we do, but that sounds weird doesn't it?). We give a 156% tip or 1.57 of the entire bill.
### When to convert 1.57 to 135/86 as a fraction
Time - spoken time is used in many forms. But we don't say It's '2.5 o'clock'. We'd say it's 'half passed two'.
### Practice Decimal Conversion with your Classroom
• If 135/86 = 1.57 what would it be as a percentage?
• What is 1 + 135/86 in decimal form?
• What is 1 - 135/86 in decimal form?
• If we switched the numerator and denominator, what would be our new fraction?
• What is 1.57 + 1/2?
|
Question Video: Finding a Side Length and an Angle Using Parallel Lines and Traversals Mathematics
Consider triangle π΄π΅πΆ and lines π΄π and πΈπ·, which are parallel to line πΆπ΅. Find the length of the line segment π΄π΅. Find the measure of β π΄π΅πΆ.
02:42
Video Transcript
Consider triangle π΄π΅πΆ and lines π΄π and πΈπ·, which are parallel to line πΆπ΅. Find the length of the line segment π΄π΅. Find the measure of angle π΄π΅πΆ.
We are given three parallel lines and two transversals of these lines. We can then recall that if a set of parallel lines divide a transversal into segments of equal length, then they divide any other transversal into segments of equal length. Since π΄πΈ is equal to πΈπΆ, the segments of the other transversal must be equal in length. So π΄π· is equal to π·π΅, which equals five millimeters. Since π΄π΅ is equal to π΄π· plus π·π΅, then π΄π΅ equals five millimeters plus five millimeters, which equals 10 millimeters. π΄π΅ is equal to 10 millimeters.
It appears in the diagram that triangle π΄π΅πΆ is a right triangle. However, we need to justify why this is the case. We can do this by recalling that if a line is perpendicular to line πΏ, then it is perpendicular to any line parallel to πΏ. Since line πΈπ· is perpendicular to line π΄πΆ and line πΈπ· is parallel to line π΅πΆ, we must have that lines π΅πΆ and π΄πΆ are perpendicular. This means the angle at πΆ has a measure of 90 degrees, so π΄π΅πΆ is a right triangle.
The sum of the measures of the interior angles in a triangle is 180 degrees. So 180 degrees equals 35 degrees plus 90 degrees plus the measure of angle π΄π΅πΆ. Rearranging the equation, we have the measure of angle π΄π΅πΆ equals 180 degrees minus 35 degrees minus 90 degrees, which equals 55 degrees. The answers to the two parts of this question are 10 millimeters and 55 degrees.
|
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# Represent and Interpret data
Sep 27, 2022
## Key Concepts
• Data and line plot
• Analyzing the line plot
## Introduction
In this chapter, we will learn about data and line plots, how to read the line plot and how to analyze the line plot.
### Data and line plot
#### What are line plots?
A line plot is a way to display data along a number line.
Line plots are also called dot plots.
Example 1:
17 turtles were walking on the beach. They walked for one hour, and each turtle covered a certain distance. The below table shows the distances covered by each turtle in meters.
Let us draw the line plot for the above data.
Step 1: Sort the data from the least to the greatest.
20, 21, 23, 24, 25, 26, 27
Step 2: Draw the number line.
Step 3: Arrange the data in a table.
Step 4: Plot the values in the number line.
Draw 3 dots above 21 because 3 turtles travelled a distance of 21 meters in an hour.
Step 5: Draw 0 dots above 22 because 0 turtles travelled a distance of 22 meters in an hour.
Similarly, draw dots for other values.
Example 2:
A class has 21 students. The following data represents the number of pets each student has.
Step 1: Sort the data from the least to the greatest.
0, 1, 2, 3, 4, 5, 6
Step 2: Draw the number line.
Step 3: Arrange the data in a table.
Step 4: Plot the values in the number line.
Example 1:
The below line plot shows the number of new necklaces made by 5 friends.
What do you understand from this line plot?
Solution:
The five friends made new necklaces of lengths 51, 52, and 54 centimetres.
### Analyzing the line plot
Example 1:
Five friends made new necklaces of the lengths 52, 51, 52, 54, and 52 centimetres.
1. How many necklaces were 51 cm long?
1. How many necklaces were 52 cm long?
1. How many necklaces were 54 cm long?
1. Which length of the necklace was made by the maximum friends?
Solution:
By observing the above line plot, we can understand the below points,
1. 1 necklace was 51 cm long.
1. 3 necklaces were 52 cm long.
1. One necklace was 54 cm long.
1. Maximum friends made a 52 cm necklace.
Example 2:
The below line plot represents the number of hours spent on reading.
1. How many students spent one hour reading?
a. 3 students
2. How many students spent three hours reading?
a. 4 students
3. How many students spent more time reading?
a. 5 students
## Exercise
1. The heights of Sabrina’s dolls are shown below. How many dolls are taller than 22 centimetres?
2. A park ranger counted stripes on each baby zebra. She is going to make a line plot of the data. Graph the measurements from the table on the line plot.
3. Suppose you count the number of students in each classroom in your school.
Draw a line plot for given data.
4. Mr Haley’s class created this line plot to show how many candy bars each student likes. How many students like three candy bars?
5. How many cars were sold in all?
6. Answer the below question using the following line plot. Each X represents one student.
Find the total number of hours spent on reading by the students.
7. The teacher wrote down how many pictures frame the students made last week. How many children made fewer than 2 picture frames?
8. Mr Jensen recorded the scores on a math quiz. How many students scored more than 7?
9. Some people took photos while visiting the zoo. How many people took at least 1 photograph?
10. Some children compared how many thank-you notes they wrote last month. How many children are there in all?
### What have we learned
• Understanding data and line plots
• How to arrange the data on a line plot
• How to read the line plots
• Understanding and analyzing the line plots
• Generating a graph based on the ordered pairs
#### Addition and Multiplication Using Counters & Bar-Diagrams
Introduction: We can find the solution to the word problem by solving it. Here, in this topic, we can use 3 methods to find the solution. 1. Add using counters 2. Use factors to get the product 3. Write equations to find the unknown. Addition Equation: 8+8+8 =? Multiplication equation: 3×8=? Example 1: Andrew has […]
#### Dilation: Definitions, Characteristics, and Similarities
Understanding Dilation A dilation is a transformation that produces an image that is of the same shape and different sizes. Dilation that creates a larger image is called enlargement. Describing Dilation Dilation of Scale Factor 2 The following figure undergoes a dilation with a scale factor of 2 giving an image A’ (2, 4), B’ […]
#### How to Write and Interpret Numerical Expressions?
Write numerical expressions What is the Meaning of Numerical Expression? A numerical expression is a combination of numbers and integers using basic operations such as addition, subtraction, multiplication, or division. The word PEMDAS stands for: P → Parentheses E → Exponents M → Multiplication D → Division A → Addition S → Subtraction Some examples […]
#### System of Linear Inequalities and Equations
Introduction: Systems of Linear Inequalities: A system of linear inequalities is a set of two or more linear inequalities in the same variables. The following example illustrates this, y < x + 2…………..Inequality 1 y ≥ 2x − 1…………Inequality 2 Solution of a System of Linear Inequalities: A solution of a system of linear inequalities […]
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Home
Linear Model - Number Lines Materials: Deck of Fraction Bars per group and one sheet of paper and pencil per person Show a bar for 2/6 and note that the rectangular shape of the bars, as compared to the circular shape of discs or pies, has the advantage that it conveniently extends to a linear model for number lines. Students often experience difficulty locating fractions on a number line. The "Forming A Number Line" activity below shows how the Fraction Bars model provides a natural connection to the number line model. Group Activity: Drawing Number Lines Each person will need a sheet of paper and one of the red bars with 1 to 5 parts shaded for drawing the beginning of a number line. Describe and carry out the following steps as each person forms a number line with their bar. 1. Use the edge of the bar to draw a line segment the length of the bar. Mark seven points on the line, one for each end of the bar and one for each of the five vertical lines of the bar. At the beginning of the bar, label the left end point on the line with 0, and at the end of the bar label the right end point on the line with 1. Point out that the length from 0 to 1 is called a unit length. 2. Each student should write the fraction for the bar beneath the division line and at the end of the shaded amount of their bar. Point out that the end of the shaded amount of the bar indicates where the fraction for the bar should be placed on the line. The use of Fraction Bars can help students understand the correct location when placing fractions on a number line. When students just have the points on the line and not the bars, a common mistake is to begin counting at the zero point, and place the fraction for 2/6 beneath the second point rather than beneath the third point. 3. Place the other red bars with different shaded amounts on the line and write their fractions on the line. 4. Use this unit line segment to measure the length of a small object such as a pencil, marker, etc., to the nearest 1/6. Discuss rounding to the nearest fraction on the line. If the length of an object is half way between two points on the number line, such as points for 4/6 and 5/6, we usually round up to the greater fraction. Discuss this activity and the process of measuring and rounding off to the nearest 1/6. To measure longer lengths, the number line can be extended from 1 to 2 by repeating the above steps, and by labeling the division points with the mixed numbers 1 1/6, 1 2/6, etc. Discuss the effect of measuring with a number line that is formed using the orange bars: The measurement will be more precise because of the greater number of division points in the unit. One activity on the Number Lines mats is to move markers along the lines to the end of the mat. The linear nature of the bars is being used on both of the mats below. The two students in the foreground are moving markers on a number lines mat according to the shaded amount of a bar, and the two students in the background are marking progress by placing the shaded amounts of their bars end- to-end on the number lines mat.
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## Friday, February 9, 2007
### Don't You Wish You Remembered Those Trig Identities. Topic: Trigonometry. Level: AMC/AIME.
Problem: (1962 IMO - #4) Solve the equation $\cos^2{(x)}+\cos^2{(2x)}+\cos^2{(3x)} = 1$.
Solution: Subtract the $\cos^2{(3x)}$ from both sides, and replace the LHS by $\sin^2{(3x)}$, so we now have
$\cos^2{(x)}+\cos^2{(2x)} = \sin^2{(3x)}$.
Recalling the sine angle addition identity, we write
$\sin{(3x)} = \sin{(2x+x)} = \sin{(2x)}\cos{(x)}+\sin{(x)}\cos{(2x)}$.
So our equation is
$\cos^2{(x)}+\cos^2{(2x)} = [\sin{(2x)}\cos{(x)}+\sin{(x)}\cos{(2x)}]^2$.
Expanding, collecting terms, and simplifying using $1-\sin^2{(2x)} = \cos^2{(2x)}$ and $1-\sin^2{(x)} = \cos^2{(x)}$, we get
$2\cos^2{(x)} \cos^2{(2x)} = 2\sin{(x)}\cos{(x)}\sin{(2x)}\cos{(2x)}$.
Divide by $2$, rearrange, and factor:
$\cos{(x)}\cos{(2x)}[\cos{(x)}\cos{(2x)}-\sin{(x)}\sin{(2x)}] = 0$.
Substitute using the double-angle identities $\cos{(2x)} = 1-2\sin^2{(x)}$ and $\sin{(2x)} = 2\sin{(x)}\cos{(x)}$ to finally find
$\cos^2{(x)} \cos{(2x)} [1-4\sin^2{(x)}] = 0$.
Solving yields
$x = \frac{\pi}{2}+k\pi \mbox{ ; } \frac{\pi}{4}+\frac{k \pi}{2} \mbox{ ; } \frac{\pi}{6}+k\pi \mbox{ ; } \frac{5 \pi}{6}+k \pi$
for all integers $k$. QED.
--------------------
Comment: Pretty standard and slightly ugly manipulation of trig identities, but overall not too bad. Considering you get like an hour or more to do this, I don't feel too bad about it. Other solutions taking into account symmetry or using DeMoivre's and stuff are also out there, but this seemed much more straightforward and easier to develop.
--------------------
Practice Problem: (2007 AMC 12A - #24) For each integer $n > 1$, let $F(n)$ be the number of solutions of the equation $\sin{(x)} = \sin{(nx)}$ on the interval $[0, \pi]$. What is $\displaystyle \sum_{n=2}^{2007} F(n)$?
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# Ratio to Percentage
Ratio to percentage conversion process helps to represent a number in ratio form in terms of percentage. You have often seen the percentage as a way to evaluate a student’s performance in examinations. Thus, the percentage is used to compare quantities. It literally means ‘per 100’, which is a number expressed as a fraction of 100. So when you say 100% of something, it means it represents the whole of it.
Similarly, when you talk about pizza slices being divided amongst 2 people, the concept of ratio comes into the picture. In simple words, the ratio is also used to compare quantities in a different manner. In this article, we will discuss how to convert ratio into percentage easily and solve a few questions to have a better understanding of the topic.
## Ratio to Percentage Conversion
Ratio to percentage conversion helps us in obtaining accuracy in mixtures of elements, or while calculating the percentage score in a test. At times you are given parts of a quantity in the form of ratios. They can also be represented in the form of percentages. Let us understand this with the help of examples.
• Example Question 1: Varun received his monthly salary. The ratio of his expenditure to savings is 7:3. What percentage of his salary, did he spend and what percentage was saved by him?
Solution:
Since the part of saving and expenditure are 3 and 7, the salary can be taken as 3 + 7 = 10 parts. This implies, 7/10 part of the salary is spent whereas 3/10 parts are saved.
Converting ratio to percentage we get,
Percentage of expenditure = 7/10 x 100% = 70%
Similarly, percentage of savings = 3/10 x 100% = 30%
Try This: Ratio To Percent Calculator
• Example Question 2: The angles of a triangle are in the ratio 1:1:2. Find the value of each angle. What will be the percentage of each angle?
Solution:
Since the angles are in ratio 1:1:2, there are 1 + 1 + 2 = 4 parts. The sum of angles in a triangle is 180 degrees.
Thus, measure of the first angle = 1/4 x 180 = 45 degrees
Measure of the second angle = 1/4 x 180 = 45 degrees
Measure of the third angle = 2/4 x 180 = 90 degrees
Similarly, converting ratio to percentage we have,
First angle = 1/4 x 100% = 25%
Second angle = 1/4 x 100% = 25%
Third angle = 2/4 x 100% = 50%
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# How to Solve an Inequality With Fractions
0
81
Inequalities with fractions are inequalities in which one or more fractions occur in the denominator or the numerator. In such cases, the unknown variable is smaller than or equal to the value of the numerator. In these cases, we need to determine the critical value of the rational expression and flip the sign.
## Problem
Inequalities with fractions are mathematical problems that involve one or more fractions. These types of equations occur when the unknown variable is in the numerator or denominator and the values of those fractions are not equal. The solution to an inequality with fractions involves reversing the sign of the inequality.
The first step to solving an inequality with fractions involves identifying the inequality. Students can do this by observing the sign. They will need to know that they need to change one side of the inequality to make the other side smaller than the left side. In order to do this, they should use the subtraction property of inequality.
The second step in solving an inequality with fractions is to identify the critical points on the number line. These are the points where the inequality has the correct form. A critical point on the number line indicates the sign of the quotient and factor in each interval. In the example, a test value for x is highlighted in yellow.
Once you have identified the negative sign, you can reverse the inequality sign by multiplying or dividing it by another negative number. After that, use the shaded part of the graph to solve the inequality.
## Test several values of n
If you want to solve an inequality with fractions, you must test several values of n to find a solution. An inequality is equivalent if its terms are like. Then, simplify the inequality by combining like terms on each side. This will create an unknown on one side and a number on the other side. You can then divide these terms by the coefficient of the unknown.
Inequalities are often written as equations whose solution sets are given by equations. A simple example is 3x – 5 or 6x – 2x. The goal is to find all values of the variable that make the inequality true. The set of these values is called the inequality solution set. Inequalities that have the same solution set are said to be equivalent. The principles for solving these equations are the same as those for solving equations.
In this exercise, students should first test several values of n to find a solution. Generally, two or three values are recommended, but students can experiment with different values to find the one that works for them. After testing different values of n, students should understand why this inequality is not true. For example, if the left side of the inequality is +1, it can never be less than the right side of the inequality. Once they understand why, they can form their own arguments for the correct answer and analyze the arguments of their classmates.
If n is an integer, the solution of the inequality is the number xy that makes the statement true when substituted into the equation. In this example, the solution of the inequality is a number greater than three. Graphing n’s solution set allows you to visually see the solution set.
## Find the critical value of the rational expression
In mathematics, the critical value of a function is a number that causes the function to equal 0 or to be undefined. The critical value can be zero on either side of the function, or a fraction. If the critical value is zero on one side, the expression is an inequality.
To solve an inequality using fractions, the first step is to find the critical value of the rational expression. This is the least common denominator. Then, take each term’s numerator and multiply it by its lowest common denominator. Next, set the remaining factors of the numerator equal to 0 and multiply both sides of the expression by the same number.
The process of finding the critical value of a rational expression is similar to solving a polynomial inequality. However, in a rational expression, there may be zeroes or undefined points. The key to solving this kind of inequality is to find these points by dividing the number line into intervals and finding the signs on each interval.
Once you have identified these values, you can simplify the rational expression. This is particularly difficult when it contains a “-” or “+” sign. You have to reverse the symbol if the value is negative.
## Flip the sign
Inequalities with fractions are easy to solve as long as you know the basic rules. You can solve inequalities by either adding the same number to both sides, dividing each side by a positive number, or reversing the inequality by dividing the sides by a negative number.
First, you must change the sign of the inequality. For example, the inequality x/9 means x divided by 9. The solution is -36. Since x is negative, you need to flip the inequality sign. This will produce the number on the other side of the inequality.
The distributive property is another common way to simplify an inequality. When using this property, you expand both sides of the inequality, which will make the equation easier to solve. Then, you can remove the parentheses. Once you’ve removed all the parentheses, you’ll be on the road to solving an inequality with fractions.
You can also include a test number in the solution. If you’re not sure that you’ve found the correct test number, you can use a critical point to check the validity of the intervals. The test number is usually highlighted in yellow. It is used to test the intervals and determine if the numerator is a valid solution. It doesn’t provide a true statement, though.
## Multiplying or dividing by a negative
A student needs to know how to use the inequality sign when multiplying or dividing by a negative fraction. There are some tricks that will help him get started. The first step is to remember to flip the inequality sign. This is similar to how you would solve an equation. Then, use the shaded part of the graph to check for an inequality.
If the inequality is in the form of a linear equation, you can solve it by multiplying or dividing by the negative number. This will solve the inequality by making the greater side larger and the lesser side smaller. In this way, you can solve any negative number and find the inequality’s solution.
Another trick for solving inequality problems involves comparing the signs of the two sides. By doing this, you can find the answer to an inequality that looks like this: a times b is greater than a. You should be able to make this work if you have a negative number and a positive number.
When you are dealing with inequalities, it is important to find a variable that is nearer to the inequality. If you find the negative side first, you’ll see a change in the inequality sign. This change will make the left-hand side of the equation x. Also, remember that negatives cancel out, so dividing by a negative number will make the equation positive.
## Order of operations
When solving an inequality with fractions, you need to know the order of operations. When multiplying or dividing by a negative number, the sign of both numbers will change. Similarly, when dividing by a positive number, the sign will change to the equals sign.
In the second step, you need to combine like terms. You can do this by multiplying the terms with the least common denominator. After you’ve combined like terms, you can then divide the terms by the coefficient of the unknown. Using the order of operations to solve an inequality with fractions can help you avoid common mistakes.
Using the correct order of operations will make solving an inequality with fractions much easier. For example, multiplying by 0.5 is the same as multiplying by 2, but you must remember to use the shaded portion of the graph when you’re solving a problem with fractions.
Using the distributive property of equality will help you solve a multistep inequality using the order of operations. This technique is also useful when you’re working with complicated terms like percentages. By using the distributive property, you can determine the best first step for a given inequality.
In multi-step equations, pay special attention to situations involving negative numbers. Consider the example below: the graph of an inequality p=12 has an open circle at 12 and an arrow stretching to the right.
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<meta http-equiv="refresh" content="1; url=/nojavascript/"> Applications Using Linear Models ( Real World ) | Algebra | CK-12 Foundation
# Applications Using Linear Models
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Practice Applications Using Linear Models
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## Real World Applications – Algebra I
### Topic
How can we represent how much a person earns making clothes in Bangladesh?
### Student Exploration
Bangladesh is one of the most popular countries for outsourcing labor because the labor is so cheap. Let’s apply our knowledge of linear equations to represent this relationship.
A child makes $20 per month for in a garment factor. Since this is a rate, this is the slope for our equation. Let’s say he/she would work every day out of the 30 days of the month. Our slope would then be represented as $\frac{\20}{30 \ days}$, or $\frac{2}{3}$. Let’s also say that this child has started saving money for his family and after working for 3 weeks, he has$24 for his family.
From this example, we have both the slope $\left ( \frac{2}{3} \right )$ and a point (21, 24). (The $x-$value in the point (21, 24) comes from the fact that there are 21 days in 3 weeks. For this relationship, we must use days, since the slope includes days, not weeks.) We can create an equation with this information. Let’s use $y = mx + b$ and follow the steps given in the concept.
$y & = mx + b\\24 & = \left ( \frac{2}{3} \right )(21) + b\\24 & = \left ( \frac{42}{3} \right ) + b\\24 & = 14 + b\\10 & = b\\y & = \left ( \frac{2}{3} \right )x + 10$
In this equation, the “10,” or the “$b$” value represents how much the child had before he started working. This equation is in slope-intercept form. If we were to represent this equation in standard form, we first want to get the $x-$term on the same side as the $y-$term.
$\left ( \frac{2}{3} \right ) x + y = 10$
We also want all whole numbers in our equation. Let’s multiply every term by “3” so the denominator disappears.
We now have $2x + 3y = 30$.
### Extension Investigation
Try finding the equation of the line in both slope-intercept form and standard form with the following information.
The child’s family found out that their child started saving money and thought it was a good idea and started saving too. One parent decided that all of her earnings are going toward their savings, and the other parent’s earnings will go toward the family’s expenses. The mother earns $36 per month. After 60 days, the family has$82.
#### ANSWERS FOR THE EXTENSION INVESTIGATION
We know that the mother makes $36 per month. This is the slope of our equation. We also know that there are 30 days (approximately) in a month, so one of our ordered pairs is (2, 82), meaning that the family had$82 in two months. We can use this information to find the equation that represents how much money the family has.
$Y & = mx + b\\82 & = 36(2) + b\\82 & = 72 + b\\10 & = b$
So, our equation is $y = 36x + 10$. But, since we want the equation in standard form, we need to get the constant by itself. So, let’s subtract $36x$ from both sides. We now have $-36x + y = 10$.
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# Three equal cubes are placed adjacently in a row. Find the ratio of total surface area of the new cuboid to that of the sum of the surface areas of the three cubes.
Given:
Three equal cubes are placed adjacently in a row.
To do:
We have to find the ratio of total surface area of the new cuboid to that of the sum of the surface areas of the three cubes.
Solution:
Let each side of the cube be $s\ cm$
This implies,
Surface area of the cube $= 6s^2\ cm^2$
Surface area of three such cubes $= 3 \times 6s^2$
$= 18s^2\ cm^2$
By placing three cubes side by side, we get a cuboid of
Length $(l) = s \times 3 = 3s$
Breadth $(b) =s$
Height $(h) = s$
Therefore,
Total surface area of the cuboid $= 2(lb + bh + lh)$
$= 2(3s \times s+s \times s+s \times 3s)$
$= 2(3s^2 + s^2 + 3s^2)$
$= 14\ s^2$
The ratio between their surface areas $= 14s^2 : 18s^2$
$= 7 : 9$
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# Divisibility Rules Explained
Related Topics:
More Lessons for Math Games and Math Trivia
Math Worksheets
Divisibility Rule for 3, 9, 7 and 11 - Why they work?
Divisibility by 3 and by 9: Why do they work?
Most everyone seems to know the rule for divisibility by three: add up the digits and see if the sum is a multiple of three.
Most seem to “know” that the same rule holds for nine.
But why do these rules work? And did you know that summing the digits tells you a bit more than just a yes/no answer?
Divisibility Rule for Seven
Many folk know divisibility rules for 3, for 5, for 9, for 11 and so on, but what about a divisibility rule for the number 7. There is one, and here it is! (Plus, you’ll learn from this how to create your own divisibility rules for 13, and 17, and 43, and 97 and the like!)
Divisibility by 11
In this video I explain the standard divisibility rule for the number eleven. (Did you know that every palindrome with an even number of digits must be a multiple of eleven?)
Dividing by Nine: Something Cute
Here is something not too serious, but cute. To divide a number by 9, just add up its digits in turn! For example:
124 / 9 = “1 and 1+2 and 1+2+4” = 13R7
Let’s explore why this little curiosity work
Dividing by 97 - and other such Numbers
Students need help “standing back” and seeking perspective on matters. It is easy to get locked into memorisation and procedure. This little video is just a tidbit that helps in this regard. It provides an easy example of the power that can result in pausing to think about things first before diving in. Dividing big numbers by 97 is actually fairly straightforward. This is not a technique to be memorised, but a statement about thinking!
Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.
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# Twenty One Cylindrical Pillars of the Parliament House Are to Be Cleaned. If the Diameter of Each Pillar is 0.50 M and Height is 4 M - Mathematics
Twenty one cylindrical pillars of the Parliament House are to be cleaned. If the diameter of each pillar is 0.50 m and height is 4 m, what will be the cost of cleaning them at the rate of Rs 2.50 per square metre?
#### Solution
$\text{ Given } :$
$\text{ Diameter of the pillars = 0 . 5 m}$
$\text{ Radius of the pillars, r = 0 . 25 m }$
$\text{ Height of the pillars, h = 4 m }$
$\text{ Number of pillars } = 21$
$\text{ Rate of cleaning = Rs 2 . 50 per square metre }$
$\text{ Curved surface area of one pillar } = 2\pi rh$
$= 2 \times \frac{22}{7} \times 0 . 25 \times 4$
$= 2 \times \frac{22}{7}$
$= \frac{44}{7} m^2$
$\therefore \text{ Curved surface area of one pillar } = \frac{44}{7} \text{ m} ^2$
$\text{ Cost of cleaning 21 pillars at the rate of Rs 2 . 50 per m}^2 = \text{ Rs } 2 . 5 \times 21 \times \frac{44}{7}$
$= 7 . 5 \times 44$
$\therefore \text{ Cost of cleaning 21 pillars at the rate of Rs 2 . 50 per m} ^2 = Rs 330$
Is there an error in this question or solution?
#### APPEARS IN
RD Sharma Class 8 Maths
Chapter 22 Mensuration - III (Surface Area and Volume of a Right Circular Cylinder)
Exercise 22.1 | Q 15 | Page 11
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# What is the equation of the line with slope m= -3/7 that passes through (17/13,14/7) ?
##### 1 Answer
Jun 20, 2018
$y = - \frac{3}{7} x + \setminus \frac{233}{91}$
#### Explanation:
When you know a given point $\left({x}_{0} , {y}_{0}\right)$ and the slope $m$, the equation of a line is
$y - {y}_{0} = m \left(x - {x}_{0}\right)$
In your case, $\left({x}_{0} , {y}_{0}\right) = \left(\setminus \frac{17}{13} , \setminus \frac{14}{7}\right) = \left(\setminus \frac{17}{13} , 2\right)$ and $m = - \frac{3}{7}$.
Let's plug these values in the formula:
$y - 2 = - \frac{3}{7} \left(x - \setminus \frac{17}{13}\right)$
Although this already is the equation of the line, you may want to write in the slope-intercept form, for example. Expanding the right hand side, we have
$y - 2 = - \frac{3}{7} x + \setminus \frac{51}{91}$
add $2$ to both sides to get
$y = - \frac{3}{7} x + \setminus \frac{233}{91}$
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How many points are there in the compass rose? Find out in this quiz.
# Position (Easy)
In this 11-plus Maths quiz, you are going to learn a little more about the compass. Did you know that the Chinese invented the compass over two thousand years ago?
Before you start this easy quiz, here's a tip: North East (NE) is midway between N and E; South East (SE) is midway between S and E; South West (SW) is midway between S and W; North West (NW) is midway between N and W. These points are called intercardinal points and they are positioned at 45° to each of the cardinal points they are positioned between.
Finally, it would be a good idea if you drew a diagram of the compass rose to help you do these questions.
Take your time, think carefully and see if you can get full marks.
Did you know...
You can play all the teacher-written quizzes on our site for just £9.95 per month. Click the button to sign up or read more.
1. If you face NE and turn through 180° anticlockwise, what direction will you be facing now?
'Anticlockwise' means that you turn (rotate) in the opposite direction to which the hands of the clock turn: the hands of the clock turn 'clockwise'. 180° = two lots of 90° or a half turn
2. If you face NW and turn through 180° anticlockwise, what direction will you be facing now?
'Anticlockwise' means that you turn (rotate) in the opposite direction to which the hands of the clock turn: the hands of the clock turn 'clockwise'. 180° = two lots of 90° or a half turn
3. If you face SE and turn through 135° clockwise, what direction will you be facing now?
135° = three lots of 45° or three-eighths of a rotation (turn): 45° is an eighth of a turn
4. If you face West and turn through 225° clockwise, what direction will you be facing now?
225° = five lots of 45° or five-eighths of a rotation (turn): 45° is an eighth of a turn
5. If you face SW and turn through 180°, what direction will you be facing now?
It doesn't make any difference whether you turn clockwise or anticlockwise because you are making a half turn; this means that you will always end up facing the same way no matter which way you turn. If you don't believe it - turn around! D'oh!
6. If you face NE, what direction is right behind you?
Look at the compass rose if you can't see this
7. If you face SE, what direction is right behind you?
Look at the compass rose if you can't see this
8. Peter and John are at the same position on the ground. Peter is facing NE. John is facing SE. If Peter turns clockwise through 135°, what angle must John turn through to face in the same direction as Peter?
Peter turns through 135° (= three lots of 35°) and ends up facing S. John will have to turn clockwise through 45° (an eighth of a turn)
9. Peter and John are at the same position on the ground again. Peter is facing NE. John is facing SE. If Peter turns clockwise through 135° what angle must John turn through to face in the same direction as Peter? This time, John is NOT allowed to turn clockwise through 45°.
John must turn in the opposite direction until he faces S again as in question 8. This shows that for every anticlockwise rotation there is also a clockwise rotation that brings you facing in the same direction. Note: the anticlockwise angle turned through + the clockwise angle turned through = 360°. This is a good check if you want to make sure that you haven't made a mistake
10. How many points are there in the compass rose?
You know eight of these points. Each point corresponds to 360 ÷ 32 = 11.25° = one thirty-second of a rotation
Quiz yourself clever - 3 free quizzes in every section
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# Empirical Formula
## Empirical Formula
The simplest formula or the empirical formula provides the lowest whole number ratio of atoms existent in a compound. The relative number of atoms of every element in the compound is provided by this formula.
Steps for Determining an Empirical Formula
• Let’s begin with what’s given in the problem, i.e., the number of grams of each element.
• We’ll assume that the total mass is 100 grams if percentages are given, so that
Each element’s mass = the given percentage
• By making use of the molar mass from the periodic table, change the mass of every element to moles.
• Divide every mole value by the lowest number of moles computed.
• Round up to the closest whole number. Â This is denoted by subscripts in the empirical formula and is the mole ratio of the elements.
Multiply each answer by the same factor to get the lowest whole number multiple, if the number is too far to round off (x.1 ~ x.9).
e.g. Â Multiply each solution in the problem by 4 to get 5, if one solution is 1.25.
e.g. Â Multiply each solution in the problem by 2 to get 3, if one solution is 1.5.
The molecular formula can be calculated for a compound if the molar mass of the compound is given when the empirical formula is found. Â To find the ratio between the molecular formula and the empirical formula. Basically, the mass of the empirical formula can be computed by dividing the molar mass of the compound by it. Â Multiply every atom (subscripts) by this ratio to compute the molecular formula.
## Solved Examples
Problem 1: A compound contains 88.79% oxygen (O) and 11.19% hydrogen (H). Compute the empirical formula of the compound.
Solution:
1. Assume 100.0g of substance. We see that the percentage of each element matches the grams of each element
11.19g H
88.79g O
1. Convert grams of each element to moles
H: (11.19/1.008) = 11.10 mol H atoms [molar mass of H=1.008g/mol]
O: (88.79/16.00) = 5.549 mol O atoms [molar mass of O= 16.00g/mol]
The formula could be articulated as H11.10O5.549. However, it’s usual to use the smallest whole number ratio of     atoms
1. By dividing the lowest number alter the numbers to whole numbers.
H =11.10/ 5.549Â = 2.000
O = 5.549/ 5.549= 1.000
The simplest ratio of H to O is 2:1
Empirical formula = H2O
Problem 2: A sulfide of iron was formed by combining 1.926g of sulfur(S) with 2.233g of iron (Fe). What is the compound’s empirical formula?
Solution:
1. As the mass of each element is known, we use them directly
2. Convert grams of each element to moles
Fe: (2.233g /55.85g) = 0.03998 mol Fe atoms [molar mass of Fe =1.008g/mol]
S: (1.926 /32.07) = 0.06006 mol S atoms [molar mass of S =32.07g/mol]
1. By dividing by the smallest number, change the numbers to whole numbers.
Fe =0.03998/0.03998 = 1.000
S = 0.06006/0.03998mol = 1.502
1. As we still have not reached a ratio that gives whole numbers in the formula we multiply by a number that will give us whole numbers
Fe: (1.000)2 = 2.000
S: (1.502)2 = 3.004
Empirical formula = Fe2S3
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### 2-digit Square
A 2-Digit number is squared. When this 2-digit number is reversed and squared, the difference between the squares is also a square. What is the 2-digit number?
### What's Possible?
Many numbers can be expressed as the difference of two perfect squares. What do you notice about the numbers you CANNOT make?
### Why 24?
Take any prime number greater than 3 , square it and subtract one. Working on the building blocks will help you to explain what is special about your results.
# Plus Minus
##### Stage: 4 Challenge Level:
Tom, from Wilson's School, found some other pairs which gave multiples of 1000 and repeated digits:
Other pairs that give multiples of $1000$ are:
$60^2 - 40^2 = 2000$
$70^2 - 30^2 = 4000$
$75^2 - 25^2 = 5000$
$80^2 - 20^2 = 6000$
$85^2 - 15^2 = 7000$
$90^2 - 10^2 = 8000$
All these pairs add up to 100 which is a factor of 1000.
Other ways to make multiples of $1000$ are $95^2 - 45^2 = 7000$ and $95^2 - 55^2 = 6000$. These numbers when added together become a factor of the answer.
Other pairs that give repeated digit answers are:
$67^2 - 34^2 = 3333$
$56^2 - 45^2 = 1111$
The numbers in the tens columns add together to make 9 while the numbers in the units columns add together to make 11.
Tabitha from The Norwood School, Hannah from Munich International School, Richard from Comberton Village College, and Paul, Dulan and Priyan from Wilson's School all explained how the diagram helped them to work out the difference between two square numbers. Here is Richard's explanation:
By drawing a $85 \times 85$ square and a superimposed $65 \times 65$ square, and subtracting the areas, a strip which covers two lengths of the larger square is formed. When segments of the strip are rotated, a rectangle is formed that has a constant width of the difference of the squares' sides and a length of the sum of the squares' sides.
So the dimensions of the purple rectangle are: width $85 - 65 = 20$, length $85 + 65 = 150$
This implies that the area of the purple strip is equal to that of the rectangle, $20 \times 150 = 3000$.
In general,by drawing a square of length $x$ with a superimposed square of length $y$, the area of the strip will be $(x+y)(x-y)$, so $x^2 - y^2 = (x+y)(x-y)$
James and Nat from Sawston both used this formula for the difference of two squares to help them to find pairs of numbers. Hannah used the formula to work out some of the other things posed in the problem:
$7778^2 - 2223^2 = (7778 - 2223)(7778+2223) = 5555 \times 10001 = 55555555$
$88889^2 - 11112^2 = (88889 - 11112)(88889 + 11112) = 77777 \times 100001 = 7777777777$
Lyman from Nanjing International School showed how the numbers could be made in lots of different ways by finding factors, including some examples using decimals. Here are some ways of making 1000:
$251^2 - 249^2$
$127^2 - 123^2$
$55^2 - 45^2$
$35^2 - 15^2$
$500.5^2 - 499.5^2$
$102.5^2 - 97.5^2$
$66.5^2 - 58.5^2$
$32.5^2 - 7.5^2$
Finally, Richard explained more generally how to make multiples of 1000 and numbers with repeated digits:
Any number of the form $n \times 10^3$ where $n$ is an integer, can be expressed as $x^2-y^2$ when $x + y = 100$ and $x - y = 10n$.
This is not the only case. More generally, it can be formed when $x + y = u \times 10^a$ and $x - y = v \times 10^b$, where $u \times v < 10$ and $a + b = 3$ or where $u \times v$ is equal to some multiple of $10$ and $a + b = 2$.
This is because $x^2 - y^2 = (x+y)(x-y) = uv \times 10^{(a+b)}$.
A repeated number can also be generated quite easily. A number of the form $mm$ can be expressed as $x^2 - y^2$ when $x + y = 11$ and $x - y$ is equal to $m$.
A number of the form $mnmn$ is expressible when $x + y = 101$ and $x - y = mn$.
$mnomno$ is expressible when $x + y$ is equal to $1001$ and $x - y = mno$
Well done to all who submitted solutions.
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There are 366 different Starters of The Day, many to choose from. You will find in the left column below some starters on the topic of Mensuration. In the right column below are links to related online activities, videos and teacher resources.
A lesson starter does not have to be on the same topic as the main part of the lesson or the topic of the previous lesson. It is often very useful to revise or explore other concepts by using a starter based on a totally different area of Mathematics.
Main Page
### Mensuration Starters:
Area Two: How many different shapes with an area of 2 square units can you make by joining dots on this grid with straight lines?
Bizarre Triangle: By how much would the area of this triangle increase if its base was enlarged to 8cm?
Christmas Tables: Which of the two shapes has the largest area? You will be surprised!
Cross Perimeter: Calculate the distance around the given shape
Goat Grazing: Find the loci of the goat's position as it eats the grass while tethered to the rope.
Missing Lengths: Introduce linear equations by solving these problems about lengths.
Missing Square Puzzle: The missing square puzzle is an optical illusion used to help students reason about geometrical figures.
Oblongs: Find the dimensions of a rectangle given the perimeter and area.
Quad Areas: Calculate the areas of all the possible quadrilaterals that can be constructed by joining together dots on this grid.
Shopping List: A quick quiz about five items on a shopping list written 40 years ago.
Stair Perimeter: Use the information implied in the diagram to calculate the perimeter of this shape.
Step Perimeter: Is it possible to work out the perimeter of this shape if not all the side lengths are given?
Average Cycling Speed: Work out the average speed of two journeys. The obvious answer is not the correct answer.
Charging Rhinos: Find the easy way to solve this kinematics problem involving a fly and two rhinos.
Cuboid: Find the dimensions of a cuboid matching the description given
Fence Optimisation: Find the length of a rectangle enclosing the largest possible area.
Hands Together: The hands of a clock are together at midnight. At what time are they next together?
Paper Ratio: Calculate the ratio of the sides of an A4 sheet of paper without any measuring.
Paper Surprising Perimeter: Find the perimeter of a folded sheet of A4 paper as described in this short video.
Piece of String: Find where a piece of string should be cut to form a circle and a square of equal areas.
Pizza Slice: A problem which can be solved by considering the areas of a triangle and a sector of a circle.
Road Connections: Design roads to connect four houses that are on the corners of a square, side of length one mile, to minimise the total length of the roads.
Rope Around The World: Imagine a long rope wrapped around Earth's equator - One metre longer than it needs to be.
Rosie's Cube: How many of the small cubes does Rosie need to complete the big cube?
Speed Circles: Find the diameters of the circles in the corners of the square.
Sphere Hole: Find the volume of the remaining part of a sphere after a 10cm cylindrical hole has been drilled through it.
Square in Rectangle: Find the area of a square drawn under the diagonal of a rectangle
#### Area and Perimeter of a Rectangle
Questions on the areas and perimeters of rectangles which will test your problem solving abilities.
Transum.org/go/?to=oblongs
### Curriculum for Mensuration:
#### Year 5
Pupils should be taught to convert between different units of metric measure (for example, kilometre and metre; centimetre and metre; centimetre and millimetre; gram and kilogram; litre and millilitre) more...
Pupils should be taught to understand and use approximate equivalences between metric units and common imperial units such as inches, pounds and pints more...
Pupils should be taught to measure and calculate the perimeter of composite rectilinear shapes in centimetres and metres more...
Pupils should be taught to calculate and compare the area of rectangles (including squares), and including using standard units, square centimetres (cm2) and square metres (m2) and estimate the area of irregular shapes more...
Pupils should be taught to estimate volume [for example, using 1 cm3 blocks to build cuboids (including cubes)] and capacity [for example, using water] more...
#### Year 6
Pupils should be taught to solve problems involving the calculation and conversion of units of measure, using decimal notation up to three decimal places where appropriate more...
Pupils should be taught to solve problems involving the calculation of percentages [for example, of measures, and such as 15% of 360] and the use of percentages for comparison more...
Pupils should be taught to use, read, write and convert between standard units, converting measurements of length, mass, volume and time from a smaller unit of measure to a larger unit, and vice versa, using decimal notation to up to three decimal places more...
Pupils should be taught to convert between miles and kilometres more...
Pupils should be taught to recognise that shapes with the same areas can have different perimeters and vice versa more...
Pupils should be taught to recognise when it is possible to use formulae for area and volume of shapes more...
Pupils should be taught to calculate the area of parallelograms and triangles more...
Pupils should be taught to calculate, estimate and compare volume of cubes and cuboids using standard units, including cubic centimetres and cubic metres, and extending to other units more...
#### Years 7 to 9
Pupils should be taught to understand and use place value for decimals, measures and integers of any size more...
Pupils should be taught to change freely between related standard units [for example time, length, area, volume/capacity, mass] more...
Pupils should be taught to derive and apply formulae to calculate and solve problems involving: perimeter and area of triangles, parallelograms, trapezia, volume of cuboids (including cubes) and other prisms (including cylinders) more...
Pupils should be taught to use scale factors, scale diagrams and maps more...
Pupils should be taught to calculate and solve problems involving: perimeters of 2-D shapes (including circles), areas of circles and composite shapes more...
Pupils should be taught to draw and measure line segments and angles in geometric figures, including interpreting scale drawings more...
Pupils should be taught to use compound units such as speed, unit pricing and density to solve problems. more...
Pupils should be taught to use standard units of mass, length, time, money and other measures, including with decimal quantities more...
Pupils should be taught to use the properties of faces, surfaces, edges and vertices of cubes, cuboids, prisms, cylinders, pyramids, cones and spheres to solve problems in 3-D more...
#### Years 10 and 11
Pupils should be taught to convert between related compound units (speed, rates of pay, prices, density, pressure) in numerical and algebraic contexts more...
Pupils should be taught to calculate arc lengths, angles and areas of sectors of circles more...
Pupils should be taught to calculate surface areas and volumes of spheres, pyramids, cones and composite solids more...
### Exam-Style Questions:
There are almost a thousand exam-style questions unique to the Transum website.
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### Notes:
Mensuration is the branch of Mathematics dealing with measurement of angles, length, area, and volume. It is linked closely to the topic of Estimation and related to the topics of Angles, Shape and Shave (3D).
It is essential for pupils to have an understanding of the units used to measure which include both the more common metric units and the Imperial units still in common usage. We have found a good teaching strategy is to ask each of the pupils to "Bring to the next Maths lesson some visual aid which will help the rest of the class remember the size of a unit of measurement". See Memorable Measures below for the printable resources. This activity provides an association with a unit, a visual aid and a known person which is a great memory enhancer.
### Mensuration Teacher Resources:
Pin Board: Rows and columns of dots that can be joined using straight lines to create shapes.
Memorable Measures: This is a visual aid and printable cards to introduce a homework activity about measures.
### Mensuration Activities:
Area and Perimeter of a Rectangle: Questions on the areas and perimeters of rectangles which will test your problem solving abilities.
Area Builder: An interactive workspace in which to make shapes using square tiles with given areas and perimeters.
Measuring Angles: Measure the size of the given angles to within two degrees of their actual value.
Area of a Triangle: Calculate the areas of the given triangles in this self marking quiz.
Area and Perimeter of a Parallelogram: Many different ways to practise your skills finding the areas and perimeters of parallelograms.
Reading Scales: A self marking exercise on the reading of scales of different types.
Area and Perimeter of a Kite: A short exercise to practise using the formulae for area and perimeter of a kite.
Area of a Trapezium: Check that you can find the area of a trapezium and use the trapezium area formula for problem solving.
Area and Perimeter: Show that you know the area and perimeter formulas of basic shapes.
Area Maze: Use your knowledge of rectangle areas to calculate the missing measurement of these composite diagrams.
Area Two: How many different shapes with an area of 2 square units can you make by joining dots on this grid with straight lines?
Mileometer: Practise converting between miles and kilometres with this self marking quiz.
Quad Areas: Calculate the areas of all the possible quadrilaterals that can be constructed by joining together dots on this grid.
Scale Drawings: Measure line segments and angles in geometric figures, including interpreting scale drawings.
Algebraic Perimeters: Questions about the perimeters and areas of polygons given as algebraic expressions.
Volume: Use formulae to solve problems involving the volumes of cuboids, cones, pyramids, prisms and composite solids.
Measuring Units: Check your knowledge of the units used for measuring with this self-marking quiz about metric and imperial units.
Sorting Units: Order the ten containers according to their value (money, length and weight)
Metric Units Pairs: Find the matching pairs of equivalent metric units in this interactive online game.
Circles: Practise using pi to calculate various circle measurements. There are six levels of difficulty.
Screen Test: Memorise the mathematical facts in the video then answer the ten quiz questions.
Volume Equals Surface Area: Find the cuboids with integer side lengths where the volume is numerically equal to the surface area.
Areas of Composite Shapes: Find the areas of combined (composite) shapes made up of one or more simple polygons and circles.
Map Scales: Test your understanding of map scales expressed as ratios with this self marking quiz.
Imperial Units: Learn about common imperial units and how they relate to other units of measurement.
Inequalities: Check that you know what inequality signs mean and how they are used to compare two quantities. Includes negative numbers, decimals, fractions and metric measures.
Imperial Units Pairs: Find the matching pairs of equivalent imperial units in this interactive online game.
Formulae to Remember: The traditional pairs or pelmanism game adapted to test recognition for formulae required to be memorised for GCSE exams.
Formulae Pairs: Find the matching pairs of diagrams and formulae for basic geometrical shapes.
Converting Standard Units: Converting measurements of length, mass, volume and time from one unit of measure to another.
Cylinders: Apply formulae for the volumes and surface areas of cylinders to answer a wide variety of questions
Bottles, Boxes and Cans: Estimate the capacity of the bottles, boxes and cans in the photograph and answer questions about volume.
Surface Area: Work out the surface areas of common solid shapes in this collection of exercises.
Compound Units: Practise using compound units such as speed, unit pricing and density to solve problems.
Finally there is Topic Test, a set of 10 randomly chosen, multiple choice questions suggested by people from around the world.
Alternatively, for the more advanced student, there is an ever-growing collection of Exam-Style Questions with worked solutions on the topic of Mensuration.
### Mensuration Investigations:
Area Builder: An interactive workspace in which to make shapes using square tiles with given areas and perimeters.
Area shapes: Investigate polygons with an area of 4 square units. This is your starting point, you can decide how to proceed.
Pin Board: Rows and columns of dots that can be joined using straight lines to create shapes.
Rectangle Perimeters: The perimeter of a rectangle is 28cm. What could its area be?
Maxvoltray: Find the maximum volume of a tray made from an A4 sheet of paper. A practical mathematical investigation.
### Mensuration Videos:
Transum's Mensuration Video
Metric System: Ted Talk: Why the metric system matters - Matt Anticole
Area And Perimeter Video: Calculate and solve problems involving perimeter and area of rectangles, triangles, parallelograms, trapezia, kites and composite shapes.
Parallelogram: Instructional video showing how the area of a parallelogram can be determined.
Volume Video: There are simple formulas that can be used to find the volumes of basic three-dimensional shapes.
Circle Facts Song: A free trial lesson from Math Upgrade dot com.
Circles Area and Circumference Video: The circumference and area of a circle can be found if the radius or diameter are known.
Pi Song: Kate Bush sings the digits of pi (audio only).
Surface Area Video: Finding the surface are of three dimensional shapes can involve some interesting formulae.
Pi and Four Fingers: Why is The Simpsons not in Base 8? In this video Simon Singh talks about Pi and Maths in The Simpsons cartoon.
Formulae for GCSE: These are the formulae candidates need to know for the GCSE(9-1) Maths exams.
### Mensuration Worksheets/Printables:
Measuring Lines and Angles: Practice using a ruler and protractor on this worksheet with answers provided.
Memorable Measures Notes: These are the printable cards to go with the activity called Memorable Measures.
Links to other websites containing resources for Mensuration are provided for those logged into 'Transum Mathematics'. Subscribing also opens up the opportunity for you to add your own links to this panel. You can sign up using one of the buttons below:
### Search
The activity you are looking for may have been classified in a different way from the way you were expecting. You can search the whole of Transum Maths by using the box below.
### Other
Is there anything you would have a regular use for that we don't feature here? Please let us know.
#### Compound Units
Practise using compound units such as speed, unit pricing and density to solve problems.
Transum.org/go/?to=compound_units
### Teaching Notes:
Many Transum activities have notes for teachers suggesting teaching methods and highlighting common misconceptions. There are also solutions to puzzles, exercises and activities available on the web pages when you are signed in to your Transum subscription account. If you do not yet have an account and you are a teacher, tutor or parent you can apply for one by completing the form on the Sign Up page.
Have today's Starter of the Day as your default homepage. Copy the URL below then select
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Set as your homepage (if you are using Internet Explorer)
CNN,
Monday, December 10, 2018
"Metric mishap caused loss of NASA orbiter. (CNN) -- NASA lost a \$125 million Mars orbiter because a Lockheed Martin engineering team used English units of measurement while the agency's team used the more conventional metric system for a key spacecraft operation, according to a review finding released Thursday.Sep 30, 1999"
For Students:
For All:
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# Michigan - Grade 1 - Math - Geometry - Composite Shapes - 1.G.2
### Description
Compose two-dimensional shapes (rectangles, squares, trapezoids, triangles, half-circles, and quarter-circles) or three-dimensional shapes (cubes, right rectangular prisms, right circular cones, and right circular cylinders) to create a composite shape, and compose new shapes from the composite shape.1
• State - Michigan
• Standard ID - 1.G.2
• Subjects - Math Common Core
• Math
• Geometry
## More Michigan Topics
Apply properties of operations as strategies to add and subtract.2 Examples: If 8 + 3 = 11 is known, then 3 + 8 = 11 is also known. (Commutative property of addition.) To add 2 + 6 + 4, the second two numbers can be added to make a ten, so 2 + 6 + 4 = 2 + 10 = 12. (Associative property of addition.)
1.OA.4 Understand subtraction as an unknown-addend problem. For example, subtract 10 – 8 by finding the number that makes 10 when added to 8. Add and subtract within 20.
1.OA.5 Relate counting to addition and subtraction (e.g., by counting on 2 to add 2).
1.OA.6 Add and subtract within 20, demonstrating fluency for addition and subtraction within 10. Use strategies such as counting on; making ten (e.g., 8 + 6 = 8 + 2 + 4 = 10 + 4 = 14); decomposing a number leading to a ten (e.g., 13 – 4 = 13 – 3 – 1 = 10 – 1 = 9); using the relationship between addition and subtraction (e.g., knowing that 8 + 4 = 12, one knows 12 – 8 = 4); and creating equivalent but easier or known sums (e.g., adding 6 + 7 by creating the known equivalent 6 + 6 + 1 = 12 + 1 = 13).
1.OA.7 Understand the meaning of the equal sign, and determine if equations involving addition and subtraction are true or false. For example, which of the following equations are true and which are false? 6 = 6, 7 = 8 – 1, 5 + 2 = 2 + 5, 4 + 1 = 5 + 2.
1.OA.8 Determine the unknown whole number in an addition or subtraction equation relating three whole numbers. For example, determine the unknown number that makes the equation true in each of the equations 8 + ? = 11, 5 = _ – 3, 6 + 6 = _.
Use addition and subtraction within 20 to solve word problems involving situations of adding to, taking from, putting together, taking apart, and comparing, with unknowns in all positions, e.g., by using objects, drawings, and equations with a symbol for the unknown number to represent the problem.
Given a two-digit number, mentally find 10 more or 10 less than the number, without having to count; explain the reasoning used.
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# Divisible by 7 | Divisibility Rule for 7 | How to Check if a Number is Divisible by 7 or Not?
Mathematics is not an easy subject until you understand the concept and compare it with the examples. Students who want to know about the divisibility rules of 7 can use this page. We help you to understand the concept of divisibility without the division method. The divisibility rule to divisibility test helps the students of 5th Grade math to check if a number is completely divisible by another number without actual division. In this article, we focus on divisible by 7 with clear-cut explanations.
Do Refer:
## Divisibility Rule of 7
The concept of divisibility is nothing but checking if a number is divisible by another number without actually dividing the number. The divisibility rule of 7 checks if a number can be completely divided by 7 without any remainder.
Usually, we perform the division arithmetic operation to know this. But divisibility rule of 7 has a shortcut method to find if a number is divisible by 7.
### Divisibility Test by 7 Examples
Have a look at the below examples to know in detail about divisibility by 7 without applying the division method. Let us check if the number is divisible by 7 or not.
Example 1.
Find whether the number 182 is divisible by 7.
Solution:
For divisible by 7 we need to double the last digit of the number and then subtract it from the remaining number. If the result is divisible by 7, then the original number will also be divisible by 7.
Here in 182
2 + 2 = 4
18 – 4 = 14
Where 14 is divisible by 7
Therefore the number 182 is also divisible by 7.
Example 2.
Find whether the number 252 is divisible by 7.
Solution:
For divisible by 7 we need to double the last digit of the number and then subtract it from the remaining number. If the result is divisible by 7, then the original number will also be divisible by 7.
Here in 252
2 + 2 = 4
25 – 4 = 21
Where 21 is divisible by 7
Therefore the number 252 is also divisible by 7.
Example 3.
Find whether the number 595 is divisible by 7
Solution:
For divisible by 7 we need to double the last digit of the number and then subtract it from the remaining number. If the result is divisible by 7, then the original number will also be divisible by 7.
Here in 595
5 + 5 = 10
59 – 10 = 49
Where 49 is divisible by 7
Therefore the number 595 is also divisible by 7.
Example 4.
Find whether the number 343 is divisible by 7.
Solution:
For divisible by 7 we need to double the last digit of the number and then subtract it from the remaining number. If the result is divisible by 7, then the original number will also be divisible by 7.
Here in 343
3 + 3 = 6
34 – 6 = 28
Where 28 is divisible by 7
Therefore the number 343 is divisible by 7.
Example 5.
Find whether the number 672 is divisible by 7.
Solution:
For divisible by 7 we need to double the last digit of the number and then subtract it from the remaining number. If the result is divisible by 7, then the original number will also be divisible by 7.
Here in 672
2 + 2 = 4
67 – 4 = 63
Where 63 is divisible by 7
Therefore the number 672 is divisible by 7.
See More Divisibility Rules
Divisible by 2 Divisible by 3 Divisible by 4 Divisible by 5 Divisible by 6 Divisible by 8 Divisible by 9 Divisible by 10
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Students can Download Chapter 2 Whole Numbers Ex 2.3 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 6 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.
## Karnataka State Syllabus Class 6 Maths Chapter 2 Whole Numbers Ex 2.3
Question 1.
Which of the following will not represent zero:
a) 1 + 0
b) 0 × 0
c) $$\frac{0}{2}$$
d) $$\frac{10-10}{2}$$
a) 1 + 0 = 1 It does not represent zero
b) 0 × 0 = 0 It represent zero
c) $$\frac{0}{2}=0$$ It represents zero
d) $$\frac{10-10}{2}=\frac{0}{2}=0$$ It represent zero
Solution:
a) 1 + 0 = 1
Question 2.
If the product of two whole numbers is zero, can we say that one or both of them will be zero ? Justify through examples.
Solution:
If the product of 2 whole numbers is zero, then one of them is definitely zero,
For example, 0 × 2 =0 and 17 × 0 = 0
It the product of 2 whole numbers is zero them both of them may be zero 0 × 0 = 0
However, 2 × 3 = 6
(Since number to be multiplied are not equal to zero, the result of the product will also be non-zero.
Question 3.
If the product of two whole numbers is 1, can we say that one or both of them will be 1? Justify through examples?
Solution:
If the product of 2 numbers is, then both the numbers have to be equal to 1
For example ,1 × 1 = 1 However, 1 × 6 = 6
Clearly, the product of two whole numbers will be 1 in the situation when both numbers to be multiplied are 1.
Question 4.
Find using distributive property:
a) 728 × 101
b) 5437 × 1001
c) 824 × 25
d) 4275 × 125
e) 504 × 35
Solution:
a) 728 × 101 = 728 × (100+1)
= 728 × 100 + 728 + 1
= 72800 + 728 = 73528
b) 5437 × 1001 = 5437 × (1000 + 1)
= 5437 × 1000 + 5437 × 1
= 5437000 + 5437 = 5442437
c) 824 × 25 = (800 + 024) × 25
= (800 + 25 – 1) × 25
= 800 × 25 + 25 × 25 – 1 × 25 = 20000 + 625 – 25
= 20000 + 600 = 20600
d) 4275 × 125
= (4000 + 200 + 100 – 25 ) × 125 = 4000 × 125 + 200 × 125 + 100 + 125 – 25 × 125
= 500000 + 25000 + 12500 – 3125
= 534375
e) 504 × 35 = ( 500 + 4) × 35
= 500 × 35 + 4 × 35
= 17500 + 140 = 17640
Question 5.
Study the pattern :
1 × 8 + 1 = 9
12 × 8 + 2 = 98
123 × 8 + 3 = 987
1234 × 8 + 4 = 9876
12345 × 8 + 5 = 98765
Write the next two steps, can you say how the pattern works?
(Hint: 12345 = 11111 + 1111 +111 +11 +1).
Solution
123456 × 8 + 6 = 987648 + 6 = 987654
1234567 × 8 + 7 = 9876536 + 7 = 9876543
Yes, the pattern works.
As 123456= 111111 + 11111 + 1111 + 111 + 11 + 1.
123456 × 8 = ( 111111 + 11111 + 1111 + 111 + 11 + 1) × 8
= 111111 × 8 + 11111 × 8 + 1111 × 8 + 111 × 8 + 11 × 8 + 1 × 8
= 888888 + 88888 + 8888 +888 + 88 + 8 = 987648
= 123456 × 8 + 6 = 987648 + 6 = 987654
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AllenHeard
March 15, 2017
110
# Big O Notation
Year 13 Lesson
March 15, 2017
## Transcript
1. ### BIG O NOTATION How well do algorithms scale as the
data size increases?
2. ### Big O Notation ▪ Big O notation is used in
Computer Science to describe the performance or complexity of an algorithm. ▪ It specifically gives an indication of how well an algorithm scales, it is not a measure of efficiency. ▪ Big O specifically describes the worst-case scenario, and can be used to describe the execution time required or the space used (e.g. in memory or on disk) by an algorithm
3. ### Big O Notation ▪ Special notation is given to describe
the performance as shown in the graph.
4. ### Example Before we look at some specific examples let’s take
a look at the following equation to get an idea of how algorithms can grow as the data set increases: 45n3 + 20n2 + 19 = ? What is the answer when n = 1?
5. ### Example What is the answer when n = 2? 45n3
+ 20n2 + 19 = ?
6. ### Example What is the answer when n = 10? 45n3
+ 20n2 + 19 = ?
7. ### Example ▪ Where n = 1 45n3 + 20n2 +
19 = 84 ▪ Where n = 2 45n3 + 20n2 + 19 = 459 When n increases from 1 to just 2, 19 becomes much less significant.
8. ### Example ▪ Where n = 10 45n3 + 20n2 +
19 = 47,019 When n increases from 2 to 10, both the 19 and the 20n2 become less significant. ▪ The real contributory factor is 45n3 = 45,000 ▪ Algorithm is said to be order n3 or O (n3)
9. ### Big O Notation ▪ Here, in order of best performance
to worst are some typical notations: – O(1) – O(log n) – O(n) – O(n log n) – O(n2) ▪ Let’s look at some examples in python to observe the difference in time taken to complete algorithms.
10. ### O(1)
▪ Executes in the same amount of time
no matter how big the data set. ▪ Irrelevant how many items are in a list, adding one more takes the same time regardless.
11. ### A note on Logarithms
▪ In computer science log
n means, the exponent you would need to raise the number 2 to to get n. ▪ So imagine, if n = 16. Our exponent would be much much smaller than the actual n value. ▪ It would be 4.
12. ### O(log n)
▪ A divide and conquer algorithm, after
each step you effectively half your problem. If you double the problem size n, your algorithm needs only a constant number of steps more. ▪ O(log n) code
13. ### O(n)
▪ O(n) describes an algorithm whose performance will
grow linearly and in direct proportion to the size of the input data set. ▪ Big O notation will always assume the upper limit where the algorithm will perform the maximum number of iterations. ▪ O(n) code
14. ### O(n log n)
▪ Time to complete grows
in direct proportion to the amount of data, best example of an algorithm of O(n log n) is merge sort. ▪ O(n log n) code
15. ### O(n 2 )
▪ Represents an algorithm whose
performance is directly proportional to the square of the size of the input data set. ▪ This is common with algorithms that involve nested iterations over the data set. Deeper nested iterations will result in O(n3), O(n4) etc. ▪ O(n2) code
16. ### Calculating Big O – Rule of thumb
▪ Simple
programs can be analysed by counting the nested loops of the program. ▪ A single loop over n items yields f(n) = n ▪ A loop within a loop yields f(n) = n2 ▪ A loop within a loop within a loop yields f(n) = n3
17. ### Calculating Big O – Nested Loops
▪ If
you have nested loops, and the outer loop iterates i times and the inner loop iterates j times, the statements inside the inner loop will execute a total of i x j times. ▪ This is because the inner loop will iterate j times for each of the i iterations of the outer loop. ▪ This means that if both the outer and inner loop are dependent on the p roblem size n, the statements in the inner loop will be executed O(n2) times.
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# How do I solve the differential equation y"-4y'-5y=0 with y(-1)=3 and y'(-1)=9?
Mar 12, 2015
As written, there is no solution to the problem.
Definitions: nth-order differential equation: a differential equation in which the highest-order derivative presented is of order n. Thus, a first-order differential equation would involve as its highest-order derivative $y '$.
Ordinary differential equation: a differential equation consisting of a function of one independent variable and its derivatives. An ordinary differential equation might involve $y \left(x\right)$, for example, but not $y \left({x}_{1} , {x}_{2}\right)$ in which ${x}_{1}$ and ${x}_{2}$ are different independent variables.
For the work shown below, we assume that $y$ is a function of $x$, and is thus denoted $y \left(x\right)$. However, this is simply a "placeholder" variable so that we can properly define our derivative. If it were instead $y \left(t\right)$, then $y ' = \frac{\mathrm{dy}}{\mathrm{dt}}$, for example.
Given the first-order ordinary differential equation above, the first thing we can do is group like terms simplify.
$y - 4 y ' - 5 y = y - 5 y - 4 y ' = - 4 \left(y ' + y\right) = 0$
Thus, dividing by -4 and moving terms to opposite sides:
$y ' = - y$.
We will first divide both sides by $y$, and then integrate both sides with respect to $x$
$y ' = - y \implies \frac{y '}{y} = - 1 \implies \int \frac{y '}{y} \mathrm{dx} = \int - 1 \mathrm{dx} \implies \int \frac{y '}{y} \mathrm{dx} = - x + {c}_{1}$, where ${c}_{1}$ is an arbitrary constant.
Recall that $\int \frac{\mathrm{du}}{u} = \ln \left(u\right)$, and that ${e}^{\ln} \left(u\right) = u \ldots$
$\ln \left(y\right) = - x + {c}_{1} \implies {e}^{\ln} \left(y\right) = {e}^{- x + {c}_{1}} = \left({e}^{-} x\right) \left({e}^{{c}_{1}}\right) \implies y = {e}^{{c}_{1}} {e}^{- x}$. Since our original ${c}_{1}$ was an arbitrary constant, we can reuse the label, declaring ${e}^{{c}_{1}}$ as the "new" ${c}_{1}$:
$y \left(x\right) = {c}_{1} {e}^{-} x$
Thus, the initial equation implies a function of the form $y = {c}_{1} {e}^{-} x$.
We would normally find the constant ${c}_{1}$ by using the boundary conditions provided. However, we already have a problem. Since $y ' = - {c}_{1} {e}^{-} x$, $y ' \left(- 1\right)$ should be equal to $- y \left(- 1\right)$. However, this is not the case, as $y ' \left(- 1\right) = 9$ while $y \left(- 1\right) = 3$.
Thus, there is no solution to the problem as presented.
|
Proof that all integers are equal June 1, 2017
Posted by Ezra Resnick in Math.
In particular, we prove that n = -1 for any integer n.
First, assume that n is odd. That means that n – 1 is even, so n – 1 = 2k for some integer k.
The following identity is trivially true for any n:
n2 – 1 = (n – 1)(n + 1)
We can substitute 2k for (n – 1):
n2 – 1 = 2k(n + 1) = 2kn + 2k
Next, we subtract (2knn – 1) from both sides of the equation, yielding:
n2 – 2kn – n = 2k – n + 1
Factoring out (n – 2k – 1) produces:
n(n – 2k – 1) = –(n – 2k – 1)
Now we simply divide both sides of the equation by (n – 2k – 1), and voila:
n = -1
Alternatively, assume that n is even. That means that n – 1 is odd, so n – 1 = 2k + 1 for some integer k.
Once again, we begin with the identity:
n2 – 1 = (n – 1)(n + 1)
This time we substitute (2k + 1) for (n – 1):
n2 – 1 = (2k + 1)(n + 1) = 2kn + 2k + n + 1
We subtract (2kn + 2n – 1) from both sides of the equation:
n2 – 2kn – 2n = 2k – n + 2
We factor out (n – 2k – 2):
n(n – 2k – 2) = –(n – 2k – 2)
And now we simply divide both sides of the equation by (n – 2k – 2) to complete the proof:
n = -1
Q.E.D.
|
### Home > AC > Chapter 16 > Lesson 16.9.1.1 > Problem9-7
9-7.
A child tosses a tennis ball into the air and it lands in the mud. The function $h\left(t\right) = -16\left(t - 0.5\right)^{2} + 8$ gives the ball’s height in feet with respect to time in seconds. (When you study physics or calculus you will learn how to find such functions yourself.)
1. Sketch a graph of the function $h\left(t\right)$. Darken the portion of the curve that fits our scenario.
2. What is happening to the height of the tennis ball with respect to the time?
First it increases, then it decreases.
3. Since speed is a function of distance traveled over time, what part of the graph represents the speed of the ball? What is happening to the speed of the tennis ball with respect to time?
The slope represents the speed of the ball. First it decreases (slope decreases). It stops momentarily (slope is at zero at the vertex) then it increases (slope gets steeper).
4. Fill in the following table of time (s) versus height (ft). Be careful. Remember that h cannot be negative in this situation.
t seconds h in feet $0$ $0.25$ $0.5$ $0.75$ $1$ $1.25$ $1.5$
t seconds h in feet $0$ $0.25$ $0.5$ $0.75$ $1$ $1.25$ $1.5$ $4$ $7$
5. Find the average velocity, $\frac { \Delta d } { \Delta t }$, in ft/s for each $0.25$ second time interval.
Here is the first interval. Complete the other $5$ intervals.
$\frac{\Delta \textit{d}}{\Delta \textit{t}} = \frac{7 - 4}{0.25 - 0} = \frac{3}{0.25} = 12$
6. Does the ball have the same velocity on any of these intervals? If so, which ones? Think carefully, this may be tricky.
No. The speeds are the same, but velocities are different (except after the ball gets stuck in the mud).
|
# Algebra for Digital Communication
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1 EPFL - Section de Mathématiques Algebra for Digital Communication Fall semester 2008 Solutions for exercise sheet 1 Exercise 1. i) We will do a proof by contradiction. Suppose 2 a 2 but 2 a. We will obtain a contradiction. Since 2 a we have that a = 2k +1 for some k Z (this follows from the Division theorem). So a 2 = (2k + 1) 2 = 4k 2 + 4k + 1 = 2(2k 2 + 2k) + 1. Hence a 2 is not divisible by 2 (its remainder on division by 2 is 1). But this is a contradiction since we assumed that 2 a 2. ii) The proof again goes by contradiction. Suppose that 2 Q. Then there exist a, b Z with 2 = a b. Without loss of generality we can assume that a and b are relatively prime (if they have a common divisor we can cancel it). Rewriting, we obtain 2 b = a, squaring both sides we obtain 2 b 2 = a 2. Hence 2 a 2 and by part (i) we conclude that 2 a. So a is of the form 2 k for some k Z. Hence 2 b 2 = a 2 = (2k) 2 = 4k 2. Dividing both sides by 2 we obtain b 2 = 2k 2. So we have 2 b 2. By using part (i) again, we obtain 2 b. This is a contradiction since a and b are relatively prime. Exercise 2. 1 of 7
2 i) a) 14 = = = = = = = = = b) The last nonzero remainder is 1, so we have gcd(14, 227) = = = = = = The last nonzero remainder is 17, so gcd(272, 1479) = 17. ii) We illustrate two methods. The first is to use the fectorization of the integers, in the second, we first find the greatest common divisor of these pairs and then use the identity gcd(a, b) lcm(a, b) = a b. a) Factoring, we obtain: 56 = = 2 2. So lcm(56, 72) = = 504. Alternatively, we use the Euclidean Algorithm: 72 = = = The last nonzero remainder is 8, so we have gcd(56, 72) = 8. Hence lcm(56, 72) = gcd(56, 72) = of 7
3 b) Factoring, we obtain: 128 = = 4. So lcm(128, 81) = = Alternatively, we use the Euclidean Algorithm: 128 = = = = = = = = = The last nonzero remainder is 1, so we have gcd(128, 81) = 1. Hence lcm(128, 81) = gcd(128, 81) = Exercise. i) Using the Euclidean Algorithm, we obtain 14 = = = = = = = = = Now substituting backwards, starting from the next to last equation (the replaced expressions are boldface), we get 1 = 7 2 = 7 (9 7 1) = = (25 9 2) 4 9 = = 25 4 ( ) 11 = = ( ) = = ( ) 7 = = ( ) = of 7
4 So we obtain 1 = ( 100). Hence one solution is given by (can you say what all solutions are?). ii) We again apply the Euclidean algorithm: x = 6, y = = = = So we have gcd(272, 119) = 17. Substituting backwards: 17 = = 119 ( ) = Hence a solution is given by x = 7, y =. iii) In this case there is no solution in Z. Suppose there would exists x, y Z such that 56x + 72y =. We have gcd(56, 72) = 8. Since 8 56 and 8 72, we would consequently have that 8 56x + 72y =, this is a contradiction. So there cannot be a solution in the integers. iv) Euclidean Algorithm: 10 = = = Hence gcd(10, 6) = 2. Substituting backwards: So we have 2 = = 6 (10 6 1) 1 = = Multiplying this equation by 2 (since we want to have 4 on the left hand side of the equation) we get 4 = Hence a solution is given by x = 4, y = 2. Can you find a general pattern, when there exists a solution in the integers and when there is no solution in the integers? In the case that there exists one solution in the integers, how can you find all solutions? Exercise 4. i) Let n be an odd integer. By the division theorem the possible remainders on division of an integer by 4 are 0, 1, 2 and. So, every integer is of the form 4k or 4k + 1 or 4k + 2 or 4k +. Since n is odd, it cannot be of the form 4k or 4k + 2 since then it would be divisible by 2 (4k = 2(2k) and 4k + 2 = 2(2k + 1)). So n is either of the form 4k + 1 or 4k +. 4 of 7
5 ii) Let n be any integer. By the division theorem the possible remainders on division by are 0, 1 and 2. So, n is either of the form m or of the form m + 1 or of the form m + 2 for some m Z. Squaring, we see that n 2 is either of the form or it is of the form (m) 2 = 9m 2 = k, where k = m 2, (m + 1) 2 = 9m 2 + 6m + 1 = (m 2 + 2m) + 1 = k + 1, where k = m 2 + 2m, or it is of the form (m + 2) 2 = 9m m + 4 = (m 2 + 4m + 1) + 1 = k + 1, where k = m 2 + 4m + 1. So n 2 is either of the form k or k + 1. Exercise 5. i) Basis: We show that the statement holds for n = 1. In this case we have so the statement is true for n = = 2 = 1 2, Induction step: We assume that the statement holds for some given n 1 and show that it also holds for n + 1. So assume that for a given n we have n (n + 1) = Adding (n + 1) (n + 2) to both sides we obtain n (n + 1) + (n + 1) (n + 2) = n(n + 1)(n + 2). n(n + 1)(n + 2) + (n + 1) (n + 2) n(n + 1)(n + 2) + (n + 1) (n + 2) (n + 1)(n + 2)(n + ) = =. But this is just the statement for n + 1, which we wanted to deduce. Hence we are done. ii) Basis: For n = 1 the statement is true since 1 1! = 1 = 2! 1. Induction step: Assume the statement for a given n 1, i.e. that 1 (1!) + 2 (2!) + (!) + + n (n!) = (n + 1)! 1. Adding (n + 1) (n + 1)! to both sides we obtain 1 (1!) + 2 (2!) + (!) + + n (n!) + (n + 1) (n + 1)! = (n + 1)! 1 + (n + 1) (n + 1)! = (n + 2) (n + 1)! 1 = (n + 2)! 1. This is just the statement for n + 1, so we are done. 5 of 7
6 Exercise 6. i) The smallest prime number which is not less than 10 is 11. Since we are looking for a composite number, the smallest possibility is = 121. ii) The Sieve of Eratosthenes is a method to find a list of all prime number up to a given number n (n = 100 in this case). It is named after the ancient greek mathematician and astronomer Eratosthenes of Cyrene. The methods works as follows: first write down all natural numbers from 2 up to n. Now cross out any multiples of 2 (except 2 itself). The smallest number (after 2) that remains uncrossed (in this case ) will be a prime number (why?). Now cross out any multiples of this number (in this case ) except the number itself. The next smallest number that remains uncrossed will again be a prime. Now cross out any all multiples of this number etc. By generalizing the reasoning in part (i), it follows easily that it is enough to start crossing out at the square of the number whose multiples you are crossing out (if you are crossing out all multiples of p, any number, which is a multiple of p and which is less than p 2 will already have been crossed out, since it has a prime divisor less than p). Clearly you can stop this process once the square of the number whose multiples you are crossing out is not contained anymore in the list (in this case for p = 11). At the end, the numbers remaining uncrossed will be the prime numbers less than n. For n = 100 the result should look as follows: /1 2 /4 5 /6 7 /8 /9 /10 11 /12 1 /14 /15 /16 17 /18 19 /20 /21 /22 2 /24 /25 /26 /27 /28 29 /0 1 /2 / /4 /5 /6 7 /8 /9 /40 41 /42 4 /44 /45 /46 47 /48 /49 /50 /51 /52 5 /54 /55 /56 /57 /58 59 /60 61 /62 /6 /64 /65 /66 67 /68 /69 /70 71 /72 7 /74 /75 /76 /77 /78 79 /80 /81 /82 8 /84 /85 /86 /87 /88 89 /90 /91 /92 /9 /94 /95 /96 97 /98 / / So the list of all prime numbers less than 100 is as follows: 2,, 5, 7, 11, 1, 17, 19, 2, 29, 1, 7, 41, 4, 47, 5, 59, 61, 67, 71, 7, 79, 8, 89, 97. Note how much the idea of the observation in part (i) speeded up the process. It is based on the fact that every composite number n has a prime divisor p such that p n. Can you prove this? iii) By the sieve above we know that neither of these numbers are prime. Since the smallest composite number now having a prime divisor less than 10 is 121, we know that these numbers must have prime divisors less than 10. So all primes dividing these numbers must be 2,, 5 or 7. By trial divison we see that 91 = 7 1, 87 = 29 and 68 = Exercise 7. i) Let d = gcd(a, b) and d = gcd(b, a b). To prove that d = d we will show that d d and d d, hence d = ±d. Since the greatest common divisor of two numbers is by definition a positive number, we will conclude that d = d. 6 of 7
7 Since d is the gcd of a and b, we know that d a and d b. So d a b and d b. Since d = gcd(b, a b), it follows that d d. Conversely, since d is the gcd of b and a b, d divides b and (a b) + b = a. But d = gcd(a, b), so d d. ii) By using part (i) we get gcd( , ) = gcd( , ) = gcd( , ). Now = = and and So gcd( , ) = 1 and consequently gcd( , ) = 1. 7 of 7
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RD Sharma Class 12 Exercise 28.8 The Plane Solutions Maths - Download PDF Free Online
# RD Sharma Class 12 Exercise 28.8 The Plane Solutions Maths - Download PDF Free Online
Edited By Lovekush kumar saini | Updated on Jan 25, 2022 11:55 AM IST
RD Sharma is a well-known name for many students in CBSE schools. Many schools widely use the RD Sharma class 12 solution of The plane exercise 28.8 for maths. Not only students, the RD Sharma class 12th exercise 28.8 is used by many teachers and experts to take in reference while giving lectures or preparing question papers. Class 12 RD Sharma chapter 28 exercise 28.7 solution is well known for its basic understanding and simple concepts which are very easy for beginners. RD Sharma solutions Their main motive is to encourage students to learn maths in the simplest possible way.
## The Plane Excercise: 28.8
The Plane exercise 28.8 question 1
Answer:- The answer of the given question is $2x-3y+z=7$.
Hint:- By find and putting the value of $k$
Given:-$2x-3y+z=0$ and point $(1,-1,2)$
Solution:- Given equation of plane is $2x-3y+z=0$ … (i)
We know that equation of a plane parallel to given plane (i) is
$2x-3y+z+k=0$ …(ii)
As given that, plane (ii) is passing through the point $(1,-1,2)$ so it satisfy the plane (ii)
$\begin{gathered} 2(1)-3(-1)+2+k=0 \\\\ k=-7 \end{gathered}$
Put the value of k in equation (ii),
$2x-3y+z-7=0$
So, equation of the required plane is, $2x-3y+z=7$
The Plane exercise 28.8 question 2
Answer:- The answer of the given question is $\vec{r} \cdot(2 \hat{\imath}-3 \hat{\jmath}+5 \hat{k})+11=0$
Hint:- By putting the value of k in equation (ii)
Given:- $\vec{r} \cdot(2 \hat{\imath}-3 \hat{\jmath}+5 \hat{k})+2=0$
Solution:- Given equation of a plane is
$\vec{r} \cdot(2 \hat{\imath}-3 \hat{\jmath}+5 \hat{k})+2=0$ …(i)
We know that the equation of a plane parallel to given plane (i) is
$\vec{r} \cdot(2 \hat{\imath}-3 \hat{\jmath}+5 \hat{k})+k=0$ … (ii)
As given that plane (ii) is passing through the point $3 \hat{\imath}+4 \hat{\jmath}-\hat{k}$ so it satisfy the equation (ii)
\begin{aligned} &(3 \hat{\imath}+4 \hat{\jmath}-\hat{k}) \cdot(2 \hat{\imath}-3 \hat{\jmath}+5 \hat{k})+k=0 \\\\ &(3)(2)+(4)(-3)+(-1)(5)+k=0 \end{aligned}
$k=11$
Put the value of k in equation (ii)
$\vec{r} \cdot(2 \hat{\imath}-3 \hat{\jmath}+5 \hat{k})+11=0$
So, the equation of the required plane is
$\vec{r} \cdot(2 \hat{\imath}-3 \hat{\jmath}+5 \hat{k})+11=0$
The Plane exercise 28.8 question 3
Answer:- The answer of the given question is $15x-47y+28z-7=0.$
Hint:- We know that, equation of a plane passing through the line of intersection of two planes $a_{1} x+b_{1} y+c_{1} z+d_{1}=0 \text { and } a_{2} x+b_{2} y+c_{2} z+d_{2}=0$ is given by
$\left(a_{1} x+b_{1} y+c_{1} z+d_{1}\right)+k\left(a_{2} x+b_{2} y+c_{2} z+d_{2}\right)=0$
Given:-$2x-7y+4z-3=0$ and $3x-5y+4z+11=0$
Solution:- So equation of plane passing through the line of intersection of given two planes is
\begin{aligned} &(2 x-7 y+4 z-3)+k(3 x-5 y+4 z+11)=0 \\\\ &2 x-7 y+4 z-3+3 k x-5 k y+4 k z+11 k=0 \\\\ &x(2+3 k)+y(-7-5 k)+z(4+4 k)-3+11 k=0 \end{aligned} … (i)
As given that, plane (i) is passing through the point $(-2,1,3)$ so it satisfy the equation (i)
$(-2)(2+3 k)+(1)(-7-5 k)+(3)(4+4 k)-3+11 k=0$
$k=\frac{1}{6}$
Put the value of k in equation (i)
$\begin{gathered} x(2+3 k)+y(-7-5 k)+z(4+4 k)-3+11 k=0 \\\\ x\left(2+\frac{3}{6}\right)+y\left(-7-\frac{5}{6}\right)+z\left(4+\frac{4}{6}\right)-3+\frac{11}{6}=0 \end{gathered}$
$\begin{gathered} x\left(\frac{12+3}{6}\right)+y\left(\frac{-42-5}{6}\right)+z\left(\frac{24+4}{6}\right)-\frac{18+11}{6}=0 \\\\ x\left(\frac{15}{6}\right)+y\left(-\frac{47}{6}\right)+z\left(\frac{28}{6}\right)-\frac{7}{6}=0 \end{gathered}$
Multiplying by 6 we get
$15 x-47 y+28 z-7=0$
The Plane exercise 28.8 question 4
Answer:- The answer of the given question is $\vec{r} \cdot(\hat{\imath}+9 \hat{\jmath}+11 \hat{k})=0$
Hint:-$\vec{r} \cdot\left(\overrightarrow{n_{1}}+k \overrightarrow{n_{2}}\right)=d_{1}+k d_{2}$
Given:-$\vec{r} \cdot(\hat{\imath}+3 \hat{\jmath}-\hat{k})=0$ and $\vec{r} \cdot(\hat{\jmath}+2 \hat{k})=0$ , Point $(2 \hat{\imath}+\hat{j}-\hat{k})$
Solution:- $\vec{r} \cdot \overrightarrow{n_{1}}=d_{1}$
$\vec{r} \cdot \overrightarrow{n_{2}}=d_{2}$ is given by
$\vec{r}\left(\overrightarrow{n_{1}}+k \overrightarrow{n_{2}}\right)=d_{1}+k d_{2}$
∴ The equation of the plane passing through the line of intersection of given two planes
$\vec{r} \cdot(\hat{\imath}+3 \hat{\jmath}-\hat{k})=0$ and $\vec{r} \cdot(\hat{j}+2 \hat{k})=0$ is given by
$\vec{r} \cdot\{(\hat{\imath}+3 \hat{\jmath}-\hat{k})+k(\hat{j}+2 \hat{k})\}=0$ … (i)
As given that, plane (i) is passing through the point $2 \hat{\imath}+\hat{\jmath}-\hat{k}$
$\begin{gathered} (2 \hat{\imath}+\hat{\jmath}-\hat{k})(\hat{\imath}+3 \hat{\jmath}-\hat{k})+k(2 \hat{\imath}+\hat{\jmath}-\hat{k})(\hat{\jmath}+2 \hat{k})=0 \\\\ (2)(1)+(1)(3)+(-1)(-1)+k[(2)(0)+(1)(1)+(-1)(2)]=0 \end{gathered}$
$\begin{gathered} (2+3+1)+k(1-2)=0 \\\\ 6-k=0 \\\\ k=6 \end{gathered}$
Putting the value of k in equation (i)
$\begin{gathered} \vec{r} \cdot\{(\hat{\imath}+3 \hat{\jmath}-\hat{k})+k(\hat{\jmath}+2 \hat{k})\}=0 \\\\ \vec{r} \cdot\{(\hat{\imath}+3 \hat{\jmath}-\hat{k})+6(\hat{\jmath}+2 \hat{k})\}=0 \\\\ \vec{r} \cdot(\hat{\imath}+9 \hat{j}+11 \hat{k})=0 \end{gathered}$
So, the equation of required plane is
$\vec{r} \cdot(\hat{\imath}+9 \hat{\jmath}+11 \hat{k})=0$
The Plane exercise 28.8 question 5
Answer:- The answer of the given question is $28x-17y+9z=0$.
Hint:- We know that, equation of a plane passing through the line of intersection of two planes $a_{1} x+b_{1} y+c_{1} z+d_{1}=0 \text { and } a_{2} x+b_{2} y+c_{2} z+d_{2}=0$ is given by
$\left(a_{1} x+b_{1} y+c_{1} z+d_{1}\right)+k\left(a_{2} x+b_{2} y+c_{2} z+d_{2}\right)=0$
Given:-$2x-y=0$ and $3z-y=0$ and plane $4x+5y-3z=8$
Solution:- So equation of plane passing through the line of intersection of given two planes is
$2x-y=0$ and $3z-y=0$ is
$\begin{gathered} (2 x-y)+k(3 z-y)=0 \\\\ 2 x-y+3 k z-k y=0 \end{gathered}$
$x(2)+y(-1-k)+z(3 k)=0$ …(i)
We know that, two planes are perpendicular if
$a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=0$ …(ii)
Given, plane (i) is perpendicular to plane
$4x+5y-3z=8$ …(iii)
Using (i) and (iii) in equation (ii)
$\begin{gathered} (2)(4)+(-1-k)(5)+(3 k)(-3)=0 \\\\ 3-14 k=0 \end{gathered}$
$k=\frac{3}{14}$
Putting value of k in equation (i)
$\begin{gathered} x(2)+y(-1-k)+z(3 k)=0 \\\\ x(2)+y\left(-1-\frac{3}{14}\right)+z\left(3\left(\frac{3}{14}\right)\right)=0 \\\\ x(2)+y\left(-\frac{17}{14}\right)+z\left(\frac{9}{14}\right)=0 \end{gathered}$
Multiplying with 14 we get
$28x-17y+9z=0$
The Plane exercise 28.8 question 6
Answer:- The answer of the given question is $33x+45y+50z-41=0$.
Hint:- We know that, equation of a plane passing through the line of intersection of two planes $a_{1} x+b_{1} y+c_{1} z+d_{1}=0 \text { and } a_{2} x+b_{2} y+c_{2} z+d_{2}=0$ is given by
$\left(a_{1} x+b_{1} y+c_{1} z+d_{1}\right)+k\left(a_{2} x+b_{2} y+c_{2} z+d_{2}\right)=0$
Given:-$x+2y+3z-4=0$ and
$2x+y-z+5=0$
Solution:- So equation of plane passing through the line of intersection of given two planes
$x+2y+3z-4=0$ and$2x+y-z+5=0$ is given by
\begin{aligned} &(x+2 y+3 z-4)+k(2 x+y-z+5)=0 \\\\ &x+2 y+3 z-4+2 k x+k y-k z+5 k=0 \end{aligned}
$x(1+2 k)+y(2+k)+z(3-k)-4+5 k=0$ … (i)
We know that, two planes are perpendicular if
$a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=0$ … (ii)
Given, plane (i) is perpendicular to plane,
$5x+3y-6z+8=0$ … (iii)
Using (i) and (iii) in equation (ii)
$\begin{gathered} 5(1+2 k)+3(2+k)+(-6)(3-k)=0 \\\\ 5+10 k+6+3 k-18+6 k=0 \\\\ k=\frac{7}{19} \end{gathered}$
Putting the value of k in equation (iii)
$\begin{gathered} x\left(1+\frac{14}{19}\right)+y\left(2+\frac{7}{19}\right)+z\left(3-\frac{7}{19}\right)-4+\frac{35}{19}=0 \\\\ x \frac{(19+14)}{19}+y\left(\frac{38+7}{19}\right)+z\left(\frac{57-7}{19}\right)+\frac{-76+35}{19}=0 \end{gathered}$
$x\left(\frac{33}{19}\right)+y\left(\frac{45}{19}\right)+z\left(\frac{50}{19}\right)-\frac{41}{19}=0$
Multiplying with 19 we get
$33x+45y+50z-41=0$
The Plane exercise 28.8 question 7
Answer:- The answer of the given question is $x-10y-5z=0$.
Hint:- We know that, equation of a plane passing through the line of intersection of two planes $a_{1} x+b_{1} y+c_{1} z+d_{1}=0 \text { and } a_{2} x+b_{2} y+c_{2} z+d_{2}=0$ is given by
$\left(a_{1} x+b_{1} y+c_{1} z+d_{1}\right)+k\left(a_{2} x+b_{2} y+c_{2} z+d_{2}\right)=0$
Given:-$x+2y+3z+4=0$ and $x-y+z+3=0$
Solution:- So equation of plane passing through the line of intersection of given two planes
$x+2y+3z+4=0$ and $x-y+z+3=0$
$\begin{array}{r} (x+2 y+3 z+4)+k(x-y+z+3)=0 \\\\ x(1+k)+y(2-k)+z(3+k)+4+3 k=0 \end{array} .$ … (i)
Equation (i) is passing through origin, so
$\begin{gathered} (1+k)+(0)(2-k)+(0)(3+k)+4+3 k=0 \\\\ 0+0+0+4+3 k=0 \\\\ k=-\frac{4}{3} \end{gathered}$
Put the value of k in equation (i)
$\begin{gathered} x(1+k)+y(2-k)+z(3+k)+4+3 k=0 \\\\ x\left(1-\frac{4}{3}\right)+y\left(2+\frac{4}{3}\right)+z\left(3-\frac{4}{3}\right)+4-\frac{12}{3}=0 \end{gathered}$
$\begin{gathered} x\left(\frac{3-4}{3}\right)+y\left(\frac{6+4}{3}\right)+z\left(\frac{9-4}{3}\right)+4-4=0 \\\\ -\frac{x}{3}+\frac{10 y}{3}+\frac{5 z}{3}=0 \end{gathered}$
Multiplying by 3, we get
$-x+10y+5z=0$
$x-10y-5z=0$
The Plane exercise 28.8 question 8
Answer:- The answer of the given question is $\vec{r}(\hat{\imath}+7 \hat{\jmath})+13=0$.
Hint:- We know that, equation of a plane passing through the line of intersection of two planes $a_{1} x+b_{1} y+c_{1} z+d_{1}=0 \text { and } a_{2} x+b_{2} y+c_{2} z+d_{2}=0$ is given by
$\left(a_{1} x+b_{1} y+c_{1} z+d_{1}\right)+k\left(a_{2} x+b_{2} y+c_{2} z+d_{2}\right)=0$
Given:-$x-3y+2z-5=0$ and $2x-y+3z-1=0$
Solution:- So equation of plane passing through the line of intersection of planes
$x-3y+2z-5=0$ and $2x-y+3z-1=0$ is given by
\begin{aligned} (x-3 y+2 z-5) &+k(2 x-y+3 z-1) &=0 & \\\\ x(1-2 k)+y(-3-k)+z(2+3 k)-5-k &=0 & \end{aligned} . … (i)
Plane (i) is passing the through the point$(1,-2,3),$ so
$\begin{gathered} 1(1+2 k)+(-2)(-3-k)+(3)(2+3 k)-5-k=0 \\\\ 1+2 k+6+2 k+6+9 k-5-k=0 \\\\ 8+12 k=0 \\\\ k=-\frac{2}{3} \end{gathered}$
Putting the value of k in equation (i)
$\begin{gathered} x(1+2 k)+y(-3-k)+z(2+3 k)-5-k=0 \\\\ x\left(1-\frac{4}{3}\right)+y\left(-3+\frac{2}{3}\right)+z\left(2-\frac{6}{3}\right)-\left(\frac{15-2}{3}\right)=0 \\\\ -\frac{1}{3} x-\frac{7}{3} y-\frac{13}{3}=0 \end{gathered}$
Multiplying by -3
$x+7y+13=0$
$\begin{gathered} (x \hat{\imath}+y \hat{\jmath}+z \hat{k})(\hat{\imath}+7 \hat{\jmath})+13=0 \\\\ \vec{r} \cdot(\hat{\imath}+7 \hat{\jmath})+13=0 \end{gathered}$
The Plane exercise 28.8 question 9
Answer:- The answer of the given question is $51x+15y-50z+173=0.$
Hints:- We know that, equation of a plane passing through the line of intersection of two planes $a_{1} x+b_{1} y+c_{1} z+d_{1}=0 \text { and } a_{2} x+b_{2} y+c_{2} z+d_{2}=0$ is given by
$\left(a_{1} x+b_{1} y+c_{1} z+d_{1}\right)+k\left(a_{2} x+b_{2} y+c_{2} z+d_{2}\right)=0$
Given:-$5x+3y+6z+8=0$
$(x +2y+3z-4)=0$ and
$x +2y+3z-4=0$
Solution:- The equation of a plane through the line of intersection of the planes $x+2y+3z-4=0$ and $2x+y-z+5=0$
\begin{aligned} (x+2 y+3 z-4) &+\lambda(2 x+y-z+5)=0 \\\\ x(1+2 \lambda)+y(2+\lambda)+z(-\lambda+3)-4+5 \lambda &=0 \end{aligned} . … (i)
Also this is perpendicular to the plane $5x+3y+6z+8=0$ …(ii)
We know that, two planes are perpendicular if $a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=0$
$\begin{gathered} 5(1+2 \lambda)+3(2+\lambda)+6(-\lambda+3)=0 \\\\ \therefore 5+10 \lambda+6+3 \lambda+18-6 \lambda=0 \\\\ \lambda=-\frac{29}{7} \end{gathered}$
∴ Putting this value of $\lambda$ in equation (i) we get equation of plane as
$51x+15y-50z+173=0$
The Plane exercise 28.8 question 10
Answer:- The answer of the given question is $\vec{r} \cdot(4 \hat{i}+2 \hat{j}-4 \hat{k})+6=0$ or
$\vec{r} \cdot(-2 \hat{i}+4 \hat{j}+4 \hat{k})+6=0$
Hint:- We know that, equation of a plane passing through the line of intersection of two planes $a_{1} x+b_{1} y+c_{1} z+d_{1}=0 \text { and } a_{2} x+b_{2} y+c_{2} z+d_{2}=0$ is given by
$\left(a_{1} x+b_{1} y+c_{1} z+d_{1}\right)+k\left(a_{2} x+b_{2} y+c_{2} z+d_{2}\right)=0$
Given:-\begin{aligned} &-\vec{r} \cdot(\hat{\imath}+3 \hat{\jmath})+6=0 \end{aligned}
$\\\\ \vec{r} \cdot(3 \hat{\imath}-\hat{\jmath}-4 \hat{k})=0$
Solution:- The equation of a plane through the line of intersection of the planes$r \cdot(\hat{\imath}+3 \hat{j})+6=0$ and $\vec{r} \cdot(3 \hat{\imath}-\hat{\jmath}-4 k)=0$
$\begin{gathered} \vec{r} \cdot(\hat{\imath}+3 \hat{\jmath})+6+\lambda[\vec{r} \cdot(3 \hat{\imath}-\hat{\jmath}-4 \hat{k})]=0 \\\\ \vec{r} \cdot[(\hat{\imath}+3 \hat{\jmath})]+6+3 \lambda \cdot \vec{r} \hat{\imath}-\vec{r} \lambda \hat{j}-4 \lambda \vec{r} \hat{k}=0 \end{gathered}$
$r(\hat{\imath}+3 \hat{\jmath}+3 \lambda \hat{\imath}-\lambda \hat{\jmath}-4 \lambda \hat{k})=-6$ ...........(i)
$\vec{r} \cdot \frac{[\hat{\imath}(1+3 \lambda)+\hat{j}(3-\lambda)+\hat{k}(-4 \lambda)]}{\sqrt{(1+3 \lambda)^{2}+(3-\lambda)^{2}+(-4 \lambda)^{2}}}=\frac{-6}{\sqrt{(1+3 \lambda)^{2}+(3-\lambda)^{2}+(-4 \lambda)^{2}}}$
The perpendicular distance from the origin is unity
\begin{aligned} &\frac{-6}{\sqrt{(1+3 \lambda)^{2}+(3-\lambda)^{2}+(-4 \lambda)^{2}}}=1 \\\\ &(1+3 \lambda)^{2}+(3-\lambda)^{2}+(-4 \lambda)^{2}=36 \end{aligned}
$\begin{gathered} 1+9 \lambda^{2}+6 \lambda+9+\lambda^{2}-6 \lambda+16 \lambda^{2}=36 \\\\ \lambda^{2}=1 \\\\ \lambda=\pm 1 \end{gathered}$
Using equation (i) the required plane is
$\vec{r} \cdot(1 \pm 3) \hat{\imath}+(3 \mp 1) \hat{\jmath}+(\mp 4) \hat{k}=-6$
$\vec{r} \cdot(4 \hat{\imath}+2 \hat{\jmath}-4 \hat{k})=-6 \text { or }\vec{r} \cdot(-2 \hat{\imath}+4 \hat{\jmath}+4 \hat{k})=-6$
$\vec{r} \cdot(4 \hat{\imath}+2 \hat{\jmath}-4 \hat{k})+6=0 \text { orr. }(-2 \hat{\imath}+4 \hat{\jmath}+4 \hat{k})+6=0$
The Plane exercise 28.8 question 11
Answer:- The answer of the given question is $7x+13y+4z=9$.
Hint:- We know that, equation of a plane passing through the line of intersection of two planes
$a_{1} x+b_{1} y+c_{1} z+d_{1}=0 \text { and } a_{2} x+b_{2} y+c_{2} z+d_{2}=0$ is given by
$\left(a_{1} x+b_{1} y+c_{1} z+d_{1}\right)+k\left(a_{2} x+b_{2} y+c_{2} z+d_{2}\right)=0$
Given:-$2x+3y-z+1=0$ and $x+y-2z+3=0$
Plane $:3x-y-2z-4=0$
Solution:- Cartesian form of equation of plane through the line of intersection of plane is
$A_{1} x+B_{1} y+C_{1} z+D_{1}+\lambda\left(A_{2} x+B_{2} y+C_{2} z+D_{2}\right)=0$ … (i)
Here the standard equation of plane is $A_{1} x+B_{1} y+C_{1} z+D_{1} \text { and } A_{2} x+B_{2} y+C_{2} z+D_{2}$
Substituting the values in equation (i) we get
$2 x+3 y-z+1+\lambda(x+y-2 z+3)=0$
$(2+\lambda) x+(3+\lambda) y+(-1-2 \lambda) z+1+3 \lambda=0$ ...........(ii)
It is given that the plane $3x-y-2z-4=0$ is perpendicular to the plane
We know that $\emptyset=90^{\circ}$ where $\cos 90^{\circ}=0$ so we get
$A_{1} A_{2}+B_{1} B_{2}+C_{1} C_{2}=0$ … (iii)
By comparing the standard equation in Cartesian form$\begin{array}{lll} A_{1}=2+\lambda, & B_{1}=3+\lambda, & C_{1}=-1-2 \lambda \\\\ A_{2}=3, & B_{2}=-1, & C_{2}=-2 \end{array}$
Substituting these values in equation (iii)
$(2+\lambda) \cdot 3+(3+\lambda)(-1)+(-1-2 \lambda)(-2)=0$
On further calculation
$\begin{gathered} 6+3 \lambda-3-\lambda+2+4 \lambda=0 \\ \lambda=-\frac{5}{6} \end{gathered}$
Substituting the value of λ in equation (ii)
$\left(2+\frac{-5}{6}\right) x+\left(3+\frac{-5}{6}\right) y+\left(-1-2 \cdot \frac{-5}{6}\right) z+1+3 \cdot \frac{-5}{6}=0$
By taking LCM
$\left(\frac{12-5}{6}\right) x+\left(\frac{18-5}{6}\right) y+\left(\frac{-6+10}{6}\right) z+\frac{6-15}{6}=0$
We get
$7x+13y+4z-9=0$
$7x+13y+4z=9$
The Plane exercise 28.8 question 12
Answer:- The answer of the given question is $33x+45y+50z-41=0$.
Hint:- We know that, equation of a plane passing through the line of intersection of two planes
$a_{1} x+b_{1} y+c_{1} z+d_{1}=0 \text { and } a_{2} x+b_{2} y+c_{2} z+d_{2}=0$ is given by
$\left(a_{1} x+b_{1} y+c_{1} z+d_{1}\right)+k\left(a_{2} x+b_{2} y+c_{2} z+d_{2}\right)=0$
Given:- $\vec{r} \cdot(\hat{\imath}+2 \hat{\jmath}+3 \hat{k})-4=0$ … (i)
$\vec{r} \cdot(2 \hat{\imath}+\hat{\jmath}-\hat{k})+5=0$ … (ii)
Solution:- The equation of the plane passing through the line of intersection of the plane given in equation (i) and equation (ii) is
$[\vec{r} \cdot(\hat{\imath}+2 \hat{\jmath}+3 \hat{k})-4]+\lambda[\vec{r} \cdot(2 \hat{\imath}+\hat{\jmath}-\hat{k})+5]=0$
$\vec{r}[(2 \lambda+1) \hat{\imath}+(\lambda+2) \hat{\jmath}+(3-\lambda) \hat{k}]+(5 \lambda-4)=0$ …(iii)
The plane in equation (iii) is perpendicular to the plane,
$\vec{r} \cdot(5 \hat{\imath}+3 \hat{\jmath}-6 \hat{k})+8=0$
We know that, two planes are perpendicular if $a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=0$
$\begin{gathered} \therefore 5(2 \lambda+1)+3(\lambda+2)-6(3-\lambda)=0 \\\\ 19 \lambda-7=0 \\\\ \lambda=\frac{7}{19} \end{gathered}$
Substituting $\lambda=\frac{7}{19}$ in equation (iii), we obtain
$\vec{r} \cdot\left[\frac{33}{19} \hat{\imath}+\frac{45}{19} \hat{\jmath}+\frac{50}{19} \hat{k}\right]-\frac{41}{19}=0$
$\vec{r} \cdot(33 \hat{\imath}+45 \hat{\jmath}+50 \hat{k})-41=0$ … (iv)
This is vector equation of the required plane
Now $(x \hat{\imath}+y \hat{\jmath}+z \hat{k}) \cdot(33 \hat{\imath}+45 \hat{\jmath}+50 \hat{k})-41=0$
$33x+45y+50z-41=0$
The Plane exercise 28.8 question 13
Answer:- The answer of the given question is $\vec{r} \cdot(20 \hat{\imath}+23 \hat{\jmath}+26 \hat{k})=69$
Hints:- We know that, equation of a plane passing through the line of intersection of two planes $a_{1} x+b_{1} y+c_{1} z+d_{1}=0 \text { and } a_{2} x+b_{2} y+c_{2} z+d_{2}=0$ is given by
$\left(a_{1} x+b_{1} y+c_{1} z+d_{1}\right)+k\left(a_{2} x+b_{2} y+c_{2} z+d_{2}\right)=0$
Given:$\vec{r} \cdot(\hat{\imath}+\hat{\jmath}+\hat{k})=6$ and
$\vec{r} \cdot\left(2 \hat{\imath}+\widehat{3}_{j}+4 \hat{k}\right)=-5$ and Point $(1,1,1)$
Solution:- The Cartesian equation of the given planes are $x+y+z-6=0$ and
$2x+3y+4z+5=0$
The family of planes is$x+y+z-6+\lambda(2 x+3 y+4 z+5)=0$ …(i)
Since it pass through $(1,1,1)$
$\begin{gathered} 1+1+1-6+\lambda(2+3+4+5)=0 \\\\ -3+\lambda(14)=0 \\ \lambda=\frac{3}{14} \end{gathered}$
$\begin{gathered} x+y+z-6+\frac{3}{14}(2 x+3 y+4 z+5)=0 \\\\ 14 x+14 y+14 z-84+6 x+9 y+12 z+15=0 \\\\ 20 x+23 y+26 z-69=0 \end{gathered}$
Vector equation is $\vec{r} \cdot(20 \hat{\imath}+23 \hat{\jmath}+26 \hat{k})=69$
The Plane exercise 28.8 question 14
Answer:- The answer of the given question is $\vec{r} \cdot(2 \hat{i}-13 \hat{j}+3 \hat{k})=0$
Hints:- We know the line of intersection of the plane $\vec{r} \cdot \vec{n}_{1}-\vec{d}_{1}=0 \text { and } \vec{r} \cdot \vec{n}_{1}-\vec{d}_{2}=0$ is
given by $\vec{r} \cdot\left(\vec{n}_{1}+k \vec{n}_{2}\right)-d_{1}+k d_{2}=0$
Given :-The intersection of the plane vector
\begin{aligned} &\vec{r} \cdot(2 \hat{i}+\hat{j}+3 \hat{k})=7 \quad \text { and } \\\\ &\vec{r} \cdot(2 \hat{i}+3 \hat{j}+3 \hat{k})=9 \end{aligned}
Solution:-We know that, the equation of a plane through the line of intersection of the plane
$\vec{r} \cdot \vec{n}_{1}-d_{1}=0 \text { and } \vec{r} \cdot \vec{n}_{1}-d_{2}=0$
Is given by $\vec{r} \cdot\left(\vec{n}_{1}+k \vec{n}_{2}\right)-d_{1}+k d_{2}=0$
So, equation of plane passing through the line of intersection of plane
\begin{aligned} &\vec{r} \cdot(2 \hat{i}+\hat{j}+3 \hat{k})-7=0 \text { and } \\ &\vec{r} \cdot(2 \hat{i}+5 \hat{j}+3 \hat{k})-9=0 \end{aligned}is given by
\begin{aligned} &{[\vec{r} \cdot(2 \hat{i}+\hat{j}+3 \hat{k})-7]+k[\vec{r} \cdot(2 \hat{i}+5 \hat{j}+3 \hat{k})-9]=0} \\\\ &\vec{r}[(2+2 k) \hat{i}+(1+5 k) \hat{j}+(3+3 k) \hat{k}]-7-9 k=0 \end{aligned}.............(1)
Given that plane (1) is passing through $2 \hat{i}+\hat{j}+3 \hat{k} \text { so }$
\begin{aligned} &(2 \hat{i}+\hat{j}+3 \hat{k})[(2+2 k) \hat{i}+(1+5 k) \hat{j}+(3+3 k) \hat{k}]-7-9 k=0 \\\\ &(2)(2+2 k)+(1)(1+5 k)+(3)(3+3 k)-7-9 k=0 \\\\ &4+4 k+1+5 k+9+9 k-7-9 k=0 \end{aligned}
\begin{aligned} &9 k=-7 \\\\ &k=\frac{-7}{9} \end{aligned}
Put the value of k in equation (1)
\begin{aligned} &\vec{r}[(2+2 k) \hat{i}+(1+5 k) \hat{j}+(3+3 k) \hat{k}]-7-9 k=0 \\\\ &\vec{r}\left[\left(2-\frac{14}{9}\right) \hat{i}+\left(1-\frac{35}{9}\right) \hat{j}+\left(3-\frac{21}{9}\right) \hat{k}\right]-7+\frac{63}{9}=0 \end{aligned}
\begin{aligned} &\vec{r}\left[\left(\frac{18-14}{9}\right) \hat{i}+\left(\frac{9-35}{9}\right) \hat{j}+\left(\frac{27-21}{9}\right) \hat{k}\right]-7+7=0 \\\\ &r\left[\left(\frac{4}{9}\right) \hat{i}-\left(\frac{26}{9}\right) \hat{j}+\left(\frac{6}{9}\right) \hat{k}\right]=0 \end{aligned}
Multiplying by $(\frac{9}{2})$,we get
$\vec{r} \cdot(2 \hat{i}-13 \hat{j}+3 \hat{k})=0$
Equation of the required plane is
$\vec{r} \cdot(2 \hat{i}-13 \hat{j}+3 \hat{k})=0$
The Plane exercise 28.8 question 15
Answer:- The answer of the given question is $7x-5y+4z-8=0.$
Hint:- We know that, equation of a plane passing through the line of intersection of two planes
$a_{1} x+b_{1} y+c_{1} z+d_{1}=0 \text { and } a_{2} x+b_{2} y+c_{2} z+d_{2}=0$ is given by
$\left(a_{1} x+b_{1} y+c_{1} z+d_{1}\right)+k\left(a_{2} x+b_{2} y+c_{2} z+d_{2}\right)=0$
Given:-$3x-y+2z=4$ and $x+y+z=2$ , Point $(2,2,1)$
Solution:- The equation of any plane through the intersection of the planes is
$3x-y+2z-4=0$ and $x+y+z-2=0$ is
$(3 x-y+2 z-4)+\alpha(x+y+z-2)=0$, where $\alpha \in R$ …(i)
The plane passes through the point $(2, 2,1).$ Therefore, this point will satisfy equation (1)
$\begin{gathered} \therefore(3 \times 2-2+2 \times 1-4)+\alpha(2+2+1-2)=0 \\\\ 2+3 \alpha=0 \\ \alpha=-\frac{2}{3} \end{gathered}$
Substituting $\alpha=-\frac{2}{3}$ in equation (i), we obtain
$\begin{gathered} (3 x-y+2 z-4)-\frac{2}{3}(x+y+z-2)=0 \\\\ 3(3 x-y+2 z-4)-2(x+y+z-2)=0 \\\\ (9 x-3 y+6 z-12)-2(x+y+z-2)=0 \\\\ 7 x-5 y+4 z-8=0 \end{gathered}$
The Plane exercise 28.8 question 16
Answer:- The answer of the given question is$\vec{r} \cdot(\hat{\imath}-\hat{k})+2=0 \text {. }$
Hints:- We know that, equation of a plane passing through the line of intersection of two planes $a_{1} x+b_{1} y+c_{1} z+d_{1}=0 \text { and } a_{2} x+b_{2} y+c_{2} z+d_{2}=0$ is given by
$\left(a_{1} x+b_{1} y+c_{1} z+d_{1}\right)+k\left(a_{2} x+b_{2} y+c_{2} z+d_{2}\right)=0$
Given:- $x+y+z=1$ ,
$2x+3y+4z=5$
$x-y+z=0$
Solution:- Equation of the plane through the intersection of plane is
\begin{aligned} &(x+y+z-1)+\lambda(2 x+3 y+4 z-5)=0 \text { or } \\\\ &(1+2 \lambda) x+(1+3 \lambda) y+(1+4 \lambda) z-(1+5 \lambda)=0 \end{aligned} …(i)
This plane is perpendicular to $x-y+z=0$
We know that, two planes are perpendicular if $a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=0$
$\begin{array}{r} 1(1+2 \lambda)-1(1+3 \lambda)+1(1+4 \lambda)=0 \text { or } \\\\ \qquad \lambda=-\frac{1}{3} \end{array}$
Equation of plane is
$\begin{gathered} (x+y+z-1)-\frac{1}{3}(2 x+3 y+4 z-5)=0 \\\\ \Rightarrow x-z+2=0 \end{gathered}$
Vector form of plane is $\vec{r} \cdot(\hat{\imath}-\hat{k})+2=0$
Yes, line lies on the plane on $(-2, 3, 0)$ satisfies$\vec{r} \cdot(\hat{\imath}-\hat{k})+2=0$
The Plane exercise 28.8 question 17
Answer:- The answer of the given question is $x+y+z=a+b+c$.
Hints:- By substituting $\vec{r}=x \hat{\imath}+y \hat{j}+z \hat{k}$ in equation (ii)
Given:-$(a, b, c)$ and parallel to plane $\vec{r} \cdot(\hat{\imath}+\hat{\jmath}+\hat{k})=2$
Solution:- Any plane passes through the point $(a, b, c)$ and parallel to plane $\vec{r} \cdot(\hat{\imath}+\hat{\jmath}+\hat{k})=2$ is given by
$\vec{r} \cdot(\hat{\imath}+\hat{\jmath}+\hat{k})=\lambda$…(i)
Here, the position vector $\vec{r}$ of this point is $\vec{r}=a \hat{\imath}+b \hat{\jmath}+c \hat{k}$
∴ Equation (i) becomes
$\begin{gathered} (a \hat{\imath}+b \hat{\jmath}+c \hat{k}) \cdot(\hat{\imath}+\hat{\jmath}+\hat{k})=\lambda \\\\ \Rightarrow a+b+c=\lambda \end{gathered}$
Substituting $\lambda =a+b+c$ in equation (i), we obtain
$\vec{r} \cdot(\hat{\imath}+\hat{\jmath}+\hat{k})=a+b+c$ … (ii)
This is vector equation of the required plane
Substituting $\vec{r}=x \hat{\imath}+y \hat{\jmath}+z \hat{k}$ in equation (ii)
$\begin{gathered} (x \hat{\imath}+y \hat{\jmath}+z \hat{k}) \cdot(\hat{\imath}+\hat{\jmath}+\hat{k})=a+b+c \\\\ x+y+z=a+b+c \end{gathered}$
The Plane exercise 28.8 question 18
1. $13 x+14 y+11 z=0,$ This plane doesn’t satisfies the given condition.
2. The equation of the required plane is, $7 x+11 y+14 z=15$.
3. The equation of the required plane is, $7 x+11 y+4 z=33$
Hint:- By using intercept form of plane
$\frac{x}{a}+\frac{y}{b}+\frac{c}{z}=1$
Given:- The equation of the family of planes passing through the intersection of the planes
$x+2 y+3 z-4=0$ and $2 x+y-z+5=0$
Solution:- The equation of the family of planes passing through the intersection of the planes
$x+2 y+3 z-4=0$ and $2 x+y-z+5=0$
$(x+2 y+3 z-4)+k(2 x+y-z+5)=0$ , where k is some constant.
\begin{aligned} &(2 k+1) x+(k+2) y+(3-k) z=4-5 k \\\\ &\frac{x}{\left(\frac{4-5 k}{2 k+1}\right)}+\frac{y}{\left(\frac{y}{k+2}\right)}+\frac{z}{\left(\frac{4-5 k}{3-k}\right)}=1 \end{aligned}
It is given that x-intercept of the required plane is twice its z intercept.
\begin{aligned} &\left(\frac{4-5 k}{2 k+1}\right)=2\left(\frac{4-5 k}{3-k}\right) \\\\ &(4-5 k)(3-k)=(4 k+2)(4-5 k) \\\\ &(4-5 k)(3-k-4 k-2)=0 \end{aligned}
$(4-5 k)(1-5 k)=0$
Either
\begin{aligned} &4-5 k=0 \quad \text { or } \quad 1-5 k=0\\\\ &k=4 / 5 \quad \text { or }\quad k=1 / 5 \end{aligned}
When $k=4 / 5$, the equation of the plane is
\begin{aligned} &\left(2 \times \frac{4}{5}+1\right) x+\left(\frac{4}{5}+2\right) y+\left(3-\frac{4}{5}\right) z=4-5 \times \frac{4}{5} \\\\ &13 x+14 y+11 z=0 \end{aligned}
This plane does not satisfies the given condition, so this is rejected
When $k=1 / 5$ , the equation of the plane is
\begin{aligned} &\left(2 \times \frac{1}{5}+1\right) x+\left(\frac{1}{5}+2\right) y+\left(3-\frac{1}{5}\right) z=4-5 \times \frac{1}{5} \\\\ &7 x+11 y+14 z=15 \end{aligned}
Thus, the equation of the required plane is $7 x+11 y+14 z=15$.
Also, the equation of the plane passing through the point $(2,3,-1)$ and parallel to the plane $7 x+11 y+14 z=15$ is
\begin{aligned} &7(x-2)+11(y-3)+14(z+1)=0 \\\\ &7 x+11 y+4 z=33 \end{aligned}
The Plane exercise 28.8 question 19
Answer:- The answer of the given question is $x+2y+3z=4.$
Hints:- We know that, equation of a plane passing through the line of intersection of two planes
$a_{1} x+b_{1} y+c_{1} z+d_{1}=0 \text { and } a_{2} x+b_{2} y+c_{2} z+d_{2}=0$ is given by
$\left(a_{1} x+b_{1} y+c_{1} z+d_{1}\right)+k\left(a_{2} x+b_{2} y+c_{2} z+d_{2}\right)=0$
Given:-$x+y+z=1$
$2x+3y+4z=5$
∴ Required equation of plane is $x+y-1+z+\lambda(2 x+3 y+4 z-5)=0$ for some λ
i.e.$(1+2 \lambda) x+(1+3 \lambda) y+(1+4 \lambda) z=(1+5 \lambda)$
According to question
$2\left(\frac{1+5 \lambda}{1+3 \lambda}\right)=3\left(\frac{1+5 \lambda}{1+4 \lambda}\right)$
Solving we get $\lambda =-1$
Thus the equation of required plane is
$-x-2y-3z=-4$
$x+2y+3z=4$
The RD Sharma class 12th exercise 28.8 covers the chapter 'The Plane.' There are about 19 questions in this exercise that are extremely basic and simple to solve if you have knowledge of the fundamentals of this chapter. The RD Sharma class 12th exercise 28.8 covers all the essential concepts of this chapter that are mentioned below,
• Equation of plane which is parallel to the line
• Equation of plane passing through points and parallel to the plane
• Equation of plane in scalar product form
• Equation of plane passing through line of intersection of the planes
Mentioned below are some benefits of the RD Sharma class 12 solution chapter 28 exercise 28.8:-
• The RD Sharma class 12th exercise 28.8 is used by teachers to take reference while teaching and also for preparing homeworks or question papers. Therefore, solving from the solution can help complete your homework on time and also easily.
• You can trust the questions provided in the RD Sharma class 12 chapter 28 exercise 28.8 as it is generously prepared by experts of maths present in the country, so it gives you the best of knowledge about the concepts.
• The RD Sharma solutions cover up the syllabus of NCERT and are updated yearly to correspond with the level of questions of NCERT so that students can also refer to the solutions to prepare for public exams.
• The RD Sharma class 12th exercise 28.8 are available online for download on the Career360 website.
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## RD Sharma Chapter-wise Solutions
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Number and Operations :
Rational Numbers
The major focus of the work on rational numbers in grade 5 is on understanding
relationships among fractions, decimals, and percents. Students make comparisons and
identify equivalent fractions, decimals, and percents, and they develop strategies for
adding and subtracting fractions and decimals.
In a study of fractions and percents, students work with halves, thirds, fourths, fifths,
sixths, eighths, tenths, and twelfths. They develop strategies for finding percent
equivalents for these fractions so that they are able to move back and forth easily between
fractions and percents and choose what is most helpful in solving a particular problem,
such as finding percentages or fractions of a group.
Students use their knowledge of fraction equivalents, fraction-percent equivalents, the
relationship of fractions to landmarks such as ½, 1, and 2, and other relationships to
decide which of two fractions is greater. They carry out addition and subtraction of
fractional amounts in ways that make sense to them by using representations such as
rectangles, rotation on a clock, and the number line to visualize and reason about fraction
equivalents and relationships.
Students continue to develop their understanding of how decimal fractions represent
quantities less than 1 and extend their work with decimals to thousandths. By
representing tenths, hundredths, and thousandths on rectangular grids, students learn
about the relationships among these numbers—for example, that one tenth is equivalent
to ten hundredths and one hundredth is equivalent to ten thousandths—and how these
numbers extend the place value structure of tens that they understand from their work
with whole numbers.
Students extend their knowledge of fraction-decimal equivalents by studying how
fractions represent division and carrying out that division to find an equivalent decimal.
They compare, order, and add decimal fractions (tenths, hundredths, and thousandths) by
carefully identifying the place value of the digits in each number and using
representations to visualize the quantities represented by these numbers.
Emphases
Rational Numbers
• Understanding the meaning of fractions and percents
• Comparing fractions
• Understanding the meaning of decimal fractions
• Comparing decimal fractions
Computation with Rational Numbers
Benchmarks
• Use fraction-percent equivalents to solve problems about the percentage of a
quantity
• Order fractions with like and unlike denominators
• Read, write, and interpret decimal fractions to thousandths
• Order decimals to the thousandths
representations
Patterns, Functions, and Change
In Grade 5, students continue their work from Grades 3 and 4 by examining,
representing, and describing situations in which the rate of change is constant . Students
create tables and graphs to represent the relationship between two variables in a variety of
contexts. They also articulate general rules for each situation. For example, consider the
perimeters of the following set of rectangles made from rows of tiles with three tiles in
each row:
If the value of one variable (the number of rows of three tiles) is known, the
corresponding value of the other variable (the perimeter of the rectangle) can be
calculated
. Students express these rules in words and then in symbolic notation. For
example:
For the first time in Grade 5, students create graphs for situations in which the rate of
change is itself changing–for example, the change in the area of a square as a side
increases by a constant increment–and consider why the shape of the graph is not a
straight line as it is for situations with a constant rate of change.
Throughout their work, students move among tables, graphs, and equations and between
those representations and the situation they represent. Their work with symbolic notation
is closely related to the context in which they are working. By moving back and forth
between the contexts, their own ways of describing general rules in words, and symbolic
notation, students learn how this notation can carry mathematical meaning .
Emphases
Using Tables and Graphs
• Using graphs to represent change
• Using tables to represent change
Linear Change
• Describing and representing a constant rate of change
Number Sequences
• Describing and representing situations in which the rate of change is not constant
Benchmarks
• Connect tables and graphs to represent the relationship between two variables
• Use tables and graphs to compare two situations with constant rates of change
• Use symbolic notation to represent the value of one variable in terms of another
variable in situations with constant rates of change
Data Analysis and Probability
Students continue to develop their understanding of data analysis in Grade 5 by
collecting, representing, describing, and interpreting numerical data . Students’ work in
this unit focuses on comparing two sets of data collected from experiments. Students
develop a question to compare two groups, objects, or conditions. (Sample questions:
Which toy car goes farthest after rolling down the ramp? Which paper bridge holds more
weight?). They consider how to ensure a consistent procedure for their experiment and
discuss the importance of multiple trials . Using representations of data, including line
plots and bar graphs, students describe the shape of the data—where the data are
concentrated, how they are spread across the range. They summarize the data for each
group or object or condition and use these summaries, including medians, to back up their
conclusions and arguments. By carrying out a complete data investigation, from
formulating a question through drawing conclusions from their data, students gain an
understanding of data analysis as a tool for learning about the world.
In their work with probability, students describe and predict the likelihood of events and
compare theoretical probabilities with actual outcomes of many trials. They use fractions
to express the probabilities of the possible outcomes (e.g., landing on the green part of the
spinner, landing on the white part of the spinner). Then they conduct experiments to see
experimental probability, for example, if half the area of a spinner is colored green and
half is colored white, why doesn’t the spinner land on green exactly half the time?
Emphases
Data Analysis
• Representing data
• Describing, summarizing, and comparing data
• Analyzing and interpreting data
• Designing and carrying out a data investigation
Probability
• Describing the probability of an event
• Describe major features of a set of data represented in a line plot or bar graph, and
quantify the description by using medians or fractional parts of the data
Benchmarks
• Draw conclusions about how 2 groups compare based on summarizing the data
for each group
• Design and carry out an experiment in order to compare two groups
• Use a decimal, fraction, or percent to describe and compare the theoretical
probabilities of events with a certain number of equally likely outcomes
Geometry and Measurement
In their work with geometry and measurement in grade 5, students further develop their
understanding of the attributes of two-dimensional (2-D) shapes, find the measure of
angles of polygons, determine the volume of three-dimensional (3-D) shapes, and work
with area and perimeter. Students examine the characteristics of polygons, including a
the classification of geometric figures, for example:
Are all squares rectangles?
Are all rectangles parallelograms?
If all squares are rhombuses, then are all rhombuses squares?
They investigate angle sizes in a set of polygons and measure angles of 30, 45, 60, 90,
120, and 150 degrees by comparing the angles of these shapes. Students also investigate
perimeter and area. They consider how changes to the shape of a rectangle can affect one
of the measures and not the other (e.g., two shapes that have the same area don’t
necessarily have the same perimeter), and examine the relationship between area and
perimeter in similar figures.
Students continue to develop their visualization skills and their understanding of the
relationship between 2-D pictures and the 3-D objects they represent. Students determine
the volume of boxes (rectangular prisms) made from 2-D patterns and create patterns for
boxes to hold a certain number of cubes. They develop strategies for determining the
number of cubes in 3-D arrays by mentally organizing the cubes—for example as a stack
of three rectangular layers, each three by four cubes. Students deepen their understanding
of the relationship between volume and the linear dimensions of length, width, and
height. Once students have developed viable strategies for finding the volume of
rectangular prisms, they extend their understanding of volume to other solids such as
pyramids, cylinders, and cones, measured in cubic units.
Emphases
Features of Shape
• Describing and classifying 2-D figures
• Describing and measuring angles
• Creating and describing similar shapes
• Translating between 2-D and 3-D shapes
Linear and Area Measurement
• Finding the perimeter and area of rectangles
Volume
• Structuring rectangular prisms and determining their volume
• Structuring prisms, pyramids, cylinders, and cones and determining their volume
Benchmarks
|
An Illustrated Proof of the First Group Isomorphism Theorem
# An Illustrated Proof of the First Group Isomorphism Theorem
Recall from The First Group Isomorphism Theorem page that The First Group Isomorphism Theorem (also known as The Fundamental Theorem of Group Homomorphisms) states that if $G$ and $H$ are groups and if $\phi : G \to H$ is a group homomorphism from $G$ to $H$ then:
(1)
\begin{align} \quad G / \ker (\phi) \cong \phi(G) \end{align}
We will now describe this very important result intuitively - using pictures, but first, we need the following proposition:
Proposition 1: Let $G$ and $H$ be groups and let $\phi : G \to H$ be a homomorphism from $G$ to $H$. Let $g \in G$. Then $x \in g \ker(\phi)$ if and only if $\phi(x) = \phi(g)$.
• Proof: $\Rightarrow$ Let $x \in g \ker (\phi)$. Then $x = gh$ for some $h \in \ker (\phi)$. Since $\phi$ is a homomorphism, we have that:
(2)
\begin{align} \quad \phi(x) = \phi(gh) = \phi(g) \phi(h) = \phi(g) e_H = \phi(g) \end{align}
• $\Leftarrow$ Let $\phi(x) = \phi(g)$. Observe that $x = gg^{-1}x$. Let $h = g^{-1}x$. Then:
(3)
\begin{align} \quad \phi(h) = \phi(g^{-1}x) = \phi(g^{-1}) \phi(x) = \phi(g)^{-1} \phi(x) = \phi(g^{-1}) \phi(g) = e_H \end{align}
• So $h \in \ker (\phi)$. Thus $x = gh \in g\ker(\phi)$. $\blacksquare$
Let $G$ and $H$ be groups and let $\phi : G \to H$ be a group homomorphism of $G$ to $H$. Recall that $\ker (\phi)$ is always a normal subgroup of $G$, and so we can construct the quotient $G / \ker (\phi)$ without any problems - that is, $G / \ker (\phi)$ is the group of all left cosets of $\ker (\phi)$ in $G$.
Now, for each $x \in G$ we have that either $x \in \ker (\phi)$ or $x \not \in \ker (\phi)$.
By Proposition 1, we see that $x \in \ker (\phi)$ ($= e_G \ker (\phi)$) if and only if $\phi(x) = e_H$ ($= \phi(e_G)$). Furthermore, Proposition 1 tells us that $x \in g \ker (\phi)$ if and only if $\phi(x) = \phi(g)$.
Now note again that $G / \ker(\phi)$ is the set of all left cosets of $\ker (\phi)$ in $G$. That is:
(4)
\begin{align} \quad G / \ker (\phi) = \{ g \ker (\phi) : g \in G \} \end{align}
So the cosets in the quotient $G / \ker (\phi)$ are really sets of the form $\{ \phi^{-1}(h) : h \in H \}$. Now suppose that we represent the group $G$ as a circle, and represent elements of $G$ as dots:
Recall that the cosets of $\ker (\phi)$ in $G$ partition $G$. This is evident as we see that each coset of $\ker(\phi)$ is the preimage of a singleton set in $\phi(G) \subseteq H$. If $x, y \in g \ker(\phi)$ then $\phi(x) = \phi(y) = \phi(g) \in \phi(G)$. So we can colour each element of $G$ such that two elements of $G$ are the same colour if and only if they are mapped to the same element in $\phi(G)$:
From the following diagram, it is clear now that the cosets of $\ker(\phi)$, i.e., $G / \ker(\phi)$ are in bijection with $\phi(G)$ (the range of $\phi$):
Furthermore, the map $\psi : G / \ker(\phi) \to \phi(G)$ defined for all $g \ker(\phi) \in G / \ker(\phi)$ by $\psi(g \ker(\phi)) = \phi(G)$ is that bijection - and is also a homomorphism, i.e., $G / \ker(\phi)$ is isomorphic to $\phi(G)$.
|
# Proportion word problems
There are lots of situations that can create proportion word problems. We will illustrate these situations with some examples
Problem # 1
Mix 3 liters of water with 4 lemons to make lemonade. How many liters of water are mixed with 8 lemons.
Set up the ratios, but make sure that the two ratios are written in the same order.
For example, all the followings can be used to solve this problem:
Let x be number of liters of water.
3 / 4 = x / 8 , 4 / 3 = 8 / x , 3 / x = 4 / 8 , x / 3 = 8 / 4 ,
3 / 4 = x / 8 , 4 / 3 = 8 / x , 3 / x = 4 / 8 , x / 3 = 8 / 4
It is very important to notice that if the ratio on the left is a ratio of number of liters of water to number of lemons, you have to do the same ratio on the right before you set them equal.
Look carefully and you will see that this is what the first proportion does.
However, the second proportion focuses on a ratio of number of lemons to number of liter of water
Number of lemons / Number of liters of water = Number of lemons / Number of liters of water
When solving proportion word problems, make sure it is set up correctly. Once you set up your proportion correctly, all you have to do if to replace values that you know and use an x or any other variable for the value you don't know
Let us solve the second proportion. I already showed you how to solve a proportion. If you do not remember, go to solving proportions
4 / 3 = 8 / x
# of lemons / # of liters of water = # of lemons / # of liters of water
When solving proportion word problems, make sure it is set up correctly. Once you set up your proportion correctly, all you have to do if to replace values that you know and use an x or any other variable for the value you don't know
Let us solve the second proportion. I already showed you how to solve a proportion. If you do not remember, go to solving proportions
4 / 3 = 8 / x
Cross product is usually used to solve proportion word problems. If you do a cross product, you will get:
4 × x = 3 × 8
4 × x = 24. Since 4 × 6 = 24, x = 6
6 liters should be mixed with 8 lemons
Problem # 2
A boy who is 3 feet tall can cast a shadow on the ground that is 7 feet long. How tall is a man who can cast a shadow that is 14 feet long?
Set up the proportion by doing ratios of height to length of shadow
Height of boy / Length of shadow = Height of man / Length of shadow
Replace the known values and use H for the unknown height of the man
3 / 7 = H / 14
Cross multiply
3 × 14 = 7 × H
42 = 7 × H
Since 7 × 6 = 42, H = 6
The man is 6 feet tall
Problem # 3
3 gallons of paint cover 900 square feet. How many gallons will cover 300 square feet?
3 / 900 = x / 300 900 / 3 = 300 / x 3 / x = 900 / 300 x / 3 = 300 / 900
We will solve the first one:
3 / 900 = x / 300
Doing cross product, will give you 3 × 300 = x × 900
900 = x × 900. Thus x = 1 because 1 × 900 = 900.
Problem # 4: Firefighter math and proportion
A firefighter truck can hold 3000 gallons of water. A firefighter can deliver 160 gallons of water every 2 minutes.
1. How much water will be delivered in 10 minutes?
2. How long will it take for the firefighter to empty the tank
1. Set up a proportion by doing ratios of number of gallons to time it takes
Number of gallons / Time it takes = Number of gallons / Time it takes
Replace the known values and use G to represent the numbers of gallons in 10 minutes
160 / 2 = G / 10
Cross multiply
160 × 10 = 2 × G
1600 = 2 × G
Since 2 × 800 = 1600, G = 800
800 gallons of water will be delivered in 10 minutes
2. Set up a proportion by doing ratios of number of gallons to time it takes
Number of gallons / Time it takes = Number of gallons / Time it takes
Replace the known values and use T to represent the time it takes to deliver 3000 gallons
160 / 2 = 3000 / T
Cross multiply
160 × T = 2 × 3000
160 × T = 6000
Divide 6000 by 160 to get T. 6000 divided by 160 = 37.5
T = 37.5 or 37 minutes and 30 seconds
I welcome any questions about these proportion word problems if you have any
Height of boy / Length of shadow = Height of man / Length of shadow
Replace the known values and use H for the unknown height of the man
3 / 7 = H / 14
Cross multiply
3 × 14 = 7 × H
42 = 7 × H
Since 7 × 6 = 42, H = 6
The man is 6 feet tall
Problem # 3
3 gallons of paint cover 900 square feet. How many gallons will cover 300 square feet?
3 / 900 = x / 300 , 900 / 3 = 300 / x , 3 / x = 900 / 300 , x / 3 = 300 / 900
We will solve the first one:
3 / 900 = x / 300
Doing cross product, will give you 3 × 300 = x × 900
900 = x × 900. Thus x = 1 because 1 × 900 = 900.
Problem # 4: Firefighter math and proportion
A firefighter truck can hold 3000 gallons of water. A firefighter can deliver 160 gallons of water every 2 minutes.
1. How much water will be delivered in 10 minutes?
2. How long will it take for the firefighter to empty the tank
1. Set up a proportion by doing ratios of number of gallons to time it takes
# of gallons / Time it takes = # of gallons / Time it takes
Replace the known values and use G to represent the numbers of gallons in 10 minutes
160 / 2 = G / 10
Cross multiply
160 × 10 = 2 × G
1600 = 2 × G
Since 2 × 800 = 1600, G = 800
800 gallons of water will be delivered in 10 minutes
2. Set up a proportion by doing ratios of number of gallons to time it takes
# of gallons / Time it takes = # of gallons / Time it takes
Replace the known values and use T to represent the time it takes to deliver 3000 gallons
160 / 2 = 3000 / T
Cross multiply
160 × T = 2 × 3000
160 × T = 6000
Divide 6000 by 160 to get T. 6000 divided by 160 = 37.5
T = 37.5 or 37 minutes and 30 seconds
I welcome any questions about these proportion word problems if you have any
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## Recent Lessons
1. ### Conjugate in Algebra
Oct 16, 17 09:34 AM
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Tough Algebra Word Problems.
If you can solve these problems with no help, you must be a genius!
Everything you need to prepare for an important exam!
K-12 tests, GED math test, basic math tests, geometry tests, algebra tests.
Real Life Math Skills
Learn about investing money, budgeting your money, paying taxes, mortgage loans, and even the math involved in playing baseball.
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# Find the amount and the compound interest on Rs 160000 for 2 years
Question:
Find the amount and the compound interest on Rs 160000 for 2 years at 10% per annum, compounded half-yearly.
Solution:
Principal, $P=$ Rs. 160000
Annual rate of interest, $R=10 \%$
Rate of interest for a half year $=\frac{10}{2} \%=5 \%$
Time, $n=2$ years $=4$ half years
Then the amount with the compound interest is given by
$A=P \times\left(1+\frac{R}{100}\right)^{n}$
$=160000 \times\left(1+\frac{5}{100}\right)^{4}$
$=160000 \times\left(\frac{100+5}{100}\right)^{4}$
$=160000 \times\left(\frac{105}{100}\right)^{4}$
$=160000 \times\left(\frac{21}{20}\right) \times\left(\frac{21}{20}\right) \times\left(\frac{21}{20}\right) \times\left(\frac{21}{20}\right)$
$=(21 \times 21 \times 21 \times 21)$
$=$ Rs 194481
Therefore, compound interest $=$ amount $-$ principal $=\mathrm{Rs}(194481-160000)=\mathrm{Rs} 34481$
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Theory:
You can use the following procedure if you know that the given number is a cube number.
Step 1: Choose any cube number.
Step 2: Start by forming groups of three digits from the number's rightmost digit. We get more than one group.
$\begin{array}{l}\underset{¯}{\phantom{\rule{0.147em}{0ex}}\mathit{abc}\phantom{\rule{0.147em}{0ex}}}\\ \phantom{\rule{0.147em}{0ex}}↓\end{array}$ $\begin{array}{l}\underset{¯}{\phantom{\rule{0.147em}{0ex}}\mathit{def}\phantom{\rule{0.147em}{0ex}}}\\ \phantom{\rule{0.147em}{0ex}}↓\end{array}$ Second group First group
Step 3: The first group will give you the ones (or unit) digit of the required cube root. You can calculate this using the ending digit of the first group.
Step 4: Now, take the second group and find which two cube numbers this group lies. Choose the smallest cube number and find the cube root of that number. This is the tens place of the required cube root.
Step 5: Join the numbers we get in the first group and second group. This is the required cube root of the given number.
Example:
Find the cube root of the cube number $$79507$$.
Solution:
Step 1: The number $$79507$$ is a cube number.
Step 2: Now, split this number into two groups.
$\begin{array}{l}\underset{¯}{\phantom{\rule{0.147em}{0ex}}79\phantom{\rule{0.147em}{0ex}}}\\ \phantom{\rule{0.147em}{0ex}}↓\end{array}$ $\begin{array}{l}\underset{¯}{\phantom{\rule{0.147em}{0ex}}507\phantom{\rule{0.147em}{0ex}}}\\ \phantom{\rule{0.147em}{0ex}}↓\end{array}$ Second group First group
Step 3: Let us take the first group, $$507$$. We know that "if a cube ends at $$7$$, then its cube root ends at $$3$$".
So, the ones place of the required number is $$3$$.
Step 4: The second group is $$79$$.
$$64 < 79 < 125$$
$$4^3 < 79 < 5^3$$
Here, the smallest cube number is $$64$$, and the cube root of $$64$$ is $$4$$.
So, the tens place of the required cube root is $$4$$.
Step 5: The required cube root is $$43$$.
Therefore, $$\sqrt[3]{79507}$$ $$=$$ $$43$$.
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# Form 1 Unit 15 Lesson 1 – Volume of Solids
### Objectives
At the end of this lesson, student should be able to:
• Recognize the need for cubic metres
• Understand that measurements of and volume are based on standard units.
• Convert between metric unit of volume
Volume is a measure of the amount of space that a three – dimensional shape occupies. Thus length × breadth × height = volume
Let’s take a look at a diagram of a prism and how volume is calculated.
Volume = length × breadth × height
= 4units × 3 units × 2 units
= 24 units3
To understand what volume means, let’s start by filling the bottom of the prism with unit cubes. This means the bottom of the prism will act as a box will hold as many as possible without stacking them on top of each other. This is what it will look like.
The diagram above does not yet represent the total volume. It only represents a partial volume, but we need to count these cubes to arrive at the total volume.
Since the length is 4 units and the breadth is 3 units, the bottom layer of cube measure rows and columns that are 3units by 4units. So, a quick way of counting the blocks would be to multiply 3 × 4 = 12 units2.
We could stack cubes on top each other until the prism is completely filled. It would be filled so that all cubes are touching each other such that no space existed between cubes. It would look like this.
To count all the cubes above, we already know there are 12 cubes on the bottom level and all levels contain the exact number of cubes. Therefore we need only to take the bottom total of 12 × 2 (since there are 2 levels).
Volume = 12 × 2 = 24 total
Note: If we review our calculations, we find that the total bottom layer of cubes was found by multiplying the prisms length by its breadth. Then we took the result and multiplied it by the prism’s height. So,
Volume of prisms = length × breadth × height
Example 1
Give the volume of the figure below in cubic unit
Solution
a)One layer of eight
Volume = 8 cubic unit
b) One layer of 5
Volume = 5 cubic unit
c) Top layer = 3
Middle layer = 2
Bottom layer = 3
Volume = 9 cubic unit
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