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Exercise: What is the Fourier transform of the function $f(x)=0$ ?
The Fourier transform of the function $f(x)=0$ is also zero. Since the function $f(x)$ is identically zero, its Fourier transform will also be zero at all frequencies. In other words, all the frequency components of the function $f(x)$ are absent, resulting in a zero Fourier transform.
(a) By expanding both sides as Fourier Sin Series, show that the solution to the equation
$$\frac{d^2 y}{d x^2}+y=2 x$$
with boundary conditions $y(x=0)=0, y(x=1)=0$ is
$$y(x)=\frac{4}{\pi} \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n\left(1-n^2 \pi^2\right)} \sin (n \pi x) .$$
(b) Show that the r.m.s. value of $y(x)$ is
$$\sqrt{\left\langle y^2(x)\right\rangle}=\frac{4}{\pi} \sqrt{\frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{n^2\left(1-n^2 \pi^2\right)^2}}$$
(a) To find the solution to the given differential equation, we expand both sides as a Fourier sine series. The general form of a Fourier sine series is given by:
$$y(x) = \sum_{n=1}^{\infty} B_n \sin(n \pi x)$$
where $B_n$ are the Fourier coefficients.
Taking the second derivative of $y(x)$ with respect to $x$, we have:
$$\frac{d^2 y}{d x^2} = -\sum_{n=1}^{\infty} B_n (n \pi)^2 \sin(n \pi x)$$
Substituting the expressions for $\frac{d^2 y}{d x^2}$ and $y$ into the differential equation, we get:
$$-\sum_{n=1}^{\infty} B_n (n \pi)^2 \sin(n \pi x) + \sum_{n=1}^{\infty} B_n \sin(n \pi x) = 2x$$
Matching the coefficients of the same sine functions on both sides, we obtain the following equation for the Fourier coefficients $B_n$:
$$-(n \pi)^2 B_n + B_n = 2 \delta_{n,1}$$
where $\delta_{n,1}$ is the Kronecker delta.
Simplifying the equation, we find:
$$B_n = \frac{4 (-1)^{n+1}}{n (1 – n^2 \pi^2)}$$
Therefore, the solution to the given differential equation with the specified boundary conditions is:
$$y(x) = \frac{4}{\pi} \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n (1 – n^2 \pi^2)} \sin(n \pi x)$$
(b) The root mean square (r.m.s.) value of $y(x)$ can be calculated as follows:
$$\sqrt{\left\langle y^2(x)\right\rangle} = \sqrt{\frac{1}{1-0} \int_{0}^{1} y^2(x) dx}$$
Substituting the expression for $y(x)$ and performing the integration, we obtain:
$$\sqrt{\left\langle y^2(x)\right\rangle} = \sqrt{\frac{4}{\pi^2} \sum_{n=1}^{\infty} \frac{1}{n^2 (1 – n^2 \pi^2)^2}}$$
Therefore, the r.m.s. value of $y(x)$ is given by the equation:
$$\sqrt{\left\langle y^2(x)\right\rangle} = \frac{4}{\pi} \sqrt{\frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{n^2 (1 – n^2 \pi^2)^2}}$$
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# Cramer's Rule
## Solving systems of equations using ratios of determinants.
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Cramer's Rule
At your school book fair, paperbacks cost one price and hardcovers cost another. You buy 3 paperbacks and 2 hardcovers. Your total comes to $54. Your best friend buys 2 paperbacks and 4 hardcovers. His total comes to$76. How could you use a matrix to find the price of each type of book?
### Cramer's Rule
We have already learned how to solve systems of linear equations using graphing, substitution and linear combinations. In this section, we will explore one way to use matrices and determinants to solve linear systems.
#### Cramer’s Rule in Two Variables
Given the system: \begin{align*}ax+by &= e\\ cx+dy &= f,\end{align*} we can set up the matrix \begin{align*}A\end{align*} and solve for \begin{align*}x\end{align*} and \begin{align*}y\end{align*} as shown below:
\begin{align*}A = \begin{bmatrix} a & b\\ c & d\end{bmatrix}, \ x=\frac{\begin{vmatrix}{\color{red}e} & d\\ {\color{red}f} & b\end{vmatrix}}{det \ A}\end{align*} and \begin{align*}y = \frac{\begin{vmatrix}a & {\color{red}e}\\c & {\color{red}f}\end{vmatrix}}{det \ A},\end{align*} provided \begin{align*}|A| \ne 0\end{align*}.
Note: When \begin{align*}|A| = 0\end{align*}, there is no unique solution. We have to investigate further to determine if there are infinite solutions or no solutions to the system. Notice that there is a pattern here. The coefficients of the variable we are trying to find are replaced with the constants.
#### Cramer’s Rule in Three Variables
Given the system: \begin{align*}ax+by+cz &= j\\ dx+ey+fz &= k\\ gx+hy+iz &= l,\end{align*} we can set up the matrix \begin{align*}A\end{align*} and solve for \begin{align*}x\end{align*} and \begin{align*}y\end{align*} as shown below:
\begin{align*}A = \begin{bmatrix} a & b & c\\ d & e & f\\ g & h & i\end{bmatrix}\end{align*}
\begin{align*}x = \frac{\begin{vmatrix}{\color{red}j} & b & c\\{\color{red}k} & e & f\\{\color{red}l} & h & i\end{vmatrix}}{det \ A}\end{align*}, \begin{align*}y = \frac{\begin{vmatrix}a & {\color{red}j} & c\\d & {\color{red}k} & f\\g & {\color{red}l} & i\end{vmatrix}}{det \ A}\end{align*} and \begin{align*}z = \frac{\begin{vmatrix}a & b & {\color{red}j}\\d & e & {\color{red}k}\\g & h & {\color{red}l}\end{vmatrix}}{det \ A}\end{align*}
provided \begin{align*}|A| \ne 0\end{align*}. Once again, if the \begin{align*}|A| = 0\end{align*}, then there is no unique solution to the system. Once again there is a pattern here. The coefficients of the variable for which we are trying to solve are replaced with the constants.
Let's use Cramer's Rule to solve the following systems of equations.
1.
\begin{align*}3x-7y &=13\\ -5x+9y &=-19\end{align*}
Matrix \begin{align*}A\end{align*} is the matrix made up of the coefficients of \begin{align*}x\end{align*} and \begin{align*}y\end{align*}:
\begin{align*}A = \begin{bmatrix} 3 & -7\\ -5 & 9\end{bmatrix}\end{align*}
Now we can find the \begin{align*}det \ A = (3\cdot9)-(-7\cdot-5) = 27-35=-8\end{align*}
So, using the formulas above:
\begin{align*}x = \frac{\begin{vmatrix}{\color{red}13} & -7\\{\color{red}-19} & 9\end{vmatrix}}{-8}=\frac{(13\cdot9)-(-7\cdot-19)}{-8}=\frac{-16}{-8}=2\end{align*}
\begin{align*}y = \frac{\begin{vmatrix}3 & {\color{red}13}\\-5 & {\color{red}-19}\end{vmatrix}}{-8}=\frac{(3\cdot-19)-(13\cdot-5)}{-8}=\frac{-57-(-65)}{-8}=\frac{8}{-8}=-1\end{align*}
Therefore, the solution is (2, -1).
1.
\begin{align*}6x+3y &=-12\\ 2x+y &=20\end{align*}
Matrix \begin{align*}A\end{align*} is the matrix made up of the coefficients of \begin{align*}x\end{align*} and \begin{align*}y\end{align*}:
\begin{align*}A = \begin{bmatrix} 6 & 3\\ 2 & 1\end{bmatrix}\end{align*}
Now we can find the \begin{align*}det \ A = (6\cdot1)-(3\cdot2) = 6-6=0\end{align*}
Because the determinant is zero, there is no unique solution and we cannot solve the system further using Cramer's Rule. Looking at the system, we see that the left-hand side of the first equation is a multiple of the second equation, by 3. The right sides are not multiples of each other, therefore there is no solution.
1.
\begin{align*}2x+2y-z &=-7\\ 5x+y-2z &=-3\\ x-3y+2z &=21\end{align*}
Matrix \begin{align*}A\end{align*} is the matrix made up of the coefficients of \begin{align*}x, y\end{align*} and \begin{align*}z\end{align*}:
\begin{align*}\begin{bmatrix} 2 & 2 & -1\\ 5 & 1 & -2\\ 1 & -3 & 2\end{bmatrix}\end{align*}
Now we can find the determinant of matrix \begin{align*}A\end{align*}:
\begin{align*}det \ A &=\begin{vmatrix}2 & 2 & -1\\ 5 & 1 & -2\\ 1 & -3 & 2\end{vmatrix}\begin{matrix} 2 & 2\\ 5 & 1\\ 1 & -3\end{matrix}\\ &= [(2)(1)(2)+(2)(-2)(1)+(-1)(5)(-3)]-[(1)(1)(-1)+(-3)(-2)(2)+(2)(5)(2)]\\ &= [4-4+15]-[-1+12+20]\\ &=15-31\\ &=-16\end{align*}
Now using the formulas above, we can find \begin{align*}x, y\end{align*} and \begin{align*}z\end{align*} as follows:
\begin{align*}x = \frac{\begin{vmatrix}{\color{red}-7} & 2 & -1\\{\color{red}-3} & 1 & -2\\{\color{red}21} & -3 & 2\end{vmatrix}}{-16}=\frac{-32}{-16}=2 \qquad y = \frac{\begin{vmatrix}2 & {\color{red}-7} & -1\\5 & {\color{red}-3} & -2\\1 & {\color{red}21} & 2\end{vmatrix}}{-16}=\frac{48}{-16}=-3 \qquad z = \frac{\begin{vmatrix}2 & 2 & {\color{red}-7}\\5 & 1 & {\color{red}-3}\\1 & -3 & {\color{red}21}\end{vmatrix}}{-16}=\frac{-80}{-16}=5\end{align*}
So the solution is (2, -3, 5).
### Examples
#### Example 1
Earlier, you were asked how you could use a matrix to find the price of each type of book.
The system of linear equation represented by this situation is:
\begin{align*}3x + 2y = 54\\ 2x + 4y = 76\end{align*}
We can now set up a matrix and apply Cramer's rule to find the price of each type of book.
\begin{align*}A = \begin{bmatrix} 3 & 2\\ 2 & 4\end{bmatrix}\end{align*}
Now we can find the \begin{align*}det \ A = (3\cdot4)-(2\cdot2) = 12-4=8\end{align*}
So, using the formulas above:
\begin{align*}x = \frac{\begin{vmatrix}{\color{red}54} & 2\\{\color{red}76} & 4\end{vmatrix}}{8}=\frac{(54\cdot4)-(76\cdot2)}{8}=\frac{216-152}{8}=8\end{align*}
\begin{align*}y = \frac{\begin{vmatrix}3 & {\color{red}54}\\2 & {\color{red}76}\end{vmatrix}}{8}=\frac{(3\cdot76)-(2\cdot54)}{8}=\frac{(228-108)}{8}=15\end{align*}
Therefore, paperbacks cost $8 and hardcovers cost$15.
#### Example 2
Use Cramer's Rule to solve the system below.
\begin{align*}2x+5y &=7\\ x+3y &=2\end{align*}
Find the \begin{align*}det \ A = \begin{vmatrix} 2 & 5\\ 1 & 3\end{vmatrix} = 1.\end{align*} Now find \begin{align*}x\end{align*} and \begin{align*}y\end{align*} as follows:
\begin{align*}x= \frac{\begin{vmatrix} 7 & 5\\ 2 & 3 \end{vmatrix}}{1} = \frac{11}{1}=11 \qquad y= \frac{\begin{vmatrix} 2 & 7\\ 1 & 2 \end{vmatrix}}{1} = \frac{-3}{1}=-3, \quad \text{solution:} \ (11,-3)\end{align*}
#### Example 3
Use Cramer's Rule to solve the system below.
\begin{align*}4x-y &=6\\ -8x+2y &=10\end{align*}
Find the
\begin{align*}det \ A= \begin{vmatrix} 4 & -1 \\ -8 & 2 \end{vmatrix} = 8-8 = 0.\end{align*}
Therefore, there is no unique solution. We must use linear combination or the substitution method to determine whether there are an infinite number of solutions or no solutions. Using linear combinations we can multiply the first equation by 2 and get the following:
\begin{align*}& \quad 8x-2y=12 \\ & \ \underline{-8x+2y=10 \; \;} \ \Rightarrow \quad \text{Therefore, there is no solution}.\\ & \qquad \quad \ \ 0=22\end{align*}
#### Example 4
Use Cramer's Rule to solve the system below.
\begin{align*}x+2y+3z &=8\\ 2x-y+4z &=3\\ -x-4y+3z &=14\end{align*}
Find the
\begin{align*}det \ A=\begin{vmatrix} 1 & 2 & 3\\ 2 & -1 & 4\\ -1 & -4 & 3 \end{vmatrix}=-34.\end{align*}
Now find \begin{align*}x, y\end{align*} and \begin{align*}z\end{align*} as follows:
\begin{align*}x= \frac{\begin{vmatrix} 8 & 2 & 3\\ 3 & -1 & 4\\ 14 & -4 & 3 \end{vmatrix}}{-34} = \frac{204}{-34}=-6 \qquad y= \frac{\begin{vmatrix} 1 & 8 & 3\\ 2 & 3 & 4 \\ -1 & 14 & 3 \end{vmatrix}}{-34} = \frac{-34}{-34}=1 \qquad z= \frac{\begin{vmatrix} 1 & 2 & 8\\ 2 & -1 & 3\\ -1 & -4 & 14 \end{vmatrix}}{-34}=\frac{-136}{-34}=4\end{align*}
Therefore, the solution is (-6, 1, 4).
### Review
Solve the systems below using Cramer’s Rule. If there is no unique solution, use an alternate method to determine whether the system has infinite solutions or no solution.
1. \begin{align*}5x-y &= 22\\ -x+6y &= -16\end{align*}
2. \begin{align*}2x+5y &= -1\\ -3x-8y &= 1\end{align*}
3. \begin{align*}4x-3y &= 0\\ -6x+9y &= 3\end{align*}
4. \begin{align*}4x-9y &= -20\\ -5x+15y &= 25\end{align*}
5. \begin{align*}2x-3y &= -4\\ 8x+12y &= -24\end{align*}
6. \begin{align*}3x-7y &= 12\\ -6x+14y &= -24\end{align*}
7. \begin{align*}x+5y &= 8\\ -2x-10y &= 16\end{align*}
8. \begin{align*}2x-y &= -5\\ -3x+2y &= 13\end{align*}
9. \begin{align*}x+y &= -1\\ -3x-2y &= 6\end{align*}
Solve the systems below using Cramer’s Rule. You may wish to use your calculator to evaluate the determinants.
1. \begin{align*}3x-y+2z &= 11\\ -2x+4y+13z &= -3\\ x+2y+9z &=5\end{align*}
2. \begin{align*}3x+2y-7z &= 1\\ -4x-3y+11z &= -2\\ x+4y-z &=7\end{align*}
3. \begin{align*}6x-9y+z &= -6\\ 4x+3y-2z &= 10\\ -2x+6y+z &=0\end{align*}
4. \begin{align*}x-2y+3z &= 5\\ 4x-y+4z &= 14\\ 5x+2y-4z &=-3\end{align*}
5. \begin{align*}-3x+y+z &= 10\\ 2x+y-2z &= 15\\ -4x-2y+4z &=-20\end{align*}
6. The Smith and Jamison families go to the county fair. The Smiths purchase 6 corndogs and 3 cotton candies for $21.75. The Jamisons purchase 3 corndogs and 4 cotton candies for$15.25. Write a system of linear equations and solve it using Cramer’s Rule to find the price of each food.
To see the Review answers, open this PDF file and look for section 4.8.
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
Cramer's rule
Cramer's rule is a formula involving ratios of determinants for the solution of a system of linear equations.
determinant
The determinant is a single number descriptor of a square matrix. The determinant is computed from the entries of the matrix, and has many properties and interpretations explored in linear algebra.
matrix equation
A matrix equation represents a system of equations by multiplying a coefficient matrix and a variable matrix to get a solution matrix.
Sarrus’ rule
Sarrus’ rule is a memorization technique that enables you to compute the determinant of matrices efficiently.
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# QuickQuiz: Four-Digit Addition With Carrying
This activity was designed to help you practice addition. You're going to see a variety of problems that ask you to add four-digit numbers. All of the questions combine random values. This quiz works with addends in a stacked (up and down) format. This quiz will also test your carrying skills. Many of the numbers you add will have values that need to be carried to the tens, hundreds, and thousands columns. When you see the answer, you will see how the carrying was supposed to work.
Good luck and have fun.
## Directions
This is another NumberNut three-choice quiz. Once you start the activity you will see a math problem. To the right side or below that problem are three possible answers. It's your job to figure out the correct answer and click the answer. The next screen will show you the correct answer. Some of these quizzes will show you how to get the right answer.
You will get a happy or sad face for every question you finish. Once you finish ten (10) questions, the quiz will be over. Take the quiz again because all of the questions are random. Chances are, you'll get a new quiz every time. It's good practice to learn these basic arithmetic operations.
RELATED LINKS LESSONS: - NumberNut.com: Addition ACTIVITIES: - QuickQuiz: Single-Digit Addition(H) - QuickQuiz: Single-Digit Addition(V) - Number Lines: One-Digit Addition - Next in Series: 1, 4, and 8 Addition - Next in Series: 3, 7, and 9 Addition - Next in Series: 2, 5, and 10 Addition - More or Less: One-Digit Addition - Pick-a-Card: Single-Digit Word Problems - Number Lines: One/Two-Digit Addition - More or Less: One and Two-Digit Addition - Pick-a-Card: One/Two-Digit Word Problems - QuickQuiz: Three Term Addition - QuickQuiz: Two-Digit Addition (No Carrying) - QuickQuiz: Two-Digit Addition (No Carrying,V) - More or Less: Two-Digit Addition (No Carry) - QuickQuiz: Two-Digit Numbers (Carrying) - More or Less: Two-Digit Addition (Carry) - QuickQuiz: Three-Digit Addition (No Carrying) - QuickQuiz: Three-Digit Addition (Carrying) - QuickQuiz: Four-Digit Addition (No Carrying) - QuickQuiz: Four-Digit Addition (Carrying)
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# Finding Unknown Side Lengths Using Trig Ratios
Lesson
As we saw in our chapter on Special Ratios, there are 3 trigonometric ratios that relate an angle and sides together. They are:
Trigonometric ratios
$\sin\theta$sinθ = $\frac{Opposite}{Hypotenuse}$OppositeHypotenuse
$\cos\theta$cosθ = $\frac{Adjacent}{Hypotenuse}$AdjacentHypotenuse
$\tan\theta$tanθ = $\frac{Opposite}{Adjacent}$OppositeAdjacent
If we know any $2$2 parts of a right-angled triangle, whether that's $2$2 side lengths, or an angle and a side length we can then find any other part of that triangle using these trigonometric ratios.
## Finding Side lengths
If we know $2$2 sides and want to find the third, we would use pythagorean formula
Pythagoras
$a^2+b^2=c^2$a2+b2=c2
If we know 1 side length and an angle, we would use one of the trigonometric ratios.
The most common mistake is when the wrong ratio is used. We have to remember the ratios and the sides that apply to those ratios. For most students the mnemonic SOHCAHTOA can be a great help.
#### Examples
##### Question 1
In the given triangle $\theta=25^\circ$θ=25° and the hypotenuse measures $12.6$12.6. Solve for the length $b$b.
Think: We need to identify the sides we have and want with respect to the angle given. Here I can see that we have the hypotenuse (H) and we want $b$b, which is opposite (O) the angle. This means I have OH - so the trig ratio I need to use here is sine.
Do:
$\sin\theta$sinθ $=$= $\frac{O}{H}$OH $\sin25^\circ$sin25° $=$= $\frac{b}{12.6}$b12.6 $b$b $=$= $12.6\times\sin25^\circ$12.6×sin25° $b$b $=$= $5.32$5.32
##### Question 2
Find the length of the hypotenuse ($c$c) in the diagram, where the angle $36^\circ$36° and the side length of $4.8$4.8 are given.
Think: We need to identify the sides we have and want with respect to the angle given. Here I can see that we want the hypotenuse (H) and we have a side length of $4.8$4.8, which is adjacent (A) the angle. This means I have AH - so the trig ratio I need to use here is cosine.
Do:
$\cos\theta$cosθ $=$= $\frac{A}{H}$AH $\cos36^\circ$cos36° $=$= $\frac{4.8}{c}$4.8c $c$c $=$= $\frac{4.8}{\cos36^\circ}$4.8cos36° $c$c $=$= $5.93$5.93 units (to 2 d.p.)
##### Question 3
Find the length of the unknown side, when the angle is $66^\circ$66° and the indicated side length is $7.3$7.3
Think: We need to identify the sides we have and want with respect to the angle given. Here I can see that we want the adjacent side (A) and we have a side length of $7.3$7.3, which is opposite (O) the angle. This means I have OA - so the trig ratio I need to use here is tangent.
Do:
$\tan\theta$tanθ $=$= $\frac{O}{A}$OA $\tan66^\circ$tan66° $=$= $\frac{7.3}{a}$7.3a $a$a $=$= $\frac{7.3}{\tan66^\circ}$7.3tan66° $a$a $=$= $3.25$3.25 units (to 2 d.p.)
##### Question 4
Find the value of $f$f, correct to two decimal places.
##### Question 5
Find the value of $h$h, correct to two decimal places.
##### Question 6
Find the value of $x$x, the side length of the parallelogram, to the nearest centimetre. |
# Areas and Perimeters
Often, you will be given a geometric figure drawn on a coordinate system and will be asked to find its area or perimeter. In these problems, use the properties of the coordinate system to deduce the dimensions of the figure and then calculate the area or perimeter. For complicated figures, you may need to divide the figure into simpler forms, such as squares and triangles.
A couple examples will illustrate:
Example
What is the area of the quadrilateral in the coordinate system?
1. 2
2. 4
3. 6
4. 8
5. 11
Solution
If the quadrilateral is divided horizontally through the line y = 2, two congruent triangles are formed.
As the figure shows, the top triangle has height 2 and base 4.
Hence, its area is .
The area of the bottom triangle is the same, so the area of the quadrilateral is 4 + 4 = 8.
Example
What is the perimeter of Triangle ABC in the figure?
Solution
Point A has coordinates (0, 4), point B has coordinates (3, 0), and point C has coordinates (5, 1).
Adding these lengths gives the perimeter of Triangle ABC:Using the distance formula to calculate the distances between points A and B, A and C, and B and C yields |
# Solutions to a Math (Choke) Lesson in Gift Planning
We left off last time with a challenge to you, the reader, to solve a problem and also to check your solution.
The problem is a real-world gift planning problem. In terms of difficulty, it’s a simple first-year algebra problem
The solution of n, the number of shares to be donated, is given to be: n = 3,000/110, rounded to the nearest whole share. The easy way to solve for n is to do what the writer did. Which is to open up an Excel spreadsheet, format the spreadsheet to calculate to the nearest whole number (0 decimal places) and to type in a cell (any cell will do):
=3,000/110,
which returns 27. This means we’ve calculated the whole number of shares to be donated to be 27.
Now, in solving math problems, it’s often advisable to check your answer. To check the answer 27, we need to grasp that the answer was obtained by setting the tax savings generated by the number of shares donated (0.4n\$200) equal to the capital gain tax generated by the number of shares sold [0.2(100 – n)(\$200 – \$50)].
If we plug in 27 for n, we get \$2,182 for the charitable deduction tax savings and \$2,182 for the capital gain tax generated.
That is what we want. Checking shows that 27 is the correct answer for the number of shares to donate. (Important Note: Depending on how you “plug in 27 for n,” you may get slightly different numbers. For consistency, the writer performed all calculations using the same Excel spreadsheet. The slight difference in numbers has to do with how Excel rounds a number with a decimal part to the nearest whole number.)
Now if this were a real, live case, we’d face the fact that the value of the donor’s stock, assumed here to be \$200 per share, likely would be bouncing around day to day. To make our lives easy, we could create an Excel spreadsheet that would perform all the calculations we want based on whatever value we entered for the stock’s value.
Hope this is helpful.
by Jon Tidd, Esq
# A Math (Choke) Lesson in Gift Planning
Here are the facts: Donor owns 100 shares of ABC stock, which is publicly traded. Each share is worth \$200 and has a \$50 cost basis.
What Donor wants to do: Donor wants to donate some of the shares and sell the remaining shares. She wants the charitable deduction for the donation to offset exactly the capital gain on the sale. So that she walks away owing net zero capital gain tax.
Here’s the question: How many shares should she donate? How many shares should she sell?
We need some more information: We need to know the marginal income tax rate against which the charitable deduction will apply. Let’s assume it’s 40 percent (0.4). We also need to know the applicable capital gain tax rate. Let’s assume it’s 20 percent (0.2).
Here we go: It’s all a matter of logic.
1. Donor is going to give n shares. She is, therefore, going to sell (100 – n) shares. We want to find n.
2. The tax savings produced by the gift will be 0.4n(\$200), which is \$80n.
3. The capital gain tax for each share sold will be 0.2(\$200 – \$50), or 0.2(\$150), or \$30.
The capital gain tax for (100 – n) shares sold will be \$30(100 – n), or (\$3,000 – \$30n).
4. Now, Donor wants the tax savings from her gift to offset (be equal to) the capital gain tax on the shares sold.
This means: \$80n = \$3,000 – 30n.
5. Solving: \$110n = \$3,000. n = 3,000/110
You have two tasks: [1] Find n to the nearest whole share. [2] Check to see that this value of n achieves Donor’s goal.
Further discussion next time.
by Jon Tidd, Esq
# Gift Substantiation: Two Recent Tax Court Cases Are Revealing
What they reveal is how tough the IRS and the Tax Court have gotten on gift substantiation . . . gift receipt and “Qualified Appraisal” rules.
The first case involves a gift of real estate to the University of Michigan. The donor was a partnership that had purchased the real estate for \$2.5 million. More than one year after purchase, the partnership gave the real estate to the university. The partnership claimed a value of \$33 million for the property.
The partnership got an appraisal and filed a Form 8283. For some unknown reason, the Form 8283 failed to state the partnership’s cost basis in the donated property, which was \$2.5 million.
Held, this omission, all by itself, was sufficient to knock out the entire \$33 million charitable contribution claim(!). RERI Holdings, 149 T.C. #1 (2017)
The second case involves some out-of-pocket cash expenditures made by a volunteer on behalf of a charity (a little more than \$1,000 in each of 2012 and 2013). The expenditures related to mileage and the purchase of T-shirts for a youth group. The volunteer had a written mileage log and also a purchase receipt for the T-shirt, but the volunteer didn’t have a gift receipt from the charity.
Held, the failure to obtain a gift receipt (which needed to state whether the charity provided any goods or services to the volunteer in consideration of the expenditures) meant the volunteer was entitled to no federal income tax charitable deduction for the expenditures. Martinez, T.C. Summ. Op. 2017-42
Important Note: Such out-of-pocket expenditures are considered cash contributions to charity and need to be acknowledged accordingly.
Charitable Deductions: Revisiting the Basics” (Give & Take, January 2018)
Taxing Matters: Token Donor Gifts” (Give & Take, October 2015)
Gift Substantiation: The Tail That Wags the Dog” (blog post, September 8, 2015)
Avoiding Charitable Tax Traps” (Give & Take, April 2015)
Or contact your Sharpe Group representative at 901-680-5300, info@SHARPEnet.com.
by Jon Tidd, Esq.
# Let’s Look at Charitable IRA Gifts
The Charitable IRA gift is likely to be the gift of choice going forward for many American 70½ and older. “Going forward” means in the wake of the 2017 tax law changes.
One doesn’t need to itemize deductions to save taxes via a Charitable IRA gift (called a “qualified charitable distribution,” or QCD) … given that the QCD isn’t taxed to the donor, and also that the donor avoids having to take a required minimum distribution from the IRA to the extent of the QCD.
There are some problems with the QCD, however. Some examples:
The feds don’t tax the QCD to the donor, but New Jersey does.
The IRS hasn’t issued guidance on when the QCD is complete for federal income tax purposes. For example, lots of IRA custodians make the QCD check payable to a charity but mail the check to the donor … requiring the donor to send the check to the charity. IRS has said this is OK but hasn’t said when the QCD is deemed to be made.
In 2017, one IRA custodian made a check payable to the donor. The donor then proposed completing a QCD by endorsing the check over to a charity. Didn’t work. The check was income to the donor.
IRA check-writing privileges also can cause problems. For example, in December 2017, a donor wrote a check on his IRA account made payable to a charity. The donor hand-delivered the check to the charity also in December 2017. The charity didn’t deposit the check until January 2018, however, because it was short-staffed due to the year-end holidays. This one should be OK, should be a 2017 QCD; but the IRA custodian is going to report it as a 2018 transaction, because that’s all the information the custodian has.
If your organization is going to promote Charitable IRA gifts this year, which may be great idea, it should start doing so well before year-end.
For professional advice on how to market this way of giving, contact your Sharpe Group representative.
By Jon Tidd, Esq
# Answers to the Planning Problem
Here’s the gift planning problem from last time.
Don wants his name on Charity’s new clinic building. Charity’s president, Ron, has told Don it will cost him a big chunk of change, \$X, in cash or securities or in a combination of both. Don is just about to sign a pledge agreement to this effect when his lawyer whispers something to him.
Don pauses and then says to Ron, “How about if I set up a \$Y charitable remainder unitrust for myself and my wife instead. You all will get the entire trust remainder.” (\$Y > \$X) Don continues, “My wife and I will take a 7% payout for 10 years. Then you’ll get the money.”
Ron says, “Let me talk to my people. I’ll get back to you.”
Ron, who is clueless about such an arrangement, goes back to his office and calls Julie, Charity’s planned giving director. Ron describes to Julie his conversation with Don and asks Julie to prepare a written briefing on the matter. Julie, by the way, is a knowledgeable and experienced gift planner.
As Ron tells Julie, Don proposes to make a pledge of \$Y, get his name on the clinic building and pay the pledge over 10 years using a 7% unitrust.
• Q. Will the gift arrangement work as Ron describes it to Julie?
• A. No. Don would be making a legally enforceable pledge and then setting up a CRT to pay the pledge with the CRT remainder. That would be prohibited self-dealing, as we’ve seen.
• Q. Could the gift arrangement work as Don describes it to Ron?
• A. Yes. It is OK to make an enforceable pledge to create a CRT. Such a pledge, however, leaves open what the charity will receive from the CRT. An unanswered question is whether Don could make an additional “backstop” pledge to ensure that Charity receives at least \$Y. I think it is possible to craft such a pledge.
• Q. In retrospect, why should Julie have joined the meeting between Don and Ron?
• A. Ron, like many charity presidents, is a “big picture” thinker. He’s not the right person to nail down a big naming pledge where tax law details are critically important. Julie, a “detail person,” should have been at the meeting, at which she would have been a much better listener than Ron.
Both charity and donor should seek advice from a specialist on complicated gift matters.
by Jon Tidd, Esq
# Planning Problem
Now that we know all about pledges (Click to read Part One, Two, Three, Four, Five, Six), a gift planning problem.
Don wants his name on Charity’s new clinic building. Charity’s president, Ron, has told Don it will cost him a big chunk of change, \$X, in cash or securities or in a combination of both. Don is just about to sign a pledge agreement to this effect when his lawyer whispers something to him.
Don pauses and then says to Ron, “How about if I set up a \$Y charitable remainder unitrust for myself and my wife instead. You all will get the entire trust remainder.” (\$Y > \$X) Don continues, “My wife and I will take a 7% payout for 10 years. Then you’ll get the money.”
Ron says, “Let me talk to my people. I’ll get back to you.”
Ron, who is clueless about such an arrangement, goes back to his office and calls Julie, Charity’s planned giving director. Ron describes to Julie his conversation with Don and asks Julie to prepare a written briefing on the matter. Julie, by the way, is a knowledgeable and experienced gift planner.
As Ron tells Julie, Don proposes to make a pledge of \$Y, get his name on the clinic building and pay the pledge over 10 years using a 7% unitrust.
Questions:
1. Will the gift arrangement work as Ron describes it to Julie?
2. Could the gift arrangement work as Don describes it to Ron?
3. In retrospect, why should Julie have joined the meeting between Don and Ron?
By: Jon Tidd, Esq
# Let’s Look at Some Key IRS Rulings, Part 6
We continue looking at IRS rulings on pledges (read Part 5 here). First up is a 1981 Revenue Ruling, Rev. Rul. 81-110.
The facts here are that Party A made a legally enforceable pledge to a charity. Subsequently, Party B paid the pledge.
Focusing on the fact that Party B’s payment relieved Party A of a legal obligation, the IRS ruled that:
• Party B’s payment was functionally a payment (a gift) to Party A.
• Therefore, Party B was not entitled to a charitable deduction for the payment.
• Party A was entitled to the charitable deduction for the payment.
Note how the IRS re-configures the transaction for federal income tax purposes. Is the IRS allowed to do this? The answer is yes. In fact, the reconfiguration of transactions according to their substance is one of the IRS’s primary tools.
This ruling, all by itself, appears to throw cold water on third-party payment of enforceable pledges.
• Yet the IRS has said it’s OK to use an “IRA rollover” distribution to pay an enforceable pledge.
• And at the very end of 2017, the IRS announced it was leaning toward allowing DAF distributions to be used to pay enforceable pledges.
Stay tuned.
by Jon Tidd, Esq
# Let’s Look at Some Key IRS Rulings, Part 5
We want now to look at some IRS rulings on pledges (Read Part 4 here). These rulings are important, as will become obvious. First, though, some background on pledges.
• A pledge is a promise to make a gift or gifts. A promise to do something else, such as to make and keep in force a specified will provision, may be a good and valuable promise, but it isn’t a pledge.
• A pledge is either enforceable against the donor or the donor’s estate or unenforceable. Enforceability depends on the law of the state that governs the pledge (enforceability varies from state to state). Enforceability has nothing to do with whether the charity would ever sue to enforce a pledge; it’s a matter of law, not policy.
Remember these points.
In Letter Ruling 8230156, an individual proposed paying an enforceable pledge with appreciated property (perhaps appreciated stock). The individual wanted assurance from the IRS that the payment wouldn’t cause him or her to realize capital gain. The individual was concerned because if one pays off a debt with appreciated property, one does realize capital gain (the debtor is deemed to sell the property). IRS ruled no gain would be realized, because according to the IRS an enforceable pledge isn’t a debt for federal income tax purposes.
That’s important and pretty interesting. It clears the way to do what many donors do—satisfy enforceable pledges with appreciated securities. It’s not end of the story on paying enforceable pledges, however, as we’ll see down the road when we consider the use of charitable remainder trusts and charitable lead trusts to pay such pledges.
We’ll dig deeper next time. Meantime, if you have a question about pledges, check with your Sharpe Group representative.
by Jon Tidd, Esq
Read Part 6 here.
# Let’s Look at Some Key IRS Rulings, Part 4
Before we leave our discussion of third-party buyers (buyers-in-the-wings), we need to look at one more IRS ruling, a private ruling from 1994 (LTR 9452026).
Read Part 1 here.
Read Part 2 here.
Read Part 3 here.
This is a good one. The facts are that an individual proposed transferring both securities and a valuable musical instrument to a charitable remainder annuity trust (CRAT). The individual wanted assurance from the IRS that if and when the CRAT sold the musical instrument, which was highly appreciated, he (the individual donor) wouldn’t be considered to have sold the instrument.
Why is this a good one? Because as a practical matter, the trustee of the CRAT is going to want to sell the musical instrument in order to meet the CRAT’s payout obligation. The trustee, however, was under no legal obligation to sell.
Given these facts, IRS ruled that the individual donor would not be deemed to sell the musical instrument when the instrument was sold by the CRAT. Meaning the donor wouldn’t realize any gain on the sale.
What we see here is important. The IRS is drawing a distinction between a legal obligation to sell and a clearly anticipated future sale. Anticipated not because the CRAT agreement requires the trustee to sell but because the trustee naturally is going to want to sell. (I understand the musical instrument was a Stradivarius violin, and the eventual buyer was a symphony orchestra.)
This ruling represents graduate-level gift planning. Such planning involves [a] attention to detail and [b] knowing where the IRS has drawn lines in the sand. As to detail, note that securities were transferred to the CRAT in addition to the musical instrument.
These were publicly traded securities—liquid assets. The securities bought the CRAT trustee time. Time to arrange for an orderly sale, time to avoid the need for a quick sale, of the musical instrument. This was a very good bit of fine tuning. It facilitated the IRS’s ruling the way it did.
Next time, we’ll switch gears and begin to look at some IRS rulings on pledges.
by Jon Tidd, Esq
Read Part 5 here.
# The Finish Line: Tax Law Update 12/22/17
Contrary to some news reports this week predicting a delay in the signing of the Tax Cuts and Jobs Act of 2017, President Trump officially signed the bill into law today. See “Trump Signs Tax Bill Into Law” from The Hill.
Now what?
Sharpe Group’s new white paper details how the new tax law will impact charitable giving. Click here to download.
We have a library of donor communication tools that are being updated with the latest tax law. Now may be a good time to refresh your Sharpe library. More info coming soon.
In addition, Sharpe Group experts are producing a new 16-page booklet with all the information your donors need to know. “Your Guide to Effective Giving After Tax Reform” will be available for orders in the coming weeks. We’ll post a preview of the content soon.
Our next Gift Planning Seminar is coming up January 22-23, 2018 in Memphis, TN. Get a first look at the impact of the new tax law on major and planned gifts at “Integrating Major and Planned Gifts”. Click here for more information and to register.
Check back here for updates on the availability of donor communication tools. |
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# 4.3: Inverse Trigonometric Properties
Difficulty Level: At Grade Created by: CK-12
## Learning Objectives
• Relate the concept of inverse functions to trigonometric functions.
• Reduce the composite function to an algebraic expression involving no trigonometric functions.
• Use the inverse reciprocal properties.
• Compose each of the six basic trigonometric functions with \begin{align*}\tan^{-1} x\end{align*}.
## Composing Trig Functions and their Inverses
In the Prerequisite Chapter, you learned that for a function \begin{align*}f(f^{-1}(x)) = x\end{align*} for all values of \begin{align*}x\end{align*} for which \begin{align*}f^{-1}(x)\end{align*} is defined. If this property is applied to the trigonometric functions, the following equations will be true whenever they are defined:
\begin{align*}\sin(\sin^{-1}(x)) = x && \cos(\cos^{-1}(x)) = x && \tan(\tan^{-1}(x)) = x\end{align*}
As well, you learned that \begin{align*}f^{-1}(f(x)) = x\end{align*} for all values of \begin{align*}x\end{align*} for which \begin{align*}f(x)\end{align*} is defined. If this property is applied to the trigonometric functions, the following equations that deal with finding an inverse trig. function of a trig. function, will only be true for values of \begin{align*}x\end{align*} within the restricted domains.
\begin{align*}\sin^{-1}(\sin(x)) = x && \cos^{-1}(\cos(x)) = x && \tan^{-1}(\tan(x)) = x\end{align*}
These equations are better known as composite functions and are composed of one trigonometric function in conjunction with another different trigonometric function. The composite functions will become algebraic functions and will not display any trigonometry. Let’s investigate this phenomenon.
Example 1: Find \begin{align*}\sin \left ( \sin^{-1} \frac{\sqrt{2}}{2} \right )\end{align*}.
Solution: We know that \begin{align*}\sin^{-1} \frac{\sqrt{2}}{2} = \frac{\pi}{4}\end{align*}, within the defined restricted domain. Then, we need to find \begin{align*}\sin \frac{\pi}{4}\end{align*}, which is \begin{align*}\frac{\sqrt{2}}{2}\end{align*}. So, the above properties allow for a short cut. \begin{align*}\sin \left ( \sin^{-1} \frac{\sqrt{2}}{2} \right ) = \frac{\sqrt{2}}{2}\end{align*}, think of it like the sine and sine inverse cancel each other out and all that is left is the \begin{align*}\frac{\sqrt{2}}{2}\end{align*}.
## Composing Trigonometric Functions
Besides composing trig functions with their own inverses, you can also compose any trig functions with any inverse. When solving these types of problems, start with the function that is composed inside of the other and work your way out. Use the following examples as a guideline.
Example 2: Without using technology, find the exact value of each of the following:
a. \begin{align*}\cos \left ( \tan^{-1} \sqrt{3} \right )\end{align*}
b. \begin{align*}\tan \left ( \sin^{-1}\left( -\frac{1}{2} \right ) \right )\end{align*}
c. \begin{align*}\cos (\tan^{-1} (-1))\end{align*}
d. \begin{align*}\sin \left ( \cos^{-1}\frac{\sqrt{2}}{2} \right )\end{align*}
Solution: For all of these types of problems, the answer is restricted to the inverse functions’ ranges.
a. \begin{align*}\cos \left ( \tan^{-1} \sqrt{3} \right )\end{align*}: First find \begin{align*}\tan^{-1} \sqrt{3}\end{align*}, which is \begin{align*}\frac{\pi}{3}\end{align*}. Then find \begin{align*}\cos \frac{\pi}{3}\end{align*}. Your final answer is \begin{align*}\frac{1}{2}\end{align*}. Therefore, \begin{align*}\cos \left ( \tan^{-1} \sqrt{3} \right ) = \frac{1}{2}\end{align*}.
b. \begin{align*}\tan \left ( \sin^{-1} \left ( -\frac{1}{2} \right ) \right ) = \tan \left ( -\frac{\pi}{6} \right ) = -\frac{\sqrt{3}}{3}\end{align*}
c. \begin{align*}\cos (\tan^{-1} (-1)) = \cos^{-1} \left ( -\frac{\pi}{4} \right ) = \frac{\sqrt{2}}{2}\end{align*}.
d. \begin{align*}\sin \left ( \cos^{-1} \frac{\sqrt{2}}{2} \right ) = \sin \frac{\pi}{4} = \frac{\sqrt{2}}{2}\end{align*}
## Inverse Reciprocal Functions
We already know that the cosecant function is the reciprocal of the sine function. This will be used to derive the reciprocal of the inverse sine function.
\begin{align*}y & = \sin^{-1} x\\ x & = \sin y\\ \frac{1}{x} & = \csc y\\ \csc^{-1} \frac{1}{x} & = y\\ \csc^{-1} \frac{1}{x} & = \sin^{-1} x\end{align*}
Because cosecant and secant are inverses, \begin{align*}\sin^{-1} \frac{1}{x} = \csc^{-1} x\end{align*} is also true.
The inverse reciprocal identity for cosine and secant can be proven by using the same process as above. However, remember that these inverse functions are defined by using restricted domains and the reciprocals of these inverses must be defined with the intervals of domain and range on which the definitions are valid.
\begin{align*}\sec^{-1} \frac{1}{x} = \cos^{-1} x \leftrightarrow \cos^{-1} \frac{1}{x} = \sec^{-1} x\end{align*}
Tangent and cotangent have a slightly different relationship. Recall that the graph of cotangent differs from tangent by a reflection over the \begin{align*}y-\end{align*}axis and a shift of \begin{align*}\frac{\pi}{2}\end{align*}. As an equation, this can be written as \begin{align*}\cot x = -\tan \left ( x-\frac{\pi}{2} \right )\end{align*}. Taking the inverse of this function will show the inverse reciprocal relationship between arccotangent and arctangent.
\begin{align*}y & = \cot^{-1} x\\ y & = -\tan^{-1} \left ( x-\frac{\pi}{2} \right )\\ x & = -\tan \left( y-\frac{\pi}{2} \right )\\ -x & = \tan \left ( y-\frac{\pi}{2} \right )\\ \tan^{-1} (-x) & = y-\frac{\pi}{2}\\ \frac{\pi}{2} + \tan^{-1} (-x) & = y\\ \frac{\pi}{2} - \tan^{-1} x & = y\end{align*}
Remember that tangent is an odd function, so that \begin{align*}\tan(-x) = -\tan(x)\end{align*}. Because tangent is odd, its inverse is also odd. So, this tells us that \begin{align*}\cot^{-1} x = \frac{\pi}{2} - \tan^{-1} x\end{align*} and \begin{align*}\tan^{-1} x =\frac{\pi}{2} - \cot^{-1} x\end{align*}. You will determine the domain and range of all of these functions when you graph them in the exercises for this section. To graph arcsecant, arccosecant, and arccotangent in your calculator you will use these conversion identities: \begin{align*}\sec^{-1} x = \cos^{-1} \frac{1}{x} , \csc^{-1} x = \sin^{-1} \frac{1}{x} , \cot^{-1} x = \frac{\pi}{2} -\tan^{-1} x\end{align*}. Note: It is also true that \begin{align*}\cot^{-1} x = \tan^{-1} \frac{1}{x}\end{align*}.
Now, let’s apply these identities to some problems that will give us an insight into how they work.
Example 3: Evaluate \begin{align*}\sec^{-1}\sqrt{2}\end{align*}
Solution: Use the inverse reciprocal property. \begin{align*}\sec^{-1} x = \cos^{-1} \frac{1}{x} \rightarrow \sec^{-1} \sqrt{2} = \cos^{-1} \frac{1}{\sqrt{2}}\end{align*}. Recall that \begin{align*}\frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2}\end{align*}. So, \begin{align*}\sec^{-1} \sqrt{2} = \cos^{-1} \frac{\sqrt{2}}{2}\end{align*}, and we know that \begin{align*}\cos^{-1} \frac{\sqrt{2}}{2} = \frac{\pi}{4}\end{align*}. Therefore, \begin{align*}\sec^{-1} \sqrt{2} = \frac{\pi}{4}\end{align*}.
Example 4: Find the exact value of each expression within the restricted domain, without a calculator.
a. \begin{align*}\sec^{-1} \sqrt{2} \end{align*}
b. \begin{align*}\cot^{-1} \left ( -\sqrt{3} \right )\end{align*}
c. \begin{align*}\csc^{-1} \frac{2\sqrt{3}}{3} \end{align*}
Solution: For each of these problems, first find the reciprocal and then determine the angle from that.
a. \begin{align*}\sec^{-1} \sqrt{2} = \cos^{-1} \frac{\sqrt{2}}{2} \end{align*} From the unit circle, we know that the answer is \begin{align*}\frac{\pi}{4}\end{align*}.
b. \begin{align*}\cot^{-1} \left ( -\sqrt{3} \right ) = \frac{\pi}{2} - \tan^{-1} \left ( -\sqrt{3} \right )\end{align*} From the unit circle, the answer is \begin{align*}\frac{5\pi}{6}\end{align*}.
c. \begin{align*}\csc^{-1} \frac{2\sqrt{3}}{3} = \sin^{-1} \frac{\sqrt{3}}{2} \end{align*} Within our interval, there are is one answer, \begin{align*}\frac{\pi}{3}\end{align*}.
Example 5: Using technology, find the value in radian measure, of each of the following:
a. \begin{align*}\arcsin 0.6384\end{align*}
b. \begin{align*}\arccos (-0.8126)\end{align*}
c. \begin{align*}\arctan (-1.9249)\end{align*}
Solution:
a.
b.
c.
## Composing Inverse Reciprocal Trig Functions
In this subsection, we will combine what was learned in the previous two sections. Here are a few examples:
Example 6: Without a calculator, find \begin{align*}\cos \left ( \cot^{-1} \sqrt{3} \right )\end{align*}.
Solution: First, find \begin{align*}\cot^{-1} \sqrt{3}\end{align*}, which is also \begin{align*}\tan^{-1}\frac{\sqrt{3}}{3}\end{align*}. This is \begin{align*}\frac{\pi}{6}\end{align*}. Now, find \begin{align*}\cos \frac{\pi}{6}\end{align*}, which is \begin{align*}\frac{\sqrt{3}}{2}\end{align*}. So, our answer is \begin{align*}\frac{\sqrt{3}}{2}\end{align*}.
Example 7: Without a calculator, find \begin{align*}\sec^{-1} \left ( \csc \frac{\pi}{3} \right )\end{align*}.
Solution: First, \begin{align*}\csc \frac{\pi}{3} = \frac{1}{\sin \frac{\pi}{3}} = \frac{1}{\frac{\sqrt{3}}{2}} = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3}\end{align*}. Then \begin{align*}\sec^{-1} \frac{2\sqrt{3}}{3} = \cos^{-1} \frac{\sqrt{3}}{2} = \frac{\pi}{6}\end{align*}.
Example 8: Evaluate \begin{align*}\cos \left ( \sin^{-1} \frac{3}{5} \right )\end{align*}.
Solution: Even though this problem is not a critical value, it can still be done without a calculator. Recall that sine is the opposite side over the hypotenuse of a triangle. So, 3 is the opposite side and 5 is the hypotenuse. This is a Pythagorean Triple, and thus, the adjacent side is 4. To continue, let \begin{align*}\theta = \sin^{-1} \frac{3}{5} \end{align*} or \begin{align*}\sin \theta = \frac{3}{5}\end{align*}, which means \begin{align*}\theta\end{align*} is in the Quadrant 1 (from our restricted domain, it cannot also be in Quadrant II). Substituting in \begin{align*}\theta\end{align*} we get \begin{align*}\cos \left ( \sin^{-1} \frac{3}{5} \right ) = \cos \theta\end{align*} and \begin{align*}\cos \theta = \frac{4}{5}\end{align*}.
Example 9: Evaluate \begin{align*}\tan \left ( \sin^{-1} \left ( -\frac{3}{4} \right ) \right )\end{align*}
Solution: Even though \begin{align*}\frac{3}{4}\end{align*} does not represent two lengths from a Pythagorean Triple, you can still use the Pythagorean Theorem to find the missing side. \begin{align*}(-3)^2 + b^2 = 4^2\end{align*}, so \begin{align*}b = \sqrt{16-9} = \sqrt{7}\end{align*}. From the restricted domain, sine inverse is negative in the \begin{align*}4^{th}\end{align*} Quadrant. To illustrate:
Let \begin{align*}\theta & = \sin^{-1} \left ( -\frac{3}{4} \right )\\ \sin \theta & = -\frac{3}{4}\\ \tan \left ( \sin^{-1} \left ( -\frac{3}{4} \right ) \right ) & = \tan \theta\\ \tan \theta & = \frac{-3}{\sqrt{7}} \ \text{or} \ \frac{-3\sqrt{7}}{7}\end{align*}
## Trigonometry in Terms of Algebra
All of the trigonometric functions can be rewritten in terms of only \begin{align*}x\end{align*}, when using one of the inverse trigonometric functions. Starting with tangent, we draw a triangle where the opposite side (from \begin{align*}\theta\end{align*}) is defined as \begin{align*}x\end{align*} and the adjacent side is 1. The hypotenuse, from the Pythagorean Theorem would be \begin{align*}\sqrt{x^2+1}\end{align*}. Substituting \begin{align*}\tan^{-1} x\end{align*} for \begin{align*}\theta\end{align*}, we get:
\begin{align*}\tan \theta & = \frac{x}{1}\\ \tan \theta & = x && hypotenuse = \sqrt{x^2+1}\\ \theta & = \tan^{-1} x\end{align*}
\begin{align*}\sin (\tan^{-1}x) & = \sin \theta = \frac{x}{\sqrt{x^2+1}} && \csc (\tan^{-1}x) = \csc \theta = \frac{\sqrt{x^2+1}}{x}\\ \cos (\tan^{-1}x) & = \cos \theta = \frac{1}{\sqrt{x^2+1}} && \sec (\tan^{-1}x) = \sec \theta = \sqrt{x^2+1}\\ \tan (\tan^{-1}x) & = \tan \theta = x && \cot (\tan^{-1}x) = \cot \theta = \frac{1}{x}\end{align*}
Example 10: Find \begin{align*}\sin (\tan^{-1} 3x)\end{align*}.
Solution: Instead of using \begin{align*}x\end{align*} in the ratios above, use \begin{align*}3x\end{align*}.
\begin{align*}\sin (\tan^{-1} 3x) = \sin \theta = \frac{3x}{\sqrt{(3x)^2 + 1}} = \frac{3x}{\sqrt{9x^2+1}}\end{align*}
Example 11: Find \begin{align*}\sec^2 (\tan^{-1} x)\end{align*}.
Solution: This problem might be better written as \begin{align*}[\sec (\tan^{-1} x)]^2\end{align*}. Therefore, all you need to do is square the ratio above.
\begin{align*}[\sec (\tan^{-1}x)]^2 = \left ( \sqrt{x^2+1} \right )^2 = x^2 + 1\end{align*}
You can also write the all of the trig functions in terms of arcsine and arccosine. However, for each inverse function, there is a different triangle. You will derive these formulas in the exercise for this section.
## Points to Consider
• Is it possible to graph these composite functions? What happens when you graph \begin{align*}y = \sin (\cos^{-1} x)\end{align*} in your calculator?
• Do exact values of functions of inverse functions exist if any value is used?
## Review Questions
1. Find the exact value of the functions, without a calculator, over their restricted domains.
1. \begin{align*}\cos^{-1} \frac{\sqrt{3}}{2} \end{align*}
2. \begin{align*}\sec^{-1} \sqrt{2} \end{align*}
3. \begin{align*}\sec^{-1} \left ( -\sqrt{2} \right )\end{align*}
4. \begin{align*}\sec^{-1} (-2)\end{align*}
5. \begin{align*}\cot^{-1} (-1)\end{align*}
6. \begin{align*}\csc^{-1} \left ( \sqrt{2} \right )\end{align*}
2. Use your calculator to find:
1. \begin{align*}\arccos (-0.923)\end{align*}
2. \begin{align*}\arcsin 0.368\end{align*}
3. \begin{align*}\arctan 5.698\end{align*}
3. Find the exact value of the functions, without a calculator, over their restricted domains.
1. \begin{align*}\csc \left ( \cos^{-1} \frac{\sqrt{3}}{2} \right )\end{align*}
2. \begin{align*}\sec^{-1} ( \tan (\cot^{-1} 1))\end{align*}
3. \begin{align*}\tan^{-1} \left ( \cos \frac{\pi}{2} \right )\end{align*}
4. \begin{align*}\cot \left ( \sec^{-1} \frac{2\sqrt{3}}{3} \right )\end{align*}
4. Using your graphing calculator, graph \begin{align*}y = \sec^{-1} x\end{align*}. Sketch this graph, determine the domain and range, \begin{align*}x-\end{align*} and/or \begin{align*}y-\end{align*}intercepts. (Your calculator knows the restriction on this function, there is no need to input it into \begin{align*}Y =\end{align*}.)
5. Using your graphing calculator, graph \begin{align*}y = \csc^{-1} x\end{align*}. Sketch this graph, determine the domain and range, \begin{align*}x-\end{align*} and/or \begin{align*}y-\end{align*}intercepts. (Your calculator knows the restriction on this function, there is no need to input it into \begin{align*}Y =\end{align*}.)
6. Using your graphing calculator, graph \begin{align*}y = \cot^{-1} x\end{align*}. Sketch this graph, determine the domain and range, \begin{align*}x-\end{align*} and/or \begin{align*}y-\end{align*}intercepts. (Your calculator knows the restriction on this function, there is no need to input it into \begin{align*}Y =\end{align*}.)
7. Evaluate:
1. \begin{align*}\sin \left ( \cos^{-1} \frac{5}{13} \right )\end{align*}
2. \begin{align*}\tan \left ( \sin^{-1} \left( -\frac{6}{11} \right) \right )\end{align*}
3. \begin{align*}\cos \left ( \csc^{-1} \frac{25}{7} \right )\end{align*}
8. Express each of the following functions as an algebraic expression involving no trigonometric functions.
1. \begin{align*}\cos^2 (\tan^{-1} x)\end{align*}
2. \begin{align*}\cot (\tan^{-1} x^2)\end{align*}
9. To find trigonometric functions in terms of sine inverse, use the following triangle.
1. Determine the sine, cosine and tangent in terms of arcsine.
2. Find \begin{align*}\tan (\sin^{-1} 2x^3)\end{align*}.
10. To find the trigonometric functions in terms of cosine inverse, use the following triangle.
1. Determine the sine, cosine and tangent in terms of arccosine.
2. Find \begin{align*}\sin^2 \left ( \cos^{-1} \frac{1}{2}x \right )\end{align*}.
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What is Arithmetic? Definition, Basic Operations, Examples
• Last Updated : 19 Oct, 2021
Arithmetic most likely has the longest history at the time. It is a calculating approach that has been used since ancient times for routine calculations such as measurements, labelling, and other day-to-day computations to achieve precise results. The name derives from the Greek word “arithmos,” which meaning “numbers.”
Arithmetic is the fundamental area of mathematics that studies numbers and the characteristics of conventional operations such as addition, subtraction, multiplication, and division.
Aside from the standard operations of addition, subtraction, multiplication, and division, arithmetic also includes sophisticated computations such as percentage, logarithm, exponentiation, and square roots, among others. Arithmetic is a field of mathematics that deals with numbers and their conventional operations.
Basic Operations of Arithmetic
Arithmetic has four basic operations that are used to perform calculations as per the statement:
2. Subtraction
3. Multiplication
4. Division
The simple definition for addition will be that it is an operation to combine two or more values or numbers into a single value. The process of adding n numbers of value is called summation.
• 0 is said to be the identity element of addition as while adding 0 to any value it gives the same result. For example, if we add 0 to 7 the result would be the same that is 7.
0 + 7= 7
• And, the inverse element includes the addition of the opposite value. The result of adding inverse elements will be an identity element that is 0. For example, if we add 4 with its opposite value -4, then the result would be:
4 + (-4) = 0
Subtraction (-)
Subtraction is the arithmetic operation that computes the difference between two values (i.e. minuend minus the subtrahend).
• In the condition where the minuend is greater than the subtrahend, the difference is positive. It is the inverse of addition.
4 – 2 = 2
• While, if the subtrahend is greater than minuend the difference between them will be negative.
2 – 4 = -2
Multiplication (×)
The two values involved in the operation of multiplication are known as multiplicand and multiplier. It combines two values that is multiplicand and multiplier to give a single product.
• The product of two values supposedly p and q is expressed in p.q or p × q form.
5 × 6 = 30
Division (÷)
The division is the operation that computes the quotient of two numbers. It is the inverse of multiplication.
• The two values involved in it are known as dividends by the divisor and if the quotient is more than 1 if the dividend is greater than the divisor the result would be a positive number.
12 ÷ 3 = 4
What is Simple Arithmetic?
Arithmetic most likely has the longest history at the time. It is a calculating approach that has been used since ancient times for routine calculations such as measurements, labelling, and other day-to-day computations to achieve precise results. The name derives from the Greek word “arithmos,” which meaning “numbers.”
Arithmetic is the fundamental area of mathematics that studies numbers and the characteristics of conventional operations such as addition, subtraction, multiplication, and division.
Aside from the standard operations of addition, subtraction, multiplication, and division, arithmetic also includes sophisticated computations such as percentage, logarithm, exponentiation, and square roots, among others. Arithmetic is a field of mathematics that deals with numbers and their conventional operations.
Sample Problems on Simple Arithmetic
Question 1: The sum of the two numbers is 30, and their difference is 20. Find the numbers.
Solution:
Let the numbers be a and b. Now, as per the situation,
a + b = 30 ………(1)
and a – b = 20 ………(2)
We can write, a = 30 – b, from equation (1),
no put the value of a in equation (2), we get,
30 – b – b = 20
30 – 2b = 20
2b = 30 – 20 = 10
b = 10/2 = 5
b = 5
And
a = 30 – b
= 30 – 5
a = 25
Therefore, the two numbers are 25 and 5.
Question 2: Solve 45 + 2(36 ÷ 3) – 9
Solution:
45 + 2(36 ÷ 3) – 9
⇒ 45 + 2(12) – 9
⇒ 45 + 24 – 9
⇒ 69 – 9 = 60
Question 3: Find the value of a in the given equation 3a – 18 = 3.
Solution:
According to the equation,
3a – 18 = 3
3a = 18 + 3
3a = 21
a = 7
Therefore, the value of a is 7.
Question 4: Solve for the value of a:
3a – 2(15 ÷ 3) – 5 × 2 = 22
Solution:
3a – 2(15 ÷ 3) – 5 x 2 = 22
3a – 2(5) – 10 = 22
3a – 10 – 10 = 22
3a – 20 = 22
3a = 22 + 20
3a = 42
a = 14
Therefore the value of a is 14.
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In mathematics, the trigonometric functions (also called circular functions, angle functions or goniometric functions) are functions of an angle. They relate the angles of a triangle to the lengths of its sides. Trigonometric functions are important in the study of triangles and modeling periodic phenomena, among many other applications.The most familiar trigonometric functions are the sine, cosine, and tangent.
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• In mathematics, the trigonometric functions (also called the circular functions) are functions of an angle. They relate the angles of a triangle to the lengths of its sides. Trigonometric functions are important in the study of triangles and modeling periodic phenomena, among many other applications.The most familiar trigonometric functions are the sine, cosine, and tangent. In the context of the standard unit circle (a circle with radius 1 unit), where a triangle is formed by a ray originating at the origin and making some angle with the x-axis, the sine of the angle gives the length of the y-component (the opposite to the angle or the rise) of the triangle, the cosine gives the length of the x-component (the adjacent of the angle or the run), and the tangent function gives the slope (y-component divided by the x-component). More precise definitions are detailed below. Trigonometric functions are commonly defined as ratios of two sides of a right triangle containing the angle, and can equivalently be defined as the lengths of various line segments from a unit circle. More modern definitions express them as infinite series or as solutions of certain differential equations, allowing their extension to arbitrary positive and negative values and even to complex numbers.Trigonometric functions have a wide range of uses including computing unknown lengths and angles in triangles (often right triangles). In this use, trigonometric functions are used, for instance, in navigation, engineering, and physics. A common use in elementary physics is resolving a vector into Cartesian coordinates. The sine and cosine functions are also commonly used to model periodic function phenomena such as sound and light waves, the position and velocity of harmonic oscillators, sunlight intensity and day length, and average temperature variations through the year.In modern usage, there are six basic trigonometric functions, tabulated here with equations that relate them to one another. Especially with the last four, these relations are often taken as the definitions of those functions, but one can define them equally well geometrically, or by other means, and then derive these relations. (en)
• In mathematics, the trigonometric functions (also called the circular functions) are functions of an angle. They relate the angles of a triangle to the lengths of its sides. Trigonometric functions are important in the study of triangles and modeling periodic phenomena, among many other applications.The most familiar trigonometric functions are the sine, cosine, and tangent. In the context of the standard unit circle (a circle with radius 1 unit), where a triangle is formed by a ray starting at the origin and making some angle with the x-axis, the sine of the angle gives the length of the y-component (the opposite to the angle or the rise) of the triangle, the cosine gives the length of the x-component (the adjacent of the angle or the run), and the tangent function gives the slope (y-component divided by the x-component). More precise definitions are detailed below. Trigonometric functions are commonly defined as ratios of two sides of a right triangle containing then angle, and can equivalently be defined as the lengths of various line segments from a unit circle. More modern definitions express them as infinite series or as solutions of certain differential equations, allowing their extension to arbitrary positive and negative values and even to complex numbers.Trigonometric functions have a wide range of uses including computing unknown lengths and angles in triangles (often right triangles). In this use, trigonometric functions are used, for instance, in navigation, engineering, and physics. A common use in elementary physics is resolving a vector into Cartesian coordinates. The sine and cosine functions are also commonly used to model periodic function phenomena such as sound and light waves, the position and velocity of harmonic oscillators, sunlight intensity and day length, and average temperature variations through the year.In modern usage, there are six basic trigonometric functions, tabulated here with equations that relate them to one another. Especially with the last four, these relations are often taken as the definitions of those functions, but one can define them equally well geometrically, or by other means, and then derive these relations. (en)
• In mathematics, the trigonometric functions (also called circular functions, angle functions or goniometric functions) are functions of an angle. They relate the angles of a triangle to the lengths of its sides. Trigonometric functions are important in the study of triangles and modeling periodic phenomena, among many other applications.The most familiar trigonometric functions are the sine, cosine, and tangent. In the context of the standard unit circle (a circle with radius 1 unit), where a triangle is formed by a ray starting at the origin and making some angle with the x-axis, the sine of the angle gives the length of the y-component (the opposite to the angle or the rise) of the triangle, the cosine gives the length of the x-component (the adjacent of the angle or the run), and the tangent function gives the slope (y-component divided by the x-component). More precise definitions are detailed below. Trigonometric functions are commonly defined as ratios of two sides of a right triangle containing the angle, and can equivalently be defined as the lengths of various line segments from a unit circle. More modern definitions express them as infinite series or as solutions of certain differential equations, allowing their extension to arbitrary positive and negative values and even to complex numbers.Trigonometric functions have a wide range of uses including computing unknown lengths and angles in triangles (often right triangles). In this use, trigonometric functions are used, for instance, in navigation, engineering, and physics. A common use in elementary physics is resolving a vector into Cartesian coordinates. The sine and cosine functions are also commonly used to model periodic function phenomena such as sound and light waves, the position and velocity of harmonic oscillators, sunlight intensity and day length, and average temperature variations through the year.In modern usage, there are six basic trigonometric functions, tabulated here with equations that relate them to one another. Especially with the last four, these relations are often taken as the definitions of those functions, but one can define them equally well geometrically, or by other means, and then derive these relations. (en)
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• In mathematics, the trigonometric functions (also called circular functions, angle functions or goniometric functions) are functions of an angle. They relate the angles of a triangle to the lengths of its sides. Trigonometric functions are important in the study of triangles and modeling periodic phenomena, among many other applications.The most familiar trigonometric functions are the sine, cosine, and tangent. (en)
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Inscribed triangle
To a circle is inscribed triangle so that the it's vertexes divide circle into 3 arcs. The length of the arcs are in the ratio 2:3:7. Determine the interior angles of a triangle.
Correct result:
α = 30 °
β = 45 °
γ = 105 °
Solution:
$\beta =3a=4{5}^{\circ }=4{5}^{\circ }$
$\gamma =7a=10{5}^{\circ }=10{5}^{\circ }$
We would be pleased if you find an error in the word problem, spelling mistakes, or inaccuracies and send it to us. Thank you!
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You need to know the following knowledge to solve this word math problem:
We encourage you to watch this tutorial video on this math problem:
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The side of the triangle inscribed in a circle is a chord passing through the circle center. What size are the internal angles of a triangle if one of them is 40°?
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# NCERT SOLUTIONS CLASS 6 MATHS CHAPTER 7 FRACTIONS
The chapter Fractions deals with concept of part of the whole. It includes presentation of fractions on number line, proper, improper and mixed fractions, equivalent fractions, the simplest form of fractions, like and unlike fractions and also comparison, addition and subtraction of like and unlike fractions. There are six exercises which cover all the topics mentioned above.
## FAQ’S about NCERT Solutions for Class 6 Maths Chapter 7 Fractions
### Q) What are the topics covered in Chapter 7 of NCERT Solutions for Class 6 Maths?
A)Chapter 7 of NCERT Solutions for Class 6 Maths has following topics:
Chapter 7 of NCERT Solutions for Class 6 Maths has following topics:
1. A Fraction
2. Fraction on the Number Line
3. Proper Fractions
4. Improper and Mixed Fractions
5. Equivalent Fractions
6. Simplest form of a Fraction
7. Like Fractions
8. Comparing Fractions
• Comparing like fractions
• Comparing unlike fractions
9. Addition and Subtraction of Fractions
• Adding or subtracting like fractions
### Q) How many exercises in NCERT Solutions for Class 6 Maths Chapter 7?
A) There are nine exercises in chapter 7 of class 6.
### Q) How many problems are there in each exercise of NCERT Solutions for Class 6 Maths Chapter 7?
A) Exercise wise count of problems of chapter 7 is as follows:
Exercise Number Count of Problems 7.1 11 7.2 3 7.3 9 7.4 10 7.5 5 7.6 9
### Q) What does Chapter 7 of NCERT Solutions for Class 6 Maths convey?
A) This chapter clear all basic concepts related to fraction such as what is a fraction, proper fraction, improper fraction, mixed fraction, equivalent fraction, the simplest form of a fraction, etc.
### Q) Which topics of chapter 7 of class 6 Maths requires more practice?
A) Topics simplest form of a fraction, comparing fractions and addition and subtraction of fractions require more practice as these concepts would be important in higher levels of classes. Students must understand these concepts and work on them on a regular basis to score well in the annual exams. |
Contents
## How To Convert Fractions To Decimals?
Arithmetic is the branch involved with the study of numbers, which is vital to understanding the more complex and advanced concepts of mathematics. Fractions and decimals play a crucial role in unveiling the importance of this branch and help understand numbers so that you have a good grasp of the real world.
We will need to understand fractions, how it is derived, and the basic concepts involved before we jump to the conversion of a fraction into a decimal. Similarly, we will also learn a bit about the study of numbers, decimals, and their importance.
### What Is a Fraction?
According to a good Maths Tutor, there are many examples to help understand a fraction. A one-line explanation states that fraction is a part of the whole.
Say there is a square. You cut it into two halves, then each half becomes a fraction of the square. They are two rectangles that are part of the entire square. When you join them, they become square again. This is a simple understanding of this term. Let’s dig a little deeper to understand this concept better using numbers.
Let’s say the square was divided into equal halves, which means the fraction would be ½ of the square. Similarly, you can have 1/8th of a whole product or, 5/8th part of the whole product too.
The fraction is divided into two parts: the numerator and the denominator. In the case of ½, 1 is the numerator & 2 is the denominator.
Similarly, in the case of 5/8- 5 is the numerator and 8 is the denominator.
### Real-World Examples
An excellent example of a fraction in the real-life is pizza. The eight pieces in which the pizza is cut is a fraction of the whole pizza.
When you need to pay money for a particular product, fractions come as a savior. They not only help you understand discounts but also works towards helping you get the right change. Fractions are used in almost all aspects of everyday life. From cooking to shopping to travel, you will find a use for fractions.
### Types of Fractions
We will introduce the different types of fractions here so that you have a basic idea.
1. Proper: In this case, the numerator is smaller as compared to the denominator. ½ is a proper fraction, and the outcome is less than one. The denominator here is the total number of parts into which the whole portion has been divided.
2. Improper: When the numerator is larger than the denominator, it is an improper fraction. The denominator is the total number of equal parts required while the numerator is the number of objects available. For instance, there are five chocolates to be divided into four people. This can lead to an improper fraction.
3. Mixed: If the chocolate example is repeated, and the four people take one chocolate each while, dividing the fifth one equally, it is a mixed fraction.
Apart from these basic fractions, you also have like fractions, unlike fractions, and equivalent fractions.
### Solving Fraction Problems
Half of the students in a particular school are girls. 3/5thof these girls are from the lower classes. What fraction of the total girls are studying in the lower classes?
Answer: ½ of the students in the school are girls.
3/5(1/2) are in the lower classes.
This means 3/5*1/2 girls study in lower classes. 3/10 fraction of girls studies in the lower classes.
A herd of cows produces 4 liters of milk on a single day. Each cow produces 1/3rdof the cow delivered. The total milk produced in six days is 24. What is the total number of cows?
Total milk produced on a single day- 4 liters
Every cow produces 1/3rdmilk
Total milk produced by each cow on a daily basis= 4/3
Total number of cows: 4 4/3= 3
The total number of cows in the herd=3
### What Is a Decimal?
If we are converting fractions to decimal, we need to know what a decimal is.
An integer is separated from its fraction using a decimal point.
435.6 is a number where 435 is the integer and 6 is 1/10th of the number, i.e. the fraction of the integer.
As we move towards the left from the decimal point, each number becomes 10 times bigger while, when we move towards the right, each number grows 10 times smaller. For instance, the number 6 is 1/10th of the number.
### Types of Decimals
• The Recurring Decimal: In this case, the digits after the decimal point repeat. In the case of 512.12121212, the number 12 repeats after the decimal point. This is an example of a recurring decimal. Another example would be 432.5555555555
• Non-recurring Decimals: The value after the decimal point could be finite or infinite in this case. The numbers are all different, and no two numbers repeat after the decimal point.
42.1234 (finite)
43.12453267.. (infinite).
• Decimal Fractions: The denominator for this fraction is in the power of 10s. 817/10=81.7, and this is a decimal fraction.
### Converting Fractions to Decimals
The division plays an important role when you want to convert fractions to decimals. The simplest method is to divide the numerator by the denominator.
1/8 is the fraction. You want to convert it into a decimal.
0.625
. 0.125
———–
8 )10 0
8.0
20
16
40
40
0
The decimal for 1/8 is 0.125
We have just seen the long-form division, which has helped us derive this result.
The alternative method for converting fractions to decimals.
1. Convert the bottom digit into 10, 100, 1000, i.e. any number with 1 followed by zeroes.
2. Next on, multiply the top and bottom digit by this number.
3. Now you will need to write the numerator you have achieved by multiplication method, and add the decimal point at the right place.
Let’s see this with an example to get a perfect understanding.
We will take the same example, 1/8.
Multiplying 8 by 125 will give 1000
We will now multiply the numerator and denominator with 125
125/1000=0.125
So, we have arrived at the same answer with fewer steps using this alternate method.
### Let’s take another example 1/7
There is no way in which you can get 10, 100, or any number in this category with seven or any odd number in the denominator.
You will need to make do with the nearest possibility for the best outcome.
Multiplying 14 with 7 will yield 98, which is closest to 100.
Let’s multiply both numerator and denominator with 14.
14/98
As the number is closest to 100, we will use two decimal spaces. The answer is 0.14
Isn’t this alternative easy? Of course, you will need to know your tables well so that you can manage to locate a number closest to 10s, 100s, and 1000s.
Let’s solve some problems to become better with this conversion?
### Examples and Solutions
Let’s do a quick revision by solving a few problems.
• Convert 10/6 to its decimal number.
We can use the long-form division or the alternate method. For the first one, we will use the long-form division.
1.66
6)10
– 6
_______
40
-36
40 and so on.
The answer to this problem is 1.66 (it is a recurring decimal)
• Convert 41/100 to its decimal form.
We already have the numerator to the nearest power of 10.
• 3/5 to the decimal form.
Let’s use the alternate method to answer this problem
Multiplying five to get the nearest digit raised to the power of 10.
20 multiplied by 5 gives 100.
We will multiply the numerator and the denominator with 20
60/100=0.60
• 6/8 to decimal form.
Multiply numerator and denominator with 125.
750/1000=0.75
This should give you an idea of converting fractions into decimals.
### Why Convert Fractions to Decimals?
When you are given a fraction, but you find it difficult to work with a fraction, it makes sense to convert it into a decimal and work on this format.
For instance, when you are calculating money or, percentages, working with fractions becomes difficult. Sometimes, you are given fraction numbers. Let’s say the results. You are graded out of 100, and you get a particular number. Knowing where you stand or, how good you have scored can be difficult if you set out to compare the fractions.
Similarly, when you need to give out change, you cannot work with fractions. If you have received input in the form of a fraction, you should know to convert it into decimal to work around the number.
### Convert Fractions To Decimals: Conclusion
A fraction is a part of the whole, while a decimal is a number represented after the decimal point.
When you divide the numerator with the denominator, you get a whole number or a decimal number. Converting the fraction to the decimal gives you an understanding of what the whole represents. There are multiple ways to convert the fraction to decimal. The alternate method needs you to multiply the denominator with a number so that you get numbers such as 10, 100, 1000, etc. The key is to understand numbers and be able to interchange them for your benefit. There are plenty of real-world benefits of being able to convert decimals to fractions.
This is just a basic insight into the world of fractions and decimals. You can know more about numbers and how it can help by taking a trial class at Cuemath. Register and get connected with our tutor right away.
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# How do you determine the remaining zeroes for g(x)=2x^5-3x^4-5x^3-15x^2-207x+108 if 3i is a zero?
Jul 4, 2015
Use conjugate zeros, factor theorem, division of polynomials, rational zeros theorem, division (again), then solve the resulting quadratic by you favorite method for quadratics.
#### Explanation:
$g \left(x\right) = 2 {x}^{5} - 3 {x}^{4} - 5 {x}^{3} - 15 {x}^{2} - 207 x + 108$
First Two Zeros
Given that $3 i$ is a zero, we observe that the coefficients of $g$ are real, so the Complex Conjugate Zero Theorem tells us that $- 3 i$ is also a zero.
The Factor Theorem tells us that $x - 3 i$ and $x + 3 i$ are factors, so their product is also a factor:
$\left(x - 3 i\right) \left(x + 3 i\right) = {x}^{2} + 9$
Divide $g \left(x\right)$ by ${x}^{2} + 9$ to get:
$2 {x}^{5} - 3 {x}^{4} - 5 {x}^{3} - 15 {x}^{2} - 207 x + 108 = \left({x}^{2} + 9\right) \left(2 {x}^{3} - 3 {x}^{2} - 23 x + 12\right)$
Third Zero
To find another zero, we need a zero of $2 {x}^{3} - 3 {x}^{2} - 23 x + 12$.
Neither of the easiest choices, $\pm 1$, are zeros. So use the Rational Zeros Theorem to search for another zero.
Possible rational zeros are: $\pm 1 , \pm 2 , \pm 3 , \pm 4 , \pm 6 , \pm 12 , \pm \frac{1}{2} , \pm \frac{3}{2}$
Test until you find a zero. (Synthetic division is recommended for this testing.)
Depending on the order in which you test, you'll find a rational zero.
I found $- 3$ is a zero first. So (or really because) $x + 3$ is a factor.
$2 {x}^{3} - 3 {x}^{2} - 23 x + 12 = \left(x + 3\right) \left(2 {x}^{2} - 9 x + 4\right)$.
Last Two Zeros
Solve $2 {x}^{2} - 9 x + 4 = 0$ by whatever method you like.
$\left(2 x - 1\right) \left(x - 4\right) = 0$
So the final two zeros of $g$ are $\frac{1}{2}$ and $4$.
List of Zeros
Zeros: $3 i , - 3 i , - 3 , 4 , \frac{1}{2}$ |
# What is 60 divided by 15%?
Nov 14, 2016
400
#### Explanation:
The % in 15% is a unit of measurement and is worth $\frac{1}{100}$
So another way of writing 15% is $\frac{15}{100}$
What is 60 divided by 15% -> 60 -:15% -> 60 -:15/100
Using the shortcut method, turn $\frac{15}{100}$ upside down and multiply.
$60 \div \frac{15}{100} \text{ " =" " 60xx100/15" "=" } \frac{60}{15} \times 100$
....................................................................................................................
I can move the 15 to be 'under' the 60 from under the 100 for the same reason that $2 \times 4 \text{ is the same as } 4 \times 2$
....................................................................................................................
But 5 will divide exactly into both 15 and 60 so we have:
$\frac{60}{5} \times 100 \text{ "=" } \frac{60 \div 5}{15 \div 5} \times 100$
$= \frac{12}{3} \times 100 \text{ }$ but 3 will divide exactly into both 3 and 12
$= \frac{12 \div 3}{3 \div 3} \times 100$
$= \frac{4}{1} \times 100 = 400$ |
# math
While adding ten two-digit numbers the digits of one of the numbers were interchanged. As a result the sum of all the ten numbers increased by a value which was four less than that number. Three times the sum of the digits of the original number is ten less than the number. What is the product of the digits of that number?
Please help how to obtain that number. and also explain (While adding ten two-digit numbers the digits of one of the numbers were interchanged. As a result the sum of all the ten numbers increased by a value which was four less than that number ) this sentence
1. 👍 0
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3. 👁 174
1. a1 ÷ a10 = your numbers.
You can write 10th digit in form:
a10 = 10 x + y
Original sum:
S = a1 + a2 +...+ a9 + 10 x + y
If 10th digit be interchanged new number will be 10 y + x
New sum wil be:
S1 = a1 + a2 +...+ a9 + 10 y + x
The sum of all the ten numbers increased by a value which was four less than that number mean:
S1 - S = 10 x + y - 4
S1 - S = a1 + a2 +...+ a9 + 10 y + x - ( a1 + a2 +...+ a9 + 10 x + y )
S1 - S = ( a1 + a2 +...+ a9 ) + 10 y + x - ( a1 + a2 +...+ a9 ) - 10 x - y
S1 - S = 10 y + x - 10 y + x
S1 - S = 9 y - 9 x
So:
S1 - S = 10 x + y - 4
9 y - 9 x = 10 x + y - 4
9 y - y = 10 x + 9 x - 4
8 y = 19 x - 4
Three times the sum of the digits of the original number is ten less than the number mean:
3 ( x + y ) = 10 x + y - 10
3 x + 3 y = 10 x + y - 10
3 y - y = 10 x - 3 x - 10
2 y = 7 x - 10
Now you must solve system:
8 y = 19 x - 4
2 y = 7 x - 10
The solutions are:
x = 4 , y = 9
The product of the digits:
x ∙ y = 4 ∙ 9 = 36
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posted by Bosnian
2. Suppose the two digits are a and b. Then the value of the number is 10a+b.
So, the value after the digit swap is 10b+a
I assume "that number" refers to the number whose digits were swapped.
The amount of increase in the sum is (10b+a)-(10a+b)=9b-9a
Now we know that
9b-9a = 10a+b-4
3(a+b) = 10a+b-10
19a-8b = 4
7a-2b = 10
a=4 b=9
ab = 36
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2. 👎 0
posted by Steve
3. Why (10b+a)-(10a+b) this step can you explain
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2. 👎 0
posted by Vipul
4. cmon. That is the difference between the original number and with its digits reversed. Better reread what I wrote.
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2. 👎 0
posted by Steve
## Similar Questions
1. ### math
While adding ten two-digit numbers the digits of one of the numbers were interchanged. As a result the sum of all the ten numbers increased by a value which was four less than that number. Three times the sum of the digits of the
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2. ### math
The digits 2, 4, 6, 8 and 0 are used to make five-digit numbers with no digits repeated. What is the probability that a number chosen at random from these numbers has the property that the digits in the thousands place and ten's
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3. ### Number Sentence
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4. ### math
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5. ### MATH Trouble
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6. ### math
given the digits 0,1,2,3,4,5,6, and 7 find the number of possibilities in each category four digit numbers, odd three digit numbers and three digit numbers without repeating digits.
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7. ### math
. The ones digit is 0. . The hundreds and tens digits are consecutive numbers. . The digits in the thousands period have a sum of 15, with the hundred thousands digit being the greatest of the threes digits and the ten thousands
asked by sofia on December 10, 2012
8. ### math
think of a five digit number composed of odd numbers. the thousands digits is two less than the ten thousands digit but two more than the hundreds digit. the tens digits is two more that ones digit which is less than the hundreds
asked by alvin on June 21, 2017
9. ### Math. Explain
From the digits 1, 2, 3, and 4, how many positive integers are less than 100,000? Consider the possibilities for 5-digit, 4-digit, 3-digit, 2-digit, and 1-digit numbers and repetition of digits. 1,364 1,024 256 A telephone dial
asked by Sebastian on March 26, 2012
10. ### 6 th grade Math
You can find the total number of different 4-digit numbers of choices for each digit (9) to the number of digits (4), or 9 to the 4th power. Based on this pattern, how many different 5-digit numbers could you make from the digits
asked by Bob on December 14, 2011
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# 9.17: Corresponding Parts of Similar Figures
Difficulty Level: At Grade Created by: CK-12
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Remember the geometric skateboard park? Well, the students are working on all kinds of designs for the sidewalk that leads into the park.
Sam is designing some figures to be painted on the sidewalk for the skateboard park.
"I am going to use all primary colors," Sam told his friend Kara at lunch.
"Let me see," Kara said looking at his drawing.
Here are Sam's designs.
"Those aren't the same," Kara said looking at the drawing.
"I know that. These are similar figures. But I think they are cool."
"They are cool, but they aren't the same," Kara said again.
"That doesn't matter with similar figures. They have corresponding sides," Sam explained.
Kara is puzzled. She isn't sure how to identify corresponding sides of similar figures.
Do you know?
In this Concept you will learn all about corresponding sides of similar figures, and you will know how to identify them too.
### Guidance
We just finished identifying the corresponding parts of congruent figures, and we can also identify the corresponding parts of similar figures. We do this in the same way.
First, notice that all of the angle measures are the same. Whether figures are similar or congruent, the angle measures are the same in both.
The side lengths are different in similar figures. The side lengths are the same in congruent figures.
Triangle ABC\begin{align*}ABC\end{align*} is similar to triangle DEF\begin{align*}DEF\end{align*}. This means that while they are the same shape, they aren’t the same size. In fact, there is a relationship between the corresponding parts of the triangle.
The side lengths are corresponding even though they aren’t congruent.
AB¯¯¯¯¯¯¯¯×DE¯¯¯¯¯¯¯¯BC¯¯¯¯¯¯¯¯×EF¯¯¯¯¯¯¯¯AC¯¯¯¯¯¯¯¯×DF¯¯¯¯¯¯¯¯\begin{align*}\overline{AB} \times \overline{DE} \\ \overline{BC} \times \overline{EF} \\ \overline{AC} \times \overline{DF}\end{align*}
We use the symbol for similar ("~") to show the relationship between the corresponding sides of the two triangles.
Use these similar figures to answer the following questions.
#### Example A
How many right angles are in the first two figures?
Solution: 4 right angles
#### Example B
What makes the two squares similar if they both have four right angles?
Solution: The side lengths are different.
#### Example C
In the triangle pair, are the two triangles similar or congruent? why?
Solution: The triangles are similar because the side lengths are different. The angle measures are the same.
Now back to Kara, Sam and the similar figures. Here is the original problem once again.
Sam is designing some figures to be painted on the sidewalk for the skateboard park.
"I am going to use all primary colors," Sam told his friend Kara at lunch.
"Let me see," Kara said looking at his drawing.
Here are Sam's designs.
"Those aren't the same," Kara said looking at the drawing.
"I know that. These are similar figures. But I think they are cool."
"They are cool, but they aren't the same," Kara said again.
"That doesn't matter with similar figures. They have corresponding sides," Sam explained.
Kara is puzzled. She isn't sure how to identify corresponding sides of similar figures.
When working with congruent figures, you had to match up the sides that were the same length. Similar figures aren't the same length, but based on the position of the figures, you can figure out which sides go together. These are the corresponding sides. Corresponding sides are in the same position on two different figures.
Here are the corresponding sides of these figures.
HG¯¯¯¯¯¯¯¯\begin{align*}\overline{HG}\end{align*} ~ XW¯¯¯¯¯¯¯¯¯¯\begin{align*}\overline{XW}\end{align*}
HI¯¯¯¯¯¯¯\begin{align*}\overline{HI}\end{align*} ~ XY¯¯¯¯¯¯¯¯\begin{align*}\overline{XY}\end{align*}
IJ¯¯¯¯¯¯\begin{align*}\overline{IJ}\end{align*} ~ XZ¯¯¯¯¯¯¯¯\begin{align*}\overline{XZ}\end{align*}
GJ¯¯¯¯¯¯¯\begin{align*}\overline{GJ}\end{align*} ~ WZ¯¯¯¯¯¯¯¯¯\begin{align*}\overline{WZ}\end{align*}
### Vocabulary
Here are the vocabulary words in this Concept.
Congruent
having the same size and shape and measurement
Similar
having the same shape, but not the same size. Similar shapes are proportional to each other.
Corresponding
matching-corresponding sides between two triangles are sides that match up
### Guided Practice
Here is one for you to try on your own.
List all of the pairs of corresponding sides in the similar figures below.
Here are the pairs of corresponding sides.
OP¯¯¯¯¯¯¯¯\begin{align*}\overline{OP}\end{align*} and RS¯¯¯¯¯¯¯\begin{align*}\overline{RS}\end{align*}
NO¯¯¯¯¯¯¯¯\begin{align*}\overline{NO}\end{align*} and QR¯¯¯¯¯¯¯¯\begin{align*}\overline{QR}\end{align*}
MP¯¯¯¯¯¯¯¯¯\begin{align*}\overline{MP}\end{align*} and TS¯¯¯¯¯¯¯\begin{align*}\overline{TS}\end{align*}
MN¯¯¯¯¯¯¯¯¯¯\begin{align*}\overline{MN}\end{align*} and TQ¯¯¯¯¯¯¯¯\begin{align*}\overline{TQ}\end{align*}
### Video Review
Here are videos for review.
### Practice
Directions: Use the following figures to answer each question.
1. Are these two triangles similar or congruent?
2. How do you know?
3. Which side is congruent to AB¯¯¯¯¯¯¯¯\begin{align*}\overline{AB}\end{align*}?
4. Which side is congruent to AC¯¯¯¯¯¯¯¯\begin{align*}\overline{AC}\end{align*}?
5. Which side is congruent to RS¯¯¯¯¯¯¯\begin{align*}\overline{RS}\end{align*}?
6. Which angle is congruent to angle A?
7. Which angle is congruent to angle B?
8. Which angle is congruent to angle C?
9. Are the two figures similar or congruent?
10. Why?
11. Which side is congruent to NO¯¯¯¯¯¯¯¯\begin{align*}\overline{NO}\end{align*}?
12. Which side is congruent to MN¯¯¯¯¯¯¯¯¯¯\begin{align*}\overline{MN}\end{align*}?
13. Which side is congruent to ST¯¯¯¯¯¯¯\begin{align*}\overline{ST}\end{align*}?
14. Which side is congruent to QT¯¯¯¯¯¯¯¯\begin{align*}\overline{QT}\end{align*}?
15. Which side is congruent to OP¯¯¯¯¯¯¯¯\begin{align*}\overline{OP}\end{align*}?
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
Congruent
Congruent figures are identical in size, shape and measure.
Corresponding
The corresponding sides between two triangles are sides in the same relative position.
Similar
Two figures are similar if they have the same shape, but not necessarily the same size.
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How do you find the distance between (3,2) and (1,-1)?
Sep 11, 2015
Distance between $\left(3 , 2\right)$ and $\left(1 , - 1\right)$ is$\sqrt{13}$
Explanation:
(see image below)
The horizontal distance between $\left(1 , - 1\right)$ and $\left(3 , 2\right)$ is
$\textcolor{w h i t e}{\text{XXX}} \Delta x = \left\mid 3 - 1 \right\mid = 2$
The vertical distance between $\left(- 1 , - 1\right)$ and $\left(3 , 2\right)$ is
$\textcolor{w h i t e}{\text{XXX}} \Delta y = \left\mid \left(- 1\right) - 2 \right\mid = 3$
By the Pythagorean Theorem the distance between $\left(- 1 , 1\right)$ and $\left(3 , 2\right)$ is
$\textcolor{w h i t e}{\text{XXX}} \sqrt{{\left(\Delta x\right)}^{2} + {\left(\Delta y\right)}^{2}}$
$\textcolor{w h i t e}{\text{XXX}} = \sqrt{{2}^{2} + {3}^{2}}$
$\textcolor{w h i t e}{\text{XXX}} = \sqrt{13}$ |
Whenever we want to add or subtract two or more fractions, first we have to check whether their denominators are same or not.
• If they have same denominators then we can put one denominator commonly and combine the numerators.
• If the denominators are not same, then we have to take L.C.M in order to make the denominators same.After making the denominators same, we can combine the numerators.
Let us see some example problems to understand the above concept.
Example 1 :
Solution :
Since the denominators are not same, we have to take L.C.M in order to make the denominators same.
L.C.M of (4, 8) = 8
Multiply the numerator and denominator of the first fraction by 2.
(1/4) x (2/2) = 2/8
(1/4) + (3/8) = (2/8) + (3/8)
= (2+3)/8
= 5/8
Example 2 :
Solution :
Since the denominators are not same, we have to take L.C.M in order to make the denominators same.
L.C.M of (4, 5) = 20
Multiply the numerator and denominator of the first fraction by 5.
Multiply the numerator and denominator of the second fraction by 4.
(3/4) x (5/5) = 15/20
(2/5) x (4/4) = 8/20
(15/20) + (8/20) = (15 + 8)/20
= 23/20
Since it is improper fraction, we have to make it is as mixed fraction.
Example 3 :
Subtract the fractions
Solution :
Since the denominators are not same, we have to take L.C.M in order to make the denominators same.
L.C.M of (8, 7) = 56
Multiply the numerator and denominator of the first fraction by 7.
Multiply the numerator and denominator of the second fraction by 8.
(5/8) x (7/7) = 35/56
(1/7) x (8/8) = 8/56
(35/56) + (8/56) = (35 + 8)/56
= 43/56
Example 4 :
Subtract the fractions
Solution :
Since the denominators are not same, we have to take L.C.M in order to make the denominators same.
L.C.M of (2, 3) = 6
Multiply the numerator and denominator of the first fraction by 3.
Multiply the numerator and denominator of the second fraction by 2.
(3/2) x (3/3) = 9/6
(4/3) x (2/2) = 8/6
(9/6) + (8/6) = (9 + 8)/6
= 17/6
Since it is improper fraction, we have to make it is as mixed fraction.
After having gone through the stuff given above, we hope that the students would have understood "Add and subtract fractions".
Apart from the stuff given in this section, if you need any other stuff in math, please use our google custom search here.
WORD PROBLEMS
HCF and LCM word problems
Word problems on simple equations
Word problems on linear equations
Algebra word problems
Word problems on trains
Area and perimeter word problems
Word problems on direct variation and inverse variation
Word problems on unit price
Word problems on unit rate
Word problems on comparing rates
Converting customary units word problems
Converting metric units word problems
Word problems on simple interest
Word problems on compound interest
Word problems on types of angles
Complementary and supplementary angles word problems
Double facts word problems
Trigonometry word problems
Percentage word problems
Profit and loss word problems
Markup and markdown word problems
Decimal word problems
Word problems on fractions
Word problems on mixed fractrions
One step equation word problems
Linear inequalities word problems
Ratio and proportion word problems
Time and work word problems
Word problems on sets and venn diagrams
Word problems on ages
Pythagorean theorem word problems
Percent of a number word problems
Word problems on constant speed
Word problems on average speed
Word problems on sum of the angles of a triangle is 180 degree
OTHER TOPICS
Profit and loss shortcuts
Percentage shortcuts
Times table shortcuts
Time, speed and distance shortcuts
Ratio and proportion shortcuts
Domain and range of rational functions
Domain and range of rational functions with holes
Graphing rational functions
Graphing rational functions with holes
Converting repeating decimals in to fractions
Decimal representation of rational numbers
Finding square root using long division
L.C.M method to solve time and work problems
Translating the word problems in to algebraic expressions
Remainder when 2 power 256 is divided by 17
Remainder when 17 power 23 is divided by 16
Sum of all three digit numbers divisible by 6
Sum of all three digit numbers divisible by 7
Sum of all three digit numbers divisible by 8
Sum of all three digit numbers formed using 1, 3, 4
Sum of all three four digit numbers formed with non zero digits
Sum of all three four digit numbers formed using 0, 1, 2, 3
Sum of all three four digit numbers formed using 1, 2, 5, 6 |
# Introduction to Polynomials
## Presentation on theme: "Introduction to Polynomials"— Presentation transcript:
Introduction to Polynomials
5.2 Introduction to Polynomials 1. Identify monomials. 2. Identify the coefficient and degree of a monomial. 3. Classify polynomials. 4. Identify the degree of a polynomial. 5. Evaluate polynomials. 6. Write polynomials in descending order of degree. 7. Combine like terms.
Objective 1 Identify monomials.
x, y a number Monomial: An expression that is a constant, a variable, or a product of a constant and variable(s) that are raised to whole number powers. Multiply a number and a variable Exponent can’t be negative! Exponent can’t have fractions!
Is the given expression a monomial?
Yes Yes When an equation in one variable is solved the answer is a point on a line. No No
Objective 2 Identify the coefficient and degree of a monomial.
Coefficient of a monomial: The numerical factor in a monomial.
Degree of a monomial: The sum of the exponents of all variables in a monomial.
Identify the coefficient and degree of each monomial:
x0 C: -6.7 D: 3 When an equation in one variable is solved the answer is a point on a line. C: 16 D: 0 C: 23 = 8 D: 5
Objective 3 Classify polynomials. Objective 4 Identify the degree of a polynomial.
Polynomial: A monomial or an expression that can be written as a sum of monomials.
Examples: 4x, 4x + 8, 2x2 - 5xy + 8y Polynomial in one variable: A polynomial in which every variable term has the same variable. Example: x2 – 5x + 2
Binomial: A polynomial containing two terms.
Trinomial: A polynomial containing three terms. Degree of a polynomial: The greatest degree of any of the terms in the polynomial.
Identify the type of polynomial and the degree:
Binomial D: 2 Monomial D: 3 When an equation in one variable is solved the answer is a point on a line. Trinomial D: 3 Polynomial D: 3 Polynomial D: 4 Not a polynomial
Objective 5 Evaluate polynomials.
Evaluate each of the following:
-2(-1)2(4) = -8 (-4)2 – (-4) – 3 = – 3 = 17 - (-1)2 (2) = - (1)(2) = -2
Objective 6 Write polynomials in descending order of degree. Objective 7 Combine like terms.
Writing a Polynomial in Descending Order of Degree
Place the highest degree term first, then the next highest degree, and so on.
Write the polynomial in descending order.
Combine like terms and write the resulting polynomial in descending order of degree.
Classify the expression a) Monomial b) Binomial c) Trinomial d) None of these Copyright © 2011 Pearson Education, Inc. 5.2
Classify the expression a) Monomial b) Binomial c) Trinomial d) None of these Copyright © 2011 Pearson Education, Inc. 5.2
Evaluate when x = –3. a) –118 b) –10 c) 10 d) 134 Copyright © 2011 Pearson Education, Inc. 5.2
Evaluate when x = –3. a) –118 b) –10 c) 10 d) 134 Copyright © 2011 Pearson Education, Inc. 5.2 |
EASIER THAN YOU THINK...
Many students struggle with analysing quadratic equations and graphing parabolas. This page is devoted to explaining how to analyse quadratic equations. Another page is devoted to the graphing of parabolas and explaining their properties.
You will find here a comprehensive series of videos and resource materials that will thoroughly explain how to analyse and factorise quadratic expressions and how to solve quadratic equations.
Simply put, a quadratic expression is one that can be resolved (simplified) into a maximum of three parts: a term consisting of a constant (number) multiplied by x², a term consisting of a constant (number) multiplied by x, and a term consisting of a number only. Such an arrangement is called a simple polynomial. Of the three terms, only the first one (containing the x²) is necessary for the expression to be called a quadratic expression. This means that quadratic expressions may have one, two or three terms depending on whether they contain an “x” term and/or a constant. Examples of quadratic expressions are: x² – 3x +5, 5x², -x² + 7, and 5x² -4x/3. Quadratic expressions that are written in a messy way may be simplified, and one of your earlier challenges will be to learn how to do that.
Note that, although other numbers are permissible, on this page we will require all the constants to be real numbers (and most will be integers … whole numbers).
Quadratic equations are obtained by letting a quadratic expression equal another quadratic expression, or a linear expression (e.f. 2x – 5), or simply a constant. Equations have an = sign, expressions do not. We solve equations, because any quadratic equation may have up to two values for x that make the equation true. All other values will fail to balance the equation.
I have said enough. It will all make sense as we proceed …
Before you learn how to factorise quadratic expressions, you will need a supply of them (and their answers). This is because constant practice can make you an accomplished 'factoriser,' and factorising is an essential skill in algebra and graphing. Avoiding this practice is a bit like saying that you want to become a good sailor, surf skier, sail boarder or surfer, but you will not invest time in learning to swim.
During the last decade or so I have developed an Excel Workbook as a resource for my students. It produces a huge variety of expressions for you to practise on. The quadratic expressions and equations are randomly generated, literally at the touch of a button (the F9 key). Also, the answers are provided on the same sheet! You can even change the level of difficulty at will.
Teachers, feel free to provide a copy of this to each of your students. It will help them develop their skills quickly. You might even consider using it in class. Parents and students are free to use and distribute it freely. The only condition is that you understand that I retain copyright (ownership) of the workbook and I give you permission to distribute freely, but not to sell it in any way.
The workbook contains no macros and is virus free. You may obtain it here.
How to Factorise Monic Quadratic Expressions Using the Cross Method
Factorising a quadratic expression is the complemetary (or reverse) process to expanding a binomial product. In fact, We use our understanding of expanding a binomial product to develop our strategies/methods to factorise quadratic expressions. Mathematicians have devised a number of methods (or algorithms) for achieving this goal.
In this video (23:59) I explain the 'conventional' technique which is often called the Cross Method.
I will show you a clear way of presenting your work, and thinking through the process for finding the factors.
If you wish to practise these skills, you may use my FREE Excel workbook to produce a huge variety of expressions for you to practise on. In each case, the answers are provided on the same sheet! Teachers, feel free to provide a copy of this to each of your students. It will help them develop their skills quickly. You might even consider using it in class. Please understand that I retain copyright and give you permission to distribute freely, but not to sell it in any way.
The above video is long (23:59). If you do not wish to learn why the cross method works, or of you have already been taught this method and simply want to refresh your memory, this video is for you. In it, I provide a short example (3:46) of factorising a quadratic expression using the cross method.
How to Factorise Monic Quadratic Expressions Using the Product-Sum (PS) Method
The Product-Sum Method underlies most methods for factorising quadratic expressions. The variations are mostly due to how the work is set out on the page.
Many students have difficulty with the concept, mainly because it is difficult to think in terms of positive and negative numbers. Here (15:18), I show a simple method to break the task into two simple steps ... first find the numbers and THEN decide whether they are positive or negative! By explaining the method first, and then rapidly solving four examples, my hope is that you will be able to understand clearly how to use this method/system.
If you wish to practise these skills, you may use my FREE Excel workbook to produce a huge variety of expressions for you to practise on. In each case, the answers are provided on the same sheet! Teachers, feel free to provide a copy of this to each of your students. It will help them develop their skills quickly. You might even consider using it in class. Please understand that I retain copyright and give you permission to distribute freely, but not to sell it in any way.
Mastering the Product-Sum Calculation for Factorising Quadratics
This is one of the most important videos in this series about factorising quadratic expressions. Regardless of the method that you use, at some stage you will need to find two numbers that multiply to give one result and add to give another result. We call this the product-sum. Many students find this process difficult, especially when the numbers required can be positive OR negative and particularly if they are weak with their number skills!
Why should you watch this video? Because, in it, I explain a method that you probably will not have seen! It allows you to find these two numbers in two simple steps instead on one, more complicated, step. I will say no more ... if you are rushed for time, start about halfway through the (15:09) video to see my demonstration (but you might find the entire video worth watching).
As I mention in the video, I have created an Excel workbook so that you can practise this skill and become very good at it! It allows you to practise ... first with positive numbers only and then with a mixture of positives and negatives. In each case, you can adjust the degree of difficulty and, by pressing the F9 key, create a randomly generated set of ten questions each time. The answers are provided on the same sheet.
You can start by working on the sheet and, as you get better and better, solve them mentally. Fold the sheet in half so that a friend can see the answers and check you as you work down the page.
Teachers, feel free to provide a copy of this to each of your students and to use the sheets in class. I have used them very successfully in classes ... pairing students up so they hold the sheet between them and quickly test each other (or getting them to work on the sheets alone for more difficult sets). It helps students develop this particular skill very quickly. Please understand that I retain copyright and give you permission to distribute the file freely, but not to sell it in any way.
Factorising Non-Monic Quadratics Using the Cross Method
Monic quadratic expressions have an x² term with a coefficient of one. If you encounter a quadratic expression where the coefficient of x² is any number other than one, it is non-monic! These quadratics are more difficult to factorise.
Again, I start by demonstrating the conventional technique which is often called the Cross Method. I also explain why it works and mention its strengths and weaknesses.
The following videos will explain how to factorise the same quadratic expressions using the PS Method.
If you wish to practise these skills, you may use my FREE Excel workbook to produce a huge variety of expressions for you to practise on. In each case, the answers are provided on the same sheet! Teachers, feel free to provide a copy of this to each of your students. It will help them develop their skills quickly. You might even consider using it in class. Please understand that I retain copyright and give you permission to distribute freely, but not to sell it in any way.
The decomposition method for factorising quadratic expressions is given this name for a reason. We begin by using the product-sum technique to separate (or decompose) the middle term into two other terms ... hence the name. E.g. we would separate 7x² +4x - 3 into 7x² -3x +7x -3 (the term +4x having been decomposed into -3x +7x).
Once the quadratic has been separated into four terms, we factorise in pairs, looking for common factors. I demonstrate the technique thoroughly and explain why some people prefer this method above others (i.e. I explain some of the benefits to you). One of the major benefits is that, about half-way through the procedure you have a very good idea whether your answer is right or wrong, without having to expand your solution to test it! The major drawback is that this technique involves more writing than almost any other technique.
I encourage you to try this technique out. Factorise a few quadratic expressions this way to 'get a feel' for the process. Even if you decide this isn't for you, you will have gained from the experience!
If you wish to practise these skills, you may use my FREE Excel workbook to produce a huge variety of expressions for you to practise on. In each case, the answers are provided on the same sheet! Teachers, feel free to provide a copy of this to each of your students. It will help them develop their skills quickly. You might even consider using it in class. Please understand that I retain copyright and give you permission to distribute freely, but not to sell it in any way.
Factorising Non-Monic Quadratics Using the Berry Method
I have used variants of this method for decades without knowing its name. If you know why it is called the Berry Method, please let me know. None of the sites that I have looked at on the Internet (including Wikipedia) seem interested in that detail.
In this video I demonstrate this clever method thoroughly using a number of examples.
Its advantages are that it requires only two lines of work (the quadratic is already factorised on the first line and the second line is basically 'tidying up') and there is no attempt to divide the coefficient of x² into factors during the more difficult first step.
I encourage you to try this technique out. Factorise a few quadratic expressions this way to 'get a feel' for the process. Even if you decide this isn't for you, you will have gained from the experience!
If you wish to practise these skills, you may use my FREE Excel workbook to produce a huge variety of expressions for you to practise on. In each case, the answers are provided on the same sheet! Teachers, feel free to provide a copy of this to each of your students. It will help them develop their skills quickly. You might even consider using it in class. Please understand that I retain copyright and give you permission to distribute freely, but not to sell it in any way.
Factorising Non-Monic Quadratics Using the Diamond Method
Here I explain the Diamond Method for factorising quadratic expressions. It is, in fact, a variant of the Berry Method (see my last video/post). This time, instead of waiting to divide the factors after they have been constructed, all the division/simplifying is completed on the right hand side of the page where all the 'working' is done!
This means that, apart from the work on the right hand side of the page, the entire factorisation is completed on the one line!
I encourage you to try this technique out. Factorise a few quadratic expressions this way to 'get a feel' for the process. Even if you decide this isn't for you, you will have gained from the experience!
If you wish to practise these skills, you may use my FREE Excel workbook to produce a huge variety of expressions for you to practise on. In each case, the answers are provided on the same sheet! Teachers, feel free to provide a copy of this to each of your students. It will help them develop their skills quickly. You might even consider using it in class. Please understand that I retain copyright and give you permission to distribute freely, but not to sell it in any way.
Completing the Square With Simple (Monic) Quadratic Equations
The second step in analysing a quadratic equation is to find its roots (or zeros). These, of course, are the x-intercepts of the parabola. These values can be found in three ways ... by factorising, by completing the square, and by using the Quadratic Formula. All three methods do essentially the same job and find the same values.
I have posted about eight videos explaining how to factorise quadratic expressions. On another page, I will explain how to use this skill to find the roots/zeros of a quadratic equation and draw its graph. You will be encouraged to know that, if you can factorise, you will have done the 'hard work.'
In this video (and the two that follow) I explain how to complete the square. I start by explaining what a perfect square looks like and how to use its pattern to solve a couple of equations.
The 'Completing the Square' videos are presented as follows:
1. Monic quadratics with an even coefficient of x (this video/post)
2. Monic quadratics with odd coefficient of x (the next videopost below this one)
3. Non-monic quadratics (i.e. I will show you how to complete the square for any quadratics expression) (the next drop-down menu item below this one)
In this video/post I explain how to complete the square of a monic quadratic equation with an odd coefficient of x. These are a little bit more difficult than the ones with an even coefficient of x (which I examined in the last video/post). This is because they require that we use fractions. Many students today try to evaluate the parts as decimals, but that involves unnecessary calculation. I show here the benefits of using fractions, and how to solve such quadratic equations with relative ease.
This is a very important skill to develop, especially if you plan to study more advanced mathematics. Therefore, I strongly encourage you not to neglect it (even though it is under-emphasised in a lot of text books)! Practise this skill until you find it easy ... that is, until you can solve a batch of them at the rate of about one every 30 seconds. Then you will/can feel confident.
Completing the Square With Non-Monic Quadratic Equations (General Solution)
After having posted two videos showing how to complete the square for simpler quadratics expressions, I am now going to show you how to use this method with any quadratic expression. As you will have seen in the previous video, this solution is best completed using fractions rather than decimals.
If you are a high school student, it is more likely that you will be asked to simplify/solve a monic quadratic by completing the square. However, I encourage you not to neglect the general solution (shown here), especially if you plan to study more advanced mathematics. Set time aside to practise simplifying/solving non-monic quadratics as well. Even with the higher degree of difficulty that they present, you should be able to simplify/solve most of them in well under 60 seconds each. Then you will/can feel confident.
How to Learn the Quadratic Formula
Many students have difficulty with learning the quadratic formula. It is the first 'complicated' formula that they learn in High School. I have found that students learn this formula best when they practise deriving it and using it.
If you wish to learn (and practise) how to derive the quadratic equation, download my PDF summary of the process and watch the video below.
If you wish to practise using the quadratic formula to solve lots of quadratic equations, I have created an Excel workbook that will allow you to generate randomised worksheets (and much more). You may download it here. Teachers will also find this workbook useful for creating practice and revision material for the classroom.
My personal preference is for students to learn by deriving and using the formula because, in this way, they know what they are learning (and why). However, if you simply want a method that 'drills the formula into your mind,' you might try one or more of the following:
• Using rhythm (rap): The only example I found with reasonable sound quality was this one.
• Using a story line with symbols: This one was particularly clever.
If you want to learn a bit about memory techniques, you might watch this TED Talk by Joshua Foer!
I encourage you to go exploring on YouTube if you wish to embark on a new adventure of developing your memory skills! Please note that I do not necessarily recommend any of these techniques. I suggest that you explore and find what works for you.
I enjoyed your presentation and no it wasn’t too long. Each subtraction algorithm has its merit as you demostrated, but after learning the “one up and one down” method, I’m employing it because of its speed and ease of usage. Even my wife, who hates mathematics with a passion, thinks it’s too easy. I look forward to your future presentations on both multiplication and number theory. I read an introduction text book some twenty five years ago on number theory by Oystein Ore who taught at Yale for better than twenty years. So in closing, please produce these lectures and the longer the better. Thanks.
Dennis Bell (on a CCM YouTube video about How to Subtract (Large) Numbers Easily)
See all Testimonials |
# What is probability weighting in prospect theory?
Table of Contents
## What is probability weighting in prospect theory?
The function is a probability weighting function and captures the idea that people tend to overreact to small probability events, but underreact to large probabilities. Let denote a prospect with outcome with probability and outcome with probability and nothing with probability .
Why is the prospect theory wrong?
2.4 Prospect theory does not apply to real-world decisions. A major limitation of prospect theory is that it is always presented as a choice between two prospects, whereas in the real world, the decision is more likely to be whether or not to adopt a single prospect.
How do you calculate weighted probabilities?
Divide the number of ways to achieve the desired outcome by the number of total possible outcomes to calculate the weighted probability. To finish the example, you would divide five by 36 to find the probability to be 0.1389, or 13.89 percent.
### What is the fourfold pattern?
The four-fold pattern risk attitude suggests that when faced with a risky choice, people will be (1) risk seeking over low-probability gains, (2) risk averse over high-probability gains, (3) risk averse over low-probability loss, and (4) risk seeking over high-probability loss.
What is probability weighted sum?
Probability-weighted sum of the squared deviations of the possible values of the random variable from its mean, or expected value of the squared deviation from the mean. Var(X) ≡ σ2.
How do I figure out a weighted grade?
How to Calculate Weighted Class Grades
1. Determine Grade and Weight. Determine your grade on each assignment and the weight of the grade.
2. Multiply Grade by Weight. Multiply the grade on the assignment by the grade weight.
3. Add together.
4. Use an online grade calculator.
## How do you find the weighted distribution?
Weighted Distribution: The percentage of the total sales volume that comes from the served outlet. Let’s clear this by an example: you have 10 outlets in a beat, now out of these 10 outlets if your product is present in 4 outlets then numeric distribution is 40%.
What is fourfold risk pattern?
Prospect Theory (PT) predicts a “Fourfold Pattern of Risk Attitudes:” risk seeking for low-probability losses/high-probability gains, and risk aversion for high-probability losses/low-probability gains. Previously, subjects primarily exhibited this pattern for willingness-to-accept tasks.
What is a four fold risk?
### What is the basic prediction of prospect theory?
Prospect theory states that decision-making depends on choosing among options that may themselves rest on biased judgments. Thus, it built on earlier work conducted by Kahneman and Tversky on judgmental heuristics and the biases that can accompany assessments of frequency and probability.
What are the odds in poker?
Poker odds refer to the probability of an event happening and allow you to work out how much money you could make if your hand wins. Good odds mean a higher chance of success are calculated based on the equity of your hand. The higher the odds against you the lower your chances of winning. What are the odds of flopping a flush draw?
How do I increase the probability of winning poker?
To increase the probability of winning poker, you must become accustomed to the odds and probabilities that are presented to you in the game. Become familiar with outs and calculating your percentages of improving and be able to quickly relate these to the pot odds you may be getting, so that you can determine if you can call profitably or not.
## What is the probability of drawing a good hand in poker?
One would then expect to draw this hand about once in every 649,740 draws, or nearly 0.000154% of the time. Cumulative probability refers to the probability of drawing a hand as good as or better than the specified one.
What is the probability of getting pocket aces in poker?
The probability of getting pocket aces in any one hand if 6/1326. When playing against nine players, the probability of winning with pocket aces is 31.36% – assuming all players stay until the end. Poker aces win 85% of the time against an opponent, although this varies depending on the other hands around the table and the number of opponents. |
# TS Inter 1st Year Maths 1B Question Paper May 2019
Access to a variety of TS Inter 1st Year Maths 1B Model Papers and TS Inter 1st Year Maths 1B Question Paper May 2019 allows students to familiarize themselves with different question patterns.
## TS Inter 1st Year Maths 1B Question Paper May 2019
Time : 3 Hours
Max. Marks : 75
Note : This question paper consists of THREE sections A, B and C.
Section – A
(10 × 2 = 20 Marks)
I. Very short answer type questions :
2. Each question carries two marks.
Question 1.
Find the equation of the straight line which makes an angle 60° with the positive x-axis measured counter-clockwise and passing through the point (1, 2).
Solution:
Slope of the line m = tan 60° = $$\sqrt{3}$$.
Required straight line equation is y-2=V3(x-1)
⇒ y – 2 = $$\sqrt{3}$$(x – 1),
⇒ y – 2 = $$\sqrt{3}$$x – $$\sqrt{3}$$
⇒ $$\sqrt{3}$$x – y + (2 – $$\sqrt{3}$$) = 0
Question 2.
Find the value of P, if the straight lines 3x + 7y – 1 = 0 and 7x – py + 3 = 0 are mutually perpendicular.
Solution:
Given straight line equations are
3x + 7y – 1 = 0 ……… (1)
7x – py + 3 = 0 ……. (2)
Since (1), (2) are mutually perpendicular.
∴ 3(7) + 7(-P) = 0
⇒ 21 – 7P = 0
⇒ 7P = 21
⇒ P = 3
Question 3.
Find the distance between the mid point of the line segment $$\overline{\mathrm{AB}}$$ and the point (3, -1, 2), where A = (6, 3, -4) and B = (-2, -1, 2).
Solution:
Given A = (6, 3, -4)
B = (-2, -1, 2)
Let P be the midpoint of the line segment $$\overline{\mathrm{AB}}$$
Question 4.
Find the angle between the planes x + 2y + 2z – 5 = 0 and 3x + 3y + 2z – 8 = 0.
Solution:
Given plane equations are
x + 2y + 2z – 5 = 0 ….. (1)
3x + 3y + 2z – 8 = 0 …… (2)
Let ‘θ’ be the angle between the planes (1) and (2)
Question 5.
Find the left hand limit and right hand limit of the function, f(x) = $$\frac{|x-2|}{x-2}$$ at x = 2.
Solution:
Given f(x) = $$\frac{|x-2|}{x-2}$$, x ≠ 2
Question 6.
Compute :
Solution:
Question 7.
Find the derivative of the function, f(x) = 5 sinx + ex Log x.
Solution:
Given f(x) = 5 sinx + ex logx.
Differentiating w.r.to. ‘x’ on both sides, we have
f'(x) = 5cosx + ex. $$\frac{1}{x}$$ + logx.ex
= 5 cosx + ex($$\frac{1}{x}$$ + logx).
Question 8.
If x = esin hy, find $$\frac{\mathrm{dy}}{\mathrm{dx}}$$.
Solution:
Given x = esin hy
Taking logarithm on both sides, we have
logx = sin hy
Differentiating w.r.to ‘x’ on both sides, we have
$$\frac{1}{x}$$ = cos hy . $$\frac{d y}{d x}$$
⇒ $$\frac{d y}{d x}$$ = $$\frac{1}{x \cosh y}$$
Question 9.
Find dy and ∆y of y = f(x) = x2 + x at x = 10. when ∆x = 0.1.
Solution:
Given f(x) = x2 + x, x = 10 and ∆x = 0.1
dy = f’(x) ∆x
= (2x + 1) ∆x
= [2(10) + 1] (0.1)
= (21) (0.1)
= 2.1
∆y = f(x + ∆x) – f(x)
= (x + ∆x)2 + (x + ∆x) – (x2 + x)
= x2 + 2x ∆x + (∆x)2 + x + ∆x – x2 – x
= 2x ∆x + (∆x)2 + ∆x
= 2(10) (0.1)+ (0.1)2 +(0.1)
= 2 + 0.01 + 0.1
= 2.11.
Question 10.
Verify Rolle’s theorem for the function y = f(x) = x2 – 1 on [-1, 1].
Solution:
Given y = f(x) = x2 – 1
since f is a second degree polynomial
∴ f is continuous on [-1, 1] and f is derivable on (-1, 1)
Also f (-1) = (-1 )2 – 1 = 1 – 1 = 0
f(1) = 12 – 1 = 1 – 1 = 0
∴ f(-1) = f(1)
∴ f statistics all the conditions of Rolle’s theorem.
∴ There exists c ∈ (-1, 1) such that f'(c) = 0.
f(x) = x2 + x
⇒ f'(x) = 2x + 1
⇒ f'(x) = 2x + 1
⇒ f'(c) = 2c + 1
⇒ 2c = – 1
⇒ c = $$\frac{-1}{2}$$ ∈ (-1, 1)
Hence Roll’s theorem is verified.
Section – B (5 × 4 = 20 Marks)
1. Attempt any five questions.
2. Each question carries four marks.
Question 11.
A(5, 3) and B(3, -2) are two fixed points. Find the equation of the locus of a P, so that the area of triangle PAB is 9.
Solution:
Given A = (5, 3)
B = (3, – 2)
Let P(x1, y1) be any point on the locus.
Given geometric condition is area of ∆PAB is 9 units,
Question 12.
When the axes are rotated through an angle $$\frac{\pi}{4}$$, find the transformed equation of 3x2 + 10xy + 3y2 = 9.
Solution:
Given equation is 3x2 + 10xy + 3y2 – 9 = 0
Question 13.
Transform the equation $$\frac{x}{a}+\frac{y}{b}$$ = 1 into the normal form when a > 0 and b > 0. If the perpendicular distance of the straight line from the origin is p, deduce that $$\frac{1}{p^2}$$ = $$\frac{1}{a^2}+\frac{1}{b^2}$$.
Solution:
Given $$\frac{x}{a}+\frac{y}{b}$$ = 1
⇒ $$\frac{b x+a y}{a b}$$ = 1
Question 14.
Check the continuity of the function f given by
at the point 3.
Solution:
Given f(x) = $$\frac{x^2-9}{x^2-2 x-3}$$, 0 < x < 5, x ≠ 3
= 1.5, x = 3
Question 15.
If y = tan-1$$\left(\frac{2 x}{1-x^2}\right)$$(| x |< 1), then find $$\frac{d y}{d x}$$.
Solution:
Given y = tan-1 $$\left(\frac{2 x}{1-x^2}\right)$$(|x| < 1)
Put x = tan θ
⇒ θ = tan-1x.
∴ y = tan-1$$\left[\frac{2 \tan \theta}{1-\tan ^2 \theta}\right]$$
⇒ y = tan-1 (tan 2θ)
⇒ y = 2θ
⇒ y = 2 tan-1x.
Differentiating with respect to ‘x’ on both sides, we have
$$\frac{d y}{d x}$$ = $$\text { 2. } \frac{1}{1+x^2}$$
$$\frac{d y}{d x}$$ = $$\frac{2}{1+x^2}$$
Question 16.
Find the equations of tangent and normal to the curve x = cost, y = sint, at t = $$\frac{d y}{d x}$$.
Solution:
Question 17.
A stone is dropped into a quiet lake and ripples move in circles at the speed of 5cm/sec. At the instant when the radius of circular ripple is 8cm, how fast is the enclosed area increases ?
Solution:
Let r, A be the radius and area of circular ripple.
Given $$\frac{\mathrm{dr}}{\mathrm{dt}}$$ = 5cm/sec and r = 8.
A = πr2
⇒ $$\frac{\mathrm{dA}}{\mathrm{dt}}$$ = $$\pi .2 \mathrm{r} . \frac{\mathrm{dr}}{\mathrm{dt}}$$
when r = 8
$$\frac{\mathrm{dA}}{\mathrm{dt}}$$ = π.2r.$$\frac{\mathrm{dr}}{\mathrm{dt}}$$
= 80 sq.cm/sec.
Section – C
1. Attempt any five questions.
2. Each question carries seven marks.
Question 18.
Find the orthocentre of the triangle whose vertices are (-2, -1) (6, -1) and (2, 5).
Solution:
Let A = (-2, -1); B = (6, -1); C = (2, 5)
Question 19.
Show that the product of the perpendicular distances from the origin to the pair of straight lines represented by ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 is $$\frac{|c|}{\sqrt{(a-b)^2+4 h^2}}$$.
Solution:
Let S ≡ ax2 + 2hxy + by2 + 2gx + 2fy + c = 0
Let s = 0 represent pain of lines be
l1x + m1y + n1 = 0 …… (1)
l2x + m2y + n2 = 0 ……. (2)
∴ s ≡ (l1x + n1y + n1) (l2x + m2y + n2)
Comparing the co-efficients of like terms on both sides, we have
l1l2 = a; l1m1 = 2h
m1m2 = b; l1n2 + l2n1 = 2g
n1n2 = c; m1n2 + m2n1 = 2f
The perpendicular distance from origin to the line (1) = $$\frac{\left|\mathrm{n}_1\right|}{\sqrt{l_1^2+\mathrm{m}_1^2}}$$
The perpendicular distance from origin to the line (2) = $$\frac{\left|n_2\right|}{\sqrt{l_2^2+m_2^2}}$$
∴ The product of the perpendicular distances from origin to the pair of straigh lines represented by s = 0 is $$\frac{\left|n_1\right|}{\sqrt{l_1^2+m_1^2}} \cdot \frac{\left|n_2\right|}{\sqrt{l_2^2+m_2^2}}$$
= $$\frac{\left|n_1 n_2\right|}{\sqrt{\left(l_1^2+m_1^2\right)\left(l_2^2+m_2^2\right)}}$$
Question 20.
Find the condition for the fines joining the origin to the points of intersection of the circle x2 + y2 = a2 and the line lx + my = 1 to coincide.
Solution:
Equation of the circle is x2 + y2 = a2 ……… (1)
Equation of the line AB is lx + my = 1 ………. 2
Homogenising (1) with the help of (2)
∴ The combined equation of OA, OB is x2 + y2 = a2 . 12
= a2 (lx + my)2
= a2 (l2x2 + 2lmxy + m2y2)
= a2l2x2 + 2a2lmxy + a2m2y2
⇒ (a2l2 – 1) x2 + 2a2 lmxy + (a2m2 – 1) = 0
Since the lines OA, OB coincide.
∴ h2 = ab
⇒ (a2 lm)2 = (a2l2 – 1) (a2m2 – 1)
⇒ a4l2m2 = a4l2m2 – a2l2 – a2m2 + 1
⇒ 0 = – a2l2 – a2m2 + 1
⇒ a2l2 + a2m2 = 1
⇒ a2(l2 + m2) = 1.
Question 21.
Find the direction cosines of two lines which are connected by the relations l – 5m + 3n = 0 and 7l2 + 5m2 – 3n2 = 0.
Solution:
Given l – 5m + 3n = 0
⇒ l = 5m – 3n ……. (1)
7l2 + 5m2 – 3n2 = 0
⇒ 7(5m – 3n)2 + 5m2 – 3n2 = 0
⇒ 7(25m2 – 30mn + 9n2) + 5m2 – 3n2 = 0
⇒ 175m2 – 210mn + 63n2 + 5m2 – 3n2 = 0
⇒ 180m2 – 210mn + 60n2 = 0
⇒ 6m2 – 7mn + 2n2 = 0
⇒ 6m2 – 4mn – 3mn + 2n2 = 0
⇒ 2m(3m – 2n) – n (3m – 2n) = 0
⇒ (2m – n) (3m – 2n) = 0
⇒ 2m – n = 0 ………. (2)
3m – 2n = 0 ….. (3)
Solving (1) and (2)
Question 22.
Find the derivative of the function (sinx)logx + xsinx.
Solution:
Given y = (sin x)log x + xsin x.
Let u = (sinx)logx
Taking logarithms on bothsides, we have
log u = log x. log (sin x)
Differentiating w.r.to ‘x’ on bothsides, we have
Differentiating w.r.to x on bothsides, we have
Question 23.
Find the angle between the curves, y2 = 4x; x2 + y2 = 5.
Solution:
Given curve equations are
y2 = 4x ……… (1)
x2 + y2 = 5 ……… (2)
From (1) and (2)
x2 + 4x = 5 ⇒ x2 + 4x – 5 = 0.
⇒ x2 + 5x – x – 5 = 0
⇒ x (x + 5) – 1 (x + 5) = 0
⇒ (x – 1) (x + 5) = 0
⇒ x = 1, x = – 5
If x = 1 then y2 = 4
⇒ y = ± 2.
∴ The points of intersection of the curves (1) and (2) are (1, 2) and (1, -2).
y2 = 4x
Differentiating w.r.to ‘x’ on bothsides, we have
Differentiating w.r.to. ‘x’ on both sides, we have
Question 24.
A window is in the shape of a rectangle surrounded by a semicircle. If the perimeter of the window is 20 ft., find the maximum area.
Solution:
Let length of the rectangle be 2x and breadth by y and radius of the semi-circle in x.
∴ Perimeter = 20
⇒ y + 2x + y + πx = 20
⇒ 2x + 2y + πx = 20
⇒ 2y = 20 – 2x – πx
⇒ y = 10 – x – $$\frac{\pi}{2}$$x.
Window area = Area of the rectangle + Area of semi-circle |
You are here: Home Interpretation of graphs
## Example 1: Determining the equation of a parabola
### Question
Use the sketch below to determine the values of $a$ and $q$ for the parabola of the form $y=ax 2 +q$.
#### Examine the sketch
From the sketch we see that the shape of the graph is a “frown”, therefore $a<0$. We also see that the graph has been shifted vertically upwards, therefore $q>0$.
#### Determine $q$ using the $y$-intercept
The $y$-intercept is the point $(0;1)$.
$y=ax 2 +q1=a(0) 2 +q∴q=1$(1)
#### Use the other given point to determine a
Substitute point $(-1;0)$ into the equation:
$y=ax 2 +q0=a(-1) 2 +1∴a=-1$(2)
$a=-1$ and $q=1$, so the equation of the parabola is $y=-x 2 +1$.
## Example 2: Determining the equation of a hyperbola
### Question
Use the sketch below to determine the values of $a$ and $q$ for the hyperbola of the form $y=a x+q$.
#### Examine the sketch
The two curves of the hyperbola lie in the second and fourth quadrant, therefore $a<0$. We also see that the graph has been shifted vertically upwards, therefore $q>0$.
#### Substitute the given points into the equation and solve
Substitute the point $(-1;2)$:
$y=a x+q2=a -1+q∴2=-a+q$(3)
Substitute the point $(1;0)$:
$y=a x+q0=a 1+q∴a=-q$(4)
#### Solve the equations simultaneously using substitution
$2=-a+q=q+q=2q∴q=1∴a=-q=-1$(5)
$a=-1$ and $q=1$, the equation of the hyperbola is $y=-1 x+1$.
## Example 3: Interpreting graphs
### Question
The graphs of $y=-x 2 +4$ and $y=x-2$ are given. Calculate the following:
1. coordinates of $A$, $B$, $C$, $D$
2. coordinates of $E$
3. distance $CD$
#### Calculate the intercepts
For the parabola, to calculate the $y$-intercept, let $x=0$:
$y=-x 2 +4=-0 2 +4=4$(6)
This gives the point $C(0;4)$.
To calculate the $x$-intercept, let $y=0$:
$y=-x 2 +40=-x 2 +4x 2 -4=0(x+2)(x-2)=0∴x=±2$(7)
This gives the points $A(-2;0)$ and $B(2;0)$.
For the straight line, to calculate the $y$-intercept, let $x=0$:
$y=x-2=0-2=-2$(8)
This gives the point $D(0;-2)$.
For the straight line, to calculate the $x$-intercept, let $y=0$:
$y=x-20=x-2x=2$(9)
This gives the point $B(2;0)$.
#### Calculate the point of intersection $E$
At $E$ the two graphs intersect so we can equate the two expressions:
$x-2=-x 2 +4∴x 2 +x-6=0∴(x-2)(x+3)=0∴x=2or-3$(10)
At $E$, $x=-3$, therefore $y=x-2=-3-2=-5$. This gives the point $E(-3;-5)$.
#### Calculate distance $CD$
$CD=CO+OD=4+2=6$(11)
Distance $CD$ is 6 units.
## Example 4: Interpreting trigonometric graphs
### Question
Use the sketch to determine the equation of the trigonometric function $f$ of the form $y=af(θ)+q$.
#### Examine the sketch
From the sketch we see that the graph is a sine graph that has been shifted vertically upwards. The general form of the equation is $y=asinθ+q$.
#### Substitute the given points into equation and solve
At $N$, $θ=210 ࢪ$ and $y=0$:
$y=asinθ+q0=asin210 ࢪ +q=a-1 2+q∴q=a 2$(12)
At $M$, $θ=90 ࢪ$ and $y=3 2$:
$3 2=asin90 ࢪ +q=a+q$(13)
#### Solve the equations simultaneously using substitution
$3 2=a+q=a+a 23=2a+a3a=3∴a=1∴q=a 2=1 2$(14)
$y=sinθ+1 2$(15) |
Edit Article
# How to Graph a Circle
A circle is a two-dimensional shape made by drawing a curve. In trigonometry and other areas of mathematics, a circle is understood to be a particular kind of line: one that forms a closed loop, with each point on the line equidistant from the fixed point in the center. Graphing a circle is simple. Just start with Step 1.
### Part 1 Understanding the Mathematical Properties of Circles
1. 1
Note the center of the circle. The center is the point inside the circle that is at an equal distance from all of the points on the line.
2. 2
Know how to find the radius of a circle. The radius is the common and constant distance from all points on the line to the center of the circle. In other words, it is any line segment that joins the center of the circle with any point on the curved line.
3. 3
Know how to find the diameter of a circle. The diameter is the length of a line segment that connects two points on a circle and passes through the center of the circle. In other words, it represents the fullest distance across the circle.
• The diameter will always be twice the radius. If you know the radius, you can multiply by 2 to get the diameter; if you know the diameter; you can divide by 2 to get the radius.
• Remember that a line that connects two points on the circle (also known as a chord) but does not pass through the center will not give you the diameter; it will have a shorter distance.
4. 4
Learn how to denote a circle. Circles are defined primarily by their centers, so in mathematics, a circle’s symbol is a circle with a dot in the center. To denote a circle at a particular location on a graph, simply put the location of the center after the symbol.
• A circle located at point 0 would look like this: ⊙O.
### Part 2 Graphing the Circle
1. 1
Know the equation of a circle. The standard form for the equation of a circle is (x – a)2 + (y – b)2 = r2. The symbols a and b represent the center of the circle as a point on an axis, with a as the horizontal displacement and b as the vertical displacement. The symbol r represents the radius.
• As an example, take the equation x2 + y2 = 16.
2. 2
Find the center of your circle. Remember that the center of the circle is shown as a and b in the circle equation. If there are no brackets – as in our example – that means that a = 0 and b = 0.
• In the example, note that you can write (x – 0)2 + (y – 0)2 = 16. You can see that a = 0 and b = 0, and the center of your circle is therefore at the origin, at point (0, 0).
3. 3
Find the radius of the circle. Recall that the r represents the radius. Be careful: if the r part of your equation does not include a square, you will have to figure out your radius.
• So, in our example, you have a 16 for r, but there is no square. To get the radius, write r2 = 16; you can then solve to see that the radius is 4. Now you can write the equation as x2 + y2 =42.
4. 4
Plot the radius points on the coordinate plane. For whatever number you have for the radius, count that number is all four directions from the center: left, right, up, and down.
• In the example, you would count 4 in all directions to plot the radius points, since our radius is 4.
5. 5
Connect the dots. To graph the circle, connect the points using a round curve.
## Article Info
Categories: Geometry
In other languages:
Español: graficar un círculo, Русский: построить окружность, Deutsch: Einen Kreis grafisch darstellen, Italiano: Rappresentare un Cerchio, Português: Representar Graficamente um Círculo, Français: dessiner un cercle, Bahasa Indonesia: Menggambarkan Grafik Lingkaran
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# How do you solve 1/2(x-4)^2+3=11?
May 7, 2017
$x = 0 \text{ or } x = 8$
#### Explanation:
$\textcolor{b l u e}{\text{Isolate " (x-4)^2" on the left side}}$
$\text{subtract 3 from both sides}$
$\frac{1}{2} {\left(x - 4\right)}^{2} \cancel{+ 3} \cancel{- 3} = 11 - 3$
$\Rightarrow \frac{1}{2} {\left(x - 4\right)}^{2} = 8$
$\text{multiply both sides by 2}$
$\cancel{2} \times \frac{1}{\cancel{2}} {\left(x - 4\right)}^{2} = 2 \times 8$
$\Rightarrow {\left(x - 4\right)}^{2} = 16 \leftarrow {\left(x - 4\right)}^{2} \text{ isolated on left}$
$\textcolor{b l u e}{\text{take the square root of both sides}}$
$\sqrt{{\left(x - 4\right)}^{2}} = \textcolor{red}{\pm} \sqrt{16} \leftarrow \text{ note plus or minus}$
$\Rightarrow x - 4 = \pm 4$
$\text{add 4 to both sides}$
$x \cancel{- 4} \cancel{+ 4} = \pm 4 + 4$
$\Rightarrow x = 4 \pm 4 \leftarrow \textcolor{red}{\text{ 2 solutions}}$
$\Rightarrow {x}_{1} = 4 - 4 = 0 \text{ or } {x}_{2} = 4 + 4 = 8$ |
in
# How To Find Asymptotes Of A Graph
Click the blue arrow to submit and see the result! An asymptote is a line that a graph approaches, but does not intersect.
Characteristics of Graphs of Rational Expressions
### Rational functions contain asymptotes, as seen in this example:
How to find asymptotes of a graph. Y = x 2 4 x 2 = 1 4. A line that can be expressed by x = a, where a is some constant. Start by graphing the equation of the asymptote on a separate expression line.
This only applies if the numerator t(x) is not zero for the same x value). The calculator can find horizontal, vertical, and slant asymptotes. The direction can also be negative:
Find the asymptotes for the function. The graph has a vertical asymptote with the equation x = 1. When you have a task to find vertical asymptote, it is important to understand the basic rules.
The vertical asymptote of this function is to be. Initially, the concept of an asymptote seems to go against our everyday experience. Remember that an asymptote is a line that the graph of a function approaches but never touches.
The asymptote calculator takes a function and calculates all asymptotes and also graphs the function. An asymptote is a line that a curve approaches, as it heads towards infinity:. Find the vertical and horizontal asymptotes of the graph of f(x) = x2 2x+ 2 x 1.
(use n as an arbitrary integer if necessary. In this example, there is a vertical asymptote at x = 3 and a horizontal asymptote at y = 1. Find any asymptotes of a function definition of asymptote:
Asymptotes an asymptote is a line that a graph approaches without touching. To find vertical asymptotes, look for any circumstance that makes the denominator of a fraction equal zero. The curves approach these asymptotes but never cross them.
An oblique or slant asymptote is, as its name suggests, a slanted line on the graph. In other words, the fact that the function's domain is restricted is reflected in the function's graph. There are two main ways to find vertical asymptotes for problems on the ap calculus ab exam, graphically (from the graph itself) and analytically (from the equation for a function).
How to find horizontal asymptotes? Finding asymptotes vertical asymptotes are holes in the graph where the function cannot have a value. X 1 = 0 x = 1 thus, the graph will have a vertical asymptote at x = 1.
As x approaches positive infinity, y gets really. Imagine a curve that comes closer and closer to a line without actually crossing it. If we find any, we set the common factor equal to 0 and solve.
Enter the function you want to find the asymptotes for into the editor. To change its styling to a dotted line, click and long hold the icon. In this lesson, we will learn how to find vertical asymptotes, horizontal asymptotes and oblique (slant) asymptotes of rational functions.
Determining vertical asymptotes from the graph. As x approaches this value, the function goes to infinity. If an answer does not exist, enter dne.)
To graph a rational function, find the asymptotes and intercepts, plot a few points on each side of each vertical asymptote and then sketch the graph. In the meantime, it's possible to create an asymptote manually. Horizontal asymptote y = 3 the following graph has a horizontal asymptote of y = 0:
The horizontal asymptote is found by dividing the leading terms: This only applies if the numerator t(x) is not zero for the same x value). It's difficult for us to automatically graph asymptotes for a variety of reasons.
Vertical asymptotes can be found by solving the equation n(x) = 0 where n(x) is the denominator of the function ( note: Learn what that is in this lesson along with the rules that horizontal asymptotes follow. Removable discontinuities of rational functions.
The curve can approach from any side (such as from above or below for a horizontal asymptote), Vertical asymptotes are unique in that a single graph can have multiple vertical asymptotes. Find the asymptotes for the function.
In the following example, a rational function consists of asymptotes. In the above example, we have a vertical asymptote at x = 3 and a horizontal asymptote at y = 1. We draw the vertical asymptotes as dashed lines to remind us not to graph there, like this:
Specifically, the denominator of a. Those are the most likely candidates, at which point you can graph the function to check, or take the limit to see how the graph behaves as it approaches the possible asymptote. Physical representations of a curve on a graph, like.
If a graph is given, then look for any breaks in the graph. Find the vertical asymptotes (if any) of the graph of the function. The function $$y=\frac{1}{x}$$ is a very simple asymptotic function.
Therefore the calculation is easy, just calculate the zero (s) of the denominator, at that point is the vertical asymptote. The graph has a vertical asymptote with the equation x = 1. The concept of an asymptote.
A graph showing a function with two asymptotes. Always go back to the fact we can find oblique asymptotes by finding the quotient of the function’s numerator and denominator. A removable discontinuity occurs in the graph of a rational function at $x=a$ if a is a zero for a factor in the denominator that is common with a factor in the numerator.we factor the numerator and denominator and check for common factors.
A straight line on a graph that represents a limit for a given function. The curves approach these asymptotes but never. How do you find all asymptotes?
Indeed, you can never get it right on asymptotes without grasping these. To recall that an asymptote is a line that the graph of a function visits but never touches. How to find asymptotes:vertical asymptote.
Vertical asymptotes can be found by solving the equation n(x) = 0 where n(x) is the denominator of the function ( note: However, we hope to have this feature in the future! The vertical asymptotes will occur at those values of x for which the denominator is equal to zero:
Conversely, a graph can only have at most one horizontal, or one oblique asymptote. A vertical asymptote is a vertical line on the graph;
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## How I won the talent show
So I don’t want to brag or anything but I won the neighborhood talent show tonight. Look, here’s my trophy:
My act was “The Great Squarerootio.” I drank beer and computed square roots in my head. So I thought it might be fun to explain how to do this! Old hat for my mathematician readers, but this is elementary enough for everyone.
Here’s how it works. Somebody in the audience said “752.” First of all, you need to know your squares. I think I know them all up to 36^2 = 1296. In this case, I can see that 752 is between 27^2 = 729 and 28^2 = 784, a little closer to 729. So my first estimate is 27. This is not too bad: 27^2 = 729 is 23 away from 752, so my error is 23.
Now here’s the rule that makes things go:
new estimate = old estimate + error/(2*old estimate).
In this case, the error is 23 and my old estimate is 27, so I should calculate 23/2*27; well, 23 is just a bit less than 27, between 10 and 20% less, so 23/2*27 is a bit less than 1/2, but should be bigger than 0.4.
So our new estimate is “27.4 something.”
Actual value of 27 + 23/54: 27.4259…
so I probably would have gotten pretty close on the hundredth place if I’d taken more care with estimating the fraction. Of course, another way to get even closer is to take 27.4 as your “old estimate” and repeat the process! But then I’d have to know the square of 27.4, which I don’t by heart, and computing it mentally would give me some trouble.
Why does this work? The two-word answer is “Newton’s method.” But let me give a more elementary answer.
Suppose x is our original estimate, and n is the number we’re trying to find the square root of, and e is the error; that is, n = x^2 + e.
We want to know how much we should change x in order to get a better estimate. So let’s say we modify x by d. Then
$(x+d)^2 = x^2 + 2xd + d^2$
Now let’s say we’ve wisely chosen x to be the closest integer to the square root of n, so d should be pretty small, less than 1 at any rate; then d^2 is really small. So small I will decide to ignore it, and estimate
$(x+d)^2 = x^2 + 2xd$.
What we want is
$(x+d)^2 = n = x^2 +e$.
For this to be the case, we need e = 2xd, which tells us we should take d = e/2x; that’s our rule above!
Here’s another way to think about it. We’re trying to compute 752^(1/2), right? And 752 is 729+23. So what is (729+23)^(1/2)?
If you remember the binomial theorem from Algebra 2, you might remember that it tells you how to compute (x+y)^n for any n. Well, any positive whole number n, right? That’s the thing! No! Any n! It works for fractions too! Your Algebra 2 teacher may have concealed this awesome fact from you! I will not be so tight-lipped.
The binomial theorem says
$(x+y)^n = x^n + {n \choose 1} x^{n-1} y + {n \choose 2} x^{n-2} y^2 + \ldots +$
Plugging in x=729,y=23,n=1/2 we get
$(729+23)^{1/2} = 729^{1/2} + {1/2 \choose 1} 729^{-1/2} \cdot 23 + {1/2 \choose 2} 729^{-3/2} \cdot 23^2 + \ldots$
Now 729^{1/2} we know; that’s 27. What is the binomial coefficient 1/2 choose 1? Is it the number of ways to choose 1 item out of 1/2 choices? No, because that makes no sense. Rather: by “n choose 1” we just mean n, by “n choose 2” we just mean (1/2)n(n-1), etc.
So we get
$(729+23)^{1/2} = 27 + (1/2) \cdot 23 / 27 + (-1/8) \cdot 23^2 / 27^3 + \ldots$
And look, that second term is 23/54 again! So the Great Squarerootio algorithm is really just “use the first two terms in the binomial expansion.”
To sum up: if you know your squares up to 1000, and you can estimate fractions reasonably fast, you can get a pretty good mental approximation to the square root of any three-digit number. Even while drinking beer! You might even be able to beat out cute kids in your neighborhood and win a big honking cup! |
Section 4.4
# Section 4.4 - Notes for Day Review : . : e Eigenvalue...
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Unformatted text preview: Notes for Day Review : . : e Eigenvalue Problem Last time, we discussed the homogeneous system of rst-order equations y = P(t)y, fundamental sets of solutions, the solution matrix, and its Wronskian. Today, we'll discuss nding solutions to rst-order systems in which P(t) is a constant matrix. Introduction Suppose we had a homogeneous system of rst-order equations y = P(t)y . We remember from our discussion of rst-order equations that the solutions tend to involve functions of the form e t , where is a constant. If we look for a solution to the homogeneous system in this form, say, y= c e t c e t c e t , we get: c e t c e t c e t = P(t) c e t c e t c e t , which is the same thing as saying P(t)y = y. But we recall that this equation is true only when is an eigenvalue of P(t) and y is an eigenvector associated with that eigenvalue. e pair of an eigenvalue and its corresponding eigenvector is called an eigenpair. Today, we'll discuss systems where the eigenpairs are real and distinct; next time, we'll discuss systems with complex eigenpairs. ere is a third case involving repeated eigenvalues that is su ciently complicated that we'll delay treatment of this case until next week. p. , Example Find the general solution of the system: y y = = y + y y +y p. , Example Find the general solution of the system: - - - y = - y In this example, there are repeated eigenvalues. However, we are able to nd distinct eigenvectors, so the overall eigenpairs are distinct. p. , Example Find the general solution of the system: y y y = y -y -y = -y + y - y = -y - y + y e Phase Plane e phase plane is a type of direction eld (see . ) which we use to visualize the solutions to systems. systems involve three variables--t, y , and y --but in the phase plane, we'll put the variables y and y on the horizontal and vertical axes, and use arrows to indicate the direction of increasing t. Like directions elds, phase planes are tedious to generate by hand, but they're easy to generate using so ware. ese p. , Example e system y y = = . y + . y . y - . y has a general solution: y(t) = c et + c e -t - . Sketch the phase plane corresponding to this solution. ...
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## This note was uploaded on 04/05/2008 for the course MATH 2214 taught by Professor Edesturler during the Spring '06 term at Virginia Tech.
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# In an AP. Sp = q, Sq = p and Sr denotes the sum of first r terms. Then, Sp+q is equal to
Question:
In an AP. $S_{p}=q, S_{q}=p$ and $S_{r}$ denotes the sum of first $r$ terms. Then, $S_{p+q}$ is equal to
(a) 0
(b) −(p + q)
(c) p + q
(d) pq
Solution:
In the given problem, we are given $S_{p}=q$ and $S_{q}=p$
We need to find $S_{p+q}$
Now, as we know,
$S_{n}=\frac{n}{2}[2 a+(n-1) d]$
So,
$S_{p}=\frac{p}{2}[2 a+(p-1) d]$
$q=\frac{p}{2}[2 a+(p-1) d]$
$2 q=2 a p+p(p-1) d$ .............(1)
Similarly,
$S_{q}=\frac{q}{2}[2 a+(q-1) d]$
$p=\frac{q}{2}[2 a+(q-1) d]$
$2 p=2 a q+q(q-1) d$...............(2)
Subtracting (2) from (1), we get
$2 q-2 p=2 a p+[p(p-1) d]-2 a q-[q(q-1) d]$
$2 q-2 p=2 a(p-q)+[p(p-1)-q(q-1)] d$
$-2(p-q)=2 a(p-q)+\left[\left(p^{2}-q^{2}\right)-(p-q)\right]$
$-2=2 a+(p+q-1) d$ .............(3)
Now,
$S_{p+q}=\frac{p+q}{2}[2 a+(p+q-1) d]$
$S_{p+0}=\frac{(p+q)}{2}(-2)$ $\ldots$ (Using 3)
$S_{p+q}=-(p+q)$
Thus, $S_{p+q}=-(p+q)$
Hence, the correct option is (b). |
Multiplying and Dividing Real Numbers Worksheet | Problems & Solutions
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# Multiplying and Dividing Real Numbers Worksheet
Multiplying and Dividing Real Numbers Worksheet
• Page 1
1.
Find the quotient: ($11\frac{1}{5}$) ÷ (- 2)
a. $\frac{1}{5}$ b. - $5\frac{3}{5}$ c. $\frac{3}{5}$ d. - $5\frac{5}{3}$
#### Solution:
(111 / 5) ÷ (- 2)
[Write the expression.]
= (56 / 5) ÷ (- 2 / 1)
[Write improper fractions.]
= (565) · (- 12)
[Multiply by the reciprocal of - 2 / 1.]
= 56(- 1)52
= - 285
[Find the products.]
= - 535
[Write as a mixed number.]
The quotient of (111 / 5) ÷ (- 2) is - 53 / 5.
Correct answer : (2)
2.
Find the quotient: $7\frac{1}{8}$ ÷ 3
a. $2\frac{8}{3}$ b. $2\frac{3}{8}$ c. $1\frac{8}{3}$ d. $\frac{3}{8}$
#### Solution:
71 / 8 ÷ 3
[Write the expression.]
= (57 / 8) ÷ (3 / 1)
[Write improper fractions.]
= (578) · (13)
[Multiply by the reciprocal of 3 / 1.]
= 57183
= 198
= 238
[Write as a mixed number.]
The quotient of 71 / 8 ÷ 3 is 23 / 8.
Correct answer : (2)
3.
Find the quotient: $13\frac{1}{2}$ ÷ 9
a. 2 b. $2\frac{1}{2}$ c. $\frac{1}{2}$ d. $1\frac{1}{2}$
#### Solution:
131 / 2 ÷ 9
[Write the expression.]
= (27 / 2) ÷ (9 / 1)
[Write improper fractions.]
= (272) · (19)
[Multiply by the reciprocal of 9 / 1.]
= 32
= 112
[Write as a mixed number.]
The quotient of 131 / 2 ÷ 3 is 11 / 2.
Correct answer : (4)
4.
Using repeated addition, find the product of 4 and 2.
a. 8 b. 6 c. 10 d. 2
#### Solution:
4 x 2 = 2 + 2 + 2 + 2
[Multiplication by a positive integer can be modeled as repeated addition.]
= 8
[Use the addition rules.]
So, 4 x 2 = 8.
Correct answer : (1)
5.
Evaluate -4(-5)($y$) for $y$ = -2.
a. -40 b. 40 c. -42 d. -20
#### Solution:
-4(-5)(y)
[Original expression.]
= 20y
[The product is positive as there are even number of negative factors.]
= 20 (-2)
[Substitute -2 for y.]
= -40
.
[Since there is one negative factor, the product is negative.]
Correct answer : (1)
6.
A kite is moving up with a velocity of 17 ft per second. Write an algebraic model for its displacement, $d$ (in feet) after $t$ seconds.
a. $d$ = -17$t$ b. $d$ = 17$t$ c. $d$ = 15$t$ d. $d$ = 19$t$
#### Solution:
Displacement = Velocity x Time
[Formula.]
d = 17 x t
[As the kite is moving up, the velocity is indicated by positive value.]
d = 17t
[Algebraic model.]
Correct answer : (2)
7.
Reduce the quotient: (2$\frac{1}{2}$) -5
a. - $\frac{1}{2}$ b. $\frac{2}{3}$ c. $\frac{1}{2}$ d. None of the above
#### Solution:
(212) - 5
[Original expression.]
= (212) ÷ - 5
[Write the fraction as a division.]
= (52) ÷ - 5
[Write in improper fraction.]
= (52) × (1- 5)
[Use reciprocal to write a division expression as a product.]
= 52 × - 15
[a- b = - a / b]
= - 12
[Since there is one negative factor, the product is negative.]
Correct answer : (1)
8.
Evaluate: - 36 ÷ ($-1\frac{6}{12}$).
a. 24 b. 34 c. -24 d. 29
#### Solution:
-36 ÷ ( -16 / 12 ) = -36 ÷ ( -18 / 12 )
[Writing in improper fraction.]
= -36 x ( 12-18)
[Use reciprocal to write a division expression as a product.]
= -36 x -1218
[a / -b = -a / b.]
= 24
[Two negative factors, so product is positive.]
Correct answer : (1)
9.
Find the product of 5 and (- 8) by using repeated addition.
a. - 43 b. - 37 c. - 40 d. 40
#### Solution:
5(- 8) = (- 8) + (- 8) + (- 8) + (- 8) + (- 8)
[Multiplication by a positive integer can be modeled as repeated addition.]
= - 40
[Use the addition rules.]
Correct answer : (3)
10.
Using the definition of opposites and repeated addition, evaluate - 2(4).
a. 8 b. -8 c. 4 d. 10
#### Solution:
-2(4) = -(2)(4)
[Use the definition of opposites.]
= -(4 + 4)
[Multiplication by a positive integer can be modeled as repeated addition.]
= - (8)
[Use the addition rules.]
= - 8
[A product is negative if it has odd number of negative factors.]
Correct answer : (2)
*AP and SAT are registered trademarks of the College Board. |
# Logan is flying a kite that is made up of three similar rectangles. The sides of the three rectangles are in the ratio 1:2:3. If the area of the smallest rectangle is 72 square inches, what are the areas of the other two rectangles?
Oct 8, 2016
$288 , 648$
#### Explanation:
Area of rectangle ${A}_{R} = a \cdot b$, where $a$=length, $b$=width.
Given that the sides of the three similar rectangles are in the ratio of $1 : 2 : 3$,
Let the sides of the three similar rectangles be $a , b , 2 a , 2 b , 3 a , 3 b$, respectively.
Given that area of the smallest rectangle $= a \cdot b = 72$
$\implies 2 a \cdot 2 b = 4 a b = 4 \cdot 72 = 288$
$\implies 3 a \cdot 3 b = 9 a b = 9 \cdot 72 = 648$
Hence, the area of the other two rectangles are $288$, and $648$, respectively.
SIde note : if the sides of three similar rectangles are in the ratio of $1 : 2 : 3$. their areas will be in the ratio of $1 : {2}^{2} : {3}^{2} = 1 : 4 : 9$. |
# If -f(x) + f(x-1) = 3x and f(1)=1000, what is f(20) ?
Jul 12, 2016
$f \left(20\right) = 373$
#### Explanation:
Given
$- f \left(x\right) + f \left(x - 1\right) = 3 x \ldots \left(1\right) \text{ And } f \left(1\right) = 1000$
$\text{Putting", x=1 " in "(1) "we have}$
$\implies - f \left(1\right) + f \left(1 - 1\right) = 3 \cdot 1$
$\implies - 1000 + f \left(0\right) = 3$
$\implies f \left(0\right) = 1003$
$\text{Now putting "x=1,2,3,...20 " in "(1)" we get}$
$\cancel{- f \left(1\right)} + f \left(0\right) = 3 \cdot 1$
$\cancel{- f \left(2\right)} + \cancel{f \left(1\right)} = 3 \cdot 2$
$\cancel{- f \left(3\right)} + \cancel{f \left(2\right)} = 3 \cdot 3$
$\cancel{- f \left(4\right)} + \cancel{f \left(3\right)} = 3 \cdot 4$
$\ldots \ldots \ldots \ldots \ldots .$
$\ldots \ldots \ldots \ldots \ldots .$
$\ldots \ldots \ldots \ldots \ldots .$
$- f \left(20\right) + \cancel{f \left(19\right)} = 3 \cdot 20$
So adding above last 20 equations we finally get
$- f \left(20\right) + f \left(0\right) = 3 \left(1 + 2 + 3 + \ldots . + 20\right)$
$\implies - f \left(20\right) + 1003 = 3 \cdot \frac{20}{2} \left(20 + 1\right)$
$\implies - f \left(20\right) + 1003 = 630$
$\implies f \left(20\right) = 1003 - 630 = 373$ |
# Math Games for Junior High Students: Fun Games to Play at Home
Are you trying to get your junior high student actively engaged in his or her math studies? Consider using the fun games below to help your child improve his or her skills and motivation.
## What Will My Junior High Student Be Learning?
At the sixth grade level, children learn to fluently add, subtract, multiply and divide multi-digit numbers. These whole numbers may or may not contain decimals. Sixth graders also determine the greatest common factor (GCF) and least common multiple (LCM) of a set of whole numbers.
In seventh grade, students solve addition, subtraction, multiplication and division problems involving rational numbers, which includes negative numbers. Finally, eigth graders are introduced to irrational numbers, such as pi.
### Find the GCF and LCM
Before beginning this activity, write a variety of numbers on small pieces of paper and place them in a bowl. To provide ample practice, include both 1-digit and 2-digit numbers. Have your child select two pieces of paper from the bowl and calculate the greatest common factor (GCF) of the two numbers. Then, have your child determine the least common multiple (LCM) of the same two numbers.
For example, if the selected numbers were three and 40, then the GCF would be one because that's the only factor that the two integers have in common. The LCM is 120, which can be calculated by multiplying 40 x 3 = 120. Continue drawing numbers from the bowl and calculating the GCF and LCM until your child feels comfortable with these skills. If your child is more advanced, feel free to have him or her draw three numbers out of the bowl at a time.
For this addition activity, you will only need a deck of cards (remove the face cards before playing). Divide the deck in half and give each player a stack of cards. Red cards will represent negative numbers and black cards will represent positive numbers. For each round, players will turn over the top two cards from their stack and add them together.
For example, if a player turns over a black seven card and a red four card then he or she would add: 7 + (-4) = 3. At the end of each round, the player with the largest sum will take possession of all the cards played. The player with the most cards at the end of the game wins!
Did you find this useful? If so, please let others know!
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# How Many Eighty Fours Are In 672?
Trying to answer the question, “How many eighty fours are in 672?” can be a difficult task for some. It requires breaking the number down into its component parts and then working out the answer. Fortunately, as long as you have some basic math skills, you should be able to figure out the answer quite quickly. In this article, we’ll look at how to calculate the answer to this question and why it’s important to understand the concept of numbers in general.
## Understanding the Concept of Numbers
Before attempting to answer the question “How many eighty fours are in 672?”, it’s important to understand the concept of numbers. Numbers are used to represent a quantity of something. For example, the number 8 represents eight items, and the number 72 represents seventy-two items. When we combine two numbers, like 8 and 72, we get the number 672. This number represents the combined amount of the two numbers. In this case, 672 represents six hundred and seventy-two items.
## Breaking Down the Problem
Now that we understand the concept of numbers, we can begin to break down the problem. To answer the question, “How many eighty fours are in 672?”, we need to first figure out how many eighty fours we can fit into 672. To do this, we can divide 672 by 84. When we do this, we get 8. This means that 8 eighty fours fit into 672. As such, the answer to the question “How many eighty fours are in 672?” is 8.
## Why It’s Important to Understand Numbers
Knowing how to answer the question, “How many eighty fours are in 672?” is an important part of understanding numbers in general. It’s important because it helps us to better understand the relationships between different numbers and how to calculate the answers more quickly and accurately. Furthermore, it helps us to understand how to break down problems into smaller, more manageable pieces.
##
In conclusion, the answer to the question “How many eighty fours are in 672?” is 8. To answer this question, we need to understand the concept of numbers and how to break down a problem into its component parts. Knowing how to answer this question is an important part of understanding numbers in general and can help us to better understand the relationships between different numbers and how to calculate the answers more quickly and accurately. |
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# What is the derivative of a constant value such as 1?
Are you curious to know what is the derivative of 1? You have come to the right place as I am going to tell you everything about the derivative of 1 in a very simple explanation. Without further discussion let’s begin to know what is the derivative of 1?
## What is the Derivative of 1?
Calculus, a branch of mathematics developed by luminaries like Isaac Newton and Gottfried Wilhelm Leibniz, is a powerful tool for understanding how things change. One of the fundamental concepts in calculus is the derivative, which measures the rate of change of a function at a particular point. However, you may wonder, what happens when we take the derivative of a constant, such as 1? The answer is both intriguing and essential in understanding the basics of calculus.
## Understanding derivatives
Before we delve into why the derivative of 1 is equal to 0, let’s refresh our understanding of derivatives. A derivative measure how a function changes as its input (usually denoted as x) changes. In simple terms, it answers the question: “How fast is a function changing at a specific point?”
## Derivative of a constant
Now, let’s consider a constant, like the number 1. When we talk about a constant in calculus, we are essentially discussing a function where the output value remains the same, regardless of changes in the input. In this case, the function can be represented as f(x) = 1 for all values of x.
So, when we take the derivative of the constant function f(x) = 1, we are essentially asking how fast it’s changing at any point. Since the function never changes, the answer is straightforward: the rate of change is zero.
## Mathematically, we express this as:
f'(x) = 0
This tells us that the derivative of a constant function is always zero. In other words, no matter where you pick on the graph of the function f(x) = 1, the slope (rate of change) at that point will be zero because the function is perfectly flat.
## Why it matters?
Understanding the derivative of a constant is fundamental in calculus and has various practical applications:
Tangent Lines: Derivatives help find the equations of tangent lines to curves. When dealing with horizontal lines (which have a slope of zero), you’re essentially dealing with the derivative of a constant.
Rate of Change: In real-world applications, we often encounter constants that represent unchanging values. Calculating their derivatives helps determine the rate of change or the lack thereof.
Calculus Foundations: Learning the derivative of a constant is a fundamental step in understanding more complex derivatives, such as those of polynomial, trigonometric, and exponential functions.
## Conclusion
In calculus, the derivative of a constant, like 1, is always equal to 0. This concept may appear simple, but it forms the foundation for understanding how functions change and how we can calculate rates of change, tangent lines, and more. The derivative of a constant reminds us that, in the world of calculus, some things stay perfectly still, while others are in a constant state of flux.
## FAQ
### What is the derivative of negative 1?
The derivative of positive and negative numbers (Constants) is zero, the rate of change or derivative does not change if the number is positive or negative.
### What derivatives equal 1?
The derivative of x will be equal to 1. Both the power rule and the first principle can be used to find the derivative of x. By using n =1 in the power given by dxn/dx = nxn-1, the derivative of x can be determined. As f(x) = x represents a straight line, hence, the derivative will be 1 at all points.
### Is the derivative of a constant 1?
The derivative (Dx) of a constant (c) is zero. Constant Coefficient Rule: The Dx of a variable with a constant coefficient is equal to the constant times the Dx. The constant can be initially removed from the derivation. Chain Rule: There is nothing new here other than the dx is now something other than 1.
### What is the derivative of 2?
Explanation: The derivative is the measure of the rate of change of a function. 2 is a constant whose value never changes. Thus, the derivative of any constant, such as 2, is 0. |
# Analytical Reasoning1(www.allexamreview.com)
```Reasoning
One of the important topic in Bank Po’s is nonverbal analytical reasoning. Now we will discuss
some examples.
1. Find the minimum number of straight lines required to make the given figure.
The figure may be labeled as shown.
The horizontal lines are IJ, AB, EF, MN, HG, DC and LK i.e. 7 in number.
The vertical lines are AD, EH, IL, FG, BC and JK i.e. 6 in number.
Thus, there are 7 + 6 = 13 straight lines in the figure.
2. Find the minimum number of straight lines required to make the given figure.
.The figure may be labelled as shown.
The horizontal lines are IK, AB, HG and DC i.e. 4 in number.
The vertical lines are AD, EH, JM, FG and BC i.e. 5 in number.
The slanting lines are IE, JE, JF, KF, DE, DH, FC and GC i.e. 8 is number.
Thus, there are 4 + 5 + 8 = 17 straight lines in the figure.
3. Find the number of triangles in the given figure.
The figure may be labelled as shown.
The simplest triangles are AHB, GHI, BJC, GFE, GIE, IJE, CEJ and CDE i.e. 8 in number.
The triangles composed of two components each are HEG, BEC, HBE, JGE and ICE i.e.
5 in number.
The triangles composed of three components each are FHE, GCE and BED i.e. 3 in number.
There is only one triangle i.e. AGC composed of four components.
There is only one triangle i.e. AFD composed of nine components.
Thus, there are 8 + 5 + 3 + 1 + 1 = 18 triangles in the given figure.
4. Find the number of triangles in the given figure.
The figure may be labelled as shown.
The simplest triangles are IJO, BCJ, CDK, KQL, MLQ, GFM, GHN and NIO i.e. 8 in number.
The triangles composed of two components each are ABO, AHO, NIJ, IGP, ICP, DEQ, FEQ,
KLM, LCP and LGP i.e.10 in number.
The triangles composed of four components each are HAB, DEF, LGI, GIC, ICL and GLG i.e.
6 in number.
Total number of triangles in the figure = 8 + 10 + 6 = 24.
5. Find the number of triangles in the given figure.
The figure may be labelled as shown.
The simplest triangles are AGE, EGC, GFC, BGF, DGB and ADG i.e. 6 in number.
The triangles composed of two components each are AGC, BGC and ABG i.e. 3 in number.
The triangles composed of three components each are AFC, BEC, BDC, ABF, ABE and DAC
i.e. 6 in number.
There is only one triangle i.e. ABC composed of six components.
Thus, there are 6 + 3 + 6 + 1 = 16 triangles in the given figure.
6. Find the number of triangles in the given figure.
The figure may be labelled as shown.
The simplest triangles are AKI, AIL, EKD, LFB, DJC, BJC, DHC and BCG i.e. 8 in number.
The triangles composed of two components each are AKL, ADJ, AJB and DBC i.e. 4 in number.
The triangles composed of the three components each are ADC and ABC i.e. 2 in number.
There is only one triangle i.e. ADB composed of four components.
Thus, there are 8+ 4 + 2 + 1= 15 triangles in the figure.
7. Find the number of triangles in the given figure.
The figure may be labelled as shown.
The simplest triangles are APQ, AEQ, QTU, QRU, BGS, BHS, RSU, SUV, TUW, UWX,
NWD, WDM, UVY, UXY, JCY and YKC i.e. 16 in number.
The triangles composed of two components each are QUW, QSU, SYU and UWY i.e. 4 in number.
The triangles composed of three components each are AOU, AFU, FBU, BIU, UIC, ULC, ULD
and OUD i.e. 8 in number.
The triangles composed of four components each are QYW, QSW, QSY and SYW i.e. 4 in number.
The triangles composed of six components each are AUD, ABU, BUC and DUC i.e. 4 in number.
The triangles composed of seven components each are QMC, ANY, EBW, PSD, CQH, AGY,
DSK and BJW i.e. 8 in number.
The triangles composed of twelve components each are ABD, ABC, BCD and ACD i.e. 4 in number.
Thus, there are 16 + 4 + 8 + 4 + 4 + 8 + 4 = 48 triangles in the figure.
8. Find the minimum number of straight lines required to make the given figure.
The figure may be labelled as shown.
The horizontal lines are DE, FH, IL and BC i.e. 4 in number.
The slanting lines are AC, DO, FN, IM, AB, EM and HN i.e. 7 in number.
Thus, there are 4 + 7 = 11 straight lines in the figure.
9. What is the number of straight lines and the number of triangles in the given figure?
The figure may be labelled as shown.
The Horizontal lines are DF and BC i.e. 2 in number.
The Vertical lines are DG, AH and FI i.e. 3 in number.
The Slanting lines are AB, AC, BF and DC i.e. 4 in number.
Thus, there are 2 + 3 + 4 = 9 straight lines in the figure.
Now, we shall count the number of triangles in the figure.
The simplest triangles are ADE, AEF, DEK, EFK, DJK, FLK, DJB, FLC, BJG and LIC i.e.
10 in number.
The triangles composed of two components each are ADF, AFK, DFK, ADK, DKB, FCK, BKH,
KHC, DGB and FIC i.e. 10 in number.
The triangles composed of three components each are DFJ and DFL i.e. 2 in number.
The triangles composed of four components each are ABK, ACK, BFI, CDG, DFB, DFC and
BKC i.e. 7 in number.
The triangles composed of six components each are ABH, ACH, ABF, ACD, BFC and CDB i.e.
6 in number.
There is only one triangle i.e. ABC composed of twelve components.
There are 10 + 10 + 2 + 7 + 6+ 1 = 36 triangles in the figure.
10. Find the number of triangles in the given figure.
The figure may be labelled as shown.
The simplest triangles are GLK, DLJ, DJM, HMN, QRE, IRA, IPA and FPO i.e. 8 in number.
The triangles having two components each are BDO, CDQ, DLM, PRA, KFI, NEI, HJI, GJI,
DKI and DNI i.e. 10 in number.
The triangles having four components each are DIE, DFI, DOA, DQA andGHI i.e. 5 in
number. The triangles having six components each are DCA and DBA i.e. 2 in number.
DEF is the only triangle having eight components.
ABC is the only triangle having twelve components.
Thus, there are 8+10+ 5 + 2+1 + 1 = 27 triangles in the figure.
11. Find the minimum number of straight lines required to make the given figure.
The figure may be labelled as shown.
The horizontal lines are AK, BJ, CI, DH and EG i.e. 5 in number.
The vertical lines are AE, LF and KG i.e. 3 in number.
The slanting lines are LC, CF, FI, LI, EK and AG i.e. 6 in number.
Thus, there are 5 + 3 + 6 = 14 straight lines in the figure.
12. Find the number of triangles in the given figure.
The figure may be labelled as shown.
The simplest triangles are AML, LRK, KWD, DWJ, JXI, IYC, CYH, HTG, GOB, BOF, FNE
and EMA i.e. 12 in number.
The triangles composed of two components each are AEL, KDJ, HIC and FBG i.e. 4 in number.
The triangles composed of three components each are APF, EQB, BQH, GVC, CVJ, IUD, DUL
and KPA i.e. 8 in number.
The triangles composed of six components each are ASB, BSG, CSD, DSA, AKF, EBH,
GGJ and IDL i.e. 8 in number.
The triangles composed of twelve components each are ADB, ABC, BCD and CDA i.e. 4 in
number. Total number of triangles in the figure = 12 + 4 + 8 + 8 + 4 = 36.
13. What is the number of triangles that can be formed whose vertices are the vertices of an
octagon but have only one side common with that of octagon?
When the triangles are drawn in an octagon with vertices same as those of the octagon and
having one side common to that of the octagon, the figure will appear as shown in (Fig. 1).
Now, we shall first consider the triangles having only one side AB common with octagon
ABCDEFGH and having vertices common with the octagon (See Fig. 2).Such triangles are
ABD, ABE, ABF and ABG i.e. 4 in number.
Similarly, the triangles having only one side BC common with the octagon and also having
vertices common with the octagon are BCE, BCF, BCG and BCH (as shown in Fig. 3). i.e.
There are 4 such triangles.
This way, we have 4 triangles for each side of the octagon. Thus, there are 8 x 4 = 32 such triangles.
14. Find the number of triangles in the given figure.
The figure may be labelled as shown.
The simplest triangles are BFG, CGH, EFM, FMG, GMN, GHN, HNI, LMK, MNK and KNJ i.e.
10 in number.
The triangles composed of three components each are FAK and HKD i.e. 2 in number.
The triangles composed of four components each are BEN, CMI, GLJ and FHK i.e. 4 in number.
The triangles composed of eight components each are BAJ and OLD i.e. 2 in number.
Thus, there are 10 + 2 + 4 + 2 = 18 triangles in the given figure.
15. Find the number of triangles in the given figure.
The figure may be labelled as shown.
The simplest triangles are AHL, LHG, GHM, HMB, GMF, BMF, BIF, CIF, FNC, CNJ, FNE,
NEJ, EKJ and JKD i.e. 14 in number.
The triangles composed of two components each are AGH, BHG, HBF, BFG, HFG, BCF, CJF,
CJE, JEF, CFE and JED i.e. 11 in number.
The triangles composed of four components each are ABG, CBG, BCE and CED i.e. 4 in number.
Total number of triangles in the given figure = 14 + 11 + 4 = 29.
16. Find the number of triangles in the given figure.
The figure may be labeled as shown.
The simplest triangles are ABJ, ACJ, BDH, DHF, CIE and GIE i.e. 6 in number.
The triangles composed of two components each are ABC, BDF, CEG, BHJ, JHK, JKI and CJI
i.e. 7 in number.
There is only one triangle JHI which is composed of four components.
Thus, there are 6 + 7 + 1 = 14 triangles in the given figure
Next issue we will discuss other analytical problem solving questions.
``` |
Proficiency Test Study Guide
MATHEMATICS / ALGEBRA OVERVIEW Key Words and Converting Words to Equations Fractions: Adding, subtracting, multiplying, dividing Simplifying Writing decimals as fractions Statistics Reading Tables and Charts Exponents Pre-Algebra and Algebra: Special notation for multiplication and division with variable Algebra word problems Order of operations Simplifying expressions Prime factorization Greatest common factor Least common multiple Factoring Sample algebra problems Coordinate System: Grid graph Slope coordinates Geometry: Basics Squares, rectangles, circles, triangles Math Definitions
ENGLISH Proof reading / spelling Reading comprehension Main theme of a paragraph Logical sequence of a paragraph Key word English grammar Basic word meanings
ABILITY TO ASSIST Worker roles and responsibilities Student discipline / behavior
WRITING Content Format Grammar Spelling Punctuation
MDUSD Proficiency Test Study Guide / Page 2
MATH
Key Words and Converting Words to Equations
Sometimes math questions use key words to indicate what operation to perform. Becoming familiar with these key words will help you determine what the question is asking for:
OPERATION OTHER WORDS WHICH INDICATE THE OPERATION Addition Increased by; more than; combined together; total of; sum; added to. Subtraction Decreased by; minus; less; difference between/of; less than; fewer than. Multiplication Of; times; multiplied by; product of (For example 4 + 4 + 4 equals 4 X3). Division Per; a; out of; ratio of; quotient of; percent (divide by 100). Equal Is; are; was; will be; gives; yields; sold for Per Divided by
Here are some examples of words converted to equations:
WORDS EQUATIONS What is the sum of 8 and y? 8 + y 4 less than y y – 4 Y multiplied by 13 13y The quotient of y and 3 y / 3 The difference of 5 and y 5 – y The ratio of 9 more than y to y (y+9) / y Nine less than the total of a number (y) and two (y+2) = 9 or y - 7
Fractions:
In order to accurately solve fraction problems it is important to distinguish between the numerator and denominator.
Numerator: Top number Denominator: Bottom number
MATHEMATICS PRACTICE QUESTIONS 1. 8 9/16 - 2 4/16 a. 10 13/16 b. 5 13/16 c. 6 5/16 d. 6 13/16 2. 4 3/4 + 6 3/5 a. 11 7/20 b. 10 2/3 c. 11 2/3 d. 10 7/20 3. 4 4/5 x 6 2/8 = a. 10 3/5 b. 30 c. 25 d. -24 6/8 4. 6 2/3 ÷ 4 2/6 = a. 2 1/13 b. 28 8/9 c. 3 2/6 d. 1 7/13 5. 80% of what is 204? a. 250 b. 240 c. 255 d. 260 6. 50 people went to a play. 3/5 of the people stayed to the end. How many people left? a. 10 b. 30 c. 20 d. 40 7. Estimate the answer: 4.88 x 6.24 a. 30 b. 24 c. 35 d. 28 8. What is the probability of rolling a 4 on a set of dice? a. 1:6 b. 2:6 c. 1:12 d. 1:4 9. How many times bigger is b than a: a. 2 times b. ½ times c. 3 times d. 1 1/2 times a b 10. Jean goes 385 miles on 14 liters of gas. What was the miles per liter? a. 25 miles / liter b. 26.3 miles / liter c. 27.5 miles / liter d. 27 miles / liter
Answers: 1. (C) 2. (A 3. (B) 4. (D) 5. (C) 6. (C) ’7. (A) 8. (A) 9. (A) 10. (C)
MDUSD Proficiency Test Study Guide / Page 3
PRE-ALGEBRA AND ALGEBRA
SPECIAL NOTATION FOR MULTIPLICATION AND DIVISION WITH VARIABLES:
Here are some examples of special notations and what they mean:
2b means 2 x b
2(a + 5) means twice the sum of a number (a) and five
bc means b x c
4bc means 4 x b x c
d/5 means d ÷ 5
ALGEBRA WORD PROBLEMS:
In algebra you solve problems by essentially making two groups, one for each side of an equation. An unknown number or value is represented by a letter (for example: x).
Basic Steps
1. Define the variable
2. Translate the problem into an equation and plug in known values
3. Solve the equation
4. Go back to the problem and plug in the new value to obtain the answer
SAMPLE A car dealership has 15 new cars and 12 used cars. How many cars to they have? Define the unknown variable Let x = Total Cars Translate the problem into an equation and plug known values in 15 + 12 = x Solve the equation 27 = x Answer There are 27 total cars
CONSECUTIVE INTEGER WORD PROBLEMS:
SAMPLE Two consecutive numbers have a sum of 71. What are the numbers? Define the unknown variables: Since two numbers are unknown, in order to solve it you must use only one variable (such as x) in the equation. Let x = The First Consecutive Number Let x + 1 = The Second Consecutive Number Translate the problem into an equation and plug known values in. x + (x + 1) = 71 Solve the equation x + (x + 1) = 71 Remove parenthesis 2x + 1 = 71 Subtract 1 from each side -1 + 2x + 1 = 71 – 1 2x = 70 Divide both sides by 2 2x = 70 2 2 x = 35 Go back to the problem and plug in new values 35 = The First Consecutive Number 35 + 1 = The Second Consecutive Number Answer 35 & 36
MDUSD Proficiency Test Study Guide / Page 4
Other tips for consecutive number word problems:
• When solving for negative consecutive numbers, ignore the negative sign and do not do anything differently. Keep the x positive and when you have obtained your answer, add the negative sign.
• When solving for even or odd consecutive numbers, add a space to the equation. For example: The next consecutive number after 15 can be found by adding 1 to it. The next even/odd number can be found by adding 2.
PRIME FACTORIZATION:
A prime number is a positive integer greater than one that can only be divided by itself and one.
Some examples are 2, 3, 5, 7, 11, 13, 17, and 19.
A composite number is a positive integer greater than one that has more than one divisor other than one and itself. Some examples are 4, 6, 9. 15, and 21
One is neither a prime nor a composite number.
Ways to obtain the prime factor
• Repeatedly divide by prime numbers.
• Choose any pair of factors and split these factors until all the factors are prime.
• Work backwards from the answers, seeing which one is BOTH only prime numbers, and produces the correct product.
ALGEBRA PRACTICE QUESTIONS 1. 15 + (- 8) = a. 23 b. 7 c. -23 d. -7 2. Select the prime factorization for 84 a. 1 x 2 x 42 b. 2 x 2 x 21 c. 3 x 2 x 8 d. 2 x 2 x 3 x 7 3. Which answer is equivalent to: 7 – 5y < 3 (4y – 2) a. 2y < 12y – 2 b. -17y < -9 c. 7-5y < 12 y - 6 d. 9 > 3 (9y) 4. Evaluate the following when r = 2, s = 8, and t = 5: t + 1 ------- + 4s r 29 35 38 44 7. Joan’s bedroom is 25.3 feet on one side and 10.5 feet on the other. What is the area of the bedroom? a. 106.26 b. 139.15 c. 255.55 d. 265.65 8. Solve for the variable: 48.3 + x = 71.2 a. 119.5 b. 23.9 c. 22.9 d. 32.9 9. Solve for the variable: 12a = 60 a. 72 b. 5 c. 48 d. 5
Answers: 1. (B) 2. (D) 3. (C) 4. (B) 5. (C) 6. (C) 7. (D) 8. (C) 9. (B)
MDUSD Proficiency Test Study Guide / Page 5
ENGLISH:
Main Theme of a Paragraph:
These questions ask you to first read a paragraph and then choose an answer based on the main idea of the paragraph. The correct answer usually restates the main idea using different wording or requires that you draw a conclusion from the contents of the paragraph.
SAMPLE
A successful weight loss program must contain a specific plan designed to achieve healthy weight loss for an individual. An appropriate plan, without necessary determination to carry it out, is useless. Determination, without a well-defined plan, will only achieve partial success.
The MAIN theme of this paragraph is
A well-defined plan will assure the success of a weight loss program.
1. A high degree of determination is necessary and sufficient for a highly successful weight loss program.
2. It’s impossible to develop a successful weight loss program.
3. Two important ingredients of a successful weight loss program are a well-defined plan and a sincere resolve to implement that plan.
Solution: To answer this question, evaluate each choice.
Choice (a) only contains one of the points: a well-defined plan; therefore, this choice is only partially correct.
Choice (b) also only lists one of the points: determination; therefore, this choice is only partially correct.
Choice (c) is not supported by evidence within the paragraph; therefore this choice is incorrect.
Choice (d) restates the idea presented in the paragraph. This choice is correct.
Logical Sequence of a Paragraph:
Some questions ask you to evaluate a paragraph for a smooth, logical progression of ideas. This is known as logical sequencing. First, it is important to know the structure of a paragraph. The topic sentence is the first sentence of a paragraph; it introduces the main theme. The supporting sentences give details and develop the main theme; they usually follow the topic sentence. The closing sentence wraps up the paragraph by restating the main theme, drawing a conclusion, or presenting a transition to another paragraph.
Some questions to ask when evaluating a paragraph’s logical sequence are:
• What is the main theme of the paragraph?
• In what order should the ideas follow?
• Are there ideas that are an extension of the main theme?
• Are there ideas that can’t be understood until other things are explained?
•
MDUSD Proficiency Test Study Guide / Page 6
ABILITY TO ASSIST:
To study for questions related to the ability to assist instruction, it is important to think about the role of a Paraprofessional / Instructional Aide and to answer questions based on this role. Paraprofessional / Instructional Aides should have knowledge of basic child guidance and development characteristics and principles and appropriate ways to manage student behavior.
Paraprofessional / Instructional Aides also need to:
• Follow instructions provided by the teacher (verbal and written).
• Be positive when interacting with students, parents, and school personnel.
• Communicate and be respectful while interacting with students and families from diverse cultures.
• Keep student information confidential (personal information, test results, medical history, etc.)
• Tutor students (individually and in small groups).
• Watch and help students in other learning environments (library, computer lab).
• Score teacher-developed tests and file information accurately.
SAMPLE
When communicating with parents from a different culture, it is most important to
1. do all of the talking so they feel more comfortable
2. be respectful of the differences between your culture and theirs
3. realize that their level of communication is not as refined as yours
4. point out your cultural differences at the beginning of the conversation
Solution: to answer this question, evaluate each choice.
Choice (a) is incorrect. Successful communication involves both speaking and listening.
Choice (b) is correct. Being respectful of cultural differences encourages open communication.
Choice (c) is incorrect. Being from another culture doesn’t mean that their level of communication is better or worse than yours.
Choice (d) is incorrect. Pointing out cultural differences may create a negative communication environment. It is best to focus on similarities between both of you, such as concern for their child.
WRITING:
Write an essay on an assigned topic. Essay will be graded on:
• Content
• Grammar
• Spelling
• Punctuation
• Penmanship |
# Texas Go Math Grade 5 Lesson 1.5 Answer Key Round Decimals
Refer to our Texas Go Math Grade 5 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 5 Lesson 1.5 Answer Key Round Decimals.
## Texas Go Math Grade 5 Lesson 1.5 Answer Key Round Decimals
Unlock the Problem
The Gold Frog of South America is one of the smallest frogs in the world. It is 0.386 of an inch long. What is this length rounded to the nearest hundredth of an inch?
• Underline the length of the Gold Frog.
• Is the frog’s length about the same as the length or the width of a large paper clip?
One Way
Use a place-value chart.
• Write the number in a place-value chart and circle the digit in the place value to which you want to round.
• In the place-value chart, underline the digit to the right of the place to which you are rounding.
• If the digit to the right is less than 5, the digit in the place value to which you are rounding stays the same.
• If the digit to the right is 5 or greater, the digit in the rounding place increases by 1.
• Drop the digits after the place to which you are rounding.
Think: Does the digit in the rounding place stay the same or increase by 1?
So, to the nearest hundredth of an inch, a Gold Frog is about 0.39 of an inch long.
Another Way
Use place value.
The Little Grass Frog is the smallest frog in North America. It is 0.437 of an inch long
A. What is the length of the frog to the nearest hundredth of an inch?
So, to the nearest hundredth of an inch,
The frog is about 0.44 of an inch long.
B. What is the length of the frog to the nearest tenth of an inch?
So, to the nearest tenth of an inch,
The frog is about 0.4 of an inch long.
Share and Show
Write the place value of the underlined digit. Round each number to the place of the underlined digit.
Question 1.
0.673
The given decimal number is: 0.673
Now,
The representation of the given number in the place value chart is:
So,
From the given place value chart,
The place value of 7 is: Hundredths
So,
The rounded number that is to the place of the underlined digit is: 0.67
Hence, from the above,
We can conclude that
The place value of 7 is: Hundredths
The rounded number that is to the place of the underlined digit is: 0.67
Question 2.
4.282
The given decimal number is: 4.282
Now,
The representation of the given number in the place value chart is:
So,
From the given place value chart,
The place value of 2 is: Tenths
So,
The rounded number that is to the place of the underlined digit is: 4.2
Hence, from the above,
We can conclude that
The place value of 2 is: Tenths
The rounded number that is to the place of the underlined digit is: 4.2
Question 3.
12.917
The given decimal number is: 12.917
Now,
The representation of the given number in the place value chart is:
So,
From the given place value chart,
The place value of 2 is: Ones
So,
The rounded number that is to the place of the underlined digit is: 13
Hence, from the above,
We can conclude that
The place value of 2 is: Ones
The rounded number that is to the place of the underlined digit is: 13
Name the place value to which each number was rounded.
Question 4.
0.982 to 0.98
The given decimal number is: 0.982
Now,
The representation of the given number in the place value chart is:
Now,
When we observe that above place value chart,
We can observe that
The thousandths place is less than 5
Hence, from the above,
We can conclude that
The place value that the given number was rounded is: Thousandths
Question 5.
3.695 to 4
The given decimal number is: 3.695
Now,
The representation of the given number in the place value chart is:
Now,
When we observe that above place value chart,
We can observe that
The tenths place is greater than 5
Hence, from the above,
We can conclude that
The place value that the given number was rounded is: Tenths
Question 6.
7.486 to 7.5
The given decimal number is: 7.486
Now,
The representation of the given number in the place value chart is:
Now,
When we observe that above place value chart,
We can observe that
The hundredths place is greater than 5
Hence, from the above,
We can conclude that
The place value that the given number was rounded is: Hundredths
Problem Solving
Write the place value of the underlined digit. Round each number to the place of the underlined digit.
Question 7.
0.592
The given decimal number is: 0.592
Now,
The representation of the given number in the place value chart is:
Now,
From the above place value chart,
The place value of 5 is: 5 tenths
So,
The rounded number that is to the place of the underlined digit for the given number is: 0.6
Hence, from the above,
We can conclude that
The place value of 5 is: 5 tenths
The rounded number that is to the place of the underlined digit for the given number is: 0.6
Question 8.
6.518
The given decimal number is: 6.518
Now,
The representation of the given number in the place value chart is:
Now,
From the above place value chart,
The place value of 6 is: 6 ones
So,
The rounded number that is to the place of the underlined digit for the given number is: 7
Hence, from the above,
We can conclude that
The place value of 6 is: 6 ones
The rounded number that is to the place of the underlined digit for the given number is: 7
Question 9.
0.809
The given decimal number is: 0.809
Now,
The representation of the given number in the place value chart is:
Now,
From the above place value chart,
The place value of 0 is: 0 hundredths
So,
The rounded number that is to the place of the underlined digit for the given number is: 0.8
Hence, from the above,
We can conclude that
The place value of 0 is: 0 hundredths
The rounded number that is to the place of the underlined digit for the given number is: 0.8
Round 16.748 to the place named.
Question 10.
tenths __________
The given number is: 16.748
Now,
The representation of the given number in the place value chart is:
Now,
From the given place value chart,
We can observe that
The tenths place value is greater than 5
Hence, from the above,
We can conclude that
The rounded number that is to the nearest tenth for the given number is: 16.8
Question 11.
hundredths __________
The given number is: 16.748
Now,
The representation of the given number in the place value chart is:
Now,
From the given place value chart,
We can observe that
The hundredths place value is less than 5
Hence, from the above,
We can conclude that
The rounded number that is to the nearest hundredth for the given number is: 16.74
Question 12.
ones _____________
The given number is: 16.748
Now,
The representation of the given number in the place value chart is:
Now,
From the given place value chart,
We can observe that
The ones place value is greater than 5
Hence, from the above,
We can conclude that
The rounded number that is to the nearest one for the given number is: 20
Question 13.
Explain why any number less than 12.5 and greater than or equal to 11.5 would round to 12 when rounded to the nearest whole number.
Let the number be x
Now,
We know that,
When the last digit of a number is greater than 5, then it will be rounded to the next whole number
When the last digit of a number is less than 5, then it will be rounded to the nearest previous whole number
So,
If x = 11.6, then
x = 12 when x > 11.5 and x < 12.5
Hence, from the above,
We can conclude that
When the last digit of a number is greater than 5, then it will be rounded to the next whole number
When the last digit of a number is less than 5, then it will be rounded to the nearest previous whole number
Question 14.
Write Math Explain what happens when you round 4.999 to the nearest tenth.
The given decimal number is: 4.999
Now,
The representation of 4.999 in the place value chart is:
Now,
When we observe the tenth’s place value in the above lace value chart
We can observe that
The number is greater than 5
Now,
We know that,
When the last digit of a number is greater than 5, then it will be rounded to the next whole number
When the last digit of a number is less than 5, then it will be rounded to the nearest previous whole number
Hence, from the above,
We can conclude that
The rounded number for 4.999 when we rounded off to the nearest tenth to the whole number is: 5
Problem Solving
Use the table for 15-18.
Question 15.
The speeds of two insects, when rounded to the nearest whole number, are the same. Which two insects are they?
The given table is:
Now,
From the given table,
We can say that
The rounded speeds nearest to the whole number of the different insects are:
Dragonfly – 7 m/s
Horsefly – 4 m/s
Bumblebee – 3 m/s
Honeybee – 3 m/s
Housefly – 2 m/s
Hence, from the above,
We can conclude that
The names of the insects that have the speeds of the two insects when rounded to the nearest to the whole number that are the same are: Bumblebee and Honeybee
Question 16.
What is the speed of the housefly rounded to the nearest hundredth?
The given table is:
Now,
From the given table,
We can observe that
The speed of Housefly is: 1.967 m/s
Now,
The representation of the speed of housefly in the place value chart is:
Now,
From the above place value chart,
We can say that
The speed of the housefly when rounded to the nearest hundredth is: 1.97 m/s
Hence, from the above,
We can conclude that
The speed of the housefly when rounded to the nearest hundredth is: 1.97 m/s
Question 17.
The speed of an insect is about 3.9 meters per second. Which insect could it be?
It is given that
The speed of an insect is about 3.9 meters per second
Now,
The given table is:
Now,
From the given table,
We can observe that
The speed of the horsefly is: 3.934 m/s
Now,
When we rounded off the speed of horsefly to the nearest hundredth,
The speed of horsefly will be: 3.9 m/s
Hence, from the above,
We can conclude that
The insect that has the speed of 3.9 m/s is: Horsefly
Question 18.
H.O.T. What’s the Error? Mark said that the speed of a dragonfly rounded to the nearest tenth was 6.9 meters per second. Is he correct? If not, what is his error?
It is given that
Mark said that the speed of a dragonfly rounded to the nearest tenth was 6.9 meters per second.
Now,
The given table is:
Now,
From the given table,
We can observe that
The speed of dragonfly is: 6.974 m/s
Now,
When we round off the speed of dragonfly to the nearest tenth,
The speed of dragonfly will be: 6.9 m/s
Hence, from the above,
We can conclude that
Mark is correct
Question 19.
H.O.T. Multi-Step A rounded number for the speed of an insect is 5.67 meters per second. What are the fastest and slowest speeds to the thousandths that could round to 5.67? Explain.
It is given that
A rounded number for the speed of an insect is 5.67 meters per second.
Now,
When we round off the given speed of insect to the nearest whole number,
The speed of insect will become: 6 m/s
When we round off the given speed of insect to the hundredth,
The speed of insect will become: 5.7 m/s
Hence, from the above,
We can conclude that
The fastest speed is: 6 m/s
The slowest speed is: 5.7 m/s
Question 20.
Chef Round uses 1.257 kilograms of meat for his famous meatballs. There are four packages at the market. Which amount is closest to 1.257 kilograms?
A. 1.26 kg
B. 1.2 kg
C. 1.3 kg
D. 1.35 kg
It is given that
Chef Round uses 1.257 kilograms of meat for his famous meatballs. There are four packages at the market.
Now,
The given number is: 1.257
When we round off the given number to the nearest thousandth,
The number will be: 1.26
Hence, from the above,
We can conclude that
The amount that is closest to 1.257 kilograms is:
Question 21.
Representations The table shows the neck lengths for four giraffes in the Long-Necked Friends Sanctuary. Which two giraffes have necks that are the same length when rounded to the nearest tenth?
A. Spot and Stretch
B. Mittens and Spot
C. Mittens and Stretch
D. Spot and Zippy
It is given that
The table shows the neck lengths for four giraffes in the Long-Necked Friends Sanctuary.
Now,
The given table is:
Now,
From the given table,
We can say that
The length of the necks when rounded off to the nearest tenth is:
Mittens – 5.6 feet
Spot – 5.7 feet
Stretch – 5.4 feet
Zippy – 5.7 feet
So,
The two giraffes that have the same neck length are: Spot and Zippy
Hence, from the above,
We can conclude that
The two giraffes that have the same neck lengths when rounded to the nearest tenth are:
Question 22.
Multi-Step Janis sells friendship bracelets at a fair. She has three bracelets that are 16.53 cm, 16.755 cm, and 16.55 cm long. She wants to label the lengths to the nearest tenth of a cm. In order, how should she label the bracelets?
A. 16.5 cm, 16.76 cm, 16.6 cm
B. 16.6 cm, 16.7 cm, 16.5 cm
C. 16.5 cm, 16.8 cm, 16.6 cm
D. 16.3 cm, 16.5 cm, 16.5 cm
It is given that
Janis sells friendship bracelets at a fair. She has three bracelets that are 16.53 cm, 16.755 cm, and 16.55 cm long
Now,
The given numbers are: 16.53, 16.755, and 16.55
So,
The given numbers when rounded to the nearest tenth will become: 16.5, 16.76, and 16.6
So,
The order of the numbers from the greatest to the least are: 16.76, 16.6, and 16.5
The order of the numbers from the least to the greatest are: 16.5, 16.6, and 16.76
Hence, from the above,
We can conclude that
The order that Janis should label the bracelets are:
Texas Test Prep
Question 23.
To which place value is the number rounded? 6.706 to 6.71
A. ones
B. tenths
C. hundredths
D. thousandths
The given decimal number is: 6.706
Now,
The representation of 6.706 in the place value chart is:
Now,
From the above place value chart,
We can say that
The given rounded can be rounded to thousandths place to become 6.71
Hence, from the above,
We can conclude that
The place value that the given number is rounded will be:
### Texas Go Math Grade 5 Lesson 1.5 Homework and Practice Answer Key
Write the place value of the underlined digit. Round each number to the place of the underlined digit.
Question 1.
0.782
The given decimal number is: 0.782
Now,
The representation of the given number in the place value chart is:
Now,
From the above place value chart,
The place value of 7 is: 7 tenths
So,
The rounded number that is to the place of the underlined digit for the given number is: 0.8
Hence, from the above,
We can conclude that
The place value of 7 is: 7 tenths
The rounded number that is to the place of the underlined digit for the given number is: 0.8
Question 2.
4.071
The given decimal number is: 4.071
Now,
The representation of the given number in the place value chart is:
Now,
From the above place value chart,
The place value of 7 is: 7 hundredths
So,
The rounded number that is to the place of the underlined digit for the given number is: 4.07
Hence, from the above,
We can conclude that
The place value of 7 is: 7 hundredths
The rounded number that is to the place of the underlined digit for the given number is: 4.07
Question 3.
21.939
The given decimal number is: 21.939
Now,
The representation of the given number in the place value chart is:
Now,
From the above place value chart,
The place value of 1 is: 1 ones
So,
The rounded number that is to the place of the underlined digit for the given number is: 22
Hence, from the above,
We can conclude that
The place value of 1 is: 1 ones
The rounded number that is to the place of the underlined digit for the given number is: 22
Question 4.
8.155
The given decimal number is: 8.155
Now,
The representation of the given number in the place value chart is:
Now,
From the above place value chart,
The place value of 1 is: 1 tenth
So,
The rounded number that is to the place of the underlined digit for the given number is: 8.16
Hence, from the above,
We can conclude that
The place value of 1 is: 1 tenth
The rounded number that is to the place of the underlined digit for the given number is: 8.16
Question 5.
34.09
The given decimal number is: 34.09
Now,
The representation of the given number in the place value chart is:
Now,
From the above place value chart,
The place value of 3 is: 3 tens
So,
The rounded number that is to the place of the underlined digit for the given number is: 34
Hence, from the above,
We can conclude that
The place value of 3 is: 3 tens
The rounded number that is to the place of the underlined digit for the given number is: 34
Question 6.
4.276
The given decimal number is: 4.276
Now,
The representation of the given number in the place value chart is:
Now,
From the above place value chart,
The place value of 7 is: 7 hundredths
So,
The rounded number that is to the place of the underlined digit for the given number is: 4.28
Hence, from the above,
We can conclude that
The place value of 7 is: 7 hundredths
The rounded number that is to the place of the underlined digit for the given number is: 4.28
Round 8.293 to the place named.
Question 7.
hundredths ___________
The given number is: 8.293
Now,
The representation of 8.293 in the place value chart is:
Now,
From the above place value chart,
We can say that
The rounded value of 8.293 nearest to the place value of hundredths is: 8.29
Hence, from the above,
We can conclude that
The rounded value of 8.293 nearest to the place value of hundredths is: 8.29
Question 8.
tenths ______________
The given number is: 8.293
Now,
The representation of 8.293 in the place value chart is:
Now,
From the above place value chart,
We can say that
The rounded value of 8.293 nearest to the place value of tenths is: 8.3
Hence, from the above,
We can conclude that
The rounded value of 8.293 nearest to the place value of tenths is: 8.3
Question 9.
ones ____________
The given number is: 8.293
Now,
The representation of 8.293 in the place value chart is:
Now,
From the above place value chart,
We can say that
The rounded value of 8.293 nearest to the place value of ones is: 8
Hence, from the above,
We can conclude that
The rounded value of 8.293 nearest to the place value of ones is: 8
Round 12.462 to the place named.
Question 10.
tenths __________
The given number is: 12.462
Now,
The representation of 8.293 in the place value chart is:
Now,
From the above place value chart,
We can say that
The rounded value of 12.462 nearest to the place value of tenths is: 12.5
Hence, from the above,
We can conclude that
The rounded value of 12.462 nearest to the place value of tenths is: 12.5
Question 11.
hundredths ___________
The given number is: 12.462
Now,
The representation of 8.293 in the place value chart is:
Now,
From the above place value chart,
We can say that
The rounded value of 12.462 nearest to the place value of hundredths is: 12.46
Hence, from the above,
We can conclude that
The rounded value of 12.462 nearest to the place value of hundredths is: 12.46
Question 12.
ones _____________
The given number is: 12.462
Now,
The representation of 8.293 in the place value chart is:
Now,
From the above place value chart,
We can say that
The rounded value of 12.462 nearest to the place value of ones is: 12
Hence, from the above,
We can conclude that
The rounded value of 12.462 nearest to the place value of ones is: 12
Problem Solving
Question 13.
What is Anita’s speed rounded to the nearest hundredth?
The given table is:
Now,
From the given table,
We can observe that
Anita’s speed is: 0.277 m/s
Now,
The value of Anita’s speed when rounded to the nearest hundredth will become: 0.28 m/s
Hence, from the above,
We can conclude that
Anita’s speed that is rounded to the nearest hundredth is: 0.28 m/s
Question 14.
The speeds of two runners when rounded to the nearest tenth are the same. Which two runners are they?
The given table is:
Now,
From the given table,
We can observe that
The values of runners speeds when rounded to the nearest tenth will be:
Luke – 0.2 m/s
Margola – 0.2 m/s
Anita – 0.28 m/s
Shateel – 0.30 m/s
So,
From the above,
We can observe that
The speed of the two runners that have the same speed when rounded to the nearest tenth are: Luke and Margola
Hence, from the above,
We can conclude that
The speed of the two runners that have the same speed when rounded to the nearest tenth are: Luke and Margola
Lesson Check
Question 15.
To which place value is the number rounded? 8.293 to 8.29
A. ones
B. tenths
C. hundredths
D. thousandths
The given decimal number is: 8.293
Now,
The representation of 8.293 in the place value chart is:
So,
From the above place value chart
We can observe that
The number is rounded at the thousandths place
Hence, from the above,
We can conclude that
The place value that the given number is rounded is:
Question 16.
To which place value is the number rounded? 0.799 to 0.8
A. ones
B. tenths
C. hundredths
D. thousandths
The given decimal number is: 0.799
Now,
The representation of 0.799 in the place value chart is:
So,
From the above place value chart
We can observe that
The number is rounded at the hundredths place
Hence, from the above,
We can conclude that
The place value that the given number is rounded is:
Question 17.
Which number is 5.389 rounded to the nearest hundredth?
A. 5.4
B. 5.38
C. 5.39
D. 5.399
The given decimal number is: 5.389
Now,
The representation of 5.389 in the place value chart is:
Now,
From the above place value chart,
We can say that
The value of 5.389 when rounded to the nearest hundredth is: 5.39
Hence, from the above,
We can conclude that
The number that is to the nearest hundredth for 5.389 is:
Question 18.
Which number is 7.323 rounded to the nearest tenth?
A. 7.3
B. 7.4
C. 7.32
D. 7.33
The given decimal number is: 7.323
Now,
The representation of 7.323 in the place value chart is:
Now,
From the above place value chart,
We can say that
The value of 7.323 when rounded to the nearest tenth is: 7.3
Hence, from the above,
We can conclude that
The number that is to the nearest tenth for 7.323 is:
Question 19.
A baker uses 2.327 kilograms of blueberries for muffins. There are four packages at the fruit market. Which amount is closest to 2.327 kilograms?
A. 2.3 kg
B. 2.42 kg
C. 2.33 kg
D. 2.4 kg
It is given that
A baker uses 2.327 kilograms of blueberries for muffins. There are four packages at the fruit market
Now,
The representation of 2.327 in the place value chart is:
Now,
From the above place value chart,
We can say that
The number that is nearest to 2.327 is: 2.33
Hence, from the above,
We can conclude that
The amount that is closest to 2.327 kilograms is:
Question 20.
Multi-Step The chart shows the length of each student’s pencil. Which person has a pencil length that can be rounded up to the nearest tenth of an inch?
A. Lori
B. Ping
C. Jason
D. Annie |
1. In the first 10 overs of a cricket game, the run rate was only 3.2. What should be the run rate in the remaining 40 overs to reach the target of 282 runs? A. 6.25 B. 5.5 C. 7.4 D. 5
Here is the answer and explanation
Explanation :
$\text{Runs scored in the first 10 overs = }10 \times 3.2 = 32$
Total runs = 282
remaining runs to be scored = 282 - 32 = 250
remaining overs = 40
$\text{Run rate needed = }\dfrac{250}{40} = 6.25$
2. A grocer has a sale of Rs. 6435, Rs. 6927, Rs. 6855, Rs. 7230 and Rs. 6562 for 5 consecutive months. How much sale must he have in the sixth month so that he gets an average sale of Rs. 6500? A. 4800 B. 4991 C. 5004 D. 5000
Here is the answer and explanation
Explanation :
Let the sale in the sixth month = x
$\text{Then }\dfrac{6435 + 6927 + 6855+ 7230 + 6562 + x}{6} = 6500$
=> 6435 + 6927 + 6855+ 7230 + 6562 + x = 6 × 6500 = 39000
=> 34009 + x = 39000
=> x = 39000 - 34009 = 4991
3. The average of 20 numbers is zero. Of them, How many of them may be greater than zero , at the most? A. 1 B. 20 C. 0 D. 19
Here is the answer and explanation
Explanation :
Average of 20 numbers = 0
$=> \dfrac{\text{Sum of 20 numbers}}{20} = 0$
=> Sum of 20 numbers = 0
Hence at the most, there can be 19 positive numbers.
(Such that if the sum of these 19 positive numbers is x, 20th number will be -x)
4. The captain of a cricket team of 11 members is 26 years old and the wicket keeper is 3 years older. If the ages of these two are excluded, the average age of the remaining players is one year less than the average age of the whole team. Find out the average age of the team. A. 23 years B. 20 years C. 24 years D. 21 years
Here is the answer and explanation
Explanation :
Number of members in the team = 11
Let the average age of of the team = x
$\text{=>}\dfrac{\text{Sum of the ages of all the 11 members of the team}}{11} = x$
=> Sum of the ages of all the 11 members of the team = 11x
Age of the captain = 26
Age of the wicket keeper = 26 + 3 = 29
Sum of the ages of 9 members of the team excluding captain and wicket keeper
= 11x - 26 - 29 = 11x - 55
Average age of 9 members of the team excluding captain and wicket keeper
$= \dfrac{11x - 55}{9}$
$\text{Given that }\dfrac{11x - 55}{9} = (x - 1)$
=> 11x - 55 = 9(x - 1)
=> 11x - 55 = 9x - 9
=> 2x = 46
$=> x = \dfrac{46}{2} = 23\text{ years}$
5. The average monthly income of A and B is Rs. 5050. The average monthly income of B and C is Rs. 6250 and the average monthly income of A and C is Rs. 5200. What is the monthly income of A? A. 2000 B. 3000 C. 4000 D. 5000
Here is the answer and explanation
Explanation :
Let the monthly income of A = a
monthly income of B = b
monthly income of C = a
\begin{align} &a + b = 2 \times 5050 ---------(Equation 1)\\\\ &b + c = 2 \times 6250 ---------(Equation 2)\\\\ &a + c = 2 \times 5200 ---------(Equation 3)\\\\ \end{align}
(Equation 1) + (Equation 3) - (Equation 2)
$=> a + b + a + c - (b + c) = \left(2 \times 5050\right) + \left(2 \times 5200\right) - \left(2 \times 6250\right)$
=> 2a = 2(5050 + 5200 - 6250)
=> a = 4000
=> Monthly income of A = 4000
6. A car owner buys diesel at Rs.7.50, Rs. 8 and Rs. 8.50 per litre for three successive years. What approximately is the average cost per litre of diesel if he spends Rs. 4000 each year? A. Rs. 8 B. Rs. 7.98 C. Rs. 6.2 D. Rs. 8.1
Here is the answer and explanation
Explanation :
\begin{align} &\text{Total Cost = }4000 \times 3\\\\ &\text{Total diesel used = }\dfrac{4000}{7.5} + \dfrac{4000}{8} + \dfrac{4000}{8.5}\\\\ &\text{average cost per litre of diesel = }\dfrac{4000 \times 3}{\left(\dfrac{4000}{7.5} + \dfrac{4000}{8} + \dfrac{4000}{8.5}\right)} = \dfrac{3}{\left(\dfrac{1}{7.5} + \dfrac{1}{8} + \dfrac{1}{8.5}\right)}\\\\\\\\ &\text{It is important how you proceed from this stage. Remember time is very important here}\\\\ &\text{and if we solve this completely in the traditional way, it may take lot of time.}\\\\ &\text{Instead, we can find out the approximate value easily and select the right answer from the}\\\\ &\text{given choices}\\\\ &\text{In this case answer = }\dfrac{3}{\left(\dfrac{1}{7.5} + \dfrac{1}{8} + \dfrac{1}{8.5}\right)}\\\\ &\approx \dfrac{3}{\left(\dfrac{1}{8} + \dfrac{1}{8} + \dfrac{1}{8}\right)} \approx \dfrac{3}{\left(\dfrac{3}{8}\right)} \approx 8 \\\\\\\\ &\text{Means we got that answer is approximately equal to 8. From the given choices, the answer}\\\\ &\text{can be 8 or 7.98 or 8.1 . But which one from these?}\\\\\\\\ &\text{It will be easy to figure out. Just see here the denominator was }\dfrac{1}{7.5} + \dfrac{1}{8} + \dfrac{1}{8.5}\\\\ &\text{and we approximated it as }\dfrac{3}{8}\text{. However}\\\\ &\dfrac{1}{7.5} + \dfrac{1}{8.5} = \dfrac{1}{8-.5}+\dfrac{1}{8+.5} = \dfrac{8+.5+8-.5}{\left(8-.5\right)\left(8+.5\right)} \\\\ &= \dfrac{16}{\left(8^2-.5^2\right)} \left[\text{because }a^2 - b^2 = \left(a-b\right) \left(a+b\right)\right] \\\\ &= \dfrac{16}{\left(64-.25\right)}\\\\ &\text{ie, }\dfrac{1}{7.5} + \dfrac{1}{8.5} = \dfrac{16}{\left(64-.25\right)}\\\\ &\text{We know that }\dfrac{1}{8} + \dfrac{1}{8} = \dfrac{1}{4} = \dfrac{16}{64}\\\\ &\text{=> }\dfrac{1}{7.5} + \dfrac{1}{8.5} > \dfrac{1}{8} + \dfrac{1}{8}\\\\\\\\ &\text{Early we had approximated the denominator as }\dfrac{3}{8}\\\\ &\text{However from the above mentioned equations, now you know that actually denominator is}\\\\ &\text{slightly greater than }\dfrac{3}{8} \\\\ \\\\ &\text{It means that answer is slightly lower that 8. Hence we can pick the choice 7.98 as}\\\\ &\text{the answer}\\\\\\\\ &\text{Try to remember the relations between numbers and which can help you to save a lot of time}\\\\ &\text{which can be very precious in competitive exams} \end{align}
7. In Kiran's opinion, his weight is greater than 65 kg but less than 72 kg. His brother does not agree with Kiran and he thinks that Kiran's weight is greater than 60 kg but less than 70 kg. His mother's view is that his weight cannot be greater than 68 kg. If all are them are correct in their estimation, what is the average of different probable weights of Kiran? A. 70 kg B. 69 kg C. 61 kg D. 67 kg
Here is the answer and explanation
Explanation :
Let Kiran's weight = x. Then
According to Kiran, 65 < x < 72 ------------(equation 1)
According to brother, 60 < x < 70 ------------(equation 2)
$\text{According to mother, }x \leq 68\text{ ------------(equation 3)}$
Given that equation 1,equation 2 and equation 3 are correct. By combining these equations,
we can write as
$65 < x \leq 68$
That is x = 66 or 67 or 68
$\text{average of different probable weights of Kiran = }\dfrac{66+67+68}{3} = 67$
8. The average weight of 16 boys in a class is 50.25 kg and that of the remaining 8 boys is 45.15 kg. Find the average weights of all the boys in the class. A. 48.55 B. 42.25 C. 50 D. 51.25
Here is the answer and explanation
Explanation :
\begin{align} &\text{Average weight of 16 boys = }50.25\\\\ &\text{Total Weight of 16 boys = }50.25 \times 16\\\\\\\\\\\\ &\text{Average weight of remaining 8 boys = }45.15\\\\ &\text{Total Weight of remaining 8 boys = }45.15 \times 8\\\\\\\\\\\\ &\text{Total weight of all boys in the class = }\left( 50.25 \times 16 \right)+\left( 45.15 \times 8\right)\\\\ &\text{Total boys = }16 + 8 = 24\\\\ &\text{Average weight of all the boys = }\dfrac{\left( 50.25 \times 16 \right)+\left( 45.15 \times 8\right)}{24}\\\\ &= \dfrac{\left( 50.25 \times 2\right)+\left( 45.15 \times 1\right)}{3} = \left( 16.75 \times 2\right)+ 15.05 = 33.5 + 15.05\\\\ &=48.55 \end{align}
9. A library has an average of 510 visitors on Sundays and 240 on other days. What is the average number of visitors per day in a month of 30 days beginning with a Sunday? A. 290 B. 304 C. 285 D. 270
Here is the answer and explanation
Explanation :
in a month of 30 days beginning with a Sunday, there will be 4 complete weeks and
another two days which will be Sunday and Monday
Hence there will be 5 Sundays and 25 other days in a month of 30 days beginning with
a Sunday
\begin{align} &\text{Average visitors on Sundays = }510\\\\ &\text{Total visitors of 5 Sundays = }510 \times 5\\\\\\\\\\\\ &\text{Average visitors on other days = }240\\\\ &\text{Total visitors of other 25 days = }240 \times 25\\\\\\\\\\\\ &\text{Total visitors = }\left( 510 \times 5 \right)+\left( 240 \times 25\right)\\\\ &\text{Total days= }30\\\\ &\text{Average number of visitors per day = }\dfrac{\left( 510 \times 5 \right)+\left( 240 \times 25\right)}{30}\\\\ &= \dfrac{\left( 51 \times 5 \right)+\left( 24 \times 25\right)}{3} = \left( 17\times 5 \right)+\left( 8\times 25\right) = 85+200=285 \\\\ \end{align}
10. A student's mark was wrongly entered as 83 instead of 63. Due to that the average marks for the class got increased by half 1/2. What is the number of students in the class? A. 45 B. 40 C. 35 D. 30
Here is the answer and explanation
Explanation :
\begin{align} &\text{Let the total number of students = x}\\\\ &\text{The average marks increased by }\dfrac{1}{2}\text{ due to an increase of 83-63=20 marks.}\\\\ &\text{But total increase in the marks = }\dfrac{1}{2}\times x = \dfrac{x}{2}\\\\ &\text{Hence we can write as}\\\\ &\dfrac{x}{2}=20\\\\ &\Rightarrow x = 20 \times 2 = 40 \end{align}
11. A family consists of two grandparents, two parents and three grandchildren. The average age of the grandparents is 67 years, that of the parents is 35 years and that of the grandchildren is 6 years. The average age of the family is A. $32 \dfrac{2}{7}$ years B. $31 \dfrac{5}{7}$ years C. $28\dfrac{1}{7}$ years D. $30 \dfrac{5}{7}$ years
Here is the answer and explanation
Explanation :
\begin{align} &\text{Total age of the grandparents = }67 \times 2\\\\ &\text{Total age of the parents = }35 \times 2\\\\ &\text{Total age of the grandchildren = }6 \times 3\\\\ &\text{Average age of the family = }\dfrac{\left(67 \times 2\right) + \left(35 \times 2\right) + \left(6 \times 3\right)}{7}\\\\ &= \dfrac{134+70+18}{7} = \dfrac{222}{7} = 31 \dfrac{5}{7} \end{align}
12. The average weight of A, B and C is 45 kg. If the average weight of A and B be 40 kg and that of B and C be 43 kg, what is the weight of B? A. $31$ kg B. $28 \dfrac{1}{2}$ kg C. $32$ kg D. $30 \dfrac{1}{2}$ kg
Here is the answer and explanation
Explanation :
\begin{align} &\text{Let the weight of A, B and C are a,b and c respectively.}\\\\\\\\\\\\ &\text{Average weight of A,B and C = 45}\\\\ &a + b + c = 45 \times 3 = 135 \text{---equation(1)}\\\\\\\\\\\\ &\text{average weight of A and B = }40\\\\ &a + b = 40 \times 2 = 80 \text{---equation(2)}\\\\\\\\\\\\ &\text{average weight of B and C = }43\\\\ & b+ c = 43 \times 2 = 86 \text{ ---equation(3)}\\\\\\\\\\\\ &\text{equation(2)+equation(3)- equation(1) }=> a + b + b+ c - (a+b+c) = 80 + 86 -135 \\\\ &=> b = 80 + 86 -135 = 166-135 = 31\\\\ &=>\text{ weight of B = 31}\\\\ \end{align}
13. If the average marks of three batches of 55, 60 and 45 students respectively is 50, 55, 60, what is the average marks of all the students? A. 53.23 B. 54.68 C. 51.33 D. 50
Here is the answer and explanation
Explanation :
Average marks of batch1 = 50
Students in batch1 = 55
$\text{Total marks of batch1} = 55 \times 50$
Average marks of batch2 = 55
Students in batch2 = 60
$\text{Total marks of batch2} = 60\times 55$
Average marks of batch3 = 60
Students in batch3 = 45
$\text{Total marks of batch3} = 45 \times 60$
Total students = 55 + 60 + 45 = 160
$\text{Average marks of all the students = }\dfrac{ \left(55 \times 50 \right)+ \left(60 \times 55 \right)+ \left( 45 \times 60 \right)}{160} \\\\= \dfrac{275+ 330+ 270}{16} = \dfrac{875}{16} = 54.68$
14. The average age of husband, wife and their child 3 years ago was 27 years and that of wife and the child 5 years ago was 20 years. What is the present age of the husband? A. 40 B. 32 C. 28 D. 30
Here is the answer and explanation
Explanation :
let the present age of the husband = h
present age of the wife = w
present age of the child = c
3 years ago, average age of husband, wife and their child = 27
$=>\text{Sum of age of husband, wife and their child before 3 years = }3 \times 27 = 81$
=> (h-3) + (w-3) + (c-3) = 81
=> h + w + c = 81+9 = 90 -------------------equation(1)
5 years ago, average age of wife and child = 20
$=>\text{Sum of age of wife and child before 5 years = }2 \times 20 = 40$
=> (w-5) + (c-5) = 40
=> w + c = 40+10 = 50 -------------------equation(2)
Substituting equation(2) in equation(1)
=> h + 50 = 90
=> h = 90 - 50 = 40
=>present age of the husband = 40
15. The average weight of 8 person's increases by 2.5 kg when a new person comes in place of one of them weighing 65 kg. What is the weight of the new person? A. 75 Kg B. 50 Kg C. 85 Kg D. 80 Kg
Here is the answer and explanation
Explanation :
$\text{Total increase in weight = }8 \times 2.5 = 20$
If x is the weight of the new person, total increase in weight = x - 65
=> 20 = x - 65
=> x = 20 + 65 = 85
16. There are two divisions A and B of a class, consisting of 36 and 44 students respectively. If the average weight of divisions A is 40 kg and that of division b is 35 kg. What is the average weight of the whole class? A. 38.25 B. 37.25 C. 38.5 D. 37
Here is the answer and explanation
Explanation :
$\text{Total weight of students in division A = }36 \times 40\\\\ \text{Total weight of students in division B = }44 \times 35\\\\ \text{Total students = }36 + 44 = 80\\\\ \text{Average weight of the whole class = }\dfrac{\left(36 \times 40 \right)+\left(44 \times 35 \right)}{80}\\\\ = \dfrac{\left(9 \times 40 \right)+\left(11\times 35 \right)}{20}=\dfrac{\left(9 \times 8\right)+\left(11\times 7\right)}{4}=\dfrac{72+77}{4}=\dfrac{149}{4}=37.25$
17. A batsman makes a score of 87 runs in the 17th inning and thus increases his averages by 3. What is his average after 17th inning? A. 39 B. 35 C. 42 D. 40.5
Here is the answer and explanation
Explanation :
Let the average after 17 innings = x
Total runs scored in 17 innings = 17x
then average after 16 innings = (x-3)
Total runs scored in 16 innings = 16(x-3)
We know that Total runs scored in 16 innings + 87 = Total runs scored in 17 innings
=> 16(x-3) + 87 = 17x
=> 16x - 48 + 87 = 17x
=> x = 39
18. A student needed to find the arithmetic mean of the numbers 3, 11, 7, 9, 15, 13, 8, 19, 17, 21, 14 and x. He found the mean to be 12. What is the value of x? A. 12 B. 5 C. 7 D. 9
Here is the answer and explanation
Explanation :
$\dfrac{3 + 11 + 7 + 9+ 15 + 13 + 8 + 19 + 17 + 21 + 14 + x}{12} = 12\\\\ => \dfrac{137 + x}{12} = 12$
=> 137 + x = 144
=> x = 144 - 137 = 7
19. Arun obtained 76, 65, 82, 67 and 85 marks (out in 100) in English, Mathematics, Chemistry, Biology and Physics. What is his average mark? A. 53 B. 54 C. 72 D. 75
Here is the answer and explanation
Explanation :
$\text{Average mark = }\dfrac{76 + 65 + 82 + 67 + 85}{5}=\dfrac{375}{5}=75$
20. Distance between two stations A and B is 778 km. A train covers the journey from A to B at 84 km per hour and returns back to A with a uniform speed of 56km per hour. Find the average speed of the train during the whole journey? A. 69.0 km /hr B. 69.2 km /hr C. 67.2 km /hr D. 67.0 km /hr
Here is the answer and explanation
Explanation :
-------------------------------------------
Solution 1 (Quick)
--------------------------------------------
\begin{align} &\text{If a car covers a certain distance at x kmph and an equal distance at y kmph. Then,}\\ &\text{the average speed of the whole journey = }\dfrac{2xy}{x+y}\text{ kmph.}\\\\\\\\\\\\ &\text{By using the same formula, we can find out the average speed quickly}\\\ &\text{average speed = }\dfrac{2 \times 84 \times 56}{84 + 56} = \dfrac{2\times84 \times 56}{140} = \dfrac{2 \times 21 \times 56}{35}\\\\ &= \dfrac{2 \times 3 \times 56}{5} = \dfrac{336}{5} = 67.2 \end{align}
-------------------------------------------
Solution 2 (Fundamentals)
--------------------------------------------
Though it is a good idea to solve the problems quickly using formulas, you should
know the fundamentals too. Let's see how we can solve this problems using basics
Train travels from A to B at 84 km per hour
Let the distance between A and B = x
$\text{Total time taken for traveling from A to B = }\dfrac{\text{distance}}{\text{speed}} = \dfrac{x}{84}$
Train travels from B to A at 56 km per hour
\begin{align} &\text{Total time taken for traveling from B to A = }\dfrac{\text{distance}}{\text{speed}} = \dfrac{x}{56}\\\\\\\\\\\\\\\ &\text{Total distance travailed = }x + x = 2x\\\\ &\text{Total time taken = }\dfrac{x}{84} + \dfrac{x}{56}\\\\ &\text{Average speed = }\dfrac{\text{Total distance traveled}}{\text{Total time taken}} = \dfrac{2x}{\dfrac{x}{84} + \dfrac{x}{56}}\\\\\\\\ &= \dfrac{2}{\dfrac{1}{84} + \dfrac{1}{56}} = \dfrac{2 \times 84 \times 56}{56 + 84}= \dfrac{2\times84 \times 56}{140} = \dfrac{2 \times 21 \times 56}{35}= \dfrac{2 \times 3 \times 56}{5}\\\\ &= \dfrac{336}{5} = 67.2 \end{align}
21. The average age of boys in a class is 16 years and that of the girls is 15 years. What is the average age for the whole class? A. 15 B. 16 C. 15.5 D. Insufficient Data
Here is the answer and explanation
Explanation :
We do not have the number of boys and girls. Hence we can not find out the answer.
22. The average age of 36 students in a group is 14 years. When teacher's age is included to it, the average increases by one. Find out the teacher's age in years? A. 51 years B. 49 years C. 53 years D. 50 years
Here is the answer and explanation
Explanation :
average age of 36 students in a group is 14
$\text{=> Sum of the ages of 36 students = }36 \times 14$
When teacher's age is included to it, the average increases by one => average = 15
\begin{align} &\text{=> Sum of the ages of 36 students and the teacher =}37 \times 15\\\\ &\text{Hence teachers age = }37 \times 15 - 36 \times 14 = 37 \times 15 - 14\left(37-1\right) = 37 \times 15 - 37 \times 14 + 14\\\\ &= 37 \left(15 - 14\right)+ 14 = 37 + 14 = 51 \end{align}
23. The average of five numbers id 27. If one number is excluded, the average becomes 25. What is the excluded number? A. 30 B. 40 C. 32.5 D. 35
Here is the answer and explanation
Explanation :
$\text{Sum of 5 numbers = }5 \times 27 \\\\ \text{Sum of 4 numbers after excluding one number = }4 \times 25\\\\ \text{Excluded number = }5 \times 27 - 4 \times 25 = 135- 100 = 35$
24. The batting average for 40 innings of a cricket player is 50 runs. His highest score exceeds his lowest score by 172 runs. If these two innings are excluded, the average of the remaining 38 innings is 48 runs. Find out the highest score of the player. A. 150 B. 174 C. 180 D. 166
Here is the answer and explanation
Explanation :
\begin{align} &\text{Total runs scored by the player in 40 innings = }40 \times 50 \\\\ &\text{Total runs scored by the player in 38 innings after excluding two innings }38 \times 48\\\\ &\text{Sum of the scores of the excluded innings = }40 \times 50 - 38 \times 48 = 2000 - 1824 = 176\\\\ &\end{align}
Given that the scores of the excluded innings differ by 172. Hence let's take
the highest score as x + 172 and lowest score as x
Now x + 172 + x = 176
=> 2x = 4
$=> x = \dfrac{4}{2} = 2$
highest score as x + 172 = 2 + 172 = 174
25. The average score of a cricketer for ten matches is 38.9 runs. If the average for the first six matches is 42, what is the average for the last four matches? A. 34.25 B. 36.4 C. 40.2 D. 32.25
Here is the answer and explanation
Explanation :
\begin{align} &\text{Total runs scored in 10 matches = }10 \times 38.9\\\\ &\text{Total runs scored in first 6 matches = }6 \times 42\\\\ &\text{Total runs scored in the last 4 matches = }10 \times 38.9 - 6 \times 42 \\\\ &\text{Average of the runs scored in the last 4 matches = }\dfrac{10 \times 38.9 - 6 \times 42}{4}\\\\ &=\dfrac{389 - 252}{4}=\dfrac{137}{4}= 34.25 \end{align}
26. The average of six numbers is x and the average of three of these is y. If the average of the remaining three is z, then A. None of these B. x = y + z C. 2x = y + z D. x = 2y + 2z
Here is the answer and explanation
Explanation :
Average of 6 numbers = x
=> Sum of 6 numbers = 6x
Average of the 3 numbers = y
=> Sum of these 3 numbers = 3y
Average of the remaining 3 numbers = z
=> Sum of the remaining 3 numbers = 3z
Now we know that 6x = 3y + 3z
=> 2x = y + z
27. Suresh drives his car to a place 150 km away at an average speed of 50 km/hr and returns at 30 km/hr. What is his average speed for the whole journey ? A. 32.5 km/hr. B. 35 km/hr. C. 37.5 km/hr D. 40 km/hr
Here is the answer and explanation
Explanation :
-------------------------------------------
Solution 1 (Quick)
--------------------------------------------
\begin{align} &\text{If a car covers a certain distance at x kmph and an equal distance at y kmph. Then,}\\ &\text{the average speed of the whole journey = }\dfrac{2xy}{x+y}\text{ kmph.}\\\\\\\\\\\\ &\text{By using the same formula, we can find out the average speed quickly}\\\ &\text{average speed = }\dfrac{2 \times 50\times 30}{50 + 30} = \dfrac{2\times 50 \times 30}{80} = \dfrac{2 \times 50 \times 3}{8}\\\\ &= \dfrac{50 \times 3}{4} = \dfrac{25\times 3}{2} = \dfrac{75}{2} = 37.5 \end{align}
-------------------------------------------
Solution 2 (Fundamentals)
--------------------------------------------
Though it is a good idea to solve the problems quickly using formulas, you should
know the fundamentals too. Let's see how we can solve this problems using basics.
\begin{align} &\text{Total time taken for traveling one side = }\dfrac{\text{distance}}{\text{speed}} = \dfrac{150}{50}\\\\ &\text{Total time taken for return journey = }\dfrac{\text{distance}}{\text{speed}} = \dfrac{150}{30}\\\\\\\\\\\\\\\ &\text{Total distance travailed = }150 + 150 = 2 \times 150\\\\ &\text{Total time taken = }\dfrac{150}{50} + \dfrac{150}{30}\\\\ &\text{Average speed = }\dfrac{\text{Total distance traveled}}{\text{Total time taken}} = \dfrac{2 \times 150}{\dfrac{150}{50} + \dfrac{150}{30}}\\\\\\\\ &= \dfrac{2}{\dfrac{1}{50} + \dfrac{1}{30}} = \dfrac{2 \times 50 \times 30}{30+ 50}\\\\ &= \dfrac{2\times 50 \times 30}{80} = \dfrac{2 \times 50 \times 3}{8}\\\\ &= \dfrac{50 \times 3}{4} = \dfrac{25\times 3}{2} = \dfrac{75}{2} = 37.5 \end{align}
28. The average age of a husband and his wife was 23 years at the time of their marriage. After five years they have a one year old child. What is the average age of the family ? A. 21 years B. 20 years C. 18 years D. 19 years
Here is the answer and explanation
Explanation :
\begin{align} &\text{Total of the age of husband and wife = }2 \times 23 = 46\\\\ &\text{Total of the age of husband and wife after 5 years + Age of the 1 year old child }\\\\ &= 46 + 5 + 5 + 1 = 57 \\\\ &\text{Average age of the family}= \dfrac{57}{3} = 19 \end{align}
29. In an examination, a student's average marks were 63. If he had obtained 20 more marks for his Geography and 2 more marks for his history, his average would have been 65. How many subjects were there in the examination? A. 12 B. 11 C. 13 D. 14
Here is the answer and explanation
Explanation :
Let the number of subjects = x
Then, Total marks he scored for all subjects = 63x
If he had obtained 20 more marks for his Geography and 2 more marks for his history,
his average would have been 65
=> Total marks he would have scored for all subjects = 65x
Now we can form the equation as 65x - 63x = the additional marks of the student = 20 + 2 = 22
=> 2x = 22
$=> x = \dfrac{22}{2} = 11$
30. The average salary of all the workers in a workshop is Rs.8000. The average salary of 7 technicians is Rs.12000 and the average salary of the rest is Rs.6000. How many workers are there in the workshop ? A. 21 B. 22 C. 23 D. 24
Here is the answer and explanation
Explanation :
Let the number of workers = x
Given that average salary of all the workers = Rs.8000
then total salary of all workers = 8000x
Given that average salary of 7 technicians is Rs.12000
$\text{=> Total salary of 7 technicians = }7 \times 12000 = 84000$
Number of the rest of the employees = (x - 7)
Average salary of the rest of the employees = Rs.6000
Total salary of the rest of the employees = (x - 7)(6000)
8000x = 84000 + (x - 7)(6000)
=> 8x = 84 + (x - 7)(6)
=> 8x = 84 + 6x - 42
=> 2x = 42
$=> x = \dfrac{42}{2} = 21$
mahender 17 Mar 2015 7:41 AM
A pool is there. In order to fill that pool, it takes 28 days.how many days does it take to fill half of the pool.day by day it will goes like twice
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Ahmad Umer Rana 14 Mar 2015 8:09 PM
More average based question
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yavanika sharma 12 Mar 2015 9:41 AM
At the end of a soccer season every player had a prime number of goals and the average of the 11 players was also a prime number. No player's individual tally was the same as of anyone else's or as the average. nobody has scored more than 45 goals.
Q1 what was the average of their goals scored?
Q2 what was the maximum goals scored by a single person?
Q3 what was the minimum number of goals scored by a single person?
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Raj 14 Mar 2015 11:22 AM
Consider all prime numbers less than 45 which is a possible super set of the 11 unique scores
{2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43}
Total of these numbers = 281
if we remove 3,5,7 (lowest three), average will slightly above 24
if we remove
37,41,43, average will be slightly above 14
So, Possible values of the average are 17, 19 and 23 (because average is also a prime)
possible set is {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43}
reminder (when divided by by 11) {2, 3, 5, 7, 0, 2, 6, 8, 1, 7, 9, 4, 8, 10
}
if the average is 17,
possible set is {2, 3, 5, 7, 11, 13, 19, 23, 29, 31, 37, 41, 43}
Total of these 13 numbers = 264 .
Total of the actual 11 numbers = 17 * 11 = 187 (because average is 17)
Remainder of
264 when divided by 11 is 0. So the two numbers which are subtracted from 264 should have the remainder sum of 11 (ie, sum of the numbers as 11, 22 etc) so that the net sum will be divisible by 11.
Even when we subtract any two numbers from 264 whose reminder sum is 11, we will not get 187. So average cannot be 17.
if the average is 19,
possible set is {2, 3, 5, 7, 11, 13, 17, 23, 29, 31, 37, 41, 43}
Total of these 13 numbers = 262
Total of the actual 11 numbers = 19 * 11 = 209
Remainder of 262 when divided by 11 is 9. so the two numbers which are subtracted from 262 should have the remainder sum of 2
(ie, sum of the numbers as 2, 13, 24 etc) so that the net sum will be divisible by 11)
Even when we subtract any two numbers from 262 whose reminder sum is 2, we will not get 209 and hence, average cannot be 19
if the average is 23,
possible set is {2, 3, 5, 7, 11, 13, 17, 19, 29, 31, 37, 41, 43}
remainder (when divided by by 11) {2, 3, 5, 7, 0, 2, 6, 8, 7, 9, 4, 8, 10
}
Total of these 13 numbers = 258(remainder when divided by 11 is 5)
Total of the actual 11 numbers = 23 * 11 = 243
we can subtract 2 and 3 (
whose reminder sum is 5) such that is sum becomes 253 which is divisible by 11
So the average is 23
The numbers are
{5, 7, 11, 13, 17, 19, 29, 31, 37, 41, 43}
Now the questions can be easily answered
1. average of their goals scored : 23
2.
maximum goals scored by a single person : 43
3. minimum number of goals scored by a single person : 5
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hyma 11 Mar 2015 1:18 PM
there are 27 students in a class whose average weight is 36kg, among these 27 students 15 are boys and 12 are girls, the average weight of 15 boys is 6.75kg more than the average weight of 12 girls. what is the average weight of 12 girls(in kg)?
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Dev 11 Mar 2015 8:11 PM
Total weight of all students = 27*36 = 972
Let average weight of the 12 girls be x kg.
Then, the average weight of 15 boys = (x+6.75)
Total weight of the 12 girls = 12x
Total weight of the 15 boys = 15(x+6.75)
12x+15(x+6.75)=972
x = 32.25 kg
average weight of 12 girls = 32.25 kg
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guru prasath 03 Mar 2015 5:08 PM
This system are very useful to my knowledge
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sudha 12 Feb 2015 6:16 PM
It,s really exellent to the all students
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ishan 20 Jan 2015 5:45 PM
the length of a rectangular plot is 4 1/2 times that of its breadth if the area of plot is 200 square metres then, then what is its length ?
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Raj 24 Jan 2015 12:29 AM
Then, the length will be 9x/2
area = x * 9x/2 = 200
9x2 = 400
3x = 20
x = 20/3
length = 9x/2 = 30 metre
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Calculus Volume 2
# 4.3Separable Equations
Calculus Volume 24.3 Separable Equations
## Learning Objectives
• 4.3.1 Use separation of variables to solve a differential equation.
• 4.3.2 Solve applications using separation of variables.
We now examine a solution technique for finding exact solutions to a class of differential equations known as separable differential equations. These equations are common in a wide variety of disciplines, including physics, chemistry, and engineering. We illustrate a few applications at the end of the section.
## Definition
A separable differential equation is any equation that can be written in the form
$y′=f(x)g(y).y′=f(x)g(y).$
(4.3)
The term ‘separable’ refers to the fact that the right-hand side of the equation can be separated into a function of $xx$ times a function of $y.y.$ Examples of separable differential equations include
$y′=(x2−4)(3y+2)y′=6x2+4xy′=secy+tanyy′=xy+3x−2y−6.y′=(x2−4)(3y+2)y′=6x2+4xy′=secy+tanyy′=xy+3x−2y−6.$
The second equation is separable with $f(x)=6x2+4xf(x)=6x2+4x$ and $g(y)=1,g(y)=1,$ the third equation is separable with $f(x)=1f(x)=1$ and $g(y)=secy+tany,g(y)=secy+tany,$ and the right-hand side of the fourth equation can be factored as $(x-2)(y+3),(x-2)(y+3),$ so it is separable as well. The third equation is also called an autonomous differential equation because the right-hand side of the equation is a function of $yy$ alone. If a differential equation is separable, then it is possible to solve the equation using the method of separation of variables.
## Problem-Solving Strategy
### Problem-Solving Strategy: Separation of Variables
1. Check for any values of $yy$ that make $g(y)=0.g(y)=0.$ These correspond to constant solutions.
2. Rewrite the differential equation in the form $dyg(y)=f(x)dx.dyg(y)=f(x)dx.$
3. Integrate both sides of the equation.
4. Solve the resulting equation for $yy$ if possible.
5. If an initial condition exists, substitute the appropriate values for $xx$ and $yy$ into the equation and solve for the constant.
Note that Step 4. states “Solve the resulting equation for $yy$ if possible.” It is not always possible to obtain $yy$ as an explicit function of $x.x.$ Quite often we have to be satisfied with finding $yy$ as an implicit function of $x.x.$
## Example 4.10
### Using Separation of Variables
Find a general solution to the differential equation $y′=(x2−4)(3y+2)y′=(x2−4)(3y+2)$ using the method of separation of variables.
## Checkpoint4.10
Use the method of separation of variables to find a general solution to the differential equation $y′=2xy+3y−4x−6.y′=2xy+3y−4x−6.$
## Example 4.11
### Solving an Initial-Value Problem
Using the method of separation of variables, solve the initial-value problem
$y′=(2x+3)(y2−4),y(0)=-1.y′=(2x+3)(y2−4),y(0)=-1.$
## Checkpoint4.11
Find the solution to the initial-value problem
$6y′=(2x+1)(y2−2y−8),y(0)=−36y′=(2x+1)(y2−2y−8),y(0)=−3$
using the method of separation of variables.
## Applications of Separation of Variables
Many interesting problems can be described by separable equations. We illustrate two types of problems: solution concentrations and Newton’s law of cooling.
### Solution concentrations
Consider a tank being filled with a salt solution. We would like to determine the amount of salt present in the tank as a function of time. We can apply the process of separation of variables to solve this problem and similar problems involving solution concentrations.
## Example 4.12
### Determining Salt Concentration over Time
A tank containing $100L100L$ of a brine solution initially has $4kg4kg$ of salt dissolved in the solution. At time $t=0,t=0,$ another brine solution flows into the tank at a rate of $2L/min.2L/min.$ This brine solution contains a concentration of $0.5kg/L0.5kg/L$ of salt. At the same time, a stopcock is opened at the bottom of the tank, allowing the combined solution to flow out at a rate of $2L/min,2L/min,$ so that the level of liquid in the tank remains constant (Figure 4.16). Find the amount of salt in the tank as a function of time (measured in minutes), and find the limiting amount of salt in the tank, assuming that the solution in the tank is well mixed at all times.
Figure 4.16 A brine tank with an initial amount of salt solution accepts an input flow and delivers an output flow. How does the amount of salt change with time?
## Checkpoint4.12
A tank contains $33$ kilograms of salt dissolved in $7575$ liters of water. A salt solution of $0.4kg salt/L0.4kg salt/L$ is pumped into the tank at a rate of $6L/min6L/min$ and is drained at the same rate. Solve for the salt concentration at time $t.t.$ Assume the tank is well mixed at all times.
### Newton’s law of cooling
Newton’s law of cooling states that the rate of change of an object’s temperature is proportional to the difference between its own temperature and the ambient temperature (i.e., the temperature of its surroundings). If we let $T(t)T(t)$ represent the temperature of an object as a function of time, then $dTdtdTdt$ represents the rate at which that temperature changes. The temperature of the object’s surroundings can be represented by $Ts.Ts.$ Then Newton’s law of cooling can be written in the form
$dTdt=k(T(t)−Ts)dTdt=k(T(t)−Ts)$
or simply
$dTdt=k(T−Ts).dTdt=k(T−Ts).$
(4.6)
The temperature of the object at the beginning of any experiment is the initial value for the initial-value problem. We call this temperature $T0.T0.$ Therefore the initial-value problem that needs to be solved takes the form
$dTdt=k(T−Ts),T(0)=T0,dTdt=k(T−Ts),T(0)=T0,$
(4.7)
where $kk$ is a constant that needs to be either given or determined in the context of the problem. We use these equations in Example 4.13.
## Example 4.13
### Waiting for a Pizza to Cool
A pizza is removed from the oven after baking thoroughly, and the temperature of the pizza when it comes out of the oven is $200°F.200°F.$ The temperature of the kitchen is $75°F,75°F,$ and after $11$ minute the temperature of the pizza is $190°F.190°F.$ We would like to wait until the temperature of the pizza reaches $150°F150°F$ before cutting and serving it (Figure 4.17). How much longer will we have to wait?
Figure 4.17 From Newton’s law of cooling, if the pizza cools $10°F10°F$ in $11$ minute, how long before it cools to $150°F?150°F?$
## Checkpoint4.13
A cake is removed from the oven after baking thoroughly, and the temperature of the cake when it comes out of the oven is $450°F.450°F.$ The temperature of the kitchen is $70°F,70°F,$ and after $1010$ minutes the temperature of the cake is $330°F.330°F.$
1. Write the appropriate initial-value problem to describe this situation.
2. Solve the initial-value problem for $T(t).T(t).$
3. How long will it take until the temperature of the cake is within $5°F5°F$ of room temperature?
## Section 4.3 Exercises
Solve the following initial-value problems with the initial condition $y0=0y0=0$ and graph the solution.
119.
$d y d t = y + 1 d y d t = y + 1$
120.
$d y d t = y − 1 d y d t = y − 1$
121.
$d y d t = –y + 1 d y d t = –y + 1$
122.
$d y d t = − y − 1 d y d t = − y − 1$
Find the general solution to the differential equation.
123.
$x 2 y ′ = ( x + 1 ) y x 2 y ′ = ( x + 1 ) y$
124.
$y ′ = tan ( y ) x y ′ = tan ( y ) x$
125.
$y ′ = 2 x y 2 y ′ = 2 x y 2$
126.
$d y d t = y cos ( 3 t + 2 ) d y d t = y cos ( 3 t + 2 )$
127.
$2 x d y d x = y 2 2 x d y d x = y 2$
128.
$y ′ = e y x 2 y ′ = e y x 2$
129.
$( 1 + x ) y ′ = ( x + 2 ) ( y − 1 ) ( 1 + x ) y ′ = ( x + 2 ) ( y − 1 )$
130.
$d x d t = 3 t 2 ( x 2 + 4 ) d x d t = 3 t 2 ( x 2 + 4 )$
131.
$t d y d t = 1 − y 2 t d y d t = 1 − y 2$
132.
$y ′ = e x e y y ′ = e x e y$
Find the solution to the initial-value problem.
133.
$y ′ = e y − x , y ( 0 ) = 0 y ′ = e y − x , y ( 0 ) = 0$
134.
$y ′ = y 2 ( x + 1 ) , y ( 0 ) = 2 y ′ = y 2 ( x + 1 ) , y ( 0 ) = 2$
135.
$d y d x = y 3 x e x 2 , y ( 0 ) = 1 d y d x = y 3 x e x 2 , y ( 0 ) = 1$
136.
$d y d t = y 2 e x sin ( 3 x ) , y ( 0 ) = 1 d y d t = y 2 e x sin ( 3 x ) , y ( 0 ) = 1$
137.
$y ′ = x sech 2 y , y ( 0 ) = 0 y ′ = x sech 2 y , y ( 0 ) = 0$
138.
$y ′ = 2 x y ( 1 + 2 y ) , y ( 0 ) = −1 y ′ = 2 x y ( 1 + 2 y ) , y ( 0 ) = −1$
139.
$d x d t = ln ( t ) 1 − x 2 , x ( 1 ) = 0 d x d t = ln ( t ) 1 − x 2 , x ( 1 ) = 0$
140.
$y ′ = 3 x 2 ( y 2 + 4 ) , y ( 0 ) = 0 y ′ = 3 x 2 ( y 2 + 4 ) , y ( 0 ) = 0$
141.
$y ′ = e y 5 x , y ( 0 ) = ln ( ln ( 5 ) ) y ′ = e y 5 x , y ( 0 ) = ln ( ln ( 5 ) )$
142.
$y ′ = −2 x tan ( y ) , y ( 0 ) = π 2 y ′ = −2 x tan ( y ) , y ( 0 ) = π 2$
For the following problems, use a software program or your calculator to generate the directional fields. Solve explicitly and draw solution curves for several initial conditions. Are there some critical initial conditions that change the behavior of the solution?
143.
[T] $y′=1−2yy′=1−2y$
144.
[T] $y′=y2x3y′=y2x3$
145.
[T] $y′=y3exy′=y3ex$
146.
[T] $y′=eyy′=ey$
147.
[T] $y′=yln(x)y′=yln(x)$
148.
Most drugs in the bloodstream decay according to the equation $y′=cy,y′=cy,$ where $yy$ is the concentration of the drug in the bloodstream. If the half-life of a drug is $22$ hours, what fraction of the initial dose remains after $66$ hours?
149.
A drug is administered intravenously to a patient at a rate $rr$ mg/h and is cleared from the body at a rate proportional to the amount of drug still present in the body, $dd$ Set up and solve the differential equation, assuming there is no drug initially present in the body.
150.
[T] How often should a drug be taken if its dose is $33$ mg, it is cleared at a rate $c=0.1c=0.1$ mg/h, and $11$ mg is required to be in the bloodstream at all times?
151.
A tank contains $11$ kilogram of salt dissolved in $100100$ liters of water. A salt solution of $0.10.1$ kg salt/L is pumped into the tank at a rate of $22$ L/min and is drained at the same rate. Solve for the salt concentration at time $t.t.$ Assume the tank is well mixed.
152.
A tank containing $1010$ kilograms of salt dissolved in $10001000$ liters of water has two salt solutions pumped in. The first solution of $0.20.2$ kg salt/L is pumped in at a rate of $2020$ L/min and the second solution of $0.050.05$ kg salt/L is pumped in at a rate of $55$ L/min. The tank drains at $2525$ L/min. Assume the tank is well mixed. Solve for the salt concentration at time $t.t.$
153.
[T] For the preceding problem, find how much salt is in the tank $11$ hour after the process begins.
154.
Torricelli’s law states that for a water tank with a hole in the bottom that has a cross-sectional area of $ATAT$ and with a height of water $hh$ above the bottom of the tank, the rate of change of volume of water flowing from the tank is proportional to the square root of the height of water, according to $dVdt=−AT2gh,dVdt=−AT2gh,$ where $gg$ is the acceleration due to gravity. Note that $dVdt=Adhdt,dVdt=Adhdt,$ where $ATAT$ is the cross-sectional area of the tank. Solve the resulting initial-value problem for the height of water, assuming a tankof radius $242242$ with a circular hole of radius $22$ ft.
155.
For the preceding problem, determine how long it takes the tank to drain.
For the following problems, use Newton’s law of cooling.
156.
The liquid base of an ice cream has an initial temperature of $200°F200°F$ before it is placed in a freezer with a constant temperature of $0°F.0°F.$ After $11$ hour, the temperature of the ice-cream base has decreased to $140°F.140°F.$ Formulate and solve the initial-value problem to determine the temperature of the ice cream.
157.
[T] The liquid base of an ice cream has an initial temperature of $210°F210°F$ before it is placed in a freezer with a constant temperature of $20°F.20°F.$ After $22$ hours, the temperature of the ice-cream base has decreased to $170°F.170°F.$ At what time will the ice cream be ready to eat? (Assume $30°F30°F$ is the optimal eating temperature.)
158.
[T] You are organizing an ice cream social. The outside temperature is $80°F80°F$ and the ice cream is at $10°F.10°F.$ After $1010$ minutes, the ice cream temperature has risen by $10°F.10°F.$ How much longer can you wait before the ice cream melts at $40°F?40°F?$
For Exercises 159—162, assume a cooling constant of $k=-0.125k=-0.125$ and assume time $tt$ is in minutes.
159.
You have a cup of coffee at temperature $70°C70°C$ and the ambient temperature in the room is $20°C.20°C.$ Assuming a cooling rate $kof0.125,kof0.125,$ write and solve the differential equation to describe the temperature of the coffee with respect to time.
160.
[T] You have a cup of coffee at temperature $70°C70°C$ that you put outside, where the ambient temperature is $0°C.0°C.$ After $55$ minutes, how much colder is the coffee?
161.
You have a cup of coffee at temperature $70°C70°C$ and you immediately pour in $11$ part milk to $55$ parts coffee. The milk is initially at temperature $1°C.1°C.$ Write and solve the differential equation that governs the temperature of this coffee.
162.
You have a cup of coffee at temperature $70°C,70°C,$ which you let cool $1010$ minutes before you pour in the same amount of milk at $1°C1°C$ as in the preceding problem. How does the temperature compare to the previous cup after $1010$ minutes?
163.
Solve the generic problem $y′=ay+by′=ay+b$ with initial condition $y(0)=c.y(0)=c.$
164.
Prove the basic continual compounded interest equation. Assuming an initial deposit of $P0P0$ and an interest rate of $r,r,$ set up and solve an equation for continually compounded interest.
165.
Assume an initial nutrient amount of $II$ kilograms in a tank with $LL$ liters. Assume a concentration of $cc$ kg/L being pumped in at a rate of $rr$ L/min. The tank is well mixed and is drained at a rate of $rr$ L/min. Find the equation describing the amount of nutrient in the tank.
166.
Leaves accumulate on the forest floor at a rate of $22$ g/cm2/yr and also decompose at a rate of $90%90%$ per year. Write a differential equation governing the number of grams of leaf litter per square centimeter of forest floor, assuming at time $00$ there is no leaf litter on the ground. Does this amount approach a steady value? What is that value?
167.
Leaves accumulate on the forest floor at a rate of $44$ g/cm2/yr. These leaves decompose at a rate of $10%10%$ per year. Write a differential equation governing the number of grams of leaf litter per square centimeter of forest floor. Does this amount approach a steady value? What is that value?
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# 6.2 Antidifferentiation by Substitution
## Presentation on theme: "6.2 Antidifferentiation by Substitution"— Presentation transcript:
6.2 Antidifferentiation by Substitution
If y = f(x) we can denote the derivative of f by either dy/dx or f’(x). What can we use to denote the antiderivative of f? We have seen that the general solution to the differential equation dy/dx = f(x) actually consists of an infinite family of functions of the form F(x) + C, where F’(x) = f(x). Both the name for this family of functions and the symbol we use to denote it are closely related to the definite integral because of the Fundamental Theorem of Calculus.
The symbol is an integral sign, the function f is
the integrand of the integral, and x is the variable of integration.
Evaluating an Indefinite Integral
Evaluate
Paying Attention to the Differential
Let f(x) = x³ + 1 and let u = x². Find each of the following antiderivatives in terms of x: a.) b.) c.)
Paying Attention to the Differential
Let f(x) = x³ + 1 and let u = x². Find each of the following antiderivatives in terms of x: a.) b.) c.)
Using Substitution Evaluate Let u = cos x du/dx = -sin x
du = - sin x dx
Using Substitution Evaluate Let u = 5 + 2x³, du = 6x² dx.
Using Substitution Evaluate
We do not recall a function whose derivative is cot 7x, but a basic trigonometric identity changes the integrand into a form that invites the substitution u = sin 7x, du = 7 cos 7x dx.
Setting Up a Substitution with a Trigonometric Identity
Find the indefinite integrals. In each case you can use a trigonometric identity to set up a substitution.
Setting Up a Substitution with a Trigonometric Identity
Find the indefinite integrals. In each case you can use a trigonometric identity to set up a substitution.
Setting Up a Substitution with a Trigonometric Identity
Find the indefinite integrals. In each case you can use a trigonometric identity to set up a substitution.
Evaluating a Definite Integral by Substitution
Evaluate Let u = tan x and du = sec²x dx.
That Absolute Value Again
Evaluate
Homework!!!!! Textbook – p # 1 – 6, 18 – 42 even, 54 – 66 even. |
gimpppa.org
You are watching: 16 divided by 9 as a fraction
Here we will show you step-by-step with thorough explanation just how to calculate 16 divided by 9 using lengthy division.Before girlfriend continue, keep in mind that in the difficulty 16 split by 9, the number are defined as follows:16 = dividend9 = divisorStep 1:Start by setting it up with the divisor 9 ~ above the left side andthe dividend 16 ~ above the best side like this:
9 ⟌ 1 6
Step 2:
The divisor (9) goes into the an initial digit of the dividend (1), 0 time(s). Therefore, put 0 on top:
0 9 ⟌ 1 6
Step 3:
Multiply the divisor by the result in the previous step (9 x 0 = 0) and also write the answer below the dividend.
0 9 ⟌ 1 6 0
Step 4:
Subtract the an outcome in the previous action from the an initial digit that the dividend (1 - 0 = 1) and also write the price below.
0 9 ⟌ 1 6 - 0 1
Step 5:
Move under the 2nd digit of the dividend (6) prefer this:
0 9 ⟌ 1 6 - 0 1 6
Step 6:
The divisor (9) goes into the bottom number (16), 1 time(s).
See more: Why Didn'T The Skeleton Cross The Road Answers, Halloween Riddles And Puns
Therefore, placed 1 on top:
0 1 9 ⟌ 1 6 - 0 1 6
Step 7:
Multiply the divisor through the result in the previous step (9 x 1 = 9) and also write that answer at the bottom:
0 1 9 ⟌ 1 6 - 0 1 6 9
Step 8:
Subtract the an outcome in the previous step from the number written above it. (16 - 9 = 7) andwrite the answer in ~ the bottom.
0 1 9 ⟌ 1 6 - 0 1 6 - 9 7
You room done, because there room no an ext digits to relocate down indigenous the dividend.The prize is the optimal number and the remainder is the bottom number.Therefore, the answer come 16 divided by 9 calculated using Long division is:1
7 RemainderLong department CalculatorEnter one more problem for united state to explain and solve:/ |
## Real Mathematics – Strange Worlds #18
Every year in December, each city changes drastically. Suddenly we find ourselves surrounded by decorations that remind us of the upcoming new year.
Steve the teacher starts to decorate his classrooms for the new year like he does every year. Though, Steve the teacher set his mind on using new year decorations for his mathematics lessons.
New Year Decorations Game (N.Y.D.G.)
Steve’s creation N.Y.D.G. is a multiplayer game. This is why the game is played in knockout stages/rounds. The winner of the game wins the new year decorations and gets to decorate the classroom as he/she wishes.
Content of N.Y.D.G.
• In each knockout round, students are given 4 decorations as follows:
• Players wind the decorations one another.
• The winding procedure should be done secretly from the opponent.
• Each player has at most four moves for winding.
Let’s use an example to explain what a “move” means during the winding procedure.
Assume that the first move is made with the red decoration as follows:
This counts as one move. The red one undergoes the blue and green decorations in this move. Let the next two moves are as follows:
In the second move, the yellow decoration undergoes the green and red ones, while the blue one passes over the green and yellow decorations. The illustration (up-right) shows us how the winding looks after these 3 moves.
In the end, winding gives us a braid.
The Goal of The Game
In any round, to knock your opponent out, you should solve the braid of your opponent faster than your opponent solves yours. (Solving a braid means, bringing the decorations to their first state. For instance, in the example given up, the first state is yellow-green-blue-red in order.)
Braids
Braids have a very important part in daily life. We encounter them not just in new year decorations, but also in a piece of cheese, a hairstyle, a basket or even in a bracelet:
In case you wish to understand what braids mean in mathematics; one can take a look at Austrian mathematician Emil Artin’s works from the 1920s.
Let’s call the following an identity braid from now on:
In Steve the teacher’s game, the ambition is to go back to the identity braid from a complex braid in the shortest amount of time. To do that, we can use Artin’s work on braids.
Example One: Solving two ropes.
Assume that we have two ropes tangles with each other as follows:
The inverse of this rope is:
If we combine these two ropes, when each rope to be stretched, the result will give us the identity braid:
Example Two: Solving three ropes.
Take three ropes and make a braid as follows:
There are three intersections in this braid:
1: Green over the blue.
2: Red over the green.
3: Blue over the red.
Now, you should repeat these steps, but from last to the first this time. Then, you should do these moves:
Move #1: Blue over the red.
Move #2: Red over the green.
Move #3: Green over the blue.
Finally, the combination will give you the identity braid. Try and see yourself.
Paper and Braids
Take an A4 paper and cut the paper using a knife like the following:
Then, hold the paper from its sides and rotate it 90 degrees to the left. You will end up with some kind of a braid:
One wonders…
• How can you use Emil Artin’s work in the game of Steve the teacher?
• In “example two”, rotate the ropes 90 degrees to the left. Start investigating the intersections from left to right. What do you notice?
• Play Steve the teacher’s game with an A4 paper. (It is more than enough to use 3 or 4 cuts on the paper.)
M. Serkan Kalaycıoğlu
## Real Mathematics – Graphs #7
Serkan’s System
Serkan the math teacher, hands out a specific number of problems to his students. Kids who can solve 1 or more of those problems would get a certain prize. At the beginning of each semester, Serkan and his students sit down and agree on what kind of prize is going to be distributed. For the current semester, oreo is chosen as the prize:
If Serkan the math teacher hands out 10 problems:
• 10 Oreos for the kids who solved 10, 9 or 8 of those problems,
• 5 Oreos for the kids who solved 7, 6 or 5 of those problems,
• 2 Oreos for the kids who solved 4, 3 or 2 of those problems,
• 1 oreo for the kids who solved 1 problem,
• Absolutely nothing for the students who solved… well… none of those problems.
If you take a careful look at the numbers, you can see that Serkan the math teacher selected those numbers with a kind of logic: 10, 5, 2 and 1.
These are the natural numbers that can divide the number of the problems (that is 10) without any remainder.
Prize Distribution Machine (P.D.M.)
One month later…
Serkan the math teacher had faced some problems 4 weeks into the semester. He realized that it took hours to distribute the prizes since he has 10 classes in total.
Serkan the math teacher had to use almost all his free time in school to distribute the Oreos. This led him to think about a machine that would help him with the distribution:
• P.D.M. will have 4 different compartments. (Because of 10, 5, 2 and 1.)
• The volumes of those compartments will be measured with Oreos. They will be 10, 5, 2 and 1 Oreo-sized.
• Oreos will enter the machine from the 10-Oreo-sized compartment. From there, Oreos will move to the other compartments using the connections that will be established.
• Golden Rule: To establish a connection between any two compartments, the size of those compartments must be factors of one another.
Connections of the compartments for 10 problems:
• For 10-Oreo-sized: 5, 2 and 1.
• For 5-Oreo-sized: 10 and 1.
• For 2-Oreo-sized: 10 and 1.
• For 1-Oreo-sized: 10, 5, and 2.
Then, the sketch of the P.D.M. would look like the following:
Is this another graph?!
If you are familiar with graph theory (or if you read the graph section of the blog) you can recognize that the sketch of Serkan the math teacher’s machine is a planar graph:
You should connect the numbers (dots) using lines (connections) according to the golden rule.
One wonders…
What if Serkan the math teacher asks 12 problems?
For 12 problems, the numbers of prizes are going to be: 12, 6, 4, 3, 2 and 1.
In such a situation, can Serkan build his machine? In other words; is it possible to connect the dots for 12-sized P.D.M.?
Hint: First, you should consider where the lines should be. Also, you can arrange the dots in any order you’d like.
M. Serkan Kalaycıoğlu
## Real Mathematics – Strange Worlds #17
Often you see me writing about changing our perspective. For example, when you encounter a baby first thing you do is to make baby sounds and try to make the baby laugh. Whereas if you’d looked carefully at baby’s hair, you could have seen a very valuable mathematical knowledge hidden on the baby’s head:
As shown above, there is a point on each baby’s head. You can see that the hair besides that point is growing in different directions. Can you tell me which direction the hair grows at that exact point?
Hairy ball theorem can give us the answer.
Hairy Ball Theorem
Hairy ball theorem asks you to comb a hairy ball towards a specific direction. The theorem states that there is always at least one point (or one hair) that doesn’t move into that direction.
You can try yourself and see it: Each time at least one hair stands high. This hair (or point) is a sort of singularity. That hair is too stubborn to bend.
Baby’s hair is some kind of a hairy ball example. (I use the expression “some kind of”because the hairy ball has hair all over its surface. Though the baby’s head is not covered with hair completely.) This is why the point on the baby’s head is a singularity. It is the hair that gives a cowlick no matter how hard you comb the baby’s hair.
Torus
Hairy ball theorem doesn’t work on a torus that is covered with hair. In other words, it is possible to comb the hair on a torus towards a single direction.
No Wind
Hairy ball theorem can be used in meteorology. The theorem states that there is a point on earth where there is no wind whatsoever.
To prove that, you can use a hairy ball. Let’s assume that there is wind all over the earth from east to west. If you comb the ball like that, you will realize that north and south poles will have no wind at all.
On Maps
The hairy ball theorem is a kind of a fixed point theorem. Actually, it is also proven by L. E. J. Brouwer in 1912.
One of the real-life examples of the fixed point theorem uses maps. For example, print the map of the country you live in, and place it on the ground:
There is a point on the printed map that is exactly the same as the map’s geographical location.
One wonders…
Assume that all the objects below are covered with hair. Which one(s) can be combed towards the same direction at all its points? Why is that?
M. Serkan Kalaycıoğlu
## Real Mathematics – Strange Worlds #16
The Walk
• Select two points in the classroom.
• Draw a line between them.
• Send a student to one of those points.
• Once the student starts his/her walk, he/she should arrive at the other point exactly 10 seconds later.
• Everybody in the classroom would count to 10 to help the walker.
Ask the student to do the same walk twice while recording the walk using a camera.
The goal of the experiment
After the experiment is done, the following question is asked to the classroom:
“Is there a moment during both walks when the student stands at the exact point?”
In other words, the student walks the same distance in the same amount of time at different speeds. The goal is to find if there is a moment in both walks when the student passes the exact point on the line.
First of all, we should give time to the students for them to think and brainstorm on the problem. Then, using the video shots, the answer is given.
The most important question comes at last: Why so?
Weeding out the stone
In my childhood, one of my duties involved weeding out the stones inside a pile of rice. To be honest, I loved weeding out. Because I was having fun with the rice as I was making different shapes with it.
Years later when I was an undergrad mathematics student I heard of a theorem that made me think of my weed out days. This theorem stated that after I finish the weed out, there should be at least one rice particle that sits in the exact point where it was before the weed out started. (Assuming that the rice particles are covering the surface completely.) In other words; no matter how hard to stir the rice particles, there should be at least one rice particle that has the exact spot where it was before stirring.
This astonishing situation was explained by a Dutch mathematician named L.E.J. Brouwer. Brouwer’s fixed point theorem is a topology subject and it is known as one of the most important theorems in mathematics.
The answer to the walking problem,
The walking problem is an example of Brouwer’s fixed point theorem. This is why the answer to the question is “yes”: There is a moment in both walks when the student stands at the exact point on the line.
I will be talking about Brouwer’s fixed point in the next article.
One wonders…
A man leaves his home at 08:00 and arrives at another city at 14:00. Next morning at 08:00 he leaves that city and arrives at his home at 14:00, using the exact roads.
Conditions
• Starting and finishing points are the same, as well as the time intervals of both trips.
• The first condition means that the man could travel in his choice of speed as long as he sticks to the first condition.
Is there a point on these trips where the man passes at the exact time during both trips?
Hint: You could assume that the distance is 600 km and the man must finish that in 6 hours. For instance, he could have been traveling 100 km/h the first day, and the next day 80 km/h in the first 2 hours; 100 km/h in the next 2 hours, and 120 km/h in the last 2 hours of the trip.
M. Serkan Kalaycıoğlu
## Real MATHEMATICS – Strange Worlds #15
Human Knot Game
Inside a classroom divide students into groups such that each group has at least 5 students. Groups should stand up and form a circle before following the upcoming instructions:
1. If there is an even number of student in a group:
• Have each student extend his/her right hand and take the right hand of another student in the group who is not adjacent to him/her.
• Repeat the same thing for the left hand.
2. If there is an odd number of student in a group:
• Have each student (except one of them) extend his/her right hand and take the right hand of another student in the group who is not adjacent to him/her.
• Then take the extra student and have him/her extend his/her right hand so that the extra student can hold the left hand of another student who is not adjacent to him/her.
• Finally, repeat the process for the students whose left hands are free.
In the end, students will be knotted.
Now, each group should find a way to untangle themselves without letting their hands go. To do that, one can use Reidemeister moves.
Reidemeister Moves
Back in 1926 Kurt Reidemeister discovered something very useful in knot theory. According to him, in the knot theory one can use three moves which we call after his name. Using these three moves we can show if two (or more) knots are the same or not.
For example, using Reidemeister moves, we can see if a knot is an unknot (in other words, if it can be untangled or not).
What are these moves?
Twist
First Reidemeister move is twist. We can twist (or untwist) a part of a knot within the knot theory rules.
Poke
Second one is poke. We can poke a part of a knot as long as we don’t break (or cut) the knot.
Slide
Final one is slide. We can slide a part of a knot according to Reidemeister.
One wonders…
After you participating in a human knot game, ask yourselves which Reidemeister moves did you use during the game?
M. Serkan Kalaycıoğlu
## Real MATHEMATICS – Strange Worlds #14
In my youth, I would never step outside without my cassette player. Though, I had two knotting problems with my cassette player: the First one was about the cassette itself. Rewinding cassettes was a big issue as sometimes the tape knotted itself. Whenever I was lucky, sticking a pencil would solve the whole problem.
The second knotting problem was about the headphone. Its cable would get knotted so bad, it would take me 10 minutes to untangle it. Most of the time I would bump into a friend of mine and the whole plan about listening to music would go down the drain.
The funny thing is I would get frustrated and chuck the headphone into my bag which would guarantee another frustration for the following day.
A similar tangling thing happens in our body, inside our cells, almost all the time.
### DNA: A self-replicating material that is present in nearly all living organisms as the main constituent of chromosomes. It is the carrier of genetic information.
DNA has a spiral curve shape called “helix”. Inside the cell, the DNA spiral sits at almost 2-meters. Let me give another sight so that you can easily picture this length in your mind: If a cell’s nucleus was at the size of a basketball, the length of the DNA spiral would be up to 200 km!
You know what happens when you chuck one-meter long headphones into your bags. Trying to squeeze a 200 km long spiral into a basketball?! Lord; knots everywhere!
Knot Theory
This chaos itself was the reason why mathematicians got involved with knots. Although, knot-mathematics relationships existed way before DNA researches started. In the 19th century, a Scottish scientist named William Thomson (a.k.a. Lord Kelvin) suggested that all atoms are shaped like knots. Though soon enough Lord Kelvin’s idea was faulted and knot theory was put aside for nearly 100 years. (At the beginning of the 20th century, Kurt Reidemeister’s work was important in knot theory. I will get back to Reidemeister in the next post.)
Is there a difference between knots and mathematical knots?
For instance the knot we do with our shoelaces is not mathematical. Because both ends of the shoelace are open. Nevertheless a knot is mathematical if only ends are connected.
###### The left one is a knot, but not mathematically. The one on the right is mathematical though.
Unknot and Trefoil
In the knot theory we call different knots different names with using the number of crosses on the knot. A knot that has zero crossing is called “unknot”, and it looks like a circle:
Check out the knots below:
They look different from each other, don’t they? The one on the left has 1 crossing as the one on the right has 2:
But, we can use manipulations on the knot without cutting (with turning aside and such) and turn one of the knots look exactly like the other one!
This means that these two knots are equivalent to each other. If you take a good look you will see that they are both unknots. For example, if you push the left side of the left knot, you will get an unknot:
Is there a knot that has 1 crossing but can’t be turned into an unknot?
The answer is: No! In fact, there are no such knots with 2 crossiongs either.
We call knots that have 3 crossings and that can’t be turned into unknots a “trefoil”.
Even though in the first glance you would think that a trefoil could be turned into an unknot, it is in fact impossible to do so. Trefoil is a special knot because (if you don’t count unknot) it has the least number of crossings (3). This is why trefoil is the basis knot for the knot theory.
One of the most important things about trefoils is that their mirror images are different knots. In the picture, knot a and knot b are not equivalent to each other: In other words, they can’t be turned into one another.
Möbius Strip and Trefoil
I talked about Möbius strips and its properties in an old post. Just to summarize what it is; take a strip of paper and tape their ends together. You will get a circle. But if you do it with twisting one of the end 180 degrees, you will get a Möbius band.
Let’s twist one end 3 times:
Then cut from the middle of the strip parallel to its length:
We will get a shape like the following:
After fixing the strip, you can see that it is a trefoil knot:
To be continued…
One wonders…
1. In order to make a trefoil knot out of a paper strip, we twist one end 3 times. Is there a difference between twisting inwards and outwards? Try and observe what you end up with.
2. If you twist the end 5 times instead of 3, what would you get? (Answer is in the next post.)
M. Serkan Kalaycıoğlu
## Real MATHEMATICS – Graphs #6
Bequeath Problem
King Serkan I decides to allocate his lands to his children. Obviously he had set up some ground rules for the allocation:
• Each child will get at least one land.
• Same child can’t have adjacent lands.
Problem: At least how many children should Serkan I has so that allocation can be done without a problem?
Map #1
Let’s start from a simple map:
In this case Serkan I can have two children:
As you can remember from the previous article a map and a graph is irreversible. If we represent lands with dots, and let two dots be connected with a line if they are adjacent, we can show maps as graphs:
Let’s add another land to this map:
Adding a land on a map is the same thing as adding a dot on a graph:
Map #2
Let’s assume there are three lands on a map:
We can convert this map into graph as follows:
As seen above, three children are needed in order to fulfill Serkan I’s rules:
Map #3
Let the third map be the following:
According to Serkan I’s rules, we will need four children for such map:
Map #3 can be shown as a graph like the following:
Map #4
For the final map, let’s assume Serkan I left a map that looks like USA’s map:
Surprisingly four children are enough in order to allocate the lands on the map of USA:
What is going on?
Careful readers already noticed that adding a dot that connects the other dots in the second map’s graph gives the graph of the third map. Same thing is true for the first and second maps:
Hence, adding a new dot to the graph means adding a new child.
Q: Is it possible to create a map that requires at least five children?
In other words: Is it possible to add a fifth dot to the graph so that it has connection to all existing four dots? (Ps: There can be no crossing in a graph as our maps are planar.)
Then, all we have to do is to add that fifth dot… Nevertheless I can’t seem to do it. When I add the fifth dot outside of the following graph:
It is impossible to connect 1 and 5 without crossing another line. No matter what I try, I can’t do it:
Four Color Theorem
“Can the areas on any map be colored with at most four colors such that no pair of neighboring areas get the same color?”
This simple problem was introduced for the first time by Francis Guthrie in 1852. Not until 1976 there was no proof for Guthrie’s conjecture. Only then with the help of computers the conjecture was proved. This proof is crucial for mathematics world as it is known as the mathematical theorem that was proven with the help of computers.
One wonders…
Add a fourth dot to the graph you see above and connect that dot to the existing dots. (You are free to place the fourth dot wherever you want on the graph.)
Now check your graph: One of those four dots is trapped inside the lines, isn’t it?
Can you fix that?
Explain how you can/can’t do it.
M. Serkan Kalaycıoğlu
## Real MATHEMATICS – Graphs #5
Cruel Traffic Light
I have to drive past the same crossroad almost every single day. Obviously I stop at the longest-lasting traffic light of the crossroad. Within time I started loving these moments because it gave me a chance to think my life over. Though, it doesn’t take too long for me to start thinking about mathematics.
One of those days I found myself questioning the traffic lights and their relationship with mathematics. (After all mathematics is everywhere; isn’t it?) Soon after I realized that there was my beloved graph theory behind traffic lights.
Light #1
Let’s assume a one-way street with two traffic lights: One for the vehicles (we’ll call it A), other for the pedestrians (we’ll call it B):
In such situation we have to avoid accidents. This means whenever A has green light, B must have red light and vice versa. We will not take account of the situation when both lights are red. Because even though there won’t be any accident, neither of the sides will be standing still (which is nonsense):
All these can be shown using graph theory: Lights will be represented by dots. Dots in a graph will be connected with lines if they are not in the same color:
Same thing can be shown with maps: If A and B are two neighboring countries, they should be colored in different colors to avoid confusion:
Light #2
This time we will assume a two-way street with three traffic lights: Two for the vehicles from opposite sides (we’ll call them A and B), and one for the pedestrians (we’ll call that C):
In such a situation when C is red, either or both of A and B should be green. When C is green, then both A and B should be red:
We’ll skip the situation where all three of them are red as no one would move in such situation.
We can show these using graph theory and map coloring as follows:
Since we set all the rules graphs make it much easier and clearer to understand the situations.
Light #3
Finally we have a two-way street (A and B), a right turn (C) and two pedestrian lights (D and E) as follows:
This time it is a much complex situation:
Although using graph theory makes it all easier for us to understand:
How Many Colors?
We will color the following graphs using the same rules we just established above: If two dots are connected with a line, then those dots must have different colors.
#### Chromatic Numbers: Whenever a graph is being colored, ambition is the use the least number of colors. This number is also known as the chromatic number of a graph.
Here we need 3 colors. Hence chromatic number of the graph is 3. Let’s add one more dot and line to the graph:
This time chromatic number of the graph becomes 2. We’ll add another dot and line:
Now chromatic number becomes 3 again. Let’s add a dot and a line for the last time:
Chromatic number is back to 2.
To be continued…
One Wonders…
What did just happen? What did you notice? Why is it happening?
How can you increase the chromatic number?
M. Serkan Kalaycıoğlu
## Real MATHEMATICS – Graphs #4
I visited the world famous Hermitage museum in Saint Petersburg (Russia) back in 2014. Hermitage is so huge that it has 1057 rooms and one would have to walk around 22 km to see all the rooms. And numbers of artistic & historical items are not that shallow either: It is believe that it would take a little over 11 years if one would spare 1 minute for each piece of art.
Now you can relate why I had trouble when I wanted to visit Hermitage for a few hours. I didn’t have enough time and I wanted to see important art works such as the Dessert: Harmony in Red by Henri Matisse. Eventually I realized that this is a problem I had faced before.
Actually, each and every one of you must have faced such problems in your daily lives. Most common one: “Which route you should take between home and work during rush hour traffic?”
Postman’s Path
Facing such problem in Saint Petersburg is a pleasing coincidence as this magnificent city once was home to one of the giants of mathematics: Leonhard Euler. If you take a look at the first article of graphs you can see that Euler is the person who initiated the discovery of Graph Theory with his solution to the famous Seven Bridges of Königsberg problem.
In 1960 (almost 230 years after the solution of Königsberg) a Chinese mathematician named Mei-Ko Kwan took a similar problem into his hands:
A postman has to deliver letters to a given neighborhood. He needs to walk through all the streets in the neighborhood and back to the post-office. How can he design his route so that he walks the shortest distance?
This problem is given the name Chinese Postman Problem as an appreciation to Kwan. There are a few things you should know about the postman problem:
• Postman must walk each street once.
• Start and finish point must be at the post-office.
• Postman must fulfill these two conditions in the shortest possible time.
Power of Graphs
Mei-Ko Kwan turned to Euler’s Königsberg solution in order to solve the postman problem. According to Kwan postman problem could have been shown as a graph: Lines represent the streets and letters represent the houses.
Example Graph:
##### A graph has an Euler circuit if and only if the degree of every vertex is even. In other words, for each vertex (point) count the number of edges (lines) it has. If that number is even for each vertex, then it is safe to say that our graph has an Euler circuit.
Solution
As you can see above, if the degree of every vertex in a graph is even, then we can conclude that the graph has an Euler circuit. Let’s assume that the following is our graph:
First of all we must determine the degree of each vertex:
As seen above all the vertices have even degrees. This is how we can conclude that the graph has an Euler circuit even without trying to find the path itself. Hence the postman can use the existent roads and finish his route in the shortest time:
When a graph has vertices with odd degrees, then we must add new line(s) to the graph in order to create an Euler circuit. These are the steps you can follow:
1. Find the vertices that have odd degrees.
2. Split these vertices into pairs.
3. Find the distance for each pair and compare them. The shortest pair(s) shows where to add new line(s).
4. Add the line(s) to the graph.
Let’s use an example and test Kwan’s algorithm for the solution. Assume that A is the starting point. First we must check the degrees of the vertices:
Unfortunately A and D have odd degrees which is why the postman can’t finish his walk:
Here we have only one pair that is A-D. There are three routes between A-D. If we find the shortest one and add that to our graph, we will have an Euler circuit:
The shortest one is the line between A-D as seen above. If we add that line, we will have the shortest route for the postman:
One wonders…
If the post office is at A, what is the route for the postman to take so that he will finish his day in the shortest way?
M. Serkan Kalaycıoğlu
## Real MATHEMATICS – Strange Worlds #13
Love of Dinosaurs
I loved reading weekly television guides when I was a child. Thanks to these booklets I knew when to catch my favorite cartoons and movies. This is why I was able to watch some great movies such as Jurassic Park more than once.
Jurassic Park (along with the famous cartoon Flintstones) was the reason why so many kids from my generation interested in genetics, paleontology and obviously dinosaurs. After 1993 when Jurassic Park was a big success, majority of the kids (including me) started learning names of the dinosaurs starting with Tyrannosaurus Rex.
Dragon Curve
In my mid 20s I was doing research about fractal geometry and I eventually found myself with Jurassic Park. Apparently in 1990 Jurassic Park novel was first published. There were strange shapes just before every chapter named as “iterations”. These iterations were actually showing some stages of a special fractal:
This fractal is known as Jurassic Park fractal or Dragon curve. I prefer using Dragon curve because let’s face it; dragons are cool!
How to construct a dragon curve?
• Draw a horizontal line.
• Take that line, spin it 90 degree clockwise. This will be the second line.
• Add second line to the first one.
• Repeat the same processes forever.
After first iteration you will end up with the following:
After second iteration:
Third and forth iterations:
Just before the first chapter of the Jurassic Park novel you can see the forth iteration named as “first iteration”:
One wonders…
You might find these ordinary. Then let me try to surprise you a bit. First of all cut a long piece of paper as shown below:
Did you do it? Well done! Now unite the right end of the paper with left end:
In other words the paper is folded in half. Now slowly unfold the paper such that two halves construct a 90-degree angle between them:
Fold the paper second time in half:
Unfold it carefully:
Do the same things for the third time:
And finally repeat the same process for the fourth time:
Conclusion: Whenever a piece of paper is folded four times in half, one would end up with the fourth iteration of the dragon curve.
M. Serkan Kalaycıoğlu |
Solving the Project Euler Problems – Problem 5
2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder.
What is the smallest positive number that is evenly divisible by all of the numbers from 1 to 20?
This one isn’t so much a problem of programming as it is mathematics, as it isn’t just a case of multiplying all the numbers from 1 to 20, that would just result in 20!, which is far bigger than our actual answer.
For reference 20! = 2,432,902,008,176,640,000, which is 2 quintillion 432 quadrillion 902 trillion 8 billion 176 million 640 thousand.
Taking into account the prime numbers up to 20 we have 2, 3, 5, 7, 11, 13, 17 and 19, so our number still needs to be divisible by 4, 6, 8, 9, 10, 12, 14, 15, 16, 18 and 20. Our number so far is 2 × 3 × 5 × 7 × 11 × 13 × 17 × 19.
By virtue of being multiples of these prime factors, our number is already divisible by 6 (2 × 3), 10 (2 × 5) and 14 (2 × 7). The numbers now remaining are 4, 8, 9, 12, 16, 18 and 20.
If we multiply our number by 2, it is now divisible by 4 (it was divisible by 2), 12 (it was divisible by 6) and 20 (it was divisible by 10). The numbers now remaining are 8, 9, 16 and 18 and our number is 2 × 2 × 3 × 5 × 7 × 11 × 13 × 17 × 19.
Multiplying by 2 again will make our number divisible by 8, and doing it again will make it divisible by 16, leaving only 9 and 18 and making our number 2 × 2 × 2 × 2 × 3 × 5 × 7 × 11 × 13 × 17 × 19.
Now multiplying our number by 3, as it was already divisible by both 3 and 6, will now make it divisible by 9 and 18.
Our number is now 2 × 2 × 2 × 2 × 3 × 3 × 5 × 7 × 11 × 13 × 17 × 19 = 116,396,280. Far smaller than our result for 20! above!
Incidentally, by virtue of being divisible by 3 and 7, our number is also divisible by 21. And as it’s divisible by 2 and 11, it is also divisible by 22.
Solving the Project Euler Problems – Problem 4
A palindromic number reads the same both ways. The largest palindrome made from the product of two 2-digit numbers is 9009 = 91 × 99.
Find the largest palindrome made from the product of two 3-digit numbers.
No gists found
Python’ s very nice way of manipulating integers into strings and string into integers was very helpful with this one, all I had to do was create a function to check if a number is a palindrome (to keep the code neat) and use this to verify our results. If this function returns true and the product of our two numbers is larger than our current result, simply overwrite our result.
Solving the Project Euler Problems – Problem 3
The prime factors of 13195 are 5, 7, 13 and 29.
What is the largest prime factor of the number 600851475143?
This problem is the first one that has taken some real effort, as no language seems to have any built-in function for determining whether or not a number is prime.
Part of my memory however is convinced that the R language, which I first worked with all the way back in university, did have a function to list all the prime factors of a number, so either my memory is corrupted or, well, chances are it is, as even after sending hopeful texts to people I’m still in touch with from that same course I came up dry. Relearning R is something for another day though, so for this problem, Python it is!
The reason for Python over PHP, as I could just as unhappily do this one in PHP, is that the point of doing these exercises is to expand my knowledge of multiple languages rather than just “solve them”.
Going forward I will likely solve these in a number of languages, but I certainly envisage the main 2 being PHP and Python.
No gists found
Trying to work out a function that would return the prime factors of a number has turned out to be a huge pain, and even then by looking at the code you’ll see I still haven’t managed it! This function does however iterate through the number, reducing it down to its highest prime factor, also covering the case where our number itself is prime.
I even tried venturing into the world of generators and using yield rather than return – definitely something to leave for another day frankly, maybe later in these problems, who knows?
Solving the Project Euler Problems – Problem 2
The main consideration here was how to calculate the Fibonacci sequence without taking up too many resources, a common danger with recursive functions which are often touted as the solution for working out the sequence and other things like factorials. It turns out the Fibonacci sequence as a recursive function is fine up until about the 10th term, but any further than that? Not advisable!
No gists found
The much quicker solution that I’ve found is to do it as an array, which takes simple values for its calculations rather than calling a recursive function again and again and again.
The second performance optimisation is checking whether our Fibonacci number is even, which when doing it a limited number of times, is fine to do just by checking if n modulo 2 is 0, but how many times will we be doing this? Off the top of your head which Fibonacci term is the largest below 4,000,000? The 10th? The 50th? The 1,000th? We don’t know how many until we actually do it. As it happens it’s the 32nd term, 3,524,578. Run that recursive function 32 times!
As division is slow and all we’re using it to do is check if a number is even, is there a better way? Well kind of, we can check if it isn’t odd. This is done by using the bitwise AND operator. This takes the binary representation of a number – 2019’s would be 11111100011 – and it’s that final bit we’re interested in, if it’s 1 then the number is odd and if it’s 0 it’s even. So using the bitwise AND operator to compare any number to 1 checks only 2 bits and avoids division.
Solving the Project Euler Problems – Problem 1
After stumbling across an article on Medium by Bennett Garner, I discovered the Project Euler problems and that it would be a good idea to showcase my skills in solving them in a variety of ways, as a good starting point to get a code portfolio going, so here goes with – as natural a starting point as any – problem 1!
All of the Project Euler Problems can be found at ProjectEuler.net.
If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.
Find the sum of all the multiples of 3 or 5 below 1000.
No gists found
There were 2 ways I considered to tackle this:
• Loop from 1 to 1000 (or as in the code, n) and check each number for dividing by 3 and 5 respectively, adding them to the result if either is true
• Reduce the number of loops by 47% and run our loop $\frac{n}{3}$ + n/5 times (for those interested, 1/3 + 1/5 = 8/15, so this reduction percentage works for any integer), by dividing n by 3, running that loop, adding 3 times each number to the result – which takes division out of the loop as well – then doing the same again, only now having to check (on a smaller loop) if our result was already added as being a multiple of 3. |
While correlation coefficients measure the strength of association between two variables, linear correlation indicates the strongest association between two variables. Visually, this represents any relationship between two variables that depicts a straight line when plotted out next to each other in a graph. Just like the visual, descriptive statistics is one area of statistical applications that uses numerical and graphical techniques to summarize the data, to look for patterns and to present the information in a useful and convenient way. Try this introduction to descriptive statistics today for a more comprehensive review.
The Pearson product-moment correlation coefficient measures the strength of the linear association between variables. The correlation coefficient of a sample is most commonly denoted by r, and the correlation coefficient of a population is denoted by ρ or R. This R is used significantly in statistics, but also in mathematics and science as a measure of the strength of the linear relationship between two variables. More simply, it expresses how much one number can be expected to be influenced by changes in another.For more study, try this course on inferential statistics to guide you through statistical tests in SPSS, including t tests, ANOVA, correlation, regression, and chi-square.
Before moving onto linear association, let’s start with how to find the correlation coefficient.
First, be sure you’re working with two different sets of data. E ach pair of which will be denoted by (xi,yi).
1. You begin with four initial calculations. The quantities from these calculations will be used in subsequent steps of our calculation of r:
1. Calculate x̄, the mean of all of the first coordinates of the data xi.
2. Calculate ȳ, the mean of all of the second coordinates of the data yi.
3. Calculate s x the sample standard deviation of all of the first coordinates of the data xi.
4. Calculate s y the sample standard deviation of all of the second coordinates of the data yi.
2. Use the formula (zx)i = (xi – x̄) / s x and calculate a standardized value for each xi.
3. Use the formula (zy)i = (yi – ȳ) / s y and calculate a standardized value for each yi.
4. Multiply corresponding standardized values: (zx)i(zy)i
5. Add the products from the last step together.
6. Divide the sum from the previous step by n – 1, where n is the total number of points in our set of paired data. The result of all of this is the correlation coefficient r.
The complete formula looks like this:
The sign and the absolute value of a correlation coefficient describe the direction and the magnitude of the relationship between two variables.
• The value of a correlation coefficient ranges between -1 and 1.
• The greater the absolute value of a correlation coefficient, the stronger the linear relationship.
• The strongest linear relationship is indicated by a correlation coefficient of -1 or 1.
• The weakest linear relationship is indicated by a correlation coefficient equal to 0.
• A positive correlation means that if one variable gets bigger, the other variable tends to get bigger.
• A negative correlation means that if one variable gets bigger, the other variable tends to get smaller.
It is important to note that just because r equals 0, it does not mean there is zero relationship between two variables. Rather, it means that there is 0 linear relationship. The Pearson product-moment correlation coefficient only measures linear relationships.
Let’s put these rules to use with a popular statistics example.
Question: A magazine reported the following correlations.
• The correlation between car weight and car reliability is -0.30.
• The correlation between car weight and annual maintenance cost is 0.20.
Which of the following statements are true?
I. Heavier cars tend to be less reliable.
II. Heavier cars tend to cost more to maintain.
III. Car weight is related more strongly to reliability than to maintenance cost.
Options:
(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I, II, and III
Answer: The correct answer is (E). The correlation between car weight and reliability is negative. This means that reliability tends to decrease as car weight increases. The correlation between car weight and maintenance cost is positive. This means that maintenance costs tend to increase as car weight increases.
The strength of a relationship between two variables is indicated by the absolute value of the correlation coefficient. The correlation between car weight and reliability has an absolute value of 0.30, meaning there is a linear correlation between the variables (strongest linear relationship is indicated by a correlation coefficient of -1 or 1) although not very strong. The correlation between car weight and maintenance cost has an absolute value of 0.20 (still a linear relationship, but slightly weaker). Therefore, the relationship between car weight and reliability is stronger than the relationship between car weight and maintenance cost.
Try using the formula to test another number of variable pairs. Need more practice on practical statistics? Try this course on approachable concepts and generating statistical solutions to common questions in user research.
Page Last Updated: February 2020
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# How to Divide by 10
Instructor: Emily Hume
Emily is a Reading Specialist and Literacy coach in a public elementary school with a Master's Degree in Elementary Education.
Dividing by ten is one of the easiest mental math problems to do! In this lesson, you will learn how to divide by 10, draw to solve, and use multiplication to divide. Let's get started!
## Share the Skittles!
Hooray! Mom has bought a giant bag of 120 Skittles for you and your friends for movie night! You have to share them with your 9 friends. So, let's figure out how we can divide the candy evenly between 10 people, your 9 friends plus yourself! When you divide the Skittles, each person will have an equal amount to eat! Yummy! So we have to figure out what is 120 divided by 10!
## How to Divide the Skittles
1. Count the number of total Skittles. We already know that the bag has 120 candies!
2. Line up 10 cups. 1 cup for each person! Begin putting 1 Skittle in each cup until each cup has 1. Then, repeat until each cup has 2, 3, 4, etc. When you have no more Skittles or you don't have enough left to give every cup one (for example, you only have 4 left) you are finished dividing.
3. How many Skittles are in each cup? This number is the answer, or quotient, to your division problem. If you were to have any left over, that number is called the remainder. So in our case, the quotient is 12 since there were 12 candies in each cup. The remainder is 0 since we do not have any pieces left over!
You just divided 120 by 10! So, 120 ÷ 10 = 12. Or, 120 Skittles divided among 10 people equals 12 Skittles per person. How cool!
## Multiplying to Divide
Did you know you could use multiplication to solve a division problem? Let's see how it's done!
You are planning a birthday party! You want each of your friends to have 10 balloons for the balloon popping game. If you already have your 10 balloons and you have 60 more balloons, how many friends can you invite to your party?
1. Count the total number of balloons: 60
2. Start putting your balloons in piles of 10 each.
3. When you have no more balloons left, count the number of groups of 10 you made. This is the number of friends you can invite, since each friend will get a group of 10 balloons.
Here's the division problem: 60 ÷ 10 = 6. You can invite 6 friends!
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Transformations on the Coordinate Plane Worksheet | Problems & Solutions
# Transformations on the Coordinate Plane Worksheet
Transformations on the Coordinate Plane Worksheet
• Page 1
1.
Find the coordinates of C′, if a square ABCD is rotated 180° clockwise around the origin.
a. (0, 6) b. (- 6, - 6) c. (0, - 6) d. (6, 0)
#### Solution:
A rotation is a transformation that turns a figure about a fixed point called the center of rotation.
The following figure is obtained when the square ABCD is rotated 180° clockwise around the origin.
The coordinates of C′ are (- 6, - 6).
2.
Find the coordinates of C′, if triangle ABC is rotated 180o counterclockwise around the origin.
a. (0, - 3) b. (- 3, 0) c. (0, 3) d. (- 3, - 2)
#### Solution:
A rotation is a transformation that turns a figure about a fixed point called the center of rotation.
The following figure is obtained when the triangle ABC is rotated 180° counterclockwise around the origin.
The coordinates of C′ are (0, - 3).
3.
The point C (4, - 3) is first reflected over $x$-axis and then reflected over $y$-axis. What are the coordinates of the final image?
a. (4, 3) b. (4, - 3) c. (3, - 4) d. (- 4, 3)
#### Solution:
When a point is reflected over a horizontal line, the x-coordinate remains the same and the sign of the y-coordinate changes.
The coordinates of the point after the reflection on x-axis is (4, 3).
When a point is reflected over a vertical line, the y-coordinate remains the same and the sign of the x-coordinate changes.
The coordinate of the point when reflected over the y-axis is (- 4, 3) as shown in the following figure.
The coordinates of the final image of point C are (- 4, 3).
4.
Point D is located at (5, 3). If you translate the point ($x$, $y$) $\to$ ($x$ - 2, $y$ - 1), then what would be the coordinates of D′?
a. (3, 4) b. (7, 2) c. (7, 4) d. (3, 2)
#### Solution:
The rule of trasformaton for the point D at (5, 3) is (x - 2, y - 1).
So, (5 - 2, 3 - 1) = (3, 2)
The coordinates of the point D′ are (3, 2).
5.
ABCD is a quadrilateral with coordinates (3, 1), (6, 1), (5, 4) and (1, 3). It is translated to points A′(- 2, 3), B′( 1, 3), C′(0, 6) and D′(- 4, 5). Find the rule for the translation.
a. 5 units right and 2 units down b. 2 units left and 2 units up c. 5 units left and 2 units up d. 2 units down and 2 units right
6.
Find the new coordinates of the figures ABCDE, if you translate 3 units left and 2 units up.
a. A′(1, 3), B′(5, 7), C′(1, 9), D′(2, 6) E′(-1, 5) b. A′(1, 3), B′(5, 7), C′(7, 9), D′(2, 6) E′(-1, 5) c. A′(-7, 3), B′(11, 7), C′(1, 9), D′(2, 6) E′(-1, 5) d. A′(-7, 3), B′(11, 7), C′(1, 9), D′(8, 6) E′(-1, 5)
7.
Triangle ABC has vertices A(4, 4), B(8, 4), C(4, 8). Which graph shows the reflect over the graph y = 3 and then the graph of y = - 2?
a. Graph 2 b. Graph 1 c. Graph 3 d. Graph 4
#### Solution:
Graph 2 shows a reflection of triangle ABC over the line y = 3 followed by another reflection over the line y = - 2.
[Only a reflection flips a figure.]
8.
Identify the rule to describe the translation of the triangle ABC to the triangle A′B′C′.
a. ($x$, $y$) $\to$ ($x$ - 4, $y$ + 3) b. ($x$, $y$) $\to$ ($x$ + 4, $y$ + 3) c. ($x$, $y$) $\to$ ($x$ + 3, $y$ + 4) d. ($x$, $y$) $\to$ ($x$ + 4, $y$ - 3)
9.
A translation is also called as _______.
a. turn b. slide c. flip d. None of the above
#### Solution:
A translation is also called as slide.
10.
What type of transformation is illustrated in the figure?
a. Symmetry b. Reflection c. Translation d. Rotation
#### Solution:
The center of the circle A is moved 5 units down to form the image of the circle with center A'
[From the figure.]
A A'
[Point A goes to point A'.]
So, it is a translation. |
# How do you find the three arithmetic means between 44 and 92?
Then teach the underlying concepts
Don't copy without citing sources
preview
?
#### Explanation
Explain in detail...
#### Explanation:
I want someone to double check my answer
5
### This answer has been featured!
Featured answers represent the very best answers the Socratic community can create.
Alan P. Share
Jan 11, 2017
$56 , 68 , 70$
#### Explanation:
Method 1: Using Arithmetic Sequence Analysis
(I did this version first since the question was asked under "Arithmetic Sequences").
If the initial value, $44$ is denoted as ${a}_{0}$
then we have:
$\textcolor{w h i t e}{\text{XXX}} {a}_{0} = 44$
color(white)("XXX")a_1=?
color(white)("XXX")a_2=?
color(white)("XXX")a_3=?
$\textcolor{w h i t e}{\text{XXX}} {a}_{4} = 92$
but we know that for an arithmetic sequence with initial value ${a}_{0}$ and arithmetic increment $d$,
the ${n}^{t h}$ value in the sequence is:
$\textcolor{w h i t e}{\text{XXX}} {a}_{n} = {a}_{0} + n \cdot d$
$\textcolor{w h i t e}{\text{XXXXXX}}$Some will denote the initial value ${a}_{1}$;
$\textcolor{w h i t e}{\text{XXXXXX}}$in this case the formula becomes:
$\textcolor{w h i t e}{\text{XXXXXXXXX}} {a}_{n} = {a}_{1} + \left(n - 1\right) \cdot d$
$\textcolor{w h i t e}{\text{XXXXXX}}$This seems unnecessarily complicated to me.
so
$\textcolor{w h i t e}{\text{XXX}} {a}_{4} = 92 = 44 + 4 d$
$\textcolor{w h i t e}{\text{XXX}} \rightarrow d = 12$
and therefore
$\textcolor{w h i t e}{\text{XXX}} {a}_{1} = 44 + 12 = 56$
$\textcolor{w h i t e}{\text{XXX}} {a}_{2} = 56 + 12 = 68$
$\textcolor{w h i t e}{\text{XXX}} {a}_{3} = 68 + 12 = 80$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Method 2: Just calculate the arithmetic averages
The primary arithmetic mean is the midpoint between $44$ and $92$
$\textcolor{w h i t e}{\text{XXX}} \frac{44 + 92}{2} = 68$
The arithmetic mean between the initial value and the midpoint is
$\textcolor{w h i t e}{\text{XXX}} \frac{44 + 68}{2} = 56$
The arithmetic mean between the midpoint and the final value is
$\textcolor{w h i t e}{\text{XXX}} \frac{68 + 92}{2} = 80$
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# Find the derivative of h(x)=(4x^3-7x+8)/x
Find the derivative of $h=\left(x\right)=\frac{4{x}^{3}-7x+8}{x}$
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Step 1
Given: $h=\left(x\right)=\frac{4{x}^{3}-7x+8}{x}$
To find- The derivative of the above function.
Identity Used- Using the derivative of the quotient function as
$\frac{d}{dx}\left(\frac{f\left(x\right)}{g\left(x\right)}\right)=\frac{g\left(x\right)×{f}^{\prime }\left(x\right)-f\left(x\right)×{g}^{\prime }\left(x\right)}{g\left(x{\right)}^{2}}$, where ,br.f(x) and g(x) are the function of x.
Step 2
Explanation- Rewrite the given expression,
$h\left(x\right)=\frac{4{x}^{3}-7x+8}{x}$
Using the derivative of the quotient function, differentiating the above expression w.r.t. x, we get,
${h}^{\prime }\left(x\right)=\frac{x\left(12{x}^{2}-7\right)-\left(4{x}^{3}-7x+8\right)\cdot 1}{{x}^{2}}$
$=\frac{12{x}^{3}-7x-4{x}^{3}+7x-8}{{x}^{2}}$
$=\frac{8{x}^{3}-8}{{x}^{2}}$
$=\frac{8\left({x}^{3}-1\right)}{{x}^{2}}$
So, the derivative of the above function is
${h}^{\prime }\left(x\right)=\frac{8\left({x}^{3}-1\right)}{{x}^{2}}$
Answer- the derivative of the function
$\frac{4{x}^{3}-7x+8}{x}$ is ${h}^{\prime }\left(x\right)=\frac{8\left({x}^{3}-1\right)}{{x}^{2}}$ |
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# Usage of Different Strategies of Multiplication
Jul 29, 2023
#### Introduction:
A scientist on a boat is studying hammerhead sharks. The length of 6 hammerhead sharks lined up nose to tail without gaps is equal to the length of the boat. How long is the boat?
#### To find the length of the boat, we have two ways/strategies.
1. Bar diagram
2. Distributive property
We solve the problem by using each method.
### 1. Bar Diagram:
A bar diagram can be defined as a pictorial representation of a number in the form of bars or boxes.
A bar diagram helps to understand the problem easily.
Length of one hammerhead shark = 5 yards
Length of 6 hammerheads sharks = ?
So, we must find 6 × 5.
6 × 5 means 6 groups of 5.
Skip count by 5s.
So, 6 × 5= 30.
The boat is 30 yards long.
### 2. Distributive Property:
The distributive property says that when you multiply a factor by two addends, you can first multiply the factor with each addend, and then add the sum.
To find the length of the boat, we must find 6 × 5.
Use 2s facts and 4s facts to help. that is (2 + 4) × 5.
2 × 5 = 10
4 × 5 = 20
Then add the two products:10 + 20 = 30.
The boat is 30 yards long.
#### To multiply any two numbers, the following are the different strategies:
1. Bar diagram
2. Distributive property
3. Array Model
4. Counters
### 1. Bar Diagram:
Example 1:
7 children shared the cost of a gift equally. Each of them paid \$5. What was the cost of the gift?
1 unit = \$5
7 units = 7 × \$5
So, the cost of the gift was \$35.
Example 2:
Roberta is planting flowers in her garden. She plants 9 rows of flowers. There are 7 flowers in each row. How many flowers does she plant in all?
Solution:
She plants 9 rows of flowers.
There are 7 flowers in each row.
So, if we find 9 × 7, we get the total number of flowers she planted. Skip count by 7s.
So, the total number of flowers she planted was 63.
### 2. Distributive Property:
Example 3:
Solve 12 × 7 by using the distributive property.
Solution:
The distributive property says you can break the problem into smaller parts and then add.
So, 12 × 7 = (7 + 5) × 7
12 × 7 = (7 + 5) × 7
= (7 × 7) + (5 × 7)
= (49 + 35)
= 84
12 × 7 = 84.
Example 4:
A soccer team traveled to a game in 11 vans. Each van has 7 players. Three of the total number of players are goalkeepers. How many players are not goalkeepers?
Solution:
Total number of players= 11 × 7
By using distributive property, we break the problem into smaller parts and then add.
11 × 7 = (5 + 6) × 7
= (5 × 7) +(6 × 7)
= 35 + 42
= 77
Total number of players = 77
The number of players that are not goalkeepers is 77 – 3 = 74.
### 3. Array Model:
Array model of multiplication
uses the number of rows and
number of columns in an array
to illustrate the product of two
numbers.
Product:
2 × 3 = 6
3 × 2 = 6
### 4. Counters:
Counters are an excellent tool that children can use in their attempts to master math skills, including counting, adding, subtracting, making patterns, and comparing numbers.
Example 5:
Share 4 counters between 2 groups.
Solution:
There are 4 altogether.
There are 2 groups in total.
There are 2 in each group.
Try it!
Example 6:
A farmer has 7 ducks. He has 5 times as many chickens as ducks. How many more chickens than ducks does he have?
Solution:
7 × 5 = 35
He has 35 chickens.
35 – 7 = 28
He has 28 chickens more than ducks.
Example 7:
Mr. Marks is studying 3 blacktip sharks and 4 tiger sharks. What is the total length of the 7 sharks? Show your strategy.
Solution:
Length of blacktip shark = 2 yards
Length of tiger shark = 4 yards
The number of blacktip sharks = 3
The number of tiger sharks = 4
Total length of the 7 sharks = (3 × 2) + (4 × 4) = 6 + 16 = 22 yards
So, the total length of the 7 sharks is 22 yards.
#### Exercise:
1. Mrs. Flu bought 3 packets of strawberries. There were 8 strawberries in each packet. How many strawberries did she buy altogether? Use the distributive property to solve.
2. Albert arranged 24 toy soldiers in 4 rows. There were an equal number of toy soldiers in each row. How many toy soldiers were there in each row? Use the counters method to solve.
3. Ben saved \$5 a week for 8 weeks. How much did he save altogether? Use a bar diagram to solve.
4. Mrs. Flu baked 6 cakes. She put 10 cherries on each cake. How many cherries did she use altogether?
5. A toy car costs \$6. A train set costs 5 times as much as the toy car. What is the cost of the train set?
6. Hassan weighs 36kg. He is 4 times as heavy as his brother. How heavy is his brother?
7. Mrs. Bob bought 4 boxes of pencils. There were 5 blue pencils and 3 red pencils in each box. How many pencils did Mrs. Bob buy?
8. Miss Li marked 5 sets of 8 books in the morning. She marked 30 books in the afternoon. How many books did she mark altogether?
9. Raj bought 18 pencils. He bought twice as many pencils as pens. How much did he pay for the pens if each pen cost \$3?
10. Minffa has 6 goldfish. He has 5 times as many guppies as goldfish. If he puts his guppies equally into 3 tanks, how many guppies are there in each tank?
#### What We Have Learned:
• Understand bar diagram.
• Understand distributive property.
• Understand counters and array model.
• Use different strategies to solve problems.
#### 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 […] |
# How do you sketch the general shape of f(x)=x^3-2x^2+1 using end behavior?
Sep 30, 2016
${x}^{3} - 2 {x}^{2} + 1 \setminus \to - \setminus \infty$ as $x \setminus \to - \setminus \infty$
${x}^{3} - 2 {x}^{2} + 1 \setminus \to + \setminus \infty$ as $x \setminus \to + \setminus \infty$
#### Explanation:
You need to know that the end behaviour of a polynomial depends on its degree:
• if the degree is even, both limits at $\setminus \pm \setminus \infty$ will be $+ \setminus \infty$
• if the degree is odd, you'll have the limit according to the direction: if $p \left(x\right)$ is your polynomial, then ${\lim}_{x \setminus \to - \setminus \infty} p \left(x\right) = - \setminus \infty$ and ${\lim}_{x \setminus \to + \setminus \infty} p \left(x\right) = + \setminus \infty$
This is easy to explain: an even degree means that you surely are the square of something: ${x}^{2}$ is the square of $x$, ${x}^{4}$ is the square of ${x}^{2}$, and so on. If $n$ is even, ${x}^{n}$ is the square of ${x}^{\frac{n}{2}}$. And since squares are always positive, the limits can only be $+ \setminus \infty$.
On the other hand, you can see an odd power as an even power of $x$ multiplied one more time by $x$. For example, see ${x}^{7}$ as ${x}^{6} \cdot x$.
We already observed that ${x}^{6}$ tends to positive infinity in both direction, so at $- \setminus \infty$ you'll have ${x}^{6} \cdot x \setminus \to \left(+ \setminus \infty\right) \left(- \setminus \infty\right) = - \setminus \infty$, while at $+ \setminus \infty$ you'll have ${x}^{6} \cdot x \setminus \to \left(+ \setminus \infty\right) \left(+ \setminus \infty\right) = + \setminus \infty$.
The reason for which the leading term is the only relevant one is simple, too: let's analyze your case: we have
${x}^{3} - 2 {x}^{2} + 1 = {x}^{3} \left(1 - \frac{2}{x} + \frac{1}{x} ^ 3\right)$
So, if we factor the greatest power of $x$, all the remaining terms will tend to zero as $x$ approaches infinity (in both direction), showing that (in this case) ${x}^{3}$ is the only relevant term to investigate the end behaviour. |
# Frank ICSE Class 8 Solutions Chapter 1
## Frank ICSE Mathematics Class 8 Solutions Chapter 1 Rational Numbers
Welcome to NCTB Solutions. Here with this post we are going to help 8th class students for the Solutions of Frank ICSE Mathematics Class 8 Math Book, Chapter 1, Rational Numbers. Here students can easily find step by step solutions of all the problems for Rational Numbers, Exercise 1.1, 1.2, 1.3 and 1.4 Also here our mathematics teacher’s are solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get chapter 1 solutions. Here in this post all the solutions are based on latest Syllabus.
Rational Numbers Exercise 1.1 Solution :
Question no – (1)
Solution :
(a) -11/16 __ 13/- 16
= – 11/16 > 13/- 16
(b) -9/-13 __ 36/52
= -9/-13 = 36/52
(c) 5/8 __ – 9/14
= 5/8 > -9/14
(d) 7/-13 __ -8/15
= 7/- 13 < – 8/15
Question no – (2)
Solution :
(a) In the given question,
= 3/8, 5/-12, -7/16, -13/18, 11/24
Now in ascending order.
= -13/118, 11/- 24, – 7/16, 5/-12, – 3/8
(b) In the question,
= 9/25, 2/5, 14/-75, – 19/10, 8/15
Now, in ascending order,
= -19/10, 14/75, 9/25, 2/5, 8/15
Question no – (3)
Solution :
(a) Given in the question,
= 7/8, 11/12, 15/-16, 1/4, 14/-15
Now in descending order,
= 11/12, 7/8, 1/4, 14/-15, 15/-16
(b) In the given question,
= -5/6, -3, -13/-4, 17/-6, -11/1, -13/9
Now, in the descending order,
= – 5/6, – 11/12, -13/9, 17/-6, -3
Question no – (4)
Solution :
(a) -5/13 + 12/13
= -5/13 + 12/13
= -5 + 12/13
= 7/13 …(simplified)
(b) -8/15 + -/15
= -8/15 + – 7/15
= – 8 – 7/15
= – 15/15
= – 1 …(simplified)
(c) 21/25 – 16/25 + 7/25
= 21/25 – 16/25 + 7/25
= 21 – 16 + 7/25
= 12/25 …(simplified)
(d) 5/12 – 7/36
= 5/12 – 7/36
= (5 × 3) – 7/36
= 15 – 7/36
= 8/36
= 2/9 …(simplified)
(e) – 8/17 + – 4/51
= – 8/17 + – 4/51
= (- 8) × 3 (- 4)/51
= – 24 – 4/51
= – 28/51 …(simplified)
(f) – 5/16 + 7/24 – 13/48
= – 5/16 + 7/24 – 13/48
= (- 5) × 3 + (7 × 2) – 13/48
= 15 + 14 – 13/48
= – 14/48
= – 7/24 …(simplified)
Question no – (5)
Solution :
(a) – 11/9 × – 81/88
= – 11/9 × – 81/88
= 9/8 …(Simplified)
(b) 8/-9 × 16/-7
= 8/- 9 × 16/-7
= 128/63 …(Simplified)
(c) 6/7 × – 49/36
= – 7/6
(d) – 5/9 × 49/56 × – 72/35
= 1
(e) – 13/9 ÷ 26/27
= – 3/2
(f) – 25/7 ÷ 15/14
= – 25/7 × 14/15
= – 10/3 …(Simplified)
Question no – (6)
Solution :
(a) All whole numbers are rational numbers → True
(b) All rational numbers are integers. → False
(c) The rational number -8/5 is in standard from. → True
(d) Integers satisfy the closure property for subtraction. → True
Question no – (7)
Solution :
(a) -165/187 = ___/- 34
= – 165/187
= – 15/17
= 30/- 34
(b) 248/- 280 = ____/105
= 248/- 280
= – 31/35
= – 93/105
(c) 275/- 330 = – 10/____
= 275/- 330
= – 5/6
= – 10/12
(d) -192/256 = 9/____
= – 192/256
= – 3/4
= 9/-12
Rational Numbers Exercise 1.2 Solution :
Question no – (1)
Solution :
(a) 11/15, – 11/15
= Yes, they are additive inverse of each other.
(b) – 18/21, 17/21
= Yes, they are additive inverse of each other.
(c) 8/-9, 8/9 /17
= Yes, they are additive inverse of each other.
(d) – 25/- 52, 25/52
= No, they are not additive inverse of each other.
Question no – (2)
Solution :
(a) 12/17, -17/-12
= Yes, they are reciprocals.
(b) -13/27, 27/13
= No, they are not reciprocals.
(c) 1/9, 9/-1
= No, they are not reciprocals.
(d) -1 3/5, 5/8
= No, they are not reciprocals.
Question no – (3)
Solution :
The multiplicative inverse of -3/5 × – 8/11 is -55/24
Question no – (4)
Solution :
(a) 5/8 × [3/4 + 7/11] = [5/8 × 3/4] + [5/8 × 7/11]
(b) (- 11/15 × 5/6) × 3 = 11/15 × (5/6 × 3)
= Associative property
(c) 12/- 23 × – 23/12 = 11
= Multiplicative identity
(d) – 9/17 × 1 = 1 × -9/17 = – 9/17
= Multiplicative identity
(e) 6/7 × – 8/9 = – 8/9 × 6/7
= Commutative property
Question no – (5)
Solution :
(a) -7/12 + 13/15
= (-7 × 5) + (13 × 4)/60
= -35 + 52/60
= -7/60 …(Simplified)
(b) -5/8 + 8/9 + – 11/12
= (5 × 9) + (8 × 8) + (- 11 × 6)/72
= -40 + 64 – 66/72
= -42/72
= -7/12 …(Simplified)
(c) (-8/15 + – 13/20) × – 5/6
= {(-8 × 4) + (-13 × 3)}/60} × – 5/6
= (32 – 39/60) × – 5/6
= – 71/60 × – 5/6
= 71/72 …(Simplified)
(d) 3 3/5 × -15/16 × 13/20
= 18/5 × -15/16 × 13/20
= – 351/160 …(Simplified)
Question no – (6)
Solution :
(a) 1/8 + – 5/12 + – 3/16
= 6 + (- 5 × 4) + (- 3 × 3)/46
= – 23/48
(b) 11/12 + – 1/3 + – 3/4 + 7/6
= 11 – 4 – 9 + 14/12
= 12/12
= 1
(c) 5/12 × – 3/8 × – 4/15
= 1/24
(d) – 6/7 × 8/- 25 × 35/- 18
= – 6/7 × 8/- 25 × 35/- 18
= – 8/15
(e) – 8/15 × – 25/16 ÷ 35/10
= 5/6 × 10/35
= 5/21
(f) 7/10 × – 15/14 ÷ 9/8
= – 3/4 × 8/9
= – 2/3
Rational Numbers Exercise 1.3 Solution :
Question no – (1)
Solution :
(a) A – 4/8, B = 7/8, C = 9/8, D = 12/8
(b) A = – 4/5, B = – 1/5, C = 2/5, D = 4/5, E = 7/5
(c) A = – 9/7, B = – 5/7, C = – 2/7, D = 3/7, ER = 6/7, F = 9/7
(d) A = -9/11, B = – 6/11, C = – 3/11, D = 2/11, E = 5/11, F = 7/11, G = 12/11
Question no – (2)
Solution :
In the given question,
Given rational numbers are – (a) 4/5 (b) -2/3 (c) -2 3/5 (d) 3 1/3
Now on the number line :
Question no – (3)
Solution :
In the question give numbers are
= -3/4, 1/4, 7/4
Now on the number line :
Question no – (4)
Solution :
Five rational numbers between -1 and +1 using the method of arithmetic mean are –
= q1 = 1/2 = (- 1 + 1) = 1/2 (0) = 0
= q2 = 1/2 (0+ 1) = 1/2 (1) = 1/2
= q3 = 1/2 (1/2 + 0) = 1/2 (1/2) = 1/4
Numbers are 0, 1/4, 1/2
Question no – (5)
Solution :
= q1 = 1/2 (2 + 3) = 1/2 950 = 5/2
= q2 = 1/2 (2 + 5/2) = 1/2 (9/2) = 9/4
= q3 = 1/2 (2 + 9/4) = 1/2 917/4) = 17/8
= q4 = 1/2 (2 + 17/80 = 1/2 (33/8) = 33/16
= q5 = 1/2 (2 + 33/16) = 1/2 (65/16) = 65/32
Numbers are = 65/32, 33/16, 17/8, 9/4, 5/2
Question no – (6)
Solution :
(a) = q1 = 1/2 (1/2 + 2/3) = 1/2 (7/6) = 7/12
= q2 = 1/2 (1/2 + 7/12) = 1/2 913/12) = 13/224
= q3 = 1/2 (1/2 + 13/24) = 1/2 (25/24) = 25/48
(b) = – 3/4 = – 3 × 50/4 × 50 = – 150/200,
= – 4/5 = – 4 × 40/5 × 40 = – 160/200
Numbers = – 151/200, – 152/200, – 153/200
(c) = 2/7 = 2 × 2/7 × 2 = 4/14
= 2 = 2 × 14/1 × 14 = 28/14
Numbers = 5/14, 8/14, 20/14
(d) = – 3 = – 3 × 5/1 × 5 = – 15/5, – 7/5
Numbers are = – 8/5, – 9/5, – 10/5
Rational Numbers Exercise 1.4 Solution :
Question no – (1)
Solution :
According to the question,
The product two rational numbers is = -8/9.
One of them is = 10/3
The other number = ?
Now, The other number,
= – 8/9 ÷ 10/3
= – 8/9 × 3/10
= – 4/15
Therefore, the other number will be – 4/15
Question no – (2)
Solution :
= (2/9 + 4/9) = 14 + 36/63 = 50/63
= (2/9 – 4/7) = 14 – 36/63 = -22/62
= 50/63 × – 63/22
= – 25/11
Question no – (3)
Solution :
In the question given,
= -9/14 and 11/8
Required number,
= (- 9/14 – 11/18)
= – 81 – 77/+ 126
= 158/126
Therefore, the required number will be 158/126
Question no – (4)
Solution :
= 5/9 + – 8/7
= 35 – 72/63
= – 37/63
= 7/8 × 3/8
= 21/40
– 37/63 × 40/21
= – 1480/1323
Question no – (5)
Solution :
According to the question,
The sum of two rational numbers is 8/25.
One of them is 3/16,
The other number.
Step by Step solution :
The other number,
= (8/25 – 3/10)
= 128 – 75/400
= 53/400
Therefore, the other number will be 53/400
Next Chapter Solution :
Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 Chapter 12 Chapter 13 Chapter 14 Chapter 15 Chapter 16 Chapter 17 Chapter 18 Chapter 19
Updated: June 26, 2023 — 5:48 am |
# Sum of $p\sum_{i = 1}^{\infty} i(1 - p)^{i - 1}$
Does anyone know how to find the exact sum of
$$p\sum_{i = 1}^{\infty} i(1 - p)^{i - 1}$$
-
Note that
$$\sum_{i=1}^{\infty} i \, r^{i-1} = \frac{d}{dr} \sum_{i=0}^{\infty} r^i = \frac{d}{dr} \frac{1}{1-r}=\frac{1}{(1-r)^2}$$
So your sum is, letting $r=1-p$,
$$p \frac1{p^2} = \frac1{p}$$
-
I'm going to do this my favorite way instead of the same way as everyone else.
Let $S$ be the sum in question. Then we can multiply $S$ by $\displaystyle \frac{1}{1-p}$ to shift the sum and then subtract, obtaining $$\frac{p}{1-p}S = \frac{S}{1-p} - S = \frac{p}{1-p} + p\sum_{i=0}^\infty (1-p)^i = \frac{p}{1-p} + p(\frac{1}{p}) = \frac{1}{1-p}.$$ Solving for $S$ we obtain $\displaystyle S = \frac{1}{p}$.
Of course, the sum only converges when $|1-p| < 1$.
-
Yes. Integrate the sum termwise with respect to $p,$ to get $$-\sum_{i=1}^\infty(1-p)^i = -(p-1) \sum_{i=0}^\infty (1-p)^i = -\frac{1}{p}.$$ Differentiate back with respect to $p,$ to get $1/p^2.$ Multiply by $p$ to finally get $1/p.$
-
Given that $|x|< 1$, $$\frac 1{(1-x)^2}=\sum_{k=0}^\infty (k+1)x^k$$
So $$p^{-1}=\frac{p}{(1-(1-p))^2}=p\sum_{k=0}^\infty (k+1)(1-p)^k$$
Reasoning: $(1-x)(1+x+x^2+\ldots)=1+x+x^2+\ldots-(x+x^2+\ldots)=\lim_{n\to\infty}1-x^n=1$ so $\frac{1}{1-x}=\sum_{k=0}^\infty x^k$. Note that $$\frac{1}{(1-x)^2}=\sum_{k=0}^\infty \frac{x^k}{1-x}=\sum_{k=0}^\infty\left(\sum_{\ell=0}^\infty x^{k+\ell}\right)$$ For any $c$, there are $c+1$ pairs $(k,\ell)$ such that $k+\ell=c$, so the sum reduces to $\sum_{k=0}^\infty (k+1)x^k$
- |
## Understand the relationship between degree and turning points
In addition to the end behavior, recall that we can analyze a polynomial function’s local behavior. It may have a turning point where the graph changes from increasing to decreasing (rising to falling) or decreasing to increasing (falling to rising). Look at the graph of the polynomial function $f\left(x\right)={x}^{4}-{x}^{3}-4{x}^{2}+4x$ in Figure 11. The graph has three turning points.
Figure 11
This function f is a 4th degree polynomial function and has 3 turning points. The maximum number of turning points of a polynomial function is always one less than the degree of the function.
### A General Note: Interpreting Turning Points
A turning point is a point of the graph where the graph changes from increasing to decreasing (rising to falling) or decreasing to increasing (falling to rising).
A polynomial of degree n will have at most n – 1 turning points.
### Example 7: Finding the Maximum Number of Turning Points Using the Degree of a Polynomial Function
Find the maximum number of turning points of each polynomial function.
1. $f\left(x\right)=-{x}^{3}+4{x}^{5}-3{x}^{2}++1$
2. $f\left(x\right)=-{\left(x - 1\right)}^{2}\left(1+2{x}^{2}\right)$
### Solution
1. $f\left(x\right)=-x{}^{3}+4{x}^{5}-3{x}^{2}++1$
First, rewrite the polynomial function in descending order: $f\left(x\right)=4{x}^{5}-{x}^{3}-3{x}^{2}++1$
Identify the degree of the polynomial function. This polynomial function is of degree 5.
The maximum number of turning points is 5 – 1 = 4.
2. $f\left(x\right)=-{\left(x - 1\right)}^{2}\left(1+2{x}^{2}\right)$
Figure 12
First, identify the leading term of the polynomial function if the function were expanded.
Then, identify the degree of the polynomial function. This polynomial function is of degree 4.
The maximum number of turning points is 4 – 1 = 3. |
# Consecutive Interior Angles
## Consecutive Interior Angles
### What do we mean by successive angles?
The following angles refer to specific pairs of angles that occur when two parallel lines are crossed by another line called transverse.
### What's the next corner of that?
Internal rear corners. When two lines are crossed by another line (called transverse): The pairs of angles on one side of the transverse but within the two lines are called consecutive internal angles.
### In addition to the above, what does it mean when successive angles are complementary?
The following set of internal angles says that if the two lines are parallel, the following internal angles are complementary to each other. It also means that the two angles add up to 180 degrees.
### How are subsequent external angles defined here?
Subsequent outer corners: Subsequent outer corners are outer corners that lie on the same side as the transverse direction.
### What does it mean to be congruent?
Compliant. Angles are congruent when they are the same size (in degrees or radians). The sides are congruent if they are the same length.
### How big is the sum of the consecutive angles?
When the two lines are parallel, a pair of continuous internal angles form 180 degrees.
### How do you find the measurements of the angles?
Using a protractor The best way to measure a protractor is to use a protractor. To do this, first install a bar along a 0 degree line on the protractor. Then align the top with the center of the protractor. Follow the second ray to determine the angle measurement to the nearest degree.
### Are the subsequent external corners complementary?
Successive external angles. When parallel lines are intersected by a transverse line, the subsequent external angles are complementary. The following external angles add up to 180 degrees.
### Are consecutive angles complementary?
The following corners are complementary, so look at the following corners (or side by side). If the figures are complementary, the shape can be a parallelogram. Complementary angles are two angles whose sum is 180 degrees.
### What is a vertical line?
In elementary geometry, the perpendicular (perpendicular) property is the ratio of two lines meeting at right angles (90 degrees). A line is perpendicular to another line if the two lines intersect at right angles.
### Are the adjacent corners congruent?
All the properties of a parallelogram apply (we are interested in parallel sides, opposite angles are congruent and subsequent angles are complementary). The definition is consistent from every point of view.
### What are consecutive angles in a square?
A square is a four-sided polygon. to make a random square! The sides, corners and adjacent corners in a polygon are said to be consecutive. (Note that consecutive sides intersect at one point.
### What are consecutive angles in a parallelogram?
The following angles in a parallelogram are always 180 degrees. This makes the angles in a parallelogram complementarily continuous. In the parallelogram above, the angles A and B, B and C, C and D, and D and A are examples of continuous angles.
### How do you find adjacent corners?
Two corners are adjacent if they have a side and vertex in common and do not overlap.
### Are the vertical angles congruent?
When two lines intersect to form an X, the angles on either side of X are called vertical angles. These angles are the same, and here is the official sentence it tells you. Vertical angles are congruent: if two angles are vertical, they are congruent (see illustration above).
### What are the consecutive angles in a diamond?
Result 4: Two consecutive corners in a diamond are complementary and the opposite sides are parallel. Proof: Through the evaluation we see that the dimension is now at the angle BDC and the dimension is at the angle DCA, i.e. the sum of these two successive angles is 180 degrees, they are additional.
### How big is the sum of the alternating external angles?
The parallel case If the transverse parallel lines intersect (normal case), the external angles are added separately (add 180 °). In the figure above, when moving point A or B, both angles are always set to 180 °.
### What is an alternative outside corner?
Alternative external angles. When two lines are crossed by another line (called transverse): the alternate external angles are a pair of external angles to each of the two lines, but on opposite sides of the transverse line. In this example there are two pairs of alternating external angles: a and h. |
# How do you simplify (x^3+8)*(x-2)/(x^2-2x+4)div(x^2-4)/(x-6)?
##### 1 Answer
Sep 27, 2016
$x - 6$
#### Explanation:
$\textcolor{red}{\left({x}^{3} + 8\right)} \cdot \frac{x - 2}{\textcolor{b l u e}{{x}^{2} - 2 x + 4}} \div \frac{\textcolor{g r e e n}{{x}^{2} - 4}}{x - 6}$
In algebraic fractions you want to factorize as much as possible.
$\frac{\textcolor{red}{\left(x + 2\right) \left({x}^{2} - 2 x + 4\right)}}{1} \times \frac{x - 2}{\textcolor{b l u e}{{x}^{2} - 2 x + 4}} \times \frac{\textcolor{g r e e n}{x - 6}}{\left(x + 2\right) \left(x - 2\right)}$
Now that everything is expressed as factors you may cancel:
$\frac{\cancel{x + 2} \cancel{{x}^{2} - 2 x + 4}}{1} \times \frac{\cancel{x - 2}}{\cancel{{x}^{2} - 2 x + 4}} \times \frac{x - 6}{\cancel{x + 2} \cancel{x - 2}}$
$= \left(x - 6\right)$ |
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# Differentiate the given function with respect to x, $y=\sqrt{{{e}^{\sqrt{x}}}}\text{ }\!\!~\!\!\text{ , }\!\!~\!\!\text{ }x>0$
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Hint: We can also write the function $\sqrt{{{e}^{\sqrt{x}}}}$ in the form ${{\text{(}{{e}^{\sqrt{x}}})}^{\dfrac{1}{2}}}$. As we can see $\sqrt{{{e}^{\sqrt{x}}}}\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }$ involves three functions, which are ${{({{e}^{\sqrt{x}}})}^{\dfrac{1}{2}}}\text{,}{{e}^{\sqrt{x}}}\text{,}\sqrt{x}\text{ }\!\!~\!\!\text{ }$. We can use the chain rule to solve this question.
The function given in the question is $y=\sqrt{{{e}^{\sqrt{x}}}}\text{ }\!\!~\!\!\text{ , }\!\!~\!\!\text{ }x>0$. As we can see, the function has multiple functions. So, for differentiating this type of functions, we can use the chain rule of differentiation.
The chain rule states that:
$\dfrac{d}{dx}(f(g(x)))=f'(g(x))g'(x)$
We know that the formula for the derivation of ${{e}^{x}}$and ${{x}^{n}}$ is given by,
$\dfrac{d}{dx}({{e}^{x}})={{e}^{x}}$ and $\dfrac{d}{dx}({{x}^{n}})=n{{x}^{n-1}}$
Let us assume ${{e}^{\sqrt{x}}}$ in the place of $x$ such that ${{({{e}^{\sqrt{x}}})}^{\dfrac{1}{2}}}$ reduces into form ${{x}^{n}}$.
Therefore, we can see that we have ${{({{e}^{\sqrt{x}}})}^{\dfrac{1}{2}}}$ as in the above form of derivative.
Now, applying the formula $\dfrac{d}{dx}({{x}^{n}})$ and chain rule, we get,
$\therefore \dfrac{d}{dx}{{({{e}^{\sqrt{x}}})}^{\dfrac{1}{2}}}=\dfrac{1}{2}{{({{e}^{\sqrt{x}}})}^{\dfrac{1}{2}-1}}\dfrac{d}{dx}({{e}^{\sqrt{x}}})$
$=\dfrac{1}{2\sqrt{{{e}^{\sqrt{x}}}}}\dfrac{d}{dx}({{e}^{\sqrt{x}}})................(i)$
Let us assume $\sqrt{x}$ in the place of $x$ such that ${{e}^{\sqrt{x}}}$ reduces into form $({{e}^{x}})$.
Therefore, we can see that we have ${{e}^{\sqrt{x}}}$ as in the above form of derivative.
Applying the formula for $\dfrac{d}{dx}({{e}^{x}})$ and chain rule we get,
$\therefore \dfrac{d}{dx}({{e}^{\sqrt{x}}})={{e}^{\sqrt{x}}}\dfrac{d}{dx}\sqrt{x}...........(ii)$
We can convert $\sqrt{x}$ in the form ${{x}^{\dfrac{1}{2}}}$.
Then, applying the formula for $\dfrac{d}{dx}({{x}^{n}})$ we get,
$\therefore \dfrac{d}{dx}\sqrt{x}=\dfrac{1}{2}{{x}^{\dfrac{1}{2}-1}}$
$=\dfrac{1}{2\sqrt{x}}...............(iii)$
From equation (i), we get:
$\dfrac{d}{dx}{{({{e}^{\sqrt{x}}})}^{\dfrac{1}{2}}}=\dfrac{1}{2\sqrt{{{e}^{\sqrt{x}}}}}\dfrac{d}{dx}{{(e)}^{\sqrt{x}}}$
From equation (ii), we get:
$\dfrac{d}{dx}({{e}^{\sqrt{x}}})={{e}^{\sqrt{x}}}\dfrac{d}{dx}(\sqrt{x})$
From equation (iii), we get:
$\dfrac{d}{dx}(\sqrt{x})=\dfrac{1}{2\sqrt{x}}$
Substituting the values from equations (i), (ii), (iii) to the question:
$\dfrac{dy}{dx}=\dfrac{d}{dx}{{({{e}^{\sqrt{x}}})}^{\dfrac{1}{2}}}$
From equation (i),
$\dfrac{d}{dx}{{({{e}^{\sqrt{x}}})}^{\dfrac{1}{2}}}=\dfrac{1}{2\sqrt{{{e}^{\sqrt{x}}}}}\dfrac{d}{dx}{{(e)}^{\sqrt{x}}}$
Substituting the value of $\dfrac{d}{dx}({{e}^{\sqrt{x}}})$ from equation (ii),
$=\dfrac{1}{2\sqrt{{{e}^{\sqrt{x}}}}}{{e}^{\sqrt{x}}}\dfrac{d}{dx}(\sqrt{x})$
Substituting the value of $\dfrac{d}{dx}(\sqrt{x})$ from equation (iii),
$=\dfrac{1}{2\sqrt{{{e}^{\sqrt{x}}}}}{{e}^{\sqrt{x}}}\dfrac{1}{2\sqrt{x}}$
$=\dfrac{{{e}^{\sqrt{x}}}}{4\sqrt{{{e}^{\sqrt{x}}}x}}$
Hence, $\dfrac{dy}{dx}=\dfrac{{{e}^{\sqrt{x}}}}{4\sqrt{{{e}^{\sqrt{x}}}x}}$
Note: In this question, x>0 is given. If it was x<0, then we would have had imaginary roots. So, in that case, we would have multiple answers and we would be asked to find which of those couldn’t be the value of x. Before applying chain rule, you have to take care that the question involves function inside the given function, only then the chain rule can be applied. If the question involves chain rule and we are comfortable using it, then we can directly differentiate each function separately and write the answer. It will take less time, for example:
Differentiate ${{e}^{{{x}^{2}}}}$
$\dfrac{d}{dx}({{e}^{{{x}^{2}}}})$
Here, we have two functions ${{e}^{{{x}^{2}}}},{{x}^{2}}$. So, we can differentiate these two functions separately and simply write the answer as ${{e}^{{{x}^{2}}}}2x$. |
# K.OA.A.4
For any number from 1 to 9, find the number that makes 10 when added to the given number, e.g., by using objects or drawings, and record the answer with a drawing or equation.
## Summary:
This standard asks students to determine what number can be added to a given number to make 10. Objects such as counters, blocks, fingers, and other manipulatives can be be used and then drawn on paper to represent this problem. Students who have mastered this skill can "make ten" as a strategy for adding; for example, to add 8 + 6, they can decompose the 6 into 2 + 4 (note that this also uses skills from K.OA.3), knowing that 8 + 2 makes ten. Then they can observe that 8 + 2 + 4 = 10 + 4 = 14.
## Understanding the Standard:
• Fact families are especially beneficial in helping students to understand the relationship between addition and subtraction. Being able to subtract from 10 might give students an advantage when determining what number can be added to a given number to make 10.
• Students should have ample opportunities to use manipulatives in constructing concrete models of missing-addend problems.
• Drawings can be used to further develop understanding the processes involved in finding the difference in a subtraction problem.
• The use of manipulatives and technologies is helpful for those students needing additional visual cues and hands-on experiences. Using a set of ten objects and decomposing it to make two smaller sets will help children visualize what they are learning.
• Starting with a random number less than 10, use the word "addition" to represent an example that includes the number 10 and a variable in the form of "?", e.g., 2 + ? = 10.
## Questions to Focus Instruction:
• When given the first addend, can students correctly solve simple addition facts in which the sum is 10 ?
• Are students able to take a drawing of a given mathematical situation that makes ten and write it in equation form?
• Are students able to represent their method of problem-solving with manipulatives or drawings using 10 objects?
## Skills
Prior to: Students have an understanding that two sets can combine to form a single set with a greater number of objects. Studnets should understand that the operation of addition gives the number of objects in the combined set. |
# HCF of Two Numbers
HCF of two numbers is the highest common number, which is available in both the numbers. Before we proceed ahead to find the HCF, let us discuss what HCF is. HCF or highest common factor is the factor of any two or more numbers, which are common among them. Sometimes, it is also called the greatest common factor (GCF) or greatest common divisor (GCD).
For example, the HCF of 2 and 4 is 2, because 2 is the number which is common to both 2 and 4. For such small numbers, finding HCF is an easy method. But for larger numbers, we need to use different techniques such as prime factorisation and long division method, to find the HCF.
Let us discuss both the techniques.
### HCF of two numbers By Prime Factorisation
Let us solve some examples here to understand this method.
Question 1: What is the HCF of 24 and 36?
Solution: By prime factorisation, we can write the two numbers;
24 = 2 x 2 x 2 x 3
36 = 2 x 2 x 3 x 3
Hence, after factoring the numbers 24 and 36, we can see, the factors 2x2x3 are common.
Therefore, the HCF (24, 36) = 2x2x3 = 12
Question 2: What is the HCF of 35 and 55?
Solution: By prime factorisaton we can write the two numbers as:
35 = 5 x 7
55 = 5 x 11
Hence, we can see the highest common factor for 35 and 55 here is 5.
Therefore, HCF (35, 55) = 5
### HCF of two numbers by Division Method
We have already understood the prime factorisation method to determine the HCF.
Steps for Division method:
If we were given two numbers, then
• First, divide the large number by a small number.
• If the remainder is left, then divide the first divisor by remainder.
• If the remainder divides the first divisor completely, then it is the HCF or highest common factor of the given two numbers.
• If the remainder does not divide the first divisor completely, then repeat the steps.
Let us now learn the division method with the help of examples
Question:1 What is the HCF of 120 and 100.
Solution: Now, let us find the HCF by using division method.
Divide 120 by 100.
120/100 → 1 and remainder is 20
Now, divide the first divisor 100 by first remainder 20.
100/20 → 5 and remainder is 0.
Therefore, 20 is the HCF of 120 and 100.
Question 2: Find the HCF of 45 and 60 by the division method.
Solution: Divide 60 by 45.
60/45 → 1 and remainder is 15
Now, divide 45 by 15
45/15 → 3
Therefore, 15 is the HCF of 45 and 60.
### HCF of Three Numbers
Let us solve an example when we need to find the HCF of three numbers.
Example: Find the HCF of 126, 162 and 180.
Solution: By prime factorisation, we can write the given numbers as;
126 = 2x3x3x7
162 = 2x3x3x3x3
180 = 2x2x3x3x5
Taking out the common factors of 126, 162 an d180, we get:
HCF(126,162,180) = 2x3x3 = 18 |
# If f(x) =-e^(2x+4) and g(x) = tan3x , what is f'(g(x)) ?
Oct 26, 2017
$f ' \left(g \left(x\right)\right) = - 2 {e}^{2 \tan \left(3 x\right) + 4}$
#### Explanation:
$f ' \left(g \left(x\right)\right)$ indicates that we need to take the derivative of $f$, and then plug in $g \left(x\right)$ for $x$.
First, let's find $f ' \left(x\right)$:
We know that $\frac{d}{\mathrm{dx}} {e}^{u} = {e}^{u} \frac{d}{\mathrm{dx}} \left(u\right)$
Therefore:
$\frac{d}{\mathrm{dx}} \left(- {e}^{2 x + 4}\right) = - {e}^{2 x + 4} \cdot \frac{d}{\mathrm{dx}} \left(2 x + 4\right)$
$= - 2 {e}^{2 x + 4}$
So $f ' \left(x\right) = - 2 {e}^{2 x + 4}$.
We also know that $g \left(x\right) = \tan \left(3 x\right)$
Therefore, $f ' \left(g \left(x\right)\right) = - 2 {e}^{2 \tan \left(3 x\right) + 4}$ |
$$\require{cancel}$$
# 16.4: Wave Speed on a Stretched String
Learning Objectives
• Determine the factors that affect the speed of a wave on a string
• Write a mathematical expression for the speed of a wave on a string and generalize these concepts for other media
The speed of a wave depends on the characteristics of the medium. For example, in the case of a guitar, the strings vibrate to produce the sound. The speed of the waves on the strings, and the wavelength, determine the frequency of the sound produced. The strings on a guitar have different thickness but may be made of similar material. They have different linear densities, where the linear density is defined as the mass per length,
$\mu = \frac{\text{mass of string}}{\text{length of string}} = \frac{m}{l} \ldotp \label{16.7}$
In this chapter, we consider only string with a constant linear density. If the linear density is constant, then the mass ($$\Delta m$$) of a small length of string ($$\Delta$$x) is $$\Delta m = \mu \Delta x$$. For example, if the string has a length of 2.00 m and a mass of 0.06 kg, then the linear density is $$\mu = \frac{0.06\; kg}{2.00\; m}$$ = 0.03 kg/m. If a 1.00-mm section is cut from the string, the mass of the 1.00-mm length is
$\Delta m = \mu \Delta x = (0.03\, kg/m)(0.001\, m) = 3.00 \times 10^{−5}\, kg. \nonumber$
The guitar also has a method to change the tension of the strings. The tension of the strings is adjusted by turning spindles, called the tuning pegs, around which the strings are wrapped. For the guitar, the linear density of the string and the tension in the string determine the speed of the waves in the string and the frequency of the sound produced is proportional to the wave speed.
## Wave Speed on a String under Tension
To see how the speed of a wave on a string depends on the tension and the linear density, consider a pulse sent down a taut string (Figure $$\PageIndex{1}$$). When the taut string is at rest at the equilibrium position, the tension in the string $$F_T$$ is constant. Consider a small element of the string with a mass equal to $$\Delta m = \mu \Delta x$$. The mass element is at rest and in equilibrium and the force of tension of either side of the mass element is equal and opposite.
If you pluck a string under tension, a transverse wave moves in the positive x-direction, as shown in Figure $$\PageIndex{2}$$. The mass element is small but is enlarged in the figure to make it visible. The small mass element oscillates perpendicular to the wave motion as a result of the restoring force provided by the string and does not move in the x-direction. The tension FT in the string, which acts in the positive and negative x-direction, is approximately constant and is independent of position and time.
Assume that the inclination of the displaced string with respect to the horizontal axis is small. The net force on the element of the string, acting parallel to the string, is the sum of the tension in the string and the restoring force. The x-components of the force of tension cancel, so the net force is equal to the sum of the y-components of the force. The magnitude of the x-component of the force is equal to the horizontal force of tension of the string $$F_T$$ as shown in Figure $$\PageIndex{2}$$. To obtain the y-components of the force, note that tan $$\theta_{1} = − \frac{F_{1}}{F_{T}}$$ and $$\tan \theta_{2} = \frac{F_{2}}{F_{T}}$$. The $$\tan \theta$$ is equal to the slope of a function at a point, which is equal to the partial derivative of y with respect to x at that point. Therefore, $$\frac{F_{1}}{F_{T}}$$ is equal to the negative slope of the string at x1 and $$\frac{F_{2}}{F_{T}}$$ is equal to the slope of the string at x2:
$\frac{F_{1}}{F_{T}} = - \left(\dfrac{\partial y}{\partial x}\right)_{x_{1}}\; and\; \frac{F_{2}}{F_{T}} = \left(\dfrac{\partial y}{\partial x}\right)_{x_{2}} \ldotp$
The net force is on the small mass element can be written as
$F_{net} = F_{1} + F_{2} = F_{T} \Bigg[ \left(\dfrac{\partial y}{\partial x}\right)_{x_{2}} - \left(\dfrac{\partial y}{\partial x}\right)_{x_{1}} \Bigg] \ldotp$
Using Newton’s second law, the net force is equal to the mass times the acceleration. The linear density of the string µ is the mass per length of the string, and the mass of the portion of the string is $$\mu \Delta$$x,
$F_{T} \Bigg[ \left(\dfrac{\partial y}{\partial x}\right)_{x_{2}} - \left(\dfrac{\partial y}{\partial x}\right)_{x_{1}} \Bigg] = \Delta ma = \mu \Delta x \left(\frac{\partial^{2} y}{\partial t^{2}}\right) \ldotp$
Dividing by FT$$\Delta$$x and taking the limit as $$\Delta$$x approaches zero,
$\begin{split} \lim_{\Delta x \rightarrow 0} \frac{\Bigg[ \left(\dfrac{\partial y}{\partial x}\right)_{x_{2}} - \left(\dfrac{\partial y}{\partial x}\right)_{x_{1}} \Bigg]}{\Delta x} & = \frac{\mu}{F_{T}} \frac{\partial^{2} y}{\partial t^{2}} \\ \frac{\partial^{2} y}{\partial x^{2}} & = \frac{\mu}{F_{T}} \frac{\partial^{2} y}{\partial t^{2}} \ldotp \end{split}$
Recall that the linear wave equation is
$\frac{\partial^{2} y(x,t)}{\partial x^{2}} = \frac{1}{v^{2}} \frac{\partial^{2} y(x,t)}{\partial t^{2}} \ldotp$
Therefore,
$\frac{1}{v^{2}} = \frac{\mu}{F_{T}} \ldotp$
Solving for $$v$$, we see that the speed of the wave on a string depends on the tension and the linear density
Speed of a Wave on a String Under Tension
The speed of a pulse or wave on a string under tension can be found with the equation
$|v| = \sqrt{\frac{F_{T}}{\mu}} \label{16.8}$
where $$F_T$$ is the tension in the string and $$µ$$ is the mass per length of the string.
Example 16.5: The Wave Speed of a Guitar Spring
On a six-string guitar, the high E string has a linear density of $$\mu_{High\; E}$$ = 3.09 x 10−4 kg/m and the low E string has a linear density of $$\mu_{Low\; E}$$ = 5.78 x 10−3 kg/m. (a) If the high E string is plucked, producing a wave in the string, what is the speed of the wave if the tension of the string is 56.40 N? (b) The linear density of the low E string is approximately 20 times greater than that of the high E string. For waves to travel through the low E string at the same wave speed as the high E, would the tension need to be larger or smaller than the high E string? What would be the approximate tension? (c) Calculate the tension of the low E string needed for the same wave speed.
Strategy
1. The speed of the wave can be found from the linear density and the tension $$v = \sqrt{\frac{F_{T}}{\mu}}$$.
2. From the equation v = $$\sqrt{\frac{F_{T}}{\mu}}$$, if the linear density is increased by a factor of almost 20, the tension would need to be increased by a factor of 20.
3. Knowing the velocity and the linear density, the velocity equation can be solved for the force of tension FT = $$\mu$$v2.
Solution
1. Use the velocity equation to find the speed: $$v = \sqrt{\frac{F_{T}}{\mu}} = \sqrt{\frac{56.40\; N}{3.09 \times 10^{-4}\; kg/m}} = 427.23\; m/s \ldotp$$
2. The tension would need to be increased by a factor of approximately 20. The tension would be slightly less than 1128 N.
3. Use the velocity equation to find the actual tension: $$F_{T} = \mu v^{2} = (5.78 \times 10^{-3}\; kg/m)(427.23\; m/s)^{2} = 1055.00\; N \ldotp$$
4. This solution is within 7% of the approximation.
Significance
The standard notes of the six string (high E, B, G, D, A, low E) are tuned to vibrate at the fundamental frequencies (329.63 Hz, 246.94 Hz, 196.00 Hz, 146.83 Hz, 110.00 Hz, and 82.41 Hz) when plucked. The frequencies depend on the speed of the waves on the string and the wavelength of the waves. The six strings have different linear densities and are “tuned” by changing the tensions in the strings. We will see in Interference of Waves that the wavelength depends on the length of the strings and the boundary conditions. To play notes other than the fundamental notes, the lengths of the strings are changed by pressing down on the strings.
Exercise 16.5
The wave speed of a wave on a string depends on the tension and the linear mass density. If the tension is doubled, what happens to the speed of the waves on the string?
## Speed of Compression Waves in a Fluid
The speed of a wave on a string depends on the square root of the tension divided by the mass per length, the linear density. In general, the speed of a wave through a medium depends on the elastic property of the medium and the inertial property of the medium.
$|v| = \sqrt{\frac{elastic\; property}{inertial\; property}}$
The elastic property describes the tendency of the particles of the medium to return to their initial position when perturbed. The inertial property describes the tendency of the particle to resist changes in velocity.
The speed of a longitudinal wave through a liquid or gas depends on the density of the fluid and the bulk modulus of the fluid,
$v = \sqrt{\frac{\beta}{\rho}} \ldotp \label{16.9}$
Here the bulk modulus is defined as Β = $$− \frac{\Delta P}{\frac{\Delta V}{V_{0}}}$$, where $$\Delta$$P is the change in the pressure and the denominator is the ratio of the change in volume to the initial volume, and $$\rho \equiv \frac{m}{V}$$ is the mass per unit volume. For example, sound is a mechanical wave that travels through a fluid or a solid. The speed of sound in air with an atmospheric pressure of 1.013 x 105 Pa and a temperature of 20°C is vs ≈ 343.00 m/s. Because the density depends on temperature, the speed of sound in air depends on the temperature of the air. This will be discussed in detail in Sound. |
# Percentage-Decimal-Fraction Relations Calculator
Choose the appropriate button below, enter your numbers, and press Calculate.
divided by equals what percent (%) and what decimal?
is percent (%) of what number?
percent (%) of equals what number?
percent (%) equals what number?
Convert percent (%) to a decimal and a fraction.
Convert the decimal to a percentage and a fraction.
Calculate the percentage change from to
What percent of is
% of % is
Convert the decimal to a fraction and a percentage:
## The decimal 88.596 is greater than 1. So we write this as a whole number and fraction:
The whole number is 88
The decimal piece is 88.596 - 88 = 0.596
For our denominator, we take 10 raised to the power of 3 = 1000.
For our numerator, we take 0.596 x 1000 = 596
596 1000
We can reduce 596 and 1000
From our GCF Calculator, we see that the highest number that reduces 596 and 1000 is 4.
Our numbers reduce as follows: 596 ÷ 4 = 149 and 1000 ÷ 4 = 250
The reduced decimal part of our fraction is:
149 250
Combining this with our whole number of 88, our decimal of 88.596 rewritten as a whole number and fraction is
88 & 149/250
Our fraction is not fully reduced. Using our GCF Calculator, we can reduce top and bottom of the fraction by 8
88.596 as a Fraction = 11 125
Now we need to calculate the percentage version of the decimal:
The Percentage version of a decimal is just the decimal multiplied by 100 with a % sign at the end.
Multiplying by 100 is the same as shifting the decimal point 2 places to the right..
Shifting 88.596 one decimal place to the right, we get --> 885.96.
Shifting 885.96 another decimal place to the right and adding a percentage (%) sign, we get --> 8859.6.
## Write this as a ratio:
Ratio = Numerator : Denominator
Ratio = 11 : 125
### How does the Percentage-Decimal-Fraction Relations Calculator work?
Calculates the relational items between a fraction, a decimal (including repeating decimal and terminating decimal), a percentage, and the numerator and denominator piece of that fraction. Also calculates the percentage change going from one number to another or the amount increase or decrease of a percentage above/below a number. Round decimals. decimals into fractions
This calculator has 1 input.
### What 3 formulas are used for the Percentage-Decimal-Fraction Relations Calculator?
1. Percent = Decimal * 100%
2. Decimal = Percent/100
3. Decimal = Numerator/Denominator
For more math formulas, check out our Formula Dossier
### What 6 concepts are covered in the Percentage-Decimal-Fraction Relations Calculator?
decimal
Number in the form a/b where the result is not a whole number
denominator
The bottom portion of a fraction. For a/b, b is the denominator
fraction
how many parts of a certain size exist
a/b where a is the numerator and b is the denominator
numerator
the number above the line in a common fraction
percent
a specified amount in or for every hundred. one part in every hundred.
n% = n/100
percentage-decimal-fraction relations |
Close
## How do you calculate square footage with multiple sides?
How to Calculate the Square Feet of Odd Shapes
1. Measure all the dimensions or sides of the area.
2. Draw the area on graph paper using the measurements you obtained.
3. Divide the drawing into shapes.
4. Figure the area of each shape.
5. Add the areas of all the individual shapes to find the total square footage.
## What is the area in square units of trapezoid ABCD?
To get the area of a trapezoid, you will do the height times the b1 plus b2 divided by 2. In this case, h = 2; b1 = 5; b2 = 20. [2 * (5 + 20)] / 2 = (2 * 25) / 2 = 50 / 2 = 25 square units.
## Which of the following is equivalent to a 1 2h b1 b2?
the formula for the area of a trapezoid is a=1/2h(b1+b2).
## What does a 1 2h b1 b2 mean in math?
| Certified Educator. A = 1/2h(b1+b2) Multiply both sides by 2. 2A = h(b1+b2) Multiply both sides by 1/h.
## What is the formula for 1/2 BH?
So, the area A of a triangle is given by the formula A=12bh where b is the base and h is the height of the triangle. Example: Find the area of the triangle. The area A of a triangle is given by the formula A=12bh where b is the base and h is the height of the triangle.
## What is a ½BH?
The formula for area of a triangle is. A = ½bh where b represents value of base and h represents value of height.
## What formula is a LW?
A formula is an algebraic relationship between two or more variable quantities. For example, A = lw is a formula for the area, A, of a rectangle of length l and width w.
## What is transposing formula?
Transposition is a method to isolate the variable to one side of the equation and everything else to the other side so that you can solve the equation.
## Which of these is an example of a literal equation?
ax-by=k is an example of literal equation.
## What is the relation between the variable in the equation a LW?
The relation between the variables in the equation A=lw is; A varies jointly as lw. Explanation: Variation problems states that involves a simple relationships or formulas, involving one variable being equal to one term.
## What is the equation of variation?
The formula y=kxn y = k x n is used for direct variation. The value k is a nonzero constant greater than zero and is called the constant of variation.
2019-05-18 |
Lesson Objectives
• Demonstrate an understanding of ratios
• Learn how to use the equality test for fractions
• Learn the definition of a proportion
• Learn how to determine if two ratios or two rates represent a proportion
## What is a Proportion
In our last lesson, we learned how to use a ratio to show the relationship between two quantities. In this lesson, we will expand on this topic and learn about proportions. Before we get into the definition of a proportion, we first need to understand how we can determine if two fractions are equal.
### Equality Test for Fractions
The equality test for fractions tells us that two fractions are equal if their cross products are equal. Cross products are formed by multiplying the denominator of one fraction by the numerator of the other. Let's take a look at a few examples.
Example 1: Replace the ? with "=" or "≠"
$$\frac{3}{5} \hspace{.25em} ? \hspace{.25em} \frac{21}{35}$$ Form the cross products: 35 x 3 = 105
5 x 21 = 105
Since the cross products are equal, the two fractions are equal. $$\frac{3}{5} \hspace{.25em} = \hspace{.25em} \frac{21}{35}$$ Example 2: Replace the ? with "=" or "≠"
$$\frac{7}{9} \hspace{.25em} ? \hspace{.25em} \frac{49}{60}$$ Form the cross products: 60 x 7 = 420
9 x 49 = 441
Since the cross products are not equal, the two fractions are not equal. $$\frac{7}{9} \hspace{.25em} ≠ \hspace{.25em} \frac{49}{60}$$
### Definition of a Proportion
When two ratios or two rates are equal, they are called a proportion. Proportions have lots of uses. One great example is to scale up or down a recipe. Let’s suppose a cake recipe calls for 5 cups of sugar and 3 cups of flour. We can write the ratio of flour to sugar as » 3:5. If we wanted to be more descriptive, we could use a fraction and write what each number represents: $$\frac{3 \hspace{.25em} cups \hspace{.25em} flour}{5 \hspace{.25em} cups \hspace{.25em} sugar }$$ Suppose we want to double the recipe. We can do this by doubling the sugar and the flour. Let's multiply each part of the ratio by 2: $$\frac{3}{5} \cdot \frac{2}{2} = \frac{6}{10}$$ Now we know we would use 6 cups of flour and 10 cups of sugar. $$\frac{6 \hspace{.25em} cups \hspace{.25em} flour}{10 \hspace{.25em} cups \hspace{.25em} sugar}$$ The two ratios here » 3:5 and 6:10 are equal and therefore represent a proportion. To determine if two ratios or two rates represent a proportion, we use the equality test for fractions. We only need to work with the number parts. Let's take a look at a few examples:
Example 3: Replace the ? with "=" or "≠" $$\frac{32 \hspace{.2em} shrubs}{50 \hspace{.2em} acres} \hspace{.25em} ? \hspace{.25em} \frac{160 \hspace{.2em} shrubs}{250 \hspace{.2em} acres}$$ We check for proportionality using the equality test for fractions. Let's form the cross products with the number parts only:
50 x 160 = 8000
250 x 32 = 8000
Since the cross products are equal, we have a proportion. $$\frac{32 \hspace{.2em} shrubs}{50 \hspace{.2em} acres} \hspace{.25em} = \hspace{.25em} \frac{160 \hspace{.2em} shrubs}{250 \hspace{.2em} acres}$$ Example 4: Replace the ? with "=" or "≠" $$\frac{15 \hspace{.2em} yen}{370 \hspace{.2em} pesos} \hspace{.25em} ? \hspace{.25em} \frac{160 \hspace{.2em} yen}{2590 \hspace{.2em} pesos}$$ We check for proportionality using the equality test for fractions. Let's form the cross products with the number parts only:
370 x 160 = 59,200
2590 x 15 = 38,850
Since the cross products are not equal, we do not have a proportion. $$\frac{15 \hspace{.2em} yen}{370 \hspace{.2em} pesos} \hspace{.25em} ≠ \hspace{.25em} \frac{160 \hspace{.2em} yen}{2590 \hspace{.2em} pesos}$$ |
# Illustrative Mathematics Unit 6.4, Lesson 9: How Much in Each Group? (Part 2)
Learning Targets:
• I can find the amount in one group in different real-world situations.
Related Pages
Illustrative Math
### Lesson 9: How Much in Each Group? (Part 2)
Let’s practice dividing fractions in different situations.
Illustrative Math Unit 6.4, Lesson 9 (printable worksheets)
### Lesson 9 Summary
The following example shows how to use diagrams and multiplication and division equations to find the amount in one group.
### Lesson 9.1 Number Talk: Greater Than 1 or Less Than 1?
Decide whether each of the following is greater than 1 or less than 1.
1. ½ ÷ ¼
2. 1 ÷ ¾
3. ⅔ ÷ ⅞
4. 2⅞ ÷ 2⅗
### Lesson 9.2 Two Water Containers
1. After looking at these pictures, Lin says, “I see the fraction ⅖.” Jada says, “I see the fraction ¾.” What quantities are Lin and Jada referring to?
2. How many liters of water fit in the water dispenser?
Write a multiplication equation and a division equation for the question, then find the answer. Draw a diagram, if needed. Check your answer using the multiplication equation.
#### Lesson 9.3 Amount in One Group
Write a multiplication equation and a division equation and draw a diagram to represent each situation and question. Then find the answer. Explain your reasoning.
1. Jada bought 3½ yards of fabric for \$21. How much did each yard cost?
2. 4/9 kilogram of baking soda costs \$2. How much does 1 kilogram of baking soda cost?
3. Diego can fill 1⅕ bottles with 3 liters of water. How many liters of water fill 1 bottle?
4. 5/4 gallons of water fill 5/6 of a bucket. How many gallons of water fill the entire bucket?
#### Are you ready for more?
The largest sandwich ever made weighed 5,440 pounds. If everyone on Earth shares the sandwich equally, how much would you get? What fraction of a regular sandwich does this represent?
#### Lesson 9 Practice Problems
1. A group of friends is sharing 3½ pounds of berries.
a. If each friend received 5/4 of a pound of berries, how many friends are sharing the berries?
b. If 5 friends are sharing the berries, how many pounds of berries does each friend receive?
b. Priya has picked 1½ cups of raspberries, which is enough for ¾ of a cake. How many cups does she need for the whole cake?
2. ⅔ kilogram of soil fills ⅓ of a container. Can 1 kilogram of soil fit in the container? Explain or show your reasoning.
3. After raining for ¾ of an hour, a rain gauge is ⅖ filled. If it continues to rain at that rate for 15 more minutes, what fraction of the rain gauge will be filled?
a. To help answer this question, Diego wrote the division equation ¾ ÷ ⅕. Explain why this equation does not represent the situation.
b. Write a multiplication equation and a division equation that does represent the situation.
4. 3 tickets to the museum cost \$12.75. At this rate, what is the cost of:
a. 1 ticket?
b. 5 tickets?
5. Elena went 60 meters in 15 seconds. Noah went 50 meters in 10 seconds. Elena and Noah both moved at a constant speed.
a. How far did Elena go in 1 second?
b. How far did Noah go in 1 second?
c. Who went faster? Explain or show your reasoning.
6. The first row in the table shows a recipe for 1 batch of trail mix. Complete the remaining rows with recipes for 2, 3, and 4 batches of the same type of trail mix.
number of batches cups of cereal cups of almonds cups of raisins 1 2 3 4
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. |
# What is the inverse equation of the function, y= 3x-6?
Nov 20, 2016
${f}^{-} 1 \left(x\right) = \frac{1}{3} x + 2$
## Concept
${f}^{- 1} \left(x\right)$ (inverse of $f \left(x\right)$) means to switch the $x$ and $y$ variables in the equation $f \left(x\right)$
This means ${f}^{- 1} \left(x\right) = f \left(y\right)$
## Solving
• Given function $f \left(x\right) = 3 x - 6$, then ${f}^{- 1} \left(x\right) \setminus \Rightarrow \setminus \textcolor{\in \mathrm{di} a n red}{x = 3 y - 6}$
• We must isolate the $y$ in ${f}^{- 1} \left(x\right)$ to get the proper form of the inverse.
$x = 3 y - 6$
$x + 6 = 3 y$
$\frac{x}{\setminus} \textcolor{red}{3} + \frac{6}{\setminus} \textcolor{red}{3} = \frac{\setminus \cancel{3} y}{\setminus} \cancel{\setminus \textcolor{red}{3}}$
$y = \frac{1}{3} x + 2$
${f}^{-} 1 \left(x\right) = \frac{1}{3} x + 2$ |
# Constructing the Circumcircle of a Triangle
Compass and straight edge constructions are of interest to mathematicians, not only in the field of geometry, but also in algebra. For thousands of years, beginning with the Ancient Babylonians, mathematicians were interested in the problem of "squaring the circle" (drawing a square with the same area as a circle) using a straight edge and compass. This problem is equivalent to finding the area of a circle. It turns out that this is impossible, but no-one managed to prove this until 1882!
However, it is possible to construct the circumcircle of a triangle using only a straight edge and compass. For this construction, you will need a straight edge (ruler - but you won't be measuring anything), a pair of compasses, a pencil and paper. I have drawn the pictures using the robocompass app. It's fun to play with, and you can use it to do all sorts of geometric constructions. There's a little bit of coding to learn, but a list of instructions is provided. Once you've written your little program, you can invite a few friends over to watch the construction. Be warned that, when you are using it to draw arcs, the robocompass compass point may appear to be a little away from the centre, but the drawing is actually accurate.
Note: these instructions assume that you are familiar with constructing the perpendicular bisector of a straight line. We are going to do that twice. If your memories of this construction are a bit hazy, it's probably a good idea to re-read the article on finding the perpendicular bisector of a line segment before trying to do this construction.
## The Construction
Step 1: Start with the triangle $ABC$ you want your circle to pass through.
Step 2: Construct the perpendicular bisector of side $AB$. Make it nice and long so that it will intersect the line segment drawn in the next step.
Step 3: Construct the perpendicular bisector of side $BC$. Make it long enough the intersect the perpendicular bisector drawn in step 2. Label the point where the two perpendicular bisectors intersect with a $J$. This will be the centre of our circle. Note that this point of intersection may lie either inside or outside of triangle $ABC$.
Step 4: Place the tip of the compasses at the point $J$. Open them out so that the pencil touches one of points $A$, $B$ or $C$. It doesn't matter which: they're all the same distance away from point $J$. Draw a circle of this radius.
Notice how it passes through all three vertices of the triangle?
The completed construction is shown on the right. Sit back and admire your art work. Tell any one who will listen that it should be in the Louvre!
What's that, Sam? Your circle doesn't pass through all three vertices? That's probably because your pencil is blunt and you didn't use a ruler to draw your line segments.
### Description
This tutorial will expose you to Rulers and how to use them. We will also have a close look as set square and compass constructions. Your feedback is important to us, if you like any other topic covered under this tutorial, please do let us know.
### Audience
Year 10 or higher
### Learning Objectives
Ruler and compass constructions and more
Author: Subject Coach
You must be logged in as Student to ask a Question. |
## Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 8 Statistics and Probability Ex 8.3
10th Maths Exercise 8.3 Samacheer Kalvi Question 1.
Write the sample space for tossing three coins using tree diagram.
Solution:
Ex 8.3 Class 10 Samacheer Question 2.
Write the sample space for selecting two balls from a bag containing 6 balls numbered 1 to 6 (using tree diagram).
Solution:
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
Exercise 8.3 Class 10 Samacheer Question 3.
If A is an event of a random experiment such that P(A) : P($$\overline{\mathbf{A}}$$) =17 : 15 and n(S) = 640 then find
(i) P($$\overline{\mathbf{A}}$$)
(ii) n(A).
Solution:
P(A): P($$\overline{\mathbf{A}}$$) = 17 : 15
10th Maths 8.3 Solutions Question 4.
A coin is tossed thrice. What is the probability of getting two consecutive tails?
Solution:
Outcomes {O}: {(HHH), (THH), (HTH), (HHT), (HTT), (THT), (TTH), (TTT)}
Two consecutive tails {F} : {(HTT), (TTH), (TTT)}
n{F} = 3
n{O} = 8
10th Maths Exercise 8.3 Question 5.
At a fete, cards bearing numbers 1 to 1000, one number on one card are put in a box. Each player selects one card at random and that card is not replaced. If the selected card has a perfect square number greater than 500, the player wins a prize. What is the probability that
(i) the first player wins a prize
(ii) the second player wins a prize if the first has won?
Sample space = {1, 2, 3,… ,1000}
n(S) = 1000
(i) Let A be the event of setting square number greater than 500
A = {529, 576, 625, 676, 729, 784, 841, 900, 961}
n(A) = 9
P(A) = $$\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{9}{1000}$$
The probability that the first player wins prize = $$\frac{9}{1000}$$
(ii) If the first player wins, the number is excluded for the second player.
n(A) = 8 and n(S) = 999
P(A) = $$\frac{n(A)}{n(S)}=\frac{8}{999}$$
Probability the second player wins a prize = $$\frac{8}{999}$$
10th Maths Probability Exercise 8.3 Question 6.
A bag contains 12 blue balls and x red balls. If one ball is drawn at random
(i) what is the probability that it will be a red ball?
(ii) If 8 more red balls are put in the bag, and if the probability of drawing a red ball will be twice that of the probability in (i), then find x.
Solution:
12 ➝ blue balls
x ➝ red balls
(i) P (red ball) = $$\frac{x}{x+12}$$
(ii) 8 red balls are added to the bag.
∴ 12 ➝ blue balls
x + 8 ➝ red balls
Given that P(ii) = 2 × P(i)
⇒ (x + 8)(x + 12) = 2x(x + 20)
⇒ (x2 + 20x + 96) = 2x2 + 40x
⇒ x2 + 20x – 96 = 0
⇒ x2 + 24x – 4x – 96 = 0
⇒ x(x + 24) – 4(x + 24) = 0
⇒ (x – 4)(x + 24) = 0
∴ x = 4 (or) x = -24
x cannot be negative ⇒ x = 4
Substituting x = 4 in (i),
10th Maths Exercise 8.3samacheer Kalvi Question 7.
Two unbiased dice are rolled once. Find the probability of getting
(i) a doublet (equal numbers on both dice)
(ii) the product as a prime number
(iii) the sum as a prime number
(iv) the sum as 1
Solution:
Doublet = {(1, 1) (2, 2) (3, 3) (4, 4) (5, 5) (6,6)}
Total number of outcomes = 6 × 6
n(S) = 36
Number of favourable outcomes = 6
(ii) Number of favourable outcomes = 6
as favourable outcomes = (1, 2), (2, 1), (1, 3), (3, 1),(1, 5),and (5, 1)
(iii) Sum as prime numbers = {(1, 1), (1, 2), (2, 3), (1, 4), (1, 6), (4, 3), (5, 6)}
Number of favourable outcomes = 7
⇒ Probability = $$\frac{7}{36}$$
(iv) With two dice, minimum sum possible = 2
∴ Prob (sum as 1) = 0 [Impossible event]
10th Maths 8.3 Question 8.
Three fair coins are tossed together. Find the probability of getting
(ii) atleast one tail
(iv) atmost two tails
Three fair coins are tossed together
Sample spade = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
n(S) = 8
(i) Let A be the event of getting all heads
A = {HHH}
n(A) = 1
$$P(A)=\frac{n(A)}{n(S)}=\frac{1}{8}$$
(ii) Let B be the event of getting atleast one tail.
B = {HHT, HTH, HTT, THH, THT, TTH, TTT}
n(B) = 7
$$P(B)=\frac{n(B)}{n(S)}=\frac{7}{8}$$
(iii) Let C be the event of getting atmost one head
C = {HTT, THT, TTH, TTT}
n(C) = 4
$$P(C)=\frac{n(C)}{n(S)}=\frac{4}{8}=\frac{1}{2}$$
(iv) Let D be the event of getting atmost two tails.
D = {HTT, TTT, TTH, THT, THH, HHT, HTH}
n(D) = 7
$$P(D)=\frac{n(D)}{n(S)}=\frac{7}{8}$$
10 Maths Exercise 8.3 Question 9.
Two dice are numbered 1,2,3,4,5,6 and 1,1,2,2,3,3 respectively. They are rolled and the sum of the numbers on them is noted. Find the probability of getting each sum from 2 to 9 separately.
Solution:
Dice 1
S = {1,2, 3, 4, 5, 6}
Dice 2
S = {1,1,2, 2, 3, 3}
Total possible outcomes when they are rolled
n(S) = 36
Event of sum (2) = A = {(1,1), (1,1)},
n(A) = 2,P(A) = $$\frac{2}{36}$$
Event of sum 3 is B = {(1, 2), (1, 2), (2, 1), (2, 1)}
Event of sum 4 is C= {(1, 3), (1, 3), (2, 2), (2, 2), (3, 1) (3, 1)}
n(C) = 6
Event of getting the sum 5 is
D = {(2, 3), (2, 3), (3, 2), (3, 2), (4, 1), (4, 1)}
n(D) = 6, P(D) = $$\frac{6}{36}$$ .
Event of getting the sum 6 is
E = {(3, 3), (3, 3), (4, 2), (4, 2), (5, 1), (5, 1)}
n(E) = 6, P(E) = $$\frac{6}{36}$$
Event of getting the sum 7 is
F = {(4, 3), (4, 3), (5, 2), (5, 2), (6, 1), (6, 1)}
n(F) = 6
P(F) = $$\frac{6}{36}$$
Event of getting the sum 8 is
G = {(5, 3), (5, 3), (6, 2), (6, 2)}
Event of getting the sum 9 is
H = {(6, 3), (6, 3), n(H) = 2
10th Standard Maths Exercise 8.3 Question 10.
A bag contains 5 red balls, 6 white balls, 7 green balls, 8 black balls. One ball is drawn at random from the bag. Find the probability that the ball is drawn
(i) white
(ii) black or red
(iii) not white
(iv) neither white nor black
Solution:
5 red 6 white 7 green 8 black total no. of balls = 5 + 6 + 7 +8= 26
Exercise 8.3 Class 10 Samacheer Kalvi Question 11.
In a box there are 20 non-defective and some defective bulbs. If the probability that a bulb selected at random from the box found to be defective is 3/8 then, find the number of defective bulbs.
Solution:
Let number of defective bulbs be ‘x’
Total number of bulbs = x + 20
⇒ 8x = 3x + 60
⇒ 5x = 60
⇒ x = 12
∴ No.of defective bulbs are = 12.
10th Maths Exercise 8.4 Samacheer Kalvi Question 12.
The king and queen of diamonds, queen and jack of hearts, jack and king of spades are removed from a deck of 52 playing cards and then well shuffled. Now one card is drawn at random from the remaining cards. Determine the probability that the card is
(i) a clavor
(ii) a queen of red card
(iii) a king of black card
Solution:
(i.e) remaining number of cards = 52 – 6 = 46 13
(i) P(a clavor) = $$\frac{13}{46}$$
(ii) P(queen of red card) = 0 as both Queen of diamond and heart have been removed.
(iii) only K of clavor is in the deck
⇒ P(king of black card) = $$\frac{1}{46}$$
10th Maths Ex 8.3 Question 13.
Some boys are playing a game, in which the stone was thrown by them landing in a circular region (given in the figure) is considered as a win and landing other than the circular region is considered as a loss. What’is the probability to win the game?
Solution:
Samacheer Kalvi 10th Maths Exercise 8.3 Question 14.
Two customers Priya and Amuthan are visiting a particular shop in the same week (Monday to Saturday). Each is equally likely to visit the shop on any one day as on another day. What is the probability that both will visit the shop on
(i) the same day
(ii) different days
(iii) consecutive days?
Solution:
10th Maths Exercise 8.3 4th Sum Question 15.
In a game, the entry fee is ₹ 150. The game consists of tossing a coin 3 times. Dhana bought a ticket for entry. If one or two heads show, she gets her entry fee back. If she throws 3 heads, she receives double the entry fees. Otherwise, she will lose. Find the probability that she
(i) gets double entry fee
(ii) just gets her entry fee
(iii) loses the entry fee.
Solution: |
Question Video: Understanding Properties of Polygons | Nagwa Question Video: Understanding Properties of Polygons | Nagwa
# Question Video: Understanding Properties of Polygons Mathematics • Second Year of Primary School
Determine whether the following statement is always, sometimes, or never true: A polygon has more vertices than sides.
03:39
### Video Transcript
Determine whether the following statement is always, sometimes, or never true. A polygon has more vertices than sides.
Let’s identify some words we need to know. A vertex, or vertices if there’s more than one, is a place where two sides meet in a flat shape. A side of a polygon is one of the line segments that make up a flat shape. Let’s look at some polygons to try and determine if this statement is true.
Here are five different polygons. Let’s count their vertices and sides and then compare those numbers. We’ll start with the square. The square has one, two, three, four sides. And it has one, two, three, four vertices.
Next up, the rectangle: one, two, three, four sides and the vertices one, two, three, four. Let’s go back and read our statement. Our statement says a polygon has more vertices than sides. In our square and in our rectangle, they both have the same number of sides and vertices. So we cross out the word “always.”
A polygon has more vertices than sides: it can’t be always true, because we’ve already found two examples where it’s not true. On to our last few polygons, this triangle has one, two, three sides and one, two, three vertices, three sides three vertices. And the pentagon one, two, three, four, five sides and one, two, three, four, five vertices.
Are we starting to see a pattern here? We’ll go ahead and check our hexagon and then make a decision about the statement. The hexagon has one, two, three, four, five, six sides and one, two, three, four, five, six vertices.
In these six examples, each polygon has the same number of sides and vertices. They do not have more vertices than sides. In fact, it is never true that a polygon will have more vertices than sides. A polygon always has the same number of vertices and sides. The statement “a polygon has more vertices than sides” is never true.
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Curvature at a Point on a Single Variable Real Valued Function
# Curvature at a Point on a Single Variable Real Valued Function
We first learned about curves in the form of a single-variable function $y = f(x)$ in $\mathbb{R}^2$, which takes every $x \in D(f)$ and maps it to an element $f(x) \in R(f) \subseteq C(f)$. We will now look at a theorem which will tell us the curvature at a point on a curve of this form, provided that the curve is twice differentiable.
Theorem 1: Let $y = f(x)$ be a real-valued twice-differentiable function that traces the curve $C$. Then the curvature at the point $x \in D(f)$ is given by the formula: $\kappa (x) = \frac{\mid f''(x) \mid}{(1 + (f'(x))^2)^{3/2}}$.
• Proof: Let $y = f(x)$ be a twice differentiable function. Then this function can rewritten as the set of parametric equations $\left\{\begin{matrix} x = x \\ y = f(x) \\ z = 0 \end{matrix}\right.$.
• So we have that $\vec{r'}(x) = (1, f'(x), 0)$ and $\vec{r''}(x) = (0, f''(x), 0)$. We now want to compute the cross product $\vec{r'}(x) \times \vec{r''}(x)$ as follows:
(1)
\begin{align} \quad \quad \vec{r'}(x) \times \vec{r''}(x) = [(f'(x))(0) - (0)(f''(x))]\vec{i} - [(1)(0) + (0)(0)] \vec{j} + [(1)(f''(x)) - 0(f'(x))] \vec{k} \\ \quad \quad \vec{r'}(x) \times \vec{r''}(x) = (0, 0, f''(x)) \end{align}
• Therefore $\| \vec{r'}(x) \times \vec{r''}(x) \| = \sqrt{(f''(x))^2} = \mid f''(x) \mid$. So we have computed the numerator to our above equation and so now we need to compute the denominator.
• We have that $\| \vec{r'}(x) \| = \sqrt{1 + (f'(x))^2}$, and so $\| \vec{r'}(x) \|^3 = (1 + (f'(x))^2)^{3/2}$. Putting this all together we see that:
(2)
\begin{align} \quad \kappa = \frac{ \| \vec{r'}(x) \times \vec{r''}(x) \|}{\| \vec{r'}(x) \|^3} = \frac{\mid f''(x) \mid}{(1 + (f'(x))^2)^{3/2}} \quad \blacksquare \end{align}
Let's look at some examples applying theorem 1.
## Example 1
Let $f(x) = 4x^2 - 2x + 1$. Compute the curvature of $f$ at $x = 2$.
Let's first compute the first and second derivatives of $f$. $f'(x) = 8x - 2$, and $f''(x) = 8$. Therefore we have that:
(3)
\begin{align} \kappa (x) = \frac{\mid 8 \mid}{(1 + (8x - 2)^2)^{3/2}} \end{align}
Plugging in $x = 2$ and:
(4)
\begin{align} \kappa (2) = \frac{8}{(197)^{3/2}} \end{align} |
# 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
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 2
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
$100x+100x+100x+...+100x=100$
$100x\times100=100$
$100x=1$
$x=0.01=\boxed{'''(C) 0.01'''}$
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 |
# Maths - Vector Calculus
In order to understand calculus on vectors we need to understand Scalar and Vector Fields.
Fist, on this page, we will look at differentiation with respect to some factor that is independant of space such as time (in classical mechanics) then we go on to look at calculus with respect to the dimensions of the space itself such as grad, div and curl.
## Vector Differentiation with respect to a scalar
Since division of one vector by another is not generally valid we can't define differentiation with respect to another vector. Differentiation with respect to a scalar is defined as follows, if:
f(x) = [a , b , c , e]
then:
d f(x) / dx = [d(a /dx) , d(b/dx) , d(c/dx) , d(e/dx)]
In other words to differentiate with respect to a scalar, we just differentiate the elements individually. So to give a more specific example if:
f(x) = [xn , sin(x) , tan(x) , ex ]
then:
d f(x) / dx = [n*xn-1 , cos(x) , sec2(x) , ex ]
So this is quite simple, provided that we can differentiate the elements of a vector, we can differentiate the whole quaternion.
Although there is not a general vector division there are specific cases which allow us to differentiate one vector with respect to another and these correspond to specific types of vector differentiation known as grad, div and curl. We will discuss these after we have reviewed scalar and vector fields.
## Scalar Field
We could represent how some scalar quantity, for example temperature, is defined for each point in the space.
We could define a function to show this, temp = f(x,y,z)
One way to illustrate this is to join up all the points with the same values. This is known as contour lines, or isothermals, or isobars, or whatever the name is that is appropriate to the scalar quantity.
## Vector Field
A vector field gives the value of a vector,for example velocity, for every position in the 2D or 3D space.
For example the following vector field might represent a velocity field for a rotating object. |
# Determine whether a) (2,13) and b) (-1,-3) lie on the graph of y = 10x - 7 . c) Sketch the...
## Question:
Determine whether
a) (2,13) and
b) (-1,-3)
lie on the graph of
{eq}y = 10x - 7 {/eq}.
c) Sketch the graph by finding {eq}x- {/eq} and {eq}y- {/eq}intercepts.
## Lines and Points on the Cartesian Plane:
From geometry, two points uniquely determine a line. This means that if we have two points, it is equivalent to knowing the line that contains them. This is why we can draw the graph of an equation for a line just by determining two of the points on it. We can determine if a point is contained on a line using the equation for the line: if the coordinates satisfy the equation, that is, if the resulting equation is true when we substitute their values, then the line contains the point.
To determine if a point is on the graph of any curve, we can check and see if the coordinates of the point satisfy the given equation of the curve. To do this, we substitute the values into their respective variables and see if we end up on a true statement. If this happens, the point satisfies the equation and therefore lies on the curve.
(a) For {eq}(2, 13) {/eq}, we sub in {eq}x = 2 {/eq} and {eq}y = 13 {/eq} to the equation and get \begin{align*} y &= 10x - 7 \\ 13 &= 10(2) - 7 \\ 13 &= 20 - 7 \\ 13 &= 13 \end{align*} and since this is true, the point {eq}(2, 13) {/eq} is on the line.
(b) We do the same for the point {eq}(-1, -3) {/eq}. We have \begin{align*} y &= 10x - 7 \\ -3 &= 10(-1) - 7 \\ -3 &= -10 - 7 \\ -3 &= -17 \end{align*} and since this is false, the point does not satisfy the equation and so the point does not lie on the line.
(c) To find the intercepts of the line, we set each of {eq}x {/eq} and {eq}y {/eq} to zero for the {eq}y {/eq}- and {eq}x {/eq}-intercepts, respectively. We have \begin{align*} x = 0: y &= 10(0) - 7 \\ y &= 3 \\ y = 0: 0 &= 10x - 7 \\ 10x &= 7 \\ x &= \frac{7}{10} \end{align*} so the intercepts are at {eq}(0, -7) {/eq} and {eq}\left( \frac{7}{10}, 0 \right) {/eq}. To sketch the graph of the line, we just have to plot these two points and draw the line connecting them: |
Subtracting Vectors Using Components
This post categorized under Vector and posted on December 3rd, 2018.
Introduction In this lesson vectors and their basic components will be defined and quantified. For this lesson we will concentrate on 2-dimensional vectors. The Lesson A vector is a quanvectory which has both magnitude and direction. For example velocity is a vector because it is both the speed at which an object is traveling and the direction. Speed is not a vector (it is called a scalar In mathematics physics and engineering a Euclidean vector (sometimes called a geometric or spatial vector oras heresimply a vector) is a geometric object that has magnitude (or vectorgth) and direction.Vectors can be added to other vectors according to vector algebra.A Euclidean vector is frequently represented by a line segment with a definite direction or graphically as an arrow When we write vectors using mathematical notation there are two different forms. If the magnitude and direction are known for a vector we write the name of the vector using a bold letter
Introduction to vectors . A vector is a quanvectory that has both a magnitude (or size) and a direction. Both of these properties must be given in order to specify a vector completely.Both addition and subtraction with vectors follows the same algebraic rules as with real numbers. The commutative rule u v v u The vectorociative rule u (v w) (u v) w The fancy terminology and symbols aside this is really quite a simple concept.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.
Vectors. Components. Vector addition and subtraction. Scalar product and vector product (dot product and cross product). Displacement velocity and acceleration. Physclips provides multimedia education in introductory physics (mechanics) at different levels. Modules may be used by teachers while students may use the whole package for self instruction or for reference.In mathematics orthogonality is the generalization of the notion of perpendicularity to the linear algebra of bilinear forms.Two elements u and v of a vector vectore with bilinear form B are orthogonal when B(u v) 0.Depending on the bilinear form the vector vectore may contain nonzero self-orthogonal vectors. In the case of function vectores families of orthogonal functions are used to form a In each of the above equations the vertical acceleration of a projectile is known to be -9.8 mss (the acceleration of gravity). As discussed earlier in Lesson 2 the v ix and v iy values in each of the above sets of kinematic equations can be determined by the use of trigonometric functions. The initial x-velocity (v ix) can be found using the equation v ix v i cosine(Theta) where
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For a real number $x$ the condition $|3x – 20| + |3x – 40| = 20$ necessarily holds if
1. $9 < x < 14$
2. $6 < x < 11$
3. $7 < x < 12$
4. $10 < x < 15$
Given that, $|3x-20| + |3x-40| = 20 ; x \in \mathbb{R} \quad \longrightarrow (1)$
We know that $,|x| = \left\{\begin{matrix} x\;;&x\geq 0 \\ -x\;; &x<0 \end{matrix}\right.$
We can open mod as positive and negative. There are four such cases.
$\textbf{Case 1:}\;\text{ Positive, Positive}$
$\Rightarrow 3x – 20 + 3x – 40 = 20$
$\Rightarrow 6x – 60 = 20$
$\Rightarrow 6x = 80$
$\Rightarrow \boxed{x = \frac{40}{3} = 13.33}$
$\textbf{Case 2:}\;\text{ Positive, Negative}$
$\Rightarrow 3x – 20 – (3x – 40) = 20$
$\Rightarrow 3x – 20 – 3x + 40 = 20$
$\Rightarrow \boxed{20 = 20\; {\color{Green} {\text{(True)}}}}$
$\textbf{Case 3:}\;\text{ Negative, Positive}$
$\Rightarrow \;– (3x – 20) + 3x – 40 = 20$
$\Rightarrow\; – 3x + 20 + 3x – 40 = 20$
$\Rightarrow \boxed{- 20 = 20\;\color{Red}{\text{(False)}}}$
$\textbf{Case 4:}\;\text{ Negative, Negative}$
$\Rightarrow \;– (3x – 20) – (3x – 40) = 20$
$\Rightarrow \;– 3x + 20 – 3x + 40 = 20$
$\Rightarrow \;– 6x =\; – 40$
$\Rightarrow \boxed{x = \frac{20}{3} = 6.66}$
$\therefore$ $\boxed{7 < x < 12}$
Correct Answer $: \text{C}$
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1
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The second derivative is the rate of change of the rate of change of a point at a graph (the “slope of the slope” if you will). This can be used to find the acceleration of an object (velocity is given by first derivative).
## What is the second derivative used for?
The second derivative is the rate of change of the rate of change of a point at a graph (the “slope of the slope” if you will). This can be used to find the acceleration of an object (velocity is given by first derivative).
## What happens when a double derivative is zero?
Since the second derivative is zero, the function is neither concave up nor concave down at x = 0. It could be still be a local maximum or a local minimum and it even could be an inflection point. Let’s test to see if it is an inflection point.
## Where is the derivative equal to zero?
It is because the derivative is a measure of the slope of the tangent line to the curve at the point, and the slope is equal to zero only when the tangent line is horizontal, at the top of a hill or the bottom of a valley, so to speak.
## What is the difference between relative maximum and absolute maximum?
A relative max/min point is a point higher or lower than the points on both of its sides while a global max/min point is a point that is highest or lowest point in the graph. In other words, there can be multiple relative max/min points while there can only be one global/absolute max/min point.
## What is first derivative used for?
The first derivative of a function is an expression which tells us the slope of a tangent line to the curve at any instant. Because of this definition, the first derivative of a function tells us much about the function. If is positive, then must be increasing.
## How do you find the maximum and minimum of a graph?
The y- coordinates (output) at the highest and lowest points are called the absolute maximum and absolute minimum, respectively. To locate absolute maxima and minima from a graph, we need to observe the graph to determine where the graph attains it highest and lowest points on the domain of the function.
## How do you find the relative maximum and minimum of a function?
Find the first derivative of a function f(x) and find the critical numbers. Then, find the second derivative of a function f(x) and put the critical numbers. If the value is negative, the function has relative maxima at that point, if the value is positive, the function has relative maxima at that point.
## What is relative extrema of a function?
5.5. 1 Relative Extrema. ¶ 🔗 A relative maximum point on a function is a point (x,y) on the graph of the function whose y -coordinate is larger than all other y -coordinates on the graph at points “close to” (x,y).
two
## How do you identify extrema?
Finding Absolute Extrema of f(x) on [a,b]
1. Verify that the function is continuous on the interval [a,b] .
2. Find all critical points of f(x) that are in the interval [a,b] .
3. Evaluate the function at the critical points found in step 1 and the end points.
4. Identify the absolute extrema.
## What Extrema means?
Extremum, plural Extrema, in calculus, any point at which the value of a function is largest (a maximum) or smallest (a minimum). There are both absolute and relative (or local) maxima and minima.
## What is a relative function?
Function achieves relative maximum or relative minimum (relative extrema) at points, at which it changes from increasing to decreasing, or vice versa. DEFINITION OF RELATIVE EXTREMA. Let f(x) be a function of x .
## What is the maximum and minimum of a parabola called?
The vertex of a parabola is the highest or lowest point, also known as the maximum or minimum of a parabola. |
## Finding the Domain of a Function Step by Step
What is the domain of a function? In what is a function we saw that we can think of a function as a rule. This rule, given an input, gives an output.
You could also say that it is a rule that "maps" the input to the output. In calculus, we often use this notation:
This means that "y" is a function of "x". Here, x is the input, often called the independent variable. And y is the output, and it is called the dependent variable.
So, back to the domain of a function. What is it? It is just the set of values that x can take. You can think of the domain as a bag. This bag contains all the x's you can choose as input for the function.
The domain of a function can be defined explicitly or implicitly, but it is always defined.
### Example 1
As an example of a domain defined explicitly, let's say I give you the expression:
Here I tell you that x must be greater than 0. You can't choose any x. In the bag you only have positive x's.
On the other hand, if I simply tell you that:
This function has an implicitly defined domain. I don't specify the valid values of x. So, it is implicit that the domain is the set of all real numbers.
### Example 2
A more interesting example of an implicitly defined domain is the function:
At first glance you may think this is the same as the previous case. However, what would happen if x=2? We'll get
And 1/0 doesn't make sense. (if you're not completely sure why division by zero doesn't make sense, here's a simple explanation).
Because f(2) doesn't make sense, we take the 2 out of the bag, and the domain is the set of all real numbers that are not 2.
More examples...
### Example 3
Let's consider the function:
In the real numbers, the squareroot of a number is defined only for positive numbers. The squareroots of negative numbers do exist, but we won't consider them here.
So, our function is only defined when there is a positive number inside (or zero!) the square root sign. That means that x-3 must be positive: |
### Mathematics Class XI
Unit-I: Sets and Functions
Chapter 1: Sets
Unit-II: Algebra
Chapter 5: Binomial Theorem
Unit-III: Coordinate Geometry
Chapter 1: Straight Lines
Chapter 2: Conic Sections
Unit-IV: Calculus
Unit-V: Mathematical Reasoning
Unit-VI: Statistics and Probability
Chapter 1: Statistics
Chapter 2: Probability
# Some various forms of the equation of a line
Lesson Progress
0% Complete
(i) If a line is at a distance and parallel to -axis, then the equation of the line is .
(ii) If a line is parallel to -axis at a distance from -axis, the the equation of the lines is
The position of points is relative to a given line:
Let and two given points .
(i) The two points are on the same side of the straight line if, and have the same sign.
(ii) The two points are on the opposite sides of the line , if and have opposite sign.
Equation of bisectors of the angle between to intersecting line:
Equation of bisectors and are
Bisectors of the angle containing a given point (h , k) its interior:
Let the equation of two-line and be given by
To obtain the equation of the bisector of the angle containing in its interior, proceed as follows:
(1) First see if and have the same or opposite signs.
(i) If they have the same sign, write
(ii) If they have the opposite sign, write
(2) Write the bisector of the angle containing in its interiro as:
Note:
Bisector of the angle, not containing in its interior, is given by ,
Provided and have the same sign.
Pair of Straight Lines:
Consider the equation which is a second degree homogenous equation in and , can be written as which is quadratic in . This gives or which represents a pair of straight lines, with slopes
Change of Axes (Shifting of Origin)
Let the origin is shifted to a point . If are coordinates of a point referred to old axes and are the coordinates of the same points referred to new axes, then
Images of a point with respect to a line:
Let the image of a point with respect to be , then
(i) The image of the point wrto -axis is
(ii) The image wrto -axis is .
(iii) The image of the point wrto mirror is .
(iv) The image of wrto the line mirror is
(v) The image of wrto origin is
(vi) The length of perpendicular from a point to a line is
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# Roots and powers Chapter 4.
## Presentation on theme: "Roots and powers Chapter 4."— Presentation transcript:
Roots and powers Chapter 4
4.1 – Estimating roots Chapter 4
Pg. 206, #1–6 Independent Practice
4.2 – irrational numbers Chapter 4
Rational and irrational numbers
Irrational numbers An irrational number cannot be written in the form m/n, where m and n are integers and n ≠ 0. The decimal representation of an irrational number neither terminates nor repeats. When an irrational number is written as a radical, the radical is the exact value of the irrational number. approximate values exact value
example Tell whether each number is rational or irrational. Explain how you know. a) b) c) –3/5 is rational, because it’s written as a fraction. In its decimal form it’s –0.6, which terminates. b) is irrational since 14 is not a perfect square. The decimal form is … which neither repeats nor terminates. c) is rational because both 8 and are perfect cubes. Its decimal form is … which is a repeating decimal.
The number system Together, the rational numbers and irrational numbers for the set of real numbers. Real numbers Rational numbers Integers Irrational numbers Whole numbers Natural Numbers
example Use a number line to order these numbers from least to greatest.
Pg , #4, 7, 8, 12, 15, 18, 2o Independent Practice
4.3 – Mixed and entire radicals
Chapter 4
Draw the following triangles on the graph paper that has been distributed, and label the sides of the hypotenuses. 1 cm 4 cm 1 cm 3 cm 3 cm 2 cm 4 cm Draw a 5 by 5 triangle. What are the two ways to write the length of the hypotenuse? 2 cm
Why? We can split a square root into its factors. The same rule applies to cube roots. Why?
where n is a natural number, and a and b are real numbers. We can use this rule to simplify radicals:
example Simplify each radical. a) b) c)
example Write each radical in simplest form, if possible. a) b) c)
Try simplifying these three:
example Write each mixed radical as an entire radical. a) b) c)
Try it:
P , #4, 5, 10 and 11(a,c,e,g,i), 14, 19, 24 Independent practice
4.4 – fractional exponents and radicals
Chapter 4
Fill out the chart using your calculator.
Fractional exponents Fill out the chart using your calculator. What do you think it means when a power has an exponent of ½? What do you think it means when a power has an exponent of 1/3? Recall the exponent law: When n is a natural number and x is a rational number:
example Evaluate each power without using a calculator. a) b) c) d)
Try it:
Powers with rational exponents
When m and n are natural numbers, and x is a rational number, Write in radical form in 2 ways. Write and in exponent form.
example Evaluate: a) b) c) d)
example Biologists use the formula b = 0.01m2/3 to estimate the brain mass, b kilograms, of a mammal with body mass m kilograms. Estimate the brain mass of each animal. A husky with a body mass of 27 kg. A polar bear with a body mass of 200 kg.
Pg , #3, 5, 10, 11, 12, 17, 20. Independent practice
4.5 – negative exponents and reciprocals
Chapter 4
challenge Factor: 5x2 + 41x – 36
Hint: try using fractions.
consider This rectangle has an area of 1 square foot. List 5 possible pairs of lengths and widths for this rectangle. (Remember, they will need to have a product of 1). Hint: try using fractions.
What is the rule for any number to the power of 0? Ex: 70?
reciprocals Two numbers with a product of 1 are reciprocals. So, what is the reciprocal of ? So, 4 and ¼ are reciprocals! What is the rule for any number to the power of 0? Ex: 70? If we have two powers with the same base, and their exponents add up to 0, then they must be reciprocals. Ex: 73 ・ 7-3 = 70
So, 73 and 7-3 are reciprocals.
73 ・ 7-3 = 70 So, 73 and 7-3 are reciprocals. What is the reciprocal of 343? 73 = 343 When x is any non-zero number and n is a rational number, x-n is the reciprocal of xn. That is,
example Evaluate each power. a) b) c) Try it:
example Evaluate each power without using a calculator. a) b) Recall:
Try it (without a calculator):
example Paleontologists use measurements from fossilized dinosaur tracks and the formula to estimate the speed at which the dinosaur travelled. In the formula, v is the speed in metres per second, s is the distance between successive footprints of the same foot, and f is the foot length in metres. Use the measurements in the diagram to estimate the speed of the dinosaur.
Pg , #3, 6, 7, 9, 13, 14, 16, 21 Independent Practice
4.6 – applying the exponent laws
Chapter 4
Exponent laws review Recall:
Try it Find the value of this expression where a = –3 and b = 2.
example Simplify by writing as a single power. a) b) c) d) Try these:
example Simplify. a) b) Try this:
example Simplify. a) b) c) d) Try these:
example A sphere has volume 425 m3.
What is the radius of the sphere to the nearest tenth of a metre?
Pg , #9, 10, 11, 12, 16, 19, 21, 22 Independent Practice |
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# Year 3 Maths Lessons – Fractions
## Lesson 1: What is a fraction?
Mathematical purpose & objective
Model and represent unit fractions including 1/2, 1/4, 1/3, 1/5 and their multiples to a
complete whole (ACMNA058) (SCSA, n.d.)
Students will recognise fractions as being part of a whole, and construct paper
models to enhance their understanding of wholes (1), ½, 1/3, ¼ and 1/5 and their
sizes.
Resources/Materials
Whiteboard/blackboard, coloured paper (strips and/or circles and rectangles), play
dough (preferably 5 different colours), pop-sticks, students’’ maths workbooks
Introduction
Students are sitting on the mat and are drawn to the sentence “A fraction is…” on the
board. Ask students to suggest possible endings to the sentence and write them
around the board. Add to students’ suggestions: “fractions are used when we count a
part of something. They tell us how many parts of a whole there are.”
Explain using visual aids: “For example, we know this is a square,” (draw on board)
“but what if we wanted to know how much this part of the square is?” (shade ¼) “We
use fractions to figure this out.”
Demonstration:
Use 5 different, equally sized balls of play dough (preferably each a different colour).
Break these down differently in front of students – one ball left as a whole (1), one
ball in halves, one in quarters, one in thirds and leave the last ball un touched (for
students to have a go dividing into fifths).
Using the pop-sticks and a piece of paper, stick a “sign” in each different sized play
dough ball, labelling it as either being 1, ½, ¼ and 1/3. Review the different play
dough partitions and their sizes “What do you notice about how the sizes of the play
dough balls change?” and ask children “What do you think fifths would look like?”
“what size do you think the play dough balls will be? Show me with your hands.” a
student volunteer will make it out of the last play dough ball and labelling with the
“1/5” pop-stick.
## Main body of lesson
Students return to their desks and are given coloured strips of paper. Students are
asked to cut each strip into different fractions and glue and label these in their maths
workbooks – strips of 1 whole, halves, thirds, quarters, fifths. (model this process
first)
Advanced students may experiment making other fractions after finishing – eights,
sixths, tenths – or the may chose to represent the same fractions of paper circles
also. Students may also be given one-on-one of group assistance if experiencing
difficulty, with the teacher giving step-by-step instructions.
Teacher makes observations of children’s independent progression throughout the
activity and collect workbooks after completion.
Conclusion/Reflection
Ask certain students to share their work with the rest of the class. “How many
halves/quarters/thirds/fifths make a whole?” “Which fraction is the largest? Which is
the smallest?”
Teacher make note of students who still have difficulty answering these questions.
Lesson 2: What do fractions look like?
Mathematical purpose & objective
Model and represent unit fractions including 1/2, 1/4, 1/3, 1/5 and their multiples to a
complete whole (ACMNA058) (SCSA, n.d.)
Students will create visual representations of fractions and connect them to their
written and symbolic form.
Resources/Materials
Smart board, fraction cards (for grouping), dice, workbooks, fraction bingo cards
Introduction
Children are on the mat and revise previous lesson. “What is a fraction? What is a
half? A quarter? A third? A fifth?”
Pause when the fractions first appear and ask children to name it. Pause when
fractions are expressed in writing and in symbolic form “Has anyone seen fractions
written using numbers before? This is what we are looking at today.” Pause after
description of symbolic representation and repeat/explain to students “the top
number is the number of pieces of the whole you have, and the bottom number is the
number of pieces that make up the whole”. Watch until end and discuss further as a
Write fractions ½, ¼, 1/3, and 1/5 symbolically as a class on the board, asking “how
many pieces do we have? How many pieces in the whole all together?” while writing.
Groups:
Hand fraction cards out to each student (each with a fraction depicted on it).
Students with matching fraction cards form a group and answer the questions: what
fraction does your piece represent? How many of these pieces are needed to make a
whole? What does it look like in symbolic form? First discuss in groups and then
share with the class.
Teacher make observations of students’ collaboration, reasoning and responses.
## Main body of lesson
Children separate from group and are given a dice each. Students roll the dice twice
– the first roll will give a numerator, the second will give the denominator. Children
will write their rolled fraction’s name and symbol, and then draw a picture (teacher
models the process first). Stronger students work independently, weaker students
work in pairs and/or with teacher assistance.
Teachers observe activity and collect workbooks.
Conclusion/Reflection
Hand out fraction bingo cards (consisting of squares with fractions in symbolic form).
Play as a class (weaker students play with a partner, advanced may have various
representation of fractions on their bingo card).
Lesson 3: How do we count fractions?
Mathematical purpose & objective
Model and represent unit fractions including 1/2, 1/4, 1/3, 1/5 and their multiples to a
complete whole (ACMNA058) (SCSA, n.d.)
Students will identify and count fractions on a number line.
Resources/Materials
Smart board, chalk, dice
Introduction
Introduce number lines using the video:
Start at 28 seconds, immediately pause and compare “whole” strip and 0-1 on the
number line. Start again and stop at 32 seconds to explain the representation of
halves. Ask the students to imagine themselves jumping along the number line in
halves, starting at 0 and draw where they’d land each time on the board. “If you kept
jumping past 1/one whole, where would your next jump land?” “How do we say and
write this number?” “What do you call these numbers with a whole number and a
fraction?”. Pause the video at 37 seconds, 40 seconds and 44 seconds and repeat
for thirds, quarters and fifths.
Precede to 1:01 and complete the 5 quiz questions as a class, pausing at each
question.
## Main body of lesson
Move students outside to chalk-draw number lines on pavement (one in halves,
quarters, sixths, thirds, fifths and tenths). Children will be in mixed ability groups and
rotate through the different number lines. Children take turns rolling the dice, moving
that number of spaces forward on the number line, and counting in their fractional
increments with each jump they take. Advanced students may roll and count
backwards when they reach the end of the number line, whilst weaker students may
start back from the beginning. Challenge: take number off the number line.
The teacher may remain at a set station, observing children’s counting as they pass
through. They may also go between groups observing and assisting where
necessary.
Conclusion/Reflection
“What did you find easy/fun?” “Which fractions did you find hard to count?”. Children play a
game of “Buzz”. Going around in a circle, students skip count by a particular fraction (that
they need practice with); whenever a whole number appears, students say “Buzz” instead of
the number. (Students are not eliminated). |
# How do you factor the trinomial 8z^2+20z-48?
Apr 22, 2018
$4 \left(z + 4\right) \left(2 z - 3\right)$
#### Explanation:
you can divide all the terms by a common factor $4$ to get
$2 {z}^{2} + 5 z - 12$
this expression can be factorised by grouping:
the first and last coefficients are $2$ and $- 12$.
the product of this is $- 24$.
then, you can find $2$ numbers that add to give the second coefficient, and multiply to give the product of the first and last coefficient.
$8 + - 3 = 8 - 3 = 5$
$8 \cdot - 3 = - 24$
the two numbers that add to give the second coefficient are $8$ and $- 3$.
therefore the second term can be expressed as the sum of two terms, $8 z$ and $- 3 z$.
this gives the expression $2 {z}^{2} - 3 z + 8 z - 12$.
then you can factorise each adjacent pair:
$2 {z}^{2} - 3 z = z \left(2 z - 3\right)$
$8 z - 12 = 4 \left(2 z - 3\right)$
therefore $2 {z}^{2} - 3 z + 8 z - 12$ can be written as $z \left(2 z - 3\right) + 4 \left(2 z - 3\right)$.
this sum is the same as $\left(z + 4\right) \left(2 z - 3\right)$.
then you have the expression $2 {z}^{2} + 5 z - 12$ grouped as $\left(z + 4\right) \left(2 z - 3\right)$.
the expression given in the question is $8 {z}^{2} + 20 z - 48$.
that is $4$ times $2 {z}^{2} + 5 z - 12$, or the grouped expression $\left(z + 4\right) \left(2 z - 3\right)$.
therefore it can be written as $4 \left(z + 4\right) \left(2 z - 3\right)$. |
# math Notes for Class 11 Quadratic Equations Download PDF
Math Notes For Class 11 Quadratic Equations Download PDF
Chapter 5 : Quadratic Equations
## Chapter 5 :- Quadratic Equations
1. Real Polynomial: Let a0, a1, a2, … , an be real numbers and x is a real variable. Then, f(x) = a0 + a1x + a2x2 + … + anxn is called a real polynomial of real variable x with real coefficients.
2. Complex Polynomial: If a0, a1, a2, … , an be complex numbers and x is a varying complex number, then f(x) = a0 + a1x + a2x2 + … + an – 1xn – 1+ anxn is called a complex polynomial or a polynomial of complex variable with complex coefficients.
3. Degree of a Polynomial: A polynomial f(x) = a0 + a1x + a2x2 + a3x3 + … + anxn , real or complex is a polynomial of degree n , if an ≠ 0.
4. Polynomial Equation: If f(x) is a polynomial, real or complex, then f(x) = 0 is called a polynomial equation. If f(x) is a polynomial of second degree, then f(x) = 0 is called a quadratic equation .
Quadratic Equation: A polynomial of second degree is called a quadratic polynomial.
Polynomials of degree three and four are known as cubic and biquadratic polynomials respectively. A quadratic polynomial f(x) when equated to zero is called quadratic equation.
i.e., ax2 + bx + c = 0 where a ≠ 0.
Roots of a Quadratic Equation: The values of variable x .which satisfy the quadratic equation is called roots of quadratic equation.
###### Important Points to be Remembered
•An equation of degree n has n roots, real or imaginary .
•Surd and imaginary roots always occur in pairs of a polynomial equation with real coefficients
i.e., if (√2 + √3i) is a root of an equation, then’ (√2 – √3i) is also its root. .
•An odd degree equation has at least one real root whose sign is opposite to that of its last’ term (constant term), provided that the coefficient of highest degree term is positive.
•Every equation of an even degree whose constant term is negative and the coefficient of highest degree term is positive has at least two real roots, one positive and one negative.
•If an equation has only one change of sign it has one positive root.
•If all the terms of an equation are positive and the equation involves odd powers of x, then all its roots are complex.
###### Solution of Quadratic Equation
1.Factorization Method: Let ax2 + bx + c = α(x – α) (x – β) = O. Then, x = α and x = β will satisfy the given equation.
2. Direct Formula: Quadratic equation ax2 + bx + c = 0 (a ≠ 0) has two roots, given by
where D = Δ = b2 – 4ac is called discriminant of the equation .
Above formulas also known as Sridharacharya formula.
#### Nature of Roots
Let quadratic equation be ax2 + bx + c = 0, whose discriminant is D.
(i) For ax2 + bx + c = 0; a, b , C ∈ R and a ≠ 0, if
(a) D < => Complex roots
(b) D > 0 => Real and distinct roots
(c) D = 0 => Real and equal roots as α = β = – b/2a
(ii) If a, b, C ∈ Q, a ≠ 0, then
(a) If D > 0 and D is a perfect square => Roots are unequal and rational.
(b) If D > 0, a = 1; b, c ∈ I and D is a perfect square. => Roots are integral. .
(c) If D > and D is not a perfect square. => Roots are irrational and unequal.
(iii) Conjugate Roots The irrational and complex roots of a quadratic equation always occur in pairs. Therefore,
(a) If one root be α + iβ, then other root will be α – iβ.
(b) If one root be α + √β, then other root will be α – √β.
(iv) If D, and D2 be the discriminants of two quadratic equations, then
(a) If D1 + D2 ≥ 0, then At least one of D1 and D2 ≥ 0 If D1 < 0, then D2 > 0 ,
(b) If D1 + D2 < 0, then At least one of D1 and D2 < 0 If D1 > 0, then D2 < 0
###### Roots Under Particular Conditions
For the quadratic equation ax2 + bx + e = 0.
(i) If b = 0 => Roots are real/complex as (c < 0/c > 0) and equal in magnitude but of opposite sign.
(ii) If c = 0 => One roots is zero, other is – b / a.
(iii) If b = C = 0 => Both roots are zero.
(iv) If a = c => Roots are reciprocal to each other.
(v) If a > 0, c < 0, a < 0, c > 0} => Roots are of opposite sign.
(vi) If a > 0, b > 0, c > 0, a < 0, b < 0, c < 0} => Both roots are negative, provided D ≥ 0
(vii) If a > 0, b < 0, c > 0, a < 0, b > 0, c < 0} => Both roots are positive, provided D ≥ 0
(viii) If sign of a = sign of b ≠ sign of c => Greater root in magnitude is negative.
(ix) If sign of b = sign of c ≠. sign of a => Greater root in magnitude is positive.
(x) If a + b + c = 0 => One root is 1 and second root is c/a.
##### Relation between Roots and Coefficients
1. Quadratic Equation: If roots of quadratic equation ax2 + bx + c = 0 (a ≠ 0) are α and β, then Sum of roost = S = α + β = -b/a = – coefficient of x / coefficient of x2 Product of roots = P = α
* β = c/a = constant term / coefficient of x2
2. Cubic Equation: If α, β and γ are the roots of cubic equation ax3 + bx2 + cx + d = 0. Then,
3. Biquadratic Equation: If α, β, γ and δ are the roots of the biquadratic equation ax4 + bx3 + cx2 + dx + e = 0, then
Symmetric Roots: If roots of quadratic equation ax2 + bx + c = 0 (a ≠ 0) are α and β, then
###### Formation of Polynomial Equation from Given Roots
If a1, a2 a3,…, an are the roots of an nth degree equation, then the equation is xn – S1X<>n – 1 + S2Xn – 2 – S3 Xn – 3 +…+( _l)n Sn = 0 where Sn denotes the sum of the products of roots taken n at a time.
##### 1. Quadratic Equation
If α and β are the roots of ‘a quadratic equation, then the equation is x2 – S1X + S2 = 0
i.e., x2 – (α + β) x + αβ = 0
###### 2. Cubic Equation
If α, β and γ are the roots of cubic equation, then the equation is
###### 3. Biquadratic Equation
If α, β, γ and δ are the roots of a biquadratic equation, then the equation is
###### Equation In Terms of the Roots of another Equation
If α, β are roots of the equation ax2 + bx + c = 0, then the equation whose roots are.
The quadratic function f(x) = ax2 + 2hxy + by2 + 2gx + 2fy + c is always resolvable into linear factor, iff abc + 2fgh – af2 – bg2 – ch2 = 0
###### 1. Only One Root is Common
If α be the common root of quadratic equations
a1x2 + b1x + C1 = 0,
and a2x2 + b2x + C2 = 0,
then a1a2 + b1α + C1 = 0,
and a2a2 + b2Mα + C2 = 0,
By Cramer’s Rule
Hence, the condition for only one root common is (c1a2 – c2a1)2 = (b1c2 – b2c1)(a1b2 – a2b1)
###### 2. Both Roots are Common
The required condition is a1 / a2 = b1 / b2 = c1 / c2
(i) To find the common root of two equations, make the coefficient of second degree term in the two equations equal and subtract. The value of x obtained is the required common root.
(ii) Two different quadratic equations with rational coefficient can not have single common root which is complex or irrational as imaginary and surd roots always occur in pair.
###### Properties of Quadratic Equation
(i) f(a) . f(b) < 0, then at least one or in general odd number of roots of the equation f(x) = 0 lies between a and b.
(ii) f( a) . f( b) > 0, then in general even number of roots of the equation f(x) = 0 lies between a and b or no root exist f(a) = f(b), then there exists a point c between a and b such that f'(c) = 0, a < c < b.
(iii) If the roots of the quadratic equation a1x2 + b1x + c1 = 0, a2x2 + b2x + c2 = 0 are in the ratio (i.e., &alpha1;/β1 = &alpha2;/β2), then b1 2 / b2 2 = a1c1 / a2c2.
(iv) If one root is k times the other root of the quadratic equation ax2 + bx + c = 0 ,then (k + 1)2 / k = b2 / ac
##### Quadratic Expression
An expression of the form ax2 + bx + c, where a, b, c ∈ R and a ≠ 0 is called a quadratic expression in x .
1. Graph of a Quadratic Expression
We have y = ax2 + bx + c = f(x)
Let y + D/4a = Y and x + D / 2a = X
Y = a * X2 => X2 = Y / a
(i) The graph of the curve y = f(x) is parabolic.
(ii) The axis of parabola is X = 0 or x + b / 2a = 0 i.e., (parallel to Y-axis).
(iii) If a > 0, then the parabola opens upward.
If a < 0, then the parabola opens downward.
2. Position of y = ax2 +bx + c with Respect to Axes.
(i) For D > 0, parabola cuts X-axis in two real and distinct points i.e, x = -b ± √D / 2a
(ii) For D = 0, parabola touch X-axis in one point, x = – b/2a.
(iii) For D < O,parabola does not cut X-axis (i.e., imaginary value of x).
###### 3. Maximum and Minimum Values of Quadratic Expression
(i) If a > 0, quadratic expression has least value at x = b / 2a. This least value is given by 4ac – b2 / 4a = – D/4a. But their is no greatest value.
(ii) If a < 0, quadratic expression has greatest value at x = – b/2a . This greatest value is given by 4ac – b2 / 4a = – D/4a. But their is no least value.
###### 4. Sign of Quadratic Expression
(i) a > 0 and D < 0, so f(x) > 0 for all x ∈ R i.e., f(x) is positive for all real values of x.
(ii) a < 0 and D < 0, so f(x) < 0 for all x ∈ R i.e., f(x) is negative for all real values of x.
(iii) a > 0 and D = 0, so f(x) ≥ 0 for all x ∈ R i.e., f(x) is positive for all real values of x except at vertex, where f(x) = 0.
(iv) a < 0 and D = 0, so f(x) ≤ 0 for all x ∈ R i.e., f(x) is negative for all real values of x except
at vertex, where f(x) = 0.
(v) a > 0 and D > 0
Let f(x) = o have two real roots α and β (α < β), then f(x) > 0 for x ∈ (- ∞, α) ∪ (β,∞) and f (x < 0 for all x ∈ (α, β).
(vi) a < 0 and D > 0
Let f(x) = 0 have two real roots α and β (α < β). Then, f(x) < 0 for all x ∈ (- ∞, α) ∪ (β,∞) and f(x) > 0 for all x ∈ (α, β).
###### 5. Intervals of Roots
In some problems, we want the roots of the equation ax2 + bx + c = 0 to lie in a given interval.
For this we impose conditions on a, b and c.
Since, a ≠ 0, we can take f(x) = x2 + b/a x + c/a.
(i) Both the roots are positive i.e., they lie in (0,∞), if and only if roots are real, the sum of the roots as well as the product of the roots is positive.
α + β = -b/a > 0 and αβ = c/a > 0 with b2 – 4ac ≥ 0
Similarly, both the roots are negative i.e., they lie in (- ∞,0) ifF roots are real, the sum of the roots is negative and the product of the roots is positive. i.e., α + β = -b/a < 0 and αβ = c/a > 0 with b2 – 4ac ≥ 0
(ii) Both the roots are greater than a given number k, iFf the following conditions are satisfied D ≥ 0, -b/2a > k and f(k) > 0
(iii) Both .the roots are less than a given number k, iff the following conditions are satisfied D ≥ 0, -b/2a > k and f(k) > 0
(iv) Both the roots lie in a’ given interval (k1, k2), iff the following conditions are satisfied D ≥ 0,k1 < -b/2a < k2 and f(k1) > 0, f(k2) > 0
(v) Exactly one of the roots lie in a given interval (k1, k2), iff f(k1) f(k2) < 0
(vi) A given number k lies between the roots iff f(k) < O. In particular, the roots of the equation will be of opposite sign, iff 0 lies between the roots. ⇒ f(0) < 0
Wavy Curve Methodd
Let f(x) = (x – a1)k 1 (x – a2)k 2(x — a3)k3 … (x – an – 1)k n – 1 (x – an)k n
where k1, k2, k3,…, kn ∈ N and a1, a2, a3,…, an are fixed natural numbers satisfying the condition. a1 < a2 < a3 < … < an – 1 < an.
First we mark the numbers a1, a2, a3,…, an on the real axis and the plus sign in the interval of the right of the largest of these numbers, i.e., on the right of an. If kn is even, we put plus sign on the left of anand if kn is odd, then we put minus sign on the left of an In the next interval we put a sign according to the following rule.
When passing through the point an – 1 the polynomial f(x) changes sign . if kn – 1 is an odd number and the polynomial f(x) has same sign if kn – 1 is an even number. Then, we consider the next interval and put a sign in it using the same rule.
Thus, we consider all the intervals. The solution of f(x) > 0 is the union of all interval in which we have put the plus sign and the solution of f(x) < 0 is the union of all intervals in which we have put the minus Sign.
###### Descarte’s Rule of Signs
The maximum number of positive real roots of a polynomial equation f(x) = 0 is the number of changes of sign from positive to negative and negative to positive in f(x) .
The maximum number of negative real roots of a polynomial equation f(x) = 0 is the number of changes of sign from positive to negative and negative to positive in f(x).
###### Rational Algebraic In equations
(i) Values of Rational Expression P(x)/Q(x) for Real Values of x, where P(x) and Q(x) are Quadratic Expressions To find the values attained by rational expression of the form a1x2 + b1x + c1 / a2x2 + b2x + c2> for real values of x.
(a) Equate the given rational expression to y.
(b) Obtain a quadratic equation in x by simplifying the expression,
(c) Obtain the discriminant of the quadratic equation.
(d) Put discriminant ≥ 0 and solve the in equation for y.
The values of y so obtained determines the set of values attained by the given rational expression.
(ii) Solution of Rational Algebraic In equation If P(x) and Q(x) are polynomial in x, then the in equation P(x) / Q(x) > 0, P(x) / Q(x) < 0, P(x) / Q(x) ≥ 0 and P(x) / Q(x) ≤ 0 are known as rational algebraic in equations.
To solve these in equations we use the sign method as
(a) Obtain P(x) and Q(x).
(b) Factorize P(x) and Q(x) into linear factors.
(c) Make the coefficient of x positive in all factors.
(d) Obtain critical points by equating all factors to zero.
(e) Plot the critical points on the number line. If these are n critical points, they divide the number line into (n + 1) regions.
(f) In the right most region the expression P(x) / Q(x) bears positive sign and in other region the
expression bears positive and negative signs depending on the exponents of the factors .
##### Lagrange’s identity
If a1, a2, a3, b1, b2, b3 ≠ R, then (a1 2 + a2 2 + a3 2) (b1 2 + b2 2 + b3 2) – (a1b1 + a2b2 + a3b3)2
= (a1b2 – a2b1)2 + (a2b3 – a3b2 )2 + (a3 b1 – a1b3)2
##### Algebraic Interpretation of Rolle’s Theorem
Let f (x) be a polynomial having α and β as its roots such that α < β, f(α) = f(β) = 0.Also, a polynomial function is everywhere continuous and differentiable, then there exist θ ∈ (α, β) such that f'(θ) = 0. Algebraically, we can say between any two zeros of a polynomial f(x) there is always a derivative f’ (x) = 0.
###### Equation and In equation Containing Absolute Value
1. Equation Containing Absolute Value By definition, |x| = x, if x ≥ 0 OR -x, if x < 0
If |f(x) + g(x)| = |f(x)| + g(x)|, then it is equivalent to the system f(x) . g(x) ≥ 0.
If |f(x) – g(x)| = |f(x)| – g(x)|, then it is equivalent to the system f(x) . g(x) ≤ 0.
###### 2.In equation Containing Absolute Value
(i) |x| < a ⇒ – a < x < a (a > 0)
(ii) |x| ≤ a ⇒ – a ≤ x ≤ a
(iii) |x| > a ⇒ x < – a or x > a
(iv) |x| ≥ a ⇒ x le; – a or x ≥ a
###### 3. Absolute Value of Real Number
|x| = -x, x < 0 OR +x, x ≥ 0
(i) |xy| = |x||y|
(ii) |x / y| = |x| / |y|
(iii) |x|2 = x2
(iv) |x| ≥ x
(v) |x + y| ≤ |x| + |y|
Equality hold when x and y same sign.
(vi) |x – y| ≥ ||x| – |y||
###### Inequalities
Let a and b be real numbers. If a – b is negative, we say that a is less than b (a < b) and if a – b is positive, then a is greater than b (a > b).
##### Important Points to be Remembered
(i) If a > b and b > c, then a > c.
Generally, if a1 > a2, a2 > a3,…., an – 1 > an, then a1 > an.
(vii) If a < x < b and a, b are positive real numbers then a2 < x2 < b2
###### Important Inequality
1. Arithmetico-Geometric and Harmonic Mean Inequality
(i) If a, b > 0 and a ≠ b, then
(ii) if ai > 0, where i = 1,2,3,…,n, then
iii) If a1, a2,…, an are n positive real numbers and m1, m2,…,mn are n positive rational numbers, then
i.e., Weighted AM > Weighted GM
(iv) If a1, a2,…, an are n positive distinct real numbers, then
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# FREE Grade 5 SBAC Math Practice Test
Welcome to our FREE SBAC Math practice test for grade 5, with answer key and answer explanations. This practice test’s realistic format and high-quality practice questions can help your student succeed on the SBAC Math test. Not only does the test closely match what students will see on the real SBAC, but it also comes with detailed answer explanations.
For this practice test, we’ve selected 20 real questions from past exams for your student’s SBAC Practice test. Your student will have the chance to try out the most common SBAC Math questions. For every question, there is an in-depth explanation of how to solve the question and how to avoid mistakes next time.
Use our free SBAC Math practice tests and study resources (updated for 2020) to help your students ace the SBAC Math test! Make sure to follow some of the related links at the bottom of this post to get a better idea of what kind of mathematics questions students need to practice.
## 10 Sample Grade 5 SBAC Math Practice Questions
1- How long is the line segment shown on the number line below?
☐A. 6
☐B. 7
☐C. 8
☐D. 9
2- If a rectangle is 30 feet by 45 feet, what is its area?
☐A. 1350
☐B. 1250
☐C. 1000
☐D. 870
3- If a vehicle is driven 32 miles on Monday, 35 miles on Tuesday, and 29 miles on Wednesday, what is the average number of miles driven each day?
☐A. 32
☐B. 33
☐C. 34
☐D. 35
4- Peter traveled 120 miles in 4 hours and Jason traveled 160 miles in 8 hours. What is the ratio of the average speed of Peter to average speed of Jason?
☐A. 3 : 2
☐B. 2 : 3
☐C. 5 : 9
☐D. 5 : 6
5- If $$x=- 8$$, which equation is true?
☐A. $$x(2x-4)=120$$
☐B. $$8 (4-x)=96$$
☐C. $$2 (4x+6)=79$$
☐D. $$6x-2=-46$$
6- A circle has a diameter of 8 inches. What is its approximate circumference?
($$π = 3.14$$)
☐A. 6.28 inches
☐B. 25.12 inches
☐C. 34.85 inches
☐D. 35.12 inches
7- A woman owns a dog walking business. If 3 workers can walk 9 dogs, how many dogs can 5 workers walk?
☐A. 13
☐B. 15
☐C. 17
☐D. 19
8- What are the coordinates of the intersection of x–axis and the y–axis on a coordinate plane?
☐A. $$(5, 5)$$
☐B. $$(1, 1)$$
☐C. $$(0, 0)$$
☐D. $$(0, 11)$$
9- Jack added 19 to the product of 16 and 26. What is this sum?
☐A. 61
☐B. 330
☐C. 435
☐D. 135
10- Joe makes $4.75 per hour at his work. If he works 8 hours, how much money will he earn? ☐A.$32.00
☐B. $34.75 ☐C.$36.50
☐D. $38.00 11- Which of the following is an obtuse angle? ☐A. 89$$^\circ$$ ☐B. 55$$^\circ$$ ☐C. 143$$^\circ$$ ☐D. 235$$^\circ$$ 12- What is the value of $$6 – 3 \frac{4}{9}$$? ☐A. $$\frac{23}{9}$$ ☐B. $$3\frac{4}{9}$$ ☐C. $$-\frac{1}{9}$$ ☐D. $$\frac{42}{9}$$ 13- The bride and groom invited 220 guests for their wedding. 190 guests arrived. What percent of the guest list was not present? ☐A. $$90\%$$ ☐B. $$20\%$$ ☐C. $$23.32\%$$ ☐D. $$13.64\%$$ 14- Frank wants to compare these two measurements. $$18.023 kg \space …….. \space 18,023 g$$ Which symbol should he use? ☐A. $$<$$ ☐B. $$>$$ ☐C. $$≠$$ ☐D. $$=$$ 15- Aria was hired to teach three identical 5th grade math courses, which entailed being present in the classroom 36 hours altogether. At$25 per class hour, how much did Aria earn for teaching one course?
☐A. $50 ☐B.$300
☐C. $600 ☐D.$1400
16- In a classroom of 60 students, 22 are male. What percentage of the class is female?
☐A. $$51\%$$
☐B. $$59\%$$
☐C. $$63\%$$
☐D. $$73\%$$
17- In a party, 6 soft drinks are required for every 9 guests. If there are 171 guests, how many soft drinks are required?
☐A. 9
☐B. 27
☐C. 114
☐D. 171
18- While at work, Emma checks her email once every 90 minutes. In 9 hours, how many times does she check her email?
☐A. 4 Times
☐B. 5 Times
☐C. 6 Times
☐D. 7 Times
19- In a classroom of 44 students, 18 are male. About what percentage of the class is female?
☐A. $$63\%$$
☐B. $$51\%$$
☐C. $$59\%$$
☐D. $$53\%$$
20- A florist has 516 flowers. How many full bouquets of 12 flowers can he make?
☐A. 40
☐B. 41
☐C. 43
☐D. 45
## Answers:
1- D
The line segment is from 1 to$$-8$$. Therefore, the line is 9 units.
$$1 –(-8)= 1+8=9$$
2- A
Use area of rectangle formula.
Area $$=$$ length $$×$$ width $$⇒ A = 30 × 45 ⇒ A = 1,350$$
3- A
$$average (mean) = \frac{sum \space of \space terms}{number \space of \space terms}⇒ average= \frac{32+35+29}{3}⇒ average = 32$$
4- A
Peter’s speed $$= \frac{120}{4}= 30$$
Jason’s speed $$= \frac{160}{8}=20$$
$$\frac{The \space average \space speed \space of \space peter}{The \space average \space speed \space of \space Jason}=\frac{30}{20}$$
equals to: $$\frac{3}{2}$$or 3 : 2
5- B
Plug in $$x=- 8$$ in each equation.
$$x(2x-4)=120→(-8)(2(-8)-4)=(-8)×(-16-4)=160$$
$$8 (4-x)=96→8(4-(-8)=8(12)=96$$
$$2 (4x+6)=79→2(4(-8)+6)=2(-32+6)=-52$$
$$6x-2=-46→6(-8)-2=-48-2=-50$$
Only option B is correct.
6- B
The diameter of the circle is 8 inches. Therefore, the radius of the circle is 4 inches.
Use circumference of circle formula.
$$C = 2πr ⇒ C = 2 × 3.14 × 4 ⇒ C = 25.12$$
7- B
3 workers can walk 9 dogs ⇒ 1 workers can walk 3 dogs.
5 workers can walk $$(5 × 3) 15$$ dogs.
8- C
The horizontal axis in the coordinate plane is called the x-axis. The vertical axis is called the y-axis. The point at which the two axes intersect is called the origin. The origin is at 0 on the x-axis and 0 on the y-axis.
9- C
$$19 + (16 × 26) = 19 + 416 = 435$$
10- D
1 hour: $$4.75$$
8 hours: $$8 × 4.75 = 38$$
11- C
An obtuse angle is an angle of greater than 90$$^\circ$$ and less than 180$$^\circ$$. From the options provided, only option C (143 degrees) is an obtuse angle.
12- A
$$6 – 3\frac{4}{9}=\frac{54}{9}-\frac{31}{9}=\frac{23}{9}$$
13- D
The number of guests that are not present are $$(220 – 190) 30$$ out of $$220 =\frac{30}{220}$$
Change the fraction to percent:
$$\frac{30}{220}×100\%=13.64\%$$
14- D
Each kilogram is 1,000 grams.
18,023 grams $$= (\frac{18,023}{1,000}) =18.023$$ kilograms.
Therefore, two amounts provided are equal.
15- B
Aria teaches 36 hours for three identical courses. Therefore, she teaches 12 hours for each course. Aria earns $25 per hour. Therefore, she earned$300 ($$12 × 25$$) for each course.
16- C
The number of female students in the class is $$(60 – 22) 38$$ out of $$60 = \frac{38}{60}$$
Change the fraction to percent:
$$\frac{38}{60} 3 ×100\%=63\%$$
17- C
Write a proportion and solve.
$$\frac{6 \space soft \space drinks}{9 \space guests}=\frac{x}{171 \space guests}$$
$$x =\frac{171×6}{9}⇒x=114$$
18- C
Every 90 minutes Emma checks her email.
In 9 hours (540 minutes), Emma checks her email $$(540 ÷ 90) 6$$ times.
19- C
There are 44 students in the class. 18 of the are male and 26 of them are female.
26 out of 44 are female. Then:
$$\frac{26}{44}=\frac{x}{100}→2,600=44x→x=2,600÷44≈59\%$$
20- C
Divide the number flowers by $$12: 516 ÷ 12 = 43$$
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# Problems on Time and Work Part 2
## Problems on Time and Work Part 2
16.
X and Y can do a piece of work in 20 days and 12 days respectively. X started the work alone and then after 4 days Y joined him till the completion of the work. How long did the work last?
A. 6 days B. 10 days C. 15 days D. 20 days
Explanation:
Work done by X in 4 days = 1 x 4 = 1 . 20 5
Remaining work = 1 - 1 = 4 . 5 5
(X + Y)'s 1 day's work = 1 + 1 = 8 = 2 . 20 12 60 15
Now, 2 work is done by X and Y in 1 day. 15
So, 4 work will be done by X and Y in 15 x 4 = 6 days. 5 2 5
Hence, total time taken = (6 + 4) days = 10 days.
17.
A is 30% more efficient than B. How much time will they, working together, take to complete a job which A alone could have done in 23 days?
A.11 days
B.13 days
C.
20 3 days 17
D.None of these
Explanation:
Ratio of times taken by A and B = 100 : 130 = 10 : 13.
Suppose B takes x days to do the work.
Then, 10 : 13 :: 23 : x x = 23 x 13 x = 299 . 10 10
A's 1 day's work = 1 ; 23
B's 1 day's work = 10 . 299
(A + B)'s 1 day's work = 1 + 10 = 23 = 1 . 23 299 299 13
Therefore, A and B together can complete the work in 13 days.
18.
Ravi and Kumar are working on an assignment. Ravi takes 6 hours to type 32 pages on a computer, while Kumar takes 5 hours to type 40 pages. How much time will they take, working together on two different computers to type an assignment of 110 pages?
A. 7 hours 30 minutes B. 8 hours C. 8 hours 15 minutes D. 8 hours 25 minutes
Explanation:
Number of pages typed by Ravi in 1 hour = 32 = 16 . 6 3
Number of pages typed by Kumar in 1 hour = 40 = 8. 5
Number of pages typed by both in 1 hour = 16 + 8 = 40 . 3 3
Time taken by both to type 110 pages = 110 x 3 hours 40
= 8 1 hours (or) 8 hours 15 minutes. 4
19.
A, B and C can complete a piece of work in 24, 6 and 12 days respectively. Working together, they will complete the same work in:
A.
1 day 24
B.
7 day 24
C.
3 3 days 7
D.4 days
Explanation:
Formula: If A can do a piece of work in n days, then A's 1 day's work = 1 . n
(A + B + C)'s 1 day's work = 1 + 1 + 1 = 7 . 24 6 12 24
Formula: If A's 1 day's work = 1 , then A can finish the work in n days. n
So, all the three together will complete the job in 24 days = 3 3 days. 7 7
20.
Sakshi can do a piece of work in 20 days. Tanya is 25% more efficient than Sakshi. The number of days taken by Tanya to do the same piece of work is:
A. 15 B. 16 C. 18 D. 25
Explanation:
Ratio of times taken by Sakshi and Tanya = 125 : 100 = 5 : 4.
Suppose Tanya takes x days to do the work.
5 : 4 :: 20 : x x = 4 x 20 5
x = 16 days.
Hence, Tanya takes 16 days to complete the work.
21.
A takes twice as much time as B or thrice as much time as C to finish a piece of work. Working together, they can finish the work in 2 days. B can do the work alone in:
A. 4 days B. 6 days C. 8 days D. 12 days
Explanation:
Suppose A, B and C take x, x and x days respectively to finish the work. 2 3
Then, 1 + 2 + 3 = 1 x x x 2
6 = 1 x 2
x = 12.
So, B takes (12/2) = 6 days to finish the work.
22.
A and B can complete a work in 15 days and 10 days respectively. They started doing the work together but after 2 days B had to leave and A alone completed the remaining work. The whole work was completed in :
A. 8 days B. 10 days C. 12 days D. 15 days
Explanation:
(A + B)'s 1 day's work = 1 + 1 = 1 . 15 10 6
Work done by A and B in 2 days = 1 x 2 = 1 . 6 3
Remaining work = 1 - 1 = 2 . 3 3
Now, 1 work is done by A in 1 day. 15
2 work will be done by a in 15 x 2 = 10 days. 3 3
Hence, the total time taken = (10 + 2) = 12 days.
23.
A and B can do a piece of work in 30 days, while B and C can do the same work in 24 days and C and A in 20 days. They all work together for 10 days when B and C leave. How many days more will A take to finish the work?
A. 18 days B. 24 days C. 30 days D. 36 days
Explanation:
2(A + B + C)'s 1 day's work = 1 + 1 + 1 = 15 = 1 . 30 24 20 120 8
Therefore, (A + B + C)'s 1 day's work = 1 = 1 . 2 x 8 16
Work done by A, B, C in 10 days = 10 = 5 . 16 8
Remaining work = 1 - 5 = 3 . 8 8
A's 1 day's work = 1 - 1 = 1 . 16 24 48
Now, 1 work is done by A in 1 day. 48
So, 3 work will be done by A in 48 x 3 = 18 days. 8 8
24.
A works twice as fast as B. If B can complete a work in 12 days independently, the number of days in which A and B can together finish the work in :
A. 4 days B. 6 days C. 8 days D. 18 days
Explanation:
Ratio of rates of working of A and B = 2 : 1.
So, ratio of times taken = 1 : 2.
B's 1 day's work = 1 . 12
A's 1 day's work = 1 ; (2 times of B's work) 6
(A + B)'s 1 day's work = 1 + 1 = 3 = 1 . 6 12 12 4
So, A and B together can finish the work in 4 days.
25.
Twenty women can do a work in sixteen days. Sixteen men can complete the same work in fifteen days. What is the ratio between the capacity of a man and a woman?
A. 3 : 4 B. 4 : 3 C. 5 : 3 D. Data inadequate
Explanation:
(20 x 16) women can complete the work in 1 day.
1 woman's 1 day's work = 1 . 320
(16 x 15) men can complete the work in 1 day.
1 man's 1 day's work = 1 240
So, required ratio
= 1 : 1 240 320
= 1 : 1 3 4
= 4 : 3 (cross multiplied)
26.
A and B can do a work in 8 days, B and C can do the same work in 12 days. A, B and C together can finish it in 6 days. A and C together will do it in :
A. 4 days B. 6 days C. 8 days D. 12 days
Explanation:
(A + B + C)'s 1 day's work = 1 ; 6
(A + B)'s 1 day's work = 1 ; 8
(B + C)'s 1 day's work = 1 . 12
(A + C)'s 1 day's work
= 2 x 1 - 1 + 1 6 8 12
= 1 - 5 3 24
= 3 24
= 1 . 8
So, A and C together will do the work in 8 days.
27.
A can finish a work in 24 days, B in 9 days and C in 12 days. B and C start the work but are forced to leave after 3 days. The remaining work was done by A in:
A.5 days
B.6 days
C.10 days
D.
10 1 days 2
Explanation:
(B + C)'s 1 day's work = 1 + 1 = 7 . 9 12 36
Work done by B and C in 3 days = 7 x 3 = 7 . 36 12
Remaining work = 1 - 7 = 5 . 12 12
Now, 1 work is done by A in 1 day. 24
So, 5 work is done by A in 24 x 5 = 10 days. 12 12
28.
X can do a piece of work in 40 days. He works at it for 8 days and then Y finished it in 16 days. How long will they together take to complete the work?
A.
13 1 days 3
B.15 days
C.20 days
D.26 days
Explanation:
Work done by X in 8 days = 1 x 8 = 1 . 40 5
Remaining work = 1 - 1 = 4 . 5 5
Now, 4 work is done by Y in 16 days. 5
Whole work will be done by Y in 16 x 5 = 20 days. 4
X's 1 day's work = 1 , Y's 1 day's work = 1 . 40 20
(X + Y)'s 1 day's work = 1 + 1 = 3 . 40 20 40
Hence, X and Y will together complete the work in 40 = 13 1 days. 3 3
29.
A and B can do a job together in 7 days. A is 1 times as efficient as B. The same job can be done by A alone in :
A.
9 1 days 3
B.11 days
C.
12 1 days 4
D.
16 1 days 3
Explanation:
(A's 1 day's work) : (B's 1 day's work) = 7 : 1 = 7 : 4. 4
Let A's and B's 1 day's work be 7x and 4x respectively.
Then, 7x + 4x = 1 11x = 1 x = 1 . 7 7 77
A's 1 day's work = 1 x 7 = 1 . 77 11
30.
A and B together can do a piece of work in 30 days. A having worked for 16 days, B finishes the remaining work alone in 44 days. In how many days shall B finish the whole work alone?
A. 30 days B. 40 days C. 60 days D. 70 days |
+0
# ///'/'/'/'/
0
244
1
+221
Forgot how to do this......
Find the fourth arithmetic means between -21 and -36
Apr 5, 2018
#1
+22500
0
Find the fourth arithmetic means between -21 and -36
$$\text{We use a_n = a_1 + (n-1)d to find the common difference d.}$$
Solution
$$\text{The first term is a_1 = - 21 and the fourth term is a_4 = - 36 .}\\ \text{We must find the common difference so that the terms }$$
$$\begin{array}{cccc} -21, & -21+d, & -21+2d, & -36 \\ \uparrow & \uparrow & \uparrow & \uparrow \\ a_1 & a_2 & a_3 & a_4 \end{array}$$
$$\text{form an arithmetic sequence.}$$
$$\text{To find the common difference d, we substitue -21 for a_1; 4 for n, and -36 for a_n \\in the formula for the 4th term:}$$
$$\begin{array}{rcll} a_4 &=& a_1 + (n-1)d \qquad & \text{This gives the 4th term of any arithmetic sequence.} \\ -36 &=& -21 + (4-1)d \qquad & \text{Substitute.} \\ -36 &=& -21 + 3d \qquad & \text{Subtract within the parentheses.}\\ -15 &=& 3d \qquad & \text{Subtract -21 from both sides.} \\ -5 &=& d \qquad & \text{To isolate d, divide both sides by 3.} \\ \end{array}$$
$$\text{To find the two arithmetic means between -21 and -36, \\we add the common difference -5, as shown:}$$
$$\begin{array}{rcll} -21+d &=& -21 +(-5) \\ &=& -26 \qquad & \text{This is a_2.} \\ \end{array}$$
$$\begin{array}{rcll} -21+2d &=& -21 +2(-5) \\ &=& -21 -10 \\ &=& -31 \qquad & \text{This is a_3.} \\ \end{array}$$
$$\text{Two arithmetic means between -21 and -36 are -26 and -31.}$$
Apr 5, 2018 |
# RS Aggarwal Class 8 Math Sixth Chapter Operation on Algebric Expressions Exercise 6E Solution
## EXERCISE 6E
### OBJECTIVE QUESTIONS
#### Tick (√) the correct answer in each of the following:
(1) The sum of (6a + 4b – c + 3), (2b – 3c + 4), (11b – 7a + 2c – 1) and (2c – 5a – 6) is
Ans: (c) (- 6a + 17b)
Solution: (6a + 4b – c + 3) + (2b – 3c + 4) + (11b – 7a + 2c – 1) + (2c – 5a – 6)
= 6a + 4b – c + 3 + 2b – 3c + 4 + 11b – 7a + 2c – 1 + 2c – 5a – 6
= – 6a + 17b
(2) (3q + 7p2 – 2r3 + 4) – (4p2 – 2q + 7r3 – 3) =?
Ans: (d) (3p2 + 5q – 9r3 + 7)
Solution: 3q + 7p2 – 2r3 + 4 – 4p2 + 2q – 7r3 + 3
= 3p2 + 5q – 9r3 + 7
(3) (x + 5) (x – 3) =?
= x2 + 5x – 3x – 15
= x2 + 2x – 15
Ans: (d)
(4) (2x+ 3) (3x – 1)
= 6x2 + 9x – 2x – 3
= 6x2 + 7x – 3
Ans: (b)
(5) (x + 4) (x + 4)
= x2 + 4x + 4x + 16
= x2 + 8x + 16
Ans: (c)
(6) (x – 6) (x – 6)
= x2 – 6x – 6x + 36
= x2 – 12x + 36
Ans: (d)
(7) (2x + 5) (2x – 5)
= 4x2 + 10x – 10x – 25
= (4x2 – 25)
Ans: (b)
(8) 8a2b3 ÷ (- 2ab)
Ans: (c)
(9) (2x2 + 3x + 1) ÷ (x + 1)
= 2x + 1
Ans: (b)
(10) (x2 – 4x + 4) ÷ (x – 2)
= (x – 2)
Ans: (a)
(11) (a + 1) (a – 1) (a2 + 1)
= [(a)2 – (1)2] (a2 + 1)
= (a2 – 1) (a2 + 1)
= (a2)2 – (1)2
= (a4 – 1)
Ans: (c)
(15) (82)2 – (18)2
= (80 + 2)2 – (20 – 2)2
= [(80)2 + (2 × 80 × 2) + (2)2] – [(20)2 – (2 × 20 × 2) + (2)2]
= (6400 + 320 + 4) – (400 – 80 + 4)
= 6724 – 324 = 6400
Ans: (c)
(16) (197 × 203)
= (200 – 3) × (200 + 3)
= (200)2 – (3)2
= 40000 – 9 = 39991
Ans: (a)
(17) If (a + b) = 12 and ab = 14, then (a2 + b2) =?
⇒ (a + b) = 12
⇒ (a + b)2 = (12)2
⇒ a2 + 2ab + b2 = 144
⇒ (a2 + b2) + (2 × 14) = 144
⇒ (a2 + b2) = 144 – 28 = 116
Ans: (b)
(18) If (a – b) = 7 and ab = 9, then, (a2 + b2) =?
⇒ (a – b) = 7
⇒ (a – b)2 = 72
⇒ a2 – 2ab + b2 = 49
⇒ (a2 + b2) – (2 × 9) = 49
⇒ (a2 + b2) = 49 + 18
⇒ (a2 + b2) = 67
Ans: (a)
(19) If x = 10, then the value of (4x2 + 20x + 25) =?
= 4x2 + 20x + 25
= [4 × (10)2] + (20 × 10) + 25
= (4 × 100) + 200 + 25
= 400 + 200 + 25
= 625
Ans: (c)
Updated: May 30, 2022 — 2:11 pm
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## Introduction
Multiplying algebraic expressions is relatively easy when one of the expressions is a monomial. Simply use the distributive property.
However, the multiplication becomes more difficult when multiplying larger expressions such as binomials, trinomials, and larger polynomials. The diagrams below give a visual example of the individual multiplications that must be done to complete these types of problems.
Two binomials are multiplied to the left, while a trinomial is multiplied by a 4-term polynomial to the right. Each arrow represents an individual multiplication that must be performed in the problem. As the two polynomials being multiplied increase in size, the number of individual multiplications also increases.
## Lesson
In this lesson, you will learn how to multiply a binomial times another binomial. This process involves taking each individual term of the first binomial and multiplying it by each term of the second binomial.
Suppose that you are multiplying the following binomials: x + 2 and x + 3. Since both terms are part of the binomial, use parenthesis for the multiplication. A good way to write the problem is by simply writing (x + 2)(x + 3) without any kind of multiplication sign. Most first time problem solvers assume that the problem is done like this:
However, this answer can be shown to be wrong by picking a number for x and seeing whether the problem value is the same as the answer value. Suppose that x = 3. The problem value and answer value are shown below.
The correct way to do this problem is to multiply each term in the first binomial times each term in the second binomial.
The method of multiplying a binomial times another binomial requires four separate multiplications. These four separate multiplications can be represented by the acronym “foil”. The problem below shows how each multiplication can be represented by a letter in the word “foil”.
Look at the above example carefully. See if you can get a good idea of what each letter in F-O-I-L stands for. Here is another example: Multiply (x + 5)(x – 2)
The above problem contains a subtraction sign. The mathematically correct way to do the problem is to consider that x – 2 really equals x + (-2). Use the (-2) in multiplication to get the correct answer.
A good way to think of the distributive property is as a “crooked rainbow” where an individual part of a rainbow connects the initial (monomial) expression to each term of the second (polynomial) expression. Each part of the rainbow represents a multiplication.
The crooked rainbow is a good way to show the foil method in one step. This time there will be two separate rainbows, one coming from each term in the first binomial. Put the second rainbow below the problem to make it easier to read. See if you can do the work and answer for each of the following problems.
When the numbers get larger, follow the same procedure. The following problem contains coefficients on the variables as well as larger numbers as the constants.
Here is one final reminder of the foil method. Scroll over the problem below to see how each letter in foil is applied in the multiplication.
## Try It
Multiply each pair of expressions:
1) (a + 3)(a + 7)
2) (2b + 5)(b – 5)
3) 2c (4c2 – 3c + 15)
4) (3d + 5)(-2d – 4)
5) (8e2 + 5)(e2 + 3)
1) a2 + 10a + 21
2) 2b2 – 5b – 25
3) 8c3 – 6c2 + 30c
4) -6d2 – 22d – 20
5) 8e4 + 29e2 + 15
Didn't find what you were looking for in this lesson? More information on polynomials can be found at the following places:
Resource Page
Related Lessons
Looking for something else? Try the buttons to the left or type your topic into the search box at the top of this page. |
# ANGLES AND THEIR MEASUREMENT
## About "Angles and their measurement"
Angles and their measurement :
An angle consists of two different rays that have the same initial point. The rays are the sides of the angles. The initial point is the vertex of the angle.
The angle that has sides AB and AC is denoted by ∠BAC, ∠CAB or ∠A. The point A is the vertex of the angle.
In this section, we are going to study two postulates about the measures of angles.
## Postulate 1 (Protractor Postulate)
Consider a point A on one side of OB. The rays of the form OA can be matched one to one with the real numbers from 0 to 180.
The measure of ∠AOB is equal to the absolute value of the difference between the real numbers for OA and OB.
## Postulate 2 (Angle Addition Postulate)
If P is in the interior of ∠RST, then we have
m∠RSP + m∠PST = m∠RST
## Classifying Angles
Angles are classified as acute, right, obtuse and straight, according to their measures. Angles have measures greater than 0°
## Angles and their measurement - Examples
Example 1 :
Name the angles in the figure given below.
Solution :
There are three different angles.
∠PQS or SQP
∠SQR or RQS
∠PQR or RQP
We should name any of the angles as ∠Q, because all three angles have Q as their vertex. The name ∠Q would not distinguish one angle from others.
Example 2 :
Each eye of a horse wearing blinkers has an angle of vision that measures 100°. The angle of vision that is seen by both eyes measures 60°.
Find the angle of vision seen by the left eye alone.
Solution :
We can use the angle addition postulate.
m∠2 + m∠3 = 100° (The total for left eye is 100°)
m∠3 = 100° - m∠2 (Subtract m∠2 from each side)
m∠3 = 100° - 60° (Substitute 60° for m∠2)
m∠3 = 40° (Subtract)
Hence, the vision for the left eye alone measures is 40°.
Example 3 :
Plot the points L (-4, 2), M(-1, -1), N (2, 2), Q(4, -1) and P(2, -4). Then, measure and classify the following angles as acute, right, obtuse or straight.
(i) m∠LMN
(ii) mLMP
(iii) mNMQ
(iv) mLMQ
Solution :
Plot the given points in xy coordinate plane.
We can use the protractor to measure and classify each angle as shown below.
MEASURE(i) m∠LMN = 90°(ii) m∠LMP = 180°(iii) m∠NMQ = 45°(iv) m∠LMQ = 135° CLASSIFICATIONRight angleStraight angleAcute angleObtuse angle
Note :
Two angles are adjacent angles, if they share a common vertex and side, but have no common interior points.
Example 4 :
Use a protractor to draw two adjacent acute angles ∠RSP and ∠PST so that ∠RST is
(a) Acute
(b) Obtuse
Solution :
Solution (a) :
Solution (b) :
After having gone through the stuff given above, we hope that the students would have understood "Angles and their measurement".
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## Inequality Symbols and Graphs
### Learning Outcomes
• Represent inequalities using an inequality symbol
• Represent inequalities on a number line
An inequality is a mathematical statement that compares two expressions using a phrase such as greater than or less than. Special symbols are used in these statements. In algebra, inequalities are used to describe sets of values, as opposed to single values, of a variable. Sometimes, several numbers will satisfy an inequality, but at other times infinitely many numbers may provide solutions. Rather than try to list a possibly infinitely large set of numbers, mathematicians have developed some efficient ways to describe such large lists.
## Inequality Symbols
One way to represent such a list of numbers, an inequality, is by using an inequality symbol:
• ${x}\lt{9}$ indicates the list of numbers that are less than $9$. Since this list is infinite, it would be impossilbe to list all numbers less than $9$.
• $-5\le{t}$ indicates all the numbers that are greater than or equal to $-5$.
If you were to read the above statement from left to right, it would translate as $-5$ is less than or equal to t. The direction of the symbol is dependent upon the statement you wish to make. For example, the following statements are equivalent. Both represent the list of all numbers less than 9. Note how the open end of the inequality symbol faces the larger value while the smaller, pointy end points to the smaller of the values:
• ${x}\lt{9}$
• ${9}\gt{x}$
Here’s another way of looking at is:
• $x\lt5$ means all the real numbers that are less than 5, whereas;
• $5\lt{x}$ means that 5 is less than x, or we could rewrite this with the x on the left: $x\gt{5}$. Note how the inequality is still pointing the same direction relative to x. This statement represents all the real numbers that are greater than 5 which is easier to interpret than 5 is less than x.
The box below shows the symbol, meaning, and an example for each inequality sign, as they would be translated reading from left to right.
Symbol Words Example
$\neq$ not equal to ${2}\neq{8}$, 2 is not equal to 8.
$\gt$ greater than ${5}\gt{1}$, 5 is greater than 1
$\lt$ less than ${2}\lt{11}$, 2 is less than 11
$\geq$ greater than or equal to ${4}\geq{ 4}$, 4 is greater than or equal to 4
$\leq$ less than or equal to ${7}\leq{9}$, 7 is less than or equal to 9
The inequality $x>y$ can also be written as ${y}<{x}$. The sides of any inequality can be switched as long as the inequality symbol between them is also reversed.
## Graphing an Inequality
Another way to represent an inequality is by graphing it on a number line:
Below are three examples of inequalities and their graphs. Graphs are often helpful for visualizing information.
$x\leq -4$. This translates to all the real numbers on a number line that are less than or equal to $4$.
${x}\geq{-3}$. This translates to all the real numbers on the number line that are greater than or equal to -3.
Each of these graphs begins with a circle—either an open or closed (shaded) circle. This point is often called the end point of the solution. A closed, or shaded, circle is used to represent the inequalities greater than or equal to $\displaystyle \left(\geq\right)$ or less than or equal to $\displaystyle \left(\leq\right)$. The end point is part of the solution. An open circle is used for greater than (>) or less than (<). The end point is not part of the solution. When the end point is not included in the solution, we often say we have strict inequality rather than inequality with equality.
The graph then extends endlessly in one direction. This is shown by a line with an arrow at the end. For example, notice that for the graph of $\displaystyle x\geq -3$ shown above, the end point is $−3$, represented with a closed circle since the inequality is greater than or equal to $−3$. The blue line is drawn to the right on the number line because the values in this area are greater than $−3$. The arrow at the end indicates that the solutions continue infinitely.
### Example
Graph the inequality $x\ge 4$
This video shows an example of how to draw the graph of an inequality.
### Example
Write an inequality describing all the real numbers on the number line that are strictly less than $2$. Then draw the corresponding graph. |
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# Selina solutions for Class 10 Mathematics chapter 13 - Section and Mid-Point Formula
## Chapter 13: Section and Mid-Point Formula
Ex. 13AEx. 13BEx. 13CEx. 13.CEx. 13.c
#### Chapter 13: Section and Mid-Point Formula Exercise 13A solutions [Page 0]
Calculate the co-ordinates of the point P which divides the line segment joining: A (1, 3) and B (5, 9) in the ratio 1 : 2
Calculate the co-ordinates of the point P which divides the line segment joining: A (-4, 6) and B(3, -5) in the ratio 3 : 2
In what ratio is the line joining (2, -3) and (5, 6) divided by the x-axis.
In what ratio is the line joining (2, -4) and (-3, 6) divided by the y – axis.
In what ratio does the point (1, a) divide the join of (-1, 4) and (4,-1)? Also, find the value of a.
In what ratio does the point (a, 6) divide the join of (-4, 3) and (2, 8)? Also, find the value of a.
In what ratio is the join of (4, 3) and (2, -6) divided by the x-axis. Also, find the co-ordinates of the point of intersection.
Find the ratio in which the join of (-4, 7) and (3, 0) is divided by the y-axis. Also, find the coordinates of the point of intersection.
Points A, B, C and D divide the line segment joining the point (5, -10) and the origin in five equal parts. Find the co-ordinates of B and D.
The line joining the points A (-3, -10) and B (-2, 6) is divided by the point P such that (PB)/(AB)=1/5 Find the co-ordinates of P.
P is a point on the line joining A(4, 3) and B(-2, 6) such that 5AP = 2BP. Find the co-ordinates of P.
Calculate the ratio in which the line joining the points (-3, -1) and (5, 7) is divided by the line x = 2. Also, find the co-ordinates of the point of intersection.
Calculate the ratio in which the line joining A(6, 5) and B(4, -3) is divided by the line y = 2.
The point P (5, -4) divides the line segment AB, as shown in the figure, in the ratio 2 : 5. Find the co-ordinates of points A and B.
Find the co-ordinates of the points of tri-section of the line joining the points (-3, 0) and (6, 6)
Show that the line segment joining the points (-5, 8) and (10, -4) is trisected by the co-ordinate axes.
Show that A (3, -2) is a point of trisection of the line segment joining the points (2, 1) and (5, -8).
Also, find the co-ordinates of the other point of trisection.
If A = (-4, 3) and B = (8, -6)
(i) Find the length of AB
(ii) In what ratio is the line joining A and B, divided by the x-axis?
The line segment joining the points M(5, 7) and N(-3, 2) is intersected by the y-axis at point L. Write down the abscissa of L. Hence, find the ratio in which L divides MN. Also, find the co-ordinates of L.
A (2, 5), B (-1, 2) and C (5, 8) are the co-ordinates of the vertices of the triangle ABC. Points P
and Q lie on AB and AC respectively,
Such that: AP : PB = AQ : QC = 1 : 2
(i) Calculate the co-ordinates of P and Q.
(ii) Show that PQ =1/3BC
A (-3, 4), B (3, -1) and C (-2, 4) are the vertices of a triangle ABC. Find the length of line segment AP, where point P lies inside BC, such that BP : PC = 2 : 3
The line segment joining A (2, 3) and B (6, -5) is intercepted by x-axis at the point K. Write down the ordinate of the point K. Hence, find the ratio in which K divides AB. Also find the coordinates of the point K.
The line segment joining A (4, 7) and B (-6, -2) is intercepted by the y – axis at the point K. write down the abscissa of the point K. hence, find the ratio in which K divides AB. Also, find the co-ordinates of the point K.
The line joining P(-4, 5) and Q(3, 2) intersects the y-axis at point R. PM and QN are perpendicular from P and Q on the x-axis Find:
(i) the ratio PR : RQ
(ii) the coordinates of R.
(iii) the area of the quadrilateral PMNQ.
In the given figure line APB meets the x-axis at point A and y-axis at point B. P is the point (-4,2) and AP : PB = 1 : 2. Find the co-ordinates of A and B.
Given a line segment AB joining the points A (-4, 6) and B (8, -3). Find:
(i) the ratio in which AB is divided by the y-axis
(ii) find the coordinates of the point of intersection
(iii) the length of AB.
#### Chapter 13: Section and Mid-Point Formula Exercise 13B solutions [Page 0]
Find the mid – point of the line segment joining the point: (-6, 7) and (3, 5)
Find the mid – point of the line segment joining the point: (5, -3) and (-1, 7)
Points A and B have co-ordinates (3, 5) and (x, y) respectively. The mid point of AB is (2, 3). Find the values of x and y.
A (5, 3), B(-1, 1) and C(7, -3) are the vertices of triangle ABC. If L is the mid-point of AB and M is the mid-point of AC, show that LM =1/2BC
Given M is the mid point of AB, find the co-ordinates of: A; if M = (1, 7) and B = (-5, 10)
Given M is the mid point of AB, find the co-ordinates of: B; if A = (3, -1) and M = (-1, 3)
P (-3, 2) is the mid-point of line segment AB as shown in the given figure. Find the co-ordinates of points A and B.
In the given figure, P (4, 2) is mid-point of line segment AB. Find the co-ordinates of A and B.
(-5, 2), (3, -6) and (7, 4) are the vertices of a triangle. Find the length of its median through the vertex (3, -6)
Given a line ABCD in which AB = BC = CD, B= (0, 3) and C = (1, 8)
Find the co-ordinates of A and D.
One end of the diameter of a circle is (-2, 5). Find the co-ordinates of the other end of it, of the centre of the circle is (2, -1)
A (2, 5), B (1, 0), C (-4, 3) and D (-3, 8) are the vertices of quadrilateral ABCD. Find the coordinates of the mid-points of AC and BD. Give a special name to the quadrilateral.
P (4, 2) and Q (-1, 5) are the vertices of parallelogram PQRS and (-3, 2) are the co-ordinates of the point of intersection of its diagonals. Find co-ordinates of R and S.
A (-1, 0), B (1, 3) and D (3, 5) are the vertices of a parallelogram ABCD. Find the co-ordinates of vertex C.
The points (2, -1), (-1, 4) and (-2, 2) are mid points of the sides of a triangle. Find its vertices
Points A (-5, x), B (y, 7) and C (1, -3) are collinear (i.e. lie on the same straight line) such that AB = BC. Calculate the values of x and y.
Points P (a, −4), Q (−2, b) and R (0, 2) are collinear. If Q lies between P and R, such that PR = 2QR, calculate the values of a and b.
Calculate the co-ordinates of the centroid of the triangle ABC, if A = (7, -2), B = (0, 1) and C =(-1, 4).
The co-ordinates of the centroid of a triangle PQR are (2, -5). If Q = (-6, 5) and R = (11, 8); calculate the co-ordinates of vertex P.
A (5, x), B (-4, 3) and C (y, -2) are the vertices of the triangle ABC whose centroid is the origin. Calculate the values of x and y.
#### Chapter 13: Section and Mid-Point Formula Exercise 13C, 13.C, 13.c solutions [Page 0]
Given a triangle ABC in which A = (4, -4), B = (0, 5) and C = (5, 10). A point P lies on BC such that BP : PC = 3 : 2, Find the length of line segment AP.
A(20, 0) and B(10, -20) are two fixed points Find the co-ordinates of the point P in AB such that : 3PB = AB, Also, find the co-ordinates of some other point Q in AB such that AB = 6 AQ.
A(-8, 0), B(0, 16) and C(0, 0) are the verticals of a triangle ABC. Point P lies on AB and Q lies on AC such that AP : PB = 3 : 5 and AQ : QC = 3 : 5
Show that : PQ =3/8 BC
Find the co-ordinates of points of trisection of the line segment joining the point (6, -9) and the origin.
A line segment joining A(-1,5/3) and B (a, 5) is divided in the ratio 1 : 3 at P, the point where the line segment AB intersects the y-axis.
(i) calculate the value of ‘a’
(ii) Calculate the co-ordinates of ‘P’.
In what ratio is the line joining A(0, 3) and B (4, -1) divided by the x-axis? Write the co-ordinates of the point where AB intersects the x-axis
The mid-point of the segment AB, as shown in diagram, is C(4, -3). Write down the co-ordinates of A and B.
AB is a diameter of a circle with centre C = (-2, 5). If A = (3, -7), find
(i) the length of radius AC
(ii) the coordinates of B.
Find the co-ordinates of the centroid of a triangle ABC whose vertices are: A(-1, 3), B(1, -1) and C(5, 1)
The mid point of the line segment joining (4a, 2b -3) and (-4, 3b) is (2, -2a). Find the values of a and b.
The mid point of the line segment joining (2a, 4) and (-2, 2b) is (1, 2a + 1). Find the values of a and b.
(i) write down the co-ordinates of the point P that divides the line joining A(-4, 1) and B(17, 10) in the ratio 1 : 2.
(ii) Calculate the distance OP, where O is the origin.
(iii) In what ratio does the Y-axis divide the line AB?
Prove that the points A(-5, 4); B(-1, -2) and C(5, 2) are the vertices of an isosceles right angled triangle. Find the co-ordinates of D so that ABCD is a square.
M is the mid-point of the line segment joining the points A(-3, 7) and B(9, -1). Find the coordinates of point M. Further, if R(2, 2) divides the line segment joining M and the origin in the ratio p : q, find the ratio p : q
Calculate the ratio in which the line joining A(-4,2) and B(3,6) is divided by point p(x,3). Also, find x
Find the ratio in which the line 2x+y=4 divides th line segment joining the point p(2,-2) and Q (3,7).
If the abscissa of point P is 2. find the ratio in which this point divides the line segment joining the point (-4,3) and (6,3). Also find the co-ordinates of point P.
The line joining the points (2,1) and (5,-8) is trisected at the point P and Q, point P lies on the line 2x-y+k=0, find the value of k Also, find the co-ordinates of point Q.
M is the mid-point of the line segment joining the points A(0,4) and B(6,0). M also divides the line segment OP in the ratio 1:3 find :
(a) co-ordinates of M
(b) Co-ordinates of P
(C) Length Of BP
Find the image of the point A(5,3) under reflection in the point P(-1,3).
A (-4,2)2, B(0,2) and C (-2,-4) are the vertices of a triangle ABC. A,Q and R are mid-Points of sides BC,CA and AB respectively.Show that the centroid of Δ PQR is the same as the centroid of Δ ABC.
## Chapter 13: Section and Mid-Point Formula
Ex. 13AEx. 13BEx. 13CEx. 13.CEx. 13.c
## Selina solutions for Class 10 Mathematics chapter 13 - Section and Mid-Point Formula
Selina solutions for Class 10 Maths chapter 13 (Section and Mid-Point Formula) include all questions with solution and detail explanation. This will clear students doubts about any question and improve application skills while preparing for board exams. The detailed, step-by-step solutions will help you understand the concepts better and clear your confusions, if any. Shaalaa.com has the CISCE Selina ICSE Concise Mathematics for Class 10 (2018-2019) solutions in a manner that help students grasp basic concepts better and faster.
Further, we at shaalaa.com are providing such solutions so that students can prepare for written exams. Selina textbook solutions can be a core help for self-study and acts as a perfect self-help guidance for students.
Concepts covered in Class 10 Mathematics chapter 13 Section and Mid-Point Formula are Mid-point Formula, Section Formula, Distance Formula, Co-ordinates Expressed as (x,y).
Using Selina Class 10 solutions Section and Mid-Point Formula exercise by students are an easy way to prepare for the exams, as they involve solutions arranged chapter-wise also page wise. The questions involved in Selina Solutions are important questions that can be asked in the final exam. Maximum students of CISCE Class 10 prefer Selina Textbook Solutions to score more in exam.
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# y=-2x+6
Understanding the Equation y = -2x + 6
Introduction: The equation y = -2x + 6 represents a linear relationship between the variables x and y. In this answer, I will provide a clear explanation of this equation and its significance. Whether you’re a beginner or have little knowledge of this topic, I’ll break it down into simple terms for easy understanding.
Explanation: Here’s what you need to know about the equation y = -2x + 6:
1. The equation is in slope-intercept form: y = mx + b. In this form, ‘m’ represents the slope of the line, and ‘b’ represents the y-intercept.
• In our equation, the slope ‘m’ is -2, indicating that for every increase of 1 unit in x, y decreases by 2 units.
• The y-intercept ‘b’ is 6, which means the line crosses the y-axis at the point (0, 6).
2. Graphical representation:
• To visualize the equation, you can plot its points on a coordinate plane.
• Start by plotting the y-intercept (0, 6). This point indicates that when x is 0, y equals 6.
• From the y-intercept, use the slope to find additional points. For example:
• When x = 1, y = -2(1) + 6 = 4. Plot the point (1, 4).
• When x = -1, y = -2(-1) + 6 = 8. Plot the point (-1, 8).
• Connect these points with a straight line. The resulting line represents the equation y = -2x + 6.
3. Meaning of the equation:
• The equation represents a linear relationship where y decreases by 2 units for every 1 unit increase in x.
• The negative slope indicates a downward trend, meaning as x increases, y decreases.
• The y-intercept at (0, 6) shows that when x is 0, y is 6. This point represents the initial value of y when x is not present.
Example applications: Understanding this equation can be helpful in various scenarios. Here are a few examples:
1. Cost and revenue: Suppose you run a business where the cost of production increases by \$2 for every additional unit produced (x). The equation y = -2x + 6 can represent your revenue (y) as a function of the number of units produced (x).
2. Temperature and time: Imagine you have a cooling process where the temperature (y) decreases by 2 degrees Celsius for every hour (x) that passes. The equation y = -2x + 6 can describe the relationship between time and temperature during this process.
3. Sports and performance: If you’re analyzing an athlete’s performance, the equation can be used to measure improvement. For instance, y could represent the number of goals scored in a game, while x represents the training time. The equation y = -2x + 6 would then depict the expected number of goals based on the time spent training.
Conclusion: In summary, the equation y = -2x + 6 represents a linear relationship with a negative slope. By understanding its slope-intercept form, graphical representation, and real-world applications, you can gain insights into how the variables x and y are related. Remember, this equation is just one example, and there are countless other equations that describe different relationships between variables.
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# How do you differentiate y = (3x - 5) / (3x^2 + 7)?
Jan 5, 2016
Use the quotient rule.
$f ' \left(x\right) = \frac{- 9 {x}^{2} + 30 x + 21}{3 {x}^{2} + 7} ^ 2$
#### Explanation:
Quotient rule
For function
$f \left(x\right) = \frac{u}{v}$
$f ' \left(x\right) = \frac{v u ' - u v '}{v} ^ 2$
So:
For $f \left(x\right) = \frac{3 x - 5}{3 {x}^{2} + 7}$
let
$u = 3 x - 5$
so
$u ' = 3$
let
$v = 3 {x}^{2} + 7$
so
$v ' = 6 x$
$f ' \left(x\right) = \frac{\left(3 {x}^{2} + 7\right) \cdot 3 - \left(3 x - 5\right) \cdot 6 x}{3 {x}^{2} + 7} ^ 2$
Expand and simplify:
$f ' \left(x\right) = \frac{9 {x}^{2} + 21 - 18 {x}^{2} + 30 x}{3 {x}^{2} + 7} ^ 2$
$f ' \left(x\right) = \frac{- 9 {x}^{2} + 30 x + 21}{3 {x}^{2} + 7} ^ 2$ |
# Absolute Value Functions
The absolute value function is explored through definitions, examples and exercises with solutions at the bottom of the page. The graphing of absolute value functions |f(x)| is discussed through examples.
## Definition of Absolute Value
Consider the number line and a point at a distance x from the origin zero of the number line.
The absolute value of x written as |x| is the distance from zero to x. Since |x| gives a distance, it is always positive or equal to zero. Hence
| -5 | = 5
| 5 | = 5
| 0 | = 0
|a - b| is the distance between points at a and b on the number line.
|5 - 15| = | - 10| = 10
definition
$|x| = \left\{ \begin{array}{ll} x & x\geq 0 \\ -x & x\lt 0 \\ \end{array} \right.$
which can be used to write
$|x - a| = \left\{ \begin{array}{ll} x - a & x\geq a \\ - (x - a) = a - x & x\lt a \\ \end{array} \right.$
## Absolute Value and Square Root
$$\sqrt { x^2} = | x |$$ ; because the square root is positive or equal zero.
$$\sqrt {(x - a)^2} = |x - a|$$
## Properties of the Absolute Value Function
1. $$| x | \ge 0$$
2. $$\sqrt{ x^2} = | x |$$
3. $$| x | = | - x |$$
4. $$| a b | = | a || b |$$
5. $$\left | \dfrac{a}{b} \right | = \dfrac{|a|}{|b|} , b \ne 0$$
## Equations and Inequalities with Absolute Value
For $$k \ge 0$$
1. $$| x | = k \iff x = k \text{ or } x = - k$$
2. $$| x | \le k \iff - k \le x \le k$$
3. $$| x | ? k \iff x \le - k \text{ or } x \ge k$$
## Graph of Absolute Value of a Function
Function f(x) used is a quadratic function of the form
$f(x) = a x^2 + b x + c$ The exploration is carried out by changing the parameters a, b and c included in $$f(x)$$ above.
Interactive Tutorial
a = 1 -10+10 b = 0 -10+10 c = 1 -10+10
>
1. click on the button above "draw" to start.
2. Use the sliders to set parameter a to zero, parameter b to zero and parameter c to a positive value; $$f(x)$$ is a constant function. Compare the graph of f(x) in blue and that of $$h(x)= |f(x)|$$ in red. Change c to a negative value and compare the graphs again. Use the definition of the absolute value functions to explain how can the graph of $$|f(x)|$$ be obtained from the graph of $$f(x)$$.
3. Keep the value of a equal to zero, select non zero values for b to obtain a linear function . How can the graph of $$h(x) = |f(x)|$$ be obtained from that of $$f(x)$$?
Hint: use the definition of the absolute value functions and reflection of a graph on the x-axis.
4. Set b and c to zero and select a positive value for a to obtain a quadratic function . Why are the two graphs the same? (Hint: use the definition of the absolute value functions).
5. Set b and c to zero and select a negative value for a to obtain a quadratic function . Why are the two graphs reflection of each other? (Hint: use the definition of the absolute value functions and reflection of a graph on the x-axis).
6. Keep the values of a and b as in 5 above and change gradually c from zero to some positive values. How can the graph of $$h(x)=|f(x)|$$ be obtained from that of $$f(x)$$?
7. Select different values for a, b and c and explore.
## Exercises
Part 1
Evaluate the expressions
1. | - 9 |
2. | 9 |
3. | 2 - 6 |
4. $$|- \dfrac{5}{2}|$$
5. $$| 11 - 4^2 |$$
6. $$2^{| 2 - 7|}$$
7. $$|2-19| - |9 - 20|$$
Part 2
Substitute and Evaluate the expressions
1. | x - 10 | + | 2 x | , x = - 7
2. | - x | , x = 8
3. $$\left |- \dfrac{|x - 2|}{|-x + 9|} \right |$$ , x = -20
4. $$| - 3 x - x^2 |$$ , x = -3
5. $$x^{| x - 7|}$$ , x = - 2
Part 3
Simplify the expressions
1. | a - b | for a = b
2. | a - b | for a > b
3. | a - b | for b > a
4. $$\sqrt {3^2}$$
5. $$\sqrt {(- 3)^2}$$
6. $$\sqrt { (12 - 20)^2 }$$
7. $$| x^2 |$$
8. $$| (-x)^2 |$$
9. $$| (x + 1)^2 |$$
10. $$\sqrt{(x - y + z)^2}$$
11. $$\sqrt{sin^2(x)}$$
Part 4
Solve the Equations
1. $$|x - 2| = 4$$
2. $$|x + 5| = 0$$
3. $$|10 - x| = - 5$$
Part 5
Solve the Inequalities
1. $$|x - 7| \gt 4$$
2. $$|x + 5| \lt 9$$
3. $$|x + 8| \lt - 5$$
4. $$|x + 8| \gt - 2$$
Part 6
Sketch the following pairs of functions in the same system of axes and explains the similarities and differences of the two graphs.
1. $$f(x) = x - 1 , h(x) = |f(x)|$$
2. $$f(x) = x^2 - 4 , h(x) = |f(x)|$$
3. $$f(x) = \sin(x) , h(x) = |\sin(x)|$$
Solutions to the Above Exercises
Part 1
Evaluate the expressions
1. | - 9 | = 9
2. | 9 | = 9
3. | 2 - 6 | = 4
4. $$|- \dfrac{5}{2}| = \dfrac{5}{2}$$
5. $$| 11 - 4^2 | = 5$$
6. $$2^{| 2 - 7|} = 32$$
7. $$|2-19| - |9 - 20| = 6$$
Part 2
Substitute and Evaluate the expressions
1. For x = - 7, | x - 10 | + | 2 x | = 31
2. For x = 8 , | - x | = 8
3. For x = - 20 , $$\left |- \dfrac{|x - 2|}{|-x + 9|} \right | = \dfrac{22}{29}$$
4. For x = - 3 , $$| - 3 x - x^2 | = 0$$
5. For x = - 2 , $$x^{| x - 7|} = - 512$$
Part 3
Simplify the expressions
1. | a - b | = 0
2. | a - b | = a - b
3. | a - b | = b - a
4. $$\sqrt {3^2} = |3| = 3$$
5. $$\sqrt {(- 3)^2} = |-3| = 3$$
6. $$\sqrt { (12 - 20)^2 = 8}$$
7. $$| x^2 | = x^2$$
8. $$| (-x)^2 | = x^2$$
9. $$| (x + 1)^2 | = (x + 1)^2$$
10. $$\sqrt{(x - y + z)^2} = |x - y + z|$$
11. $$\sqrt{sin^2(x)} = |sin(x)|$$
Part 4
Solve the Equations
1. $$|x - 2| = 4$$ , two solutions: x = - 2 and x = 6
2. $$|x + 5| = 0$$ , one solution: x = - 5
3. $$|10 - x| = - 5$$ , no real solutions.
Part 5
Solve the Inequalities
1. $$|x - 7| \gt 4$$ , solution set: x < 3 ? x > 11
2. $$|x + 5| \lt 9$$ , solution set: -14 < x < 4
3. $$|x + 8| \lt - 5$$ , solution set: {?}
4. $$|x + 8| \gt - 2$$ , solution set: {all real numbers}
Part 6
1. $$f(x) = x - 1 , h(x) = |f(x)| = |x - 1|$$
For x - 1 ? 0 or x ? 1, h(x) = | x - 1 | = x - 1 and we therefore have h(x) = f(x) which we can be clearly seen from the graphs.
For x - 1 < 0 or x < 1 , h(x) = |x - 1| = - (x - 1) = - f(x).
Therefore for x < 1 h(x) = - f(x) which means that the graph of h is a reflection of the graph of f on the x axis and this is can be clearly seen on the graphs.
2. $$f(x) = x^2 - 4 , h(x) = |f(x)| = |x^2 - 4|$$
For x2 - 4 ? 0 or -2 ? x ? 2 , h(x) = |(x2 - 4)| = - (x2 - 4) = - f(x) . Therefore, on the interval -2 ? x ? 2, the graph of h is a reflection of the graph of f. Outside this interval, the two graphs coincide.
3. $$f(x) = \sin(x) , h(x) = |\sin(x)|$$
Over intervals -2? to -? , 0 to ? , 2? to 3?, ... sin(x) ? 0 and h(x) = sin(x) = f(x) therefore the graphs of h and f coincide.
For intervals - ? to 0 , ? to 2? ... sin(x) < 0 , h(x) = | sin(x) | = - sin(x) and the graph of h is a reflection of the graph of f on the x axis.
Conclusion: In all three exercises above, the graph of h is on or above the x axis because h(x) = |f(x)| and the absolute value is positive or equal to 0. |
Numbers, units and arithmetic
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# 3.21 Division rules
Lastly consider division. Dividing 8 by 2 means ‘How many times does 2 go into 8?’ or ‘What must you multiply 2 by to get 8?’. The answer is 4.
So to find 8 ÷ 2, you need to ask ‘What do I have to multiply 2 by to get 8?’. The answer is 4, since 2 × 4 = 8.
So 8 ÷ 2 = 4.
Similarly, to find 8 ÷ 2 you need to ask ‘what do I have to multiply 2 by to get 8?’ and the answer is 4, since 2 × 4 = 8.
So 8 ÷ 2 = 4.
Using this sort of argument you can work out the rules for division of and by negative numbers. They are remarkably similar to those for multiplication (as you might expect, since they are inverse processes).
Multiplying or dividing a positive number by a positive number gives a positive answer.
Multiplying or dividing a negative number by a positive number gives a negative answer.
Multiplying or dividing a positive number by a negative number gives a negative answer.
Multiplying or dividing a negative number by a negative number gives a positive answer.
The last one is the most difficult to remember. If you are in doubt, you can always use your calculator to check.
These rules may be summarised as follows.
Multiplying or dividing two numbers of the same sign gives a positive answer.
Multiplying or dividing two numbers of different signs gives a negative answer.
## Try some yourself
### Activity 62
Give examples of all the rules for the division of and by positive and negative numbers.
Your list should look something like this.
Dividing a positive number by a positive number gives a positive answer.
4 ÷ 2 = 2.
Dividing a negative number by a positive number gives a negative answer.
4 ÷ 2 = 2.
Dividing a positive number by a negative number gives a negative answer.
4 ÷ 2 = 2.
Dividing a negative number by a negative number gives a positive answer.
4 ÷ 2 = 2.
### Activity 63
Evaluate each of the following and where possible give an example from everyday life to illustrate the calculation.
• (a) 24 ÷ 4
• (b) 40 ÷ 4
• (c) 45 ÷ 15
• (a) 24 ÷ 4 = 6
• Thomas has lots of £4 IOUs. A gift of £24 will subtract 6 of them.
• (b) 40 ÷ 4 = 10
• An IOU of £40 divided into smaller IOUs of £4 each (which could be paid off more easily from Thomas's pocket money) gives 10 smaller IOUs.
• (c) 45 ÷ 15 = 3
• A debt of £45 shared between 15 people means each owes £3.
### Activity 64
Evaluate each of the following.
• (a) 7 × 6
• (b) 56 ÷ 7
Think of a financial context where each might be an appropriate calculation (bear in mind that negative numbers can represent debts).
• (a) 7 × 6 = 42
• (7 debts of £6 result in a debt of £42).
• (b) 56 ÷ 7 = 8
• (a £56 debt divided into £7 debts gives eight £7 debts).
If you have a calculator check that it gives the same answers as yours.
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# Construct the Following Quadrilaterals. Quadrilateral Abcd Ab = 4.5 Cm, Bc = 5.5 Cm - Mathematics
Course
ConceptConstructing a Quadrilateral When the Lengths of Four Sides and a Diagonal Are Given
AB = 4.5 cm
BC = 5.5 cm
CD = 4 cm
AC = 7 cm
#### Solution
Firstly, a rough sketch of this quadrilateral can be drawn as follows.
1) ΔABC can be constructed by using the given measurements as follows.
2) Vertex D is 6 cm away from vertex A. Therefore, while taking A as centre, draw an arc of radius 6 cm.
3) Taking C as centre, draw an arc of radius 4 cm, cutting the previous arc at point D. Join D to A and C.
Is there an error in this question or solution?
#### APPEARS IN
NCERT Solution for Mathematics Textbook for Class 8 (2018 to Current)
Chapter 4: Practical Geometry
Ex. 4.1 | Q: 1.1 | Page no. 60
#### Video TutorialsVIEW ALL [1]
Solution Construct the Following Quadrilaterals. Quadrilateral Abcd Ab = 4.5 Cm, Bc = 5.5 Cm Concept: Constructing a Quadrilateral - When the Lengths of Four Sides and a Diagonal Are Given.
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Stage 2
# Ordering and counting with fractions
Lesson
You may have already learned about representing fractions as areas of shapes.
When representing fractions the:
• denominator (the bottom number) represents how many equal parts make the whole
• numerator (the top number) represents how many equal parts are selected.
Fractions can be written in:
• words, e.g. six eighths
• pictures, e.g.
• symbols, e.g. $\frac{6}{8}$68
Watch this video to learn about counting and ordering fractions in words, symbols and pictures.
Try these questions for yourself.
#### Worked examples
##### Question 1
Danielle was counting fractions in the eighths.
Fill in the gaps for her.
1. $\frac{0}{8}$08 $\frac{\editable{}}{8}$8 $\frac{2}{8}$28
$\frac{3}{\editable{}}$3 $\frac{4}{8}$48 $\frac{5}{\editable{}}$5
$\frac{6}{8}$68 $\frac{\editable{}}{8}$8 $\frac{8}{8}$88
##### Question 2
Order these fractions from smallest to largest.
1. $\frac{1}{5}$15, , $2$2 fifths
$\frac{1}{5}$15, $2$2 fifths,
A
, $2$2 fifths, $\frac{1}{5}$15
B
$2$2 fifths, , $\frac{1}{5}$15
C
## Improper fractions and mixed fractions
Improper fractions have a numerator that is greater than or equal to the denominator, for example, $\frac{8}{5}$85.
Mixed fractions have a whole number and a fraction part, for example, $4$4$\frac{5}{6}$56.
Watch this video to learn about improper and mixed fractions.
Try these questions for yourself.
#### Worked examples
##### Question 3
Fill in the gaps in table by completing the conversions.
1. Mixed Number Improper Fraction
$1\frac{2}{3}$123 $\frac{5}{3}$53
$\editable{}$$\frac{\editable{}}{\editable{}} \frac{3}{2}32 1\frac{3}{4}134 \frac{\editable{}}{\editable{}} 2\frac{2}{3}223 \frac{\editable{}}{\editable{}} \editable{}$$\frac{\editable{}}{\editable{}}$
$\frac{9}{4}$94
##### Question 4
Kenneth has two whole pies.
He cuts each pie into $6$6 slices.
1. How many slices of pie does Kenneth have in total?
He has $\editable{}$ slices.
2. Write a fraction representing one slice of one pie.
One slice is $\frac{\editable{}}{\editable{}}$ of a whole pie.
3. Write an improper fraction which represents one whole pie.
A whole pie is $\frac{\editable{}}{\editable{}}$ of a whole pie.
4. Fill in the gap in the table below.
One Slice One Pie One Pie + $\editable{}$ slice
$\frac{1}{6}$16 $\frac{6}{6}$66 $\frac{7}{6}$ |
# Measuring Human Rights: High School Mathematics Unit
### Learning Objectives
Students will be able to:
• Compare and Contrast Three Methods of Representing Data
• Represent data appropriately using frequency tables, histograms, and box plots.
• Draw conclusions from the data displayed.
• Analyze data provided in various plot forms and draw conclusions from the data displayed
• Identify similarities and differences in shape, center and spread of various data sets across and within specific representations.
Math Content Standards
• S.ID.1. Represent data with plots on the real number line (dot plots, histograms, and box plots).
• S.ID.2. Use statistics appropriate to the shape of the data distribution to compare center (median, mean) and spread (interquartile range, Standard deviation) for two or more data sets.
Math Practice Standards
• MP.1 Make sense of problems and persevere in solving them.
• MP.4 Model with mathematics.
• MP.5 Use appropriate tools strategically.
## Instructional Approach
### Teacher Introduction:
• Ask students to imagine a class of 400 students doing the same exercise (of putting up post its on the number line). What do they think would happen? (Total chaos)
• Remind students that one way to organize this data is by creating “bins” (also known as ‘class’). In our case, we will take a range and divide it into equal intervals. In our case, we will divide the number line into 6 equal intervals (in millions) 1-25, 26-50, 51-75, 76-100, 101-125, 126-150.
• Distribute Handout #3
• Project (or draw) a frequency distribution table with 5 population bins (classes) on the board and ask students to raise their hands if their estimate falls within the first bin (class). Make a tally mark for each student who raises his or her hand. Repeat this process for the rest of the ranges. Ask students to copy the results on the corresponding table in Handout #3.
Handout #3 With Example (Red)
Handout #3 With Example (Red and Green)
### Individual Work
Begin filling the cumulative column on the board and ask students to complete the cumulative frequency column in Handout #3.
• Once students complete the table ask them to individually complete the questions in the handouts (15 minutes), and then review answers as a class before going to next process (10 minutes).
1. Use the table and create a histogram that matches the frequency table. (If needed, give students a refresher handout on how to create a histogram)
2. Which of these representation(s) (e.g. frequency table, histogram, and box plot) shows the most popular estimation in the class? Why?
3. Which representation do you think shows students’ estimations most accurately? Why?
4. Optional if time permits: Can you make up data for another class, which will be different from your class’s data set but result in the same box plot and histogram? If your answer is ‘yes’, show the data set. If your answer is ‘no’ explain why.
### Small Group Work
• Divide students into pairs or trios, hand out Class Mix Up cards to each group
• Tell students that four (optional: six) more classes participated in this activity. Each class produced a card with a histogram and a card with a box plot that represent their classroom data. Unfortunately you mixed up the cards. The group’s task is to match each histogram card with the appropriate box plot card. Tell students that each group will be asked to justify their matching choices to the rest of the class. (Give each group set of 8 cards containing 4 box plots and 4 histograms, or a set of 12 cards with 6 histograms and 6 box plots)
• Once all group members agree on the 4 matches (pairs), ask them to glue matching solutions on a poster.
• Post and conduct a gallery walk and ask students to look for commonalities and differences across the posters.
• Once the class reconvenes, using a projector, present the histograms and box plots and ask volunteers from various groups to explain why they matched their cards as they did.
• Encourage the rest of the students to respectfully challenge and counter answers.
• Ask students if they can tell the number of students for each match. Do they have enough information? How do you know?
Handout #4
Class Mix Up Cards: Match Box Plots with Histograms
### Teacher Explanation
According to the World Health Organization 15.7% of all children who are under the age of 5 in the world are underweight. As we discussed in the previous lesson, in 2012 it was estimated there were 642 million children in the world under the age of 5. Given this number, ask students to calculate the number of children under 5 in the world that are underweight. (Approximately 101 million)
### Conclusion
Conclude the class with discussion of the following questions:
• Which representation would you use to determine the number of individual students who had the closest estimation? (box plot, frequency table, or histogram)
• How does our class’ estimation compare the other (6) classes? Which class came the closest (within the bin) to the actual number of children <5 in the world who are underweight (101 million)?
Homework: The focus of the next class is on the first indicator: Prevalence of underweight and stunted children under five years of age. Ask students to think about 3 ways to determine if children are underweight.
### Differentiation and Supports
• Struggling: During small group work students can work with the first 3 pairs of cards only.
• Struggling: During the matching card game, direct students to focus on one aspect at a time. Begin with students figuring out how many students are in each class.
• Advanced: Can you make up data for another class, which will be different from your class’s data set but result in the same box plot and histogram? If your answer is ‘yes’, show the data set. If your answer is ‘no’ explain why.
• Advanced: Create additional set of cards for the card game and share it with class.
Supports
• While students engage in the group task, teacher can work with a small group of students on the same task. Explicitly demonstrating how he/she goes about marching the cards.
Note: The lesson is dealing with the notion of body weight (e.g. prevalence of underweight in children and BMI in adults). Be aware of many students who may have low self-esteem due to body weight. In the first part of the lesson we are focusing on children under five years of age (the UN indicator). In the second part we are looking at adults BMI and the focus is on BMI<18.5 as a function of underweight. The notion of BMI has several limitations, which we will discuss in the lesson. In addition, both, the notion of standard weight for children and the classification for BMI are useful as indicators for evaluating populations. NOT individuals.
### Assessment
Handout #3 will provide assessment data on student learning.
Observations of Class Mix Up Activity. |
Exponential Growth, its properties, how graph relates to the equation and formula--Visual Lesson | Math Warehouse
# Exponential Growth Equations and Graphs
Formula and graph for exponential growth equations
### How Equation relates to Graph
Graph 1
In the general example shown in Graph 1, 'a' stands for the initial amount, and 'b' is any real number that is greater than 1.
The graph's asymptote at x -axis
What about when b is exactly equal to 1?
In this case, you are not dealing with an exponential equation, but rather a linear equation. (Link) $y = ab^x \\ y = a(1)^x \\ y = a$
As you can see from the work above, and the graph, when b is 1, you end up with the equation of a horizontal line ( Link ) .
#### How does this compare to other graphs/functions?
As the graph below shows, exponential growth
• at first, has a lower rate of growth than the linear equation f(x) =50x
• at first, has a slower rate of growth than a cubic function like f(x) = x3, but eventually the growth rate of an exponential function f(x) = 2 x , increases more and more -- until the exponential growth function has the greatest value and rate of growth!
Moral of the story: Exponential growth eventually grows at massive rates, even though it starts out growing slowly. This 'trend' is true and, when compared to other graphs, exponential growth eventually outpaces most other functions's rate of increase.
### Role of a in $$y=ab^x$$
For $$y = \color{red}{a} \cdot b^x$$, a determines the y-intercept.
### Some Examples
Three different exponential growth functions are graphed in the diagram below.
The graph on the left helps show the role of 'b' in the $$y = a \cdot b^x$$
Namely, the greater the value of b
• the 'steeper' the curve looks
• the greater the rate of growth
#### What would you do?
Take the Poll! (and think about exponential growth)
(This poll has its own page)
Option 1
You can have $1000 a year for twenty years Option 2 You can get$1 for the first year, $2 for the second,$4 for the 3rd, doubling the amount very year -- for twenty years.
### Closer Look at Graph/Equation
Below is a picture of the most commonly studied example of exponential growth, the equation $$y = 2^x$$ |
1. If three fair coins are tossed randomly 175 times and it is found that three heads appeared 21 times, two heads appeared 56 times, one head appeared 63 times and zero head appeared 35 times. What is the probability of getting?
```Answer
Answer: Option C Explanation:Total number of trials = 175.
Number of times three heads appeared = 21.
Number of times two heads appeared = 56.
Number of times one head appeared = 63.
Number of times zero head appeared = 35.
Number of times three heads appeared
= P(E1) = total number of trials
= 21/175
= 0.12
Number of times two heads appeared
= P(E2) = total number of trials
= 56/175
= 0.32
Number of times one head appeared
= P(E3) = total number of trials
= 63/175
= 0.36
Number of times zero head appeared
= P(E4) = total number of trials
= 35/175
= 0.20
Note: Remember when 3 coins are tossed randomly, the only possible outcomes
are E2, E3, E4 and
P(E1) + P(E2) + P(E3) + P(E4)
= (0.12 + 0.32 + 0.36 + 0.20)
= 1 ```
Report
2. Two dice are rolled, find the probability that the sum is?
equal to 1
equal to 4
less than 13
None of these
```Answer
Answer: Option A Explanation: a) The sample space S of two dice is shown below.
S = { (1,1),(1,2),(1,3),(1,4),(1,5),(1,6)
(2,1),(2,2),(2,3),(2,4),(2,5),(2,6)
(3,1),(3,2),(3,3),(3,4),(3,5),(3,6)
(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)
(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)
(6,1),(6,2),(6,3),(6,4),(6,5),(6,6) }
Let E be the event "sum equal to 1". There are no outcomes which correspond to a sum equal to 1, hence
P(E) = n(E) / n(S) = 0 / 36 = 0
b) Three possible outcomes give a sum equal to 4: E = {(1,3),(2,2),(3,1)}, hence.
P(E) = n(E) / n(S) = 3 / 36 = 1 / 12
c) All possible outcomes, E = S, give a sum less than 13, hence.
P(E) = n(E) / n(S) = 36 / 36 = 1
```
Report
3. Which of these numbers cannot be a probability?
a) -0.00001
0.5
1.001
1
```Answer
Answer: Option D Explanation:A probability is always greater than or equal to 0 and less than or equal to 1, hence only a) and c) above cannot represent probabilities: -0.00010 is less than 0 and 1.001 is greater than 1.
```
Report
4. In how many different ways the word FAVOUR can be arranged?
820
530
560
720
```Answer
Answer: Option D Explanation:6! = 6 * 5 * 4 * 3 * 2 * 1 = 720 ```
Report
5. The probability of two persons of passing the interview are 1/3 and 3/5. Then calculate the probability that only one of them pass the interview?
7/15
8/15
11/15
13/15
```Answer
Answer: Option B Explanation:PF + FP = (P = pass, F = fail)
1/3 * 2/5 + 2/3*3/5 = 8/15 ```
Report |
# Multiplication Table Up To 15 Printable
Learning multiplication right after counting, addition, as well as subtraction is good. Young children learn arithmetic using a natural progression. This progression of studying arithmetic is generally the subsequent: counting, addition, subtraction, multiplication, and ultimately division. This declaration results in the question why discover arithmetic in this particular series? Furthermore, why find out multiplication following counting, addition, and subtraction but before department?
## These details respond to these inquiries:
1. Youngsters learn counting very first by associating visible items making use of their fingers. A perceptible illustration: Just how many apples are available inside the basket? More abstract illustration is when aged are you?
2. From counting phone numbers, the following reasonable stage is addition followed by subtraction. Addition and subtraction tables can be quite helpful training aids for children because they are graphic instruments creating the transition from counting much easier.
3. Which will be acquired up coming, multiplication or division? Multiplication is shorthand for addition. At this moment, young children possess a company understanding of addition. Consequently, multiplication is the following plausible method of arithmetic to discover.
## Assess basics of multiplication. Also, assess the essentials the way you use a multiplication table.
Let us review a multiplication case in point. Using a Multiplication Table, grow four times about three and obtain an answer a dozen: 4 x 3 = 12. The intersection of row three and line 4 of any Multiplication Table is 12; 12 may be the answer. For kids starting to find out multiplication, this can be effortless. They can use addition to resolve the trouble therefore affirming that multiplication is shorthand for addition. Illustration: 4 by 3 = 4 4 4 = 12. It is really an superb overview of the Multiplication Table. The added gain, the Multiplication Table is graphic and reflects to understanding addition.
## In which can we start discovering multiplication using the Multiplication Table?
1. Very first, get knowledgeable about the table.
2. Get started with multiplying by a single. Start off at row number one. Move to line number 1. The intersection of row a single and column the first is the solution: one.
3. Repeat these steps for multiplying by 1. Grow row a single by columns one through twelve. The responses are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12 respectively.
4. Recurring these techniques for multiplying by two. Grow row two by posts one by means of several. The solutions are 2, 4, 6, 8, and 10 respectively.
5. Let us leap in advance. Replicate these techniques for multiplying by five. Multiply row several by columns 1 by means of 12. The replies are 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, and 60 correspondingly.
6. Now let us improve the degree of problems. Repeat these techniques for multiplying by three. Multiply row about three by posts one particular by way of a dozen. The responses are 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, and 36 correspondingly.
7. In case you are confident with multiplication so far, consider using a test. Remedy the next multiplication difficulties in your mind and after that assess your answers to the Multiplication Table: flourish 6 and two, increase nine and a few, multiply 1 and 11, increase 4 and four, and increase several as well as 2. The situation responses are 12, 27, 11, 16, and 14 correspondingly.
Should you obtained several out of 5 difficulties correct, create your personal multiplication checks. Estimate the responses in your thoughts, and check them while using Multiplication Table. |
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