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# Divide a Number into Three Parts in a Given Ratio
To divide a number into three parts in a given ratio
Let the number be p. It is to be divided into three parts in the ratio a : b : c.
Let the parts be x, y and z. Then, x + y + z = p .................... (i)
and x = ak, y =bk, z = ck.................... (ii)
Substituting in (i), ak + bk + ck = p
⟹ k(a + b + c) = p
Therefore, k = $$\frac{p}{a + b + c}$$
Therefore, x = ak = $$\frac{ap}{a+ b + c}$$, y = bk = $$\frac{bp}{a+ b + c}$$, z = ck = $$\frac{cp}{a+ b + c}$$.
The three parts of p in the ratio a : b : c are
$$\frac{ap}{a+ b + c}$$, $$\frac{bp}{a+ b + c}$$, $$\frac{cp}{a+ b + c}$$.
Solved examples on dividing a number into three parts in a given ratio:
1. Divide 297 into three parts that are in the ratio 5 : 13 : 15
Solution:
The three parts are $$\frac{5}{5 + 13 + 15}$$ ∙ 297, $$\frac{13}{5 + 13 + 15}$$ ∙ 297 and $$\frac{15}{5 + 13 + 15}$$ ∙ 297
i.e., $$\frac{5}{33}$$ ∙ 297, $$\frac{13}{33}$$ ∙ 297 and $$\frac{15}{33}$$ ∙ 297 i.e., 45, 117 and 135.
2. Divide 432 into three parts that are in the ratio 1 : 2 : 3
Solution:
The three parts are $$\frac{1}{1 + 2 + 3}$$ ∙ 432, $$\frac{2}{1 + 2 + 3}$$ ∙ 432 and $$\frac{3}{1 + 2 + 3}$$ ∙ 432
i.e., $$\frac{1}{6}$$ ∙ 432, $$\frac{2}{6}$$ ∙ 432 and $$\frac{3}{6}$$ ∙ 432
i.e., 72, 144 and 216.
3. Divide 80 into three parts that are in the ratio 1 : 3 : 4.
Solution:
The three parts are $$\frac{1}{1 + 3 + 4}$$ ∙ 80, $$\frac{3}{1 + 3 + 4}$$ ∙ 80 and $$\frac{4}{1 + 3 + 4}$$ ∙ 80
i.e., $$\frac{1}{8}$$ ∙ 80, $$\frac{3}{8}$$ ∙ 80 and $$\frac{4}{8}$$ ∙ 80
i.e., 10, 30 and 40.
● Ratio and proportion |
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Speed and distance problems are among the most complained about problems on the GMAT. Numerous clients come to us and say they have difficulty with speed and distance problems, word problems, or work rate problems. So we’re going to look at a particularly difficult one and see just how easy it can be with the right approach.
## 1. The Two Cars Problem
Two cars are travelling in the same direction. Car A is travelling at a speed of 58 miles per hour and car B is travelling at 50 miles per hour. If car A is 20 miles behind car B, how long will it take car A to pass car B by 8 miles?
A) 2 hours
B) 1½ hours
C) 4 hours
D) 5½ hours
E) 3½ hours
In this problem, we have two cars – car A and B. Car A begins 20 miles behind car B and needs to catch up. Our immediate DSM (Default Solving Mechanism) is to dive in and create an equation for this and that’s exactly what we don’t want to do.
These types of problems are notorious for being algebraically complex, while conceptually simple. If you hold on to the algebra, rather than getting rid of it, you’re going to have a hard time.
## 2. Speed and Distance Problems – Solution Paths
In this problem, we’re going to build up solution paths. We’re gonna skip the entirely. We’re going to take a look at an iterative way to get to the answer and then do a conceptual scenario, where we literally put ourselves in the driver’s seat to understand how this problem works. So if we want to take the iterative process we can simply drive the process hour-by-hour until we get to the answer.
## 3. Iterative Solution Path
We can imagine this on a number line or just do it in a chart with numbers. Car A starts 20 miles behind car B – so let’s say ‘A’ starts at mile marker zero and ‘B’ starts at 20. After one hour ‘A’ is at 58, ‘B’ is at 70 and the differential is now -12 and not -20. After the second hour ‘A’ is at 116, ‘B’ is at 120. ‘A’ is just four behind ‘B’. After the third hour ‘A’ has caught up! Now it’s 4 miles ahead. At the fourth hour it’s not only caught up but it’s actually +12, so we’ve gone too far. We can see that the correct answer is between three and four and our answer is three and a half.
Now let’s take a look at this at a higher level. If we take a look at what we’ve just done we can notice a pattern with the catching up: -20 to -12 to -4 to +4. We’re catching up by 8 miles per hour. And if you’re self-prepping and don’t know what to do with this information, this is exactly the pattern that you want to hinge on in order to find a better solution path.
You can also observe (and this is how you want to do it on the exam) that if ‘A’ is going 8 miles an hour faster than ‘B’, then it’s catching up by 8 miles per hour. What we care about here is the rate of catching up, not the actual speed. The 50 and 58 are no different than 20 and 28 or a million and a million and eight. That is, the speed doesn’t matter. Only the relative distance between the cars and that it changes at 8 miles per hour.
Now the question becomes starkly simple. We want to catch up 20 miles and then exceed 8 miles, so we want to have a 28-mile shift and we’re doing so at 8 miles an hour. 28 divided by 8 is 3.5.
## 4. Speed and Distance Problems – Conceptual Scenario Solution Path
You might ask yourself what to do if you are unable to see those details. The hallmark of good scenarios is making them personal. Imagine you’re driving and your friend is in the car in front of you. He’s 20 miles away. You guys are both driving and you’re trying to catch up. If you drive at the same speed as him you’re never going to get there. If you drive one mile per hour faster than him you’ll catch up by a mile each hour. It would take you 20 hours to catch up. This framework of imagining yourself driving and your friend in the other car, or even two people walking down the street, is all it takes to demystify this problem. Make it personal and the scenarios will take you there.
Thanks for the time! For other solutions to GMAT problems and general advice for the exam check out the links below. Hope this helped and good luck! |
# How do you solve 2<=4-3x<7?
Feb 20, 2017
See the entire solution process below:
#### Explanation:
First, subtract $\textcolor{red}{4}$ from each segment of the inequality to isolate the $x$ term while keeping the system of inequalities balanced:
$2 - \textcolor{red}{4} \le 4 - 3 x - \textcolor{red}{4} < 7 - \textcolor{red}{4}$
$- 2 \le 4 - \textcolor{red}{4} - 3 x < 3$
$- 2 \le 0 - 3 x < 3$
$- 2 \le - 3 x < 3$
Now, divide each segment of the inequality system by $\textcolor{b l u e}{- 3}$ to solve for $x$ while keeping the system of inequalities balanced. However, remember, when dividing or multiplying an inequality by a negative number you must reverse the inequality terms:
$\frac{- 2}{\textcolor{b l u e}{- 3}} \textcolor{red}{\ge} \frac{- 3 x}{\textcolor{b l u e}{- 3}} \textcolor{red}{>} \frac{3}{\textcolor{b l u e}{- 3}}$
$\frac{2}{3} \textcolor{red}{\ge} \frac{\textcolor{b l u e}{\cancel{\textcolor{b l a c k}{- 3}}} x}{\cancel{\textcolor{b l u e}{- 3}}} \textcolor{red}{>} - 1$
$\frac{2}{3} \textcolor{red}{\ge} x \textcolor{red}{>} - 1$ |
# How you simplify (Square root of 75) +(square root of 48)?
Jul 27, 2015
= color(blue)(9sqrt3
#### Explanation:
• Square root of a number can be simplified by prime factorising the number.
• Prime factorisation involves expressing a number as a product of prime numbers.
($75$ has $3$ and $5$ as its prime factors and $48$ has $2$ and $3$ as its prime factors)
So,
$\sqrt{75} = \sqrt{5 \cdot 5 \cdot 3} = \sqrt{{5}^{2} \cdot 3} = 5 \sqrt{3}$
$\sqrt{48} = \sqrt{2 \cdot 2 \cdot 2 \cdot 2 \cdot 3} = \sqrt{{2}^{2} \cdot {2}^{2} \cdot 3} = 4 \sqrt{3}$
Now,
sqrt75 +sqrt48 = 5sqrt3+4sqrt3 = color(blue)(9sqrt3 |
Question:
Two sides of a rectangle differ by 3.5 cm. Find the dimensions of the rectangle if its perimeter is 67 cm.
Difficulty: Easy
Solution:
Let the length of the rectangle is $x$ cm and width is $y$ cm.
According to the given conditions:
The two sides $x$ and $y$ of the rectangle differ by $3.5$.
$x-y=3.5$
$10x-10y=35 \quad\quad\quad\quad\quad...~(1)$
Perimeter = 67 cm
$2(x + y) = 67$
$2x + 2y = 67 \quad\quad\quad\quad\quad...~(2)$
Method 1: Matrix Inversion Method
Write the linear equations $(1)$ and $(2)$ in matrix form.
$AX = B$
$\left[ \begin{matrix} 10 & -10 \\ 2 & 2 \\ \end{matrix} \right]\left[ \begin{matrix} x \\ y \\ \end{matrix} \right]$ = $\left[ \begin{matrix} 35 \\ 67 \\ \end{matrix} \right]$
According to matrix inversion method, solution $X$ can be found by the following formula:
$X=A^{-1}B$
$\left| \text{A} \right|=\left| \begin{matrix} 10 & -10 \\ 2 & 2 \\ \end{matrix} \right|=10\times 2-\left( -10 \right)\times 2=20+20=40\ne 0$
The coefficient matrix $A$ is non-singular therefore $A^{-1}$ exists.
$\left[ \begin{matrix} x \\ y \\ \end{matrix} \right]$ = ${{\text{A}}^{-1}}\left[ \begin{matrix} 35 \\ 67 \\ \end{matrix} \right]$
= $\frac{Adj~\text{A}}{\left| \text{A} \right|}~\left[ \begin{matrix} 35 \\ 67 \\ \end{matrix} \right]$
= $\frac{1}{40}\left[ \begin{matrix} 2 & 10 \\ -2 & 10 \\ \end{matrix} \right]\left[ \begin{matrix} 35 \\ 67 \\ \end{matrix} \right]$
= $\frac{1}{40}\left[ \begin{matrix} 2~\times 35+10\times 67 \\ -2~\times 35+10\times 67 \\ \end{matrix} \right]$
= $\frac{1}{40}\left[ \begin{matrix} 70+670 \\ -70+670 \\ \end{matrix} \right]$
= $\frac{1}{40}\left[ \begin{matrix} 740 \\ 600 \\ \end{matrix} \right]$
$\left[ \begin{matrix} x \\ y \\ \end{matrix} \right]$= $\left[ \begin{matrix} 18.5 \\ 15 \\ \end{matrix} \right]$
Therefore $x=18.5$ and $y=15$
Length of the Rectangle = $x$ = 18.5cm
Width of the Rectangle = $y$ = 15 cm
Method 2: Cramer’s Rule
Write the linear equations $(1)$ and $(2)$ in matrix form.
$AX = B$
$\left[ \begin{matrix} 10 & -10 \\ 2 & 2 \\ \end{matrix} \right]\left[ \begin{matrix} x \\ y \\ \end{matrix} \right]$ = $\left[ \begin{matrix} 35 \\ 67 \\ \end{matrix} \right]$
According to Cramer's Rule:
$x=\frac{\left| {{A}_{x}} \right|}{\left| A \right|}$
$y=\frac{\left| {{A}_{y}} \right|}{\left| A \right|}$
$\left| \text{A} \right|=\left| \begin{matrix} 10 & -10 \\ 2 & 2 \\ \end{matrix} \right|=10\times 2-\left( -10 \right)\times 2=20+20=40\ne 0$
The coefficient matrix $A$ is non-singular therefore solution exists.
For ${{A}_{x}}$, the $x$ column is replaced with the constant column $B$
${{A}_{x}}=\left[ \begin{matrix} 35 & -10 \\ 67 & 2 \\ \end{matrix} \right]$
$\left| {{A}_{x}} \right|=\left| \begin{matrix} 35 & -10 \\ 67 & 2 \\ \end{matrix} \right|=35\times 2-\left( -10 \right)\times 67=70+670=740$
$x=\frac{\left| {{A}_{x}} \right|}{\left| A \right|}=\frac{740}{40}=18.5$
For ${{A}_{y}}$, the $y$ column is replaced with the constant column $B$
${{A}_{y}}=\left[ \begin{matrix} 10 & 35 \\ 2 & 67 \\ \end{matrix} \right]$
$\left| {{A}_{y}} \right|=\left| \begin{matrix} 10 & 35 \\ 2 & 67 \\ \end{matrix} \right|=10\times 67-35\times 2=670-70=600$
$y=\frac{\left| {{A}_{y}} \right|}{\left| A \right|}=\frac{600}{40}=15$
Length of the Rectangle = $x$ = 18.5 cm
Width of the Rectangle = $y$ = 15 cm |
# Eureka Math Grade 7 Module 6 Mid Module Assessment Answer Key
## Engage NY Eureka Math 7th Grade Module 6 Mid Module Assessment Answer Key
### Eureka Math Grade 7 Module 6 Mid Module Assessment Task Answer Key
Question 1.
In each problem, set up and solve an equation for the unknown angles.
a. Four lines meet at a point. Find the measures m° and n°.
b. Two lines meet at the vertex of two rays. Find the measures m° and n°.
c. Two lines meet at a point that is the vertex of two rays. Find the measures m° and n°.
d. Three rays have a common vertex on a line. Find the measures m° and n°.
a. n° = 90° , vertical angles
25° + (90°) + 40° + m° = 180°
155° + m° = 180°
155° – 155° + m° = 180° – 155°
m° = 25°
b. 50° + 90° + n° = 180°
140° + n° = 180°
140° – 140° + n° = 180° – 140°
n° = 40°
m° + 50° = 90°
m° + 50° – 50° = 90° – 50°
m° = 40°
c. m° + 52° = 90°
m° + 52° – 52° = 90° – 52°
m° = 38°
40 + 52 + (38) + n° = 180
130 + n° = 180
130 – 130 + n° = 180 – 130
n° = 50°
d. n° + 62° = 90°
n° + 62° – 62° = 90° – 62°
n° = 28°”
m° + 62° + (28°) + 27° = 180°
m° + 117° = 180
m° + 117° – 117° = 180° – 117°
m° = 63°
Question 2.
Use tools to construct a triangle based on the following given conditions.
a. If possible, use your tools to construct a triangle with angle measurements 20°, 55°, and 105°, and leave evidence of your construction. If it is not possible, explain why.
b. Is it possible to construct two different triangles that have the same angle measurements? If it is, construct examples that demonstrate this condition, and label all angle and length measurements. If it is not possible, explain why.
a. Solutions will vary. An example of a correctly constructed triangle is shown here.
b. Solutions will vary; refer to the rubric.
Question 3.
In each of the following problems, two triangles are given. For each: (1) state if there are sufficient or insufficient conditions to show the triangles are identical, and (2) explain your reasoning.
a. The triangles are identical by the two angles and included side condition. The marked side is between the given angles.
△ABC ↔ △YXZ
b. There is insufficient evidence to determine that the triangles are identical. In △DEF , the marked side is between the marked angles, but in △ABC , the marked side is not between the marked angles.
c. The triangles are identical by the two sides and included angle condition. △DEF ↔ △GIH
d. The triangles are not identical. In △ABC , the marked side is opposite ∠B . In △WXY , the marked side is opposite ∠W . ∠B and ∠W are not necessarily equal in measure.
Question 4.
Use tools to draw rectangle ABCD with AB = 2 cm and BC = 6 cm. Label all vertices and measurements.
Question 5.
The measures of two complementary angles have a ratio of 3:7. Set up and solve an equation to determine the measurements of the two angles.
3x + 7x = 90
10x = 90
($$\frac{1}{10}$$)10x = ($$\frac{1}{10}$$)90
x = 9
Measure of Angle 1: 3(9) = 27 . The measure of the first angle is 27° .
Measure of Angle 2: 7(9) = 63 . The measure of the second angle is 63° .
Question 6.
The measure of the supplement of an angle is 12° less than the measure of the angle. Set up and solve an equation to determine the measurements of the angle and its supplement.
Let y° be the number of degrees in the angle.
y + (y – 12) = 180
2y – 12 = 180
2y – 12 + 12 = 180 + 12
2y = 192
($$\frac{1}{2}$$)2y = ($$\frac{1}{2}$$)192
y = 96
Measure of the angle: 96°
Measure of its supplement: (96)° – 12° = 84°
Question 7.
Three angles are at a point. The ratio of two of the angles is 2:3, and the remaining angle is 32° more than the larger of the first two angles. Set up and solve an equation to determine the measures of all three angles.
2x + 3x + (3x + 32) = 360
8x + 32 = 360
8x + 32 – 32 = 360 – 32
8x = 328
($$\frac{1}{8}$$)8x = ($$\frac{1}{8}$$)328
x = 41
Measure of Angle 1: 2(41)° = 82°
Measure of Angle 2: 3(41)° = 123°
Measure of Angle 3: 3(41)° + 32° = 155°
Question 8.
Draw a right triangle according to the following conditions, and label the provided information. If it is not possible to draw the triangle according to the conditions, explain why. Include a description of the kind of figure the current measurements allow. Provide a change to the conditions that makes the drawing feasible.
a. Construct a right triangle ABC so that AB = 3 cm, BC = 4 cm, and CA = 5 cm; the measure of angle B is 90°.
b. Construct triangle DEF so that DE = 4 cm, EF = 5 cm, and FD = 11 cm; the measure of angle D is 50°.
b. It is not possible to draw this triangle because the lengths of the two shorter sides do not sum to be greater than the longest side. In this situation, the total lengths of $$\overline{D E}$$ and $$\overline{E F}$$ are less than the length of $$\overline{F D}$$; there is no way to arrange $$\overline{D E}$$ and $$\overline{E F}$$ so that they meet. If they do not meet, there is no arrangement of three non-collinear vertices of a triangle; therefore, a triangle cannot be formed. I would change $$\overline{E F}$$to 9 cm instead of 5 cm so that the three sides would form a triangle. |
# Multivariable calculus
## Introduction
So far in A-level maths you've dealt mainly with functions like $$y=f(x)$$
where the variable $y$ is equal to some function of one variable $x$. Such expressions can be differentiated with respect to this one independent variable $$\frac{dy}{dx} = f'(x)$$
In FP3 we deal with functions of multiple variables, sketch what they look like in three dimensions, and analyse their gradients and behaviours.
## Three-dimensional surfaces
The function $y=f(x)$ represents a curve in two dimensions.
Geometrically a function expressed by $z=f(x,y)$ represents a surface in three dimensions, where $z$ is the third dimension in Cartesian three-dimensional space.
This is the surface $z = 2x + 3y$, a flat plane.
You can take sections and contours of this surface to see how 2D and 3D space relate to each other. For example if you set $z=2$, then the surface becomes $$2 = 2x + 3y \Rightarrow y = \frac{2}{3}\left(1-x\right)$$
which looks like
Indeed if you compare this to the 3D graph it has similar decreasing behaviour as $x$ increases while $z$ is held constant.
This is a contour plot of the same surface. Since the $x$ and $y$ in $f(x,y)$ are linear the contours aren't particularly interesting. They're just straight lines.
In two dimensions the contour plot gives another perspective.
A much more interesting surface is $z=x^2-y^3$. It is non-linear and therefore has a much 'curvier' shape.
Its contour plot shows the changes in gradient across the surface.
In two dimensions the contour plot looks like this
## Partial derivatives
A partial derivative of a multivariable function is a derivative with respect to one of its variables, holding the others constant.
For a function $z=f(x,y)$ the partial derivative of $z$ with respect to $x$ is denoted $$\frac{\partial z}{\partial x}$$
A partial derivative of $f(x,y)$ is often denoted in several other ways too $$\frac{\partial f}{\partial x} = f_x = \partial_x f = \frac{\partial}{\partial x}f$$
Take the function $z=xy-\frac{2}{x+y}$. This function is too complicated to differentiate using the conventional methods you learned so far in A-level maths, but you can analyse its behaviour as well as the 3D surface it represents using partial derivatives.
Every point on the surface has infinitely many tangent lines. When you take a partial derivative, you treat all the other variables as constant. That means you treat then just as if they were numbers. So let's the find the partial derivatives
$$\frac{\partial z}{\partial x} = y+\frac{2}{\left(x+y\right)^2}$$
and similarly
$$\frac{\partial z}{\partial y} = x+\frac{2}{\left(x+y\right)^2}$$
### Example
Q) Find the partial derivatives of $z=\sin(x^2 +y)$ with respect to $x$ and $y$.
A) By the chain rule $$\frac{\partial z}{\partial x} = 2x\cos(x^2 +y)$$ and $$\frac{\partial z}{\partial y} = \cos(x^2 +y)$$
### Example
Q) Find the partial derivatives of $z=x\left(x-y\right)^3+3x$ with respect to $x$ and $y$.
A) Using the product rule and the chain rule \begin{align} \frac{\partial z}{\partial x} &= 3x\left(x-y\right)^2+\left(x-y\right)^3+3 \\ &= \left(x-y\right)^2\left(3x+x-y\right)+3 \\ &= \left(x-y\right)^2\left(4x-y\right)+3 \end{align}
and using the chain rule $$\frac{\partial z}{\partial y} = -3xy\left(x-y\right)^2$$
## Stationary points in three dimensions
In C1 and C2 you learned that to find the stationary points of a function $y=f(x)$ you needed to set $$\frac{dy}{dx} = 0$$
and solve for $x$. This told you the points where the gradient was zero. We can do the same thing for surfaces in three dimensions. However instead of the above condition, we need to find points $(x,y,z)$ such that $$\frac{\partial z}{\partial x} = \frac{\partial z}{\partial y} = 0$$
which makes sense since we're looking for points that are 'flat' in three dimensions. Remember that each stationary point must satisfy both $\frac{\partial z}{\partial x}=0$ and $\frac{\partial z}{\partial y}=0$ at the same time.
### Example
Q) Find the stationary points of the surface $z=\sin(x^2 +y)$.
A) We found that the partial derivatives were \begin{align} \frac{\partial z}{\partial x} &= 2x\cos(x^2 +y) \\[0.7em] \frac{\partial z}{\partial y} &= \cos(x^2 +y) \end{align}
Equating both of them to zero and subtracting one from the other $$\frac{\partial z}{\partial x} - \frac{\partial z}{\partial y} = \left(2x-1\right)\cos(x^2 +y) = 0$$
Therefore either $x=\frac{1}{2}$ or $\cos(x^2 +y) = 0$. This means that stationary points occur along the line $x=\frac{1}{2}$ and also on the lines $x^2+y=n\frac{\pi}{2}$ for $n\in\mathbb{Z}$.
### Example
Q) Find the stationary points of the surface $z=x^2+3y$.
A) The partial derivatives are $$\frac{\partial z}{\partial x} = 2x ~\textrm{ and }~ \frac{\partial z}{\partial y} = 3$$
Equating both of them to zero and subtracting one from the other $$\frac{\partial z}{\partial x} - \frac{\partial z}{\partial y} = 2x-3 = 0$$
Therefore stationary points occur along the line $x=\frac{3}{2}$.
Surfaces in three dimensions can also be expressed in the form $$f(x,y,z)=c$$ where $c$ is a constant. Here's an example, $-4x-y^2+z^2=3$
Then $\textbf{grad}f$ or $\nabla f$ is a vector function that always points in the direction of greatest increase of $f$ at each particular point. In the context of surfaces it is also called the gradient. The magnitude of $\nabla f$ is the value of the steepest slope. It is calculated as $$\nabla f = \left(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial f}{\partial z}\right)$$
In the special case where $f(x,y,z)=c$ can be expressed in the form $z=g(x,y)$ then $$\nabla f = \left(\frac{\partial g}{\partial x},\frac{\partial g}{\partial y},-1\right)$$
For the above example $$\nabla f = \left(-4,-2y,2z\right)$$
and the value of the steepest slope at each point is the norm of the gradient, since it's a vector $$\left\Vert\nabla f\right\Vert = \sqrt{\left(-4\right)^2+\left(-2y\right)^2+\left(2z\right)^2} = 2\sqrt{4+y^2+z^2}$$
### Tangent planes and normal lines
One application of partial derivatives and the surface gradient is finding tangent planes and normal lines to points on surfaces. Recall from AS-level maths that to find a tangent line to a point $(x_0,y_0)$ on a function in two dimensions, where the gradient at that point is $m$, sub the values into $$y - y_0 = m\left(x-x_0\right)$$
In three dimensions, the equation for a plane tangent to a point $(x_0,y_0,z_0)$ on a surface $f(x,y,z)=c$ is made up of the values of the partial derivatives at the point $$f_x(x_0,y_0,z_0)\left(x-x_0\right)+f_y(x_0,y_0,z_0)\left(y-y_0\right)+f_z(x_0,y_0,z_0)\left(z-z_0\right) = 0$$
### Example
Q) Find the tangent plane to the point $(1,3,1)$ on the surface $(x+z)^2+y+\sqrt{z}=2$.
A) Let $f(x,y,z)=(x+z)^2+y+\sqrt{z}$. The partial derivatives are
\begin{align} f_x &= 2x+2z \Rightarrow f_x(1,3,1) = 4 \\ f_y &= 1 \\ f_z &= 2x+2z+\frac{1}{2\sqrt{z}} \Rightarrow f_z(1,3,1) = \frac{9}{2} \\ \end{align}
Therefore the equation of the tangent plane is $$4\left(x-1\right)+\left(y-3\right)+\frac{9}{2}\left(z-1\right) = 0$$ $$\Rightarrow 4x+y+\frac{9}{2}z=\frac{23}{2}$$
The normal line to a point $(x_0,y_0,z_0)$ on a surface $f(x,y,z)=c$ is a straight line starting from the point and going in the same direction as $\nabla f$ at that point. The line is expressed as $$\mathbf{r}(\lambda) = (x_0,y_0,z_0)+ \lambda\nabla f(x_0,y_0,z_0)$$
### Example
Q) Find the equation of the normal line to the point $(2,0,-1)$ on the surface $\left(\frac{x+y}{z}\right)^2 = 0$.
A) Let $f(x,y,z)=\left(\frac{x+y}{z}\right)^2$. The partial derivatives are
\begin{align} f_x &= \frac{2x\left(x+y\right)}{z^2} \Rightarrow f_x(2,0,-1) = 8 \\ f_y &= \frac{2y\left(x+y\right)}{z^2} \Rightarrow f_x(2,0,-1) = 0 \\ f_z &= -\frac{2\left(x+y\right)}{z^3} \Rightarrow f_z(2,0,-1) = 4 \\ \end{align}
Therefore the equation of the normal line is $$\mathbf{r}(\lambda) = (2,0,-1) + (8,0,4)\lambda = (2+8\lambda,0,4\lambda-1)$$
### Example
Q) Find the equations of the tangent plane and the normal line to the point $(-1,-1,\frac{\pi}{3})$ on the surface $\sin(xyz)=1$.
A) Let $f(x,y,z)=\sin(xyz)$. The partial derivatives are
\begin{align} f_x &= yz\cos(xyz) \Rightarrow f_x(-1,-1,\frac{\pi}{3}) = -\frac{\pi}{6} \\ f_y &= xz\cos(xyz) \Rightarrow f_x(-1,-1,\frac{\pi}{3}) = -\frac{\pi}{6} \\ f_z &= xy\cos(xyz) \Rightarrow f_z(-1,-1,\frac{\pi}{3}) = \frac{1}{2} \\ \end{align}
Therefore the equation of the tangent plane is $$-\frac{\pi}{6}\left(x+1\right)-\frac{\pi}{6}\left(y+1\right)+\frac{1}{2}\left(z-\frac{\pi}{3}\right) = 0$$ $$\Rightarrow -\frac{\pi}{6}x-\frac{\pi}{6}y+\frac{1}{2}z = -\frac{\pi}{2}$$ $$\Rightarrow \pi x+\pi y-3z = 3\pi$$
and the equation of the normal line is $$\mathbf{r}(\lambda) = \left(-1,-1,\frac{\pi}{3}\right) + \left(-\frac{\pi}{6},-\frac{\pi}{6},\frac{1}{2}\right)\lambda = \left(-1-\frac{\pi}{6}\lambda,-1-\frac{\pi}{6}\lambda,\frac{\pi}{3}+\frac{1}{2}\lambda\right)$$
## The total derivative and its application to error estimates
You'll see further on in your calculus studies at university that there are many relations between the single-variable derivatives you're used to and partial derivatives. In FP3 you need to know a relation that can be applied to estimating the overall error in a calculation.
The total derivative of a function with multiple variables with respect to one of its variables, uses all of the partial derivatives and regular derivatives to find out the overall dependency of the function on the single variable.
Take a function $f(x,y,z)$, then its total derivative with respect to $x$ is $$\frac{df}{dx} = \frac{\partial f}{\partial x}\frac{dx}{dx}+\frac{\partial f}{\partial y}\frac{dy}{dx}+\frac{\partial f}{\partial z}\frac{dz}{dx}$$
As you can spot from the first term the total derivative simplifies to $$\frac{df}{dx} = \frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}\frac{dy}{dx}+\frac{\partial f}{\partial z}\frac{dz}{dx}$$
Then multiply through by the differential $\mathop{dx}$ $$\mathop{df} = \frac{\partial f}{\partial x}\mathop{dx}+\frac{\partial f}{\partial y}\mathop{dy}+\frac{\partial f}{\partial z}\mathop{dz}$$
This equation shows that any change in $f$ depends on changes that happen in all of the variables, and also the partial derivatives with respect to each variable.
For functions that can be expressed as $z=f(x,y)$ a shorter form for $z$ is $$\mathop{dz} = \frac{\partial z}{\partial x}\mathop{dx}+\frac{\partial z}{\partial y}\mathop{dy}$$
This can be applied to approximation errors in calculations. If you know that one variable may be off by some measure, then you can predict how much that will affect the dependent variable $z$.
### Tangent planes as an approximation
For 3D surfaces a tangent plane to a point on it provides a limited approximation of the surface around the neighbourhood of the point.
In FP3 you need to understand that a small uniform change in the variables away from the tangent point will cause an overall change in the value of the function.
### Example
Q) (Exam question, June 2012) Let $g(x,y,z)=x^2+2y^2-z^2+2xz+2yz+4z-3$ and $S$ be a surface defined by $g(x,y,z)=0$. $P(-2,-1,1)$ is a point on $S$. A point $Q$ lies on the normal line to $S$ at $P$. At $Q$, $g(x,y,z)=h$ where $h$ is small. Find the constant $c$ such that $PQ\approx c|h|$.
A) First find all the partial derivatives of $g$
\begin{align} g_x &= 2x+2z \Rightarrow g_x(-2,-1,1)=-2 \\ g_y &= 4y+2z \Rightarrow g_y(-2,-1,1)=-2 \\ g_z &= 2z+2x+2y+4 \Rightarrow g_z(-2,-1,1)=-4 \\ \end{align}
The equation of the normal line to $P$ on $S$ is therefore $$\mathbf{r}(\lambda)=(-2,-1,1)+(1,1,2)\lambda = (-2+\lambda,-1+\lambda,1+2\lambda)$$
We were able to divide the derivative values by -2 there because we were forming a direction vector of the normal line, so only the direction matters. You can keep the direction vector as $(-2,-2,-4)$ if you like, and do the below calculations again - you'll get the same overall answer.
The coordinates of $Q$ are $(-2+\lambda,-1+\lambda,1+2\lambda)$, then sub each partial derivative into the following total derivative approximation as well as each $\lambda$ from the direction of the normal line.
In such cases replace $d$s with $\delta$, lower case delta, to show that the total derivative approximation uses very small but measurable perturbations (i.e. not infinitesimally small) in each variable.
\begin{align} h &= \mathop{\delta g} \approx \frac{\partial g}{\partial x}\mathop{\delta x}+\frac{\partial g}{\partial y}\mathop{\delta y}+\frac{\partial g}{\partial z}\mathop{\delta z} \\ &= \left(-2\right)\left(\lambda\right)+\left(-2\right)\left(\lambda\right)+\left(-4\right)\left(2\lambda\right) \\ &= -12\lambda \end{align}
So we've found the actual value of $h$. But we need to keep going and calculate $PQ$. It is along the normal line, and the direction vector is $(\lambda,\lambda,2\lambda)$. Therefore the length of $PQ$ is $$\sqrt{\lambda^2+\lambda^2+\left(2\lambda^2\right)} = \sqrt{6}|\lambda|$$
Therefore because we were asked for $c$ such that $PQ\approx c|h|$ $$\sqrt{6}|\lambda| = c\left|-12\lambda\right| \Rightarrow c=\frac{\sqrt{6}}{12}$$ |
# Percent of a Number
## Use multiplication and proportions to find the percent of a number.
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Taylor wants to buy a new pair of shoes. She goes to the shoe store and discovers a shelf on which all of the shoes are 60% off. She finds a pair of shoes on the shelf that are ticketed at $50. “I wonder how much money I will get back if I pay for the shoes with a$50 bill,” Taylor muses. At the checkout register, how much change will Taylor get back?
In this concept, you will learn to find the percent of a number using fraction multiplication.
### Finding the Percent of a Number
To work with percents you have to understand how they relate to parts and fractions.
The table below shows the fractional equivalents for common percents.
5% 10% 20% 25% 30% 40% 50% 60% 70% 75% 80% 90%
The word “of” in a percent problem means to multiply. If you know the fractional equivalents for common percents, you can use this information to find the percent of a number by multiplying the fraction by that number. If you want to find a part of a whole using a percent, you use multiplication to solve.
Let’s look at an example.
Find 40% of 45.
First, 40% of 45 means .
Next, look at the chart. The fraction is equivalent to 40%.
Then perform the multiplication, simplifying as you go.
The answer is that 40% of 45 is 18.
Now, let’s look at another example using an alternate way to solve.
What is 18% of 50?
First, to figure this out, you can change the percent to a fraction and then create a proportion.
Next, you can cross multiply and solve for .
The answer is 18% of 50 is 9.
### Examples
#### Example 1
Earlier, you were given a problem about Taylor and her shoes.
They were ticketed at $50 but there was a 60% off sale on them. Taylor wanted to know how much money she would get back if she paid for the shoes with a$50 bill.
First, 60% of 50 means .
Next, change the percent to a fraction in simplest form.
Then perform the multiplication, simplifying as you go.
The answer is that the shoes are marked down by $30. If Taylor pays with a$50 bill she will get \$30 back.
#### Example 2
Change the percent to a fraction in simplest form.
Find 85% of 20.
First, 85% of 20 means .
Next, change the percent to a fraction in simplest form.
Then perform the multiplication, simplifying as you go.
The answer is that 85% of 20 is 17.
#### Example 3
What is 10% of 50?
First, 10% of 50 means .
Next, look at the chart. The fraction is equivalent to 10%.
Then perform the multiplication, simplifying as you go.
The answer is that 10% of 50 is 5.
#### Example 4
What is 25% of 80?
First, 25% of 80 means .
Next, look at the chart. The fraction is equivalent to 25%.
Then perform the multiplication, simplifying as you go.
The answer is that 25% of 80 is 20.
#### Example 5
What is 22% of 100?
First, 22% of 100 means .
Next, change the percent to a fraction in simplest form.
Then perform the multiplication, simplifying as you go.
The answer is that 22% of 100 is 22.
### Review
Use fraction multiplication to find each percent of the number.
1. 10% of 25
2. 20% of 30
3. 25% of 80
4. 30% of 90
5. 75% of 200
6. 8% of 10
7. 10% of 100
8. 19% of 20
9. 15% of 30
10. 12% of 30
11. 15% of 45
12. 25% of 85
13. 45% of 60
14. 50% of 200
15. 55% of 300
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
TermDefinition
improper fraction An improper fraction is a fraction in which the absolute value of the numerator is greater than the absolute value of the denominator. |
## Sunday, April 19, 2009
### Mathematics - Quadratic Equation (Part 1)
Picture is obtained from http://foldspace.files.wordpress.com/2008/06/c4743r.jpg
The mathematics question is,
Find the value of k of the quadratic equation
Such that one of its roots is thrice the value of the other.
Solution:
Let α and β be the roots. We have,
Sum of roots:
Product of roots:
Given that,
Substituting into eq. (1), we get
Therefore the other root is,
What to do next?
The roots,
α and β are found but in terms of k.
How to find the value of k?
Note that equation (2) is not yet used.
Substituting α and β into eq. (2), we have,
How to obtain the Quadratic Formula
In mathematics, a quadratic equation is a polynomial equation of the second degree. The general form is
The quadratic coefficients are the following:
a is the coefficient of x2
b is the coefficient of x
c is the constant coefficient (also called the free term or constant term).
The roots x or the quadratic formula can be derived by the method of completing the square. Dividing the quadratic equation
by a (which is allowed because a is non-zero), gives:
or
The quadratic equation is now in a form to which the method of completing the square can be applied.
To "complete the square" add the constant,
, to both sides,
The left side is now a perfect square because
Substituting the left side with the perfect square we have,
The right side can be written as a single fraction, with common denominator 4a2. This gives
Taking the square root of both sides yields
Isolating x, gives
This equation is known as the quadratic formula.
How to obtain Sum of roots and Product of roots
The general form is of a quadratic equation is
Dividing the quadratic equation by coefficient, a, we have
If a quadratic equation has roots, α and β, the equation can be written as
That is,
Comparing eq.(1) and eq.(2),
Reference |
### Archive for the ‘Uncategorized’ Category
Given parallel lines, there are four main relationships that angles can have to one another: vertical angles, interior angles relationships, exterior angle relationships, and corresponding angles. These relationships can all be derived from one another, and in what follows, we include some of those derivations so that, if you forget one of them on exam day, you can recover it from knowledge of the others.
Vertical Angles: Two angles formed by the same lines are vertical angles if they are on opposite sides of the point of intersection between the two lines.
This is a fundamental fact that is often critical in finding an unknown angle. For example:
Example 1:
Interior Angles:
In the above diagram,
Example 2:
In the following diagram, let and let What is the value of ?
Exterior Angles:
In the above diagram,
Example 3:
In the below diagram, let and What is the value of ?
Corresponding Angles: Two angles (which are in the same relative position) formed by a single line cutting across two parallel lines. Such angles are equal.
For example:
In this diagram, and are corresponding angles and thus equal in degree.
Practice Problems:
1. Find the value of .
2. Find the value of .
3. Find the value of
4. Find the value of
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Many geometry problems either directly ask about the degree of certain angles or will require you to figure out the degree of an angle in the course of solving some problem. In successive posts, we will cover the basics of what an angle is, how we can use parallel lines to find unknown angles, and how we can use triangles/circles/polygons to find unknown angles.
Angles: Some Preliminaries
To understand angles, we need to begin with the idea of a line:
Line: a straight line that continues in both directions without end.
Now, an angle appears where two lines intersect. More formally,
Angle: a measure of how much you would need to turn one line to make them part of the same line
For example,
And as you can see in the above diagram, there are actually four angles formed when two lines intersect. So there is a convention for naming each of the four angles:
[Practice problems with matching angles: label points on the lines and degree measures and have them match the two]
Now, here are two important facts about angles:
Lines Have 180 Degrees: The angle formed by a single line has 180 degrees.
Circles Have 360 Degrees: The angle formed by a circle is 360 degrees.
Now, we also have two relationships among lines which are very important:
Parallel Lines: Parallel lines go in the same direction and therefore never intersect. We often denote that lines and are parallel by writing , as in the below diagram:
Perpendicular Lines: Perpendicular lines intersect at a 90 degree angle. We denote two lines as perpendicular by writing :
In our next post, we will talk about some of the relationships that exist between angles and lines and shapes.
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Below, we include some additional geometry practice problems.
1. What is the area of ? Of ?
2. In the following diagram, What is the area of ?
3. What is the area of ?
4. The radius of the below circle is 3. Find the area of the shaded region.
5. The circle below has a radius of . Find the ratio of the shaded region's area to the area of the square below.
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The next 3D shape we shall look at is the circular cylinder (also called the “right circular cylinder” by the GRE for its right angles):
Circular Cylinder: A 3D shape that has a circle for both bases and a perpendicular line connecting the center of those bases.
Now, it follows from the definition that a circular cylinder’s top and bottom circles will be the same size. So we use to denote the radius of either circle, and we use to denote the height of the cylinder. Without further ado, we present:
Volume of a Circular Cylinder: The volume of a right circular cylinder with a radius of and a height of is
Surface Area of a Circular Cylinder: The surface area of a circular cylinder with a radius of and a height of is
Practice Problems:
1. The cylinder below has a radius of and a height of 6. Find .
2. The surface area of a right circular cylinder is and its radius is 2. What is the height of the cylinder?
3. The surface area of a cylinder is twice its volume. Its radius equals its height. What is the volume of the cylinder?
Comment on this
One important kind of 3D shape is the rectangular solid:
which is defined as follows:
Rectangular Solid: A 3D shape with six faces which are all rectangles placed perpendicularly to one another.
The fact that the faces are all perpendicular to each other helps a great deal in calculating the volume, for we simply have:
Volume of a Rectangular Solid: For the following rectangular solid,
its volume is equal to
Example 1
Find the area of a rectangular solid whose dimensions are .
Example 2
A rectangular solid has two sides that are 3 and 8 units long. It has a volume of 48 units. What is the length of its remaining side?
And, to find the surface area of any rectangular solid, we simply need to add up the area of each of the 6 rectangles which surround it.
Surface Area of a Rectangular Solid: For the following rectangular solid,
its surface area is equal to
How did we get this formula? Well, since each of the faces are perpendicular to each other, we know that there are two rectangles with sides of and units long, one in front and one at the back:
and there are two with sides of units long:
and there are two with sides of units long:
Thus, adding up the area of all of those rectangles, we get: which is the same as what we wrote above.
Now, one common 3D shape is the cube, and the cube is just a special kind of rectangular solid:
Cube: a cube is a rectangular solid whose six sides are all squares.
Our formulas become especially simple for a cube. Let be the length of any side of the cube:
Volume of a Cube: The volume of a cube is .
Surface Area of a Cube: The surface area of a cube is .
These formulas follow directly from just applying our above formulas for the volume/surface area of any rectangular solid.
Finally, it is worth noting how we can use the Pythagorean Theorem to find certain distances on a rectangular solid. Consider the following problem:
Example 3
For a rectangular solid with a length of , a width of and a height of , find the length of :
Practice Problems
1. Find the length of
2. Find the length of .
3. Find the area of the shaded plane below:
Comment on this
Finally, we turn to the kinds of shapes we see in our daily lives: 3D shapes. Now, by contrast to 2D shapes, 3D shapes appear on relatively few GRE problems, so it may help your scores more to really master the different aspects of 2D shapes.
There are two main 3D shapes that the GRE tests on: the "rectangular solid" (also known as the rectangular prism) and the cylinder (also called the "right circular cylinder").
And there are two main properties of such objects that we will care about:
1. Surface Area: this is the total area of all the surfaces of a 3D shape
So, in a rectangular solid, we add up the area of each of the 6 sides. And, in a cylinder, we add up the area of the circle on the top, the circle on the bottom, and the curved portion in between the two.
2. Volume: the space enclosed by a 3D shape
Imagine filling up the 3D shape with water. The amount of space that the water takes up, that’s the volume of the object.
By analogy, you could think of the area of a 2D shape as capturing how much space the lines enclosed. Similarly, volume is a measure of how much space the surfaces of a 3D shape enclose.
In subsequent posts, we will talk about the rectangular solid and cylinder in greater detail and show how to calculate the surface area and volume for each.
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The perimeter is just the distance it would take if you were to walk the outside edges of a given shape. So, for the following triangle:
the perimeter is just
Finding the perimeter of a square or a rectangle is similarly straightforward:
Perimeter of a Square or Rectangle: In the following diagrams, the perimeter of the square is and the perimeter of the rectangle is
Perimeter of a Rhombus: Since a rhombus is defined by having four sides of equal length, the perimeter of the below rhombus is just
Perimeter of a Trapezoid:
The perimeter of a trapezoid is a little trickier, but recall that we can break a trapezoid up into two right triangles and a rectangle:
And then, if we have the bases of the right triangles, we can use the Pythagorean Theorem to find the length of the diagonal bits:
Thus, we get that the perimeter is:
Perimeter of a Parallelogram
Remember that the parallel sides of a parallelogram have the same length. Thus, for the below parallelogram:
the perimeter is just .
include some simpler problems here for finding the perimeter of when they give you basically all the right information
Sometimes, the perimeter figures in a word problem. So, for example, you might have:
Example 1
A farmer has a square field whose perimeter is twice its area. What is the area of the field?
Practice Problems
1. The outer rectangle below is units apart from the smaller rectangle on the top and bottom sides, and units apart on the left and right sides. The outer rectangle has a base of 8 units and a height of 12 units. What is the perimeter of the inner rectangle?
2. The perimeter of a regular -sided shape is Find the length of a side.
3. The length of a side of a regular hexagon is What is the perimeter?
Comment on this
Many geometry problems ask you to find the area of a shape or to compare the area of two shapes or to use the area of a shape to deduce some other value. In previous posts, we have already given the standard formulas for finding the area of such shapes (see our appendix for a full list of such formulas). But the GRE will often give you the relevant information in subtle ways, or ask you to do surprising things with the information they’ve given. So in what follows, we give some practice problems intended to help you become used to using these standard formulas in less standard ways.
But first, here are some general steps to follow in trying to solve a geometry problem:
1. Record what the problem tells you.
- The problem may give you certain lengths or angle measurements or areas or perimeters; keep track of this. It is best to write it down somewhere, either on a diagram or just in a list.
2. If you’re stuck, write down some formulas you think may be relevant.
- Thus, if the problem asks you to find the area of some shape, write down the area formula for that shape! Or if the problem gives you a 45-45-90 right triangle, write down the ratio of the side lengths. Sometimes writing such stuff down can help spark connections in your mind.
3. If you’re stuck, look back carefully at the problem and see if there’s any information you missed.
- Generally, these problems are quite parsimonious: they give you exactly what is needed to solve the problem, and no more. So if you find yourself with some unused fact, try and fit it into the problem somewhere. It’s unlikely they would’ve included something totally irrelevant to your problem.
4. If you don’t think you can get the solution quickly, just move on!
- Remember that the math section gives you 35 minutes for 20 questions, so you can only afford to spend an average of one minute and 45 seconds per question. There’s no shame in flagging a question that looks tough so that you can come back to it if time permits. Look at it this way: if you don’t get through the test, there may be easy questions down the road that you’re effectively giving up on. By flagging it and coming back later, you help ensure that you get all the low-hanging fruit in the test.
Practice Problems
1. In the following diagram, and are both squares with side lengths of 2 and area of area of area of Find the area of the shaded rhombus.
2. In the following diagram, , and are equally spaced. How many scalene triangles can be formed with vertices at those points?
3. The following points are equally spaced. How many equilateral triangles can be formed? If three distinct points are randomly chosen, what is the likelihood that they form an equilateral triangle?
4. The outer rectangle below is units apart from the smaller rectangle on the top and bottom sides, and unit apart on the left and right sides. The inner rectangle has a height of 4 units and a base of 2 units. What is the area of the outer rectangle?
5. The radius of both circles below is 1. What is the area of triangle ?
6. What is the area of ?
7. In the following diagram, What is the area of ?
Challenge Problem: You are not likely to see anything as convoluted as the following on the GRE. Still, it might be a good way to push your understanding of the above material:
8. The radius of the following circle is 5. Find the shaded area.
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Rectangles: four sides connected by four 90-degree angles
Rectangle Area Formula:
Squares: four sides of equal length connected by four 90-degree angles.
Square Area Formula:
Parallelograms: four-sided figure where opposite sides are parallel
Parallelogram Area Formula:
Parallelograms also have the nice property that opposite angles are of the same degree, and opposite sides are the same length.
Rhombus: a parallelogram with equal sides
Rhombus Area Formula:
Trapezoids: four-sided figure with one pair of parallel opposing sides
Trapezoid Area Formula:
Now a lot of these definitions overlap with others, so here is a diagram that explains how all of these shapes relate to one another:
Finally, one fact that is sometimes helpful in solving problems with quadrilaterals is:
Quadrilaterals Have 360 Degrees: The sum of the angles in a quadrilateral is 360.
This fact will come up again when we talk about finding the values of unknown angles.
Practice Problems:
1. Find the area of the entire figure below.
2. Find the area of the entire figure below.
3. The area of the below parallelogram is 30, is perpendicular to , and the ratio of the area of to the area of is 1 to 3. What is the value of ?
4. The dotted lines below meet at a right angle. Find
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Not all triangles are equally loved. Right triangles, for example, were adored by early mathematicians, and even now, are cherished by standardized-test makers. These are the triangles that have a ninety-degree angle, like so
It is tradition to assign to the shortest side, to the second shortest, and to the longest side. We also call the side opposite from the right angle (in the above diagram, ) the hypotenuse of the right triangle, and we call the other two sides the legs of the right triangle.
These figures are so well-beloved partly because they have many interesting properties. The most famous property is:
Pythagorean Theorem: If and are the legs of a right triangle and is the hypotenuse of that right triangle, then
And even among the right triangles, there are two kinds that are especially cherished:
30-60-90 Right Triangle: Any right triangle with angles of 30, 60, and 90 degrees will have the following side lengths (where is some fixed number):
45-45-90 Right Triangle: Any right triangle with angles of 45, 45 and 90 degrees will have the following side lengths (where is some fixed number):
Knowing the ratios of the side lengths of such triangles can help (and may sometimes be crucial) to solving a geometry problem on the GRE. For example:
Example 1
Find the area of the following right triangle:
Example 2
In the following diagram, the area of triangle A is 18. Find . Also, find the area of triangle B.
Now, for the 30-60-90 right triangle, one just has to memorize the side ratios. But you can easily derive the side ratios of the 45-45-90 right triangle:
Finally, some problems may give you the side lengths of a triangle and you will have to infer that the triangle in question is actually a right triangle. In other words, we have:
Reverse Pythagorean Theorem: If a triangle has side lengths such that , then the triangle is a right triangle.
In applying the Reverse Pythagorean Theorem, here are some common side-ratios to look out for:
3-4-5
5-12-13
So if you see a triangle like:
You should note that its side lengths have the ratio 3:4:5 (simply divide all the side lengths by 3) and thus it is a right triangle.
Practice Problems
1. Find the value of
2. Find the value of
3. Find the length of AC.
4. In the following diagram, What is the ratio of the area of to the area of ?
5. What is the length of ?
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# 14.4 Electric power and energy (Page 2/7)
Page 2 / 7
Different insights can be gained from the three different expressions for electric power. For example, $P={V}^{2}/R$ implies that the lower the resistance connected to a given voltage source, the greater the power delivered. Furthermore, since voltage is squared in $P={V}^{2}/R$ , the effect of applying a higher voltage is perhaps greater than expected. Thus, when the voltage is doubled to a 25-W bulb, its power nearly quadruples to about 100 W, burning it out. If the bulb’s resistance remained constant, its power would be exactly 100 W, but at the higher temperature its resistance is higher, too.
## Calculating power dissipation and current: hot and cold power
(a) Consider the examples given in Ohm’s Law: Resistance and Simple Circuits and Resistance and Resistivity . Then find the power dissipated by the car headlight in these examples, both when it is hot and when it is cold. (b) What current does it draw when cold?
Strategy for (a)
For the hot headlight, we know voltage and current, so we can use $P=\text{IV}$ to find the power. For the cold headlight, we know the voltage and resistance, so we can use $P={V}^{2}/R$ to find the power.
Solution for (a)
Entering the known values of current and voltage for the hot headlight, we obtain
$P=\text{IV}=\left(2\text{.}\text{50 A}\right)\left(\text{12}\text{.}\text{0 V}\right)=\text{30}\text{.}\text{0 W.}$
The cold resistance was $0\text{.}\text{350}\phantom{\rule{0.25em}{0ex}}\Omega$ , and so the power it uses when first switched on is
$P=\frac{{V}^{2}}{R}=\frac{\left(\text{12}\text{.}\text{0 V}{\right)}^{2}}{0\text{.}\text{350}\phantom{\rule{0.25em}{0ex}}\Omega }=\text{411 W.}$
Discussion for (a)
The 30 W dissipated by the hot headlight is typical. But the 411 W when cold is surprisingly higher. The initial power quickly decreases as the bulb’s temperature increases and its resistance increases.
Strategy and Solution for (b)
The current when the bulb is cold can be found several different ways. We rearrange one of the power equations, $P={I}^{2}R$ , and enter known values, obtaining
$I=\sqrt{\frac{P}{R}}=\sqrt{\frac{\text{411 W}}{0\text{.}\text{350}\phantom{\rule{0.25em}{0ex}}\Omega }}=\text{34}\text{.}\text{3 A.}$
Discussion for (b)
The cold current is remarkably higher than the steady-state value of 2.50 A, but the current will quickly decline to that value as the bulb’s temperature increases. Most fuses and circuit breakers (used to limit the current in a circuit) are designed to tolerate very high currents briefly as a device comes on. In some cases, such as with electric motors, the current remains high for several seconds, necessitating special “slow blow” fuses.
## The cost of electricity
The more electric appliances you use and the longer they are left on, the higher your electric bill. This familiar fact is based on the relationship between energy and power. You pay for the energy used. Since $P=E/t$ , we see that
$E=\text{Pt}$
is the energy used by a device using power $P$ for a time interval $t$ . For example, the more lightbulbs burning, the greater $P$ used; the longer they are on, the greater $t$ is. The energy unit on electric bills is the kilowatt-hour ( $\text{kW}\cdot \text{h}$ ), consistent with the relationship $E=\text{Pt}$ . It is easy to estimate the cost of operating electric appliances if you have some idea of their power consumption rate in watts or kilowatts, the time they are on in hours, and the cost per kilowatt-hour for your electric utility. Kilowatt-hours, like all other specialized energy units such as food calories, can be converted to joules. You can prove to yourself that $\text{1 kW}\cdot \text{h = 3}\text{.}6×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}\text{J}$ .
anyone know any internet site where one can find nanotechnology papers?
research.net
kanaga
Introduction about quantum dots in nanotechnology
what does nano mean?
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
Abigail
for teaching engĺish at school how nano technology help us
Anassong
Do somebody tell me a best nano engineering book for beginners?
there is no specific books for beginners but there is book called principle of nanotechnology
NANO
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc
NANO
so some one know about replacing silicon atom with phosphorous in semiconductors device?
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
for screen printed electrodes ?
SUYASH
What is lattice structure?
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials and their applications of sensors.
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
Good
Berger describes sociologists as concerned with
How we can toraidal magnetic field
How we can create polaidal magnetic field
4
Because I'm writing a report and I would like to be really precise for the references
where did you find the research and the first image (ECG and Blood pressure synchronized)? Thank you!! |
# 1.16: Coterminal Angles
Difficulty Level: At Grade Created by: CK-12
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While playing a game with friends, you use a spinner that looks like this:
As you can see, the angle that the spinner makes with the horizontal is 60\begin{align*}60^\circ\end{align*}. Is it possible to represent the angle any other way?
### Coterminal Angles
Consider the angle 30\begin{align*}30^\circ\end{align*}, in standard position.
Now consider the angle 390\begin{align*}390^\circ\end{align*}. We can think of this angle as a full rotation (360)\begin{align*}(360^\circ)\end{align*}, plus an additional 30 degrees.
Notice that 390\begin{align*}390^\circ\end{align*} looks the same as 30\begin{align*}30^\circ\end{align*}. Formally, we say that the angles share the same terminal side. Therefore we call the angles co-terminal. Not only are these two angles co-terminal, but there are infinitely many angles that are co-terminal with these two angles. For example, if we rotate another 360\begin{align*}360^\circ\end{align*}, we get the angle 750\begin{align*}750^\circ\end{align*}. Or, if we create the angle in the negative direction (clockwise), we get the angle 330\begin{align*}-330^\circ\end{align*}. Because we can rotate in either direction, and we can rotate as many times as we want, we can continuously generate angles that are co-terminal with 30\begin{align*}30^\circ\end{align*}.
#### Identifying Co-Terminal Angles
For the following questions, determine if the angle is co-terminal with 45\begin{align*}45^\circ\end{align*}
1. 45\begin{align*}-45^\circ\end{align*}
No, it is not co-terminal with 45\begin{align*}45^\circ\end{align*}
2. 405\begin{align*}405^\circ\end{align*}
Yes, 405\begin{align*}405^\circ\end{align*} is co-terminal with 45\begin{align*}45^\circ\end{align*}.
3. 315\begin{align*}-315^\circ\end{align*}
Yes, 315\begin{align*}-315^\circ\end{align*} is co-terminal with 45\begin{align*}45^\circ\end{align*}.
### Examples
#### Example 1
Earlier, you were asked if it is possible to represent the angle any other way.
You can either think of 60\begin{align*}60^\circ\end{align*} as 420\begin{align*}420^\circ\end{align*} if you rotate all the way around the circle once and continue the rotation to where the spinner has stopped, or as 300\begin{align*}-300^\circ\end{align*} if you rotate clockwise around the circle instead of counterclockwise to where the spinner has stopped.
#### Example 2
Find a coterminal angle to 23\begin{align*}23^\circ\end{align*}
A coterminal angle would be an angle that is at the same terminal place as 23\begin{align*}23^\circ\end{align*} but has a different value. In this case, 337\begin{align*}-337^\circ\end{align*} is a coterminal angle.
#### Example 3
Find a coterminal angle to 90\begin{align*}-90^\circ\end{align*}
A coterminal angle would be an angle that is at the same terminal place as 90\begin{align*}-90^\circ\end{align*} but has a different value. In this case, 270\begin{align*}270^\circ\end{align*} is a coterminal angle.
#### Example 4
Find two coterminal angles to 70\begin{align*}70^\circ\end{align*} by rotating in the positive direction around the circle.
Rotating once around the circle gives a coterminal angle of 430\begin{align*}430^\circ\end{align*}. Rotating again around the circle gives a coterminal angle of 790\begin{align*}790^\circ\end{align*}.
### Review
1. Is 315\begin{align*}315^\circ\end{align*} co-terminal with 45\begin{align*}-45^\circ\end{align*}?
2. Is 90\begin{align*}90^\circ\end{align*} co-terminal with 90\begin{align*}-90^\circ\end{align*}?
3. Is 350\begin{align*}350^\circ\end{align*} co-terminal with 370\begin{align*}-370^\circ\end{align*}?
4. Is 15\begin{align*}15^\circ\end{align*} co-terminal with 1095\begin{align*}1095^\circ\end{align*}?
5. Is 85\begin{align*}85^\circ\end{align*} co-terminal with 1880\begin{align*}1880^\circ\end{align*}?
For each diagram, name the angle in 3 ways. At least one way should use negative degrees.
1. Name the angle of the 8 on a standard clock two different ways.
2. Name the angle of the 11 on a standard clock two different ways.
3. Name the angle of the 4 on a standard clock two different ways.
4. Explain how to determine whether or not two angles are co-terminal.
5. How many rotations is 4680\begin{align*}4680^\circ\end{align*}?
To see the Review answers, open this PDF file and look for section 1.16.
### Notes/Highlights Having trouble? Report an issue.
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### Vocabulary Language: English
TermDefinition
Coterminal Angles A set of coterminal angles are angles with the same terminal side but expressed differently, such as a different number of complete rotations around the unit circle or angles being expressed as positive versus negative angle measurements.
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# Congruence And Similarity Of Triangles : Download Maths Notes For CTET Exam Free PDF
Mathematics is an equally important section for CTET, MPTET, KVS & DSSSB Exams and has even more abundant importance in some other exams conducted by central or state govt. Generally, there are questions asked related to basic concepts and Facts of the Congruence And Similarity Of Triangles.
To let you make the most of Mathematics section, we are providing important facts related to the Congruence And Similarity Of Triangles. At least 2-3 questions are asked from this topic in most of the teaching exams. We wish you all the best of luck to come over the fear of the Mathematics section.
How to Overcome Exam Fever, Especially When You Fear Maths
## Congruence and Similarity of Triangles
Congruent Triangles: Triangles are congruent when they have exactly the same three sides and exactly the same three angles.
What is “Congruent”?
It means that one shape can become another using turns, flips and/or sliders.
The equal sides and angles may not be in the same position (if there is a turn or a flip), but they are there.
Same Sides: When the sides are same then the triangles are congruent.
For example:
Same Angles: Two triangles with the same angles might be congruent.
1. SSS congruent
Side – Side – Side congruence. When two triangles have corresponding sides equal that are congruent as shows below, the triangles are congruent.
1. SAS Congruence
Side – Angle – Side Congruence. When two triangles have corresponding angles and sides equal that are congruent as shown below, the triangles are congruent.
1. ASA Congruence
Angle – Side – Angle Congruence. When two triangles have corresponding angles and sides equal that are congruent as shown below, the triangles themselves are congruent.
Mathematics Study Notes For All Teaching Exams
1. AAS Congruence OR SAA Congruence
Angle – Angle – Side congruence. When two triangles have corresponding angles and sides equal that are congruent as shown below, the triangles are congruent.
1. HL Congruence
Hypotenuse – leg congruence. When two triangles have corresponding sides equal that are congruent as shown below, the triangles are congruent.
Similarity of Triangles
Two triangles are similar if and only if the corresponding sides are in proportion and the corresponding angles are congruent.
How To Score 25+ Marks In Mathematics For CTET Exam?
Methods of proving triangles similar:
• If two angles of one triangle are congruent to two angles of another triangle, the triangles are similar.
If ∠ A = ∠ D
∠ B = ∠ E
Then : ∆ ABC ∼ ∆ DEF
• If the three sets of corresponding sides of two triangle are in proportion, the triangles are similar.
Then : ∆ ABC ∼ ∆ DEF
Practice More Mathematics Quiz Here
• If an angle of one triangle is congruent to the corresponding angle of another triangle and the lengths of the sides including these angles are in proportion, the triangles are similar.
If ∠ A = ∠ D
Then : ∆ ABC ∼ ∆ DEF
• If a line is parallel to one side of a triangle and intersects the other two sides of the triangle, the line divides these two sides proportionally.
Properties of Similar triangles
If the two triangles are similar, then for the proportional / corresponding sides have the following results.
= Ratio of medians
= Ratio of angles bisectors
1. Ratio of areas = Ratio of squares of corresponding sides.
i.e. If ∆ ABC ∼ ∆ PQR,
Then,
1. In a right-angled triangle, the triangles on each side of the altitude drawn from the vertex of the right angle to the hypotenuse are similar to the original triangle and to each other too.
i.e., ∆ ABC ∼ ∆ BDC ∼∆ CDA.
### Some facts on Right angle triangle
• CD² = BD × DA
• BC × CA = BA × CD
• BC² = BD × BA
• AC² = AD × BA
Area Based question when two triangles are similar
If ∆ ABC ∼ ∆ DEF Then
The Ratio of their area is the square of their sides, Medians, Altitudes, Perimeters, Angle Bisectors.
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Grade 1 Module 1 Eureka Vocabulary
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Growing set of math vocabulary cards used in grade 1 module 1 Eureka math lessons.
Vocabulary Words:
-Numerals
-5 Group Cards
-Number Bond
-Number Sentence
-Expression
-Equality
-Doubles
-Doubles + 1
-Subtraction
Determine the unknown whole number in an addition or subtraction equation relating three whole numbers. For example, determine the unknown number that makes the equation true in each of the equations 8 + ? = 11, 5 = ▯ - 3, 6 + 6 = ▯.
Understand the meaning of the equal sign, and determine if equations involving addition and subtraction are true or false. For example, which of the following equations are true and which are false? 6 = 6, 7 = 8 - 1, 5 + 2 = 2 + 5, 4 + 1 = 5 + 2.
Add and subtract within 20, demonstrating fluency for addition and subtraction within 10. Use strategies such as counting on; making ten (e.g., 8 + 6 = 8 + 2 + 4 = 10 + 4 = 14); decomposing a number leading to a ten (e.g., 13 - 4 = 13 - 3 - 1 = 10 - 1 = 9); using the relationship between addition and subtraction (e.g., knowing that 8 + 4 = 12, one knows 12 - 8 = 4); and creating equivalent but easier or known sums (e.g., adding 6 + 7 by creating the known equivalent 6 + 6 + 1 = 12 + 1 = 13).
Relate counting to addition and subtraction (e.g., by counting on 2 to add 2).
Apply properties of operations as strategies to add and subtract. If 8 + 3 = 11 is known, then 3 + 8 = 11 is also known. (Commutative property of addition.) To add 2 + 6 + 4, the second two numbers can be added to make a ten, so 2 + 6 + 4 = 2 + 10 = 12. (Associative property of addition.)
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# Vector Fundamentals 4
$$\newcommand{\bv}[1]{{\bf #1}}$$ $$\newcommand{\bperp}{{\bf perp}}$$ $$\newcommand{\bproj}{{\bf proj}}$$
Geometry plays a central role in game programming, underlying not only graphics, but also physics and spatial AI. A lot has happened since geometry was first formalised by the Greeks more than two thousand years ago, most notably the invention of coordinates and vectors. These inventions are indispensible for doing geometry on machines (at least, the kind of geometry that is useful for games).
This article introduces the basic concepts of vectors. In a following article, we will look at how vectors can be used to solve geometric problems that arise in computer graphics, physics and AI.
# Introduction
(I am assuming here that you already know about coordinates, axes and points; if you don’t, check out this link.)
A vector specifies the relationship between two points; in essence, it says how to get from point A to point B.
Suppose point A is given by the coordinates (ax,ay), and B is given by the coordinates (bx, by). Then the vector from A to B is defined as [bx – ax,by – ay]. The two values are called the components or sometimes the coordinates of the vector. We can use a variable $$\bv{v}$$ to denote a vector; vectors are sometimes printed with an over-bar $$\bar{v}$$ or arrow $$\vec{v}$$.
Examples
Notice that even though the first two examples have different pairs of points, the same vector results. Vectors at different positions are the same as long as the vector components are the same. This is convenient when we draw pictures – we can translate the vector everywhere, without changing it.
In the last example, the vector from the origin has the same coordinates as the point B. This useful fact makes it possible to drop the distinction between points and vectors – we always think of a point as a vector with the same coordinates. For the remainder of this article, when I say point A, I really mean the vector from the origin to point A.
When we operate with vectors, an operand or result that is not a vector is referred to as a scalar. A scalar is simply a real number; the terminology is useful to distinguish it from a vector component.
• u, v, and w for vectors (with components vx , vy , etc.);
• a and b for scalars; and
• $$\alpha$$ for angles.
Suppose v is a vector from point A to point B, and w is a vector from point B to point C. Then the sum of the vectors v and w gives a vector from point A to point C. Here is how to calculate the sum of two vectors:
If v = [vx,vy] and w = [wx,wy], then
$$\bv{v} + \bv{w} = [v_x + w_x, v_y + w_y].$$
Examples
Notice in the figures that the second vector starts where the first vector ends. In this configuration, the sum starts where the first vector starts, and ends where the second vector ends.
We can find a vector in the opposite direction (with the same size) by inverting the signs of the components. We can also write a minus in front of a vector to show that the components should be inverted.
If v = [vx ,vy], then
$$-\bv v = [-v_x, -v_y] .$$
Examples
Subtraction is performed by subtracting vector components.
If v = [vx,vy] and w = [wx,wy], then
$$\bv v – \bv w = [v_x – w_x, v_y – w_y].$$
Examples
In the figures, both vectors are drawn to start from the same point. In this configuration, the difference is from the endpoint of the second vector to the endpoint of the first vector.
The vector with all components equal to 0 is called the zero vector, and we denote it by 0.
Vector addition and subtraction have properties that mimic addition and subtraction of real numbers. These properties, and those listed for other vector operations, can be exploited for two purposes.
Firstly, when tackling a geometric problem, they are used to manipulate the equations that you write. Typically, the problem is to express some unknown quantity in terms of known quantities; properties of vector operations make this possible.
Secondly, when implementing geometric code on a computer, they can be used to implement algorithms more efficiently and with less code. In particular, they can help reduce the number of cases to handle.
Many properties have names (given in the right-hand columns of property tables). The names are not so important for understanding and using vector operation properties, but as always names are useful when talking about stuff. In particular, they can help make comments that state your assumptions and reasoning precise. Even if a fellow coder (such as your future self!) does not know what the term means, using standard terms will serve as helpful keywords that can be searched on the Internet.
## Scalar Multiplication
We can multiply vectors with real numbers by multiplying each of the components with a real number.
If v = [vx,vy], then
$$a \bv v = [av_x, av_y].$$
The multiplication does not affect the direction in which the vector is pointing, only its length.
Examples
Division by a scalar is defined as multiplication with the scalar’s reciprocal, that is
$$v / a = (1 / a) v.$$
When a = 0, v/a is undefined.
Properties of scalar multiplication
## Unary Vector Operations
### Vector Length
The length of a vector is given by Pythagoras’s Theorem. We denote the length of a vector v by |v|.
If v = [vx,vy], then
$$|\bv v| = \sqrt{v_x^2 + v_y^2}.$$
Vector length is also called the Euclidean norm (yes, there are many other norms as well).
Example
Properties of vector length
### Normalisation
It is sometimes necessary to work with a vector that points in the same direction as a given vector, but is of unit length. If the given vector is v, then the unit vector in the same direction is denoted by v* , and is called the normalised vector of v:
$$\bv{v}^* = \bv v / |\bv v|$$
Example
Many formulas simplify drastically if the vectors in them are normalised; this can be exploited to gain a performance benefit in code.
### Perpendicular Operator
The perpendicular operator gives a vector of the same length, rotated 90º counter clockwise, and is denoted by $$\bv v ^ \perp$$:
$$\bv v ^ \perp = [-v_y, v_x].$$
This operator is useful for doing some cheap rotations of special angles. For example, $$(\bv v + \bv v^\perp) / \sqrt{2}$$gives a vector of the same size as v, rotated 45º counter clockwise.
Example
Properties of the perpendicular operator
## Vector Products
There are at least six ways to define useful vector products. Of these the dot product, perp dot product and cross product are the most useful in game programming. The dot and perp dot products are defined in this section; the cross product only applies to 3D, and is defined in the next section.
### Dot Product
The dot product of two vectors is defined as follows:
If v = [vx,vy] and w = [wx,wy], then
$$\bv v \cdot \bv w = v_xw_x + v_yw_y.$$
The most useful property of the dot product is that it gives us an easy way to calculate the cosine of the (smaller) angle between two vectors:
$$\bv v \cdot \bv w = |\bv v||\bv w|\cos \alpha,$$
where $$\alpha[\latex] is the smaller of the angles between two vectors. This formula is the key to many geometric algorithms, and you should memorise it. Examples Properties of the dot product Perp dot product The 2D perp dot product is defined as follows: If v = [vx,vy] and w = [wx,wy], then \bv v * \bv w = v_xw_y – v_yw_x. (Note: the notation for the perp dot product used here is not conventional. It is usually written $\bv v^\perp \cdot \bv w[\latex]. The star notation is used to emphasise that the perp dot product is an operation in its own right, and not merely a derivative of the dot product.) The perp dot product’s most useful property is that it gives us an easy way to compute the sine of the smaller angle between two vectors: \bv v * \bv w| = |v||w|\sin \alpha. Note the absolute value marks in the formula above. The product v*w can be either positive or negative, but [latex]|\v v| |\bv w| \sin \alpha$$ is always positive or zero (since $$0^\circ \leq \alpha < 180^\circ$$). If we work with directed angles, so that clockwise angles are negative, and counter clockwise angles are positive, we can drop the absolute value markers. Then the perp dot product also gives us a convenient way to determine in which direction something was rotated. Examples Properties of the perp dot product Relationships between the dot and perp dot product The dot and perp dot products are closely related. The table below shows those relationships. Relationships between the dot and perp dot product ### Projections and Perpendiculars Any vector v can be written as the sum of a vector in the same direction as another vector w, and a vector perpendicular to w (as long as w is not the zero vector). The first vector is called the projection ofv onto w; the second is the perpendicular of v on w. We denote these vectors by projw v and perpw vrespectively. A little bit of trigonometry help us find projections and perpendiculars in terms of dot and perp dot products: \bproj _\bv w \bv v = \frac{\bv v \cdot \bv w}{|\bv w|^2 }\bv w and \bperp _\bv w \bv v = \frac{\bv v * \bv w}{|\bv w|^2 }\bv w^\perp Examples Properties of proj and perp ### Three Dimensions In three dimensions, we use three components to denote a vector: [x,y,z]. Most of the definitions in 3D are analogous to the 2D case, and all properties in the tables hold in 3D as well. Here is a quick summary of definitions for vector operations in 3D: • $$\bv v + \bv w = [v_x + w_x, v_y + w_y, v_z + w_z]$$ • $$– \bv v = [-v_x, -v_y, -v_z]$$ • $$\bv v \cdot \bv w = v_xw_x + v_yw_y + v_zw_z$$ • $$|\bv v| = \sqrt{v_x^2 + v_y^2 + v_z^2}$$ The exception is the perp dot product, which is only defined for 2D. In 3D, we make use of the cross product (see the next section). This also means that we cannot calculate the perpendicular of a vector on another vector in terms of the perp dot product; instead, we use the identity \bperp_\bv w \bv v = \bv v – \bproj_\bv w \bv v. ### Cross Product The cross product replaces the 2D perp dot product in 3D. It differs from other vector products defined so far in that the result is a vector, not a scalar. The somewhat complicated definition looks like this: \bv v \times \bv w = [v_yw_z – v_zw_y,v_zw_y – v_yw_z,v_xw_y – v_yw_x]. The usual trick to remember this formula requires a bit a math that is not covered here. I will ignore the mathematical aspects, and show the basic trick in any case. The mnemonic is this diagram: To calculate the x component, cross out the row and column containing x. Now subtract the product of the orange components from the product of the blue components. Do the same for y: cross out the row and column containing y, and subtract orange from blue (note, the blue and orange components are reversed). And finally, do the same with z: cross out the row and column containing z, and subtract blue from orange (here the blue and orange components are as they were for x). If this confuses you, do not worry. You can look for a better trick, or just learn the formula straight. The cross product in 3D gives a vector with the property that it is perpendicular to both operands. There are two such vectors, the right one is given by the right hand rule: when the fingers of the right hand curl from v to w, the thumb points to v × w. Like the 2D perp dot product, we can use the cross product to calculate the sine of the angle between two vectors: |\bv v \times \bv w| = |\bv v||\bv w|\sin \alpha. When efficiency is a concern, a problem in 3D that requires the sine of an angle is converted to a problem that requires the cosine instead – this is computed efficiently using the dot product. Properties of the cross product ## Relationships between the dot and cross product The dot and cross products are closely related. The table below shows those relationships. Relationships between the dot and cross products $$|\bv v \times \bv w| / \bv v \cdot \bv w = \tan \alpha$$ This gives us a way to caclulate the tangent of the angle between two vectors. $$|\bv v \times \bv w|^2 + (\bv v \cdot \bv w)^2 = (|\bv v||\bv w|)$$ This is the 3D vector translation of the familiar trigonometric identity [\latex]\sin^2 \alpha + \cos^2 \alpha = 1$.
Herman Tulleken is a game developer co-founder of Plinq. He thinks computers are a necessary evil to make games, and sees frying a CPU or two in the process as a worthy sacrifice.
## 4 thoughts on “Vector Fundamentals”
• gregersn
Seems like there’s a lot of problems with rendering the formulas. At least with Chrome as I am using here.
• QueryVillain
Excellent article as always.
There’s a small mistake in first table, third vector’s B coordinate. It should be (3, -1)
• wofwof
Why in cross product equation for y component you use wy and vy? I thought that for y component we cross out those. |
From math^2
The Breakdown
A quadrilateral is a four-sided, two-dimensional enclosed figure (quad means four, lateral means side). Squares and rectangles are quadrilaterals, but quadrilaterals are not always square or rectangular. The sides do not have to be the same length, and can meet at any angle.
This section will look at angle relationships within quadrilaterals. Here are a few shapes that meet the requirements:
Here are a few that don’t. None of the following shapes are quadrilaterals. If you don’t understand why, slow down and have a think about each one. The first sentence in this article ought to help you out.
Figure 2: Examples of shapes that are not quadrilaterals
Sum of all angles in a quadrilateral
What do you think the angles of different quadrilaterals add up to? Why? Do you think quadrilaterals all have different angle sums, or does it depend on the length of their sides? (An angle sum is the total value of all interior angles in a shape.) In the following diagram (see Figure 3), does A have the same angle sum as B?
Figure 3: Totally different quadrilaterals A and B but their angle sums are just the same
Think about cutting a square or rectangle in half diagonally…
If you already know that the angles of any and every triangle add to 180°, it should make sense to you that the angles of a quadrilateral sum to 360° (two times 180°). This is easy to do with square and rectangular quadrilaterals, but harder when the sides are not all equal.
Later on in this article we’ll have a look at trapezoids. For now, let’s examine some principles related to quadrilaterals of a general nature.
Principle 2:
If two opposite angles in a quadrilateral are right angles, the other two opposite angles are supplementary (they add to 180°).
In the following diagram, angles a and b must add to 180° for the total enclosure to equal 360°.
Figure 4: Since quadrilaterals have sum of its angles not more than 360 degrees, angles a and b must add up to 180.
Principle 3:
Quadrilateral edges drawn to extend past corners will create supplementary interior/exterior angles.
In the following diagram, angles a and c are supplementary, and angles b and d are supplementary.
Figure 4: Quadrilaterals with extending sides.
What are trapezoids?
A trapezoid is a quadrilateral that has exactly two parallel sides. In other words, it has two parallel sides and two non-parallel sides. The parallel sides are called bases, the non-parallel sides connecting the bases are called legs. The median is the finite line segment connecting the two midpoints of the legs. In the following diagram showing Trapezoid Y, \bar{AC} and \bar{BD} are the legs, \bar{AB} and \bar{CD} are the bases, and \bar{EF} is the median.
Figure 6: A diagram of Trapeziod Y.
Trapezoid Y is an isosceles trapezoid because its legs are congruent (\bar{AC} = \bar{BD}). A trapezoid without congruent legs is simply called a trapezoid.
An isosceles trapezoid is symmetrical when cut at a right angle to the midpoint of the median.
The base angles of a trapezoid are the angles at the ends of its longer base, in Trapezoid Y’s case, \angle{C} and \angle{D} .
Principles related to trapezoids
Let’s look at some basic principles pertaining trapezoids.
Principle 1:
The base angles of an isosceles trapezoid are congruent.
Figure 7: Other form of trapezoid.
In the above diagram \angle{a} \propto \angle{b}.
Principle 2:
Any trapezoid with congruent base angles is an isosceles trapezoid.
Thus, in the diagram used above, if \angle{A} \propto \angle{B}, then \angle{A} \propto \angle{B} .
Applying Algebra to Trapezoid Figures
In the following diagram, use your knowledge of geometry to find values for all four angles of each trapezoid by first finding values for x and y. These shouldn’t be too hard for you to solve, just use a little logic and you’ll be fine.
Figure 8
Example #1
ABCD is an isosceles trapezoid.
Please note that a lower case x means ‘letter x’ which replaces a number, while upper case X denotes multiplication.
The statement that ABCD is isosceles has been given to us. From this we can deduce two truths:
1. \angle{C}= 5x + 60°. 2. 5x = 2x + y.
Therefore 5x – y = 2x. Therefore y = 3x. Working out the values for x and y is not too difficult from here. Because and are parallel, we know that = 180°. In other words 10x + 60 = 180°. Therefore 10x = 120°. (Deduct 60° from 180°.) Therefore x = 120°/10 x = 12°. Finding values for angles is as simple as substituting this value into x in the diagram. = 5x = 5 X 12 = 60° = 5x + 60 = (5 X 12) + 60 = 120° Because ABCD is isosceles, we know and .
Figure 9
Example #2
As before, work out the angle values for each corner.
Solving the second diagram is a little more interesting. The fact that alone tells us that ABCD is not an isosceles trapezoid, so we’ll have to work out at least three angles in order to deduce the fourth. We know that because and are parallel, we know that = 180°, and that = 180°. We’ll work with and first. To put this mathematically, 12x – 15 + 15 = 180°. Simplified, 12x = 180°. Therefore x = 180/12 = 15°. = 5 X 15 = 75°. = 7 X 15 = 105°. (A possibility to consider: These two values are equal to 180, therefore the lines are parallel, ABCD is indeed a trapezoid. If 180°, and are not parallel… and ABCD is not a trapezoid.) = 9 X 15 = 135°. There are two possible ways to find a value for . The first is by deductive reasoning: = 360 – ( + + ) = 360 – (75 + 105 + 135) = 45°. To put this in words, must equal 360 minus the sum of the other three angles. The second is by algebra: We know that there are 24 fifteens in 360 ( 360/15 = 24). Because the other three angles sum to 21x, we know = 3x. We’ve been given a value of x + 2y for . From this we can conclude x = y because x + 2y = 3x. = 3x = 3 X 15 = 45°. |
Mastering Equation Solving with Fractions: A Step-by-Step Guide
Welcome to Warren Institute! In this article, we will dive into the topic of how to solve equations with fractions. Equations with fractions can seem daunting, but fear not! We will guide you through step-by-step strategies to simplify and solve these equations with ease. From multiplying both sides by a common denominator to clearing fractions, we will equip you with the necessary tools to tackle any equation involving fractions. By the end of this article, you'll be confident in your ability to solve equations with fractions like a pro!
Understanding fractions in equations
In this section, we will explore the basics of working with fractions in equations. We will discuss how to identify different types of fractions and how to handle them when solving equations.
Applying the concept of least common denominator
The concept of least common denominator is essential when solving equations with fractions. In this section, we will explain how to find the least common denominator and utilize it to simplify equations and facilitate the solving process.
Clearing fractions from equations
Sometimes, it is necessary to clear the fractions from equations to obtain a more manageable form. We will demonstrate various methods for eliminating fractions and transforming the equation into an equivalent form that is easier to solve.
Verifying solutions involving fractions
Once we solve an equation with fractions, it is crucial to verify our solutions to ensure they are valid. This section will guide you through the process of substituting the obtained solution back into the original equation and confirming its accuracy.
How do you solve equations with fractions step by step?
To solve equations with fractions step by step in Mathematics education:
1. Multiply both sides of the equation by the least common denominator (LCD) to eliminate the fractions.
2. Simplify the resulting equation by distributing and combining like terms, if necessary.
3. Isolate the variable on one side of the equation by performing addition, subtraction, multiplication, or division operations.
4. Continue simplifying until the variable is isolated.
5. Check the solution by substituting it back into the original equation and ensuring both sides are equal.
What are some strategies for simplifying equations with fractions before solving?
Some strategies for simplifying equations with fractions before solving include:
1. Clearing the fractions: Multiply both sides of the equation by the common denominator of all the fractions involved to eliminate them. This will result in an equation with whole numbers.
2. Simplifying fractions: Simplify any fractions in the equation by canceling out common factors in the numerator and denominator.
3. Combining like terms: Combine any like terms on both sides of the equation to simplify the expression further.
4. Using the least common denominator: If there are multiple fractions with different denominators, find the least common denominator and convert all the fractions to have that denominator for easier simplification.
By simplifying equations with fractions before solving, it becomes easier to manipulate and solve the equation accurately.
Can you provide examples of equations with fractions and their solutions?
Sure! Here are a few examples of equations with fractions and their solutions:
1. Example: Solve for x: (frac{2}{3}x = 4)
Solution: Multiply both sides of the equation by (frac{3}{2}) to isolate x. This gives us (x = frac{4}{frac{2}{3}} = frac{4}{1} cdot frac{3}{2} = 6).
2. Example: Solve for x: (frac{5}{x} = 2)
Solution: Multiply both sides of the equation by x to get rid of the fraction. This gives us (5 = 2x). Then, divide both sides by 2 to isolate x. So, (x = frac{5}{2}).
3. Example: Solve for x: (frac{3}{4}x + frac{1}{2} = 2)
Solution: Subtract (frac{1}{2}) from both sides of the equation to isolate the fraction term. This gives us (frac{3}{4}x = 2 - frac{1}{2} = frac{3}{2}). Then, multiply both sides by (frac{4}{3}) to solve for x. So, (x = frac{4}{3} cdot frac{3}{2} = 2).
These are just a few examples of equations with fractions and their solutions.
Are there any specific rules or formulas to follow when solving equations with fractions?
Yes, there are specific rules and formulas to follow when solving equations with fractions. One important rule is to clear the equation of fractions by multiplying both sides by the least common denominator (LCD) of all the fractions involved. This helps eliminate the fractions and allows you to work with whole numbers or simpler expressions. Another important rule is to remember that when multiplying or dividing fractions, you can simplify by canceling out common factors in the numerator and denominator. Additionally, when adding or subtracting fractions, you need to find a common denominator to combine the fractions. These rules and formulas help ensure accurate solutions when dealing with equations involving fractions.
What are common mistakes to avoid when solving equations with fractions?
One common mistake to avoid when solving equations with fractions is forgetting to find a common denominator before adding or subtracting fractions. This can lead to incorrect answers and confusion. It is important to ensure that all fractions have the same denominator before performing any operations.
In conclusion, understanding how to solve equations with fractions is a crucial skill in mathematics education. By following a systematic approach and applying the necessary techniques, students can confidently tackle equations involving fractions. Remember to clear denominators, combine like terms, and isolate the variable. Keep in mind that practice is key to mastering this concept. With perseverance and determination, students can overcome any challenges they may face when solving these types of equations. So, let's continue to embrace the beauty of mathematics and empower ourselves with the knowledge and skills needed to excel in this subject. |
## Selina Concise Mathematics Class 9 ICSE Solutions Compound Interest (Using Formula)
APlusTopper.com provides step by step solutions for Selina Concise Mathematics Class 9 ICSE Solutions Chapter 3 Compound Interest (Using Formula). You can download the Selina Concise Mathematics ICSE Solutions for Class 9 with Free PDF download option. Selina Publishers Concise Mathematics for Class 9 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.
Selina ICSE Solutions for Class 9 Maths Chapter 3 Compound Interest (Using Formula)
Exercise 3(A)
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At 5% per annum the sum of Rs.6,000 amounts to Rs.6,615 in 2 years when the interest is compounded annually.
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Let Rs.x and Rs.y be the money invested by Pramod and Rohit respectively such that they will get the same sum on attaining the age of 25 years.
Pramod will attain the age of 25 years after 25 – 16 = 9 years
Rohit will attain the age of 25 years after 25 -18 = 7 years
Pramod and Rohit should invest in 400:441 ratio respectively such that they will get the same sum on attaining the age of 25 years.
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More Resources for Selina Concise Class 9 ICSE Solutions
## Selina Concise Mathematics Class 9 ICSE Solutions Expansions (Including Substitution)
APlusTopper.com provides step by step solutions for Selina Concise Mathematics Class 9 ICSE Solutions Chapter 4 Expansions (Including Substitution). You can download the Selina Concise Mathematics ICSE Solutions for Class 9 with Free PDF download option. Selina Publishers Concise Mathematics for Class 9 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.
Selina ICSE Solutions for Class 9 Maths – Chapter 4 – Expansions (Including Substitution)
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# NCERT Solutions for Class 9 Maths Chapter 9 Circles
## NCERT Solutions for Class 9 Maths Chapter 9 Circles
NCERT solutions Chapter 9 Circles is a resource that has been thoughtfully prepared to help students ace a very integral topic. Class 9 maths chapter 9 Circles deals with numerous essential topics like the properties of the circles and manipulating lines and angles within the circles. There are numerous questions based on these topics that can pose a serious obstacle for anyone studying them so it is for the benefit of such students that we have compiled all the answers of each exercise in Class 9 maths chapter 9 Circles.
Students will be exposed to a broad spectrum of concepts in Class 9 maths chapter 9 Circles which may be tough to learn initially. The easiest way to grasp these ideas would be via understanding the questions given in the exercises of chapter 9. For this purpose students will find NCERT solutions class 9 maths chapter to be of great utility.
## Theorems included in NCERT Class 9 Chapter 9 Solutions
Theorem 1: Equal chords of a circle subtend equal angles at the centre.
Theorem 2: If the angles subtended by the chords of a circle at the centre are equal, then the chords are equal.
Theorem 3: The perpendicular from the centre of a circle to a chord bisects the chord.
Theorem 4: The line drawn through the centre of a circle to bisect a chord is perpendicular to the chord.
Theorem 5: Equal chords of a circle (or of congruent circles) are equidistant from the centre (or centres)
Theorem 6: Chords equidistant from the centre of a circle are equal in length.
Theorem 7: The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
Theorem 8: Angles in the same segment of a circle are equal.
Theorem 9: f a line segment joining two points subtends equal angles at two other points lying on the same side of the line containing the line segment, the four points lie on a circle (i.e. they are concyclic).
Theorem 10: The sum of either pair of opposite angles of a cyclic quadrilateral is 180º.
Theorem 11: If the sum of a pair of opposite angles of a quadrilateral is 180º, the quadrilateral is cyclic.
### What is the Importance of NCERT Solutions for Class 9 Maths Chapter 9 Circles?
Students can expect questions along the lines of finding unknown angles, proving the relations given in the question and even word problems based on the concepts. The possible questions that students can encounter in an exam have immense variation and this may also intimidate them while studying the chapter. NCERT Solutions for Class 9 Maths Chapter 9 have been designed to systematically guide the students to the answer to each problem in the NCERT textbook which will make them much more at ease with the topic and reduce their fears surrounding the content. These solutions also promote active recall which is the most efficient strategy for revising any topic.
### How are NCERT Solutions for Class 9 Maths Chapter 9 helpful for Class 9 students?
Students can expect questions along the lines of finding unknown angles, proving the relations given in the question and even word problems based on the concepts. The possible questions that students can encounter in an exam have immense variation and this may also intimidate them while studying the chapter. NCERT Solutions for Class 9 Maths Chapter 9 have been designed to systematically guide the students to the answer to each problem in the NCERT textbook which will make them much more at ease with the topic and reduce their fears surrounding the content. These solutions also promote active recall which is the most efficient strategy for revising any topic.
### Why should we follow NCERT Solutions for Class 9 Maths Chapter 9?
NCERT Solutions Class 9 Maths Chapter 9 Circles are very comprehensive, covering every exercise and every sub-part found in the chapter. Additionally these solutions have been prepared by skilled and qualified individuals who prioritise the understanding and learning of the students. With these two aims in mind the class 9 maths NCERT solutions chapter 9 becomes the best way to become familiar with the concepts present in the NCERT curriculum. Furthermore these solutions have been designed to be in line with the updated NCERT book content ensuring the accuracy of the information provided.
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# How do you solve$\ln x - \ln 3 = 2$?
Last updated date: 16th Jun 2024
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Hint: An exponent that is written in a special way is known as a logarithm. Logarithm functions are just opposite or inverse of exponential functions. We can easily express any exponential function in a logarithm form. Similarly, all the logarithm functions can be easily rewritten in exponential form. In order to solve this equation, we have to use some of the logarithm function properties.
Complete step by step solution:
Here, in this question we have to solve $\ln x - \ln 3 = 2$ for the value of $x$. This question deals with logarithm functions, which are just the inverse of exponential functions. In order to solve this question, we will have to make use of logarithm function properties.Given is, $\ln x - \ln 3 = 2$.
We know that one of the logarithm function properties is, logarithm quotient rule. The logarithm quotient rule says that if ${\log _b}\left( {\dfrac{x}{y}} \right) = {\log _b}\left( x \right) - {\log _b}\left( y \right)$. By making use of the same property in the given equation we get,
$\ln x - \ln 3 = 2 \\ \Rightarrow \ln \left( {\dfrac{x}{3}} \right) = 2 \\$
Same as, ${\ln _e}\left( {\dfrac{x}{3}} \right) = 2$.
Now, we take exponential on both the sides of the equation and we get,
$\Rightarrow \dfrac{x}{3} = {e^2} \\ \therefore x = 3{e^2} \\$
Hence, the value of $x$ in $\ln x - \ln 3 = 2$ is $3{e^2}$.
Note: This problem and similar to these can very easily be solved by making use of different logarithm properties. Students should keep in mind the properties of logarithmic functions. Logarithms are useful when we want to work with large numbers. Logarithm has many uses in real life, such as in electronics, acoustics, earthquake analysis and population prediction. When the base of common logarithm is $10$ then, the base of a natural logarithm is number $e$. |
# How do you write the equation y+2=-2(x-5) in slope intercept form?
##### 1 Answer
May 24, 2017
The slope intercept form is $y = - 2 x + 8$.
#### Explanation:
The slope intercept form for a linear equation is $y = m x + b$, where $m$ is the slope, and $b$ is the y-intercept. To change the equation to slope intercept form, solve for $y$.
$y + 2 = - 2 \left(x - 5\right)$
Expand.
$y + 2 = - 2 x + 10$
Subtract $2$ from both sides.
$y + \textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} - \textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} = - 2 x + 10 - 2$
Simplify.
$y = - 2 x + 8$ $\Leftarrow$ Slope intercept form
graph{y=-2x+8 [-15.82, 16.2, -5.45, 10.57]} |
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Home / Geometry / Circle| Part – 2
# Circle| Part – 2
Exercise : Two chords AB and AC of a circle make circle make equal angles with the radius through A. Prove that AB=AC.
Particular enunciation: Let O is the centre of the circle ABC. The two chords AB and AC has made equal angles ∠BAO and ∠CAO with the radius OA through A.
It is required to prove that AB=AC.
Construction: from the centre O we draw the perpendicular OM and ON respectively to AB and AC.
Proof: O is the centre and OM⊥AB
The perpendicular from the centre of a circle to a chord bisects the chord.
∴AM= ½ AB Again ON⊥AC
∴AN= ½ AC —————————– (1)
In ∆AOM and ∆AON
∠AMO=∠ANO [1 right angle]
∠MAO=∠NAO [supposition]
And AO=AO [common side]
So, ∆AOM≌∆AON [ASA theorem]
∴AM=AN
Or, ½ AB= ½ AC [from 1]
Or, AB=AC [proved]
Exercise: In the figure , O is the centre of circle and the chord AB = chord AC . Prove that ∠BAO=∠CAO.
Particular enunciation: Given that O is the centre of the circle and the chord AB=AC. We join O and A.
We have to prove that ∠BAO=∠CAO
Construction: We join O, B and O, C
Proof: In ∆AOB and ∆AOC
AB=AC [given]
OA=OA [common side]
So ∆AOB≌∆AOC [SSS theorem]
∴∠BAO=∠CAO. [proved]
Ex: A circle passes through the vertices of a right – angled triangle. Show that the centre lies on the midpoint of the hypotenuse.
General enunciation: A circle passes through the vertices of a right-angled triangle. It is required to show the centre lies on the midpoint of the hypotenuse.
Particular enunciation: Let ABC is a right – angled triangle. Its ∠ABC= 1 right angle and AC is the hypotenuse, The circle passes through the vertices A,B and C.
Construction: From O we draw the perpendicular OD to AB and the perpendicular OE to BC. Again we join O, B.
Proof: O is the centre of the circle and OD⊥AB
The perpendicular from the centre of a circle to a chord bisects the chord]
In ∆AOD and ∆BOD
OD=OD [common side]
And ∠ADO=∠BDO= 1 right angle [OD⊥AB]
So ∆AOD≌∆BOD [SAS theorem]
∴AO=BO
Similarly from ∆BOE and ∆COE it is proved that CO=BO
So AO=BO=CO ——————————-(2)
Therefore O is the centre of the circle.
Hypotenuse AC=AO+CO
=CO+CO [from 2]
=2CO
Or, 2CO=AC
Or, CO= ½ AC
Therefore O lies on the midpoint of the hypotenuse AC. (proved)
Exercise: A chord AB of one of two concentric circles intersect the other at C and D. Prove that AC=BD.
General enunciation: A chord AB of one of two concentric circles intersects the other at C and D. We have to prove that AC=BD.
Particular enunciation: Let O be the two circles ABE and CDF. The chord AB of the circle ABE intersects the circle CDF at C and D.
It is required to prove that AC=BD.
Construction: From O we draw the perpendicular OP on AB or CD.
Proof: O is the centre and OP⊥CD, OP⊥AB.
The perpendicular from the centre of a circle to a chord bisects the chord.
∴CP=PD and AP=BP ————————- (1)
Again, AP=AC+CP
And BP=PD+BD
Since AP=BP [from 1]
So AC+CP=PD+BD
Or, AC+CP=CP+BD [CP=PD]
Or, AC+CP=BD+CP
Or, AC=BD. (proved)
## If two triangles have the three sides of the one equal to the three sides of the other, each to each, then they are equal in all respects.
If two triangles have the three sides of the one equal to the three sides ... |
# What is the Simplest Form of a Ratio? | Don't Memorise | Summary and Q&A
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December 17, 2014
by
Infinity Learn NEET
What is the Simplest Form of a Ratio? | Don't Memorise
## TL;DR
Learn how to simplify ratios to their simplest form by dividing the numerator and denominator by their highest common factor (HCF).
## Key Insights
• 😑 Ratios can be expressed in different ways by multiplying or dividing the numerator and denominator by the same value.
• 🥳 Dividing the numerator and denominator by their HCF simplifies the ratio to its simplest form.
• 💁 Common factors help identify the values to divide the numerator and denominator by to obtain the simplest form.
• ✋ The HCF is the highest number that divides both the numerator and denominator evenly.
## Transcript
ratio of the number of chocolates gorov received to the number of chocolates J received is 2 is to 3 and that can also be written as 2x3 we have seen that 2x3 can be written in different ways we can multiply the numerator as well as the denominator by two and get 4X 6 or we can also multiply the numerator and the denominator by 3 to get 6 by 9 all ... Read More
### Q: How can we write the ratio 2:3 in different ways?
The ratio 2:3 can also be written as 4:6 or 6:9 by multiplying the numerator and denominator by 2 or 3, respectively. These ratios have the same value as 2:3.
### Q: How can we simplify the ratio 12:18?
To simplify 12:18, we divide the numerator and denominator by their highest common factor (HCF), which is 6. The simplified ratio is 2:3.
### Q: What if we have the ratio 30:50 and don't know the HCF?
In such cases, we look for the common factors of the numerator and denominator. Dividing both by the common factor(s) will simplify the ratio. In this case, dividing by 2 gives 15:25. Then, dividing by 5 gives the simplest form of 3:5.
### Q: How do we determine the HCF of a ratio?
The HCF is the highest number that divides both the numerator and denominator evenly. In the ratio 30:50, the HCF is 10 (2 * 5), which is the product of all the numbers we divided the numerator and denominator by.
## Summary & Key Takeaways
• Ratios express the relationship between two quantities and can be written in multiple ways.
• By multiplying or dividing the numerator and denominator by the same value, the ratio remains unchanged.
• To simplify a ratio, divide the numerator and denominator by their highest common factor (HCF). |
SOLUTION: The relationship between the number of calculators x that a company can sell per month and the price of each calculator p is given by x = 1700 - 100p. Find the price at which a c
Algebra -> -> SOLUTION: The relationship between the number of calculators x that a company can sell per month and the price of each calculator p is given by x = 1700 - 100p. Find the price at which a c Log On
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Click here to see ALL problems on Polynomials-and-rational-expressions Question 342259: The relationship between the number of calculators x that a company can sell per month and the price of each calculator p is given by x = 1700 - 100p. Find the price at which a calculator should be sold to produce a monthly revenue of \$7000. (Hint: Revenue = xp.)Answer by jsmallt9(3438) (Show Source): You can put this solution on YOUR website!With and we have a system of two equations with two variables. Since the second equation is of degree 2 (because of the x*p term), we will use the Substitution Method. (With second degree equations other methods are often either not possible or not practical.) Since we already have the first equation solved for x, we will use that and substitute into the second equation: To solve this we will start by simplifying: We can simplify the equation by dividing both sides by 100: Now we can factor (or use the Quadratic Formula): From the Zero Product property we know that: or solving these we get: p = 7 or p = 10 Now we find the x for each of these p's. We'll use the first equation and substitute in the value for p: For p = 7: For p = 10: So we have two solutions which generate \$7000 revenue: When the price is \$7, 1000 calculators will be sold. When the price is \$10, 700 calculators will be sold. If it is cheaper to build 700 calculators than it would be to build 1000 calculators, the greatest profit would come from making 700 calculators and selling them for \$10 each. |
# Composite Functions
Lesson
The idea behind the composition of functions is best explained with an example.
Suppose we think about the function given by $f\left(x\right)=2x+1$f(x)=2x+1. We understand that the function takes values of $x$x in the domain and maps them to values, say $y=2x+1$y=2x+1, in the range.
Suppose however that this is only the first part of a two-stage treatment of $x$x. Suppose we now take these function values and map them using another function, say $g\left(x\right)=x^2$g(x)=x2. This means that the $y$y values given by $\left(2x+1\right)$(2x+1) become the squared values $\left(2x+1\right)^2$(2x+1)2. The diagram below captures the idea.
The function values $f\left(x\right)$f(x) have become the domain values of $g\left(x\right)$g(x). Thus we could describe the complete two-stage process by the expression $g\left(f\left(x\right)\right)$g(f(x)), sometimes written $g\left[f\left(x\right)\right]$g[f(x)] and spoken of as the "gof" of $x$x
Algebraically, we can write $g\left(f\left(x\right)\right)=g\left[2x+1\right]=\left(2x+1\right)^2$g(f(x))=g[2x+1]=(2x+1)2
Note that if we reversed the order of the two-stage processing, we would, in this instance, develop a different composite function.
Thus $f\left(g\left(x\right)\right)=f\left(x^2\right)=2\left(x^2\right)+1=2x^2+1$f(g(x))=f(x2)=2(x2)+1=2x2+1. This is known as the "fog" of $x$x.
## The domain of Composite functions
One very important point needs to be made here in terms of the domain of composite functions.
If we consider, say $g\left(f\left(x\right)\right)$g(f(x)), then range of the function $f\left(x\right)$f(x) (which is the function applied first) must be a subset of the domain of the function $g\left(x\right)$g(x) (the function that is applied second).
That is to say, the domain of $g\left(f\left(x\right)\right)$g(f(x)) must only consist of elements that can be mapped by both $f\left(x\right)$f(x) and then by $g\left(x\right)$g(x) without causing an issue in either function.
A similar situation applies for $f\left(g\left(x\right)\right)$f(g(x)).
For example, if $f\left(x\right)=\sqrt{x}$f(x)=x and $g\left(x\right)=\frac{1}{x}$g(x)=1x, then both $f\left(g\left(x\right)\right)$f(g(x)) and $g\left(f\left(x\right)\right)$g(f(x)) have the restricted domain $x\in R^+$xR+. Lets explain why.
If we consider $g\left(f\left(x\right)\right)$g(f(x)), the range of $f\left(x\right)$f(x) includes non-negative real numbers, but we can't use all of these in $g\left(x\right)$g(x). The number zero needs to be deleted.
If we consider $f\left(g\left(x\right)\right)$f(g(x)), the range of $g\left(x\right)$g(x) includes all real numbers other than zero, but only positive reals can be considered for $f\left(x\right)$f(x)
Taking this last example, we have $f\left(g\left(x\right)\right)=f\left(\frac{1}{x}\right)=\sqrt{\frac{1}{x}}=\frac{1}{\sqrt{x}}$f(g(x))=f(1x)=1x=1x and $g\left(f\left(x\right)\right)=g\left(\sqrt{x}\right)=\frac{1}{\sqrt{x}}$g(f(x))=g(x)=1x and so in this instance the fog and the gof are equal.
The graph of the composition is shown here. Note that the composition is graphed for positive reals only.
#### Worked examples
##### Question 1
If $f\left(x\right)=4x+4$f(x)=4x+4,
1. find $f\left(2\right)$f(2).
2. find $f\left(-5\right)$f(5).
##### Question 2
Consider the functions $f\left(x\right)=-2x-3$f(x)=2x3 and $g\left(x\right)=-2x-6$g(x)=2x6.
1. Find $f\left(7\right)$f(7).
2. Hence, or otherwise, evaluate $g\left(f\left(7\right)\right)$g(f(7)).
3. Now find $g\left(7\right)$g(7).
4. Hence, evaluate $f\left(g\left(7\right)\right)$f(g(7)).
5. Is it true that $f\left(g\left(x\right)\right)=g\left(f\left(x\right)\right)$f(g(x))=g(f(x)) for all $x$x?
Yes
A
No
B
Yes
A
No
B
##### Question 3
Consider the functions $f\left(x\right)=-2x+6$f(x)=2x+6 and $g\left(x\right)=3x+1$g(x)=3x+1.
1. The function $r\left(x\right)$r(x) is defined as $r\left(x\right)=f\left(x^2\right)$r(x)=f(x2). Define $r\left(x\right)$r(x).
2. Using the results of the previous part, define $q\left(x\right)$q(x), which is $g\left(f\left(x^2\right)\right)$g(f(x2)). |
# Mastering the Art of Dividing Improper Fractions Unlock the secrets of dividing improper fractions with our expert guidance. Learn foolproof techniques, avoid common pitfalls, and boost your math confidence through clear explanations and practice exercises.
Now Playing:Divide fractions and mixed numbers – Example 0a
Intros
1. Simplify fractions: Method A - By using greatest common factors
2. Simplify fractions: Method B - By using common factors
Examples
1. Dividing Fractions and Mixed Numbers Using Diagrams
Find each of the following quotients by using diagrams.
1. $\frac{5}{6} \div \frac{3}{4}$
2. $2\frac{1}{3} \div \frac{1}{8}$
3. $\frac{7}{5} \div \frac{1}{2}$
Practice
What are fractions?
Notes
In this section, we will use diagrams to divide fractions and mixed numbers (a.k.a. compound fractions) for the purpose of helping you getting a better understanding about the concepts of fraction division. We will also teach you how to divide fractions by using multiplications.
In this lesson, we will learn:
• Dividing Fractions and Mixed Numbers Using Diagrams
• Dividing Fractions and Mixed Numbers Algebraically
• Word Problems: Dividing Fractions and Mixed Numbers
• Dividing Fractions and Mixed Numbers Involving Multiple-digit and Negative Numbers
Concept
## Introduction to Dividing Fractions and Mixed Numbers
Dividing fractions and mixed numbers is a crucial mathematical skill that builds upon students' understanding of basic fraction operations. Our introduction video serves as an essential tool in helping students visualize this complex concept, making it more accessible and easier to grasp. The video breaks down the process step-by-step, illustrating how to convert mixed numbers to improper fractions and apply the reciprocal method for division. Many students often struggle with the counterintuitive nature of fraction division, where multiplying by the reciprocal is key. Common challenges include forgetting to flip the second fraction or mishandling whole numbers in mixed fractions. By providing clear visual representations and real-world examples, the video aims to demystify these concepts and build confidence in students' ability to tackle more advanced fraction problems. Understanding division of fractions and mixed numbers is fundamental for success in higher-level mathematics and practical applications in everyday life.
Example
Dividing Fractions and Mixed Numbers Using Diagrams
Find each of the following quotients by using diagrams. $\frac{5}{6} \div \frac{3}{4}$
#### Step 1: Understanding the Problem
To solve the problem $\frac{5}{6} \div \frac{3}{4}$ using diagrams, we need to understand what the division of fractions means. Essentially, we are trying to determine how many times $\frac{3}{4}$ fits into $\frac{5}{6}$. This can be visualized using diagrams to make the concept clearer.
#### Step 2: Drawing the Diagrams
First, we need to represent both fractions using diagrams. We will use rectangles to represent these fractions.
• Draw a rectangle and divide it into 6 equal parts to represent $\frac{5}{6}$. Shade 5 out of the 6 parts to show $\frac{5}{6}$.
• Draw another rectangle of the same size and divide it into 4 equal parts to represent $\frac{3}{4}$. Shade 3 out of the 4 parts to show $\frac{3}{4}$.
#### Step 3: Finding a Common Denominator
To compare the two fractions accurately, we need to find a common denominator. The least common multiple of 6 and 4 is 12. Therefore, we will convert both fractions to have a denominator of 12.
• Convert $\frac{5}{6}$ to $\frac{10}{12}$ by multiplying both the numerator and the denominator by 2.
• Convert $\frac{3}{4}$ to $\frac{9}{12}$ by multiplying both the numerator and the denominator by 3.
#### Step 4: Redrawing the Diagrams
Now, redraw the rectangles to reflect the new fractions with a common denominator of 12.
• For $\frac{10}{12}$, divide the rectangle into 12 equal parts and shade 10 of them.
• For $\frac{9}{12}$, divide the rectangle into 12 equal parts and shade 9 of them.
#### Step 5: Comparing the Fractions
Now, we need to determine how many times $\frac{9}{12}$ fits into $\frac{10}{12}$. This can be visualized by comparing the shaded parts of the two rectangles.
• Count how many full $\frac{9}{12}$ sections fit into the $\frac{10}{12}$ section. You will find that one full $\frac{9}{12}$ section fits, with some leftover.
• Determine the leftover part. The leftover part is $\frac{1}{12}$ of the original rectangle.
#### Step 6: Calculating the Final Answer
To find the final answer, we need to express the leftover part as a fraction of $\frac{9}{12}$.
• The leftover part is $\frac{1}{12}$, and since $\frac{9}{12}$ is the fraction we are dividing by, we need to determine what fraction of $\frac{9}{12}$ is $\frac{1}{12}$.
• $\frac{1}{12}$ is $\frac{1}{9}$ of $\frac{9}{12}$.
Therefore, the final answer is 1 full $\frac{9}{12}$ section plus $\frac{1}{9}$ of another $\frac{9}{12}$ section, which can be written as $1 \frac{1}{9}$.
FAQs
Here are some frequently asked questions about dividing fractions and mixed numbers:
#### 1. How do you divide by an improper fraction?
To divide by an improper fraction, follow these steps: 1. Convert the improper fraction to a mixed number if desired. 2. Flip the improper fraction (find its reciprocal). 3. Multiply the first fraction by the reciprocal of the second fraction. 4. Simplify the result if possible.
#### 2. What are the 3 rules for dividing fractions?
The three main rules for dividing fractions are: 1. Keep the first fraction as it is. 2. Change the division sign to multiplication. 3. Flip the second fraction (find its reciprocal).
#### 3. How do you divide improper fractions in algebra?
To divide improper fractions in algebra: 1. Write the division as a fraction. 2. Multiply the first fraction by the reciprocal of the second fraction. 3. Simplify the resulting fraction if possible. For example: (5/3) ÷ (7/4) = (5/3) × (4/7) = 20/21
#### 4. What are the steps in dividing proper fractions?
To divide proper fractions: 1. Keep the first fraction unchanged. 2. Change the division sign to multiplication. 3. Flip the second fraction (find its reciprocal). 4. Multiply the numerators and denominators. 5. Simplify the result if possible.
#### 5. How do you divide mixed numbers?
To divide mixed numbers: 1. Convert both mixed numbers to improper fractions. 2. Follow the steps for dividing fractions (keep, change, flip). 3. Multiply the numerators and denominators. 4. Simplify the result and convert back to a mixed number if needed.
Prerequisites
Understanding the fundamentals of fractions and mixed numbers is crucial before diving into the complex world of dividing them. A solid grasp of basic fraction operations forms the foundation for more advanced calculations. These operations include addition, subtraction, multiplication, and division, which are essential skills for manipulating fractions effectively.
Before tackling division, it's important to master multiplying improper fractions and mixed numbers. This skill is closely related to division, as division of fractions often involves multiplication by the reciprocal. Additionally, understanding integer division provides a crucial stepping stone, as it introduces the concept of division and its properties.
While it may seem unrelated at first, proficiency in adding and subtracting mixed numbers is also important. These skills help in simplifying expressions and finding common denominators, which can be useful when dividing fractions and mixed numbers. Moreover, the ability to convert between improper fractions and mixed numbers is often necessary during division problems.
As you progress, you'll find that solving linear equations using multiplication and division becomes relevant. This algebraic skill helps in understanding how division of fractions can be applied in more complex mathematical scenarios. It also reinforces the inverse relationship between multiplication and division.
Although it might not seem directly related, understanding proportional reasoning through graphs of linear relationships can provide valuable insights. This concept helps in visualizing how fractions relate to each other and how division affects these relationships.
Ultimately, the goal is to become proficient in solving problems with rational numbers in fraction form. This skill encompasses all the previous topics and directly applies to dividing fractions and mixed numbers. It involves interpreting word problems, setting up appropriate equations, and using division of fractions to find solutions.
By mastering these prerequisite topics, you'll build a strong foundation for understanding and applying the division of fractions and mixed numbers. Each concept contributes to your overall comprehension, making the learning process smoother and more intuitive. Remember, mathematics is a cumulative subject, and each new skill builds upon previously learned concepts. Take the time to thoroughly understand these prerequisites, and you'll find that dividing fractions and mixed numbers becomes a manageable and even enjoyable challenge. |
# Question d4a35
Apr 14, 2016
$\text{12.8 kJ}$
#### Explanation:
The key to this problem lies with the value of water's specific heat, which is said to be equal to
${c}_{w} = {\text{4.184 J g"^(-1)""^@"C}}^{- 1}$
Now, a substance's specific heat tells you how much heat is needed in order to increase the temperature of $\text{1 g}$ of that substance by ${1}^{\circ} \text{C}$.
In your case, water's specific heat tells you that you need $\text{4.184 J}$ of energy in the form of heat in order to increase the temperature of $\text{1 g}$ of water by ${1}^{\circ} \text{C}$.
Notice that specific heat is expressed in joules per gram per degree Celsius. This means that you can break up the calculation into two parts
• determine how much heat you need to add in order to increase the temperature of $\text{45.0 g}$ of water by ${1}^{\circ} \text{C}$
• use this value to determine how much heat would be needed in order to increase the sample's temperature by $\Delta T$
In this case, $\Delta T$, which represents change in temperature, will be equal to
$\Delta T = {T}_{\text{final" - T_"initial}}$
$\Delta T = {74.0}^{\circ} \text{C" - 6.2^@"C" = 67.8^@"C}$
So, to increase the temperature of $\text{45.0 g}$ of water by ${1}^{\circ} \text{C}$, you'd need
45.0 color(red)(cancel(color(black)("g"))) * overbrace(("4.184 J"^@"C"^(-1))/(1color(red)(cancel(color(black)("g")))))^(color(purple)("heat needed per 1"^@"C")) = "188.28 J"^@"C"^(-1)
If this is how much heat you need to produce a ${1}^{\circ} \text{C}$ increase in temperature for a $\text{45.0-g}$ sample of water, it follows that a ${67.8}^{\circ} \text{C}$ increase in temperature would require
67.8 color(red)(cancel(color(black)(""^@"C"))) * "188.28 J"/(1color(red)(cancel(color(black)(""^@"C")))) = "12,765.4 J"
I'll leave the answer rounded to three sig figs and express it in kilojoules
"heat needed" = color(green)(|bar(ul(color(white)(a/a)"12.8 kJ"color(white)(a/a)|)))#
ALTERNATIVE APPROACH
You can also solve this problem by using the following equation
$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} q = m \cdot c \cdot \Delta T \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$, where
$q$ - the amount of heat gained / lost
$m$ - the mass of the sample
$c$ - the specific heat of the substance
$\Delta T$ - the change in temperature, defined as the difference between the final temperature and the initial temperature
Plug in your values to get
$q = 45.0 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{g"))) * "4.184 J" color(red)(cancel(color(black)("g"^(-1)))) color(red)(cancel(color(black)(""^@"C"^(-1)))) * 67.8color(red)(cancel(color(black)(""^@"C}}}}$
$q = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \text{12.8 kJ} \textcolor{w h i t e}{\frac{a}{a}} |}}}$ |
# Prealgebra: Variables
### Contents
page 1 of 2
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#### Solutions to Algebraic Equations
When we solve an algebraic equation, instead of plugging in a given number for the variable, we find a number that, when plugged in for the variable, would make the equation true. Such a number is called a solution to an equation. 58 is a solution to the equation h + 2 = 60, because 58 + 2 = 60. 46 is not a solution to h + 2 = 60, because 46 + 2 does not equal 60.
Some equations have more than one solution. For example, 4 and -4 are both solutions to r2 = 16. Most of the equations we will deal with, however, have only one solution.
#### Fundamentals of Equations
The goal in solving an equation is to get the variable by itself on one side of the equation and a number on the other side of the equation.
Generally, the variable will start on one side with operations being performed on it. We must reverse these operations by performing the inverse of each operation. However, we cannot just perform the inverse operation on on e side, because that would change the equation. However, if you perform the same operation on both sides of an equation the equation will not change.
Performing an operation on one side of an equation will change the equation and make it false.
Given, 5×6 = 30
5×6 = 30×3; 5×6 = 30 while 30×3 = 90
5×6 = 30 + 18; 5×6 = 30 while 30 + 18 = 48
5×6 = 30/10; 5×6 = 30 while 30/10 = 3
Performing the same operation on each side of an equation won't change the equation:
Given, 7 + 4 = 11
(7 + 4)×12 = 11×12; both sides equal 132
(7 + 4) + 3 = 11 + 3; both sides equal 14
- (7 + 4) = - 11; both sides equal -11
Herein lies a vital role of solving algebraic equations: whatever operation is carried out on one side of the equal sign in an equation must be carried out on the other side as well.
#### Solving Algebraic Equations
To solve an algebraic equation, reverse all the operations on the variable side of the equation by performing their inverse operations on both sides of the equation.
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# 0.3 Equations and inequalities: linear simultaneous equations
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## Equations and inequalities: linear simultaneous equations
Thus far, all equations that have been encountered have one unknown variable that must be solved for. When two unknown variables need to be solved for, two equations are required and these equations are known as simultaneous equations. The solutions to the system of simultaneous equations are the values of the unknown variables which satisfy the system of equations simultaneously, that means at the same time. In general, if there are $n$ unknown variables, then $n$ equations are required to obtain a solution for each of the $n$ variables.
An example of a system of simultaneous equations is:
$\begin{array}{c}\hfill 2x+2y=1\\ \hfill \frac{2-x}{3y+1}=2\end{array}$
## Finding solutions
In order to find a numerical value for an unknown variable, one must have at least as many independent equations as variables. We solve simultaneous equations graphically and algebraically.
## Graphical solution
Simultaneous equations can be solved graphically. If the graph corresponding to each equation is drawn, then the solution to the system of simultaneous equations is the co-ordinate of the point at which both graphs intersect.
$\begin{array}{c}\hfill x=2y\\ \hfill y=2x-3\end{array}$
Draw the graphs of the two equations in [link] .
The intersection of the two graphs is $\left(2,1\right)$ . So the solution to the system of simultaneous equations in [link] is $y=1$ and $x=2$ .
This can be shown algebraically as:
$\begin{array}{ccc}\hfill x& =& 2y\hfill \\ \hfill ∴\phantom{\rule{1.em}{0ex}}y& =& 2\left(2y\right)-3\hfill \\ \hfill y-4y& =& -3\hfill \\ \hfill -3y& =& -3\hfill \\ \hfill y& =& 1\hfill \\ \hfill \mathrm{Substitute into the first equation:}\phantom{\rule{2.em}{0ex}}\mathrm{x}& =& 2\left(1\right)\hfill \\ & =& 2\hfill \end{array}$
Solve the following system of simultaneous equations graphically.
$\begin{array}{ccc}\hfill 4y+3x& =& 100\hfill \\ \hfill 4y-19x& =& 12\hfill \end{array}$
1. For the first equation:
$\begin{array}{ccc}\hfill 4y+3x& =& 100\hfill \\ \hfill 4y& =& 100-3x\hfill \\ \hfill y& =& 25-\frac{3}{4}x\hfill \end{array}$
and for the second equation:
$\begin{array}{ccc}\hfill 4y-19x& =& 12\hfill \\ \hfill 4y& =& 19x+12\hfill \\ \hfill y& =& \frac{19}{4}x+3\hfill \end{array}$
2. The graphs intersect at $\left(4,22\right)$ .
3. $\begin{array}{ccc}\hfill x& =& 4\hfill \\ \hfill y& =& 22\hfill \end{array}$
## Solution by substitution
A common algebraic technique is the substitution method: try to solve one of the equations for one of the variables and substitute the result into the other equations, thereby reducing the number of equations and the number of variables by 1. Continue until you reach a single equation with a single variable, which (hopefully) can be solved; back substitution then allows checking the values for the other variables.
In the example [link] , we first solve the first equation for $x$ :
$x=\frac{1}{2}-y$
and substitute this result into the second equation:
$\begin{array}{ccc}\hfill \frac{2-x}{3y+1}& =& 2\hfill \\ \hfill \frac{2-\left(\frac{1}{2}-y\right)}{3y+1}& =& 2\hfill \\ \hfill 2-\left(\frac{1}{2}-y\right)& =& 2\left(3y+1\right)\hfill \\ \hfill 2-\frac{1}{2}+y& =& 6y+2\hfill \\ \hfill y-6y& =& -2+\frac{1}{2}+2\hfill \\ \hfill -5y& =& \frac{1}{2}\hfill \\ \hfill y& =& -\frac{1}{10}\hfill \end{array}$
$\begin{array}{ccc}\hfill ∴\phantom{\rule{1.em}{0ex}}x& =& \frac{1}{2}-y\hfill \\ & =& \frac{1}{2}-\left(-\frac{1}{10}\right)\hfill \\ & =& \frac{6}{10}\hfill \\ & =& \frac{3}{5}\hfill \end{array}$
The solution for the system of simultaneous equations [link] is:
$\begin{array}{ccc}\hfill x& =& \frac{3}{5}\hfill \\ \hfill y& =& -\frac{1}{10}\hfill \end{array}$
Solve the following system of simultaneous equations:
$\begin{array}{ccc}\hfill 4y+3x& =& 100\hfill \\ \hfill 4y-19x& =& 12\hfill \end{array}$
1. If the question does not explicitly ask for a graphical solution, then the system of equations should be solved algebraically.
2. $\begin{array}{ccc}\hfill 4y+3x& =& 100\hfill \\ \hfill 3x& =& 100-4y\hfill \\ \hfill x& =& \frac{100-4y}{3}\hfill \end{array}$
3. $\begin{array}{ccc}\hfill 4y-19\left(\frac{100-4y}{3}\right)& =& 12\hfill \\ \hfill 12y-19\left(100-4y\right)& =& 36\hfill \\ \hfill 12y-1900+76y& =& 36\hfill \\ \hfill 88y& =& 1936\hfill \\ \hfill y& =& 22\hfill \end{array}$
4. $\begin{array}{ccc}\hfill x& =& \frac{100-4\left(22\right)}{3}\hfill \\ & =& \frac{100-88}{3}\hfill \\ & =& \frac{12}{3}\hfill \\ & =& 4\hfill \end{array}$
5. $\begin{array}{ccc}\hfill 4\left(22\right)+3\left(4\right)=88+12& =& 100\hfill \\ \hfill 4\left(22\right)-19\left(4\right)=88-76& =& 12\hfill \end{array}$
A shop sells bicycles and tricycles. In total there are 7 cycles (cycles includes both bicycles and tricycles) and 19 wheels. Determine how many of each there are, if a bicycle has two wheels and a tricycle has three wheels.
1. The number of bicycles and the number of tricycles are required.
2. If $b$ is the number of bicycles and $t$ is the number of tricycles, then:
$\begin{array}{ccc}\hfill b+t& =& 7\hfill \\ \hfill 2b+3t& =& 19\hfill \end{array}$
3. $\begin{array}{ccc}\hfill b& =& 7-t\hfill \\ \hfill \mathrm{Into second equation:}\phantom{\rule{1.em}{0ex}}2\left(7-\mathrm{t}\right)+3\mathrm{t}& =& 19\hfill \\ \hfill 14-2t+3t& =& 19\hfill \\ \hfill t& =& 5\hfill \\ \hfill \mathrm{Into first equation:}:\phantom{\rule{1.em}{0ex}}\mathrm{b}& =& 7-5\hfill \\ & =& 2\hfill \end{array}$
4. $\begin{array}{ccc}\hfill 2+5& =& 7\hfill \\ \hfill 2\left(2\right)+3\left(5\right)=4+15& =& 19\hfill \end{array}$
## Simultaneous equations
1. Solve graphically and confirm your answer algebraically: $3a-2b7=0$ , $a-4b+1=0$
2. Solve algebraically: $15c+11d-132=0$ , $2c+3d-59=0$
3. Solve algebraically: $-18e-18+3f=0$ , $e-4f+47=0$
4. Solve graphically: $x+2y=7$ , $x+y=0$
what is the stm
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
anyone know any internet site where one can find nanotechnology papers?
research.net
kanaga
sciencedirect big data base
Ernesto
Introduction about quantum dots in nanotechnology
what does nano mean?
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
Abigail
for teaching engĺish at school how nano technology help us
Anassong
How can I make nanorobot?
Lily
Do somebody tell me a best nano engineering book for beginners?
there is no specific books for beginners but there is book called principle of nanotechnology
NANO
how can I make nanorobot?
Lily
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
how did you get the value of 2000N.What calculations are needed to arrive at it
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# 10.3: Triangle Area
Difficulty Level: At Grade Created by: CK-12
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Have you ever tried to figure out the area of a triangle?
Now that Jillian has figured out the parallelograms, she is on to the triangles. There are right triangles in the quilt block that she is working on with her grandmother. Jillian needs to figure out the area of each triangle so that she can calculate the size of the triangle and the amount of fabric that she will need to make them all.
Here is the quilt block once again.
There are 16 right triangles in the quilt block. The good news for Jillian is that they all have the same dimensions. Here are the dimensions of the triangle.
Each side of the triangle is three inches. Given these measurements, what is the area of one of the triangles? What is the area for 16 triangles?
Jillian is puzzled. She just figured out how to find the area of a parallelogram and now she is on to triangles. Jillian knows that the triangle and the square are related, she just isn’t sure how.
Use the information in this Concept to learn about finding the area of a triangle!
### Guidance
Think back to the dilemma you just read. When Jillian looked at the triangles, she could see that they were related to squares. In fact, triangles are related to parallelograms, and a square is a type of parallelogram.
How is a triangle related to a parallelogram?
Let’s look at a parallelogram and see if we can figure out the connection.
Here is a parallelogram. If you look at it carefully, you will notice that we can divide the parallelogram into two triangles.
A rectangle is a type of parallelogram. We can divide a rectangle into two triangles also.
Notice that a rectangle is divided into two right triangles.
A square is a type of rectangle. We can divide a square into two triangles also.
We have two right triangles here too.
If a parallelogram can be divided into two triangles, then what can we say about the area of a triangle?
Based on this information, we could say that the area of a triangle is one-half the area of a parallelogram. Let’s look at how this works.
What is the area of this parallelogram?
We find the area of the parallelogram by multiplying the base times the height.
\begin{align*}A & = bh\\ A & = 2(5)\\ A & = 10 \ sq. \ inches.\end{align*}
A parallelogram can be divided into two triangles.
We can divide the area of the parallelogram in half and that will give us the area of one of the triangles.
10 \begin{align*}\div\end{align*} 2 \begin{align*}=\end{align*} 5 sq. inches
Based on this information, we can write the following formula for finding the area of a triangle.
\begin{align*}A = \frac{1}{2}bh\end{align*}
If you think about this it makes perfect sense. A triangle is one-half of a parallelogram, so the formula for the parallelogram multiplied by one-half is the formula for finding the area of a triangle. Said another way, the area of the parallelogram is divided in half to find the area of a triangle.
Now put this information into practice. Take the area of the following parallelograms and find the area of one of the triangles inside the parallelogram.
#### Example A
Area of a rectangle is 12 sq. inches
Solution: 6 sq. inches
#### Example B
Area of a parallelogram is 24 sq. feet
Solution: 12 sq. feet
#### Example C
Area of a parallelogram is 18 sq. feet
Solution: 9 sq. feet
Here is the original problem once again. Use what you learned about finding the area of a triangle to help Jillian with her quilt block.
Now that Jillian has figured out the parallelograms, she is on to the triangles. There are right triangles in the quilt block that she is working on with her grandmother. Jillian needs to figure out the area of each triangle so that she can calculate the size of the triangle and the amount of fabric that she will need to make them all.
Here is the quilt block once again.
There are 16 right triangles in the quilt block. The good news for Jillian is that they all have the same dimensions. Here are the dimensions of the triangle.
Each side of the triangle is three inches. Given these measurements, what is the area of one of the triangles? What is the area for 16 triangles?
Jillian is puzzled. She just figured out how to find the area of a parallelogram and now she is on to triangles. Jillian knows that the triangle and the square are related, she just isn’t sure how.
First, Jillian needs to find the area for one of the triangles. To do this, she can use the formula for finding the area of a triangle.
\begin{align*}A & = \frac{1}{2}bh\\ A & = \frac{1}{2}(3)(3)\\ A & = 4.5 \ square \ inches\end{align*}
Wow! Jillian is a bit nervous about every triangle having an area of 4.5 inches. That might be hard to manage. However, Jillian has another idea. If each triangle has an area of 4.5 inches, then each square has an area of 9 inches.
Think about this-a nine inch square will be easier to cut in half and get two equal triangles.
If Jillian needs 16 triangles, then she can cut 8 nine inch squares, then she will have enough because each square can be cut into two triangles.
How much material will she need?
If each square is 3 \begin{align*}\times\end{align*} 3 or has an area of 9” and Jillian needs eight squares, then the area of material is 9 \begin{align*}\times\end{align*} 8.
Jillian will need 72 square inches of material for the triangles.
### Vocabulary
Here are the vocabulary words in this Concept.
Triangle
a polygon with three sides.
Parallelogram
a four sided figure with opposite sides congruent and parallel.
Rectangle
a parallelogram with opposite sides congruent and parallel and with four right angles.
Square
a parallelogram with four congruent, parallel sides and four congruent right angles.
### Guided Practice
Here is one for you to try on your own.
Sometimes, you will have a figure that uses both triangles and another figure like a parallelogram. We can find the area of the individual parts, add them together and find the total area of the entire figure. Here are a few problems that use these skills.
Here are a series of triangles that line the center median of a city street. The triangles are overlapping.
Let’s say that that the first triangle has a base of 6 feet and a height of 5 feet. What is the area of the triangle?
\begin{align*}A & = \frac{1}{2}bh\\ A & = \frac{1}{2}(6)(5)\\ A & = \frac{1}{2}(30)\\ A & = 15 \ square \ feet\end{align*}
Let’s say that there are eight triangles in this strip of median. We can take the area of one of the triangles and multiply it by 8.
15(8) = 120 square feet
This is the area of the median. We were able to use the area of each triangle to find the total area of the median.
Here we have a rectangle and a triangle together. If we want to find the total area of the figure, we need to find the area of the rectangle and the area of the triangle and then find the total sum.
Area of the rectangle
\begin{align*}A & = lw\\ A & = 3(5)\\ A & = 15 \ sq. \ meters\end{align*}
Area of the triangle
\begin{align*}A & = \frac{1}{2}bh\\ A & = \frac{1}{2}(3)(1.5)\\ A & = 2.25 \ sq. \ meters\end{align*}
Now we add the two areas together.
Area of rectangle + area of triangle = total area of figure
15 sq. meters + 2.25 sq. meters = 17.25 sq. meters
The total area is 17.25 square meters.
### Video Review
Here is a video for review.
### Practice
Directions: Find the area of each triangle.
1.
Base = 10 in, Height = 4 in
2.
Base = 16 meters, Height = 10 meters
3.
Base = 8 in, Height = 6.5 in
4.
Base = 10 cm, Height = 7 cm
5.
Base = 5 ft, Height = 8.5 feet
Directions: Find the area of each triangle given the base and height.
6. Base = 4 in
Height = 5 in
7. Base = 6 in
Height = 4 in
8. Base = 8 ft
Height = 7 ft
9. Base = 10 meters
Height = 8 meters
10. Base = 10 meters
Height = 5 meters
11. Base = 12 feet
Height = 14 feet
12. Base = 11 feet
Height = 6 feet
13. Base = 14 inches
Height = 8 inches
14. Base = 22 feet
Height = 19 feet
15. Base = 30 cm
Height = 28 cm
16. Base = 18 inches
Height = 16 inches
17. Base = 13 meters
Height = 10 meters
18. Base = 18 meters
Height = 5.5 meters
19. Base = 12.5 feet
Height = 2.5 feet
20. Base = 13.75 inches
Height = 1.5 inches
### Notes/Highlights Having trouble? Report an issue.
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### Vocabulary Language: English
TermDefinition
Parallelogram A parallelogram is a quadrilateral with two pairs of parallel sides.
Rectangle A rectangle is a quadrilateral with four right angles.
Square A square is a polygon with four congruent sides and four right angles.
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# Determinants. Chapter Properties of the Determinant
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1 Chapter 4 Determinants Chapter 3 entailed a discussion of linear transformations and how to identify them with matrices. When we study a particular linear transformation we would like its matrix representation to be simple, diagonal if possible. We therefore need some way of deciding if we can simplify the matrix representation and then how to do so. This problem has a solution, and in order to implement it, we need to talk about something called the determinant of a matrix. The determinant of a square matrix is a number. It turns out that this number is nonzero if and only if the matrix is invertible. In the first section of this chapter, different ways of computing the determinant of a matrix are presented. Few proofs are given; in fact no attempt has been made to even give a precise definition of a determinant. Those readers interested in a more rigorous discussion are encouraged to read Appendices C and D. 4.1 Properties of the Determinant The first thing to note is that the determinant of a matrix is defined only if the matrix is square. Thus, if A is a 2 2 matrix, it has a determinant, but if A is a 2 3 matrix it does not. The determinant of a 2 2 matrix is now defined. Definition 4.1. Determinant(A) = det(a) a b = det = ad bc. c d Example 1. Compute the determinants of the following matrices: 1 6 a. det = 3 ( 12) = b. det = 8 0 = c. det = 6 6 =
2 150 CHAPTER 4. DETERMINANTS To compute the determinant of a 3 3 or n n matrix, we need to introduce some notation. Definition 4.2. Let A = [a jk ] be an n n matrix. Let M jk be that (n 1) (n 1) matrix obtained from A by deleting its jth row and kth column. This submatrix of A is referred to as the j,k minor of A. Example 2. Let A = Find M 11, M 23, and M M 11 = M 23 = M 32 = Using minors we demonstrate one way to compute the determinant of a 3 3 matrix. The technique is called expansion by cofactors. Let A be any 3 3 matrix: A = a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 Then det(a) = a 11 det(m 11 ) a 12 det(m 12 )+a 13 det(m 13 ) Notethat anyminorofa3 3matrixisa2 2matrix, and henceitsdeterminant is defined. We also wish to stress that we did not have to expand across the first row. We could have used any row or column. Example 3. Compute the determinant of the matrix below by expanding across the first row and also by expanding down the second column. A = Expanding across the first row we have det(a) = a 11 det(m 11 ) a 12 det(m 12 )+a 13 det(m 13 ) = ( 1)det (2)det +(4)det = ( 35) 2(15)+4(42+9) = Expanding down the second column we have det(a) = a 12 det(m 12 )+a 22 det(m 22 ) a 32 det(m 32 ) = (2)det +(3)det (7)det = ( 2)(15)+3(12) 7( 29) = 209
3 4.1. PROPERTIES OF THE DETERMINANT 151 It seems from the above two computations that minus signs creep in at random. That is not true. There is a rule for deciding whether or not a minus sign should appear, and it is given in the following theorem. Theorem 4.1. Let A = [a jk ] be any n n matrix. Then det(a) = det(a) = n ( 1) j+k a jk det(m jk ) k = 1,2...,n j=1 expansion down the kth column n ( 1) j+k a jk det(m jk ) j = 1,2...,n k=1 expansion across the jth row (4.1) Figure 4.1 should clarify whether or not a minussign precedes the term det(m jk ). Notice that the 1,1 entry has a + sign, and whenever we move one space horizontally or vertically the sign changes. Since it takes three moves to go from 1, 1 to 2, 3, the coefficient of a 23 in (4.1) equals det(m 23 ). The terms ( 1) j+k det(m jk ) arecalled the cofactorsof a jk, hence the phrase expansion by cofactors. Notice that this theorem reduces the problem of computing the determinant of an n n matrix to the problem of calculating the determinant of an (n 1) (n 1) matrix. Continued use of this procedure will reduce the original problem to one of calculating the determinants of 2 2 matrices Figure 4.1 It is clear that computing the determinant of a matrix, especially a large one, is painful. It s also clear that the more zeros in a matrix the easier the chore. The following theorems enable us to increase the number of zeros in a matrix and at the same time keep track of how the value of the determinant changes. Theorem 4.2. Let A be a square matrix. If A 1 is a matrix obtained from A by interchanging any two rows or columns, then det(a 1 ) = det(a).
4 152 CHAPTER 4. DETERMINANTS Example 4. det = det : rows one and two interchanged det = det : columns two and three interchanged Corollary 4.1. If an n n matrix has two identical rows or columns, its determinant must equal zero. Proof. The preceding theorem says that if you interchange any two rows or columns, the determinant changes sign. But if the two rows interchanged are identical, the determinant must remain unchanged. Since zero is the only number equal to its negative, we have det(a) = 0. Example a. det = 2 2 = b. det = 0: rows one and three are identical Theorem 4.3. If any row or column of a square matrix A is multiplied by a constant c to get a matrix A 1, then det(a 1 ) = c[det(a)]. Corollary 4.2. If a square matrix A has a row or column of zeros, then det(a) = 0. Example 6. det (1) = det 3(2) = 3det (3) = 3det (1) 2(8) 2(15) = 6det 1 4 3(4) 1 8 3(5) = 18det 1 4(1) 4 1 4(2) (7) 3 4(2) 7 = 72det
5 4.1. PROPERTIES OF THE DETERMINANT 153 Theorem 4.4. If any multiple of a row (column) is added to another row (column) of a square matrix A to get another matrix A 1, then det(a 1 ) = det(a). Example 7. det = det = det = det ] = (1)det[ = This last example illustrates perhaps the easiest way to evaluate the determinant of a matrix. That is, use the elementary row or column operations to get a row or column with at most one nonzero entry and then use Theorem 4.1. Our next example also demonstrates this idea. The reader might try computing the determinant in this example by using Theorem 4.1 directly and then comparing the two techniques. Example det = det = det = det ] = det[ = Theorem 4.5. Let A and B be two n n matrices. Then Example 9. Let A = two matrices. det(ab) = det(a) det(b) 3 4, B = Verify Theorem 4.5 for these 2 8 Solution det(a) = det = 10 det(b) = det = det(ab) = det = 140 = ( 10)(14) = det(a)det(b) 5 13 Theorem 4.6. Let A be any square matrix and let A T be its transpose, then det(a) = det(a T )
6 154 CHAPTER 4. DETERMINANTS Example 10. Verify Theorem 4.6 for the matrix A = Solution det(a) = det = 18 ( 8) = det(a T 6 2 ) = det = 18 ( 8) = 26 = det(a) 4 3 Theorem 4.7. A square matrix A is invertible if and only if det(a) is nonzero. This last theorem is one that we use repeatedly in the remainder of this text. For example, in the next section we discuss how to compute the inverse of a matrix in terms of the determinants of its minors, and in Chapter 5 we use an equivalent version of Theorem 4.7 that says, if ker(a) has nonzero vectors in it, then det(a) = 0. Problem Set Compute the determinants of each of the following matrices: a. b. c d Compute the determinants of each of the following matrices: a. b. c d a 1 a 2 a 3 3. a. det b. det a 1 +k a 2 +k a 3 +k a 1 +2k a 2 +2k a 3 +2k 4. a. det b. det If E is an elementary row matrix associated with adding a multiple of one row to another, then det(e) =? 6. Let A be any upper or lower triangular matrix., Show that det(a) = a 11 a 22...a nn ; that is, the determinant of A equals the product of the diagonal entries of A.
7 4.1. PROPERTIES OF THE DETERMINANT Let a j,j = 1,...,n be arbitrary numbers and let k be any number. We assume below that n > 2. a. Show that a 1 a 2... a n a 1 +k a 2 +k... a n +k det a 1 +2k a 2 +2k... a n +2k... = 0 a 1 +(n 2)k a 2 +(n 2)k... a n +(n 2)k a 1 +(n 1)k a 2 +(n 1)k... a n +(n 1)k b. Show that det n n+1 n n 2n+1 2n n+n... n(n 1)+1... n 2 = 0 c. What happens if n = 2? 0 a2 8. Let A 2 =. Let A a = 0 0 a 3 0 a 2 0, and a a n a n 1 0 let A n =... 0 a Thus, A n is an n n matrix whose a only possible nonzero entries occur in the (n i+1)st column of the ith row. Show that det(a n ) = ( 1) n(n 1)/2 a 1 a 2...a n. 9. Let x 1,x 2,...,x n be any n numbers. Let 1 1 = A x 1 x x 1 x 2 x 3 = A 2 2 x 2 1 x 2 2 x x 1 x 2... x n... = A n x1 n 1... xn n 1 The reader should note that these matrices (actually their transposes) appeared in Section 3.4, when we discussed fitting polynomials to prescribed data. The determinants in this problem are called Vandermonde determinants.
8 156 CHAPTER 4. DETERMINANTS a. Show that det(a 2 ) = x 2 x 1. b. det(a 3 ) = (x 3 x 2 )(x 3 x 1 )(x 2 x 1 ). c. Find a similar formula for det(a n ). 10. Let A be any n n matrix. Show that det(a) equals zero if and only if the rows (columns) of A form a linearly dependent set of vectors in R n. (Hint: Use Theorems 4.4 and 4.7.) 11. Let A be a nonsingular matrix. Show that det(a 1 ) = [det(a)] 1. (Hint: AA 1 = I n. Now use Theorem 4.5.) 12. Let S be any scalar matrix, that is, S = ci n for some number c. Show that det(s) = c n. 13. Let A and B be any two similar matrices; that is, there is a nonsingular matrix P such that A = PBP 1. Show that det(a) = det(b). 14. Compute the determinants of the following matrices: a b A matrix is said to be skew symmetric if A = A T. Thus 0 1 skew symmetric, while is not is 1 0 a. Show that all the diagonal elements of a skew symmetric matrix are equal to zero. b. Let A be a 3 3 skew symmetric matrix. Show that det(a) = 0. c. If A is a 2 2 skew symmetric matrix, must det(a) = 0? d. What happens if A is an n n skew symmetric matrix? 16. Let P be an n n permutation matrix. Show that det(p) = ±1. For the definition of a permutation matrix see problem 13 in Problem Set The Adjoint Matrix and A 1 Theorem 4.7 states that A is invertible if and only if det(a) is nonzero. In this section we show how to compute the inverse of a matrix by using the determinants of its (n 1) (n 1) minors (cf. Definition 4.2). Since these determinants will appear quite frequently, we introduce a special notation for them.
9 4.2. THE ADJOINT MATRIX AND A Definition 4.3. Let A = [a jk ] be any n n matrix. The cofactor of a jk, denoted by A jk, is defined to be where M jk is the j,k minor ofa. Example 1. Let A = A jk = ( 1) j+k det(m jk ) A 11 = ( 1) 2 ( 72) A 12 = ( 1) 3 (14 36) A 13 = ( 1) 4 (16) A 21 = (42 24) A 22 = ( 7 12) A 23 = ( 8 24) A 31 = (54) A 32 = ( 9 6) A 33 = ( 12) Using the cofactors of a matrix, we construct its adjoint matrix. Definition 4.4. Let A = [a jk ] be an n n matrix. The adjoint matrix of A is the following n n matrix: where A jk are the cofactors of A. adj(a) = [A jk ] T Note that the transpose of the matrix [A jk ] must be taken. Example 2. Let A = From Example 1 and the definition of adj(a) we have adj(a) = = Example 3. Let A be the matrix in the preceding example. Compute det(a) and A[adj(A)]. det(a) = det = det = A[adj(A)] = = = 252I 3 = det(a)i 3 T
12 160 CHAPTER 4. DETERMINANTS 4.3 Cramer s Rule Cramer s rule is a formula for computing the solution to a system of linear equations when the coefficient matrix A is nonsingular. This formula is just the component verseion of the equation A 1 = (det(a)) 1 adj(a) (4.2) WederiveCramer sruleforasystemoftwoequationswith twounknownsbefore stating the rule for n equations with n unknowns. Example 1. Find the solution to Assuming that det(a) = a 11 a 22 a 12 a 21 0, a 11 x 1 +a 12 x 2 = b 1 a 21 x 1 +a 22 x 2 = b 2 (4.3) A 1 = (det(a)) 1 adj(a) = (deta) 1 a22 a 22 a 21 a 11 Thus, and Thus, [ x1 x 2 ] = A 1 [ b1 b 2 ] [ x1 ] = x 2 = (deta) 1 a22 a 22 b1 a 21 a 11 b 2 b1 a 22 b 2 a 12 (deta) b 2 a 11 b 1 a 1 21 x 1 = b 1a 22 b 2 a 12 det(a) x 2 = b 2a 11 b 1 a 21 det(a) b1 a det 12 b 2 a 22 = det(a) a11 b det 1 a 21 b 2 = det(a) Thus we see that x 1 can be expressed as the ratio of two determinants. The matrix in the numerator is obtained by replacing the first column of A with the right-hand side of (4.3). To calculate x 2, the matrix in the numerator is obtained by replacing the second column of A with the right-hand side of (4.3). Consider the general system with n equations and n unknowns Ax = b (4.4) where A = [a jk ] is an n n nonsingular matrix x = [x 1,...,x n ] T, and b = [b 1,...,b n ] T. We know that x = A 1 b = (deta) 1 adj(a)b (4.5)
13 4.3. CRAMER S RULE 161 Using (4.5), we write x k in terms of b: x k = (deta) 1 (A 1k b 1 +A 2k b 2 + +A nk b n ) (4.6) remember adj(a) = [A jk ] T, where A jk is the cofactor of a jk. The last factor in (4.6) can be thought of as the expansion by minors (going down the kth column) of the determinant of the matrix obtained by replacing the kth column of A with the column [b 1,...,b n ] T. That is, a a 1(k 1) b 1 a 1(k+1)... a 1n... det a j1... b j... a jn... a n1... a n(k 1) b n a n(k+1)... a nn x k = det(a) (4.7) Formula (4.7) is Cramer s rule. Note that it is valid only for systems whose coefficient matrix is nonsingular. Example 2. Find the value of x 4 for which x 1,x 2,x 3,x 4, and x 5 solves the following system: 2x 1 + x 2 +2x 5 = 2 x 1 + x 4 = 2 3x 2 +x 3 +2x 4 = 1 x 1 + x 2 + x 4 + x 5 = 1 x 3 + x 4 + x 5 = 1 An easy computation shows that the coefficient matrix A has a determinant equal to 7. Thus, A is nonsingular and we have det ( ) x 4 = = det(a) 7 We now have three techniques for solving systems of equations: Gaussian elimination, which always works; inversion of the coefficient matrix A, i.e., compute A 1 ; and Cramer s rule. The last two methods assume, of course, that A is invertible. For small systems, n = 2,3, or 4, any of these methods would be fine, at least if A is invertible. For large n, however, the most efficient method to use in solving a system of equations is usually Gaussian elimination. It is for this reason that Cramer s rule is a theoretical rather than a problem-solving tool.
14 162 CHAPTER 4. DETERMINANTS Problem Set Use Cramer s rule to solve each of the following systems of equations: a. 2x 1 +2x 2 = 7 b. 8x 1 +6x 2 = 4 8x 1 + x 2 = 2 3x 1 +2x 2 = 6 2. Use Gaussian elimination to solve each of the systems in problem Use Cramer s rule to solve the following system: 2x 1 6x 2 +x 3 = 2 x 2 +x 3 = 1 x 1 x 2 x 3 = 0 4. Consider the following system of equations: 2x 1 +x 2 +x 3 x 4 = 1 3x 1 x 2 +x 3 +x 4 = 0 x 1 +x 2 x 3 +x 4 = 0 6x 1 x 2 +x 4 = 0 a. Use Cramer s rule to solve for x 1. b. Use Gaussian elimination to solve for x Consider the following system of equations: x 1 +3x 2 + x 3 = 7 x 1 x 2 x 3 = 2 x 1 +6x 2 +2x 3 = 3 a. Solve this systsem using Cramer s rule. b. Solve using Gaussian elimination. c. Solve by finding the inverse of the coefficient matrix. 6. Solve the following system by using Cramer s rule. 7. Solve using Cramers rule. 8. Solve using Cramer s rule. 2x 1 x 3 = 1 2x 1 +4x 2 x 3 = 0 x 1 8x 2 3x 3 = 2 x 1 + x 2 + x 3 = a x 1 +(1+a)x 2 + x 3 = 2a x 1 + x 2 +(1+a)x 3 = 0 x 1 + x 2 + x 3 + x 4 = 4 x 1 +2x 2 +2x 3 +2x 4 = 1 x 1 + x 2 +2x 3 +2x 4 = 2 x 1 + x 2 + x 3 +2x 4 = 3
15 4.4. AREA AND VOLUME Area and Volume Given any two vectors in R 2 that are not parallel, they determine a parallelogram; cf. problems 9 and 10 in Section 2.1. With a few manipulations it is possible to express the area of this parallelogram as the absolute values of the determinant of a matrix constructed from the coordinates of these vectors. Example 1. Compute the area of the parallelogram determined by the vectors (1,2) and (8,0) (see Figure 4.2). The area of this parallelogram is the base length 8 times the height Area = 8 2 = det (4.8) where the columns of the matrix are the coordinates of the two vectors 0 2 (8,0) and (1,2), with respect to the standard basis of R 2. x 2 (1,2) (8,0) x 1 Figure 4.2 We derive a similar formula for any two vectors (a 1,a 2 ) and (b 1,b 2 ). Our proof and picture assume that both vectors lie in the first quadrant, but the other cases can be handled in a similar manner. Considertheparallelogramdeterminedbythetwovectors(a 1,a 2 )and(b 1,b 2 ) (see Figure 4.3). Area of parallelogram OBCA = area of triangle(obp 2 ) +area of trapezoid(p 2 BCP 3 ) area of triangle(oap 1 ) area of trapezoid(p 1 ACP 3 ) = 1 2 (b 1b 2 )+ 1 2 (a 1)(b 2 +a 2 +b 2 ) 1 2 (a 1a 2 ) 1 2 (b 1)(a 2 +a 2 +b 2 ) a1 b = a 1 b 2 a 2 b 1 = det 1 a 2 b 2
16 164 CHAPTER 4. DETERMINANTS Since we do not wish to worry about the order in which we list the columns of the above matrix, we write the area as Area = det a1 b 1 a 2 b 2 We formally state this in the following theorem. B(b 1,b 2) C(a 1 +b 1,a 2 +b 2) A(a 1,a 2) 0 P 1 P 2 P 3 Figure 4.3 Theorem 4.9. Let a = (a 1,a 2 ) and b = (b 1,b 2 ) be any two vectors in R 2. Let P be the parallelogram generated by these two vectors, i.e., P = {x: x = t 1 a+t 2 b,t 1 and t 2 arbitrary scalars between 0 and 1}. Then Area(P) = det a1 b 1 a 2 b 2 Example 2. a. Sketch the parallelogram determined by the vectors ( 6, 4) and (3, 2) and calculate its areas. We see from Figure 4.4a that the parallelogram is just the straight-line segment joining the two points ( 6,4) and (3, 2). Thus the area should be zero. Indeed, we have Area(P) = 6 3 det = = b. Repeattheabove,usingthevectors(1, 2)and( 6, 3)(seeFigure4.4b). Area(P) = 1 6 det = ( 3 12) =
17 4.4. AREA AND VOLUME 165 ( 6, 4) (3, 2) (a) ( 6, 3) (1, 2) (b) Figure 4.4 As one would expect, there is a generalization of this formula to higher dimensions, and we state it in the next theorem. Theorem Let {x 1,...,x n } be any n vectors in R n. Let P be the n- dimensional parallelepiped generated by these vectors, that is, P = {y: y = t 1 x 1 + +t n x n, 0 t j 1} Then the n-dimensional volume of P equals x 1 x 2 Vol(P) = det x n.. (4.9) The matrix in (4.9) is obtained by using the coordinates of the vector x j (with respect to the standard basis) as the jth column. Thus, if we had the four
18 166 CHAPTER 4. DETERMINANTS vectors x 1 = (1,1,2,1) x 2 = ( 1, 1,3,4) x 3 = (8,9,1,1) x 4 = (10,11,1,0) then the matrix would be Example 3. Sketch the parallelepiped generated by the three vectors a = (1,1,0),b = (0,1,1),andc = (1,0,1)(seeFigure4.5), anddetermineitsvolume. Vol(P) = det [ = det det ] = 1+1 = 2 (1,0,1) (0,1,1) Figure 4.5 (1,1,0) We remark that these determinants will be zero if and only if the n vectors used to form the matrices are linearly dependent. In that case, the solid they generate
19 4.4. AREA AND VOLUME 167 will lie in an (n 1)-dimensional plane, and hence should have n-dimensional volume equal to zero. See Example 2a. If L is a linear transformation from R 2 to R 2, it has a matrix representation A, where a11 a A = 12 a 21 a 22 As we saw in Chapter 3, L(e 1 ) = (a 11,a 21 ) and L(e 2 ) = (a 12,a 22 ). Thus, geometrically we can picture L as transforming the parallelogram generated by e 1 and e 2 into the parallelogram generated by (a 11,a 21 ) and (a 12,a 22 ) (see Figure 4.6). Moreover, we have Area(L(P)) = det a11 a 12 = det(a) a 21 a 22 Since in this particular case we have area(p) = 1, we may rewrite this formula as Area(L(P)) = det(a) area(p) To see that this formula is true in general let x = (x 1,x 2 ) and y = (y 1,y 2 ). Let P be the parallelogram generated by x and y, that is, P = {t 1 x+t 2 y: 0 t 1 1,0 t 2 1} L[e 2] e 2 P L[P] L[e 1] e 1 Figure 4.6 Then where and L(P) = {t 1 L(x)+t 2 L(y): 0 t 1 1,0 t 2 1} L(x) = (a 11 x 1 +a 12 x 2,a 21 x 1 +a 22 x 2 ) L(y) = (a 11 y 1 +a 12 y 2,a 21 y 1 +a 22 y 2 ).
20 168 CHAPTER 4. DETERMINANTS We have from (4.9) that Area(L(P)) = det a11 x 1 +a 12 x 2 a 11 y 1 +a 12 y 2 a 21 x 1 +a 22 x 2 a 21 y 1 +a 22 y 2 () = det a11 a 12 x1 y 1 a 21 a 22 x 2 y 2 = det a11 a 12 x1 y det 1 a 21 a 22 y 2 = det(a) [area(p)] Thus, a linear transformation from R 2 to R 2 maps parallelograms into parallelograms (rank A = 2), or line segments (rank A = 1), or a point (rank A = 0). Moreover, the change in area depends only on the linear transformation and not on the particular parallelogram. Naturally there is a generalization of this to higher dimensions which we state below. Theorem Let A be the n n matrix representation of L: R n R n, with respect to the standard basis of R n. Let P be an n-dimensional parallelogram generated by the n vectors {x 1,...,x n }. Then L(P) is generated by the n vectors {L(x 1 ),...,L(x n )} and their volumes are related by the formula x 2 Vol(L(P)) = det(a) Vol(P) (4.10) Problem Set Calculate the area of the triangles whose vertices are: a. (0,0), (1,6), ( 2,3) b. (8,17), (9,2), (4,6) 2. Calculate the volume of the tetrahedron whose vertices are: a. (0,0,0), (1, 1,2), ( 3,6,7), (1,1,1) b. (1,1,1), ( 1, 1, 1), (0,4,8), ( 3,0,2) 3. Find the area of the parallelograms determined by the following vectors; cf. problem 1. a. (1,6), ( 2,3) b. (1, 15),( 4, 11) 4. Sketch the parallelograms P generated by the following pairs of vectors. Let O be the parallelogram generated by the standard basis vectors. For each of the parallelograms P find a linear transformation L such that P = L(O). a. (1,0), (1,2) b. (1, 1), (3,6)
21 4.4. AREA AND VOLUME In R 2, show that the straight line passing through the two points (a 1,a 2 ) and (b 1,b 2 ) has equation det x 1 x 2 1 a 1 a 2 1 = 0 b 1 b Let L, a linear transformation from R 2 to R 2, have matrix representation 1 2 A = 1 2 Let P be the parallelogram generated by the vectors ( 1,2) and (1,3). Sketch P and L(P) and compute their areas. Then verify (4.10) LetA = be the matrix representation of a linear transformation 2 1 L. Let P be the parallelogram generated by the following pairs of vectors: a. (1,0), (0,1) b. (1,1), ( 1,1) In each case sketch P,L(P), and verify (4.10). 8. Let A = be the matrix representation of a linear transformation L. Let P be the solid generated by the following vectors: a. (1,0,0), (0,1,0), (0,0,1) b. (1,1,0), (1,0,1), (0,1,1) In each case sketch P,L(P), and verify (4.10). Supplementary Problems 1. Let A = [a jk ] be an n n matrix. Define each of the following: a. M jk minor of A b. Cofactor of a jk c. Adj(A) 2. Compute the determinant and adjoint of each of the following matrices: a
22 170 CHAPTER 4. DETERMINANTS b Let A = 2 x 1 x 1 1. Find all values of x for which det(a) = Find all values of λ for which the following system of equations has a nontrivial solution: 2x 1 7x 2 = λx 1 4x 1 + x 2 = λx 2 a(t) b(t) 5. Let f(t) = det, which a(t), b(t), c(t), and d(t) are differential c(t) d(t) functions of t. a a. Show that f b (t) = det a b +det c d c d. b. Derive a similar formula for the determinant of a 3 3 matrix of functions. 6. Let y 1 (t) and y 2 (t) be two solutions of the differential equation ay +by +cy = 0 Define the Wronskian W(t) of the solutions y j (t) by y1 (t) y W(t) = det 2 (t) y 1(t) y 2(t) Show that the Wronskian satisfies the differential equation aw +bw = 0 7. Using the fact that a matrix has nonzero determinant if and only if its rows (columns) are linearly independent, determine which of the following sets are linearly independent: a. (2,1), ( 3,4) b. (2,1, 2,3), (0,1,0,1), (1,1,1,0), (3,3, 1,4) 8. If A = (a 1,a 2,a 3 ) and B = (b 1,b 2,b 3 ), then the cross product of A and B is defined as A B = det i j k a 1 a 2 a 3 b 1 b 2 b 3 = (a 2 b 3 a 3 b 2,a 3 b 1 a 1 b 3,a 1 b 2 a 2 b 1 ) The reader should realize that the term involving the determinant is merely a mnemonic, useful in remembering the last expression. Show that A B = B A and the A A = 0.
23 4.4. AREA AND VOLUME Verify the following formulas: a. det a b 0 ] c d 0 a b = det[ e c d 0 0 e a b 0 0 b. det c d 0 0 a b e f 0 0 e f = det det c d g h 0 0 g h A Let A = be an n n matrix, where A 0 A 1 is an m m matrix and 2 A 2 is an (n m) (n m) matrix, 1 m n 1. Show that det(a) = det(a 1 )det(a 2 ) 11. A group (G, ) is a mathematical system that consists of a collection of objects G and an operation between two elements of G, which gives an element of G. Thus, if x and y are elements of G, then x y (we drop the dot in the future) is also in G. We also suppose that this operation, called multiplication, satisfies the following properties: 1. (xy)z = x(yz). 2. There is an identity element e in G such that ex = xe = x for every x in G. 3. For each x in G, there is an x 1 in G such that xx 1 = x 1 x = e a. Show that the set of vectors in a vector space forms a group, the group operation being vector addition. b. Let GL(n) denote the set of invertible n n matrices, with the group operation being matrix multiplication. Show that GL(n) forms a group. This group is called the general linear group of order n. c. Let SL(n) denote the set of invertible n n matrices with determinant equal to 1. Show that SL(n), with the group operation being matrix multiplication, is a group. This group is called the special linear group of order n. d. Show that the set of all n n matrices under matrix multiplication does not form a group. 12. An n n matrix P is said to be orthogonal if P 1 = P T. a. Show that if P is orthogonal, then det(p) = ±1. b. Deduce from part a that if P is orthogonal and if K is some n- dimensional parallelepiped, then vol(p K) = vol(k).
24 172 CHAPTER 4. DETERMINANTS c. Find a 2 2 matrix P that is not orthogonal, and for which det(p) equals 1. d. Show that O(n), the set of n n orthogonal matrices, forms a group under the operation of matrix multiplication; cf. problem A matrix B is said to be similar to the matrix A if there is a nonsingular matrix P such that B = PAP 1. a. Show that if B is similar to A, then det(b) = det(a). b. Find a pair of 2 2 matrices A and B for which det(a) = det(b) and B is not similar to A. Hint: If B is similar to a scalar matrix, then B = A. 14. Find two 2 2 matrices A and B for which det(a+b) det(a)+det(b).
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# Math Expressions Grade 2 Unit 1 Lesson 21 Answer Key Focus on Mathematical Practices
## Math Expressions Common Core Grade 2 Unit 1 Lesson 21 Answer Key Focus on Mathematical Practices
Math Expressions Grade 2 Unit 1 Lesson 21 Homework
Mrs. Wise and her three children went to the apple orchard. The table shows the number of apples each picked.
Apples Picked
Use the table to solve each story problem. Show your work.
Lesson 21 Homework Math Expressions Grade 2 Unit 1 Question 1.
What was the total number of apples they picked?
17 apples
Explanation:
Mrs. Wise has 6 apples
Michelle has 4 apples
George has 3 apples
Jen has 4 apples
Total apples = 6 + 4 + 3+ 4 = 17
Math Expressions Grade 2 Math Expressions Lesson 21 Question 2.
Two children picked the same number of apples. Who were the children?
How many apples did those two children pick in all?
Michelle and Jen picked same number of apples 8
Explanation:
Michelle has 4 apples
Jen has 4 apples
Total apples = 4 + 4 = 8
Use the information in the table to write your own problem. Solve the problem.
Who was collected more apples in the group and how many apples collected by Jen. what is the difference between them?
Explanation:
Mrs. Wise = 6 apples
Jen collected = 3 apples
The difference between them is 6 – 3 = 3 apples
Math Expressions Grade 2 Unit 1 Lesson 21 Remembering
Write two equations for each Math Mountain.
7 + 9 = 16 13 – 5 = 8 15 – 7 = 8
9 + 7 = 16 5 + 8 = 13 7 + 8 = 15
Explanation:
In the Math mountain, the top part of the mountain to called Total or Sum and it consists of two parts called Partners or Addends
The addition of two whole numbers results in the total amount or sum of those values combined.
Question 2.
6 + ________ = 11
18 – 9 = ________
5 + ________ = 13
6 + 5 = 11
18 – 9 = 9
5 + 8 = 13
Explanation:
In the Math mountain, the top part of the mountain to called Total or Sum and it consists of two parts called Partners or Addends
The addition of two whole numbers results in the total amount or sum of those values combined.
Solve the word problem. Show your work.
Question 3.
Don has 5 more pencils than crayons. He has 3 more markers than pencils. Don has 7 crayons. How many markers does Don have?
Done has 15 markers.
Explanation:
Done has 7 crayons
Done has 5 more pencils than crayons 7 + 5 = 12
Done has 12 pencils
Done has 3 more markers than pencils 12 + 3 =15
Total markers = 15
Question 4.
Stretch Your Thinking Fifteen children voted for their favorite color. The votes for red and blue together were double the votes for green and yellow together. How did the children vote? |
# Second Derivative Calculator
Instructions: Use second derivative calculator to compute the second derivative (this is, the derivative of the derivative) of any differentiable function that you provide, showing all the steps. Please type the function in the form box below.
Enter the function $$f(x)$$ you want to compute the second derivative (Ex: f(x) = x*sin(x), etc.)
## More on Second Derivatives
This calculator can help you to compute the second derivative of any valid function you provide, showing all the steps of the process. All you need to do is to provide a valid, differentiable function.
A valid function could be f(x) = x*tan(x), or f(x) = 3x^3 + 2x - 1, etc. It could be any valid function, and it does not have to necessarily come simplified, since the calculator will simplify it, in case it is needed.
Once you provide a valid function has been provided, you can click on the "Calculate" button, in order to get all the calculations and steps shown.
Second derivatives are tremendously practical in many applications, especially in Calculus, with the second derivative test for maximization and minimization, to assess whether a critical point is a maximum, minimum or none.
## What is the second derivative
In very simple terms, a second derivative is just the derivative of the derivative. So the process of computing a second derivative involves computing a derivative one time, and then another time, using the common derivative rules. The second derivative of a function $$f(x)$$ is usually written as $$f''(x)$$.
The idea of second derivative also applies to partial derivatives, and it corresponds to the derivative twice, but in this case, it can be computed with respect to different variables.
## Steps for calculating the second derivative
• Step 1: Identify the function f(x) you want to differentiate twice, and simplify as much as possible first
• Step 2: Differentiate one time to get the derivative f'(x). Simplify the derivative obtained if needed
• Step 3: Differentiate now f'(x), to get the second derivative f''(x)
The steps appear to be easy, but depending on the given function, the amount of algebraic calculations could be large.
## Second derivative notation
The most common notation for the second derivative is $$f''(x)$$, which reflects well the fact that the derivative operation, denoted by ', is applied twice to the function.
There is another notation for the second derivative, which is particularly useful when the function $$f(x)$$ is referred as 'y = y(x)'. Then, we use the following notation for the second derivative.
$\displaystyle \frac{d^2y}{dx^2} = \displaystyle \frac{d}{dx} \left(\frac{dy}{dx}\right)$
## Steps for computing second derivatives for implicit functions
• Step 1: Identify the equation involving x and y
• Step 2: Differentiate both sides of the equality. Each side could potentially depend on x, y and y'. Simplify obvious terms, but it is not strictly necessary
• Step 3: Differentiate again both sides of the equality. Each side could potentially depend on x, y, y' and y''. Then, solve for y''
It is usually much easier to compute the second derivative by implicit differentiation than by solving y in terms of x first and then differentiating, in case that x and y are defined implicitly by an equation, like $$x^2 + y^2 = 1$$.
## Second derivative at a point
Same as the derivative, the second derivative is a function defined point by point. Notice that a common error students make is thinking, since I want to differentiate at a point, and the function evaluated at a point is constant, its derivative must be constant. WRONG. You first compute the derivative, and THEN you evaluate.
### Example: Second derivative calculation
Calculate the second derivative of : $$f(x) = \cos(x^2)$$
Solution: In this example, we will compute the second derivative of the function $$\displaystyle f(x)=\cos\left(x^2\right)$$.
$$\displaystyle \frac{d}{dx}\left(\cos\left(x^2\right)\right)$$
By using the Chain Rule: $$\frac{d}{dx}\left( \cos\left(x^2\right) \right) = \frac{d}{dx}\left(x^2\right)\cdot \left(-\sin\left(x^2\right)\right)$$
$$\displaystyle = \,\,$$
$$\displaystyle \frac{d}{dx}\left(x^2\right)\cdot \left(-\sin\left(x^2\right)\right)$$
We use the Power Rule for polynomial terms: $$\frac{d}{dx}\left( x^2 \right) = 2x$$
$$\displaystyle = \,\,$$
$$\displaystyle \left(2x\right) \left(-\sin\left(x^2\right)\right)$$
$$\displaystyle = \,\,$$
$$\displaystyle 2x\cdot \left(-\sin\left(x^2\right)\right)$$
Finally, the following is obtained
$$\displaystyle = \,\,$$
$$\displaystyle -2x\sin\left(x^2\right)$$
Second Derivative: Now, we differentiate the derivative obtained so to get the second derivative:
$$\displaystyle \frac{d^2f}{dx^2} = \frac{d}{dx}\left(-2x\sin\left(x^2\right)\right)$$
By using the Product Rule: $$\frac{d}{dx}\left( \left(-1\right)\times 2x\sin\left(x^2\right) \right) = \frac{d}{dx}\left(-2x\right) \cdot \sin\left(x^2\right)+\left(-1\right)\times 2x \cdot \frac{d}{dx}\left(\sin\left(x^2\right)\right)$$
$$\displaystyle = \,\,$$
$$\displaystyle \frac{d}{dx}\left(-2x\right) \cdot \sin\left(x^2\right)+\left(-1\right)\times 2x \cdot \frac{d}{dx}\left(\sin\left(x^2\right)\right)$$
By linearity, we know $$\frac{d}{dx}\left( (-1)\times 2x \right) = \left(-1 \right) \cdot \frac{d}{dx}\left(2x\right)$$, so plugging that in:
$$\displaystyle = \,\,$$
$$\displaystyle \left(\left(-1 \right) \cdot \frac{d}{dx}\left(2x\right)\right) \sin\left(x^2\right)+\left(-1\right)\times 2x \cdot \frac{d}{dx}\left(\sin\left(x^2\right)\right)$$
Using the Chain Rule: $$\frac{d}{dx}\left( \sin\left(x^2\right) \right) = \frac{d}{dx}\left(x^2\right)\cdot \cos\left(x^2\right)$$ and directly we get: $$\frac{d}{dx}\left( 2x \right) = 2$$
$$\displaystyle = \,\,$$
$$\displaystyle \left(\left(-1 \right) \cdot 2\right) \sin\left(x^2\right)+\left(-1\right)\times 2x \cdot \frac{d}{dx}\left(x^2\right)\cdot \cos\left(x^2\right)$$
In this case we use the Power Rule for polynomial terms: $$\frac{d}{dx}\left( x^2 \right) = 2x$$
$$\displaystyle = \,\,$$
$$\displaystyle \left(\left(-1 \right) \cdot 2\right) \sin\left(x^2\right)+\left(-1\right)\times 2x \cdot 2x\cdot \cos\left(x^2\right)$$
and then we find
$$\displaystyle = \,\,$$
$$\displaystyle -2x\cdot 2x\cos\left(x^2\right)+\left(-2\right)\sin\left(x^2\right)$$
Putting together the numerical values, reducing the ones in $$-2x\cdot 2x\cos\left(x^2\right) = -4x^2\cos\left(x^2\right)$$ and grouping the terms with $$x$$ in the term $$-2x\cdot 2x\cos\left(x^2\right)$$
$$\displaystyle = \,\,$$
$$\displaystyle -2\cdot 2x^2\cos\left(x^2\right)-2\sin\left(x^2\right)$$
Simplifying the integers that can be multiplied together: $$\displaystyle -2\times2 = -4$$
$$\displaystyle = \,\,$$
$$\displaystyle -4x^2\cos\left(x^2\right)-2\sin\left(x^2\right)$$
Final Conclusion: We find that the second derivative we are looking for is:
$f''(x) = -4x^2\cos\left(x^2\right)-2\sin\left(x^2\right)$
### Example: More Second derivatives
For the following function : $$f(x) = x \cos(x)$$, compute its second derivative
Solution: Now, we do the same in tis $$\displaystyle f(x)=x\cos\left(x\right)$$, for which we need to compute its derivative.
The function came already simplified, so we can proceed directly to compute its derivative:
$$\displaystyle \frac{d}{dx}\left(x\cos\left(x\right)\right)$$
Using the Product Rule: $$\frac{d}{dx}\left( x\cos\left(x\right) \right) = \frac{d}{dx}\left(x\right) \cdot \cos\left(x\right)+x \cdot \frac{d}{dx}\left(\cos\left(x\right)\right)$$
$$\displaystyle = \,\,$$
$$\displaystyle \frac{d}{dx}\left(x\right) \cdot \cos\left(x\right)+x \cdot \frac{d}{dx}\left(\cos\left(x\right)\right)$$
Directly differentiating: $$\frac{d}{dx}\left( \cos\left(x\right) \right) = -\sin\left(x\right)$$
$$\displaystyle = \,\,$$
$$\displaystyle \frac{d}{dx}\left(x\right) \cdot \cos\left(x\right)+x \left(-\sin\left(x\right)\right)$$
$$\displaystyle = \,\,$$
$$\displaystyle x\cdot \left(-\sin\left(x\right)\right)+\cos\left(x\right)$$
By reorganizing/simplifying/expanding the terms that are amenable to
$$\displaystyle = \,\,$$
$$\displaystyle -x\sin\left(x\right)+\cos\left(x\right)$$
Second Derivative Calculation: The next step is to differentiate the derivative obtained in the previous steps:
$$\displaystyle \frac{d^2f}{dx^2} = \frac{d}{dx}\left(-x\sin\left(x\right)+\cos\left(x\right)\right)$$
By linearity, we know $$\frac{d}{dx}\left( (-1)x\sin(x)+\cos(x) \right) = \frac{d}{dx}\left((-1)x\sin(x)\right)+\frac{d}{dx}\left(\cos(x)\right)$$, so plugging that in:
$$\displaystyle = \,\,$$
$$\displaystyle \frac{d}{dx}\left(\left(-1\right)x\sin\left(x\right)\right)+\frac{d}{dx}\left(\cos\left(x\right)\right)$$
Directly differentiating: $$\frac{d}{dx}\left( \cos\left(x\right) \right) = -\sin\left(x\right)$$ and we can use the Product Rule: $$\frac{d}{dx}\left( \left(-1\right)x\sin\left(x\right) \right) = \frac{d}{dx}\left(\left(-1\right)x\right) \cdot \sin\left(x\right)+\left(-1\right)x \cdot \frac{d}{dx}\left(\sin\left(x\right)\right)$$
$$\displaystyle = \,\,$$
$$\displaystyle \frac{d}{dx}\left(\left(-1\right)x\right) \cdot \sin\left(x\right)+\left(-1\right)x \cdot \frac{d}{dx}\left(\sin\left(x\right)\right)-\sin\left(x\right)$$
Directly differentiating: $$\frac{d}{dx}\left( \sin\left(x\right) \right) = \cos\left(x\right)$$ and directly we get: $$\frac{d}{dx}\left( \left(-1\right)x \right) = -1$$
$$\displaystyle = \,\,$$
$$\displaystyle \left(-1\right) \sin\left(x\right)+\left(-1\right)x \cdot \cos\left(x\right)-\sin\left(x\right)$$
and then we get
$$\displaystyle = \,\,$$
$$\displaystyle \left(-1\right)x\cos\left(x\right)+\left(-1\right)\sin\left(x\right)+\left(-\sin\left(x\right)\right)$$
Reducing the multiplication by ones in $$\left(-1\right)x\cos\left(x\right) = \left(-1\right)x\cos\left(x\right)$$ and
$$\displaystyle = \,\,$$
$$\displaystyle \left(-1\right)x\cos\left(x\right)-\sin\left(x\right)+\left(-\sin\left(x\right)\right)$$
Simplifying:
$$\displaystyle = \,\,$$
$$\displaystyle -x\cos\left(x\right)-2\sin\left(x\right)$$
Second Derivative Conclusion: We conclude that the second derivative of the given function isa:
$f''(x) = -x\cos\left(x\right)-2\sin\left(x\right)$
### Example: Second derivative and implicit differentiation
Using implicit differentiation, calculate the second derivative of y with respect to x, for $$x^2 + y^2 = 1$$.
Solution:We apply implicit differentiation, assuming that y depends on x, and we differentiate both sides of the equality:
$\frac{d}{dx}\left(x^2 + y^2\right) = \frac{d}{dx} (1)$ $\Rightarrow 2x + 2yy' = 0$
Now, applying implicit differentiation again:
$\frac{d}{dx}\left( 2x + 2yy' \right) = \frac{d}{dx} 0$ $\Rightarrow 2 + 2y'^2+2yy'' = 0$ $\Rightarrow 2y'^2 + 2yy'' = -2$ $\Rightarrow yy'' = -1 - y'^2$ $\Rightarrow y'' = \frac{-1 - y'^2}{y}$
which concludes the calculation.
## More derivative calculators
When finding the derivative of a function, it is natural to think about doing the process again, which is finding the derivative of the derivative, and that is precisely what this second derivative calculator does.
The concept of second derivative is quite useful in Calculus, especially at the time of maximize or minimize functions. The second derivative gives you information about the concavity of a function, which is also crucial at the time to understand the shape of the graph of the function.
Second derivatives can be calculated both for regular derivatives and for implicit differentiation, in which you calculate implicit differentiation rule two times. |
# Standard Form To Slope Intercept Worksheet
## The Definition, Formula, and Problem Example of the Slope-Intercept Form
Standard Form To Slope Intercept Worksheet – One of the numerous forms that are used to illustrate a linear equation among the ones most frequently found is the slope intercept form. You may use the formula of the slope-intercept find a line equation assuming you have the straight line’s slope as well as the y-intercept. This is the y-coordinate of the point at the y-axis meets the line. Learn more about this particular linear equation form below.
## What Is The Slope Intercept Form?
There are three primary forms of linear equations: the standard slope-intercept, the point-slope, and the standard. Although they may not yield identical results when utilized in conjunction, you can obtain the information line that is produced more quickly using an equation that uses the slope-intercept form. As the name implies, this form employs an inclined line where you can determine the “steepness” of the line is a reflection of its worth.
This formula is able to determine the slope of a straight line. It is also known as the y-intercept, also known as x-intercept where you can utilize a variety formulas that are available. The line equation in this formula is y = mx + b. The straight line’s slope is symbolized in the form of “m”, while its y-intercept is represented through “b”. Every point on the straight line is represented as an (x, y). Note that in the y = mx + b equation formula the “x” and the “y” have to remain as variables.
## An Example of Applied Slope Intercept Form in Problems
In the real world in the real world, the slope-intercept form is often utilized to illustrate how an item or problem changes in it’s course. The value of the vertical axis represents how the equation tackles the extent of changes over the amount of time indicated via the horizontal axis (typically in the form of time).
An easy example of using this formula is to find out how many people live in a certain area as the years go by. In the event that the area’s population grows annually by a certain amount, the point value of the horizontal axis will increase by one point with each passing year and the point values of the vertical axis is increased in proportion to the population growth by the fixed amount.
You can also note the starting value of a challenge. The beginning value is at the y value in the yintercept. The Y-intercept represents the point where x is zero. If we take the example of a previous problem the beginning point could be when the population reading starts or when the time tracking starts along with the associated changes.
This is the location that the population begins to be tracked by the researcher. Let’s say that the researcher begins to do the calculation or the measurement in the year 1995. This year will be”the “base” year, and the x = 0 point will occur in 1995. This means that the 1995 population corresponds to the y-intercept.
Linear equations that use straight-line formulas can be solved this way. The starting value is depicted by the y-intercept and the change rate is expressed in the form of the slope. The main issue with the slope intercept form typically lies in the horizontal interpretation of the variable in particular when the variable is associated with one particular year (or any other type number of units). The first step to solve them is to ensure that you understand the meaning of the variables. |
# Math Snap
## $\int \frac{1}{1+\cos x} d x$ Select one: a. $\cot x-\csc x+c$ b. $\cot x+\csc x+c$ c. $\frac{1}{1+\sin x}+c$ d. $-\cot x+\csc x+c$
#### STEP 1
Assumptions 1. We are given the integral $\int \frac{1}{1+\cos x} \, dx$. 2. We need to find the antiderivative of the given function. 3. We will use trigonometric identities to simplify the integrand.
#### STEP 2
Recall the trigonometric identity for cosine in terms of tangent half-angle substitution: $\cos x = \frac{1 - \tan^2\left(\frac{x}{2}\right)}{1 + \tan^2\left(\frac{x}{2}\right)}$
#### STEP 3
Let $t = \tan\left(\frac{x}{2}\right)$. Then, $dx = \frac{2 \, dt}{1 + t^2}$.
#### STEP 4
Rewrite $\cos x$ in terms of $t$: $\cos x = \frac{1 - t^2}{1 + t^2}$
#### STEP 5
Substitute $\cos x$ into the integrand: $\int \frac{1}{1 + \frac{1 - t^2}{1 + t^2}} \cdot \frac{2 \, dt}{1 + t^2}$
#### STEP 6
Simplify the integrand: $\int \frac{1}{\frac{1 + t^2 + 1 - t^2}{1 + t^2}} \cdot \frac{2 \, dt}{1 + t^2}$
#### STEP 7
Combine the terms in the denominator: $\int \frac{1}{\frac{2}{1 + t^2}} \cdot \frac{2 \, dt}{1 + t^2}$
#### STEP 8
Simplify the fraction: $\int \frac{(1 + t^2)}{2} \cdot \frac{2 \, dt}{1 + t^2}$
#### STEP 9
Cancel out the common terms: $\int dt$
#### STEP 10
Integrate with respect to $t$: $t + C$
#### STEP 11
Substitute back $t = \tan\left(\frac{x}{2}\right)$: $\tan\left(\frac{x}{2}\right) + C$
#### STEP 12
Recognize that the given options are in terms of $\cot x$ and $\csc x$. Use the identity $\tan\left(\frac{x}{2}\right) = \frac{\sin x}{1 + \cos x}$.
#### STEP 13
Rewrite $\tan\left(\frac{x}{2}\right)$ in terms of $\cot x$ and $\csc x$: $\tan\left(\frac{x}{2}\right) = \frac{\sin x}{1 + \cos x} = \cot x - \csc x$
#### STEP 14
Thus, the integral $\int \frac{1}{1+\cos x} \, dx$ simplifies to: $\cot x - \csc x + C$
##### SOLUTION
The correct answer is: $\boxed{\cot x - \csc x + C}$ Therefore, the correct option is: $\text{d. } -\cot x + \csc x + c$ |
# Question 4 and 5 Exercise 3.3
Solutions of Question 4 and 5 of Exercise 3.3 of Unit 03: Vectors. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.
Show that the vector $\hat{i}+7 \hat{j} + 3 \hat{k}$ is perpendicular to both $\hat{i}-\hat{j}+2 \hat{k}$ and $2 \hat{i}-$ $\hat{j}+3 \hat{k}$.
Let $\vec{a}=\hat{i}+7 \hat{j}+3 \hat{k}$, $\vec{b}=\hat{i}-\hat{j}+2 \hat{k}$ and $\vec{c} = 2 \hat{i}-\hat{j}-3 \hat{k}$. Then \begin{align}\vec{a} \cdot \vec{b}&=(\hat{i}+7 \hat{j}+3 \hat{k}) \cdot(\hat{i}-\hat{j}+2 \hat{k}) . \\ \Rightarrow \vec{a} \cdot \vec{b}&=1(1)+7(-1)+3(2) \\ \Rightarrow \vec{a} \cdot \vec{b}&=1-7+6=0 \\ \Rightarrow \vec{a} \perp \vec{b} \cdot \text { Now } \\ \vec{a} \cdot \vec{c}&=(\hat{i}+7 \hat{j}+3 \hat{k}) \cdot(2 \hat{i}+\hat{j}-3 \hat{k}) \\ \Rightarrow \vec{a} \cdot \vec{c}&=1(2)+7(1)+3(-3) \\ \Rightarrow \vec{a} \cdot \vec{c}&=2+7-9=0 . \\ \Rightarrow \vec{a} \perp \vec{c}\end{align}
Let $\vec{a}=\hat{i}+2 \hat{j}+\hat{k}$ and $\vec{b}=2 \hat{i}+\hat{j}-\hat{k}$. Find a vector that is orthogonal to both $\vec{a}$ and $\vec{b}$.
We know that $\vec{c}=\vec{a} \times \vec{b}$ is a vector that is orthogonal to both $\vec{a}$ and $\vec{b}$. Therefore, \begin{align}\vec{c}&=\vec{a} \times \vec{b}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 1 \\ 2 & 1 & -1 \end{array}\right| \end{align} expanding by $R_1$ we have \begin{align} & \vec{c}=(-2-1) \hat{i}+(-1-2) \hat{j}+(1-4) \hat{k} \\ & \Rightarrow \vec{c}=-3 \hat{i}-3 \hat{j}-3 \hat{k}\end{align} is the desired vector perpendicular to both $\vec{a}$ and $\vec{b}$. |
x2 x = 0
a = 2, b = 3, c = -5
D = (3)2 - 4⋅2⋅(-5) = 49
D = 72
The solutions are:
x1 = (-3 + √49)/(2⋅2) = (-3 + 7)/4 = 1
x2 = (-3 - √49)/(2⋅2) = (-3 - 7)/4 = -2.5
Quadratic equations looks like: ax2 + bx + c = 0
where a,b,c are real numbers, and a ≠ 0. Every quadratic equation can have 0, 1 or 2 real decidions derived by the formula:
The number D = b2 - 4ac is called the "discriminant".
If D < 0 then the quadratic equation has no real solutions(it has 2 complex solutions).
If D = 0 then the quadratic equation has 1 solution that is given by the formula: x = - b/2a.
If D > 0 then the quadratic equation has 2 distinct solutions.
Example:
Let's solve the quadratic equation: x2 + 3x - 4 = 0
a = 1, b = 3, c = -4
#### Parabola
The graph of a quadratic equatin is called a parabola.
If a > 0 then graph horns pointing down:
if a < 0 then graph horns pointing up:
The midpoint of any parabola is the point x = -b/2a.
#### Vieta's formulas
If x1 and x2 are the roots of the quadratic equation ax2 + bx + c = 0 then:
$x_1 + x_2 = -\frac{b}{a} \qquad \qquad \qquad x_1x_2 = \frac{c}{a}$
These formulas are called Vieta's formulas.
We can find the roots x1 and x2 of a quadratic equation by solving the system above.
1) x2 - 4 = 0; x = ?
Solution: x2 - 4 = (x - 2)(x + 2)
x = 2 or x = -2
2) 3x2 + 4x + 5 = 0; x = ?
Solution: the discriminant is 42 - 4.3.5 = 16 - 60 = -44 < 0 So the quadratic equation have no real decisions.
3) x2 + 4x - 5 = 0; x = ?
Solution: The discriminant is 42 - (-4.1.5) = 16 + 20 = 36 > 0 So there are two real solutions: 1/2(-4 ± √36)
x = 1 or x = -5
4) x2 + 4x + 4 = 0; x = ?
Solution: The discriminant is 42 - (4.1.4) = 16 - 16 = 0 So there is one real solution: 1/2(-4)
x = -2
5) x2 - 13x + 12 = 0
Solution: 1, 12
Draw the graph of the function: f(x) = x2 - 13x + 12
6) 8x2 - 30x + 7 = 0
Solution: 3.5, 0.25
Draw the graph of the function: f(x) = 8x2 - 30x + 7 |
# A 7.5 kg block rests on a table
Sure, here's an introduction for your blog post:
Exploring Weight and Balance: In the world of Mathematics education, understanding the principles of weight and balance is crucial. In this article, we'll delve into the scenario of a 7.5 kg block placed on a table, exploring the mathematical concepts at play. From calculating forces to analyzing equilibrium, we'll uncover the mathematical intricacies behind this seemingly simple scenario. Join us as we unravel the mathematical mysteries of weight and balance!
## The concept of weight and mass
In the context of Mathematics education, it's important to understand the difference between weight and mass. Weight is the force exerted on an object due to gravity, while mass is the amount of matter in an object. When a 7.5 kg block is placed on a table, its mass remains constant, but its weight depends on the strength of the gravitational field.
## Calculating the force of gravity
When the 7.5 kg block is placed on a table, it experiences a force due to gravity pulling it towards the center of the Earth. The force of gravity can be calculated using the formula F = mg, where F is the force of gravity, m is the mass of the object, and g is the acceleration due to gravity (approximately 9.81 m/s^2 on the surface of the Earth).
## Understanding normal force
The table exerts a force on the block, perpendicular to the surface, to support the weight of the block. This force is known as the normal force. In this case, the normal force exerted by the table on the block is equal in magnitude and opposite in direction to the force of gravity acting on the block.
## Equilibrium and net force
When the 7.5 kg block is placed on the table, it remains at rest, indicating that the forces acting on it are balanced. The net force on the block is zero, resulting in a state of equilibrium. This concept is crucial in understanding the dynamics of objects in a stable condition.
### How can the concept of force be taught using the scenario of a
The concept of force can be taught using the scenario of a rocket launch in Mathematics education.
### 5 kg block placed on a table in a mathematics education setting?
In a mathematics education setting, a 5 kg block placed on a table can be used to demonstrate concepts such as weight, mass, and force.
### What mathematical principles and formulas are involved in calculating the force exerted by a
The mathematical principles involved in calculating the force exerted by a are Newton's second law and the formula F = ma, where F is the force, m is the mass, and a is the acceleration.
### 5 kg block placed on a table?
The 5 kg block is placed on a table.
### In a mathematics classroom, how can the scenario of a
In a mathematics classroom, the scenario of a real-world application can be used to engage students and demonstrate the practical relevance of mathematical concepts.
### 5 kg block on a table be used to illustrate the concept of weight and mass?
Yes, a 5 kg block on a table can be used to illustrate the concept of weight and mass in Mathematics education.
Building 3D shapes with modeling clay, using measuring tapes to understand geometry, and creating graphs with real-world data are practical hands-on activities that can help students understand mathematical concepts in Mathematics education.
### 5 kg block being placed on a table?
The 5 kg block is being placed on a table.
### How can the scenario of a
In Mathematics education, the scenario of a real-world problem can help students understand the practical application of mathematical concepts.
### 5 kg block on a table be used to introduce the concept of gravitational force and its mathematical representation to students in a mathematics education setting?
Yes, a 5 kg block on a table can be used to introduce the concept of gravitational force and its mathematical representation to students in a mathematics education setting.
In conclusion, the 7.5 kg block placed on the table serves as an excellent example for students to understand the concepts of weight, mass, and force in the context of Mathematics education. By exploring the interaction between the block and the table, students can gain a deeper understanding of the principles of Newton's laws of motion and static equilibrium. This practical application of mathematical concepts allows students to bridge the gap between theory and real-world scenarios, fostering a more comprehensive grasp of mathematical principles. |
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# Seating Arrangement | Aptitude
• Difficulty Level : Easy
• Last Updated : 10 May, 2021
Overview :
The seating arrangement is the arrangement of people/objects logically. It can be a linear arrangement or circular arrangement. In circular seating arrangement, we arrange people around a circle while in linear seating arrangement, we arrange people in a line. Some detail regarding the objects/person and how they are seated is given, one should arrange either in a Linear or circular as mentioned. This concept involves the arrangement of people in many possible ways. In these type of questions, you will have to arrange a group of persons satisfying certain conditions. The questions on this topic can be asked in any sequence (linear arrangement, circular arrangement). By applying the logical analysis, we can perform the logical arrangement to answer the questions or decode.
Rules for Seating arrangement :
Here, we will discuss the rules as follows.
1. Identifying the right and left in a given seating arrangement
2. If it is not mentioned in the question we cannot assume left an immediate left.
3. Always start the arrangement with complete fixed details.
4. In the case of a circular arrangement, if nothing is mention regarding the direction they are facing then by default take it as facing Centre.
Features :
The questions on Seating arrangement are the most common types of questions asked in reasoning in all entrance exams. There are different types of problems with the concept of seating arrangement as follows.
1. Linear Arrangement
2. Square/Rectangular Arrangement
3. Circular Arrangement
4. Triangular arrangement
5. Hexagonal arrangement
6. Pentagon arrangement.
7. Combination of the above.
Examples :
Following are some questions on seating arrangement as follows.
Example-1 :
B, L, M, N, P, and Q are in a row. P and Q are in the center, B and L are at the ends. M is sitting to the left of B. Who is to the right of L?
Solution –
The seating arrangement is as follows.
`L--N--P--Q--M--B`
Therefore, the right of L is N.
Example-2 :
Five boys are sitting to be photographed. S is to the left of R and to the right of B. M is to the right of R. E is between R and M.
1. Who is sitting immediate right to E?
2. Who is second from the right?
Solution –
The seating arrangement is as follows.
`B--S--R--E--M`
Example-3 :
Five girls are standing in a line. One of the two girls at extreme end is P and the other is B. A is to the right of S. C is to the left of the B. What is the position of A from the left?
Solution –
The seating arrangement is as follows.
```P--S--A--C--B |
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# In a class test, the sum of Shefali’s marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product of their marks would have been 210. Find her marks in the two subjects.
Let Shefali’s marks in Mathematics = x
Let Shefali’s marks in English = 30 - x
If, she had got 2 marks more in Mathematics, her marks would be = x + 2
If, she had got 3 marks less in English, her marks in English would be = 30 – x - 3 = 27 - x
According to given condition:
$$\Rightarrow$$ $$(x + 2) (27 - x) = 210$$
$$\Rightarrow$$ $$27x - x^2 + 54 - 2x = 210$$
$$\Rightarrow$$ $$x^2 - 25x + 156 = 0$$
Comparing quadratic equation $$x^2 - 25x + 156 = 0$$ with general form $$ax^2 + bx + c = 0$$, We get a = 1, b = -25 and c = 156
Applying Quadratic Formula = $$x = {{-b ± \sqrt{b^2 - 4ac}} \over {2a}}$$
$$\Rightarrow$$ $$x = {{25 ± \sqrt{(25)^2 - 4(1)(156)}} \over {(2)(1)}}$$
$$\Rightarrow$$ $$x = {{25 ± \sqrt{625 - 624}} \over {2}}$$
$$\Rightarrow$$ $$x = {{25 + 1} \over {2}} , {{25 - 1} \over {2}}$$
$$\Rightarrow$$ $$x = 13, 12$$
Therefore, Shefali’s marks in Mathematics = 13 or 12
Shefali’s marks in English = 30 – x = 30 – 13 = 17
Or Shefali’s marks in English = 30 – x = 30 – 12 = 18
Therefore, her marks in Mathematics and English are (13, 17) or (12, 18). |
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Studying for a test? Prepare with these 14 lessons on Limits and continuity.
See 14 lessons
# Limits of combined functions
Video transskription
- [Voiceover] So let's find the limit of f of x times h of x, as x approaches zero. Alright, we have graphical depictions of the graphs, y equals f of x and y equals h of x. And we know from our limit properties, that this is gonna be the same thing as the limit, as x approaches zero of f of x, times, times the limit as x approaches zero of h of x. And let's think about what each of these are, so let's first think about f of x, right over here. So in f of x, as x approaches zero, notice, the function itself isn't defined there, but we see when we approach from the left, we are approaching, the function seems to be approaching the value of negative one, right over here. And as we approach from the right, the function seems to be approaching the value of negative one. So the limit here, this limit here, is negative one. As we approach from the left, we're approaching negative one, as we approach from the right, the value of the function seems to be approaching negative one. Now what about h of x? Well, h of x we have down here. As x approaches zero, as x approaches zero, the function is defined at x equals zero, it looks like it is equal to one, and you could, and the limit is also equal to one, we could see that as we approach it from the left, we are approaching one, as we approach from the right, we are approaching one. As we approach x equals zero from the left, we approach, the function approaches one. As we approach x equals zero from the right, the function itself is approaching one. And it makes sense that the function is defined there, and is defined at x equals zero, and the limit as x approaches zero is equal to the same as, is equal to the value of the function at that point, because this is a continuous function. So this is, this is one, and so negative one times one is going to be equal to, is equal to, negative one. So that is equal to negative one. Let's do a few more of these. So we have the limit of negative two times f of x plus three times h of x, as x approaches negative three. Well, once again, we can use our limit properties. We know that this is the same thing as the limit, as x approaches negative three of negative two f of x, plus the limit as x approaches negative three of three times h of x. And this is the same thing as the limit, or I should say this is equal to negative two times the limit as x approaches negative three, the limit of f of x, as x approaches negative three, plus, we can take this scaler out, three times the limit of h of x, as x approaches negative three. And so we just have to figure out what the limit of f of x is as x approaches negative three, and the limit of h of x as x approaches negative three. So I'll first do f of x, so f of x right over here, limit is x approaches negative three, when we approach negative three from the left-hand side, it seems the value of the function is approaching three, and as we approach it from the right-hand side, it seems like the value of the function is zero. So our left-handed and right-handed limits are approaching different things, so this limit actually does not exist. Does not exist. This limit actually here does exist. But since this limit doesn't exist, and we need to figure out this limit in order to figure out this entire limit, this whole thing does not exist. Does not, does not exist. The limits that it's made up of need to exist, this is a combined limit, so each of the pieces need to exist in order for the, in this case, the scaled up sum, to actually exist. Let's do one more. Alright. So these are both, looks like, continuous functions. So we have the limit is x approaches zero h of x over g of x. So once again, using our limit properties, this is going to be the same thing as the limit of h of x as x approaches zero over the limit of g of x as x approaches zero. Now what's the limit of h of x as x approaches zero? This is, let's see, as we approach zero from the left, as we approach x equals zero from the left, our function seems to be approaching four, and as we approach x equals zero from the right, our function seems to be approaching four, and that's also what the value of the function is at x equals zero, that makes sense, because this is a continuous function, so the limit as we approach x equals zero should be the same as the value of the function at x equals zero. So this top, this is going to be four, now let's think about the limit of g of x as x approaches zero. So from the left it looks like, as x approaches zero, the value of the function is approaching zero. And as x approaches zero from the right, the value of the function is also approaching zero, which happens to also be, which also happens to be, g of zero, g of zero is also zero. And that makes sense that the limit and the actual value of the function at that point is the same, because it's continuous. So this also is zero. But now we're in a strange situation. We have to take four and divide it by zero. So this limit will not exist, cause we can't take four and divide it by zero. So even though the limit of h of x is x equals, as x approaches zero exists, and the limit of g of x as x approaches zero exists, we can't divide four by zero, so this whole entire limit does not exist. Does, does not exist. And actually, if you were to plot h of x over g of x, if you were to plot that graph, you would see it even clearer that that limit does not exist, you would actually be able to see it graphically. |
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# A man running on a horizontal road at $8m{s^{ - 1}}$ finds rain falling vertically. If he increases his speed to $12m{s^{ - 1}}$ , he finds that drops make ${30^ \circ }$ angle with the vertical. Find velocity of rain with respect to the road.(A) $4\sqrt 7 m{s^{ - 1}}$ (B) $8\sqrt 2 m{s^{ - 1}}$ (C) $7\sqrt 3 m{s^{ - 1}}$ (D) $8.32m{s^{ - 1}}$
Last updated date: 13th Jun 2024
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Hint: This is an example of the motion that occurs in two dimensions. Hence, when we consider each element, we have to look into its horizontal and vertical components as well. Further we have to frame the connection between both the velocities by involving the concept of relative velocity. Using these methods we will be able to solve the problem.
Complete Step by Step Solution
From the question we are supposed to consider three main entities; velocity of the man, velocity of the rain and the velocity of the rain with respect to the man.
Velocity of rain: ${\vec v_r} = a\hat i + b\hat j$ . Let the velocity of man be ${\vec v_m}$ .
Given that, a man is running on a horizontal road at $8m{s^{ - 1}}$ , thus ${\vec v_m} = 8\hat i$ .
According to the first condition, when the man runs with a speed of $8m{s^{ - 1}}$ , he sees the rain falling vertically. It is represented by, ${\vec v_{rm}} = 0\hat i + b\hat j$ where ${\vec v_{rm}}$ is the relative velocity of the rain and the man, and it’s value is given by, ${\vec v_{rm}} = {\vec v_r} - {\vec v_m}$ .
Thus, ${\vec v_{rm}} = {\vec v_r} - {\vec v_m} = b\hat j$ .
We can write, $a\hat i + b\hat j - 8\hat i = b\hat j$ . Simplifying, we get,
$\left( {a - 8} \right)\hat i + b\hat j = b\hat j$ .
Thus, from this, it can be written,
$\left( {a - 8} \right) = 0$
$\Rightarrow a = 8$
According to the second condition, when the speed of the man is increased to $12m{s^{ - 1}}$ , he finds that drops make ${30^ \circ }$ angle with the vertical.
${\vec v_m} = 12\hat i$
${\vec v_{rm}} = {\vec v_r} - {\vec v_m} = \left( {a\hat i + b\hat j} \right) - 12\hat i$
Simplifying, we get,
$\left( {a - 12} \right)\hat i + b\hat j = b\hat j$ .
Thus, from this, it can be written,
$\left( {a - 12} \right) = 0$
$\Rightarrow a = 12$
Since the angle with the vertical is ${30^ \circ }$ , we can write,
$\dfrac{b}{{a - 12}} = \tan {30^ \circ }$
Since $a = 4$ , $\Rightarrow \dfrac{1}{{\sqrt 3 }} = \dfrac{b}{{4 - 12}}$ .
Simplifying, $b = - \dfrac{4}{{\sqrt 3 }}$ .
As we’ve got the values of $a$ and $b$ , the velocity of rain with respect to road is,
${\vec v_r} = 8\hat i - \dfrac{4}{{\sqrt 3 }}\hat j$
$\Rightarrow \left| {{v_r}} \right| = \sqrt {{8^2} + {{\left( { - \dfrac{4}{{\sqrt 3 }}} \right)}^2}} = 8.32m{s^{ - 1}}.$
Hence, option D is the correct answer among the given options.
Note
The sign of the relative velocity shows variations in accordance to the direction of the motion. If both the bodies are in the same directions, we get a positive value. Else it is a negative value. The students should read the question well and get an idea about the sign of the relative velocity. |
# Law of Sines: How to Use & Practice Examples
The Law of Sines (or the Sine Rule) tells us that the ratio of a side length to the sine of its opposite angle is the same for all three sides.
As a formula, it looks like this:
$\frac{a}{sin\alpha}=\frac{b}{sin\beta}=\frac{c}{sin\gamma}$
or
$\frac{sin\alpha}{a}=\frac{sin\beta}{b}=\frac{sin\gamma}{c}$
In the formula,
• a is the side and α is its opposite angle,
• b is the side and β is its opposite angle,
• c is the side and γ is its opposite angle
So the ratio between each of these sides and the sine of their opposite angle is the same.
The law of sines works for any triangle, unlike Sohcahtoa which works for the right-angled ones only.
Sides and angles are called the six measures of a triangle and to solve the triangle means to find all of these six measures. If we are given only some of them, we can use The Law of Sines to find the rest of them. Precisely, if we are given:
1. One side and its opposite angle
2. One or more other sides or angles
When solving problems, you won’t have to use all three ratios at the same time. Which ones you’ll take depends on what information is given about the triangle in your problem.
## Example I
Find the measure of side c.
Let’s first check if we are given all that we need for The Law of Sines.
Side a and its opposite angle are there, so yes, we can apply the rule.
Looking at the information given, we’ll use the first and the last ratio from the formula:
$\frac{a}{sin\alpha}=\frac{c}{sin\gamma}$
Plug in the values we have:
$\frac{4}{(sin50^\circ)}=\frac{c}{(sin20^\circ)}$
Swap the sides (we want the unknown measure to be on the left):
$\frac{c}{(sin20^\circ)}=\frac{4}{(sin50^\circ)}$
Multiply both sides with sin20°:
$c=\frac{4}{(sin50\circ)}*sin20^\circ$
And now calculate using your calculator:
$c=\frac{4}{0.766}*0.342$
$c=1.786$
$c=1.9 \text{(to one decimal place)}$
## Example II
Find the missing angle β.
We are given a side c and its opposite angle, so we can apply the Sine Rule. When you need to find the missing angle, it’s best to use the second formula:
$\frac{sin\gamma}{c}=\frac{sin\beta}{b}$
Plug in the values we have:
$\frac{(sin110^\circ)}{78}=\frac{sin\beta}{60}$
Swap the sides to get the unknown angle to the left side:
$\frac{sin\beta}{60}=\frac{(sin110^\circ)}{78}$
Multiply both sides with 60:
$sin\beta=\frac{(sin110^\circ)}{78}*60$
And now use the calculator to solve the right side of this equation:
$sin\beta=\frac{0.940}{78}*60$
$sin\beta=0.806$
Finally, find the inverse of the sine:
$\beta=sin^{-1}$ $0.806$
$\beta=53.7^\circ$
## How would you find the third side?
Let’s solve this triangle completely, i.e. let’s find the remaining side and angle.
To find the angle we’ll use the property that the sum of interior angles in a triangle is 180°:
$\alpha+\beta+\gamma=180^\circ$
$\alpha+53.7^\circ+110^\circ=180^\circ$
$\alpha+163.7^\circ=180^\circ$
$\alpha=180^\circ-163.7^\circ$
$\alpha=16.3^\circ$
Now we can set up the equation to find the side a, using the Sine Rule:
$\frac{a}{sin\alpha}=\frac{b}{sin\beta}$
$\frac{a}{(sin16.3^\circ)}=\frac{60}{(sin53.7^\circ)}$
$a=\frac{60}{(sin53.7^\circ)}*sin16.3^\circ$
$a=\frac{60}{0.806}*0.281$
$a=20.918$
$a=20.9$
## Example III
Solve the triangle below.
Let’s first check if we can apply the Sine Rule here.
We are given side b but not its opposite angle. Also, we are given side c but not its opposite angle. Therefore, we can’t apply the formula. This triangle is not suited for The Law of Sines (but it is for The Law of Cosines).
## The Ambiguous Case
The ambiguous case comes from the fact that an acute angle and an obtuse angle have the same sine (check the Unit Circle). For example, sin30°=1/2 and sin150°=1/2.
Let us use the Law of Sines to find angle γ.
$\frac{sin\beta}{b}=\frac{sin\gamma}{c}$
$\frac{sin30^\circ}{15}=\frac{sin\gamma}{20}$
$\frac{sin\gamma}{20}=\frac{sin30^\circ}{15}$
$sin\gamma=\frac{sin30^\circ}{15}*20$
The sine of 30° is 1/2:
$sin\gamma=\frac{\frac{1}{2}}{15}*20$
$sin\gamma=\frac{10}{15}$
$sin\gamma=\frac{2}{3}$
$sin\gamma=0.667$
$\gamma=sin^{-1}$ $0.667$
$\gamma\approx 42^\circ$
However, angle 138° has the same sine as angle 42°, i.e. sin138°=0.667, so this problem has two solutions.
Not only is angle ∠BCA a solution, but so is angle ∠BDA.
However, this won’t always be the case. Sometimes you’ll have two possible solutions and sometimes one, so you should always check if the other solution makes sense.
## The Circumcircle Connection
The Law of Sines tells us that the ratio between the sides and their opposite angles is the same. Additionally, that ratio is also the diameter of the circumcircle, which is the circle that passes through all three vertices of the triangle:
$\frac{a}{sin\alpha}=\frac{b}{sin\beta}=\frac{c}{sin\gamma}=2r$
where r is the radius of the triangle’s circumcircle.
## Summary
If we are given one side and its opposite angle we can apply the Law of Sines formula and then, using other given elements, we can solve the triangle.
Did this help? You can view more examples in Trigonometry + 100 other subjects at Educator.com, the best place to get in-depth and instant homework help online. |
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# Ratio
In this article, we are going to learn about the ratios and the fractions as per the Primary 6 Math level.
The learning objectives are
1. Relationship between ratio and fraction
2. Simple word problems involving ratios
This article can serve as a refresher for the ratios and fractions concepts as taught in P5.
## What is a Ratio?
A ratio tells us the relationship between two or more quantities. It may not represent the actual number.
For quantities with units (e.g. kg, m), we must remember to express them in the same units before expressing them as ratios. There are no units in the ratio.
Example 1:
Two brothers have some money. Tom has $10 and Donnie has$15. Express the amount of money Tom has to the amount of money Donnie has as a ratio in its simplest form.
Solution:
\begin{align} \text{Tom} &: \text{Donnie}\\[2ex] 10 &: 15\\[2ex] 2 &: 3\\[2ex] \end{align}
$$2 : 3$$
## Equivalent Ratio
Similar to fractions, there are equivalent ratios. We express ratios in their simplest form by dividing each term by a common factor.
$$24 : 16$$, $$12 : 8$$, and $$3 : 2$$ are equivalent ratios. The ratio $$3 : 2$$ is in its simplest form.
## Fraction And Ratio
Other than using a ratio to compare two quantities, we can also express one quantity as a fraction of another.
A ratio can be expressed as a fraction and a fraction can be expressed as a ratio.
Example 2:
Tom has $5 and Donnie has$2.
1. Express the amount of money Tom has to the amount of money Donnie has as a ratio.
Solution:
\begin{align} \text{Tom} &: \text{Donnie}\\[2ex] 5 &: 2 \end{align}
1. Express the amount of money Tom has as a fraction of the amount of money Donnie.
Solution:
\begin{align} \frac{\text{Tom}}{\text{Donnie}} = \frac{5}{2} \end{align}
### Practice Questions on Ratios
Now let us try some simple questions to understand the concept.
Question 1:
See the picture given below.
1. The ratio of the number of cups to the number of saucers is __________ : __________.
Solution:
Number of cups $$= 3$$
Number of saucers $$= 7$$
\begin{align} \text{Cups} &: \text{Saucers}\\[2ex] 3 &: 7 \end{align}
$$3 : 7$$
1. The ratio of the number of saucers to the number of cups is __________: __________.
Solution:
Number of saucers $$= 7$$
Number of cups $$= 3$$
\begin{align} \text{Saucers} &: \text{Cups}\\[2ex] 7 &: 3 \end{align}
$$7 : 3$$
1. The ratio of the number of cups to the total number of cups and saucers is __________ : __________.
Solution:
Number of cups $$= 3$$
Total number of cups and saucers\begin{align}\\[2ex] &= 3 + 7\\[2ex] &= 10 \end{align}
\begin{align} \text{Cups} &: \text{Total number of cups and saucers}\\[2ex] 3 &: 10 \end{align}
$$3 : 10$$
Question 2:
Solve the following,
$$24 : 18 = 4 :$$__________
Solution:
3
Question 3:
Solve the following,
__________ $$: 15 = 35 : 21$$
Solution:
Step 1:
Find the simplest form of $$35 : 21$$.
$$35 : 21 = 5 : 3$$
Step 2:
Find the equivalent ratio of $$5 : 3$$.
We get
$$25 : 15 = 5 : 3$$
$$25$$
Question 4:
Chris has $$\displaystyle{\frac{2}{5}}$$ as many marbles as John.
1. The ratio of the number of marbles Chris had to the number of marbles John had is __________ : __________.
2. The ratio of the number of marbles John had to the number of marbles Chris had is __________ : __________.
3. The ratio of the number of marbles Chris had to the total number of marbles the two boys had is __________ : __________.
4. The number of marbles John had is __________ of the number of marbles Chris had.
5. The number of marbles John had is __________ of the total number of marbles the two boys had.
Solution:
Number of marbles Chris had $$\displaystyle{=\frac{2}{5}}$$ number of marbles John had
1. The ratio of the number of marbles Chris had to the number of marbles John had is 2 : 5.
2. The ratio of the number of marbles John had to the number of marbles Chris had is 5 : 2.
3. The ratio of the number of marbles Chris had to the total number of marbles is 2 : 7.
4. The number of marbles John had $$\displaystyle{\frac{5}{2}}$$ of the number of marbles Chris had.
5. The number of marbles John had is $$\displaystyle{\frac{5}{7}}$$ of the total number of marbles the two boys had.
Question 5:
Cookies were sold in packs of 5 each. Mrs Tong bought 8 packs of chocolate cookies, 2 packs of butter cookies and 6 packs of strawberry cookies.
1. Express the ratio of the number of chocolate cookies to the number of butter cookies to the number of strawberry cookies Mrs Tong bought in its simplest form.
2. Express the number of chocolate cookies as a fraction of the total number of cookies Mrs Tong bought in its simplest form.
Solution:
A.
Number of cookies in each pack $$= 5$$
Number of packs of chocolate cookie $$= 8$$
Total number of chocolate cookies\begin{align}\\[2ex] \displaystyle &= 8 \times 5\\[2ex] &=40 \text{ cookies} \end{align}
Number of packs of butter cookie $$= 2$$
Total number of butter cookies\begin{align}\\[2ex] \displaystyle &= 2 \times 5\\[2ex] &=10 \text{ cookies} \end{align}
Number of packs of strawberry cookie $$= 6$$
Total number of strawberry cookies\begin{align}\displaystyle\\[2ex] &= 6 \times 5\\[2ex] &=30 \end{align}
$$4 : 1 : 3$$
B.
Number of packs of chocolate cookie $$= 8$$ packs
Total number of packs of cookie\begin{align}\\[2ex] \displaystyle &= 8 + 2 + 6\\[2ex] &=16 \text{ packs} \end{align}
\begin{align} \frac{\text{ Number Of Chocolate Cookies }}{\text{ Total Number Of Cookies }} &= \frac{40}{80} \\[2ex] &= \frac{8}{16} \\[2ex] &= \frac{1}{2} \end{align}
$$\displaystyle{\frac{1}{2}}$$
Question 6:
A sum of money is shared among Matthew, Paul and Luke in the ratio $$9 : 3 : 4$$.
1. Express Matthew’s share as a fraction of Luke’s share.
2. What fraction of Paul’s share is Mathew’s share?
3. What fraction of the total share is Luke’s share in its simplest form?
Solution:
A.
Matthew : Paul : Luke = $$9 : 3 : 4$$
$$\displaystyle{\frac{\text{Matthew}}{\text{Luke}} = \frac{9}{4} = 2\frac{1}{4}}$$
$$\displaystyle{2\frac{1}{4}}$$
B.
Paul’s share $$= 3$$ units
Mathew’s share $$= 9$$ units
$$\displaystyle{\frac{\text{Matthew}}{\text{Paul}} = \frac{9}{3} = 3}$$
3
C.
Luke’s share $$= 4$$ units
Total share $$= 9$$ units $$+ \;3$$ units $$+ \;4$$ units
$$= 16$$ units
$$\displaystyle{\frac{\text{Luke}}{\text{Total}} = \frac{4}{16} = \frac{1}{4}}$$
$$\displaystyle\frac{1}{4}$$
Question 7:
Ali’s savings is $$\displaystyle 1\frac{2}{5}$$ times that of Rose’s.
1. Find the ratio of Ali’s savings to their total savings.
2. Find the ratio of Ali’s savings to the difference between Ali’s and Rose’s savings.
Solution:
A.
Ali’s savings $$\displaystyle =1\frac{2}{5}$$ times Rose’s savings
Ali’s savings $$\displaystyle =\frac{7}{5}$$ times Rose’s savings
Total savings $$= 7$$ units $$+ \;5$$ units
$$= 12$$ units
The ratio of Ali’s savings to their total savings is $$7 : 12$$.
$$7 : 12$$
B.
Ali’s savings $$= 7$$ units
Rose’s savings $$= 5$$ units
Difference between Ali’s and Rose’s savings
$$= 7$$ units $$- \;5$$ units
$$= 2$$ units
The ratio of Ali’s savings to the difference between Ali’s and Rose’s savings is $$7 : 2$$.
$$7 : 2$$
Question 8:
On the bookshelf, for every 3 fiction books, there are 5 non-fiction books. If there were a total of 432 books, how many fiction books are there?
Solution:
\begin{align} \text{Fiction} &: \text{Non-Fiction}\\[2ex] 3 &: 5 \end{align}
Total number of books (in units) $$= 3$$ units $$+ \;5$$ units
$$= 8$$ units
Since there were a total of 432 books, 8 units is 432.
\begin{align} 8 \text{ units} &= 432\\[2ex] 1 \text{ unit} &= 432 \div 8\\[2ex] &= 54 \end{align}
Number of fiction books
$$= 3$$ units
$$= 3 \times 54$$
$$= 162$$
$$162$$ books
Question 9:
The length of a rectangle is 4 times its breadth. Its perimeter is 30 cm. Find its area.
Solution:
\begin{align} \text{Length (L)} &: \text{Breadth (B)}\\[2ex] 4 &: 1 \end{align}
Perimeter of rectangle\begin{align}\\\\[4ex] &= 2 \times \text{ (L + B)}\\[2ex] &= 2 \times (4 \text{ units} + 1 \text{ unit})\\[2ex] &= 10 \text{ units} \end{align}
\begin{align} 10 \text{ units} &= 30 \text{ cm}\\[2ex] 1 \text{ unit} &= 30 \text{ cm} \div 10\\[2ex] &= 3 \text{ cm} \end{align}
Length (L)\begin{align}\\\\[4ex] &= 4 \text{ units}\\[2ex] &= 4 \times 3 \text{ cm}\\[2ex] &= 12 \text{ cm} \end{align}
Breadth (B)\begin{align}\\[2ex] &= 1 {\text{ unit}}\\[2ex] &= 3 \text{ cm} \end{align}a
Area of rectangle\begin{align}\\\\[4ex] &= \text{L} \times \text{B}\\[2ex] &= 12\text{ cm} \times 3\text{ cm}\\[2ex] &= 36 \text{ cm}^2 \end{align}
$$36 \text{ cm}^2$$
Question 10:
At a party, the number of children was $$\displaystyle\frac{7}{3}$$ the number of adults. If there were 225 adults, how many children were at the party?
Solution:
\begin{align} \text{Children} &: \text{Adults}\\[2ex] 7 &: 3 \end{align}
Number of children $$= 7$$ units
Number of adults $$= 3$$ units
\begin{align} 3 \text{ units} &= 225\\[2ex] 1 \text{ unit} &= 225 \div 3\\[2ex] &= 75 \end{align}a
Number of children\begin{align}\\\\[4ex] &= 7 \text{ units}\\[2ex] &= 7 \times 75\\[2ex] &= 525 \end{align}
$$525$$ children
Question 11:
$$\displaystyle\frac{1}{4}$$ of May’s savings was equal to $$\displaystyle\frac{3}{7}$$ of Rena’s savings. Find the ratio of May’s savings to Rena’s savings.
Solution:
$$\displaystyle{\frac{1}{4}}$$ of May’s savings $$\displaystyle{=\frac{3}{7}}$$ of Rena’s savings
$$\displaystyle{\frac{3}{12}}$$ of May’s savings $$\displaystyle{=\frac{3}{7}}$$ of Rena’s savings
Once the numerator is made the same, we can refer to the denominator to get the total number of units representing May’s savings and Rena’s savings.
\begin{align} \text{May’s savings} &: \text{Rena’s savings}\\[2ex] 12 &: 7 \end{align}
$$12 : 7$$
Question 12:
$$\displaystyle{\frac{6}{7}}$$ of Peter’s savings is thrice of Jon’s savings. Find the ratio of Peter’s savings to Jon’s savings.
Solution:
$$\displaystyle{\frac{6}{7}}$$ of Peter’s savings is thrice of Jon’s savings.
$$\displaystyle{\frac{2}{7}}$$ of Peter’s savings is equal to Jon’s savings.
So, Peter’s savings is represented by 7 units and Jon’s savings is represented by 2 units.
\begin{align} \text{Peter’s savings} &: \text{Jon’s savings}\\[2ex] 7 &: 2 \end{align}
7 : 2
## Conclusion
In this article, we learnt about the relationship between the ratios and fractions. We learnt that ratios can be expressed as fractions and fractions can be expressed as ratios.
We also solved simple word problems involving ratio.
Hope you have understood the concepts well. Keep practicing.
Continue Learning Algebra Distance, Speed and Time Volume of Cubes and Cuboid Fundamentals Of Pie Chart Finding Unknown Angles Number Patterns: Grouping & Common Difference Fractions Of Remainder Fractions - Division Ratio Repeated Identity: Ratio Strategies
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## ML Aggarwal Class 6 Solutions for ICSE Maths Chapter 3 Integers Ex 3.3
Question 1.
Evaluate the following, using the number line:
(i) 4 – (-2)
(ii) -4 – (-2)
(iii) 3 – 6
(iv) -3 – (-5)
Solution:
(i) Start from 4 on the number line.
Move 2 units to the digits we reach at 6
∴ 4 – (-2) = 4 + 2 = 6
(ii) Start from -4 on the number line.
Move 2 units to the right, we reach at -2
∴ -4 – (-2) = —4 + 2 = -2
(iii) Start from 3 on the number line.
Move 6 units to the left, we reach at -3
3 – 6 = -3
(iv) Start from -3 on the number line.
Move 5 units to the right, we reach at 2
-3 – (-5) = -3 + 5 = 2
Question 2.
Subtract :
(i) -6 from 9
(ii) 6 from -9
(iii) -6 from -9
(iv) -725 from -63
(v) -376 from 10
(vi) 92 from -620
Solution:
(i) 9 – (-6) = 9 + 6 = 15
(ii) -9 – 6 = -15
(iii) -9 – (-6) = -9 + 6 = -3
(iv) -63 – (-725) = -63 + 725 = +662
(v) 10 – (-376) = 10 + 376 = 386
(vi) -620 – 92 = -712
Question 3.
Evaluate the following:
(i) -237 – (+ 1884)
(ii) -346 – (- 1275)
(iii) -190 – (-3512)
(iv) -2718 – (+ 6827)
Solution:
(i) -237 – (+ 1884)
= -237 – 1884
= -(237 + 1884) = -2121
(ii) -346 -(- 1275)
= -346 + 1275
= 1275 – 346 = 929
(iii) -190 – (-3512)
= -190 + 3512
= 3512- 190 = 3322
(iv) 2718 – (+ 6827)
= -2718 – 6827
= -(2718 + 6827) = -9545
Question 4.
The sum of two integers is 17. If one of them is -35, find the other.
Solution:
One number = -35
Sum of two integers =17
Second number = Sum of integers – (The given number)
= 17 – (-35)
= 17 + 35 = 52
Question 5.
What must be added to -23 to get -9?
Solution:
Let the number to be added = x
∴ -23 + x = -9
∴ The required number = -9 – (-23)
= -9 + 23 = 14
Question 6.
Find the predecessor of 0.
Solution:
Predecessor of 0 = 0 – 1 = -1
Question 7.
Find the successor and the predecessor of the following integers:
(i) -31
(ii) -735
(iii) -240
Solution:
(i) Successor of -31 = -31 + 1 = -30
Predecessor of -31 = -31 – 1 = -32
(ii) Successor of -735 = -735 + 1 = -734
Predecessor of-735 = -735 – 1 = -736
(iii) Successor of -240 = -240 + 1 = -239
Predecessor of -240 = -240 – 1 = -241 |
# What Is A Mixed Number: Explained for Primary School
Mixed numbers and improper fractions are introduced once children are secure in their understanding of proper fractions (unit- and non-unit) in Upper KS2.
### What is a mixed number?
A mixed number, sometimes also called a mixed fraction, is a combination of an integer (whole number) and fraction (part of a whole number).
### Examples of mixed numbers
Mixed numbers can be written with or without ‘and’, e.g. 5 and ¾ or 5¾. The fractional part of the mixed number must be a proper fraction (less than one whole). In a proper fraction, the numerator (top number) is less than the denominator (bottom number), such as 3⁄7, or 11⁄15. A mixed number cannot be composed of an integer and an improper fraction (more than one whole), such as 5 and 5⁄4. This would have to be corrected to a mixed number – in this case, it would be 6 and ¼.
### When do children learn about mixed numbers in the national curriculum?
Children first encounter mixed numbers in UKS2. In Year 5, pupils must “recognise mixed numbers and improper fractions and convert from one form to the other and write mathematical statements > 1 as a mixed number (for example, 2⁄5 + ⅘ = 6⁄5 = 1 and ⅕)” and “multiply proper fractions and mixed numbers by whole numbers, supported by materials and diagrams”.
Finally, in Year 6 maths lessons, pupils will be “adding and subtracting fractions with different denominators and mixed numbers, using the concept of equivalent fractions”.
### How do mixed numbers relate to other areas of maths?
Mixed numbers often appear in measurement topics, requiring children to convert between units of measure. For example, children would be expected to know that 1½ litres is equivalent to 1,500ml, or that 2¾ hours = 165 minutes. Some mixed numbers may require simplifying, e.g. 4 2⁄4 = 4½. These types of measures will also sometimes be represented as decimals, such as 1.5 or 2.75, as children are expected to know fraction-decimal equivalents in UKS2.
### How are mixed numbers used in real life?
Children should be shown how the maths they are learning is applicable in real-life contexts. Therefore, as previously mentioned, mixed numbers can be most commonly found in real-life when referring to units of measure, e.g. 1½ tablespoons, 1¾ hours, 5½ pizzas, etc.
Fractions Lessons Resource Pack
Download a free pack of worksheet on fractions, including questions on mixed numbers and word problems.
### 3 worked examples for mixed number
1) Convert between mixed numbers and improper fractions
To convert an improper fraction to a mixed number, you’ll need to understand how to divide fractions. The process will involve dividing the numerator (in a division, this is also known as the dividend) by the denominator (also known as the divisor). The answer to this (also known as the quotient) becomes the whole number part; the remainder (if there is one) becomes the numerator; the denominator (which was the divisor) remains the same.
For example, to convert 23⁄5 to a mixed number, step-by-step:
1. Divide 23 by 5.
2. 5 fits into 23 4 whole times, so the whole number is 4.
3. There is a remainder of 3, so the new numerator is 3 in the fraction part of the mixed number (the denominator remains the same as the original improper fraction). The answer is therefore 4⅗.
This can be much more clearly visualised with a diagram such as a bar model:
As shown above, twenty-three fifths can also be written as four wholes and three-fifths.
To convert a mixed number to an improper fraction, multiply the whole number by the denominator and add the numerator. The answer to this becomes the new numerator; the denominator remains the same.
For example, to convert 2⅔ to an improper fraction step-by-step:
1. 2 (whole number) x 3 (denominator) = 6
2. 6 + 2 (numerator) = 8 (the new numerator). The answer is therefore 8⁄3.
As shown above, two wholes and two-thirds can also be written as eight-thirds.
2) Add or subtract mixed numbers
Example 1: 1¼ + 1½ Example 2: 2⅓ – 1⅖
An addition or subtraction such as these can be approached in one of two ways:
• by first partitioning the mixed numbers into integers and proper fractions, calculating and then recombining, such as example 1 above: 1 + 1 = 2, ¼ + ½ = ¾, then 2 + ¾ = 2¾
• by first converting the mixed numbers into improper fractions, calculating and then converting back into mixed numbers if necessary, such as example 2 above: 2⅓ – 1⅖ = 7⁄3 – 7⁄5 = 35⁄15 – 21⁄15 = 14⁄15
3) Multiply mixed numbers by whole numbers
1⅓ x 5
As above, this can approached in one of two ways:
• by first partitioning the mixed number into an integer and proper fraction, calculating and then recombining: 1 x 5 = 5, ⅓ x 5 = 1⅔, 5 + 1⅔ = 6⅔
• by first converting the mixed number into an improper fraction, calculating and then converting back into a mixed number if necessary: 1⅓ = 4⁄3, 4⁄3 x 5 = 20⁄3 = 6⅔
### 5 mixed number practice questions and answers
1. 2½ + 1⅗
Answer = 5⁄2 + 8⁄5 = 25⁄10 + 16⁄10 = 41⁄10 = 4 and 1⁄10
2. 3 x 2⅖
Answer = 3 x 12⁄5 = 3⁄5 = 7⅕
3. The length of a day on Earth is 24 hours. The length of a day on Mercury is 58⅔ times the length of a day on Earth. What is the length of a day on Mercury, in hours? (2018 SATs)
Answer: 24 x 58⅔ = 1,408 hours
4. Which improper fraction is equivalent to 6⅞?
67⁄8 48⁄8 62⁄8 55⁄8 76⁄8 (2018 SATs)
5. Potatoes cost £1.50 per kg and carrots cost £1.80 per kg. Jack buys 1½kg of potatoes and ½kg of carrots. How much change does he get from £5? (2019 SATs)
Answer: £1.50 x 1½ = £2.25, ½ of £1.80 = 90p, £2.25 + 90p = £3.15
Read more about adding, subtracting and and learning how to multiply fractions in this fractions for kids article.
What is the difference between an improper fraction and a mixed number?
Both are larger than one whole but are represented differently: an improper fraction has only a numerator and a denominator (the former of which is larger than the latter, e.g. 5⁄3, the equivalent of 1⅔); a mixed number has a whole number and a proper fraction (e.g. 1⅔, the equivalent of 5⁄3).
How do I convert a mixed number to an improper fraction?
See the ‘worked examples’ section above
Read more: How to Teach Fractions: adding, subtracting, multiplying and dividing fractions.
For more explanations teaching primary maths topics, see our Primary Maths Dictionary
x
#### FREE Percentages & Decimals All Kinds of Word Problems (Years 5 & 6)
20+ worded problems to extend your pupils’ reasoning skills using percentages and decimals, from Years 3-6.
10 questions for each year group of ascending difficulty, and a challenge question to stretch pupils beyond the core understanding of percentages and decimals and on to mastery. |
# 4.1 Related rates (Page 3/7)
Page 3 / 7
What is the speed of the plane if the distance between the person and the plane is increasing at the rate of $300\phantom{\rule{0.2em}{0ex}}\text{ft/sec}?$
$500\phantom{\rule{0.2em}{0ex}}\text{ft/sec}$
We now return to the problem involving the rocket launch from the beginning of the chapter.
## Chapter opener: a rocket launch
A rocket is launched so that it rises vertically. A camera is positioned $5000\phantom{\rule{0.2em}{0ex}}\text{ft}$ from the launch pad. When the rocket is $1000\phantom{\rule{0.2em}{0ex}}\text{ft}$ above the launch pad, its velocity is $600\phantom{\rule{0.2em}{0ex}}\text{ft/sec}.$ Find the necessary rate of change of the camera’s angle as a function of time so that it stays focused on the rocket.
Step 1. Draw a picture introducing the variables.
Let $h$ denote the height of the rocket above the launch pad and $\theta$ be the angle between the camera lens and the ground.
Step 2. We are trying to find the rate of change in the angle of the camera with respect to time when the rocket is 1000 ft off the ground. That is, we need to find $\frac{d\theta }{dt}$ when $h=1000\phantom{\rule{0.2em}{0ex}}\text{ft}.$ At that time, we know the velocity of the rocket is $\frac{dh}{dt}=600\phantom{\rule{0.2em}{0ex}}\text{ft/sec}.$
Step 3. Now we need to find an equation relating the two quantities that are changing with respect to time: $h$ and $\theta .$ How can we create such an equation? Using the fact that we have drawn a right triangle, it is natural to think about trigonometric functions. Recall that $\text{tan}\phantom{\rule{0.1em}{0ex}}\theta$ is the ratio of the length of the opposite side of the triangle to the length of the adjacent side. Thus, we have
$\text{tan}\phantom{\rule{0.1em}{0ex}}\theta =\frac{h}{5000}.$
This gives us the equation
$h=5000\phantom{\rule{0.1em}{0ex}}\text{tan}\phantom{\rule{0.1em}{0ex}}\theta .$
Step 4. Differentiating this equation with respect to time $t,$ we obtain
$\frac{dh}{dt}=5000\phantom{\rule{0.1em}{0ex}}{\text{sec}}^{2}\theta \phantom{\rule{0.1em}{0ex}}\frac{d\theta }{dt}.$
Step 5. We want to find $\frac{d\theta }{dt}$ when $h=1000\phantom{\rule{0.2em}{0ex}}\text{ft}.$ At this time, we know that $\frac{dh}{dt}=600\phantom{\rule{0.2em}{0ex}}\text{ft/sec}.$ We need to determine ${\text{sec}}^{2}\theta .$ Recall that $\text{sec}\phantom{\rule{0.1em}{0ex}}\theta$ is the ratio of the length of the hypotenuse to the length of the adjacent side. We know the length of the adjacent side is $5000\phantom{\rule{0.2em}{0ex}}\text{ft}.$ To determine the length of the hypotenuse, we use the Pythagorean theorem, where the length of one leg is $5000\phantom{\rule{0.2em}{0ex}}\text{ft},$ the length of the other leg is $h=1000\phantom{\rule{0.2em}{0ex}}\text{ft},$ and the length of the hypotenuse is $c$ feet as shown in the following figure.
We see that
${1000}^{2}+{5000}^{2}={c}^{2}$
and we conclude that the hypotenuse is
$c=1000\sqrt{26}\phantom{\rule{0.2em}{0ex}}\text{ft}.$
Therefore, when $h=1000,$ we have
${\text{sec}}^{2}\theta ={\left(\frac{1000\sqrt{26}}{5000}\right)}^{2}=\frac{26}{25}.$
Recall from step 4 that the equation relating $\frac{d\theta }{dt}$ to our known values is
$\frac{dh}{dt}=5000\phantom{\rule{0.1em}{0ex}}{\text{sec}}^{2}\theta \phantom{\rule{0.1em}{0ex}}\frac{d\theta }{dt}.$
When $h=1000\phantom{\rule{0.2em}{0ex}}\text{ft},$ we know that $\frac{dh}{dt}=600\phantom{\rule{0.2em}{0ex}}\text{ft/sec}$ and ${\text{sec}}^{2}\theta =\frac{26}{25}.$ Substituting these values into the previous equation, we arrive at the equation
$600=5000\left(\frac{26}{25}\right)\frac{d\theta }{dt}\text{.}$
Therefore, $\frac{d\theta }{dt}=\frac{3}{26}\phantom{\rule{0.2em}{0ex}}\text{rad/sec}.$
What rate of change is necessary for the elevation angle of the camera if the camera is placed on the ground at a distance of $4000\phantom{\rule{0.2em}{0ex}}\text{ft}$ from the launch pad and the velocity of the rocket is 500 ft/sec when the rocket is $2000\phantom{\rule{0.2em}{0ex}}\text{ft}$ off the ground?
$\frac{1}{10}\phantom{\rule{0.2em}{0ex}}\text{rad/sec}$
In the next example, we consider water draining from a cone-shaped funnel. We compare the rate at which the level of water in the cone is decreasing with the rate at which the volume of water is decreasing.
What is derivative of antilog x dx ?
what's the meaning of removable discontinuity
what's continuous
Brian
an area under a curve is continuous because you are looking at an area that covers a range of numbers, it is over an interval, such as 0 to 4
Lauren
using product rule x^3,x^5
Kabiru
may god be with you
Sunny
Luke 17:21 nor will they say, See here or See there For indeed, the kingdom of God is within you. You've never 'touched' anything. The e-energy field created by your body has pushed other electricfields. even our religions tell us we're the gods. We live in energies connecting us all. Doa/higgsfield
Scott
if you have any calculus questions many of us would be happy to help and you can always learn or even invent your own theories and proofs. math is the laws of logic and reality. its rules are permanent and absolute. you can absolutely learn calculus and through it better understand our existence.
Scott
ya doubtless
Bilal
help the integral of x^2/lnxdx
Levis
also find the value of "X" from the equation that follow (x-1/x)^4 +4(x^2-1/x^2) -6=0 please guy help
Levis
Use integration by parts. Let u=lnx and dv=x2dx Then du=1xdx and v=13x3. ∫x2lnxdx=13x3lnx−∫(13x3⋅1x)dx ∫x2lnxdx=13x3lnx−∫13x2dx ∫x2lnxdx=13x3lnx−19x3+C
Bilal
itz 1/3 and 1/9
Bilal
now you can find the value of X from the above equation easily
Bilal
Pls i need more explanation on this calculus
usman
usman from where do you need help?
Levis
thanks Bilal
Levis
Do we ask only math question? or ANY of the question?
yh
Gbesemete
How do i differentiate between substitution method, partial fraction and algebraic function in integration?
usman
you just have to recognize the problem. there can be multiple ways to solve 1 problem. that's the hardest part about integration
Lauren
test
we asking the question cause only the question will tell us the right answer
Sunny
find integral of sin8xcos12xdx
don't share these childish questions
Bilal
well find the integral of x^x
Levis
bilal kumhar you are so biased if you are an expert what are you doing here lol😎😎😂😂 we are here to learn and beside there are many questions on this chat which you didn't attempt we are helping each other stop being naive and arrogance so give me the integral of x^x
Levis
Levis I am sorry
Bilal
Bilal it okay buddy honestly i am pleasured to meet you
Levis
x^x ... no anti derivative for this function... but we can find definte integral numerically.
Bilal
thank you Bilal Kumhar then how we may find definite integral let say x^x,3,5?
Levis
evaluate 5-×square divided by x+2 find x as limit approaches infinity
i have not understood
Leo
Michael
welcome
Sunny
I just dont get it at all...not understanding
Michagaye
0 baby
Sunny
The denominator is the aggressive one
Sunny
wouldn't be any prime number for x instead ?
Harold
or should I say any prime number greater then 11 ?
Harold
just wondering
Harold
I think as limit Approach infinity then X=0
Levis
ha hakdog hahhahahaha
ha hamburger
Leonito
Fond the value of the six trigonometric function of an angle theta, which terminal side passes through the points(2x½-y)²,4
What's f(x) ^x^x
What's F(x) =x^x^x
Emeka
are you asking for the derivative
Leo
that's means more power for all points
rd
Levis
iam sorry f(x)=x^x it means the output(range ) depends to input(domain) value of x by the power of x that is to say if x=2 then x^x would be 2^2=4 f(x) is the product of X to the power of X its derivatives is found by using product rule y=x^x introduce ln each side we have lny=lnx^x =lny=xlnx
Levis
the derivatives of f(x)=x^x IS (1+lnx)*x^x
Levis
what is a maximax
A maxima in a curve refers to the maximum point said curve. The maxima is a point where the gradient of the curve is equal to 0 (dy/dx = 0) and its second derivative value is a negative (d²y/dx² = -ve).
Viewer
what is the limit of x^2+x when x approaches 0
it is 0 because 0 squared Is 0
Leo
0+0=0
Leo
simply put the value of 0 in places of x.....
Tonu
the limit is 2x + 1
Nicholas
the limit is 0
Muzamil
limit s x
Bilal
The limit is 3
Levis
Leo we don't just do like that buddy!!! use first principle y+∆y=x+∆x ∆y=x+∆x-y ∆y=(x+∆x)^2+(x+∆x)-x^2+x on solving it become ∆y=3∆x+∆x^2 as ∆x_>0 limit=3 if you do by calculator say plugging any value of x=0.000005 which approach 0 you get 3
Levis
find derivatives 3√x²+√3x²
3 + 3=6
mujahid
How to do basic integrals
the formula is simple x^n+1/n+1 where n IS NOT EQUAL TO 1 And n stands for power eg integral of x^2 x^2+1/2+1 =X^3/3
Levis
write something lmit
find the integral of tan tanxdx
-ln|cosx| + C
Jug
lnSecx+c
Levis |
# Geometry 8 5 The Tangent Ratio Trigonometry The
• Slides: 18
Geometry 8. 5 The Tangent Ratio
Trigonometry The word trigonometry comes from the Greek words that mean “triangle measurement. ” In this course we will introduce the trigonometry of right triangles.
Tangent ratio = The tangent ratio is the ratio of the length of the legs in a Rt. ∆ Tangent of <A: opposite leg B Tangent of <B: adjacent leg hy pot enu se adjacent leg hy pot en use opposite leg A
I. Express tan X and tan Y as ratios. 1. 2. Z 9 X Y 3. 40 41 Y 21 4. Z X 29 4 Z 20 X Z 2 Y 8 X 8 Y
Exercises: tan = 1. Express tan X and tan Y as ratios. Z 40 9 X 41 Y
Exercises: tan = 3. Express tan X and tan Y as ratios. 2√ 3 X 4 Z 2 Y
Table of Trigonometric Ratios • Turn to page 311 in your textbook. • This table gives approximate decimal values of the tangent ratio for some angles. • The table also can be used to find an angle measure when given a tangent value.
5. tan 15 o = ______ 6. tan 43 o = ______ 7. tan 82 o = ______ 8. tan ____ =. 4663 9. tan _____ = 1. 5399 10. tan _____ = 2. 7475
Exercises. 2679 5. Complete: tan 15˚ ≈ _____ Step 1. Locate 15˚ in the angle column. Step 2. Go across to the tangent column. Step 3. Read the value: . 2679 25˚ ≈. 4663 8. Complete: tan _____ Step 1. Locate. 4663 in the tangent column. Step 2. Go across to the angle column. Step 3. Read the value: 25˚
Answers to Exercises 6. . 9325 7. 7. 1154 9. 57˚ 10. 70˚
II. Find the value of x to the nearest tenth. 12. 11. 13. 17 25 x 14. x x 12 51 37 7. 4 x 72 3 15. 16. 17 17. A freeway ramp has a 10% grade. What angle does the ramp have with the ground? [Note: Grade(%) = y x 8 200 x 60 z
Exercises 11. Find the value of x to the nearest tenth. x 25 37˚
Exercises 13. Find the value of x to the nearest tenth. 17 x 7. 4
Exercises 15. Find the value of x and y to the nearest tenth. 17 y x 8
Exercises 17. A freeway ramp has a 10% grade. What angle does the ramp have with the ground? ramp ground 1 10
Answers to Exercises 12. 9. 2 14. 9. 7 16. x ≈ 115. 5, z ≈ 230. 9
Homework pg. 308 #1 -24 Add hypotenuse, the three proportions, Pythag and Converse, multiply fractions vs. proportions, 45 -45 -90 and 30 -60 -90 formulas, pythagorean triples to vocab list, tangent |
Question Video: Finding the Dot Product between Vectors | Nagwa Question Video: Finding the Dot Product between Vectors | Nagwa
# Question Video: Finding the Dot Product between Vectors Mathematics • Third Year of Secondary School
## Join Nagwa Classes
The angle between π and π is 22Β°. If |π| = 3|π| = 25.2, find π β
π to the nearest hundredth.
02:11
### Video Transcript
The angle between vector π and vector π is 22 degrees. If the magnitude of vector π is equal to three times the magnitude of vector π is equal to 25.2, find the dot product between π and π to the nearest hundredth.
In this question, weβre given some information about two vectors π and π. First, weβre told the angle between these two vectors is equal to 22 degrees. Next, weβre also told information about their magnitudes. We know the magnitude of π is equal to 25.2, and we know that three times the magnitude of π is also equal to 25.2. So the magnitude of π is three times bigger than the magnitude of π. We need to use this to find the dot product of π and π. And we need to give our answer to the nearest hundredth.
To answer this question, we need to notice that we know a formula which connects the angle between two vectors with their dot product. We recall if π is the angle between two vectors π and π, then we know that the cos of π must be equal to the dot product between π and π divided by the magnitude of π times the magnitude of π. And in this question, we already know some of these values. For example, weβre told the angle between our two vectors is 22 degrees. Next, weβre also told the magnitude of π is equal to 25.2.
And we could also find the magnitude of π using the information given to us in the question. One way of doing this is to notice that three times the magnitude of π is equal to 25.2. We can then solve this to find the magnitude of π by dividing both sides of our equation through by three. And calculating this, we get the magnitude of π is 8.4. So, in fact, the only unknown in this equation is the dot product between π and π. And thatβs exactly what weβre asked to calculate.
So weβll substitute the angle of π equal to 22 degrees, the magnitude of π equal to 25.2, and the magnitude of π equal to 8.4 into our equation. This gives us the cos of 22 degrees should be equal to the dot product between π and π divided by 25.2 times 8.4. And now we can just rearrange this equation for the dot product between π and π. We multiply through by 25.2 times 8.4. This gives us π dot π is 25.2 times 8.4 times the cos of 22 degrees. And we can calculate this to the nearest hundredth or to two decimal places. Itβs equal to 196.27.
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# Math Mammoth Fractions 2 - 5th grade fractions workbook
## Math Mammoth
\$5.75
• Subjects: Math
• Product Type: Printable, Workbook
## Product Description
Math Mammoth Fractions 2 is a workbook (actually worktext) for 5th grade, covering the following topics: simplifying; multiplication of fractions (and of mixed numbers); division of fractions (and of mixed numbers); converting fractions to decimals.
Math Mammoth Fractions 2 continues the study of fraction topics after Math Mammoth Fractions 1. I sincerely recommend that the student study the Fractions 1 book prior to studying this book, if he has not already done so.
Sample pages (PDF) Contents and Introduction Multiplying Fractions by Whole Numbers 1 Multiplying Fractions by Fractions Multiplication and Area Dividing Fractions 2: Fitting the Divisor
This book is meant for fifth or sixth grade, and deals in-depth with the following topics:
• simplifying; including simplifying before multiplying
• multiplication of fractions (and of mixed numbers);
• division of fractions (and of mixed numbers);
• converting fractions to decimals.
We start out by simplifying fractions. Since this process is the opposite of making equivalent fractions, studied in Math Mammoth Fractions 1, it should be relatively simple for students to understand. We also use the same visual model, just backwards: This time the pie pieces are joined together instead of split apart.
Next comes multiplying a fraction by a whole number. Since this can be solved by repeated addition, it is not a difficult concept at all.
Multiplying a fraction by a fraction is first explained as taking a certain part of a fraction, in order to teach the concept. After that, students are shown the usual shortcut for the multiplication of fractions.
Then, we find the area of a rectangle with fractional side lengths, and show that the area is the same as it would be found by multiplying the side lengths. Students multiply fractional side lengths to find areas of rectangles, and represent fraction products as rectangular areas.
Simplifying before multiplying is a process that is not absolutely necessary for fifth graders. I have included it here because it prepares students for the same process in future algebra studies and because it makes fraction multiplication easier. I have also tried to include explanations of why we are allowed to simplify before multiplying. These explanations are actually proofs. I feel it is a great advantage for students to get used to mathematical reasoning and proof methods well before they start high school geometry.
Students also multiply mixed numbers, and study how multiplication can be seen as resizing or scaling. This means, for example, that the multiplication (2/3) × 18 km can be thought of as finding two-thirds of 18 km.
Next, we study division of fractions in special cases. The first one is seeing fractions as divisions; in other words recognizing that 5/3 is the same as 5 ÷ 3. This of course gives us a means of dividing whole numbers and getting fractional answers (for example, 20 ÷ 6 = 3 2/6).
Then students encounter sharing divisions with fractions. For example, if two people share equally 4/5 of a pizza, how much will each person get? This is represented by the division (4/5) ÷ 2 = 2/5. Another case we study is dividing unit fractions by whole numbers (such as (1/2) ÷ 4). We also divide whole numbers by unit fractions, such as 6 ÷ (1/3). Students will solve these thinking how many times the divisor "fits into" the dividend.
After these types of divisions, students learn the “shortcut” for fraction division, that is, the usual rule for dividing any fraction by any fraction (the rule of “invert and multiply”). We also study dividing mixed numbers.
The lesson on introduction to ratios is optional. Ratios will be studied a lot in 6th and 7th grades, especially in connection with proportions. We are laying the groundwork for that.
The last major topic is converting fractions to decimals. Problems accompanied by a small picture of a calculator are meant to be solved with the help of a calculator. Otherwise, a calculator should not be allowed.
The book contains 119 pages, which includes the answers.
This book is enabled for annotation, which means you can fill it in on a computer (with Adobe Reader 9 or higher) or on a tablet using a PDF app with annotation tools.
## Reviews
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Tutoring Precalculus Polar Coordinates Intro 9 Examples.
Why do we need them?
What are polar coordinates?
When you think of polar coordinates, something about the North Pole, polar bears and the arctic come to mind. They are not the coordinates dedicated to finding polar bears! However the idea for polar coordinates can be thought of as the coordinates “surrounding” the pole or the origin.
What do we need them for?
Some of the real-life uses of polar coordinates include avoiding collisions between vessels and other ships or natural obstructions, guiding industrial robots in various production applications and calculating groundwater flow in radially symmetric wells. Polar coordinates can also be used to determine the best audio pickup patterns for cardioid microphones. Calculations involving aircraft navigation, gravitational fields and radio antennae are additional applications in which polar coordinates are used.
The first number u is taken to be a distance from the origin and the second number v is taken to be an angle (usually in radians). To be explicit about this, we will denote the pair as (r, θ) instead of (u, v). The numbers r and θ can be positive, negative or zero.
The pole (or origin)
and is labeled O.
Plotting Polar Coordinates.
EXAMPLE1
a) (-2, π/2) b) (3, –) c) (-1, )
a) START:
Find the location of the angle: if the angle is positive use counterclockwise orientation, if the angle is negative use clockwise.
π/2 is 90 degrees. 90 degrees is the top half of the 180 x axis. To find r=-2 we travel the opposite direction from the origin and end up in the lower half of the coordinate system.
b) START:
Find the location of the angle: the angle is negative which mean we need to use clockwise orientation.
Lets divide pi=180=line into six equal parts and locate –pi/6= -30 degrees in the clockwise fashion(going underneath the x axis.
If the radius r was negative, r =-3 we would travel in the opposite direction after we locate the angle. But since the radius r=3 we do not travel in the opposite direction
.
c) START:
Let’s start with the angle, is it positive or negative? It is negative , which means we go underneath the x axis two divisions. pi is divided by 3 times in this case. R=-1 is negative which means we travel in the opposite direction to find the point in Quadrant I.
Exceptions:
• There is a possibility of a two or more distinct points (polar coordinates) to be on the same spot.
(r1,θ) , (r2, θ) If r1=r2.
Example:
(2, pi/6) is on the same spot as (2, 13pi/6).
The reason this is true is that adding a full circle rotation to the angle will result in the same spot.
Try to put your finger on the number 2 and circulate your finger until it reaches 2 again (this is a full circle). Now add 30 degrees to that and you will end up on the same spot.
2) For negative angles add a pi=180 degree, if r is the opposite sign.
• Coordinate (0, θ) all map to zero.
Converting from Rectangular Coordinates to polar Coordinates.
Dropping the perpendicular line to an x axis in any quadrant will always create a 90 degree angle. Using the Pythagorean theorem r is the hypotenuse, Y is the side opposite the , X is the adjacent side to the .
We use the convention that an angle is:
• Positive—if measured in the counterclockwise
direction from the polar axis.
• Negative—if measured inthe clockwise
direction from the polar axis.
• We agree that, as shown, the points (–r,θ)
and (r,θ) lie on the same line through O
and at the samedistance | r | fromO,but
on oppositesides of O.
EXAMPLE 1:
Convert the given rectangular coordinates into polar coordinates.
1. (0, 10) b. (6, -8) c.(-4, 2)
Things to remember: Theta -the angle is always from the x axis.
Use the right triangle formulas, Pythagorean Theorem and trig function tan(theta). Since the x and y will be known always.
Locate the Point first on the coordinate axis. Plot the point in question.
Polar coordinates are expressed as a radius and an angle if you know both that means you solved the question.
a.
.
b.
Here the angle can be expressed as a positive angle by subtracting it from 360 degrees.
1. Tan (8/6) = theta.Use the tangent formula to find the angle because opposite side 8 is known and adjacent side 6 is known tangent (theta) = opposite /adjacent .
The answer theta is positive but notice that the angle is underneath the x axis which means it is negative in respect to the x axis.
c.
Converting from Polar Coordinates to Rectangular Coordinates.
To convert from polar to rectangular simply use the formulas for x and y.
1. (20, 150)
Using a calculator: This is the quickest and simplest method of getting the answer, although it will not yield the exact answer in cases with square roots. For this example especially since it contains square root of 3.
Using parenthesis around the values is essential. MODE must be set to radians when using pi, degrees if using numbers. In this example we are using pi and the MODE is set to radians. To set the Mode on your calculator:
MODE make sure the word RADIAN is highlighted black. Use the up and down arrow to reach the third line. If you want to select RADIAN move curser to the word RADIAN then hit enter, if you want to select the word DEGREE move move over to the word degree and hit enter. 2ndQUIT to escape.
If you are using 180 instead of pi which you can do, make sure you set your calculator to degree mode first. Notice how the answer in the second picture produces the right result only after we change the radian mode to degree mode.
Online Ti-84 Plus can be found here:
https://www.cemetech.net/projects/jstified/
Using mental Math:
An exact answer can be found for this question, since the angle is a multiple of 30 degrees.
(See the chart below).
Sine function is negative in the quadrants where the y values are negative (Quadrants III, Quadrant IV). Cosine function is positive in Quadrants I,IV
Cosine of 150 is negative because the angle 150 is in the second quadrant where the cosine is negative. Using the value of 30 as a reference angle (see lesson on reference angles), the value of 30 is the same as the value of 150 except the sign is negative.
b. (75, -π)
Ref:s
https://www.reference.com/geography/examples-real-life-uses-polar-coordinates-42f7df9b107292ae
http://noodle.med.yale.edu/hdtag/notes/coord.pdf |
## Engage NY Eureka Math Precalculus Module 1 Lesson 11 Answer Key
### Eureka Math Precalculus Module 1 Lesson 11 Exercise Answer Key
Opening Exercise
a. Plot the complex number z=2+3i on the complex plane. Plot the ordered pair (2,3) on the coordinate plane.
Answer:
b. In what way are complex numbers points?
Answer:
When a complex number is plotted on a complex plane, it looks just like the corresponding ordered pair plotted on a coordinate plane. For example, when 2+3i is plotted on the complex plane, it looks exactly the same as when the ordered pair (2,3) is plotted on a coordinate plane. We can interchangeably think of a complex number x+yi in the complex plane as a point (x,y) in the coordinate plane, and vice versa.
c. What point on the coordinate plane corresponds to the complex number -1+8i?
Answer:
(-1,8)
d. What complex number corresponds to the point located at coordinate (0,-9)?
Answer:
0-9i or -9i
Exercise 1.
The endpoints of $$\overline{\boldsymbol{AB}}$$ are A(1,8) and B(-5,3). What is the midpoint of $$\overline{\boldsymbol{AB}}$$?
Answer:
The midpoint of $$\overline{\boldsymbol{AB}}$$ is (-2,$$\frac{11}{2}$$).
→ How do you find the midpoint of $$\overline{\boldsymbol{AB}}$$?
→ You find the average of the x-coordinates and the y-coordinates to find the halfway point.
→ In Geometry, we learned that for two points A(x1,y1) and B(x2, y2), the midpoint of $$\overline{\boldsymbol{AB}}$$ is ($$\frac{x_{1}+x_{2}}{2}$$, $$\frac{y_{1}+y_{2}}{2}$$).
→ Now, view these points as complex numbers: A=x1+y1i and B=x2 + y2i.
Exercise 2.
What is the midpoint of A=1+8i and B=-5+3i?
Answer:
M=$$\frac{1+-5}{2}$$ + $$\frac{(8+3)}{2}$$i = -2 + $$\frac{11}{2}$$i
b. Using A=x1+y1i and B=x2+y2i, show that, in general, the midpoint of points A and B is $$\frac{\boldsymbol{A}+\boldsymbol{B}}{2}$$, the arithmetic average of the two numbers.
Answer:
M=$$\frac{x_{1}+x_{2}}{2}$$+$$\frac{y_{1}+y_{2}}{2}$$i
= $$\frac{x_{1}+y_{1} i+x_{2}+y_{2} i}{2}$$
= $$\frac{\boldsymbol{A}+\boldsymbol{B}}{2}$$
Exercise 3.
The endpoints of $$\overline{\boldsymbol{AB}}$$ are A(1,8) and B(-5,3). What is the length of $$\overline{\boldsymbol{AB}}$$?
Answer:
The length of $$\overline{\boldsymbol{AB}}$$ is √61.
→ How do you find the length of $$\overline{\boldsymbol{AB}}$$?
→ You use the Pythagorean theorem, which can be written as AB=$$\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$$ for two points A(x1, y1) and B(x2, y2).
→ As we did previously, view these points as complex numbers: A=x1+y1i and B=x2+y2i.
Exercise 4.
a. What is the distance between A=1+8i and B=-5+3i?
Answer:
d=$$\sqrt{(1-(-5))^{2}+(8-3)^{2}}$$=$$\sqrt{61}$$
b. Show that, in general, the distance between A=x1+y1i and B=x2+y2i is the modulus of A-B.
Answer:
A-B=(x1 – x2)+(y1 – y2)i
|A-B|=$$\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$$= distance between A and B
Exercise 5.
Suppose z=2+7i and w=-3+i.
a. Find the midpoint m of z and w.
Answer:
m=-$$\frac{1}{2}$$+4i
b. Verify that |z-m|=|w-m|.
Answer:
### Eureka Math Precalculus Module 1 Lesson 11 Problem Set Answer Key
Question 1.
Find the midpoint between the two given points in the rectangular coordinate plane.
a. 2+4i and 4+8i
Answer:
M=$$\frac{2+4}{2}$$+$$\frac{4+8}{2}$$i=3+6i
b. -3+7i and 5-i
Answer:
M=$$\frac{-3+5}{2}$$+$$\frac{7-1}{2}$$i=1+3i
c. -4+3i and 9-4i
Answer:
M=$$\frac{-4+9}{2}$$+$$\frac{3-4}{2}$$i=$$\frac{5}{2}$$–$$\frac{1}{2}$$ i
d. 4+i and -12-7i
Answer:
M=$$\frac{4-12}{2}$$+$$\frac{1-7}{2}$$i=-4-3i
e. -8-3i and 3-4i
Answer:
M=$$\frac{-8+3}{2}$$+$$\frac{-3-4}{2}$$i=-$$\frac{5}{2}$$ – $$\frac{7}{2}$$i
f. $$\frac{2}{3}$$–$$\frac{5}{2}$$i and -0.2+0.4i
Answer:
Question 2.
Let A=2+4i, B=14+8i, and suppose that C is the midpoint of A and B and that D is the midpoint of A and C.
a. Find points C and D.
Answer:
b. Find the distance between A and B.
Answer:
|A-B|=|2+4i-14-8i|
=|-12-4i|
=$$\sqrt{(-12)^{2}+(-4)^{2}}$$
=$$\sqrt{144+16}$$
=$$\sqrt{160}$$
=$$4 \sqrt{10}$$
c. Find the distance between A and C.
Answer:
|A-C|=|2+4i-8-6i|
=|-6-2i|
=$$\sqrt{(-6)^{2}+(-2)^{2}}$$
=$$\sqrt{40}$$
=$$2 \sqrt{10}$$
d. Find the distance between C and D.
Answer:
|C-D|=|8+6i-5-5i|
=|3+i|
=$$\sqrt{(3)^{2}+(1)^{2}}$$
=$$\sqrt{10}$$
e. Find the distance between D and B.
Answer:
|D-B|=|5+5i-14-8i|
=|-9-3i|
=$$\sqrt{(-9)^{2}+(-3)^{2}}$$
=$$\sqrt{90}$$
=3$$\sqrt{10}$$
f. Find a point one-quarter of the way along the line segment connecting segment A and B, closer to A than
to B.
Answer:
The point is D=5+5i.
g. Terrence thinks the distance from B to C is the same as the distance from A to B. Is he correct? Explain why or why not.
Answer:
The distance from B to C is 2$$\sqrt{10}$$, and the distance from A to B is 4$$\sqrt{10}$$. The distances are not the same.
h. Using your answer from part (g), if E is the midpoint of C and B, can you find the distance from E to C? Explain.
Answer:
The distance from B to C is 2$$\sqrt{10}$$, and the distance from E to C should be half of this value, $$\sqrt{10}$$.
i. Without doing any more work, can you find point E? Explain.
Answer:
B is 5+5i, which is 3 units to the right of A in the real direction and 1 unit up in the imaginary direction. From C, you should move the same amount to get to E, so E would be 11+7i.
### Eureka Math Precalculus Module 1 Lesson 11 Exit Ticket Answer Key
Question 1.
Kishore said that he can add two points in the coordinate plane like adding complex numbers in the complex plane. For example, for point A(2,3) and point B(5,1), he will get A+B=(7,4). Is he correct? Explain your reasoning.
Answer:
No. Kishore is not correct because we cannot add two points in the rectangular plane. However, we can add two complex numbers in the complex plane, which has the geometric effect of performing a translation to points in complex numbers.
Question 2.
Consider two complex numbers A=-4+5i and B=4-10i.
a. Find the midpoint of A and B.
Answer:
M=$$\frac{A+B}{2}$$
=$$\frac{-4+5 i+4-10 i}{2}$$
=$$\frac{-5 i}{2}$$
=$$-\frac{5}{2} i$$ or 0-$$\frac{5}{2}$$i
b. Find the distance between A and B.
Answer:
d=$$\sqrt{(4-(-4))^{2}+(-10-5)^{2}}$$=17 |
# Cardinal and Ordinal Numbers: Do these confuse your child?
Counting of numbers (also known as, enumeration) is considered by many parents to be the beginning of mathematics education. The concept of counting (and numbers) is usually introduced to children when they are in the pre-primary and kindergarten age group. It is important for parents of young children to understand that in mathematics, numbers are classified into two categories: a. Cardinal Numbers and b. Ordinal Numbers. In this blog, we explain the difference between the two types. We have avoided using technical language as much as possible.
##### Cardinal Numbers
These are numbers that represent quantity. When we say there are five apples in the basket, the number 5 (five) represents the quantity of apples in the basket. If we ask a child to count the number of balls that he has, we are asking the child to represent the quantity of balls that he has, using a number.
If you wish to know more about cardinal numbers, please read this blog.
##### Ordinal Numbers
Ordinal numbers represent position. They do NOT represent quantity. In an elevator, when we press 3 to get to the third floor of an apartment, the number 3 represents the position of the desired floor with reference to the other floors. When a child tells you that she came second in a quiz competition, she again is telling you about her position in the competition with reference to other children, using the number 2.
If you wish to know more about ordinal numbers, please read this blog.
##### Cardinal vs Ordinal: What should I teach my child first?
Parents need to be careful about not confusing the child by switching frequently between the two systems. Especially if you are teaching numbers to young children who are in the pre-primary or kindergarten age group. Adults are so used to using these two systems of numbers that they often fail to realise that differences between cardinal and ordinal numbers could be confusing from the child’s perspective.
In general, it is easier for a young child to understand the concept of quantity rather than position. Further, operations such as addition or subtraction or multiplication are relevant to the cardinal system and not the ordinal system. We advise parents to try and use the cardinal system (quantity) as the basis for teaching numbers to young children, at least till the age of 4. Please note that each child is unique and some kids can grasp and be comfortable with the differences from a very early age.
###### Summary:
• Counting (enumeration) is an important aspect of learning mathematics.
• Cardinal and ordinal numbers are two distinct categories. Cardinal numbers represent quantity while ordinal numbers represent position.
• Start with cardinal numbers, when the child is in the 2-4 years of age group (pre-primary and lower kindergarten).
• Move on to ordinal numbers after 4 years (upper kindergarten and above), depending on the child. |
Quadratic transformations are a fundamental part of algebra, and they are essential to understanding how to solve quadratic equations. By understanding how to transform a quadratic equation into a different form, you can then solve the equation and find the unknown variables. The answers to quadratic transformation worksheets are often not as straightforward as they seem, and it is important to understand how to work through each transformation step-by-step.
General Transformations
The most common types of quadratic transformations are horizontal shifts, vertical shifts, and stretching or shrinking the graph. For example, if the equation is y=x^2+2, the horizontal shift would be y= (x-2)^2+2. This shifts the graph 2 units to the left. The vertical shift would be y=(x-2)^2+4, which shifts the graph 2 units up. The stretching or shrinking of the graph can be done by multiplying the x-term by a constant. For example, y=(2x-2)^2+4 stretches the graph by a factor of 2.
Inverse Transformations
Inverse transformations are the opposite of general transformations, and they are used to transform the graph back to its original form. To do this, you need to start with the original equation and work backward. For example, if the equation is y=(2x-2)^2+4, the inverse transformation would be y=x^2+2. This brings the graph back to its original form. The inverse transformations for horizontal and vertical shifts are the opposite of the general transformations. For example, if the equation is y=(x-2)^2+4, the inverse transformation would be y=x^2+2. This brings the graph back to its original form.
Transformations to Standard Form
To solve a quadratic equation, it is often helpful to transform it into standard form, which is y=ax^2 + bx + c. To do this, the equation must first be rewritten in the form y=a(x-h)^2+k. Then, the constants a, h, and k can be used to solve for b and c. For example, if the equation is y=(x-2)^2+4, then a=1, h=2, and k=4. Plugging these values into the equation y=ax^2 + bx + c results in y=x^2 + 2x + 0. So, the equation has been transformed into standard form.
Finding Solutions
Once the equation is in standard form, it can then be solved using the quadratic formula. The quadratic formula is used to find the solutions of a quadratic equation and is given by x= [-b +/- sqrt(b^2-4ac)]/2a. In this equation, a, b, and c are the constants from the standard form equation. For example, if the equation is y=x^2+2x+0, then a=1, b=2, and c=0. Plugging these values into the quadratic formula results in x = [-2 +/- sqrt(4)]/2, which simplifies to x= -1 or 1. So, the solutions of this equation are x=-1 and x=1.
Quadratic transformations are an essential part of solving quadratic equations and finding unknown variables. General transformations involve shifting the graph horizontally and vertically, as well as stretching or shrinking it. Inverse transformations are used to transform the graph back to its original form. To solve a quadratic equation, it must first be rewritten in standard form. Then, the solutions can be found using the quadratic formula. With the right steps, you can find the answers to any quadratic transformation worksheet. |
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## Section7.2Geometric Series
###### Motivating Questions
• What is a geometric series?
• What is a partial sum of a geometric series? What is a simplified form of the $n$th partial sum of a geometric series?
• Under what conditions does a geometric series converge? What is the sum of a convergent geometric series?
Many important sequences are generated by addition. In Example7.12, we see an example of a sequence that is connected to a sum.
###### Example7.12
Warfarin is an anticoagulant that prevents blood clotting; often it is prescribed to stroke victims in order to help ensure blood flow. The level of warfarin has to reach a certain concentration in the blood in order to be effective.
Suppose warfarin is taken by a particular patient in a 5 mg dose each day. The drug is absorbed by the body and some is excreted from the system between doses. Assume that at the end of a 24 hour period, 8% of the drug remains in the body. Let $Q(n)$ be the amount (in mg) of warfarin in the body before the $(n+1)$st dose of the drug is administered.
1. Explain why $Q(1) = 5 \times 0.08$ mg.
2. Explain why $Q(2) = (5+Q(1)) \times 0.08$ mg. Then show that
\begin{equation*} Q(2) = (5 \times 0.08)\left(1+0.08\right) \text{mg}\text{.} \end{equation*}
3. Explain why $Q(3) = (5+Q(2)) \times 0.08$ mg. Then show that
\begin{equation*} Q(3) = (5 \times 0.08)\left(1+0.08+0.08^2\right) \text{mg}\text{.} \end{equation*}
4. Explain why $Q(4) = (5+Q(3)) \times 0.08$ mg. Then show that
\begin{equation*} Q(4) = (5 \times 0.08)\left(1+0.08+0.08^2+0.08^3\right) \text{mg}\text{.} \end{equation*}
5. There is a pattern that you should see emerging. Use this pattern to find a formula for $Q(n)\text{,}$ where $n$ is an arbitrary positive integer.
6. Complete Table7.13 with values of $Q(n)$ for the provided $n$-values (reporting $Q(n)$ to 10 decimal places). What appears to be happening to the sequence $Q(n)$ as $n$ increases?
Solution
1. After the first dose and just before the second dose, $8\%$ of the first 5 mg dose remains. Thus, $Q(1)=5 \rtimes 0.08$ mg.
2. When the second dose is adminstered, the amount of drug in the body is the new 5 mg in addition to the residual amount remaining from the first dose, or $5 + Q(1)$ mg. After 24 hours, the remaining amount is $8\%$ of this total amount. So
\begin{equation*} Q(2)=(5+Q(1))(0.08) \end{equation*}
Now plugging in $Q(1)$ from part a., we have
\begin{align*} Q(2) \amp =(5+Q(1))(0.08) \\ \amp = (5+5(0.08))(0.08) \\ \amp = 5(0.08)+5(0.08)^2 \\ \amp = 5(0.08)(1+0.08) \end{align*}
where in the last line, we factor out 5(0.08) from each term.
3. Similarly, just before the third dose, the amount in the body is $8/%$ of the total from the previous day, which is $Q(2)+5\text{.}$ We can use similar algebra to the previous part:
\begin{align*} Q(3)\amp=(5+Q(2))(0.08) \\ \amp=(5+5(0.08)(1+0.08))(0.08)\\ \amp=5(0.08) + 5(0.08)(0.08+0.08^2)\\ \amp=5(0.08)(1+0.08+0.08^2) \end{align*}
4. Use a similar process again to the previous part.
5. Using the pattern, we can see that in general,
\begin{equation*} Q(n)=5(0.08)(1+0.08+0.08^2+0.08^3+\cdots+0.08^{n-1}) \end{equation*}
6. It appears that the sequence $Q(n)$ is getting closer and closer to a specific number.
### SubsectionGeometric Series
In Example7.12 we encountered the sum
\begin{equation*} (5 \times 0.08)\left(1+0.08+0.08^2+0.08^3+ \cdots + 0.08^{n-1}\right) \end{equation*}
for the long-term level of Warfarin in the patient's system. This sum has the form
$$a+ar+ar^2+ \cdots + ar^{n-1}\label{eq-8-2-part-sum-geometric-1}\tag{7.1}$$
where $a=5 \times 0.08$ and $r=0.08\text{.}$ Such a sum is called a finite geometric series with ratio $r\text{.}$
###### Example7.15
Let $a$ and $r$ be real numbers (with $r \ne 1$) and let
\begin{equation*} S_n = a+ar+ar^2 + \cdots + ar^{n-1}\text{.} \end{equation*}
In this example we will find a shortcut formula for $S_n$ that does not involve a sum of $n$ terms.
1. Multiply $S_n$ by $r\text{.}$ What does the resulting sum look like?
2. Subtract $rS_n$ from $S_n$ and explain why
$$S_n - rS_n = a - ar^n\text{.}\label{eq-8-2-1-partial-geometric-sum}\tag{7.2}$$
3. Solve equation (7.2) for $S_n$ to find a simple formula for $S_n$ that does not involve adding $n$ terms.
Hint
1. Distribute the factor of $r\text{.}$
2. Look for common terms in the two expressions being subtracted.
3. Observe that you can remove a factor of $S_n$ from $S_n - rS_n\text{.}$
1. $rS_n = ar+ar^2+ar^3 + \cdots + ar^n \text{.}$
2. $S_n - rS_n = a - ar^n\text{.}$
3. $S_n = a\frac{1-r^n}{1-r}\text{.}$
Solution
1. Note that
\begin{equation*} rS_n = ar+ar^2+ar^3 + \cdots + ar^n\text{.} \end{equation*}
2. When we subtract $rS_n$ from $S_n$ the middle terms all cancel and we are left with
\begin{align*} S_n - rS_n \amp = \left(a+ar+ar^2 + \cdots + ar^{n-1}\right)\\ \amp\phantom{={}}{}- \left(ar+ar^2+ar^3 + \cdots + ar^n\right)\\ \amp =a + (ar-ar) + \left(ar^2-ar^2\right) + \cdots + \left(ar^{n-1}-ar^{n-1}\right) - ar^n\\ \amp = a - ar^n\text{.} \end{align*}
3. Factoring $S_n$ from left hand side and dividing gives us a formula for $S_n\text{:}$
\begin{align*} S_n(1-r) \amp = a - ar^n\\ S_n \amp = a\frac{1-r^n}{1-r}\text{.} \end{align*}
The sum of the terms of a sequence is called a series. We summarize the result of Example7.15 in the following way.
###### Finite Geometric Series
A finite geometric series $S_n$ is a sum of the form
$$S_n = a + ar + ar^2 + \cdots + ar^{n-1}\text{,}\label{eq-8-2-geometric-sum}\tag{7.3}$$
where $a$ and $r$ are real numbers such that $r \ne 1\text{.}$ The finite geometric series $S_n$ can be written more simply as
$$S_n = a+ar+ar^2+ \cdots + ar^{n-1} = \frac{a(1-r^n)}{1-r}\text{.}\label{eq-8-2-part-sum-geometric}\tag{7.4}$$
We now apply Equation (7.4) to the example involving Warfarin from Example7.12. Recall that
\begin{equation*} Q(n)=(5 \times 0.08)\left(1+0.08+0.08^2+0.08^3+ \cdots + 0.08^{n-1}\right) \text{mg}\text{,} \end{equation*}
so $Q(n)$ is a geometric series with $a=5 \times 0.08 = 0.4$ and $r = 0.08\text{.}$ Thus,
\begin{equation*} Q(n) = 0.4\left(\frac{1-0.08^n}{1-0.08}\right) = \frac{1}{2.3} \left(1-0.08^n\right)\text{.} \end{equation*}
Notice that as $n$ goes to infinity, the value of $0.08^n$ goes to 0. So,
\begin{equation*} \lim_{n \to \infty} Q(n) = \lim_{n \to \infty} \frac{1}{2.3} \left(1-0.08^n\right) = \frac{1}{2.3} \approx 0.435\text{.} \end{equation*}
Therefore, the long-term level of Warfarin in the blood under these conditions is $\frac{1}{2.3}\text{,}$ which is approximately 0.435 mg.
To determine the long-term effect of Warfarin, we considered a finite geometric series of $n$ terms, and then considered what happened as $n$ was allowed to grow without bound. In this sense, we were actually interested in an infinite geometric series (the result of letting $n$ go to infinity in the finite sum).
###### Definition7.16
An infinite geometric series is an infinite sum of the form
$$a + ar + ar^2 + \cdots = \sum_{n=0}^{\infty} ar^n\text{.}\label{eq-8-2-geometric-series}\tag{7.5}$$
The value of $r$ in the geometric series (7.5) is called the common ratio of the series because the ratio of the ($n+1$)st term, $ar^n\text{,}$ to the $n$th term, $ar^{n-1}\text{,}$ is always $r\text{:}$
\begin{equation*} \frac{ar^n}{ar^{n-1}} = r\text{.} \end{equation*}
Geometric series are common in mathematics and arise naturally in many different situations. As a familiar example, suppose we want to write the number with repeating decimal expansion
\begin{equation*} N=0.1212\overline{12} \end{equation*}
as a rational number. Observe that
\begin{align*} N \amp = 0.12 + 0.0012 + 0.000012 + \cdots\\ \amp = \left(\frac{12}{100}\right) + \left(\frac{12}{100}\right)\left(\frac{1}{100}\right) + \left(\frac{12}{100}\right)\left(\frac{1}{100}\right)^2 + \cdots\text{.} \end{align*}
This is an infinite geometric series with $a=\frac{12}{100}$ and $r = \frac{1}{100}\text{.}$
By using the formula for the value of a finite geometric sum, we can also develop a formula for the value of an infinite geometric series. We explore this idea in the following example.
###### Example7.17
Let $r \ne 1$ and $a$ be real numbers and let
\begin{equation*} S = a+ar+ar^2 + \cdots ar^{n-1} + \cdots \end{equation*}
be an infinite geometric series. For each positive integer $n\text{,}$ let
\begin{equation*} S_n = a+ar+ar^2 + \cdots + ar^{n-1}\text{.} \end{equation*}
Recall that
\begin{equation*} S_n = a\frac{1-r^n}{1-r}\text{.} \end{equation*}
1. What should we allow $n$ to approach in order to have $S_n$ approach $S\text{?}$
2. What is the value of $\lim_{n \to \infty} r^n$ for $|r| \gt 1\text{?}$ for $|r| \lt 1\text{?}$ Explain.
3. If $|r| \lt 1\text{,}$ use the formula for $S_n$ and your observations in (a) and (b) to explain why $S$ is finite and find a resulting formula for $S\text{.}$
Hint
1. Let $n$ increase without bound.
2. Think about what happens to powers of numbers that are less than or greater than 1.
3. Consider $\frac{1-r^n}{1-r}$ and how the numerator tends to 1 as $n \to \infty$ for certain values of $r\text{.}$
1. Observe that
\begin{equation*} S = \lim_{n \to \infty} S_n\text{.} \end{equation*}
2. If $r \gt 1\text{,}$ then $\lim_{n \to \infty} r^n = \infty\text{.}$ If $0 \lt r \lt 1\text{,}$ then $\lim_{n \to \infty} r^n = 0\text{.}$
3. Since
\begin{equation*} S = \lim_{n \to \infty} S_n = \lim_{n \to \infty} a\frac{1-r^n}{1-r} \end{equation*}
and
\begin{equation*} \lim_{n \to \infty} r^n = 0 \end{equation*}
for $0 \lt r \lt 1\text{,}$ we conclude that
\begin{equation*} S = \lim_{n \to \infty} a\frac{1-r^n}{1-r} = \frac{a}{1-r} \end{equation*}
when $0 \lt r \lt 1\text{.}$
Solution
1. Observe that
\begin{equation*} S = \lim_{n \to \infty} S_n\text{.} \end{equation*}
2. If $r \gt 1\text{,}$ then $\lim_{n \to \infty} r^n = \infty\text{.}$ If $0 \lt r \lt 1\text{,}$ then $\lim_{n \to \infty} r^n = 0\text{.}$
3. Since
\begin{equation*} S = \lim_{n \to \infty} S_n = \lim_{n \to \infty} a\frac{1-r^n}{1-r} \end{equation*}
and
\begin{equation*} \lim_{n \to \infty} r^n = 0 \end{equation*}
for $0 \lt r \lt 1\text{,}$ we conclude that
\begin{equation*} S = \lim_{n \to \infty} a\frac{1-r^n}{1-r} = \frac{a}{1-r} \end{equation*}
when $0 \lt r \lt 1\text{.}$
We can now find the value of the geometric series
\begin{equation*} N = \left(\frac{12}{100}\right) + \left(\frac{12}{100}\right)\left(\frac{1}{100}\right) + \left(\frac{12}{100}\right)\left(\frac{1}{100}\right)^2 + \cdots\text{.} \end{equation*}
Using $a = \frac{12}{100}$ and $r = \frac{1}{100}\text{,}$ we see that
\begin{equation*} N = \frac{12}{100} \left(\frac{1}{1-\frac{1}{100}}\right) = \frac{12}{100} \left(\frac{100}{99}\right) = \frac{4}{33}\text{.} \end{equation*}
The sum of a finite number of terms of an infinite geometric series is often called a partial sum of the series. Thus,
\begin{equation*} S_n = a+ar+ar^2 + \cdots + ar^{n-1} = \sum_{k=0}^{n-1} ar^k\text{.} \end{equation*}
is called the $n$th partial sum of the series $\sum_{k=0}^{\infty} ar^k\text{.}$ We summarize our recent work with geometric series as follows.
###### Infinite Geometric Series
• An infinite geometric series is an infinite sum of the form
$$a + ar + ar^2 + \cdots = \sum_{n=0}^{\infty} ar^n\text{,}\label{eq-8-2-geometric-series-2}\tag{7.6}$$
where $a$ and $r$ are real numbers such that $r \ne 0\text{.}$
• The $n$th partial sum $S_n$ of an infinite geometric series is
\begin{equation*} S_n = a+ar+ar^2+ \cdots + ar^{n-1}\text{.} \end{equation*}
• If $|r| \lt 1\text{,}$ then using the fact that $S_n = a\frac{1-r^n}{1-r}\text{,}$ it follows that the sum $S$ of the infinite geometric series (7.6) is
\begin{equation*} S = \lim_{n \to \infty} S_n = \lim_{n \to \infty} a\frac{1-r^n}{1-r} = \frac{a}{1-r}. \end{equation*}
In this case we say that the geometric series converges.
• If $|r|\geq1\text{,}$ the sequence $S_n$ does not converge, so we can't assign a numerical value to the expression $\sum_{n=0}^{\infty} ar^n\text{.}$ In this case we say that the geometric series does not converge.
###### Example7.18
The formulas we have derived for an infinite geometric series and its partial sum have assumed that we begin indexing the sums at $n=0\text{.}$ If instead we have a sum that does not begin at $n=0\text{,}$ we can factor out common terms and then use the established formulas. This process is illustrated in this example.
1. Consider the sum
\begin{equation*} \sum_{k=1}^{\infty} (2)\left(\frac{1}{3}\right)^k = (2)\left(\frac{1}{3}\right) + (2)\left(\frac{1}{3}\right)^2 + (2)\left(\frac{1}{3}\right)^3 + \cdots\text{.} \end{equation*}
Remove the common factor of $(2)\left(\frac{1}{3}\right)$ from each term and hence find the sum of the series.
2. Next let $a$ and $r$ be real numbers with $-1\lt r\lt 1\text{.}$ Consider the sum
\begin{equation*} \sum_{k=3}^{\infty} ar^k = ar^3+ar^4+ar^5 + \cdots\text{.} \end{equation*}
Remove the common factor of $ar^3$ from each term and find the sum of the series.
3. Finally, we consider the most general case. Let $a$ and $r$ be real numbers with $-1\lt r\lt 1\text{,}$ let $n$ be a positive integer, and consider the sum
\begin{equation*} \sum_{k=n}^{\infty} ar^k = ar^n+ar^{n+1}+ar^{n+2} + \cdots\text{.} \end{equation*}
Remove the common factor of $ar^n$ from each term to find the sum of the series.
Hint
1. Think about how $r = \frac{1}{3}\text{.}$
2. Note that $ar^3+ar^4+ar^5 + \cdots = ar^3(1 + r + r^2 + \cdots)\text{.}$
3. Compare your work in (b).
1. $(2)\left(\frac{1}{3}\right) \left[1 + \left(\frac{1}{3}\right) + \left(\frac{1}{3}\right)^2 + \cdots \right] \text{.}$
2. $ar^3+ar^4+ar^5 + \cdots = \frac{ar^3}{1-r}\text{.}$
3. $r^n+ar^{n+1}+ar^{n+2} + \cdots = \frac{ar^n}{1-r}\text{.}$
Solution
1. Factoring out $(2)\left(\frac{1}{3}\right)$ gives us
\begin{align*} (2)\left(\frac{1}{3}\right) \left[1 + \left(\frac{1}{3}\right) + \left(\frac{1}{3}\right)^2 + \cdots \right] \amp = (2)\left(\frac{1}{3}\right)\sum_{k=0}^{\infty} \left(\frac{1}{3}\right)^k\\ \amp = (2)\left(\frac{1}{3}\right) \left(\frac{1}{1-\frac{1}{3}}\right)\\ \amp = (2)\left(\frac{1}{3}\right) \left(\frac{3}{2}\right)\\ \amp = 4\text{.} \end{align*}
2. Factoring out $ar^3$ gives us
\begin{align*} ar^3+ar^4+ar^5 + \cdots \amp = ar^3\left(1+r+r^2+ \cdots \right)\\ \amp = ar^3\sum_{k=0}^{\infty} r^k\\ \amp = ar^3 \left(\frac{1}{1-r}\right)\\ \amp = \frac{ar^3}{1-r}\text{.} \end{align*}
3. Factoring out $ar^n$ gives us
\begin{align*} ar^n+ar^{n+1}+ar^{n+2} + \cdots \amp = ar^n\left(1+r+r^2+ \cdots \right)\\ \amp = ar^n\sum_{k=0}^{\infty} r^k\\ \amp = ar^n \left(\frac{1}{1-r}\right)\\ \amp = \frac{ar^n}{1-r}\text{.} \end{align*}
### SubsectionSummary
• An infinite geometric series is an infinite sum of the form
\begin{equation*} \sum_{k=0}^{\infty} ar^k \end{equation*}
where $a$ and $r$ are real numbers and $r \neq 0\text{.}$
• The $n$th partial sum of the geometric series $\sum_{k=0}^{\infty} ar^k$ is
\begin{equation*} S_n = \sum_{k=0}^{n-1} ar^k\text{.} \end{equation*}
A formula for the $n$th partial sum of a geometric series is
\begin{equation*} S_n = a \frac{1-r^n}{1-r}\text{.} \end{equation*}
If $|r| \lt 1\text{,}$ the infinite geometric series $\sum_{k=0}^{\infty} ar^k$ converges, meaning it has the finite sum $\frac{a}{1-r}\text{.}$ If $|r| \geq 1\text{,}$ the series does not converge.
### SubsectionExercises
We will often encounter geometric series that contain a variable such as $x\text{.}$ One such series is
\begin{equation*} 3x+ 6x^2+ 12x^3 +24x^4+... \end{equation*}
1. Find the initial term and common ratio for this geometric series, just as you would for a geometric series containing numbers.
2. Recall that a geometric series converges if and only if its common ratio $r$ satisfies $|r|\lt 1\text{.}$ If $x=1$ does this geometric series converge? What if $x=-0.1, 0.2, 0.5\text{?}$
3. Write down the entire interval of $x$ values for which this geometric series converges.
4. For $x$-values that allow the series to converge, what does the series converge to? (Your answer will be an expression in terms of $x\text{.}$)
There is an old question that is often used to introduce the power of geometric growth. Here is one version. Suppose you are hired for a one month job (30 days, working every day) and are given two options to be paid.
Option 1.
You can be paid $500 per day or Option 2. You can be paid 1 cent the first day, 2 cents the second day, 4 cents the third day, 8 cents the fourth day, and so on, doubling the amount you are paid each day. 1. How much will you be paid for the job in total under Option 1? 2. Complete Table7.19 to determine the pay you will receive under Option 2 for the first 10 days. 3. Find a formula for the amount paid on day $n\text{,}$ as well as for the total amount paid by day $n\text{.}$ Use this formula to determine which option (1 or 2) you should take. Suppose you drop a golf ball onto a hard surface from a height $h\text{.}$ The collision with the ground causes the ball to lose energy and so it will not bounce back to its original height. The ball will then fall again to the ground, bounce back up, and continue. Assume that at each bounce the ball rises back to a height $\frac{3}{4}$ of the height from which it dropped. Let $h_n$ be the height of the ball on the $n$th bounce, with $h_0 = h\text{.}$ In this exercise we will determine the distance traveled by the ball and the time it takes to travel that distance. 1. Determine a formula for $h_1$ in terms of $h\text{.}$ 2. Determine a formula for $h_2$ in terms of $h\text{.}$ 3. Determine a formula for $h_3$ in terms of $h\text{.}$ 4. Determine a formula for $h_n$ in terms of $h\text{.}$ 5. Write an infinite series that represents the total distance traveled by the ball. Then determine the sum of this series. 6. Next, let's determine the total amount of time the ball is in the air. 1. When the ball is dropped from a height $H\text{,}$ if we assume the only force acting on it is the acceleration due to gravity, then the height of the ball at time $t$ is given by \begin{equation*} H - \frac{1}{2}gt^2\text{.} \end{equation*} Use this formula to determine the time it takes for the ball to hit the ground after being dropped from height $H\text{.}$ 2. Use your work in the preceding item, along with that in (a)-(e) above to determine the total amount of time the ball is in the air. Suppose you play a game with a friend that involves rolling a standard six-sided die. Two players alternate rolling the die, and the play to first roll a six wins. Assume that you roll first. In this exercise we will determine the probability that you roll the first six and win the game. 1. Explain why the probability of rolling a six on any single roll (including your first turn) is $\frac{1}{6}\text{.}$ 2. If you don't roll a six on your first turn, then in order for you to roll the first six on your second turn, both you and your friend had to fail to roll a six on your first turns, and then you had to succeed in rolling a six on your second turn. Explain why the probability of this event is \begin{equation*} \left(\frac{5}{6}\right)\left(\frac{5}{6}\right)\left(\frac{1}{6}\right) = \left(\frac{5}{6}\right)^2\left(\frac{1}{6}\right)\text{.} \end{equation*} 3. Now suppose you fail to roll the first six on your second turn. Explain why the probability is \begin{equation*} \left(\frac{5}{6}\right)\left(\frac{5}{6}\right)\left(\frac{5}{6}\right)\left(\frac{5}{6}\right)\left(\frac{1}{6}\right) = \left(\frac{5}{6}\right)^4\left(\frac{1}{6}\right) \end{equation*} that you to roll the first six on your third turn. 4. The probability of you rolling the first six is the probability that you roll the first six on your first turn plus the probability that you roll the first six on your second turn plus the probability that your roll the first six on your third turn, and so on. Explain why this probability is \begin{equation*} \frac{1}{6} + \left(\frac{5}{6}\right)^2\left(\frac{1}{6}\right) + \left(\frac{5}{6}\right)^4\left(\frac{1}{6}\right) + \cdots\text{.} \end{equation*} Find the sum of this series and determine the probability that you roll the first six. The goal of a federal government stimulus package is to positively affect the economy. Economists and politicians quote numbers like $k$ million jobs and a net stimulus to the economy of $n$ billion of dollars. Where do they get these numbers? Let's consider one aspect of a stimulus package: tax cuts. Economists understand that tax cuts or rebates can result in long-term spending that is many times the amount of the rebate. For example, assume that for a typical person, 75% of her entire income is spent (that is, put back into the economy). Further, assume the government provides a tax cut or rebate that totals $P$ dollars for each person. 1. The tax cut of $P$ dollars is income for its recipient. How much of this tax cut will be spent? 2. In this simple model, we will say that the spent portion of the tax cut/rebate from part (a) then becomes income for another person who, in turn, spends 75% of this income. After this second round" of spent income, how many total dollars have been added to the economy as a result of the original tax cut/rebate? 3. This second round of spending becomes income for another group who spend 75% of this income, and so on. In economics this is called the multiplier effect. Explain why an original tax cut/rebate of $P$ dollars will result in multiplied spending of \begin{equation*} 0.75P(1+0.75+0.75^2+ \cdots )\text{.} \end{equation*} dollars. 4. Based on these assumptions, and assuming consumer spending remains consistent forever, how much stimulus will a 200 billion dollar tax cut/rebate to consumers add to the economy? Like stimulus packages, home mortgages and foreclosures also impact the economy. A problem for many borrowers is the adjustable rate mortgage, in which the interest rate can change (and usually increases) over the duration of the loan, causing the monthly payments to increase beyond the ability of the borrower to pay. Most financial analysts recommend fixed rate loans, ones for which the monthly payments remain constant throughout the term of the loan. In this exercise we will analyze fixed rate loans. When most people buy a large ticket item like car or a house, they have to take out a loan to make the purchase. The loan is paid back in monthly installments until the entire amount of the loan, plus interest, is paid. With a loan, we borrow money, say $P$ dollars (called the principal), and pay off the loan at an interest rate of $r$%. To pay back the loan we make regular monthly payments, some of which goes to pay off the principal and some of which is charged as interest. In most cases, the interest is computed based on the amount of principal that remains at the beginning of the month. We assume a fixed rate loan, that is one in which we make a constant monthly payment $M$ on our loan, beginning in the original month of the loan. Suppose you want to buy a house. You have a certain amount of money saved to make a down payment, and you will borrow the rest to pay for the house. Of course, for the privilege of loaning you the money, the bank will charge you interest on this loan, so the amount you pay back to the bank is more than the amount you borrow. In fact, the amount you ultimately pay depends on three things: the amount you borrow (called the principal), the interest rate, and the length of time you have to pay off the loan plus interest (called the duration of the loan). For this example, we assume that the interest rate is fixed at $r$%. To pay off the loan, each month you make a payment of the same amount (called installments). Suppose we borrow $P$ dollars (our principal) and pay off the loan at an interest rate of $r$% with regular monthly installment payments of $M$ dollars. So in month 1 of the loan, before we make any payments, our principal is $P$ dollars. Our goal in this exercise is to find a formula that relates these three parameters to the time duration of the loan. We are charged interest every month at an annual rate of $r$%, so each month we pay $\frac{r}{12}$% interest on the principal that remains. Given that the original principal is $P$ dollars, we will pay $\left(\frac{0.01r}{12}\right)P$ dollars in interest on our first payment. Since we paid $M$ dollars in total for our first payment, the remainder of the payment ($M-\left(\frac{r}{12}\right)P$) goes to pay down the principal. So the principal remaining after the first payment (let's call it $P_1$) is the original principal minus what we paid on the principal, or \begin{equation*} P_1 = P - \left( M - \left(\frac{r}{12}\right)P\right) = \left(1 + \frac{r}{12}\right)P - M\text{.} \end{equation*} As long as $P_1$ is positive, we still have to keep making payments to pay off the loan. 1. Recall that the amount of interest we pay each time depends on the principal that remains. How much interest, in terms of $P_1$ and $r\text{,}$ do we pay in the second installment? 2. How much of our second monthly installment goes to pay off the principal? What is the principal $P_2\text{,}$ or the balance of the loan, that we still have to pay off after making the second installment of the loan? Write your response in the form $P_2 = ( \ )P_1 - ( \ )M\text{,}$ where you fill in the parentheses. 3. Show that $P_2 = \left(1 + \frac{r}{12}\right)^2P - \left[1 + \left(1+\frac{r}{12}\right)\right] M\text{.}$ 4. Let $P_3$ be the amount of principal that remains after the third installment. Show that \begin{equation*} P_3 = \left(1 + \frac{r}{12}\right)^3P - \left[1 + \left(1+\frac{r}{12}\right) + \left(1+\frac{r}{12}\right)^2 \right] M\text{.} \end{equation*} 5. If we continue in the manner described in the problems above, then the remaining principal of our loan after $n$ installments is $$P_n = \left(1 + \frac{r}{12}\right)^nP - \left[\displaystyle \sum_{k=0}^{n-1} \left(1+\frac{r}{12}\right)^k \right] M\text{.}\label{eq-loan-1}\tag{7.7}$$ This is a rather complicated formula and one that is difficult to use. However, we can simplify the sum if we recognize part of it as a partial sum of a geometric series. Find a formula for the sum $$\displaystyle \sum_{k=0}^{n-1} \left(1+\frac{r}{12}\right)^k\text{.}\label{eq-part-sum}\tag{7.8}$$ and then a general formula for $P_n$ that does not involve a sum. 6. It is usually more convenient to write our formula for $P_n$ in terms of years rather than months. Show that $P(t)\text{,}$ the principal remaining after $t$ years, can be written as $$P(t) = \left(P - \frac{12M}{r}\right)\left(1+\frac{r}{12}\right)^{12t} + \frac{12M}{r}\text{.}\label{eq-loan2}\tag{7.9}$$ 7. Now that we have analyzed the general loan situation, we apply formula (7.9) to an actual loan. Suppose we charge$1,000 on a credit card for holiday expenses. If our credit card charges 20% interest and we pay only the minimum payment of $25 each month, how long will it take us to pay off the$1,000 charge? How much in total will we have paid on this $1,000 charge? How much total interest will we pay on this loan? 8. Now we consider larger loans, e.g., automobile loans or mortgages, in which we borrow a specified amount of money over a specified period of time. In this situation, we need to determine the amount of the monthly payment we need to make to pay off the loan in the specified amount of time. In this situation, we need to find the monthly payment $M$ that will take our outstanding principal to $0$ in the specified amount of time. To do so, we want to know the value of $M$ that makes $P(t) = 0$ in formula(7.9). If we set $P(t) = 0$ and solve for $M\text{,}$ it follows that \begin{equation*} M = \frac{rP \left(1+\frac{r}{12}\right)^{12t}}{12\left(\left(1+\frac{r}{12}\right)^{12t} - 1 \right)}\text{.} \end{equation*} 1. Suppose we want to borrow$15,000 to buy a car. We take out a 5 year loan at 6.25%. What will our monthly payments be? How much in total will we have paid for this $15,000 car? How much total interest will we pay on this loan? 2. Suppose you charge your books for winter semester on your credit card. The total charge comes to$525. If your credit card has an interest rate of 18% and you pay $20 per month on the card, how long will it take before you pay off this debt? How much total interest will you pay? 3. Say you need to borrow$100,000 to buy a house. You have several options on the loan:
• 30 years at 6.5%
• 25 years at 7.5%
• 15 years at 8.25%.
1. What are the monthly payments for each loan?
2. Which mortgage is ultimately the best deal (assuming you can afford the monthly payments)? In other words, for which loan do you pay the least amount of total interest? |
What is 84/100 as a decimal?
There are a lot of different ways you can convert a number into (decimals, percentages, and more) but in this guide, we’ll demonstrate how to use the division method to convert a fraction to a decimal.
## Solution: 84/100 as a decimal is 0.84
Methods
### Explanation using the division method:
Put in a nutshell, a fraction is written in terms of two parts separated by a line in between: the number above the line is called the numerator and the number below the line is called the denominator. To solve this question, we can use the division method to get a decimal: simply divide the numerator 84 by the denominator 100 to get the decimal:
84 (numerator) ÷ 100 (denominator) = 0.84
That’s it! When you convert 84/100 to a decimal, 0.84 is your answer.
### Method 2 – Explanation using the factors of 10 method:
Another method is to change the denominator to a power of 10 (like 10, 100, 1000 etc.). This way, the fraction will be coming simpler to compute.
Even if our denominator is not a power of 10, we can make it one. If the denominator is a factor of a power of 10, like 2, 4, 5, 8, 10, 16, 20, 25, 40, 50 and so on, then we can convert them into a power of 10. Since our denominator is 100, we can convert it because it is a factor of a power of 10.
To do that, we multiply 100 by 1 to make the denominator 100. And since we multiplied the denominator by 1, we also need to multiply the numerator by 1 too:
$\frac{84 * 1}{100 * 1} = \frac{84}{100} = 0.84$
And finally, you get 0.84 as your answer when you convert 84/100 to a decimal.
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ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 8 Algebraic Expressions Ex 8.1
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 8 Algebraic Expressions Ex 8.1
Question 1.
From the algebraic expressions using variables, constants, and arithmetic operations:
(i) 6 more than thrice a number x.
(ii) 5 times x is subtracted from 13.
(iii) The numbers x and y both squared and added.
(iv) Number 7 is added to 3 times the product of p and q.
(v) Three times of x is subtracted from the product of x with itself.
(vi) Sum of the numbers m and n is subtracted from their product.
Solution:
Question 2.
A taxi charges ₹ 9 per km and a fixed charge of ₹ 50. If the taxi is hired for x km, write an algebraic expression for this situation.
Solution:
Question 3.
Write down the algebraic expression whose terms are:
(i) 5a, -3b, c
(ii) x2, -5x, 6
(iii) x2y, xy, -xy2
Solution:
Question 4.
Write all the terms of each of the following algebraic expressions:
(i) 3 – 7x
(ii) 2 – 5a + $$\frac { 1 }{ 2 }$$ b
(iii) 3x5 + 4y3 – 7xy2 + 3
Solution:
Question 5.
Identify the terms and their factors in the algebraic expressions given below:
(i) -4x + 5y
(ii) xy + 2x2y2
(iii) 1.2ab – 2.4b + 3.6a
Solution:
Question 6.
Show the terms and their factors by tree diagrams of the following algebraic expressions:
(i) 8x + 3y2
(ii) y – y3
(iii) 5xy2 + 7x2y
(iv) -ab + 2b2 – 3a2
Solution:
Question 7.
Write down the numerical coefficient of each of the following:
(i) -7x
(ii) -2x3y2
(iii) 6abcd2
(iv) $$\frac { 2 }{ 3 }$$ pq2
Solution:
Question 8.
Write down the coefficient of x in the following:
(i) -4bx
(ii) 5xyz
(iii) -x
(iv) -3x2y
Solution:
Question 9.
In -7xy2z3, write down the coefficient of:
(i) 7x
(ii) -xy2
(iii) xyz
(iv) 7yz2
Solution:
Question 10.
Identify the terms (other than constants) and write their numerical coefficients in each of the following algebraic expressions:
(i) 3 – 7x
(ii) 1 + 2x – 3x2
(iii) 1.2a + 0.8b
Solution:
Question 11.
Identify the terms which contain x and write the coefficient of x in each of the following expressions:
(i) 13y2 – 8xy
(ii) 7x – xy2
(iii) 5 – 7xyz + 4x2y
Solution:
Question 12.
Identify the term which contain y2 and write the coefficient of y2 in each of the following expressions:
(i) 8 – xy2
(ii) 5y2 + 7x – 3xy2
(iii) 2x2y – 15xy2 + 7y2
Solution:
Question 13.
Classify into monomials, binomials and trinomials:
(i) 4y – 7z
(ii) -5xy2
(iii) x + y – xy
(iv) ab2 – 5b -3a
(v) 4p2q – 5pq2
(vi) 2017
(vii) 1 + x + x2
(viii) 5x2 – 7 + 3x + 4
Solution:
Question 14.
State whether the given pair of terms is of like or unlike terms:
(i) -7x, $$\frac { 5 }{ 2 }$$ x
(ii) -29x, -29y
(iii) 2xy, 2xyz
(iv) 4m2p, 4mp2
(v) 12xz, 12x2z2
(vi) -5pq, 7qp
Solution:
Question 15.
Identify like terms in the following:
(i) x2y, 3xy2, -2x2y, 4x2y2
(ii) 3a2b, 2abc, -6a2b, 4abc
(iii) 10pq, 7p, 8q – p2q2, -7qp, -100q, -23, 12q2p2, -5p2, 41, 2405p, 78qp, 13p2q, qp2, 701p2
Solution:
Question 16.
Write down the degree of following polynomials in x:
(i) x2 – 6x7 + x8
(ii) 3 – 2x
(iii) -2
(iv) 1 – x2
Solution:
Question 17.
Write the degree of the following polynomials:
(i) 3x2 – 5xy2 + 7
(ii) xy2 – y3 + 3y4 – 2
(iii) 7 – 2x3 – 5xy3 + 9y5
Solution:
Question 18.
State true or false:
(i) If 5 is constant andy is variable, then 5y and 5 + y are variables
(ii) 7x has two terms, 7 and x
(iii) 5 + xy is a trinomial
(iv) 7a × bc is a binomial
(v) 7x3 + 2x2 + 3x – 5 is a polynomial
(vi) 2x2 – $$\frac { 3 }{ x }$$ is a polynomial
(vii) Coefficient of x in -3xy is -3
Solution: |
# How do you differentiate f(x)=e^(4x)*sin(5-x^2) using the product rule?
Dec 4, 2015
$f ' \left(x\right) = 2 {e}^{4 x} \left(2 \sin \left(5 - {x}^{2}\right) - x \cos \left(5 - {x}^{2}\right)\right)$
#### Explanation:
$f ' \left(x\right) = \sin \left(5 - {x}^{2}\right) \frac{d}{\mathrm{dx}} \left[{e}^{4 x}\right] + {e}^{4 x} \frac{d}{\mathrm{dx}} \left[\sin \left(5 - {x}^{2}\right)\right]$
Find each individual derivative, both of which use the chain rule:
$\frac{d}{\mathrm{dx}} \left[{e}^{4 x}\right] = {e}^{4 x} \frac{d}{\mathrm{dx}} \left[4 x\right] = 4 {e}^{4 x}$
$\frac{d}{\mathrm{dx}} \left[\sin \left(5 - {x}^{2}\right)\right] = \cos \left(5 - {x}^{2}\right) \frac{d}{\mathrm{dx}} \left[5 - {x}^{2}\right] = - 2 x \cos \left(5 - {x}^{2}\right)$
Plug back in:
$f ' \left(x\right) = 4 {e}^{4 x} \sin \left(5 - {x}^{2}\right) - 2 x {e}^{4 x} \cos \left(5 - {x}^{2}\right)$
$f ' \left(x\right) = 2 {e}^{4 x} \left(2 \sin \left(5 - {x}^{2}\right) - x \cos \left(5 - {x}^{2}\right)\right)$ |
# Chapter 1 Functions and Operations
Topics Dimensionality; Interval Notation for $${\bf R}^1$$; Neighborhoods: Intervals, Disks, and Balls; Introduction to Functions; Domain and Range; Some General Types of Functions; $$\log$$, $$\ln$$, and $$\exp$$; Other Useful Functions; Graphing Functions; Solving for Variables; Finding Roots; Limit of a Function; Continuity; Sets, Sets, and More Sets.
## 1.1 Summation Operators $$\sum$$ and $$\prod$$
Addition (+), Subtraction (-), multiplication and division are basic operations of arithmetic -- combining numbers. In statistics and calculus, we want to add a sequence of numbers that can be expressed as a pattern without needing to write down all its components. For example, how would we express the sum of all numbers from 1 to 100 without writing a hundred numbers?
For this we use the summation operator $$\sum$$ and the product operator $$\prod$$.
Summation:
$\sum\limits_{i=1}^{100} x_i = x_1+x_2+x_3+\cdots+x_{100}$
The bottom of the $$\sum$$ symbol indicates an index (here, $$i$$), and its start value $$1$$. At the top is where the index ends. The notion of "addition" is part of the $$\sum$$ symbol. The content to the right of the summation is the meat of what we add. While you can pick your favorite index, start, and end values, the content must also have the index.
• $$\sum\limits_{i=1}^n c x_i = c \sum\limits_{i=1}^n x_i$$
• $$\sum\limits_{i=1}^n (x_i + y_i) = \sum\limits_{i=1}^n x_i + \sum\limits_{i=1}^n y_i$$
• $$\sum\limits_{i=1}^n c = n c$$
Product:
$\prod\limits_{i=1}^n x_i = x_1 x_2 x_3 \cdots x_n$
Properties:
• $$\prod\limits_{i=1}^n c x_i = c^n \prod\limits_{i=1}^n x_i$$
• $$\prod\limits_{i=k}^n c x_i = c^{n-k+1} \prod\limits_{i=k}^n x_i$$
• $$\prod\limits_{i=1}^n (x_i + y_i) =$$ a total mess
• $$\prod\limits_{i=1}^n c = c^n$$
Other Useful Functions
Factorials!:
$x! = x\cdot (x-1) \cdot (x-2) \cdots (1)$
Modulo: Tells you the remainder when you divide the first number by the second.
• $$17 \mod 3 = 2$$
• $$100 \ \% \ 30 = 10$$
Example 1.1 (Operators)
1. $$\sum\limits_{i=1}^{5} i =$$
2. $$\prod\limits_{i=1}^{5} i =$$
3. $$14 \mod 4 =$$
4. $$4! =$$
Exercise 1.1 (Operators) Let $$x_1 = 4, x_2 = 3, x_3 = 7, x_4 = 11, x_5 = 2$$
1. $$\sum\limits_{i=1}^{3} (7)x_i$$
2. $$\sum\limits_{i=1}^{5} 2$$
3. $$\prod\limits_{i=3}^{5} (2)x_i$$
## 1.2 Introduction to Functions
A function (in $${\bf R}^1$$) is a mapping, or transformation, that relates members of one set to members of another set. For instance, if you have two sets: set $$A$$ and set $$B$$, a function from $$A$$ to $$B$$ maps every value $$a$$ in set $$A$$ such that $$f(a) \in B$$. Functions can be "many-to-one", where many values or combinations of values from set $$A$$ produce a single output in set $$B$$, or they can be "one-to-one", where each value in set $$A$$ corresponds to a single value in set $$B$$. A function by definition has a single function value for each element of its domain. This means, there cannot be "one-to-many" mapping.
Dimensionality: $${\bf R}^1$$ is the set of all real numbers extending from $$-\infty$$ to $$+\infty$$ --- i.e., the real number line. $${\bf R}^n$$ is an $$n$$-dimensional space, where each of the $$n$$ axes extends from $$-\infty$$ to $$+\infty$$.
• $${\bf R}^1$$ is a one dimensional line.
• $${\bf R}^2$$ is a two dimensional plane.
• $${\bf R}^3$$ is a three dimensional space.
Points in $${\bf R}^n$$ are ordered $$n$$-tuples (just means an combination of $$n$$ elements where order matters), where each element of the $$n$$-tuple represents the coordinate along that dimension.
For example:
• $${\bf R}^1$$: (3)
• $${\bf R}^2$$: (-15, 5)
• $${\bf R}^3$$: (86, 4, 0)
Examples of mapping notation:
Function of one variable: $$f:{\bf R}^1\to{\bf R}^1$$
• $$f(x)=x+1$$. For each $$x$$ in $${\bf R}^1$$, $$f(x)$$ assigns the number $$x+1$$.
Function of two variables: $$f: {\bf R}^2\to{\bf R}^1$$.
• $$f(x,y)=x^2+y^2$$. For each ordered pair $$(x,y)$$ in $${\bf R}^2$$, $$f(x,y)$$ assigns the number $$x^2+y^2$$.
We often use variable $$x$$ as input and another $$y$$ as output, e.g. $$y=x+1$$
Example 1.2 (Functions) For each of the following, state whether they are one-to-one or many-to-one functions.
1. For $$x \in [0,\infty]$$, $$f : x \rightarrow x^2$$ (this could also be written as $$f(x) = x^2$$).
2. For $$x \in [-\infty, \infty]$$, $$f: x \rightarrow x^2$$.
Exercise 1.2 (Functions) For each of the following, state whether they are one-to-one or many-to-one functions.
1. For $$x \in [-3, \infty]$$, $$f: x \rightarrow x^2$$.
2. For $$x \in [0, \infty]$$, $$f: x \rightarrow \sqrt{x}$$
Some functions are defined only on proper subsets of $${\bf R}^n$$.
• Domain: the set of numbers in $$X$$ at which $$f(x)$$ is defined.
• Range: elements of $$Y$$ assigned by $$f(x)$$ to elements of $$X$$, or $f(X)=\{ y : y=f(x), x\in X\}$ Most often used when talking about a function $$f:{\bf R}^1\to{\bf R}^1$$.
• Image: same as range, but more often used when talking about a function $$f:{\bf R}^n\to{\bf R}^1$$.
Some General Types of Functions
Monomials: $$f(x)=a x^k$$
$$a$$ is the coefficient. $$k$$ is the degree.
Examples: $$y=x^2$$, $$y=-\frac{1}{2}x^3$$
Polynomials: sum of monomials.
Examples: $$y=-\frac{1}{2}x^3+x^2$$, $$y=3x+5$$
The degree of a polynomial is the highest degree of its monomial terms. Also, it's often a good idea to write polynomials with terms in decreasing degree.
Exponential Functions: Example: $$y=2^x$$
## 1.3$$\log$$ and $$\exp$$
Relationship of logarithmic and exponential functions: $y=\log_a(x) \iff a^y=x$
The log function can be thought of as an inverse for exponential functions. $$a$$ is referred to as the "base" of the logarithm.
Common Bases: The two most common logarithms are base 10 and base $$e$$.
1. Base 10: $$\quad y=\log_{10}(x) \iff 10^y=x$$. The base 10 logarithm is often simply written as "$$\log(x)$$" with no base denoted.
2. Base $$e$$: $$\quad y=\log_e(x) \iff e^y=x$$. The base $$e$$ logarithm is referred to as the "natural" logarithm and is written as $$\ln(x)$$".
Properties of exponential functions:
• $$a^x a^y = a^{x+y}$$
• $$a^{-x} = 1/a^x$$
• $$a^x/a^y = a^{x-y}$$
• $$(a^x)^y = a^{x y}$$
• $$a^0 = 1$$
Properties of logarithmic functions (any base):
Generally, when statisticians or social scientists write $$\log(x)$$ they mean $$\log_e(x)$$. In other words: $$\log_e(x) \equiv \ln(x) \equiv \log(x)$$
$\log_a(a^x)=x$ and $a^{\log_a(x)}=x$
• $$\log(x y)=\log(x)+\log(y)$$
• $$\log(x^y)=y\log(x)$$
• $$\log(1/x)=\log(x^{-1})=-\log(x)$$
• $$\log(x/y)=\log(x\cdot y^{-1})=\log(x)+\log(y^{-1})=\log(x)-\log(y)$$
• $$\log(1)=\log(e^0)=0$$
Change of Base Formula: Use the change of base formula to switch bases as necessary: $\log_b(x) = \frac{\log_a(x)}{\log_a(b)}$
Example: $\log_{10}(x) = \frac{\ln(x)}{\ln(10)}$
You can use logs to go between sum and product notation. This will be particularly important when you're learning maximum likelihood estimation.
$\begin{eqnarray*} \log \bigg(\prod\limits_{i=1}^n x_i \bigg) &=& \log(x_1 \cdot x_2 \cdot x_3 \cdots \cdot x_n)\\ &=& \log(x_1) + \log(x_2) + \log(x_3) + \cdots + \log(x_n)\\ &=& \sum\limits_{i=1}^n \log (x_i) \end{eqnarray*}$
Therefore, you can see that the log of a product is equal to the sum of the logs. We can write this more generally by adding in a constant, $$c$$:
$\begin{eqnarray*} \log \bigg(\prod\limits_{i=1}^n c x_i\bigg) &=& \log(cx_1 \cdot cx_2 \cdots cx_n)\\ &=& \log(c^n \cdot x_1 \cdot x_2 \cdots x_n)\\ &=& \log(c^n) + \log(x_1) + \log(x_2) + \cdots + \log(x_n)\\\\ &=& n \log(c) + \sum\limits_{i=1}^n \log (x_i)\\ \end{eqnarray*}$
Example 1.3 (Logarithmic Functions) Evaluate each of the following logarithms
1. $$\log_4(16)$$
2. $$\log_2(16)$$
Simplify the following logarithm. By "simplify", we actually really mean - use as many of the logarithmic properties as you can.
1. $$\log_4(x^3y^5)$$
Exercise 1.3 (Logarithmic Functions)
Evaluate each of the following logarithms
1. $$\log_\frac{3}{2}(\frac{27}{8})$$
Simplify each of the following logarithms. By "simplify", we actually really mean - use as many of the logarithmic properties as you can.
1. $$\log(\frac{x^9y^5}{z^3})$$
2. $$\ln{\sqrt{xy}}$$
## 1.4 Graphing Functions
What can a graph tell you about a function?
• Is the function increasing or decreasing? Over what part of the domain?
• How fast" does it increase or decrease?
• Are there global or local maxima and minima? Where?
• Are there inflection points?
• Is the function continuous?
• Is the function differentiable?
• Does the function tend to some limit?
• Other questions related to the substance of the problem at hand.
## 1.5 Solving for Variables and Finding Roots
Sometimes we're given a function $$y=f(x)$$ and we want to find how $$x$$ varies as a function of $$y$$. Use algebra to move $$x$$ to the left hand side (LHS) of the equation and so that the right hand side (RHS) is only a function of $$y$$.
Example 1.4 (Solving for Variables)
Solve for x:
1. $$y=3x+2$$
2. $$y=e^x$$
Solving for variables is especially important when we want to find the roots of an equation: those values of variables that cause an equation to equal zero. Especially important in finding equilibria and in doing maximum likelihood estimation.
Procedure: Given $$y=f(x)$$, set $$f(x)=0$$. Solve for $$x$$.
Multiple Roots: $f(x)=x^2 - 9 \quad\Longrightarrow\quad 0=x^2 - 9 \quad\Longrightarrow\quad 9=x^2 \quad\Longrightarrow\quad \pm \sqrt{9}=\sqrt{x^2} \quad\Longrightarrow\quad \pm 3=x$
Quadratic Formula: For quadratic equations $$ax^2+bx+c=0$$, use the quadratic formula: $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
Exercise 1.4 (Finding Roots) Solve for x:
1. $$f(x)=3x+2 = 0$$
2. $$f(x)=x^2+3x-4=0$$
3. $$f(x)=e^{-x}-10 = 0$$
## 1.6 Sets
Interior Point: The point $$\bf x$$ is an interior point of the set $$S$$ if $$\bf x$$ is in $$S$$ and if there is some $$\epsilon$$-ball around $$\bf x$$ that contains only points in $$S$$. The interior of $$S$$ is the collection of all interior points in $$S$$. The interior can also be defined as the union of all open sets in $$S$$.
• If the set $$S$$ is circular, the interior points are everything inside of the circle, but not on the circle's rim.
• Example: The interior of the set $$\{ (x,y) : x^2+y^2\le 4 \}$$ is $$\{ (x,y) : x^2+y^2< 4 \}$$ .
Boundary Point: The point $$\bf x$$ is a boundary point of the set $$S$$ if every $$\epsilon$$-ball around $$\bf x$$ contains both points that are in $$S$$ and points that are outside $$S$$. The boundary is the collection of all boundary points.
• If the set $$S$$ is circular, the boundary points are everything on the circle's rim.
• Example: The boundary of $$\{ (x,y) : x^2+y^2\le 4 \}$$ is $$\{ (x,y) : x^2+y^2 = 4 \}$$.
Open: A set $$S$$ is open if for each point $$\bf x$$ in $$S$$, there exists an open $$\epsilon$$-ball around $$\bf x$$ completely contained in $$S$$.
• If the set $$S$$ is circular and open, the points contained within the set get infinitely close to the circle's rim, but do not touch it.
• Example: $$\{ (x,y) : x^2+y^2<4 \}$$
Closed: A set $$S$$ is closed if it contains all of its boundary points.
• Alternatively: A set is closed if its complement is open.
• If the set $$S$$ is circular and closed, the set contains all points within the rim as well as the rim itself.
• Example: $$\{ (x,y) : x^2+y^2\le 4 \}$$
• Note: a set may be neither open nor closed. Example: $$\{ (x,y) : 2 < x^2+y^2\le 4 \}$$
Complement: The complement of set $$S$$ is everything outside of $$S$$.
• If the set $$S$$ is circular, the complement of $$S$$ is everything outside of the circle.
• Example: The complement of $$\{ (x,y) : x^2+y^2\le 4 \}$$ is $$\{ (x,y) : x^2+y^2 > 4 \}$$.
Empty: The empty (or null) set is a unique set that has no elements, denoted by {} or $$\emptyset$$.
• The empty set is an example of a set that is open and closed, or a "clopen" set.
• Examples: The set of squares with 5 sides; the set of countries south of the South Pole.
## Answers to Examples and Exercises
1. 1 + 2 + 3 + 4 + 5 = 15
2. 1 * 2 * 3 * 4 * 5 = 120
3. 2
4. 4 * 3 * 2 * 1 = 24
1. 7(4 + 3 + 7) = 98
2. 2 + 2 + 2 + 2 + 2 = 10
3. $$2^3(7)(11)(2)$$ = 1232
1. one-to-one
2. many-to-one
1. many-to-one
2. one-to-one
1. 2
2. 4
3. $$3\log_4(x) + 5\log_4(y)$$
1. 3
2. $$9\log(x) + 5\log(y) - 3\log(z)$$
3. $$\frac{1}{2}(\ln{x} + \ln{y})$$
1. $$y=3x+2 \quad\Longrightarrow\quad -3x=2-y \quad\Longrightarrow\quad 3x=y-2 \quad\Longrightarrow\quad x=\frac{1}{3}(y-2)$$
2. $$x = \ln{y}$$
1. $$\frac{-2}{3}$$
3. x = - $$\ln10$$ |
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# 6.2: Graphs of the Sine and Cosine Functions
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Learning Objectives
• Graph variations of $$y=\sin( x )$$ and $$y=\cos( x )$$.
• Use phase shifts of sine and cosine curves.
White light, such as the light from the sun, is not actually white at all. Instead, it is a composition of all the colors of the rainbow in the form of waves. The individual colors can be seen only when white light passes through an optical prism that separates the waves according to their wavelengths to form a rainbow.
Light waves can be represented graphically by the sine function. In the chapter on Trigonometric Functions, we examined trigonometric functions such as the sine function. In this section, we will interpret and create graphs of sine and cosine functions.
## Graphing Sine and Cosine Functions
Recall that the sine and cosine functions relate real number values to the $$x$$- and $$y$$-coordinates of a point on the unit circle. So what do they look like on a graph on a coordinate plane? Let’s start with the sine function. We can create a table of values and use them to sketch a graph. Table $$\PageIndex{1}$$ lists some of the values for the sine function on a unit circle.
$$x$$ $$\sin(x)$$ $$0$$ $$\frac{\pi}{6}$$ $$\frac{\pi}{4}$$ $$\frac{\pi}{3}$$ $$\frac{\pi}{2}$$ $$\dfrac{2\pi}{3}$$ $$\dfrac{3\pi}{4}$$ $$\dfrac{5\pi}{6}$$ $$\pi$$ $$0$$ $$\frac{1}{2}$$ $$\frac{\sqrt{2}}{2}$$ $$\frac{\sqrt{3}}{2}$$ $$1$$ $$\dfrac{\sqrt{3}}{2}$$ $$\dfrac{\sqrt{2}}{2}$$ $$\dfrac{1}{2}$$ $$0$$
Plotting the points from the table and continuing along the x-axis gives the shape of the sine function. See Figure $$\PageIndex{2}$$.
Notice how the sine values are positive between $$0$$ and $$\pi$$, which correspond to the values of the sine function in quadrants I and II on the unit circle, and the sine values are negative between $$\pi$$ and $$2\pi$$, which correspond to the values of the sine function in quadrants III and IV on the unit circle. See Figure $$\PageIndex{3}$$.
Now let’s take a similar look at the cosine function. Again, we can create a table of values and use them to sketch a graph. Table $$\PageIndex{2}$$ lists some of the values for the cosine function on a unit circle.
$$x$$ $$\cos(x)$$ $$0$$ $$\frac{\pi}{6}$$ $$\frac{\pi}{4}$$ $$\frac{\pi}{3}$$ $$\frac{\pi}{2}$$ $$\frac{2\pi}{3}$$ $$\dfrac{3\pi}{4}$$ $$\dfrac{5\pi}{6}$$ $$\pi$$ $$1$$ $$\frac{\sqrt{3}}{2}$$ $$\frac{\sqrt{2}}{2}$$ $$\dfrac{1}{2}$$ $$0$$ $$-\dfrac{1}{2}$$ $$-\dfrac{\sqrt{2}}{2}$$ $$-\dfrac{\sqrt{3}}{2}$$ $$-1$$
As with the sine function, we can plots points to create a graph of the cosine function as in Figure $$\PageIndex{4}$$.
Because we can evaluate the sine and cosine of any real number, both of these functions are defined for all real numbers. By thinking of the sine and cosine values as coordinates of points on a unit circle, it becomes clear that the range of both functions must be the interval $$[ −1,1 ]$$.
In both graphs, the shape of the graph repeats after $$2\pi$$, which means the functions are periodic with a period of $$2\pi$$. A periodic function is a function for which a specific horizontal shift, $$P$$, results in a function equal to the original function: $$f(x+P)=f(x)$$ for all values of $$x$$ in the domain of $$f$$. When this occurs, we call the smallest such horizontal shift with $$P>0$$ the period of the function. Figure $$\PageIndex{5}$$ shows several periods of the sine and cosine functions.
Looking again at the sine and cosine functions on a domain centered at the $$y$$-axis helps reveal symmetries. As we can see in Figure $$\PageIndex{6}$$, the sine function is symmetric about the origin. Recall from The Other Trigonometric Functions that we determined from the unit circle that the sine function is an odd function because $$\sin(−x)=−\sin\space x$$. Now we can clearly see this property from the graph.
Figure $$\PageIndex{7}$$ shows that the cosine function is symmetric about the $$y$$-axis. Again, we determined that the cosine function is an even function. Now we can see from the graph that $$\cos(−x)=\cos\space x$$.
CHARACTERISTICS OF SINE AND COSINE FUNCTIONS
The sine and cosine functions have several distinct characteristics:
• They are periodic functions with a period of $$2\pi$$.
• The domain of each function is $$(−\infty,\infty)$$ and the range is $$[ −1,1 ]$$.
• The graph of $$y=\sin\space x$$ is symmetric about the origin, because it is an odd function.
• The graph of $$y=\cos\space x$$ is symmetric about they- $$y$$-axis, because it is an even function.
## Investigating Sinusoidal Functions
As we can see, sine and cosine functions have a regular period and range. If we watch ocean waves or ripples on a pond, we will see that they resemble the sine or cosine functions. However, they are not necessarily identical. Some are taller or longer than others. A function that has the same general shape as a sine or cosine function is known as a sinusoidal function. The general forms of sinusoidal functions are
$y=A\sin(Bx−C)+D$
and
$y=A\cos(Bx−C)+D$
## Determining the Period of Sinusoidal Functions
Looking at the forms of sinusoidal functions, we can see that they are transformations of the sine and cosine functions. We can use what we know about transformations to determine the period.
In the general formula, $$B$$ is related to the period by $$P=\dfrac{2\pi}{|B|}$$. If $$|B|>1$$, then the period is less than $$2\pi$$ and the function undergoes a horizontal compression, whereas if $$| B |<1$$, then the period is greater than $$2\pi$$ and the function undergoes a horizontal stretch. For example, $$f(x)=\sin(x)$$, $$B=1$$, so the period is $$2\pi$$,which we knew. If $$f(x)=\sin(2x)$$, then $$B=2$$, so the period is $$\pi$$ and the graph is compressed. If $$f(x)=\sin\left(\dfrac{x}{2}\right)$$, then $$B=\dfrac{1}{2}$$, so the period is $$4\pi$$ and the graph is stretched. Notice in Figure $$\PageIndex{8}$$ how the period is indirectly related to $$|B|$$.
PERIOD OF SINUSOIDAL FUNCTIONS
If we let $$C=0$$ and $$D=0$$ in the general form equations of the sine and cosine functions, we obtain the forms
• $$y=A\sin(Bx)$$
• $$y=A\cos(Bx)$$
The period is $$\dfrac{2\pi}{|B|}$$.
Example $$\PageIndex{1}$$: Identifying the Period of a Sine or Cosine Function
Determine the period of the function $$f(x)=\sin\left(\dfrac{\pi}{6}x\right)$$.
Solution
Let’s begin by comparing the equation to the general form $$y=A\sin(Bx)$$.
In the given equation, $$B=\dfrac{\pi}{6}$$, so the period will be
\begin{align*} P&=\dfrac{2\pi}{|B|} \\[4pt] &=\dfrac{2\pi}{\dfrac{\pi}{6}} \\ &=2\pi ⋅ \dfrac{6}{\pi} \\[4pt] &=12 \end{align*}
Exercise $$\PageIndex{1}$$
Determine the period of the function $$g(x)=\cos(\frac{x}{3})$$.
$$6\pi$$
## Determining Amplitude
Returning to the general formula for a sinusoidal function, we have analyzed how the variable $$B$$ relates to the period. Now let’s turn to the variable $$A$$ so we can analyze how it is related to the amplitude, or greatest distance from rest. $$A$$ represents the vertical stretch factor, and its absolute value $$|A|$$ is the amplitude. The local maxima will be a distance $$|A|$$ above the vertical midline of the graph, which is the line $$x=D$$; because $$D=0$$ in this case, the midline is the x-axis. The local minima will be the same distance below the midline. If $$| A |>1$$, the function is stretched. For example, the amplitude of $$f(x)=4 sin x$$ is twice the amplitude of
$$f(x)=2\sin x$$
If $$| A |<1$$, the function is compressed. Figure $$\PageIndex{9}$$ compares several sine functions with different amplitudes.
AMPLITUDE OF SINUSOIDAL FUNCTIONS
If we let $$C=0$$ and $$D=0$$ in the general form equations of the sine and cosine functions, we obtain the forms
\begin{align} y=A\sin(Bx)\text { and } y=A\cos(Bx) \end{align}
The amplitude is $$A$$, and the vertical height from the midline is $$|A|$$. In addition, notice in the example that
$|A| = amplitude = \dfrac{1}{2}∣maximum − minimum|$
Example $$\PageIndex{2}$$: Identifying the Amplitude of a Sine or Cosine Function
What is the amplitude of the sinusoidal function $$f(x)=−4\sin(x)$$? Is the function stretched or compressed vertically?
Solution
Let’s begin by comparing the function to the simplified form $$y=A\sin(Bx)$$.
In the given function, $$A=−4$$, so the amplitude is $$| A |=| −4 |=4$$. The function is stretched.
Analysis
The negative value of $$A$$ results in a reflection across the $$x$$-axis of the sine function, as shown in Figure $$\PageIndex{10}$$.
Exercise $$\PageIndex{2}$$
What is the amplitude of the sinusoidal function $$f(x)=\frac{1}{2}\sin(x)$$? Is the function stretched or compressed vertically?
$$\frac{1}{2}$$ compressed
## Analyzing Graphs of Variations of $$y = \sin\space x$$ and $$y = \cos\space x$$
Now that we understand how $$A$$ and $$B$$ relate to the general form equation for the sine and cosine functions, we will explore the variables $$C$$ and $$D$$. Recall the general form:
$y=A\sin(Bx-C)+D\qquad \text{ and } \qquad y=A\cos(Bx-C)+D$
or
$y=A\sin\left (B\left (x-\dfrac{C}{B} \right ) \right )+D \qquad \text{ and } \qquad y=A\cos\left (B\left (x-\dfrac{C}{B} \right ) \right )+D$
The value $$\frac{C}{B}$$ for a sinusoidal function is called the phase shift, or the horizontal displacement of the basic sine or cosine function. If $$C>0$$, the graph shifts to the right. If $$C<0$$, the graph shifts to the left. The greater the value of $$| C |$$, the more the graph is shifted. Figure $$\PageIndex{11}$$ shows that the graph of $$f(x)=\sin(x−\pi)$$ shifts to the right by $$\pi$$ units, which is more than we see in the graph of $$f(x)=\sin\left(x−\frac{\pi}{4}\right)$$, which shifts to the right by $$\frac{\pi}{4}$$ units.
While $$C$$ relates to the horizontal shift, $$D$$ indicates the vertical shift from the midline in the general formula for a sinusoidal function. See Figure $$\PageIndex{12}$$. The function $$y=\cos(x)+D$$ has its midline at $$y=D$$.
Any value of $$D$$ other than zero shifts the graph up or down. Figure $$\PageIndex{13}$$ compares $$f(x)=\sin x$$ with $$f(x)=\sin x+2$$, which is shifted $$2$$ units up on a graph.
VARIATIONS OF SINE AND COSINE FUNCTIONS
Given an equation in the form $$f(x)=A \sin (Bx−C)+D$$ or $$f(x)=A \cos (Bx−C)+D$$, $$\frac{C}{D}$$ is the phase shift and $$D$$ is the vertical shift.
Example $$\PageIndex{3}$$: Identifying the Phase Shift of a Function
Determine the direction and magnitude of the phase shift for $$f(x)=\sin\left(x+\frac{\pi}{6}\right)−2$$.
Solution
Let’s begin by comparing the equation to the general form $$y=A\sin(Bx−C)+D$$.
In the given equation, notice that $$B=1$$ and $$C=−\frac{\pi}{6}$$. So the phase shift is
\begin{align*} \dfrac{C}{B}&= -\frac{\frac{\pi}{6}}{1}\\ &= -\frac{\pi}{6} \end{align*}
or $$\frac{\pi}{6}$$ units to the left.
Analysis
We must pay attention to the sign in the equation for the general form of a sinusoidal function. The equation shows a minus sign before $$C$$. Therefore $$f(x)=\sin(x+\frac{\pi}{6})−2$$ can be rewritten as $$f(x)=\sin\left(x−\left(−\frac{\pi}{6}\right)\right)−2$$. If the value of $$C$$ is negative, the shift is to the left.
Exercise $$\PageIndex{3}$$
Determine the direction and magnitude of the phase shift for $$f(x)=3\cos\left(x−\frac{\pi}{2}\right)$$.
$$\frac{\pi}{2}$$; right
Example $$\PageIndex{4}$$: Identifying the Vertical Shift of a Function
Determine the direction and magnitude of the vertical shift for $$f(x)=\cos(x)−3$$.
Solution
Let’s begin by comparing the equation to the general form $$y=A\cos(Bx−C)+D$$.
In the given equation, $$D=−3$$ so the shift is $$3$$ units downward.
Exercise $$\PageIndex{4}$$
Determine the direction and magnitude of the vertical shift for $$f(x)=3\sin(x)+2$$.
$$2$$ units up
How to: Given a sinusoidal function in the form $$f(x)=A\sin(Bx−C)+D$$,identify the midline, amplitude, period, and phase shift
1. Determine the amplitude as $$| A |$$.
2. Determine the period as $$P=\frac{2\pi}{| B |}$$.
3. Determine the phase shift as $$\frac{C}{B}$$.
4. Determine the midline as $$y=D$$.
Example $$\PageIndex{5}$$: Identifying the Variations of a Sinusoidal Function from an Equation
Determine the midline, amplitude, period, and phase shift of the function $$y=3\sin (2x)+1$$.
Solution
Let’s begin by comparing the equation to the general form $$y=A\sin (Bx−C)+D$$.
$$A=3$$, so the amplitude is $$| A |=3$$.
Next, $$B=2$$, so the period is $$P=\dfrac{2\pi}{| B |}=\dfrac{2\pi}{2}=\pi$$.
There is no added constant inside the parentheses, so $$C=0$$ and the phase shift is $$\dfrac{C}{B}=\dfrac{0}{2}=0$$.
Finally, $$D=1$$, so the midline is $$y=1$$.
Analysis
Inspecting the graph, we can determine that the period is $$\pi$$, the midline is $$y=1$$, and the amplitude is $$3$$. See Figure $$\PageIndex{14}$$.
Exercise $$\PageIndex{5}$$
Determine the midline, amplitude, period, and phase shift of the function $$y=\frac{1}{2}\cos \left(\frac{x}{3}−\frac{\pi}{3}\right)$$.
midline: $$y=0$$; amplitude: $$| A |=\frac{1}{2}$$; period: $$P=\frac{2\pi}{| B |}=6π$$; phase shift: $$\frac{C}{B}=\pi$$
Example $$\PageIndex{6}$$: Identifying the Equation for a Sinusoidal Function from a Graph
Determine the formula for the cosine function in Figure $$\PageIndex{15}$$.
Solution
To determine the equation, we need to identify each value in the general form of a sinusoidal function.
$$y=A\sin (Bx−C)+D$$
$$y=A\cos (Bx−C)+D$$
The graph could represent either a sine or a cosine function that is shifted and/or reflected. When $$x=0$$, the graph has an extreme point, $$(0,0)$$. Since the cosine function has an extreme point for $$x=0$$, let us write our equation in terms of a cosine function.
Let’s start with the midline. We can see that the graph rises and falls an equal distance above and below $$y=0.5$$. This value, which is the midline, is $$D$$ in the equation, so $$D=0.5$$.
The greatest distance above and below the midline is the amplitude. The maxima are $$0.5$$ units above the midline and the minima are $$0.5$$ units below the midline. So $$| A |=0.5$$. Another way we could have determined the amplitude is by recognizing that the difference between the height of local maxima and minima is $$1$$, so $$| A |=\frac{1}{2}=0.5$$. Also, the graph is reflected about the $$x$$-axis so that $$A=−0.5$$.
The graph is not horizontally stretched or compressed, so $$B=1$$; and the graph is not shifted horizontally, so $$C=0$$.
Putting this all together,
$$g(x)=−0.5\cos (x)+0.5$$
Exercise $$\PageIndex{6}$$
Determine the formula for the sine function in Figure $$\PageIndex{16}$$.
$$f(x)=\sin(x)+2$$
Example $$\PageIndex{7}$$: Identifying the Equation for a Sinusoidal Function from a Graph
Determine the equation for the sinusoidal function in Figure $$\PageIndex{17}$$.
Solution
With the highest value at $$1$$ and the lowest value at $$−5$$, the midline will be halfway between at $$−2$$. So $$D=−2$$.
The distance from the midline to the highest or lowest value gives an amplitude of $$| A |=3$$.
The period of the graph is $$6$$, which can be measured from the peak at $$x=1$$ to the next peak at $$x=7$$,or from the distance between the lowest points. Therefore, $$P=\dfrac{2\pi}{| B |}=6$$. Using the positive value for $$B$$,we find that
\begin{align*} B&=\dfrac{2\pi}{P}\\ &=\dfrac{2\pi}{6}\\ &=\dfrac{\pi}{3} \end{align*}
So far, our equation is either $$y=3\sin\left(\dfrac{\pi}{3}x−C\right)−2$$ or $$y=3\cos\left(\dfrac{\pi}{3}x−C\right)−2$$.For the shape and shift, we have more than one option. We could write this as any one of the following:
• a cosine shifted to the right
• a negative cosine shifted to the left
• a sine shifted to the left
• a negative sine shifted to the right
While any of these would be correct, the cosine shifts are easier to work with than the sine shifts in this case because they involve integer values. So our function becomes
\begin{align*} y&=3\cos \left (\frac{\pi}{3}x-\dfrac{\pi}{3} \right )-2 \qquad \text{or} \\ y&=-3\cos \left (\dfrac{\pi}{3}x+\dfrac{2\pi}{3} \right )-2 \end{align*}
Again, these functions are equivalent, so both yield the same graph.
Exercise $$\PageIndex{7}$$
Write a formula for the function graphed in Figure $$\PageIndex{18}$$.
two possibilities: $$y=4\sin\left(\dfrac{\pi}{5}x−\dfrac{\pi}{5}\right)+4$$ or $$y=−4\sin\left(\dfrac{\pi}{5}x+\dfrac{4\pi}{5}\right)+4$$
## Graphing Variations of $$y = \sin\space x$$ and $$y = \cos\space x$$
Throughout this section, we have learned about types of variations of sine and cosine functions and used that information to write equations from graphs. Now we can use the same information to create graphs from equations.
Instead of focusing on the general form equations
$$y=A\sin(Bx-C)+D \text{ and } y=A\cos(Bx-C)+D$$
we will let $$C=0$$ and $$D=0$$ and work with a simplified form of the equations in the following examples.
How to: Given the function $$y=A\sin(Bx)$$, sketch its graph.
1. Identify the amplitude, $$| A |$$.
2. Identify the period, $$P=\dfrac{2\pi}{| B |}$$.
3. Start at the origin, with the function increasing to the right if $$A$$ is positive or decreasing if $$A$$ is negative.
4. At $$x=\dfrac{\pi}{2| B |}$$ there is a local maximum for $$A>0$$ or a minimum for $$A<0$$, with $$y=A$$.
5. The curve returns to the x-axis at $$x=\dfrac{\pi}{| B |}$$.
6. There is a local minimum for $$A>0$$ (maximum for $$A<0$$) at $$x=\dfrac{3\pi}{2| B |}$$ with $$y=–A$$.
7. The curve returns again to the x-axis at $$x=\dfrac{\pi}{2| B |}$$.
Example $$\PageIndex{8}$$: Graphing a Function and Identifying the Amplitude and Period
Sketch a graph of $$f(x)=−2\sin\left(\dfrac{\pi x}{2}\right)$$.
Solution
Let’s begin by comparing the equation to the form $$y=A\sin(Bx)$$.
• Step 1. We can see from the equation that $$A=−2$$, so the amplitude is 2.
$$|A|=2$$
• Step 2. The equation shows that $$B=\dfrac{\pi}{2}$$, so the period is
\begin{align*} P&=\dfrac{2\pi}{\dfrac{\pi}{2}}\\ &=2\pi \cdot \dfrac{2}{\pi}\\ &=4 \end{align*}
• Step 3. Because $$A$$ is negative, the graph descends as we move to the right of the origin.
• Step 4. The $$x$$-intercepts are at the beginning of one period, $$x=0$$, the horizontal midpoints are at $$x=2$$ and at the end of one period at $$x=4$$.
The quarter points include the minimum at $$x=1$$ and the maximum at $$x=3$$. A local minimum will occur $$2$$ units below the midline, at $$x=1$$, and a local maximum will occur at $$2$$ units above the midline, at $$x=3$$. Figure $$\PageIndex{19}$$ shows the graph of the function.
Exercise $$\PageIndex{8}$$
Sketch a graph of $$g(x)=−0.8\cos(2x)$$. Determine the midline, amplitude, period, and phase shift.
midline: $$y=0$$; amplitude: $$| A |=0.8$$; period: $$P=\dfrac{2\pi}{| B |}=\pi$$; phase shift: $$\dfrac{C}{B}=0$$ or none
How to: Given a sinusoidal function with a phase shift and a vertical shift, sketch its graph
1. Express the function in the general form $$y=A\sin(Bx−C)+D$$ or $$y=A\cos(Bx−C)+D$$.
2. Identify the amplitude, $$| A |$$.
3. Identify the period, $$P=\dfrac{2\pi}{| B |}$$.
4. Identify the phase shift, $$\dfrac{C}{B}$$.
5. Draw the graph of $$f(x)=A\sin(Bx)$$ shifted to the right or left by $$\dfrac{C}{B}$$ and up or down by $$D$$.
Example $$\PageIndex{9}$$: Graphing a Transformed Sinusoid
Sketch a graph of $$f(x)=3\sin\left(\dfrac{\pi}{4x}−\dfrac{\pi}{4}\right)$$.
Solution
• Step 1. The function is already written in general form: $$f(x)=3\sin\left(\dfrac{\pi}{4x}−\dfrac{\pi}{4}\right)$$.This graph will have the shape of a sine function, starting at the midline and increasing to the right.
• Step 2. $$| A |=| 3 |=3$$. The amplitude is $$3$$.
• Step 3. Since $$| B |=| \dfrac{\pi}{4} |=\dfrac{\pi}{4}$$, we determine the period as follows.
\begin{align*} P&=\dfrac{2\pi}{|B|}\\ &=\dfrac{2\pi}{\dfrac{\pi}{4}}\\ &=2\pi \cdot \dfrac{4}{\pi}\\ &=8 \end{align*}
The period is $$8$$.
• Step 4. Since $$C=\dfrac{\pi}{4}$$, the phase shift is
$\dfrac{C}{B}=\dfrac{\dfrac{\pi}{4}}{\dfrac{\pi}{4}}=1$.
The phase shift is $$1$$ unit.
• Step 5. Figure $$\PageIndex{21}$$ shows the graph of the function.
Exercise $$\PageIndex{9}$$
Draw a graph of $$g(x)=−2\cos\left(\dfrac{\pi}{3}x+\dfrac{\pi}{6}\right)$$. Determine the midline, amplitude, period, and phase shift.
midline: $$y=0$$; amplitude: $$| A |=2$$; period: $$P=\dfrac{2\pi}{| B |}=6$$; phase shift: $$\dfrac{C}{B}=−\dfrac{1}{2}$$
Example $$\PageIndex{10}$$: Identifying the Properties of a Sinusoidal Function
Given $$y=−2cos\left(\dfrac{\pi}{2}x+\pi\right)+3$$, determine the amplitude, period, phase shift, and horizontal shift. Then graph the function.
Solution
Begin by comparing the equation to the general form.
$$y=A\cos(Bx−C)+D$$
• Step 1. The function is already written in general form.
• Step 2. Since $$A=−2$$, the amplitude is $$| A |=2$$.
• Step 3. $$| B |=\dfrac{\pi}{2}$$, so the period is $$P=\dfrac{2\pi}{| B |}=\dfrac{2\pi}{\dfrac{pi}{2}}=2\pi⋅\dfrac{2}{\pi}=4$$. The period is 4.
• Step 4. $$C=−\pi$$,so we calculate the phase shift as $$\dfrac{C}{B}=−\dfrac{\pi}{\dfrac{\pi}{2}}=−\pi⋅\dfrac{2}{\pi}=−2$$. The phase shift is $$−2$$.
• Step 5. $$D=3$$,so the midline is $$y=3$$, and the vertical shift is up $$3$$.
Since $$A$$ is negative, the graph of the cosine function has been reflected about the $$x$$-axis.
Figure $$\PageIndex{23}$$ shows one cycle of the graph of the function.
## Using Transformations of Sine and Cosine Functions
We can use the transformations of sine and cosine functions in numerous applications. As mentioned at the beginning of the chapter,circular motion can be modeled using either the sine or cosine function.
Example $$\PageIndex{11}$$: Finding the Vertical Component of Circular Motion
A point rotates around a circle of radius $$3$$ centered at the origin. Sketch a graph of the $$y$$-coordinate of the point as a function of the angle of rotation.
Solution
Recall that, for a point on a circle of radius $$r$$, the $$y$$-coordinate of the point is $$y=r \sin(x)$$, so in this case, we get the equation $$y(x)=3 \sin(x)$$. The constant $$3$$ causes a vertical stretch of the $$y$$-values of the function by a factor of $$3$$, which we can see in the graph in Figure $$\PageIndex{24}$$.
Analysis
Notice that the period of the function is still $$2\pi$$; as we travel around the circle, we return to the point $$(3,0)$$ for $$x=2\pi,4\pi,6\pi,$$....Because the outputs of the graph will now oscillate between $$–3$$ and $$3$$, the amplitude of the sine wave is $$3$$.
Exercise $$\PageIndex{10}$$
What is the amplitude of the function $$f(x)=7\cos(x)$$? Sketch a graph of this function.
Example $$\PageIndex{12}$$: Finding the Vertical Component of Circular Motion
A circle with radius $$3$$ ft is mounted with its center $$4$$ ft off the ground. The point closest to the ground is labeled $$P$$, as shown in Figure $$\PageIndex{26}$$. Sketch a graph of the height above the ground of the point $$P$$ as the circle is rotated; then find a function that gives the height in terms of the angle of rotation.
Solution
Sketching the height, we note that it will start $$1$$ ft above the ground, then increase up to $$7$$ ft above the ground, and continue to oscillate $$3$$ ft above and below the center value of $$4$$ ft, as shown in Figure $$\PageIndex{27}$$.
Although we could use a transformation of either the sine or cosine function, we start by looking for characteristics that would make one function easier to use than the other. Let’s use a cosine function because it starts at the highest or lowest value, while a sine function starts at the middle value. A standard cosine starts at the highest value, and this graph starts at the lowest value, so we need to incorporate a vertical reflection.
Second, we see that the graph oscillates $$3$$ above and below the center, while a basic cosine has an amplitude of $$1$$, so this graph has been vertically stretched by $$3$$, as in the last example.
Finally, to move the center of the circle up to a height of $$4$$, the graph has been vertically shifted up by $$4$$. Putting these transformations together, we find that
$$y=−3\cos(x)+4$$
Exercise $$\PageIndex{11}$$
A weight is attached to a spring that is then hung from a board, as shown in Figure $$\PageIndex{28}$$. As the spring oscillates up and down, the position $$y$$ of the weight relative to the board ranges from $$–1$$ in. (at time $$x=0$$) to $$–7$$ in. (at time $$x=π$$) below the board. Assume the position of $$y$$ is given as a sinusoidal function of $$x$$. Sketch a graph of the function, and then find a cosine function that gives the position $$y$$ in terms of $$x$$.
$$y=3\cos(x)−4$$
Example $$\PageIndex{13}$$: Determining a Rider’s Height on a Ferris Wheel
The London Eye is a huge Ferris wheel with a diameter of $$135$$ meters ($$443$$ feet). It completes one rotation every $$30$$ minutes. Riders board from a platform $$2$$ meters above the ground. Express a rider’s height above ground as a function of time in minutes.
Solution
With a diameter of $$135$$ m, the wheel has a radius of $$67.5$$ m. The height will oscillate with amplitude $$67.5$$ m above and below the center.
Passengers board $$2$$ m above ground level, so the center of the wheel must be located $$67.5+2=69.5$$ m above ground level. The midline of the oscillation will be at $$69.5$$ m.
The wheel takes $$30$$ minutes to complete $$1$$ revolution, so the height will oscillate with a period of $$30$$ minutes.
Lastly, because the rider boards at the lowest point, the height will start at the smallest value and increase, following the shape of a vertically reflected cosine curve.
• Amplitude: $$67.5$$, so $$A=67.5$$
• Midline: $$69.5$$, so $$D=69.5$$
• Period:$$30$$, so $$B=\dfrac{2\pi}{30}=\dfrac{\pi}{15}$$
• Shape: $$−\cos(t)$$
An equation for the rider’s height would be
$$y=−67.5\cos\left(\dfrac{\pi}{15}t\right)+69.5$$
where $$t$$ is in minutes and $$y$$ is measured in meters.
Media
Access these online resources for additional instruction and practice with graphs of sine and cosine functions.
## Key Equations
Sinusoidal functions $$f(x)=A\sin(Bx−C)+D$$ $$f(x)=A\cos(Bx−C)+D$$
## Key Concepts
• Periodic functions repeat after a given value. The smallest such value is the period. The basic sine and cosine functions have a period of $$2\pi$$.
• The function $$\sin x$$ is odd, so its graph is symmetric about the origin. The function $$\cos x$$ is even, so its graph is symmetric about the y-axis.
• The graph of a sinusoidal function has the same general shape as a sine or cosine function.
• In the general formula for a sinusoidal function, the period is $$P=\dfrac{2\pi}{| B |}$$. See Example $$\PageIndex{1}$$.
• In the general formula for a sinusoidal function, $$| A |$$ represents amplitude. If $$| A |>1$$, the function is stretched, whereas if $$| A |<1$$, the function is compressed. See Example $$\PageIndex{2}$$.
• The value $$\dfrac{C}{B}$$ in the general formula for a sinusoidal function indicates the phase shift. See Example $$\PageIndex{3}$$.
• The value $$D$$ in the general formula for a sinusoidal function indicates the vertical shift from the midline. See Example $$\PageIndex{4}$$.
• Combinations of variations of sinusoidal functions can be detected from an equation. See Example $$\PageIndex{5}$$.
• The equation for a sinusoidal function can be determined from a graph. See Example $$\PageIndex{6}$$ and Example $$\PageIndex{7}$$.
• A function can be graphed by identifying its amplitude and period. See Example $$\PageIndex{8}$$ and Example $$\PageIndex{9}$$.
• A function can also be graphed by identifying its amplitude, period, phase shift, and horizontal shift. See Example $$\PageIndex{10}$$.
• Sinusoidal functions can be used to solve real-world problems. See Example $$\PageIndex{11}$$, Example $$\PageIndex{12}$$, and Example $$\PageIndex{13}$$. |
# Arithmetic Sequences and Arithmetic Series
### Arithmetic Sequences / Progressions
The terms “sequence” and “progression” are interchangeable. An “arithmetic sequence” is the same thing as an “arithmetic progression”. This post uses the term “sequence”… but if you live in a place that tends to use the word “progression” instead, it means exactly the same thing. So, let’s investigate how to create an arithmetic sequence (also known as an arithmetic progression).
Pick a number, any number, and write it down. For example:
$5$
Now pick a second number, any number (I’ll choose 3), which we will call the common difference. Now add the common difference to the first number, then write their sum down to the right of the first number:
$5,~8$
Now, continue adding the common difference to the sum and writing the result down… over, and over, and over:
$5,~8,~11,~14,~17,~20,~23,~26,~29, ...$
By following this process, you have created an “Arithmetic Sequence”, a sequence of numbers that are all the same distance apart when graphed on a number line:
### Vocabulary and Notation
In the example above 5 is the first term, or starting term, of the sequence. To refer to the starting term of a sequence in a generic way that applies to any sequence, mathematicians use the notation
$a_1$
This notation is read as “A sub one” and means: the 1st value of the sequence represented by “a”. The one is a “subscript” (value written slightly below the line of text), and indicates the position of the term within the sequence. So $a_1$ represents the value of the first term of the sequence (5 in the example above), and $a_7$ represents the value of the seventh term of the sequence (23 in the example above).
### The Common Difference
Since all of the terms in an Arithmetic Sequence must be the same distance apart by definition (3 apart in the example above), the magnitude of this distance is given a formal name (the common difference) and is often referred to using the variable $d$ (for Difference). If you add this value to any term, you end up with the value of the next term. It can be calculated by subtracting the previous term from any term: $(a_2 - a_1)$ or $(a_7 - a_6)$, etc.
Every Arithmetic Sequence has a common difference between consecutive terms. Examples include:
$1,~3,~5,~7,~9,~...$
$10,~9,~8~,7,~6,~...$
$0,~0.4,~0.8,~0.12,~...$
The common difference can be positive or negative. It can be a whole number, a fraction, or even an irrational number. No matter what value it has, it will be the difference between all consecutive terms in that Arithmetic Sequence.
### Algebraic Description Of An Arithmetic Sequence
The existence of a common difference ($d$) allows us to calculate terms in a generic way:
$a_2~=~a_1+d$
$a_3~=~a_2+d$
$~~~~etc.$
Since every line above follows the same pattern, the whole process can be described a bit more generally and compactly by using a variable as a subscript:
$a_n~=~a_{n-1}+d$
This would be read as “A sub N is equal to A sub N-1 plus the common difference d”. If $a_n$ refers to the “Nth” term, then $a_{n-1}$ has a subscript that is one less than N, and therefore refers to the term that immediately precedes $a_n$. A more intuitive way of reading this equation is “Any term may be calculated by adding the common difference to the previous term”.
These insights allow a complete description of an Arithmetic Sequence to take a number of forms:
$5,~8,~11,~14, ...$
Specifying the first three or four terms is enough to demonstrate the common difference
$a_1~=~5,~~d~=~3$
Specify the first term and the common difference
$a_1~=~5,~~a_n~=~a_{n-1}+3$
Specify the first term, with a rule to get you from each term to the next
This is a “Recursive” definition (you must know the previous term)
$a_n~=~5+(n-1)(3)$
Specify a rule (based on the term number) for calculating the “Nth” term
This is a “Closed Form” definition (you only need to know the term number)
Note that if a sequence starts with a 5 then increases by 3 from one term to the next, this situation can be modeled using a linear equation with 5 as its y-intercept and 3 as its slope (with the domain restriction that “n” must be a positive integer). The last equation above uses this linear model, and provides the fastest way to calculate the Nth term of the sequence. Generalizing this linear equation approach leads to a description that applies to any Arithmetic Sequence:
$a_n~=~a_1+(n-1)d$
### Why Use (n-1) In The Equation?
If you know the first term of a sequence ($a_1$), how many common differences do you need to add to it to get to the second term of the sequence ($a_2$)? Since you seek the very next term, only one difference is needed:
$a_2~=~a_1+d$
How many common differences are needed to get from the first to the third term?
$a_3~=~a_1+d+d$
Now generalize the situation based on these two examples. When the term numbers were one apart (2 – 1 = 1), one common difference was needed to get from one to the other. When the term numbers were two apart (3 – 1 = 2), two common differences were needed to get from one to the other. We will always need as many common differences as the difference between the two term numbers. For the general case, to get from $a_1$ to $a_n$, what is the difference between the two term numbers? One less than the value of “n”, or “n – 1”.
Thus “n-1” values of “d” must be added to the value of $a_1$ in order to arrive at the Nth term.
### Solving Arithmetic Sequence Problems
How many possible “unknowns” does the equation for $a_n$ have?
$a_n~=~a_1+(n-1)d$
Four: $a_n,~a_1,~n,~and~d$. Therefore problems involving Arithmetic Sequences typically ask one of four questions:
What is the value of the Nth term? (Calculate the value for $a_n$)
What is the value of first term? (Solve for $a_1$)
Given a value, what term number must it be? (Solve for “n”)
What is the common difference? (Solve for “d”)
To answer one of the above questions, you must know (or be given enough information to determine) values for three of the “unknowns” in the equation above. For example, if you are told that $a_{12}=24$, you can conclude that when “n” is 12, $a_n$ is 24, so you know two of the three bits of information you would need to answer a question about this sequence. Most Arithmetic Sequence problems can by solved by:
1. Determining the values for three of the four unknowns in the equation for $a_n$
2. Substituting those values into the equation above
3. Solving for the only variable remaining
Some problems will be a little more complex, but you should still be able to use the information provided to determine values for three of the four unknowns.
For example, suppose the only information that a problem provides are values for the 10th and 15th terms. You can find “d” either by a) taking the difference between the two terms and dividing it by 5 (the number of common differences needed to get from the 10th to the 15th term), or b) treating the 10th term as $a_1$, and the 15th term as $a_6$, then using the equation for $a_n$ to find “d”
### Applications of Arithmetic Sequences in “Real Life”
Arithmetic Sequences can be thought of as linear equations with their domains restricted to integers. So they can model any situation that includes a constant rate of change, but where the only inputs that make sense are integers. Examples include:
• Manufacturing situations, where the total quantity of finished product produced depends on the number of machine cycles completed
• Product pricing, where the total price equals a fixed amount for shipping and handling plus an amount per unit ordered
• Game scores for games with a fixed point value per goal scored
### Arithmetic Series
The Nth term of a “series” is the sum of the first N terms of its underlying sequence.
Consider the Arithmetic Sequence described at the beginning of this post:
$5,~8,~11,~14,~...$
The series based on this sequence is:
$5,~13,~24,~38,~...$
The 3rd term of the Series is the sum of the first three terms of the underlying sequence, and is typically described using Sigma Notation with the formula for the Nth term of an Arithmetic Sequence (as derived above):
$S_3=\displaystyle\sum_{i=1}^{3}(5+(i-1)3)\\*~\\*~\\*=5+8+11\\*~\\*=24$
### Formula for the Nth Term
Just it is sometimes useful to have a formula for the Nth term of an Arithmetic Sequence, it is also useful to have a formula for the Nth term of an Arithmetic Series, which avoids having to a add up a long list of terms.
Arithmetic Sequences have a useful pattern to them that leads to the formula we seek. Suppose we need to find the sum of the first six terms of this Arithmetic Sequence:
$5,~8,~11,~14,~17,~20,~...$
The difference between successive terms in this sequence (“d”) is always 3 . What happens when, instead of adding the first two terms together, we add the pair of terms on outside first, then the next pair of “in” from them, etc.:
$5+20=25\\*~\\*8+17=25\\*~\\*11+14=25$
### How Does This Pattern Lead To A Formula?
Three insights lead us to a formula for the sum of an Arithmetic Sequence.
1) Why do the above pairs always add to the same number? Think about the second pair that was added above: $(a_2+a_5)$. $a_2$ is 3 greater than $a_1$, while $a_5$ is 3 less than $a_6$. So
$a_2+a_5\\*~\\*=(a_1+3)+(a_6-3)\\*~\\*=a_1+a_6+3-3\\*~\\*=a_1+a_6$
The two differences of three (one positive and one negative) cancel each other out, and the sum of the second pair will be the same as the sum of the first pair. This will be true for any Arithmetic Sequence (if you substitute “d” for “3” above).
2) What is the significance of the number that these number pairs add to (25 in the example above $=a_1+a_6$)? Consider what would happen if we were looking at the first seven (an odd number) terms of the sequence… there would be a term “left over” after we have paired off as many terms as we could, with each pair having the same sum as all the other pairs (28 in this case):
5, 8, 11, 14, 17, 20, 23
That remaining term will be the “middle” term in the sequence. Both members of each number pair (for example, 5 and 23) will be equally distant from that middle term, one to the left and one to the right. That makes the middle term the average of each of these “outer” pairs. The middle term will also be the average of the first “n” terms:
$\dfrac{(a_1+a_n)}{2}$ is the average of the first “n” terms
If there are an even number of terms, there won’t be a “middle” term but every pair will still have the same sum, and therefore the same average (equal to half their sum). Therefore the sum of each pair, the quantity that ended up always being 25 or 28 above, will always be twice the average of the first “n” terms in the sequence.
3) How can knowing the average of the first “n” terms help? Recall that the average of a set of terms is equal to the sum of all terms, divided by the total number of terms. If you know the average of a sequence, and you know the number of terms in the sequence, then you can easily calculate the sum of all terms:
$Average=\dfrac{(Sum~of~Terms)}{(Number~of~Terms)}$
Multiplying both sides of this equation by the Number of Terms produces:
$(Number~of~Terms)(Average)=(Sum~of~Terms)$
This provides an easy way to calculate the sum of the first “n” terms of an Arithmetic Sequence. Since we now know that
Average of first n terms $=\dfrac{(a_1+a_n)}{2}$
we can multiply it by “n” to arrive at a simple formula for the sum of the first “n” terms of any Arithmetic Sequence:
Sum of first n terms $=\displaystyle\sum_{i=1}^{n}a_i=n\cdot \dfrac{(a_1+a_n)}{2}$
Therefore, the formula that describes the Nth term of an Arithmetic Series (where the Nth term is the sum of the first N terms of the underlying Arithmetic Sequence) is:
$S_n=n\cdot \dfrac{(a_1+a_n)}{2}$
Note that $a_n$ is usually used to represent the value of the Nth term of an Arithmetic Sequence, while $S_n$ (S for Sum) is usually used to represent the value of the Nth term of an Arithmetic Series.
### Solving Arithmetic Series Problems
Just as with Arithmetic Sequence problems, there are four possible unknowns in an Arithmetic Series problem: $S_n$$n$, $a_1$, and $a_n$. So the four types of questions that are typically asked are:
What is the value of the Nth term? (Calculate the value for $S_n$)
What is the value of first term? (Solve for $a_1$)
Given a value, what term number must it be? (Solve for “n”)
What is the Nth term of the underlying Arithmetic Sequence? (Solve for $a_n$)
To answer one of the above questions, you must know (or be given enough information to determine) values for three of the “unknowns” in the equation above. From there, algebra skills should get you to an answer for the question.
### Applications of Arithmetic Series in “Real Life”
An Arithmetic Sequence describes something that is periodically growing in a linear fashion (by the same amount each time), and an Arithmetic Series describes the sum of the periodic values. Examples of Arithmetic Series include:
– The total number of seats in a fan-shaped auditorium, where each row has a two more seats than the previous one
– The total amount of grain produced over 10 years if farmers produce 50 tons more than the previous year every year
More things in our world seem to grow exponentially than linearly, so you will probably run into more applications of Geometric Series than Arithmetic Series. However, the mathematical approaches learned in working with Arithmetic Sequences and Series will serve as a good background for working with Geometric Sequences and Series.
### Whit Ford
Math tutor since 1992. Former math teacher, product manager, software developer, research analyst, etc.
## 3 thoughts on “Arithmetic Sequences and Arithmetic Series”
1. Anonymous says: |
Overview: Seating Arrangement
### What is Seating Arrangement
The seating arrangement is the logical arrangement of either objects or people in a logical manner. One has to either perform the arrangement to answer the questions or decode the predefined arrangement by applying the logical analysis.
• Seating arrangement questions are the most common problem types in all entrance exams containing the Logical Reasoning section.
• These concepts involve the arrangement of objects/ people in many possible ways.
The question based on sitting arrangement can be classified into the following types:
### Type 1: Person Sitting in a Circle / Around a Table
• In the questions of Type 1, the persons are sitting either around a table or circle. In either of the condition, the persons are facing the centre.
• The important point to be remembered is that the left side of the persons who is facing North, is just opposite of one, sitting opposite to him, who is facing South.
• From the diagram above, we observe that A is facing North and B is facing South. Also, the left side of A is just opposite to Right side of B. Similarly right side of A is just opposite of left side of B.
• Whenever we are presented with this kind of problem, the first step should be to locate the “Fulcrum,” i.e. the position around which we can locate the other positions.
• The next step is to draw the circle or the table and start the process of allocating the positions.
• In almost all the questions, the position of one person in relation to two other persons is given. We find that two different positions are possible.
• Let us say, we are given that A is sitting between G and H, just opposite to B.
In such a case, following are the two possibilities:
In Case 1, G is to the left of A and in Case 2, G is to the right of A
• If we are further provided with the information that C is sitting second to the right of H.
• We note that this information can only be true in case 1, because in case 2, G is already second to the right of H. So we can reject case 2 and continue with case 1 only.
• In the next step, the other positions given can be allocated up very easily and we can answer the questions given to us.
• Generally based on one diagram, four to five questions are asked in the examinations. Once we have completed the diagram, all the questions can be answered.
### Type 2: Linear Arrangement
• Any logical arrangement of objects or people either horizontally or vertically or diagonally is known as Linear arrangement.Example: The row of the travellers in a train, students in a prayer hall, etc.
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# How to Find Direction Ratios and Direction Cosines?
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Last updated date: 20th Sep 2024
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## Vector: Direction Cosines
Before discussing the directional cosines of a vector, let us discuss the position vector. Just like the name suggests, a position vector indicates the position of any point relative with respect to any reference origin.
Consider any arbitrary point in three-dimensional space having the coordinates (x,y,z) as concerning the origin O (0,0,0).
The position vector in the above figure is given as |PQ| = r = √(x - 0)2 + (y - 0)2 + (z - 0)2
|PQ| = r = √x2 + y2 + z2
## Direction Cosines
Consider the following figure that represents a vector P in space with variable O being the reference origin of the vector P. Let the position vector make a positive angle (anticlockwise direction) of α, β, and γ with the positive x, y, and z-axis respectively. These angles are known as direction angles. When we take the cosine of these angles, we can find out the direction cosines. Taking direction cosines makes it easy to represent the direction of a vector in terms of angles for reference.
$cos \alpha = \frac{x}{|\overline{r}|}$
$cos \beta = \frac{y}{|\overline{r}|}$
$cos \gamma = \frac{z}{|\overline{r}|}$
This is how the cosines of the directions of a vector are represented mathematically.
$cos \alpha = \frac{x}{\sqrt{x^{2} + y^{2} + z^{2}}}$
$cos \beta = \frac{y}{\sqrt{x^{2} + y^{2} + z^{2}}}$
$cos \gamma = \frac{z}{\sqrt{x^{2} + y^{2} + z^{2}}}$
## Direction Ratios
The product of the magnitude of any given vector can be represented with point P, and the cosines of direction on the three axes, i.e.
a = lr
b = mr
c = nr
Where,
l = direction of the cosine on the axis X.
m = direction of the cosine on the axis Y.
n = direction of the cosine on the axis Z.
This helps to understand that lr, mr, and nr are in proportion to direction cosines. Hence, they are called direction ratios and are represented by the variables a, b and c.
Where the axes l, m, n represent the respective direction cosines of any given vector on the axes X, Y, Z respectively. We can see that lr, mr, nr are in proportion to the direction cosines and these are called the direction ratios and they are denoted by a, b, c.
$\frac{l}{a} = \frac{1}{\sqrt{x^{2} + y^{2} + z^{2}}}$
$\frac{m}{b} = \frac{1}{\sqrt{x^{2} + y^{2} + z^{2}}}$
$\frac{n}{c} = \frac{1}{\sqrt{x^{2} + y^{2} + z^{2}}}$
From the above theory, we have learned about direction ratios and direction cosines. Let us apply this knowledge and solve some problems.
### Solved Examples
1. Consider the point A(x, y, z) having coordinates (3, 4, 5). Find the direction ratios and the direction cosines with the origin point being O(0, 0, 0).
We have learned that,
$cos \alpha = \frac{x}{|\overline{r}|}$
$cos \beta = \frac{y}{|\overline{r}|}$
$cos \gamma = \frac{z}{|\overline{r}|}$
$\Rightarrow \sqrt{x^{2} + y^{2} + z^{2}} = \sqrt{3^{2} + 4^{2} + 5^{2}}$
$\Rightarrow \sqrt{x^{2} + y^{2} + z^{2}} = \sqrt{50} = 5\sqrt{2}$
Hence, we can conclude that:
$l = cos \alpha = \frac{3}{5\sqrt{2}}$
$m = cos \beta = \frac{4}{5\sqrt{2}}$
$n = cos \gamma = \frac{5}{5\sqrt{2}} = \frac{1}{\sqrt{2}}$
The direction ratio of the given point A(x, y, z) will be 3:4:5.
x = 3
y = 4
z = 5 |
# Triangle Inequality: Activity-Triangle Inequality Theorem and Triangle Inequality Theorem
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A triangle is a three-sided polygon. It has three sides and three angles. The three sides and three angles share an important relationship. Term “inequality” represents the meaning “not equal”.
## What is Triangle Inequality?
“Triangle inequality” is meant for any triangles. Let us take a, b, and c are the lengths of the three sides of a triangle, in which no side is being greater than the side c, then the triangle inequality states that,
This states that the sum of any two sides of a triangle is greater than or equal to the third side of a triangle.
## Activity - Triangle Inequality Theorem
Activity 1: On a paper mark two points Y and Z and join them to form a straight line. Mark another point X outside the line lying on the same plane of the paper. Join XY as shown.
Now mark another point X’ on the line segment XY, join X’Z. Similarly mark X’’ and join X’’Z with dotted lines as shown.
From the above figure we can easily deduce that if we keep on decreasing the length of side XY such that XY> X’Y> X’’Y> X’’’Y the angle opposite to side XY also decreases i.e. ∠XZY >∠X’ZY >∠X’’ZY >∠X’’’ZY. Thus, from the above activity we can infer that if we keep on increasing one side of a triangle then the angle opposite to it increases.
Now let us try out another activity.
Activity: Draw 3 scalene triangles on a sheet of paper as shown.
Let us consider fig. (i). In is the longest side and is the shortest. We observe that is the largest in measure and is the smallest. Similarly, in is the largest side and is the smallest and is the largest in measure and is the smallest. In the last figure also the same kind of pattern is followed i.e. side is largest and so is the opposite to it.
## Triangle Inequality Theorem
Let us consider the triangle. The following are the triangle inequality theorems.
Theorem 1: In a triangle, the side opposite to the largest side is greatest in measure.
The converse of the above theorem is also true according to which in a triangle the side opposite to a greater angle is the longest side of the triangle.
In the above fig., since is the longest side, the largest angle in the triangle is
Another theorem which follows can be stated as:
Theorem 2: The sum of lengths of any two sides of a triangle is greater than the length of its third side.
According to triangle inequality,
Example: In and , which side of the triangle is the longest?
Solution:
• Here given,
• In
• For the finding of other triangle, using angle sum property of triangle.
Put the value of in above equation.
Sum of the angle P and angle Q.
• So, is the largest angle.
• According to Theorem 1, In a triangle, the side opposite to the largest side is greatest in measure.
• Here largest angle is , opposite side of the triangle is the largest side.
• is the longest side. |
Diagonally dominant matrix
In mathematics, a square matrix is said to be diagonally dominant if, for every row of the matrix, the magnitude of the diagonal entry in a row is greater than or equal to the sum of the magnitudes of all the other (off-diagonal) entries in that row. More precisely, the matrix ${\displaystyle A}$ is diagonally dominant if
${\displaystyle |a_{ii}|\geq \sum _{j\neq i}|a_{ij}|\ \ \forall \ i}$
where ${\displaystyle a_{ij}}$ denotes the entry in the ${\displaystyle i}$th row and ${\displaystyle j}$th column.
This definition uses a weak inequality, and is therefore sometimes called weak diagonal dominance. If a strict inequality (>) is used, this is called strict diagonal dominance. The unqualified term diagonal dominance can mean both strict and weak diagonal dominance, depending on the context.[1]
Variations
The definition in the first paragraph sums entries across each row. It is therefore sometimes called row diagonal dominance. If one changes the definition to sum down each column, this is called column diagonal dominance.
Any strictly diagonally dominant matrix is trivially a weakly chained diagonally dominant matrix. Weakly chained diagonally dominant matrices are non-singular and include the family of irreducibly diagonally dominant matrices. These are irreducible matrices that are weakly diagonally dominant, but strictly diagonally dominant in at least one row.
Examples
The matrix
${\displaystyle A={\begin{bmatrix}3&-2&1\\1&3&2\\-1&2&4\end{bmatrix}}}$
is weakly diagonally dominant because
${\displaystyle |a_{11}|\geq |a_{12}|+|a_{13}|}$ since ${\displaystyle |+3|\geq |-2|+|+1|}$
${\displaystyle |a_{22}|\geq |a_{21}|+|a_{23}|}$ since ${\displaystyle |3|\geq |+1|+|+2|}$
${\displaystyle |a_{33}|\geq |a_{31}|+|a_{32}|}$ since ${\displaystyle |+4|\geq |-1|+|+2|}$.
The matrix
${\displaystyle B={\begin{bmatrix}-2&2&1\\1&3&2\\1&-2&0\end{bmatrix}}}$
is not diagonally dominant because
${\displaystyle |b_{11}|<|b_{12}|+|b_{13}|}$ since ${\displaystyle |-2|<|+2|+|+1|}$
${\displaystyle |b_{22}|\geq |b_{21}|+|b_{23}|}$ since ${\displaystyle |+3|\geq |+1|+|+2|}$
${\displaystyle |b_{33}|<|b_{31}|+|b_{32}|}$ since ${\displaystyle |+0|<|+1|+|-2|}$.
That is, the first and third rows fail to satisfy the diagonal dominance condition.
The matrix
${\displaystyle C={\begin{bmatrix}-4&2&1\\1&6&2\\1&-2&5\end{bmatrix}}}$
is strictly diagonally dominant because
${\displaystyle |c_{11}|>|c_{12}|+|c_{13}|}$ since ${\displaystyle |-4|>|+2|+|+1|}$
${\displaystyle |c_{22}|>|c_{21}|+|c_{23}|}$ since ${\displaystyle |+6|>|+1|+|+2|}$
${\displaystyle |c_{33}|>|c_{31}|+|c_{32}|}$ since ${\displaystyle |+5|>|+1|+|-2|}$.
Applications and properties
The following results can be proved trivially from Gershgorin's circle theorem. Gershgorin's circle theorem itself has a very short proof.
A strictly diagonally dominant matrix (or an irreducibly diagonally dominant matrix[2]) is non-singular.
A Hermitian diagonally dominant matrix ${\displaystyle A}$ with real non-negative diagonal entries is positive semidefinite. This follows from the eigenvalues being real, and Gershgorin's circle theorem. If the symmetry requirement is eliminated, such a matrix is not necessarily positive semidefinite. For example, consider
${\displaystyle {\begin{pmatrix}-2&2&1\end{pmatrix}}{\begin{pmatrix}1&1&0\\1&1&0\\1&0&1\end{pmatrix}}{\begin{pmatrix}-2\\2\\1\end{pmatrix}}<0.}$
However, the real parts of its eigenvalues remain non-negative by Gershgorin's circle theorem.
Similarly, a Hermitian strictly diagonally dominant matrix with real positive diagonal entries is positive definite.
No (partial) pivoting is necessary for a strictly column diagonally dominant matrix when performing Gaussian elimination (LU factorization).
The Jacobi and Gauss–Seidel methods for solving a linear system converge if the matrix is strictly (or irreducibly) diagonally dominant.
Many matrices that arise in finite element methods are diagonally dominant.
A slight variation on the idea of diagonal dominance is used to prove that the pairing on diagrams without loops in the Temperley–Lieb algebra is non-degenerate.[3] For a matrix with polynomial entries, one sensible definition of diagonal dominance is if the highest power of ${\displaystyle q}$ appearing in each row appears only on the diagonal. (The evaluations of such a matrix at large values of ${\displaystyle q}$ are diagonally dominant in the above sense.)
Notes
1. ^ For instance, Horn and Johnson (1985, p. 349) use it to mean weak diagonal dominance.
2. ^ Horn and Johnson, Thm 6.2.27.
3. ^ K.H. Ko and L. Smolinski (1991). "A combinatorial matrix in 3-manifold theory". Pacific J. Math. 149: 319–336. |
# Laws of Exponents
Exponents are also called Powers or Indices
The exponent of a number says how many times to use the number in a multiplication.
In this example: 82 = 8 × 8 = 64
In words: 82 could be called "8 to the second power", "8 to the power 2" or simply "8 squared"
Try it yourself:
So an Exponent saves us writing out lots of multiplies!
### Example: a7
a7 = a × a × a × a × a × a × a = aaaaaaa
Notice how we wrote the letters together to mean multiply? We will do that a lot here.
## The Key to the Laws
Writing all the letters down is the key to understanding the Laws
### Example: x2x3 = (xx)(xxx) = xxxxx = x5
Which shows that x2x3 = x5, but more on that later!
So, when in doubt, just remember to write down all the letters (as many as the exponent tells you to) and see if you can make sense of it.
## All you need to know ...
The "Laws of Exponents" (also called "Rules of Exponents") come from three ideas:
The exponent says how many times to use the number in a multiplication.
A negative exponent means divide, because the opposite of multiplying is dividing
A fractional exponent like 1/n means to take the nth root:
If you understand those, then you understand exponents!
And all the laws below are based on those ideas.
## Laws of Exponents
Here are the Laws (explanations follow):
Law Example
x1 = x 61 = 6
x0 = 1 70 = 1
x-1 = 1/x 4-1 = 1/4
xmxn = xm+n x2x3 = x2+3 = x5
xm/xn = xm-n x6/x2 = x6-2 = x4
(xm)n = xmn (x2)3 = x2×3 = x6
(xy)n = xnyn (xy)3 = x3y3
(x/y)n = xn/yn (x/y)2 = x2 / y2
x-n = 1/xn x-3 = 1/x3
And the law about Fractional Exponents:
## Laws Explained
The first three laws above (x1 = x, x0 = 1 and x-1 = 1/x) are just part of the natural sequence of exponents. Have a look at this:
Example: Powers of 5
.. etc..
52 1 × 5 × 5 25
51 1 × 5 5
50 1 1
5-1 1 ÷ 5 0.2
5-2 1 ÷ 5 ÷ 5 0.04
.. etc..
Look at that table for a while ... notice that positive, zero or negative exponents are really part of the same pattern, i.e. 5 times larger (or 5 times smaller) depending on whether the exponent gets larger (or smaller).
## The law that xmxn = xm+n
With xmxn, how many times do we end up multiplying "x"? Answer: first "m" times, then by another "n" times, for a total of "m+n" times.
### Example: x2x3 = (xx)(xxx) = xxxxx = x5
So, x2x3 = x(2+3) = x5
## The law that xm/xn = xm-n
Like the previous example, how many times do we end up multiplying "x"? Answer: "m" times, then reduce that by "n" times (because we are dividing), for a total of "m-n" times.
### Example: x4/x2 = (xxxx) / (xx) = xx = x2
So, x4/x2 = x(4-2) = x2
(Remember that x/x = 1, so every time you see an x "above the line" and one "below the line" you can cancel them out.)
This law can also show you why x0=1 :
## The law that (xm)n = xmn
First you multiply "m" times. Then you have to do that "n" times, for a total of m×n times.
### Example: (x3)4 = (xxx)4 = (xxx)(xxx)(xxx)(xxx) = xxxxxxxxxxxx = x12
So (x3)4 = x3×4 = x12
## The law that (xy)n = xnyn
To show how this one works, just think of re-arranging all the "x"s and "y"s as in this example:
## The law that (x/y)n = xn/yn
Similar to the previous example, just re-arrange the "x"s and "y"s
## The law that
OK, this one is a little more complicated!
I suggest you read Fractional Exponents first, or this may not make sense.
Anyway, the important idea is that:
x1/n = The n-th Root of x
And so a fractional exponent like 43/2 is really saying to do a cube (3) and a square root (1/2), in any order.
Just remember from fractions that m/n = m × (1/n):
### Example:
The order does not matter, so it also works for m/n = (1/n) × m:
## Exponents of Exponents ...
432
We do the exponent at the top first, so we calculate it this way:
## And That Is It!
If you find it hard to remember all these rules, then remember this:
you can work them out when you understand the
### Oh, One More Thing ... What if x = 0?
Positive Exponent (n>0) 0n = 0 Negative Exponent (n<0) 0-n is undefined (because dividing by 0 is undefined) Exponent = 0 00 ... ummm ... see below!
### The Strange Case of 00
There are different arguments for the correct value of 00
00 could be 1, or possibly 0, so some people say it is really "indeterminate":
x0 = 1, so ... 00 = 1 0n = 0, so ... 00 = 0 When in doubt ... 00 = "indeterminate"
Hard: |
# How do you solve Log_2 x + log_4 x + log_8 x = 11 ?
Mar 8, 2018
$x = 64$
#### Explanation:
As ${\log}_{a} b = \log \frac{b}{\log} a$, we can write ${\log}_{2} x + {\log}_{4} x + {\log}_{8} x = 11$ as
$\log \frac{x}{\log} 2 + \log \frac{x}{\log} 4 + \log \frac{x}{\log} 8 = 11$
or $\log \frac{x}{\log} 2 + \log \frac{x}{2 \log 2} + \log \frac{x}{3 \log 2} = 11$
or $\log \frac{x}{\log} 2 \left(1 + \frac{1}{2} + \frac{1}{3}\right) = 11$
or $\log \frac{x}{\log} 2 \left(\frac{6 + 3 + 2}{6}\right) = 11$
or $\log \frac{x}{\log} 2 \left(\frac{11}{6}\right) = 11$
or $\log \frac{x}{\log} 2 = 11 \times \frac{6}{11} = 6$
i.e. $\log x = 6 \log 2 = \log {2}^{6}$
and $x = {2}^{6} = 64$ |
# The sum of the lengths of all sides of the quadrilateral is 66 dm. Find the length of each side, knowing
The sum of the lengths of all sides of the quadrilateral is 66 dm. Find the length of each side, knowing that these lengths are equal to consecutive numbers.
If the lengths of the sides are consecutive numbers. so they differ from each other by 1, for example: 10, 11 = 10 +1, 12 = 10 + 2, 13 = 10 + 3.
Let the length of the first side be x dm, then the length of the second side is (x + 1) dm, the length of the third side is (x + 2) dm and the fourth is (x + 3). The sum of all sides of the quadrangle is (x + (x + 1) + (x + 2) + (x + 3)) dm or 66 dm. Let’s make an equation and solve it.
x + (x + 1) + (x + 2) + (x + 3) = 66;
x + x + 1 + x + 2 + x + 3 = 66;
4x + 6 = 66;
4x = 66 – 6;
4x = 60;
x = 60: 4;
x = 15 (dm) – the first side;
15 + 1 = 16 (dm) – the second side;
15 + 2 = 17 (dm) – third party;
15 + 3 = 18 (dm) – the fourth side.
Answer. 15 dm, 16 dm, 17 dm, 18 dm.
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Which square, which circle, and which triangle has the closest area to the doughnut shape on the left? The drawings are to scale, so you might be able to judge it, as well as working the actual areas out.
Circle: 3.5.
Square: 3.1.
Triangle: 4.7.
Doughnut
The area of a circle is π x Radius2.
The larger circle has diameter = 4, therefore the radius is 2, and the area is π x 22 = 4π.
The smaller circle has diameter = 2, therefore the radius is 1, and the area is π x 12 = π.
Therefore the shaded area is 4π - π = 3π ≈ 9.42.
Circle
The area of a circle is π x Radius2.
The circle with diameter 3.1 has a radius of 1.55 and an area of π x 1.552 = 7.55.
The circle with diameter 3.5 has a radius of 1.75 and an area of π x 1.752 = 9.62 (** closest match).
The circle with diameter 3.9 has a radius of 1.95 and an area of π x 1.952 = 11.95.
Square
The area of a square is Side x Side.
The square with side 3.1 has an area of 3.1 x 3.1 = 9.61 (** closest match).
The square with side 3.5 has an area of 3.5 x 3.5 = 12.25.
The square with side 3.9 has an area of 3.9 x 3.9 = 15.21.
Triangle
The area of a triangle is ½ x Base x Height. Using Pythagoras' theorem it can be shown that the area of an equilateral triangle is Sqrt(3) x Base2 ÷ 4.
The triangle with side 3.9 has an area of Sqrt(3) x 3.9 x 3.9 ÷ 4 = 6.59.
The triangle with side 4.3 has an area of Sqrt(3) x 4.3 x 4.3 ÷ 4 = 8.01.
The triangle with side 4.7 has an area of Sqrt(3) x 4.7 x 4.7 ÷ 4 = 9.57 (** closest match).
Back to the puzzles...
Top 10 Illusions
Shadow Illusion Level Balance? Same Length Red Lines? Spinning Dancer Purple Nurple |
1. ## Derivative and graphing!!!
The graphs of the function (given in blue) and (given in red) are plotted above. Find each of the following:
= ?
= ?
2. If
find the values of and that make differentiable everywhere. m= ? and b=?
2. Hello, Kayla_N!
The graphs of the function $\displaystyle f$ (in blue) and $\displaystyle g$ (in red) are plotted above.
Find the following: .$\displaystyle (1)\;(fg)'(1)\qquad (2)\;\left(\frac{f}{g}\right)\!'(1)$
We have piecewise functions . . .
$\displaystyle f(x) \;=\;\bigg\{ \begin{array}{cccc}\frac{3}{2}x & & 0 < x < 3 \\ \frac{9}{2}-x & & x > 3 \end{array}$
$\displaystyle g(x) \;=\;\bigg\{\begin{array}{cccc}4-x & & x < 2 \\ \frac{1}{2}x + 2 & & x > 2 \end{array}$
If $\displaystyle x = 1$, we have: .$\displaystyle \begin{array}{ccc}f(x) \;=\; \frac{3}{2}x \\ g(x) \;=\; 4 - x \end{array}$
$\displaystyle (1)\;\;(fg) \;=\;\tfrac{3}{2}x(4 - x) \;=\;6x - \tfrac{3}{2}x^2$
Then:. . $\displaystyle (fg)' \;=\;6 - 3x$
Therefore: .$\displaystyle (fg)'(1) \;=\;6-3(1) \;=\;\boxed{3}$
$\displaystyle (2)\;\;\left(\frac{f}{g}\right) \;=\;\frac{\frac{3}{2}x}{4-x} \;=\;\frac{3}{2}\cdot\frac{x}{4-x}$
Then: .$\displaystyle \left(\frac{f}{g}\right)\!' \;=\;\frac{3}{2}\cdot\frac{(4-x)\!\cdot\!1 - x(-1)}{(4-x)^2} \;=\;\frac{6}{(4-x)^2}$
Therefore: .$\displaystyle \left(\frac{f}{g}\right)\!'(1) \;=\;\frac{6}{(4-1)^2} \;=\;\frac{6}{9} \;=\;\boxed{\frac{2}{3}}$
3. Thanks for explaining steps by steps. Thats really helped and i understand it now. |
June 27, 2022
All cards are divided into 4 fits. There are 2 black fits– spades (♠) and clubs (♣) and 2 red fits– hearts (♥) and diamonds() In each fit there are 13 cards consisting of a 2, 3, 4, 5, 6, 7, 8, 9, 10, a jack, a queen, a king and an ace.
Consequently, one may likewise ask, how many sevens are in a deck of 52 playing cards?
There are 4 cards of each worth in a requirement deck of playing cards, one of each fit: spades, hearts, clubs, and diamonds.
How many jacks are in a deck of 52 cards?
4 Jacks
How many sixes are in a deck of 52 cards?
A basic deck has 52 cards (4 fits of 13). Each fit (clubs ♣, diamonds., hearts ♥, or spades ♠) includes an ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, jack, queen, and king.
1
4 kings 2
## How many Queen of Hearts is in a deck of cards?
A basic deck of 52 cards consist of 4 fit, clubs (♣), diamonds (), hearts (♥) and spades (♠). Each fit includes one Queen So, overall there are 4 queen in a deck of 52 cards. 3
## How many black court card are in a deck of 52 cards?
There are 12 court card in a deck of 52 cards (a jack, queen and king of all 4 fits). Half of the 12 court card are black, and half are red. So we can divide your issue into 2 cases: 1) The court card you choose is black. 4
## How many red faces are in a deck of cards?
There are 12 court card in thedeck 6 of the court card are red (diamonds or hearts) and 6 of the court card are black (areas or clubs). let A be the occasion that you have a red card. Considering that there are 26 red cards in the deck, then p( A) = 26/52 5
## How many fits are in a deck of cards?
2.1 In this research, we will think about “regular” decks of playing cards which have 52 cards, with 13 of each of the 4 fits (Hearts, Spades, Diamonds and Clubs), with each fit having the 13 ranks (Ace, 2, 3,, 9, 10, Jack, Queen, King).
6
## How many even numbers are there in a deck of cards?
If you disregard the court card (the Jack, Queen, and King) then you have 5 even-numbered cards in each fit (the 2, 4, 6, 8, and 10). There are 4 fits in a basic deck, so you have a overall of 20 even numbered cards in a deck of 52. 7
## How many deals with are in a deck of cards?
A basic deck of playing cards has 52 cards in it burglarized 4 fits of 13 cards each. Each fit has 3 face cards (a King, a Queen, and a Jack) so 4 fits increased by 3 face cards equates to 12 face cards in a basic deck of cards. 8
## Is ace card is a court card?
Jacks, queens and kings are the court card in a deck of playing cards due to the fact that these cards have photos. As revealed in image listed below: Nevertheless, the Ace card is rewarded with a duality of being utilized as a high or low card by some video games. 9
## What is a fit in a deck of cards?
5 ° Measurement is an 80-card deck presented in 2007. The 5 fits are Hearts (red), Spades (black), Clubs (green), Diamonds (yellow) and Stars (blue). Each fit has 16 cards: 1 to 10, King, Queen, Jack, Princess, Ace (unique from 1) and a Joker. 10
## Why are there 52 cards in a regular pack of cards?
Why Are There 52 Cards? The 4 fits represent the 4 seasons, while the 52 cards represent the 52 cards in a year. The thirteen cards per fit represent the thirteen lunar cycles, and even the 365 days of the year are marked by the number of overall points (fit signs) in a total deck. 11
## What is the spade a sign of?
The spade represents a leaf of the “cosmic” tree, and hence life. Together with its buddy fit, clubs, spades represent fall and winter season and the power of darkness. In the Tarot, they represent intelligence, action, air, and death 12
## Why is the ace of spades so unique?
Apparently, United States soldiers thought that Vietnamese customs held the importance of the spade to imply death and ill-fortune and in a quote to scare and demoralize Viet Cong soldiers, it prevailed practice to mockingly leave an ace of spades on the bodies of eliminated Vietnamese and even to litter the forested premises and 13
## What do the 52 cards in a deck represent?
The 4 fits– hearts, clubs, spades and diamonds– represent the 4 seasons. On the other hand, the 13 cards in each fit represent the 13 stages of the lunar cycle. And did you ever see that there are 52 cards in a deck, simply as there are 52 weeks in a year? 14
## What is the significance of cards in a deck?
The deck is divided into 4 fits: the Spades, the Diamonds, the Clubs, and Hearts. Likewise, each fit is divided into thirteen ranks: Ace through King. Each Card has a various significance in ARCANA ARCANORUM. The Ace of a fit has the significance of that fit– for example, the Ace of Diamonds indicates materiality. 15
## How many hearts are there in a deck of 52 cards?
A single card is drawn from a basic 52-deck of cards with 4 fits: hearts, clubs, diamonds, and spades; there are 13 cards per fit. 16
## How many 10s are in a deck of 52 cards?
A basic deck has 52 cards (4 fits of 13). Each fit (clubs ♣, diamonds., hearts ♥, or spades ♠) includes an ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, jack, queen, and king. 17
## How many red cards are in a deck of 52 cards?
There are 12 court card in thedeck 6 of the court card are red (diamonds or hearts) and 6 of the court card are black (areas or clubs). let A be the occasion that you have a red card. Considering that there are 26 Red Cards in the deck, then p( A) = 26/52 18
## How many aces are in a deck of 52 cards?
The likelihood of getting an ace or a king is 8/52 because in a deck of 52 cards there are 4 aces and 4 kings, which amounts to 8. The likelihood of not getting an ace or king is merely 52/52 – 8/52 = 44/52 19
4 Jacks |
GreeneMath.com - Factoring Trinomials with a Leading Coefficient that is not 1 using the Reverse FOIL Method Lesson
# In this Section:
In this lesson, we review how to factor a trinomial into the product of two binomials when the leading coefficient is not one. For this scenario, the process is much more tedious. We generally can use two different methods: factoring by grouping, or reverse FOIL. In order to factor a trinomial with grouping, we first re-write the trinomial as a four term polynomial. We do this by finding two integers whose product is a • c and whose sum is b. We use those two integers to expand the middle term; we can then use factoring by grouping to attain the product of two binomials. The alternative method uses reverse FOIL. When we use reverse FOIL, we must undo the FOIL process. In most cases, this process is more tedious than using the factoring by grouping method. We will also look at some special case scenarios that require us to factor out the GCF before we begin or factor when two variables are involved.
Sections:
# In this Section:
In this lesson, we review how to factor a trinomial into the product of two binomials when the leading coefficient is not one. For this scenario, the process is much more tedious. We generally can use two different methods: factoring by grouping, or reverse FOIL. In order to factor a trinomial with grouping, we first re-write the trinomial as a four term polynomial. We do this by finding two integers whose product is a • c and whose sum is b. We use those two integers to expand the middle term; we can then use factoring by grouping to attain the product of two binomials. The alternative method uses reverse FOIL. When we use reverse FOIL, we must undo the FOIL process. In most cases, this process is more tedious than using the factoring by grouping method. We will also look at some special case scenarios that require us to factor out the GCF before we begin or factor when two variables are involved. |
# Proof that There Are Only Three Regular Tessellations
Tessellation is the process of covering the plane with shapes without gaps and overlaps. Tiles on walls and floors are the most common tessellations.
In this post, we are going to prove that the there are only three regular polygons that can tile the plane. Tiling the plane with regular polygons is called regular tessellation. These are equilateral triangles, squares, and regular hexagons. This proof only requires simple algebra. We will use the notation $\{a,b \}$, where $a$ is the number angles of the polygon and $b$ is the number of vertices that meets at a point. For example, a square has four angles, and at every point on the tessellation, four vertices meet at a point. So, we can represent square as $\{ 4,4 \}$. Another example is the regular hexagon. A hexagon has 6 sides, and at every point 3 vertices meet (see figure below), so a hexagon can be represented as $\{ 6,3 \}$
Theorem: There are only three polygons that can tessellate the plane: equilateral triangles, squares, and regular hexagons.
Proof:
We know that the angle sum of a polygon is $180(a - 2)$ where $a$ is the number of angles (or sides). This means that each angle has measure $\frac{180(a-2)}{a}$.
For example, pentagon has 5 angles. Then, the angle sum is $180(5-2) = 180(3) = 540$. Now, to get the measure of each angle, we divide $540$ by $5$ (the number of sides/angles) and this gives us $108$ degrees.
Now, notice that to be able to tessellate without gaps or overlaps, the measure of the angle of each vertex times the number of vertices that meet at a point is equal degrees 360 (Can you see why?).
But we know from above that the measure of each angle of a regular polygon is
$\frac{180(a-2)}{a}$ and the number of vertices that meet at a point is $b$.
Therefore,
$\frac{180(a - 2)}{a}(b) = 360$
Dividing both sides by 180 we have
$\frac{(a - 2)}{a}(b) = 2$
Multiplying both sides by $a$,
$(a - 2)(b) = 2a$
$ab - 2b = 2a$
Now, this seems not going anywhere, but adding 4 to both sides we have
$ab - 2b + 4 = 2a + 4$
$ab - 2b + 4 - 2a = 4$
Factoring, we have
$b(a - 2) - 2(a - 2) = 4$
$(a - 2)(b-2) = 4$
Notice that the only possible values to get a 4 is (4)(1), (2)(2), and (1)(4).
Case 1
$a - 2 = 4$, $a = 6$
$b - 2 = 1$, $b = 3$
Therefore, $\{a,b \} = \{ 6,3 \}$.
Case 2
$a - 2 = 2$, $a = 4$
$b - 2 = 2$, $b = 4$
Therefore, ${a, b} = \{ 4,4 \}$
Case 3
$a - 2 = 1$, $a = 3$
$b - 2 = 4$, $b = 6$
Therefore, $\{a,b\} = \{3,6 \}$
Using the representation above, Case 1 which is $\{ 6,3 \}$ is an equilateral triangle, Case 2 which is $\{ 4,4 \}$ is a square, and Case 3 which is $\{ 3,6 \}$ is a regular hexagon. |
# Math Worksheets Land
Math Worksheets For All Ages
# Math Worksheets Land
Math Worksheets For All Ages
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# Ratios and Proportion Worksheets
Many students and even adults that have not been around math for a while often get these two distinct concepts confused. Ratios are used to compare two values. The integers that are used tell us how much of one thing we have compared to another. There are several different ways in which they are stated. Over the series of these topics, we go over each of them. Proportions is a math statement that indicates that two ratios are equal. Know that these things are equal allows us to scale things by making them bigger or smaller quickly and easily. The worksheets and lessons that you will find below will not only learn skills of these topic, but also how they can be applied to the real world.
### What are Ratios?
To compare values, we use the concept of ratios. When we use the term, "to," write two numbers as a fraction, or with a colon between them, we are representing a ratio. It is a comparison of the quantities of two things. It is a measure of how much of thing is there, in comparison to another thing.
A ratio can be used to represent a comparison between two things, and we call it part-to-part ratios. There are cases when you have to compare a part to a whole lot, and we call these ratios part-to-whole. The concept of ratios is very commonly used in writing down recipes.
Example: A delegation comprising of five pupils was sent to XYZ college to represent a school. Out of these five, three were female, and two were male pupils.
We can represent this information in the form of two ratios; part-to-part and whole-to-part.
Part-to-Part: - The ratio of females to male can be written as 3:2 or 3/2 The ratio of males to females can be written as 2:3 or 2/3
Whole-to-Part: - The ratio of females to the whole delegation can be written as 3:5 or 3/5 The ratio of males to the whole delegation can be written as 2:5 or 2/5
This really gets hot right around the middle grade levels. These skills are used endless throughout life, so it is important for students to grasp this.
### What are Proportions?
Proportions are equations that state that two ratios are equivalent. If we know that we have a equivalent ratios it allows us to scale things up in size or quantity very quickly. There will be times where you will need to evaluate the truth of proportions. If they are equal ratios, they are true. If they are not equal, they are false. We can use proportions to help solve all types of unit rate based problems.
Example: Jennifer travels in a car at a constant speed of 60 miles per hour from Boston to Quebec City. The distance between the two cities is 300 miles. How long does it take her?
Solution: We know that we have a proportion of 60 miles per 1 hour. We want to know the equivalent proportion that would travel 300 miles. The math would look like this:
We would then cross multiply to rearrange the portion as: 300 = 60x
We would divide both sides by 60 and be left with 5 = x.
This means it would take 5 hours to travel that distance.
### What is The Difference Between a Ratio and a Proportion?
Ratios are used to compare two quantities. This comparison is made by using the division operation. They are written in form a/b. Proportions are equations that we use to explain that two ratios are equal or equivalent. They are presented in the form: a/b = c/d. The division operator is sometimes removed or replaced with the symbol (:). Proportions are often given with unknown values. These unknown or missing values are easy to calculate by working off of the other three values that you are given. The unknown value would just need to satisfy the equivalence of proportions. They each serve their own based on what measures you working with and the nature of the data that you are exploring. Ratios are often given to explain unit rates and a wide variety of measures. Proportions are often used to compare the overall value of these unit rates and measures. Both of these have a wide array of applications, but you will use both any time you go grocery shopping.
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# Find the equation of the hyperbola with vertices at
Question:
Find the equation of the hyperbola with vertices at $(0, \pm 5)$ and foci at $(0, \pm 8)$.
Solution:
Given: Vertices at $(0, \pm 5)$ and foci at $(0, \pm 8)$
Need to find: The equation of the hyperbola.
Let, the equation of the parabola be:
$\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1$
Vertices of the parabola are at $(0, \pm 5)$
That means $\mathrm{a}=5$
The foci are given at $(0, \pm 8)$
$A$ and $B$ are the vertices. $C$ and $D$ are the foci.
That means, ae $=8$, where $e$ is the eccentricity.
$\Rightarrow 5 \mathrm{e}=8[$ As $\mathrm{a}=5]$
$\Rightarrow \mathrm{e}=\frac{8}{5}$
We know that, $e=\sqrt{1+\frac{b^{2}}{a^{2}}}$
Therefore,
$\Rightarrow \sqrt{1+\frac{b^{2}}{a^{2}}}=\frac{8}{5}$
$\Rightarrow 1+\frac{b^{2}}{a^{2}}=\frac{64}{25}$ [Squaring both sides]
$\Rightarrow \frac{b^{2}}{a^{2}}=\frac{64}{25}-1=\frac{39}{25}$
$\Rightarrow b^{2}=a^{2} \frac{39}{25}$
$\Rightarrow \mathrm{b}^{2}=25 \times \frac{39}{25}=39[$ As $\mathrm{a}=5]$
So, the equation of the hyperbola is,
$\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1 \Rightarrow \frac{y^{2}}{25}-\frac{x^{2}}{39}=1$ [Answer] |
# Class 8 NCERT Solutions – Chapter 9 Algebraic Expressions and Identities – Exercise 9.5 | Set 1
Last Updated : 05 Apr, 2021
### Question 1. Use a suitable identity to get each of the following products.
(i) (x + 3) (x + 3)
Solution:
(x + 3) (x + 3)
Putting formula (a + b)2 = a2 + b2+ 2ab
Put a = x & b = 3
(x + 3) (x +3 ) = (x + 3)2
= x2 + 6x + 9
(ii) (2y + 5) (2y + 5)
Solution:
(2y + 5) (2y + 5)
Putting formula (a + b)2 = a2 + b2 + 2ab
Put a = 2y & b = 5
(2y + 5) (2y + 5) = (2y + 5)2
= 4y2 + 20y + 25
(iii) (2a – 7) (2a – 7)
Solution:
(2a – 7) (2a – 7)
Putting formula (x – y)2 = x2 + y2 – 2xy
Put x = 2a & y = 7
(2a – 7) (2a – 7) = (2a – 7)2
= 4a2 – 28a + 49
(iv) (3a – 1/2) (3a – 1/2)
Solution:
(3a – 1/2) (3a – 1/2)
Putting formula (x – y)2 = x2 + y2 – 2xy
Put x = 3a & y = 1/2
= (3a)2 + (1/2)2 – 2 (3a) (1/2)
= 32a2 + 1/4 – 3a
= 9a2 + 1/4 – 3a
(v) (1.1m – 0.4) (1.1m + 0.4)
Solution:
(1.1m – 0.4) (1.1m + 0.4)
Putting formula (a + b) (a – b) = a2– b2
Put a = 1.1m & b = 0.4
(1.1m – 0.4) (1.1m + 0.4) = (1.1m)2 – (0.4)2
= 1.21m2 – 0.16
(vi) (a2 + b2) (– a2 + b2)
Solution:
(a2 + b2) (– a2 + b2)
(a2 + b2) (– a2 + b2) = (b2 + a2) (b2 – a2)
Putting formula (x + y) (x – y) = x2 – y2
Put x = b2 & y = a2
(b2 + a2) (b2 – a2) = b2×2 – a2×2
= b4 – a4
(vii) (6x – 7) (6x + 7)
Solution:
(6x – 7) (6x + 7)
Putting formula (a – b) (a + b) = a2 – b2
Put a = 6x & b = 7
(6x – 7) (6x + 7) = (6x)2 – 72
= 36x2 – 49
(viii) (– a + c) (– a + c)
Solution:
(– a + c) (– a + c)
(– a + c) (– a + c) = (c – a) (c – a)
Putting formula (x – y)2 = x2 + y2 – 2xy
Put x = c & y = a
(c – a) (c – a) = c2 + a2 – 2ca
= a2 + c2 – 2ac
(ix) (x/2 + 3y/4) (x/2 + 3y/4)
Solution:
(x/2 + 3y/4) (x/2 + 3y/4)
Putting formula (a + b)2 = a2 + b2 + 2ab
put a = x/2 & b = 3y/4
= (x/2)2 + (3y/4)2 + 2 (x/2) (3y/4)
= x2/4 + 9y2/16 + 3xy/4
(x) (7a – 9b) (7a – 9b)
Solution:
(7a – 9b) (7a – 9b)
Putting formula (x – y)2 = x2+ y2– 2xy
Put x = 7a & y = 9b
(7a – 9b) (7a – 9b) = (7a)2 + (9b)2 – 2(7a)(9b)
= 49a2 + 81b2 – 126ab
### Question 2. Use the identity (x + a) (x + b) = x2 + (a + b) x + ab to find the following products.
(i) (x + 3) (x + 7)
Solution:
(x + 3) (x + 7)
Formula (x + a) (x + b) = x2 + (a + b) x + ab
Put a = 3 & b = 7
= x2 + (3 + 7) x + (3 * 7)
= x2 +10x + 21
(ii) (4x + 5) (4x + 1)
Solution:
(4x + 5) (4x + 1)
Formula (y + a) (y + b) = y2 + (a + b) y + ab
Put y = 4x , a = 5 & b = 1
= (4x)2 + (5 + 1) 4x + (5 * 1)
= 16x2 + 24x + 5
(iii) (4x – 5) (4x – 1)
Solution:
(4x – 5) (4x – 1)
Formula (y + a) (y + b) = y2 + (a + b) y + ab
Put y = 4x , a = -5 & b = -1
= (4x)2 + (-5 – 1) 4x + (-5 * -1)
= 16x2 – 24x + 5
(iv) (4x + 5) (4x – 1)
Solution:
(4x + 5) (4x – 1)
Formula (y + a) (y + b) = y2 + (a + b) y + ab
Put y = 4x , a = 5 & b = -1
= (4x)2 + (5 – 1) 4x + (5 * -1)
= 16x2 + 16x – 5
(v) (2x + 5y) (2x + 3y)
Solution:
(2x + 5y) (2x + 3y)
Formula (t + a) (t + b) = t2 + (a + b) t + ab
Put t = 2x , a = 5y & b = 3y
= (2x)2 + ( 5y + 3y) 2x + (5y * 3y)
= 4x2 + 16xy + 15y2
(vi) (2a2 + 9) (2a2 + 5)
Solution:
(2a2 + 9) (2a2 + 5)
Formula (x + y) (x + z) = x2 + (y + z) x + yz
Put x = 2a2 , y = 9 & z = 5
= (2a2)2 + (9 + 5) 2a2 + (9 * 5)
= 4a4 + 28a2 + 45
(vii) (xyz – 4) (xyz – 2)
Solution:
(xyz – 4) (xyz – 2)
Formula (t + a) (t + b) = t2 + (a + b) t + ab
Put t = xyz , a = -4 & b = -2
= (xyz)2 + (-4 + (-2)) xyz + ((-4) * (-2))
= x2y2z2 – 6xyz + 8
### Question 3. Find the following squares by using the identities.
(i) (b – 7)2
Solution:
(b – 7)2
Using Formula (x – y)2 = x2 + y2 – 2xy
Putting x = b & y = 7
= b2 + 72 – 2(b)(7)
= b2 – 14b + 49
(ii) (xy + 3z)2
Solution:
(xy + 3z)2
Using Formula (a + b)2 = a2 + b2 + 2ab
Putting a = xy & b = 3z
= x2y2 + 6xyz + 9z2
(iii) (6x2 – 5y)2
Solution:
(6x2 – 5y)2
Using Formula (a – b) 2 = a2 + b2 – 2ab
Putting a = 6x2 & b = 5y
= 36x4 – 60x2y + 25y2
(iv) [(2m/3) + (3n/2)]2
Solution:
[(2m/3) + (3n/2)]2
Using Formula (a + b)2 = a2 + b2 + 2ab
Putting a = 2m/3 & b = 3n/2
= (2m/3)2 + (3n/2)2 + 2 (2m/3) (3n/2)
= (4m2/9) + (9n2/4) + 2mn
(v) (0.4p – 0.5q)2
Solution:
(0.4p – 0.5q)2
Using Formula (a – b)2 = a2 + b2 – 2ab
Putting a = 0.4p & b = 0.5q
= 0.16p2 – 0.4pq + 0.25q2
(vi) (2xy + 5y)2
Solution:
(2xy + 5y)2
Using Formula (a + b)2 = a2 + b2 + 2ab
Putting a = 2xy & b = 5y
= (2xy)2 + (5y)2 + 2 (2xy) (5y)
= 4x2y2 + 20xy2 + 25y2
### Question 4. Simplify.
(i) (a2 – b2)2
Solution:
(a2 – b2)2
Putting formula (x – y)2 = x2 + y2 – 2xy
Put x = a2 & y = b2
= a4 + b4 – 2a2b2
(ii) (2x + 5)2 – (2x – 5)2
Solution:
(2x + 5)2 – (2x – 5)2
Putting formula (a + b)2 = a2 + b2+ 2ab & (a – b)2 = a2 + b2 – 2ab
= 4x2 + 20x + 25 – (4x2 – 20x + 25)
= 4x2 + 20x + 25 – 4x2 + 20x – 25
= 40x
(iii) (7m – 8n)2 + (7m + 8n)2
Solution:
(7m – 8n)2 + (7m + 8n)2
Putting formula (a – b)2 = a2 + b2 – 2ab & (a + b)2 = a2 + b2+ 2ab
= (49m2 – 112mn + 64n2 ) + (49m2 + 112mn + 49n2)
= 98m2 + 128n2
(iv) (4m + 5n)2 + (5m + 4n)2
Solution:
(4m + 5n)2 + (5m + 4n)2
Putting formula (a + b)2 = a2 + b2+ 2ab
= (16m2 + 40mn + 25n2) + (25m2 + 40mn + 16n2)
= 41m2 + 80mn + 41n2
(v) (2.5p – 1.5q)2 – (1.5p – 2.5q)2
Solution:
(2.5p – 1.5q)2 – (1.5p – 2.5q)2
Putting formula (a – b)2 = a2 + b2 – 2ab
= (6.25p2 – 7.5pq + 2.25q2) – (2.25p2 + 7.5pq – 6.25q2)
= 4p2 – 4q2
(vi) (ab + bc)2 – 2ab2c
Solution:
(ab + bc)2 – 2ab²c
Putting formula (a + b)2 = a2 + b2 + 2ab
= (a2b2 + 2ab2c + b2c2) – 2ab2c
= a2b2 + b2c2
(vii) (m2 – n2m)2 + 2m3n2
Solution:
(m2 – n2m)2 + 2m3n2
Putting formula (a – b)2 = a2 + b2 – 2ab
= (m4 – 2m3n2 + m2n4) + 2m3n2
= m4 + m2n4
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# How do you solve 5y+13=18?
Dec 28, 2016
$\textcolor{p u r p \le}{\text{y = 1}}$
#### Explanation:
$\textcolor{\tan}{\text{Subtract 13 on both sides:}}$
$5 y + \cancel{13} = 18$
$\textcolor{w h i t e}{a a a} - \cancel{13}$ $\textcolor{w h i t e}{} - 13$
13 cancels out on the left side, and when you subtract 13 from 18 on the right you get 5:
$5 y = 5$
$\textcolor{red}{\text{Divide both sides by 5 to get y by itself.}}$
(cancel"5"y)/cancel"5"=(5)/5
Now you are just left with $y$ on the left side and 1 on the right side.
Therefore,
$\textcolor{m a \ge n t a}{\text{y = 1}}$ |
# Square Root Of 42: Calculation Methods, Properties, And Applications
//
Thomas
Discover the square root of 42 and explore its , , and real-world in geometry, engineering, and physics.
## What is the Square Root of 42?
The square root of 42 is a mathematical value that represents the number which, when multiplied by itself, gives a product of 42. In other words, it is the number that, when squared, equals 42. The square root of 42 is an irrational number, meaning it cannot be expressed as a fraction or a finite decimal.
### Calculation Methods
Calculating the square root of 42 can be done using various methods. Here are a few common techniques:
1. Prime Factorization Method: This method involves finding the prime factors of the number and grouping them in pairs. Since 42 can be expressed as 2 * 3 * 7, we can pair the prime factors as (2 * 3) * 7. Taking the square root of each pair gives us √(2 * 3) * √7, which simplifies to √6 * √7. The square root of 6 is approximately 2.449 and the square root of 7 is approximately 2.646. Multiplying these values together gives us an approximate square root of 42 as 6.480.
2. Estimation Method: Another way to calculate the square root of 42 is through estimation. By considering the nearest perfect squares, we can make an educated guess. The nearest perfect squares to 42 are 36 (6 * 6) and 49 (7 * 7). Since 42 is closer to 49, we can estimate that the square root of 42 is closer to 7. This estimation method provides a quick approximation but may not give an exact value.
### Approximation Techniques
When an exact calculation is not required, approximation techniques can be used to find a close approximation of the square root of 42. Here are a couple of common approximation techniques:
1. Decimal Approximation: Using a calculator or mathematical software, the square root of 42 can be expressed as a decimal approximation. The square root of 42 is approximately 6.4807407. This value can be rounded to the desired level of precision based on the requirements of the problem.
2. Iterative Approximation: Iterative methods involve repeatedly refining an initial guess until a desired level of accuracy is achieved. One such method is the Newton-Raphson method, which uses calculus to iteratively improve the approximation. While this method may not be necessary for simple calculations like the square root of 42, it can be used for more complex square roots.
## Properties of the Square Root of 42
### Irrationality
The square root of 42 is an example of an irrational number, which means it cannot be expressed as a fraction or a terminating decimal. When we calculate the square root of 42, we get a decimal that goes on forever without repeating. In fact, the decimal representation of the square root of 42 is approximately 6.48074. This implies that the square root of 42 cannot be expressed as a simple fraction or a ratio of two integers.
### Positive and Negative Roots
Like any other square root, the square root of 42 has both a positive and a negative value. The positive square root of 42 is approximately 6.48074, as mentioned earlier. On the other hand, the negative square root of 42 is simply the opposite of the positive square root, which means it is approximately -6.48074. The positive and negative roots of the square root of 42 have the same magnitude but differ in their signs.
To better understand the concept of positive and negative roots, let’s think of a simple analogy. Imagine you have a rectangular garden with an area of 42 square units. The square root of 42 represents the length of one side of a square garden with the same area. Now, since a square has four equal sides, there are two possible lengths for each side – a positive length and a negative length. Similarly, the square root of 42 has two possible values, one positive and one negative.
In summary, the square root of 42 is an irrational number that cannot be expressed as a fraction or a terminating decimal. It has both a positive and a negative value, with approximately 6.48074 and -6.48074 respectively. Understanding the of the square root of 42 can help us in various mathematical and problem-solving scenarios.
## Applications of the Square Root of 42
### Geometry and Measurement
When it comes to geometry and measurement, the square root of 42 plays a significant role. This mathematical concept is particularly useful in calculating the length of the sides of a square with an area of 42. By taking the square root of 42, we can determine the exact length of each side, enabling us to construct a square with the desired area.
Furthermore, the square root of 42 is also utilized in determining the diagonal of a square with sides measuring 42. This is particularly useful in various fields such as architecture and construction, where accurately calculating dimensions is crucial.
### Engineering and Physics
In the fields of engineering and physics, the square root of 42 finds application in various calculations and problem-solving scenarios. One example is in electrical engineering, where the square root of 42 is used to determine the magnitude of alternating current (AC) or voltage in certain circuits.
Moreover, in physics, the square root of 42 is often employed in equations related to waveforms and oscillations. For instance, when analyzing simple harmonic motion, which is prevalent in fields like mechanical engineering and physics, the square root of 42 is utilized to calculate the angular frequency and period of the oscillating system.
In both engineering and physics, the square root of 42 serves as a fundamental tool in solving complex problems and understanding the underlying principles that govern these disciplines.
By incorporating the square root of 42 into various equations and calculations, professionals in these fields can unlock new insights, make accurate predictions, and develop innovative solutions to real-world challenges.
Overall, whether it’s in geometry and measurement or in the realms of engineering and physics, the square root of 42 proves to be an invaluable mathematical concept with a wide range of practical applications. Its ability to provide precise measurements and aid in problem-solving makes it an essential tool for professionals across different disciplines.
## Interesting Facts about the Square Root of 42
### Prime Factorization
The prime factorization of a number is the expression of that number as a product of its prime factors. When we calculate the square root of 42, we can break it down into its prime factors to get a better understanding of its .
To find the prime factorization of 42, we can start by dividing it by the smallest prime number, which is 2. We see that 42 divided by 2 equals 21. Now, we continue dividing 21 by 2, but it does not evenly divide. So, we move on to the next prime number, which is 3. Dividing 21 by 3 gives us 7. Finally, 7 is a prime number itself, so the prime factorization of 42 is 2 x 3 x 7.
The prime factorization of 42 as 2 x 3 x 7 reveals that the square root of 42 is an irrational number. This means that it cannot be expressed as a simple fraction or a finite decimal. Instead, it goes on infinitely without repeating. The irrationality of the square root of 42 adds to its uniqueness and mathematical intrigue.
### Decimal Representation
When we calculate the square root of 42, we can express it as a decimal to provide a more precise approximation. The square root of 42 is approximately equal to 6.48074. However, it’s important to note that this decimal representation is an approximation, as the square root of 42 is an irrational number.
To understand the decimal representation of the square root of 42, we can use a calculator or a computer program to calculate it to a desired level of accuracy. The more decimal places we calculate, the closer we get to the true value of the square root of 42. However, due to its irrationality, we can never find the exact decimal representation of the square root of 42.
In real-world , the decimal approximation of the square root of 42 is often sufficient for practical purposes. Engineers, scientists, and mathematicians use these approximations to make calculations and solve problems in various fields. It allows them to work with the square root of 42 without needing to deal with its irrational nature directly.
In summary, the prime factorization of 42 as 2 x 3 x 7 reveals the square root of 42 to be an irrational number. Its decimal representation, approximately equal to 6.48074, provides a useful approximation for practical . Understanding these interesting facts about the square root of 42 helps deepen our knowledge of mathematics and its applications in various fields.
## Related Mathematical Concepts
### Perfect Squares
Have you ever wondered what makes a number a perfect square? Well, a perfect square is a number that can be obtained by multiplying an integer by itself. In other words, it is the square of an integer. For example, 4 is a perfect square because it can be written as 2 * 2, where 2 is an integer. Similarly, 9 is a perfect square because it can be written as 3 * 3.
Perfect squares have some interesting that make them worth exploring. Let’s take a closer look at a few of them:
1. Pattern of Perfect Squares: If you look at a sequence of perfect squares, you’ll notice a pattern. The squares of consecutive integers form a sequence of odd numbers. For instance, the first few perfect squares are 1, 4, 9, 16, 25, and so on. Notice how the difference between consecutive squares is always an odd number.
2. Square Roots: The concept of perfect squares is closely related to square roots. The square root of a number is the value that, when multiplied by itself, gives the original number. For example, the square root of 16 is 4, because 4 * 4 = 16. Similarly, the square root of 9 is 3, because 3 * 3 = 9. In the case of perfect squares, the square root is always an integer.
3. Applications: Perfect squares find applications in various fields. In geometry, they are used to calculate the areas of squares and rectangles. In algebra, they help in simplifying expressions and solving equations. In computer science, they are used in algorithms and data structures. Understanding perfect squares is fundamental to many mathematical concepts.
Another important mathematical concept closely related to the square root is radicals and exponents. Radicals and exponents provide a way to express numbers in a concise and powerful manner. Let’s delve into these concepts:
1. Radicals: Radicals are mathematical expressions that involve a root, such as square root, cube root, etc. The square root is the most commonly encountered radical. It is denoted by the symbol √. For example, the square root of 9 is written as √9. Radicals are used to represent numbers that are not perfect squares, as they cannot be expressed as the product of two identical integers.
2. Exponents: Exponents, also known as powers, are a way to represent repeated multiplication of a number. The exponent tells us how many times the base number should be multiplied by itself. For instance, 2^3 is read as “2 raised to the power of 3” and means 2 * 2 * 2, which equals 8. Exponents can be used to represent perfect squares as well. The square of a number can be written as the base number raised to the power of 2.
Understanding radicals and exponents is crucial in various mathematical . They are used in algebraic expressions, scientific notation, logarithms, and even in advanced calculus. Mastering these concepts will enhance your problem-solving skills and help you navigate the world of mathematics more efficiently.
In summary, perfect squares, radicals, and exponents are integral to understanding the square root of 42. By exploring these related mathematical concepts, we gain a deeper appreciation for the and of the square root. So, let’s continue our journey as we uncover more fascinating aspects of the square root of 42.
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# BPS - 5th Ed. Chapter 21 Describing Distributions with Numbers.
## Presentation on theme: "BPS - 5th Ed. Chapter 21 Describing Distributions with Numbers."— Presentation transcript:
BPS - 5th Ed. Chapter 21 Describing Distributions with Numbers
BPS - 5th Ed. Chapter 22 Numerical Summaries u Center of the data –mean –median u Variation –range –quartiles (interquartile range) –variance –standard deviation
BPS - 5th Ed. Chapter 23 Mean or Average u Traditional measure of center u Sum the values and divide by the number of values
BPS - 5th Ed. Chapter 24 Median (M) u A resistant measure of the data’s center u At least half of the ordered values are less than or equal to the median value u At least half of the ordered values are greater than or equal to the median value u If n is odd, the median is the middle ordered value u If n is even, the median is the average of the two middle ordered values
BPS - 5th Ed. Chapter 25 Median (M) Location of the median: L(M) = (n+1)/2, where n = sample size. Example: If 25 data values are recorded, the Median would be the (25+1)/2 = 13 th ordered value.
BPS - 5th Ed. Chapter 26 Median u Example 1 data: 2 4 6 Median (M) = 4 u Example 2 data: 2 4 6 8 Median = 5 (ave. of 4 and 6) u Example 3 data: 6 2 4 Median 2 (order the values: 2 4 6, so Median = 4)
BPS - 5th Ed. Chapter 27 Comparing the Mean & Median u The mean and median of data from a symmetric distribution should be close together. The actual (true) mean and median of a symmetric distribution are exactly the same. u In a skewed distribution, the mean is farther out in the long tail than is the median [the mean is ‘pulled’ in the direction of the possible outlier(s)].
BPS - 5th Ed. Chapter 28 Question A recent newspaper article in California said that the median price of single-family homes sold in the past year in the local area was \$136,000 and the mean price was \$149,160. Which do you think is more useful to someone considering the purchase of a home, the median or the mean?
BPS - 5th Ed. Chapter 29 Case Study Airline fares appeared in the New York Times on November 5, 1995 “...about 60% of airline passengers ‘pay less than the average fare’ for their specific flight.” u How can this be? 13% of passengers pay more than 1.5 times the average fare for their flight
BPS - 5th Ed. Chapter 210 Spread, or Variability u If all values are the same, then they all equal the mean. There is no variability. u Variability exists when some values are different from (above or below) the mean. u We will discuss the following measures of spread: range, quartiles, variance, and standard deviation
BPS - 5th Ed. Chapter 211 Range u One way to measure spread is to give the smallest (minimum) and largest (maximum) values in the data set; Range = max min u The range is strongly affected by outliers
BPS - 5th Ed. Chapter 212 Quartiles u Three numbers which divide the ordered data into four equal sized groups. u Q 1 has 25% of the data below it. u Q 2 has 50% of the data below it. (Median) u Q 3 has 75% of the data below it.
BPS - 5th Ed. Chapter 213 Quartiles Uniform Distribution Q1Q1 Q2Q2 Q3Q3
BPS - 5th Ed. Chapter 214 Obtaining the Quartiles u Order the data. u For Q 2, just find the median. u For Q 1, look at the lower half of the data values, those to the left of the median location; find the median of this lower half. u For Q 3, look at the upper half of the data values, those to the right of the median location; find the median of this upper half.
BPS - 5th Ed. Chapter 215 Weight Data: Sorted L(M)=(53+1)/2=27 L(Q 1 )=(26+1)/2=13.5
BPS - 5th Ed. Chapter 216 Weight Data: Quartiles u Q 1 = 127.5 u Q 2 = 165 (Median) u Q 3 = 185
BPS - 5th Ed. Chapter 217 10 0166 11 009 12 0034578 13 00359 14 08 15 00257 16 555 17 000255 18 000055567 19 245 20 3 21 025 22 0 23 24 25 26 0 third quartile median or second quartile first quartile Weight Data: Quartiles
BPS - 5th Ed. Chapter 218 Five-Number Summary u minimum = 100 u Q 1 = 127.5 u M = 165 u Q 3 = 185 u maximum = 260 Interquartile Range (IQR) = Q 3 Q 1 = 57.5 IQR gives spread of middle 50% of the data
BPS - 5th Ed. Chapter 219 Boxplot u Central box spans Q 1 and Q 3. u A line in the box marks the median M. u Lines extend from the box out to the minimum and maximum.
BPS - 5th Ed. Chapter 220 M Weight Data: Boxplot Q1Q1 Q3Q3 minmax 100 125 150 175 200 225 250 275 Weight
BPS - 5th Ed. Chapter 221 Example from Text: Boxplots
BPS - 5th Ed. Chapter 222 Identifying Outliers u The central box of a boxplot spans Q 1 and Q 3 ; recall that this distance is the Interquartile Range (IQR). u We call an observation a suspected outlier if it falls more than 1.5 IQR above the third quartile or below the first quartile.
BPS - 5th Ed. Chapter 223 Variance and Standard Deviation u Recall that variability exists when some values are different from (above or below) the mean. u Each data value has an associated deviation from the mean:
BPS - 5th Ed. Chapter 224 Deviations u what is a typical deviation from the mean? (standard deviation) u small values of this typical deviation indicate small variability in the data u large values of this typical deviation indicate large variability in the data
BPS - 5th Ed. Chapter 225 Variance u Find the mean u Find the deviation of each value from the mean u Square the deviations u Sum the squared deviations u Divide the sum by n-1 (gives typical squared deviation from mean)
BPS - 5th Ed. Chapter 226 Variance Formula
BPS - 5th Ed. Chapter 227 Standard Deviation Formula typical deviation from the mean [ standard deviation = square root of the variance ]
BPS - 5th Ed. Chapter 228 Variance and Standard Deviation Example from Text Metabolic rates of 7 men (cal./24hr.) : 1792 1666 1362 1614 1460 1867 1439
BPS - 5th Ed. Chapter 229 Variance and Standard Deviation Example from Text ObservationsDeviationsSquared deviations 1792 1792 1600 = 192 (192) 2 = 36,864 1666 1666 1600 = 66 (66) 2 = 4,356 1362 1362 1600 = -238 (-238) 2 = 56,644 1614 1614 1600 = 14 (14) 2 = 196 1460 1460 1600 = -140 (-140) 2 = 19,600 1867 1867 1600 = 267 (267) 2 = 71,289 1439 1439 1600 = -161 (-161) 2 = 25,921 sum = 0sum = 214,870
BPS - 5th Ed. Chapter 230 Variance and Standard Deviation Example from Text
BPS - 5th Ed. Chapter 231 Choosing a Summary u Outliers affect the values of the mean and standard deviation. u The five-number summary should be used to describe center and spread for skewed distributions, or when outliers are present. u Use the mean and standard deviation for reasonably symmetric distributions that are free of outliers.
BPS - 5th Ed. Chapter 232 Number of Books Read for Pleasure: Sorted M L(M)=(52+1)/2=26.5
BPS - 5th Ed. Chapter 233 Five-Number Summary: Boxplot Median = 3 interquartile range (iqr) = 5.5-1.0 = 4.5 range = 99-0 = 99 Mean = 7.06 s.d. = 14.43 0 10 20 30 40 50 60 70 80 90 100 Number of books
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# Solve the following $x^{2}\large\frac{dy}{dx}$ =$y^{2}+2xy$ given that $y=1,$ when $x=1$
Toolbox:
• First order , first degree DE
• Variable separable : Variables of a DE are rearranged to separate then, ie
• $f_1(x)g_2(y)dx+f_2(x)g_1(y)dy=0$
• Can be written as $\large\frac{g_1 (y)}{g_2(y)}$$dy=-\large\frac{f_1(x)}{f_2(x)}$$dx$
• The solution is therefore $\int \large\frac{g_1(y)}{g_2(y)}$$dy=-\int \large\frac{f_1(x)}{f_2(x)}$$dx+c$
• A D.E of first order and first degree is said to be homogeneous if it can be put in the form $\large\frac{dy}{dx}=f\bigg(\large\frac{y}{x}\bigg)$$\;or\; \large\frac{dy}{dx}=\frac{f_1(x,y)}{f_2(x,y)} Where f_1 and f_2 are homogeneous functions in x and y. • To solve we put y=vx and proceed. Step 1: x^2 \large\frac{dy}{dx}$$=y^2+2xy$ given $y=1$ when $x=1$
$\large\frac{dy}{dx}=\frac{y^2+2xy}{x^2}$ is homogeneous in x,y substitute $y=vx$=> $\large\frac{dy}{dx}$$=v+x \large\frac{dv}{dx} Step 2: The DE becomes v+ x \large\frac{dv}{dx}=\large\frac{v^2x^2+2x.vx}{x^2} \qquad= v^2 +2v \therefore x \large\frac{dv}{dx}$$=v^2+v$
$\large\frac{dv}{v(1+v)}=\large\frac{dx}{x}$
Step 3:
The variables are separated
$\int \large\frac{dv}{v(1+v)}$$=\int \large\frac{dx}{x}$$+\log c$
$\int \bigg(\large\frac{1}{v}-\frac{1}{1+v}\bigg)$$dv=\int \large\frac{dx}{x}$$+\log c$
$\log v -\log (1+v)=\log x +\log c$
$\log \large\frac{v}{x(1+v)}$$=\log c => v=x(1+v)c Substitute v=\large\frac{y}{x} \large\frac{y}{x}$$=x\bigg(1+\large\frac{y}{x}\bigg)c$
When $x=1,y=1$
$\therefore$ substitute $x=1,y=1$
$1=(1+1)c=>c=\large\frac{1}{2}$ |
4.1 - Sampling Distribution of the Sample Mean
Let’s put some numbers into Ellie’s example.
Note! The sampling method is done without replacement.
Sample Means with a Small Population: Runner’s MIleage Section
In this example, the population is the mileage of six runners. Ellie is going to try to guess the true average mileage of the six runners by taking a random sample without replacement from the population.
Mileage A B C D E F
19 14 15 9 10 17
Since we know the miles from the population, we can find the population mean.
$$\mu=\dfrac{19+14+15+9+10+17}{6}=14$$ miles
To demonstrate the sampling distribution, let’s start with obtaining all of the possible samples of size $$n=2$$ from the populations, sampling without replacement. The table below show all the possible samples, the weights for the chosen runners the sample mean and the probability of obtaining each sample. Since we are drawing at random, each sample will have the same probability of being chosen.
View Full Table
Sample Mileage $$\boldsymbol{\bar{x}}$$ Probability A, B 19, 14 16.5 $$\frac{1}{15}$$ A, C 19, 15 17.0 $$\frac{1}{15}$$ A, D 19, 9 14.0 $$\frac{1}{15}$$ A, E 19, 10 14.5 $$\frac{1}{15}$$ A, F 19, 17 18.0 $$\frac{1}{15}$$ B, C 14, 15 14.5 $$\frac{1}{15}$$ B, D 14, 9 11.5 $$\frac{1}{15}$$ B, E 14, 10 12.0 $$\frac{1}{15}$$ B, F 14, 17 15.5 $$\frac{1}{15}$$ C, D 15, 9 12.0 $$\frac{1}{15}$$ C, E 15, 10 12.5 $$\frac{1}{15}$$ C, F 15, 17 16.0 $$\frac{1}{15}$$ D, E 9, 10 9.5 $$\frac{1}{15}$$ D, F 9, 17 13.0 $$\frac{1}{15}$$ E, F 10, 17 13.5 $$\frac{1}{15}$$
We can combine all of the values and create a table of the possible values and their respective probabilities.
$$\boldsymbol{\bar{x}}$$ 9.5 11.5 12.0 12.5 13.0 13.5 14.0 14.5 15.5 16.0 16.5 17.0 18.0 Probability $$\frac{1}{15}$$ $$\frac{1}{15}$$ $$\frac{2}{15}$$ $$\frac{1}{15}$$ $$\frac{1}{15}$$ $$\frac{1}{15}$$ $$\frac{1}{15}$$ $$\frac{2}{15}$$ $$\frac{1}{15}$$ $$\frac{1}{15}$$ $$\frac{1}{15}$$ $$\frac{1}{15}$$ $$\frac{1}{15}$$
The table is the probability table for the sample mean and it is the sampling distribution of the sample mean mileage of the runners when the sample size is 2. It is also worth noting that the sum of all the probabilities equals 1. It might be helpful to graph these values.
One can see that the chance that the sample mean is exactly the population mean is only 1 in 15, very small. (In some other examples, it may happen that the sample mean can never be the same value as the population mean.) When using the sample mean to estimate the population mean, some possible error will be involved since the sample mean is random.
Now that we have the sampling distribution of the sample mean, we can calculate the mean of all the sample means. In other words, we can find the mean (or expected value) of all the possible $$\bar{x}$$’s.
The mean of the sample means is
$$\mu_\bar{x}=\sum \bar{x}_{i}f(\bar{x}_i)=9.5\left(\frac{1}{15}\right)+11.5\left(\frac{1}{15}\right)+12\left(\frac{2}{15}\right)\\+12.5\left(\frac{1}{15}\right)+13\left(\frac{1}{15}\right)+13.5\left(\frac{1}{15}\right)+14\left(\frac{1}{15}\right)\\+14.5\left(\frac{2}{15}\right)+15.5\left(\frac{1}{15}\right)+16\left(\frac{1}{15}\right)+16.5\left(\frac{1}{15}\right)\\+17\left(\frac{1}{15}\right)+18\left(\frac{1}{15}\right)=14$$
Even though each sample may give you an answer involving some error, the expected value is right at the target: exactly the population mean. In other words, if one does the experiment over and over again, the overall average of the sample mean is exactly the population mean.
Now, let's do the same thing as above but with sample size $$n=5$$
Sample Mileage $$\boldsymbol{\bar{x}}$$ Probability A, B, C, D, E 19, 14, 15, 9, 10 13.4 1/6 A, B, C, D, F 19, 14, 15, 9, 17 14.8 1/6 A, B, C, E, F 19, 14, 15, 10, 17 15.0 1/6 A, B, D, E, F 19, 14, 9, 10, 17 13.8 1/6 A, C, D, E, F 19, 15, 9, 10, 17 14.0 1/6 B, C, D, E, F 14, 15, 9, 10, 17 13.0 1/6
The sampling distribution is:
$$\boldsymbol{\bar{x}}$$ 13.0 13.4 13.8 14.0 14.8 15.0 Probability 1/6 1/6 1/6 1/6 1/6 1/6
The mean of the sample means is...
$$\mu=(\dfrac{1}{6})(13+13.4+13.8+14.0+14.8+15.0)=14$$ miles
The following dot plots show the distribution of the sample means corresponding to sample sizes of $$n=2$$ and of $$n=5$$.
Again, we see that using the sample mean to estimate population mean involves sampling error. However, the error with a sample of size $$n=5$$ is on the average smaller than with a sample of size$$n= 2$$.
Sampling Error and Size Section
Sampling Error
The error resulting from using a sample characteristic to estimate a population characteristic.
Sample size and sampling error: As the dot plots above show, the possible sample means cluster more closely around the population mean as the sample size increases. Thus, the possible sampling error decreases as sample size increases.
What happens when the population is not small?
Sample Means with Large Samples Section
An instructor of an introduction to statistics course has 200 students. The scores out of 100 points are shown in the histogram.
The population mean is $$\mu=69.77$$ and the population standard deviation is $$\sigma=10.9$$.
Let's demonstrate the samping distribution of the sample means using the StatKey website. The first video will demonstrate the sampling distribution of the sample mean when n = 10 for the exam scores data. The second video will show the same data but with samples of n = 30.
You should start to see some patterns. The mean of the sampling distribution is very close to the population mean. The standard deviation of the sampling distribution is smaller than the standard deviation of the population.
In the examples so far, we were given the population and sampled from that population.
What happens when we do not have the population to sample from? What happens when all that we are given is the sample? Fortunately, we can use some theory to help us. The mathematical details of the theory are beyond the scope of this course but the results are presented in this lesson.
In the next two sections, we will discuss the sampling distribution of the sample mean when the population is Normally distributed and when it is not. |
# NCERT Solutions for Class 4 Maths Fields And Fences
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NCERT solutions for Class 4 Mathematics
## NCERT Solutions for Class 4 Maths Fields And Fences
Chapter -13 Class 4 Maths Fields And Fences
### 1. I need a fence around my field. How much wire should I buy.
Ans. The length of the wire required for fencing the field is equal to the boundary of the field.
### 2. Can you find it from this picture?
Ans. Yes, I can find its boundary. It is equal to 21m + 15m + 9m + 9m = 54m of the field.
### 3. How much wire did Rahmat give Ganpat?
Ans. Rahmat gave Ganpat a wire of length 70m – 54 m = 16m.
### 4. How long is the boundary of Ganpat’s field?
Ans. Boundary of the Ganpat’s field = 18m + 9m + 15m + 15m + 9m = 66 metres.
### 5. How much wire will Ganpat need for his field?
Ans. Ganpat’s total requirement of wire = 66 metres
Length of wire given by Rahmat = 16 metres
Therefore, the length of more wire needed by Ganpat = (66-16) = 50 metres.
### 6. Here are the picture of some more fields. Find out which one has the biggest boundary.
(a)
Ans. Boundary = (24 + 15 + 6 + 15) metres= 60 metres
(b)
Ans. Boundary = (12 + 6 +6 + 3 +6 + 9) metres = 42 metres.
(c)
Ans. Boundary = (15 + 12 + 9) metres = 36 metres.
(d)
Ans. (d) Boundary = (9 + 15 + 15 + 9 + 15 + 15) metres = 78 metres.
### 7. Chandu’s father is called the “young old man” in his village. At 70 years age, he is fully fit. Do you know his secret? He goes for a walk around the field every morning. Everyday he takes four rounds of Chandu’s fields. What is the total distance he covers.
4 ___ = _____ m = ________ km
Ans. Boundary of Chandu’s field = 100 m + 150 m + 100 m + 150 m = 500m.
Total Distance covered by Chandu’s father = 4 Boundary of Chandu’s field
= 4 500m = 2000 m = 2 km.
### 8. Look at the picture of the table cloth and tell how much is used for one table cloth.
Ans. Length of lace = 2 (1m 50 cm + 50 cm) = 2 2m = 4m.
### 9. How much lace will be used in 3 such table clothes? How much lace will be left in the roll?
Ans. Lace used for one table cloth = 1m 50 cm + 50 cm + 1m 50 cm + 50 cm = 4metres
Lace used for 3 such table cloths = 3 4 Lace used for on one table cloth
= 3 4 metres = 12 metres.
Lace left in the roll = Total Lace – Lace used = 100m – 12m = 88m.
### 10. Find out the length of the boundary of these shapes.
Ans. To find the length of the boundary, take a long piece of thread and carefully place it along the boundary of the shape. Cut out the exact length of the thread needed for covering the shape, starting from one point and coming back to the same point. On measuring the length of this thread, we obtain the length of this thread, we obtain boundary of the shape. Repeat the process for each shape.
### 11. Now count the square and find out:
(a) How many squares are there in each shape?
Ans. The number of complete squares in shape A is 1, in shape B is 2, in shape C is 3 and shape D is 2.
(b) Which shape covers the least number of squares?
Ans. The least number of complete squares are in shape A.
(c) Which shape covers the most number of squares?
Ans. Shape C has the most number of complete squares.
### 12. A square has a boundary of 12 cm.
(a) From the corner of this square, a small square of side 1 cm is cut off. Will the boundary of B be less or more? Find its length.
Ans. Boundary of B = 3cm + 2 cm + 1 cm + 2 cm +3 cm = 12cm
Since the boundary of A is also 12 cm.
So, the boundary of B is neither less nor more than that of A. But their boundaries are equal.
(b) If you cut 1 cm square to get shape C what will be the length of the boundary of C?
Ans. Boundary of C = 3cm + 3cm + 1cm + 1cm + 1 cm +1cm + 1cm + 3 cm = 14 cm.
### 13. (a) Find the length of the boundary of square D.
Ans. Length of the boundary of square D = 5cm + 5cm + 5cm + 5cm = 20cm
(b) The boundary of this square (1cm) is ___ + ____ + ____ + ____.
Ans. The boundary of this square (1cm) is 1cm + 1cm+ 1cm + 1cm = 4cm.
Yes, we can also say that the boundary is 4 1 cm.
### 14. A hockey field is 91 metres 40cm and 55 metres wide. How long is the boundary of the field?
Ans. Length of the boundary of a hockey field = 91m + 40cm +55m + 91m 40 cm +55m
= 292 m 80 cm
Class 4 Maths Fields And Fences
### 15. Usha and Valsamma are running a race. Usha is running on the inner circle. Valsamma is running on the outer circle. Valsamma runs after than Usha. But still loses the race. Can you guess why?
Ans. Since inner boundary is smaller than the outer boundary and as such Valsamma has to run for more distance, therefore, she loses the race.
### 16.Have you seen any race where runners start from different places – like in this picture? Guess why?
Ans. In order to make their running distances equal.
### 17. How will Neetu find out if the two gardens are equally big?
Ans. To find out if the two gardens are equally big:
Let us cover each garden with cardboards of same size without overlapping. We see that the same number of cardboards cover each garden. Therefore, we can say that the two gardens are equally big.
### 18. (a) How many small squares of size 1cm are there in this big green square?
Ans. There are 36 small squares of size 1 cm in the big green square.
(b) Can you think of a faster way to know the total number of small square without counting it?
Ans. Yes, there is a faster way to find the total number of small square. Just find: 6 6=36.
Class 4 Maths Fields And Fences
### 19. Guess how many squares of one centimetre can fill this rectangle? Checking your guess by completing the grid.
Ans. By completing the grid with squares of one centimetre, we find the number of such squares is 32.
## NCERT solutions for Class 4 Mathematics Chapter 13 Fields And Fences
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The Way the World LooksView Solutions
The Junk SellerView Solutions
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NCERT Solutions Class 4 Maths Fields And Fences PDF (Download) Free from myCBSEguide app and myCBSEguide website. Ncert solution class 4 Maths includes text book solutions from Class 4 Maths Book . NCERT Solutions for CBSE Class Maths have total 14 chapters. 4 Maths NCERT Solutions in PDF for free Download on our website. Ncert Maths class 4 solutions PDF and Maths ncert class 4 PDF solutions with latest modifications and as per the latest CBSE syllabus are only available in myCBSEguide.
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To download NCERT Solutions for Class 4 Maths, EVS Hindi English, Maths Science do check myCBSEguide app or website. myCBSEguide provides sample papers with solution, test papers for chapter-wise practice, NCERT solutions, NCERT Exemplar solutions, quick revision notes for ready reference, CBSE guess papers and CBSE important question papers. Sample Paper all are made available through the best app for CBSE students and myCBSEguide website.
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## 17Calculus Precalculus - Trig Identities
##### 17Calculus
Trig Identities
Here is a list of the trig identities you will use most in calculus.
Set 1 - Basic Identities
$$\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }$$
$$\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }$$
$$\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }$$
$$\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }$$
Set 2 - Squared Identities
$$\sin^2t + \cos^2t = 1$$
$$1 + \tan^2t = \sec^2t$$
$$1 + \cot^2t = \csc^2t$$
Set 3 - Double-Angle Formulas
$$\sin(2t) = 2\sin(t)\cos(t)$$
$$\cos(2t) = \cos^2(t) - \sin^2(t)$$
Set 4 - Half-Angle Formulas
$$\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }$$
$$\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }$$
Remembering trig identities can be difficult. However, there are techniques to help you learn and memorize them. The main way to remember anything is to use it. It also helps to see how other people remember them. Here are some videos where the instructor explains how he remembers them. You may be able to pick up some techniques and new ideas from these videos.
### Trig Identities - Derive and Remember
Okay, so these identities may be a bit overwhelming to learn and remember. Here is a fun video that shows the geometric interpretation of all 6 trig functions. He goes through them pretty fast, so you may want to take notes but this gives you another way to remember the trig identities.
### 3Blue1Brown - Tattoos on Math [8min-14secs]
video by 3Blue1Brown
When using the material on this site, check with your instructor to see what they require. Their requirements come first, so make sure your notation and work follow their specifications.
DISCLAIMER - 17Calculus owners and contributors are not responsible for how the material, videos, practice problems, exams, links or anything on this site are used or how they affect the grades or projects of any individual or organization. We have worked, to the best of our ability, to ensure accurate and correct information on each page and solutions to practice problems and exams. However, we do not guarantee 100% accuracy. It is each individual's responsibility to verify correctness and to determine what different instructors and organizations expect. How each person chooses to use the material on this site is up to that person as well as the responsibility for how it impacts grades, projects and understanding of calculus, math or any other subject. In short, use this site wisely by questioning and verifying everything. If you see something that is incorrect, contact us right away so that we can correct it. |
# Solving algebra word problems calculator
In this blog post, we will explore one method of Solving algebra word problems calculator. Keep reading to learn more!
## Solve algebra word problems calculator
In algebra, one of the most important concepts is Solving algebra word problems calculator. Algebra is the branch of mathematics that deals with the rules of operations and relations, and the study of quantities which may be either constant or variable. Factoring is a technique used to simplify algebraic expressions. When an expression is factored, it is rewritten as a product of simpler factors. This can be helpful in solving equations and graphing functions. In general, factoring is the process of multiplying two or more numbers to get a product. For example, 6 can be factored as 2 times 3, since 2 times 3 equals 6. In algebra, factoring is often used to simplify equations or to find solutions. For example, the equation x^2+5x+6 can be simplified by factoring it as (x+3)(x+2). This can be helpful in solving the equation, since now it can be seen that the solution is x=-3 or x=-2. Factoring can also be used to find zeroes of polynomials, which are important in graphing functions. In general, polynomials can be factored into linear factors, which correspond to zeroes of the function. For example, the function f(x)=x^2-4x+4 has zeroes at x=2 and x=4. These zeroes can be found by factoring the polynomial as (x-2)(x-4). As a result,factoring is a powerful tool that can be used to simplify expressions and solve equations.
Polynomials are equations that contain variables with exponents. The simplest type of polynomial is a linear equation, which has only one variable. To solve a linear equation, you need to find the value of the variable that makes the equation true. For example, the equation 2x + 5 = 0 can be solved by setting each side of the equation equal to zero and then solving for x. This gives you the equation 2x = -5, which can be simplified to x = -5/2. In other words, the value of x that makes the equation true is -5/2. polynomials can be more difficult to solve, but there are still some general strategies that you can use. One strategy is to factor the equation into a product of two or more linear factors. For example, the equation x2 + 6x + 9 can be factored into (x + 3)(x + 3). This gives you the equation (x + 3)(x + 3) = 0, which can be solved by setting each factor equal to zero and solving for x. This gives you the equations x + 3 = 0 and x + 3 = 0, which both have solutions of x = -3. Therefore, the solutions to the original equation are x = -3 and x = -3. Another strategy for solving polynomials is to use algebraic methods such as completing the square or using synthetic division. These methods are usually best used when you have a high-degree polynomial with coefficients that are not easily factored. In general, however, polynomials can be solved using a variety of different methods depending on their specific form. With some practice and patience, you should be able to solve any type of polynomial equation.
There are a variety of websites that offer help with math word problems. Some of these sites provide step-by-step solutions, while others simply give the answer. However, there are a few things to keep in mind when using these websites. First, make sure that the site you're using is reputable. There are many fake sites out there that will give you incorrect answers. Second, be sure to read the instructions carefully. Many sites require you to input specific information, such as the type of problem and the variables involved. Finally, take your time and double-check your work. With a little patience and effort, you should be able to find a website that will help you solve even the most difficult math word problem.
For example, the equation 2 + 2 = 4 states that two plus two equals four. To solve an equation means to find the value of the unknown variable that makes the equation true. For example, in the equation 2x + 3 = 7, the unknown variable is x. To solve this equation, we would need to figure out what value of x would make the equation true. In this case, it would be x = 2, since 2(2) + 3 = 7. Solving equations is a vital skill in mathematics, and one that can be used in everyday life. For example, when baking a cake, we might need to figure out how many eggs to use based on the number of people we are serving. Or we might need to calculate how much money we need to save up for a new car. In both cases, solving equations can help us to get the answers we need.
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# Polynomials and Factoring
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1 Lesson 2 Polynomials and Factoring A polynomial function is a power function or the sum of two or more power functions, each of which has a nonnegative integer power. Because polynomial functions are built from power functions, their domains are all real numbers. The graph of a polynomial function is smooth and unbroken. This important property of polynomial functions makes them particularly useful for transforming one image into another during special effects in movies, for creating fonts for computers, and for approximating solutions to otherwise uncomputable problems. In this lesson, you will explore some of the advantages of expressing polynomials in different equivalent forms. Think About This Situation Polynomials are used to morph one image into another on Scholastic Production s Animorphs. Polynomials can be written symbolically in several different forms. Consider the three forms of one polynomial shown below. Standard Form: P(x) = 5x x 3 65x 2 70x Factored Form: P(x) = 5(x 1)(x + 2)(x 3)(x + 4) Nested Multiplication Form: P(x) = (((5x + 10)x 65)x 70)x a b What mathematical questions could be most easily answered using: the standard form? the factored form? the nested form? Can every polynomial be written in factored form? In nested multiplication form? 382 UNIT 6 POLYNOMIAL AND RATIONAL FUNCTIONS
2 INVESTIGATION 1 Need for Speed In the previous lesson, you may have noticed that the time needed to produce a curve on a calculator depends upon the complexity of the equation involved. This fact is very important to computer scientists, mathematicians, and engineers, for whom computation time is often a critical component of any project. Saving computer time saves money. Super computers such as the SGI Origin 2000 at the National Center for Super-computing Applications at the University of Illinois at Champaign Urbana can cost between 2 and 3 million dollars a year to operate. Kray super computer 1. The form of a function rule can greatly influence the time it takes to calculate function values or produce its graph. In this activity, you will use your calculator to measure computation time for polynomial functions. a. Working in pairs, measure and record the time required to graph the function y = x 5 15x x 3 225x x 120 with the viewing window set to 0 x 6 and 10 y 10. One person should watch the graph and the other should watch the time. Be sure to turn off any other functions or plots before doing this experiment. b. Next measure and record the time it takes to graph the function y = (x 1)(x 2)(x 3)(x 4)(x 5). Be sure to turn off any other functions. c. Verify that the function rules in Parts a and b are equivalent. d. Which form produces the graph more quickly? What percentage of time is saved using this form? e. Computation time is measured in units called cycles. Suppose computing any power, such as (1.2) 4, uses twice as much computation time as does any multiplication, subtraction, or addition, which each require 1 cycle. How many cycles are required when y = x 5 15x x 3 225x x 120 is evaluated for a particular value of x? How many cycles are required when the equivalent rule y = (x 1)(x 2)(x 3)(x 4)(x 5) is evaluated? How does this computation time for the factored form compare to the computation time for the standard polynomial form? LESSON 2 POLYNOMIALS AND FACTORING 383
3 2. The factored form of a polynomial can reduce the computation time, but not all polynomials can be written as a product of linear factors. Another way a polynomial can be rewritten to avoid exponents is to use nested multiplication. a. Verify that the polynomial 5x 5 + 4x 4 + 3x 3 + 2x 2 + x + 1 can be rewritten in nested form as ((((5x + 4)x + 3)x + 2)x + 1)x + 1. b. Determine the percentage of computation time (in seconds) saved in graphing the function y = 5x 5 + 4x 4 + 3x 3 + 2x 2 + x + 1 expressed in nested multiplication form rather than in standard polynomial form. Use the window 2 x 2 and 40 y 40. c. How many cycles are required to compute each form of the polynomial? How is the cycle count related to your answer in Part b? d. Investigate other viewing windows for the graph of the polynomial function. Explain why this polynomial cannot be written as a product of linear factors. Into how many linear factors can this polynomial be factored? e. Rewrite the polynomial of Activity 1 using nested multiplication. How does the computation time for nested multiplication compare to the computation time for the factored form? Checkpoint The form of a polynomial influences computation speed as well as the information that can be determined by examining the form. a b Write the polynomial function y = (x + 2)(x 5)(x 2 + 1) in standard polynomial form; in nested multiplication form. What information about the graph of a polynomial function can you get from the factored form? From the expanded or standard form? From the nested multiplication form? Be prepared to discuss the importance of each form of a polynomial with your classmates. Evaluating polynomials using nested multiplication is frequently called Horner s Method after the English mathematician William George Horner ( ). However, evidence was published in 1911 that Paolo Ruffini ( ) had used the method 15 years before Horner. More recently, the method has been found over 500 years before either Ruffini or Horner in the works of Chinese mathematicians during the late Sung Dynasty: Li Chih ( ), Chu Shih-chieh ( ), Ch in Chiu-shao (c ) and Yang Hui (c ). 384 UNIT 6 POLYNOMIAL AND RATIONAL FUNCTIONS
4 On Your Own Because evaluation of polynomial functions only requires addition, subtraction, and multiplication, they are often used to estimate nonpolynomial functions. Graph the function y = sin x in the window 4 x 4 and 2 y 2. Then graph each of the following polynomial functions in the same viewing window. i. y = x ii. y = x x 3 3! iii. y = x x 3 3! + x 5 5! iv. y = x x 3 3! + x 5 5! x 7 7! a. Compare the graphs of the polynomial functions to the graph of the sine function. b. Determine the zeroes of each polynomial function. How do these values compare to the zeroes of the sine function? c. Write the next polynomial function approximation in this sequence. d. Compare the time it takes your calculator to graph the fourth polynomial function above written in standard form and in nested multiplication form. INVESTIGATION 2 Strategic Factors Throughout your work with quadratic and higher degree polynomial functions, the factored form of the polynomials has played an important role. Historically, writing polynomials in factored form was important for finding solutions to equations. Problems involving the solution of polynomial equations date back to the time of the ancient Egyptians, Babylonians, and Greeks. The first systematic presentation of the solution of polynomial equations is attributed by some historians to a group of Arab mathematicians starting with al-kwarizmi (c ). 1. In her article The Art of Algebra (History of Science, June 1988, pp ), Karen Parshall describes the evolution of solving polynomial equations. a. She reports that a systematic solution of the equation x x = 39 was described by al-kwarizmi in the early 800s. Describe all the methods you know for solving this equation. Karen Parshall Solve this equation by reasoning with the symbolic form itself. b. Several centuries earlier around 250 A.D., Diophantus of Alexandria solved the equation 630x x = 6. Using a method of your choice, find the solutions to this equation. LESSON 2 POLYNOMIALS AND FACTORING 385
5 2. Until the sixteenth century, the solution of polynomial equations involved systematic guessing-and-testing for possible linear factors involving integers. a. Expand the polynomial P(x) = (x a)(x b)(x c), where a, b, and c are given constants. Write the polynomial in standard form where the coefficients of each power of x are expressions involving a, b, and c. b. What do you observe about the constant term? How could the constant term be used to identify possible factors of a polynomial? c. What do you observe about the coefficient of the x 2 term? d. How might you generalize your observations in Parts b and c so that they apply to nth-degree polynomials? Compare your generalization with those of other groups. Resolve any differences. e. Use your generalization to solve the following equations. i. x 3 10x x 18 = 0 ii. x 3 28x = 48 iii. x 3 4x = 0 3. Rewrite each polynomial function in factored form using information from its graph and your generalization in Activity 2. a. p(x) = x 3 + 9x x 21 b. y = x 4 + 2x 3 13x 2 14x + 24 c. A = w 4 3w 3 + 2w 2 d. s = t 4 + 2t 3 11t 2 12t + 36 While searching for factors of a given polynomial, you probably used a fundamental relationship between the factors and the zeroes of the polynomial function. Factor Theorem The linear expression (x b) is a factor of a polynomial function f(x) if and only if f(b) = 0. You will be asked to complete a proof of this theorem in Organizing Task 1 (page 400). 4. The method of Activity 2 can be generalized to find factors of the form (ax b). a. Without expanding completely, mentally determine the leading coefficient and constant term of the polynomial f(x) = ( 1x + 1)(3x 2)( 5x 4). b. What are the zeroes of the polynomial function in Part a? How are the zeroes related to divisors of the leading coefficient and of the constant term of the polynomial? 386 UNIT 6 POLYNOMIAL AND RATIONAL FUNCTIONS
6 c. Use your observations from Parts a and b to factor each of the following polynomials: i. 4x 3 + 9x 2 4x 9 ii. 3x x x + 4 iii. x 3 2x 2 4x + 8 Your work in Activities 2 and 4 suggests the following useful theorem about possible rational zeroes of a polynomial function with integer coefficients. Rational Zeroes Theorem For a polynomial function with integer coefficients, if a is a divisor of the coefficient of the term of highest degree and b is a divisor of the constant term, then b a is a possible zero of the function, with a corresponding possible factor of the form (ax b). 5. You now have several ways to determine if a polynomial of the form (x a) is a factor of a polynomial p(x). In this activity, you will examine a method for finding the polynomial q(x) such that p(x) = (x a)q(x). a. Describe two ways of verifying that (x + 1) is a factor of 2x 3 + x 2 4x 3. b. To find the polynomial which when multiplied by (x + 1) gives 2x 3 + x 2 4x 3, you can use a procedure similar to long division. Polynomial division is illustrated below. Explain how the term in bold below is chosen at each step. 2x 2 x 3 x x 3 + x 2 4 x 3 2x 3 + 2x 2 Multiply (x + 1) by 2x 2 x 2 4x 3 Subtract x 2 x Multiply (x + 1) by x 3x 3 Subtract 3x 3 Multiply (x + 1) by 3 0 Subtract Therefore, 2x 3 + x 2 4x 3 = (x + 1)(2x 2 x 3). c. Verify that 2x 3 + x 2 4x 3 = (x + 1)(2x 2 x 3). d. Show that 2x 3 + x 2 4x 3 can be written as a product of three linear factors. e. The polynomial x 3 5x 12 is equivalent to x 3 0x 2 5x 12. Use polynomial division to show that x 3 5x 12 = (x 3)(x 2 + 3x + 4). Can x 3 5x 12 be written as a product of three linear factors? Explain your reasoning. LESSON 2 POLYNOMIALS AND FACTORING 387
7 6. Factor each polynomial into polynomials of the smallest possible degrees. a. f(x) = x 3 2x 2 4x + 8 b. g(x) = x 4 3x 3 + 2x 2 c. h(x) = 2x 3 5x 2 28x + 15 d. p(x) = x 3 + 3x 2 + 4x + 2 e. q(x) = 8x x x x In Lesson 1, you discovered that the multiplicity of a zero of a polynomial function was related to the shape of its graph near the zero. y y f (x) = (x 3) 2 x x g(x) = (x + 3) 3 If the graph of a polynomial function touches, but does not intersect, the x-axis at x = a, then the zero a is repeated an even number of times. If the graph crosses the x-axis at x = a and has a flattened appearance at this x-intercept, then the zero a is repeated an odd number of times. a. Compare the graphs of the polynomial functions in Activity 6 with their factorizations. Is the expected relationship between the shape of the graph, multiple zeroes, and repeated factors confirmed? b. Find a polynomial function whose graph matches the graph shown below. y x 20 c. Use the connection between graphs, multiple zeroes, and repeated factors to help factor each of these polynomials into polynomials of the smallest degrees possible. i. y = x 3 7x 2 5x + 75 ii. y = x 5 + 4x 4 + x 3 10x 2 4x + 8 iii. y = x 5 + x 4 2x 3 5x 2 5x 2 iv. y = x 4 4x x 2 36x UNIT 6 POLYNOMIAL AND RATIONAL FUNCTIONS
8 Checkpoint Writing and interpreting polynomials in factored form has a long history as a central idea in algebraic thinking. a b c d How is the solution of a polynomial equation p(x) = 0 related to the factored form of the polynomial p(x)? What strategies can you use to find the factored form of a polynomial? How does a repeated linear factor reveal itself in the graph of a polynomial function? Explain why a polynomial of degree 7 cannot have just four linear factors (counting the repeated factors). Be prepared to explain your strategies for, and thinking about, factoring polynomials to the class. The quadratic formula can be used to solve second-degree polynomial equations. More complicated formulas exist for solving third- and fourth-degree polynomial equations. In the early 1800s, Paolo Ruffini, Neils Abel, and Evariste Galois found ways to show that it is not possible to solve fifth-degree equations by a formula. Since no general formula exists for finding roots of polynomial equations, mathematicians have developed systematic methods for estimating roots. One of these methods is investigated in the MORE set. Neils Abel On Your Own As you complete these tasks, think about the relationships among zeroes, factors, and graphs of polynomial functions. a. Factor each polynomial into polynomials of the smallest degrees possible. f(x) = x 3 + 4x 2 11x + 6 g(x) = x 3 3x 2 10x h(x) = x 4 + 5x 3 + 7x 2 + 5x + 6 LESSON 2 POLYNOMIALS AND FACTORING 389
9 b. Sketch the graph of a fifth-degree polynomial function that has one factor repeated twice and a second factor repeated three times. Write a symbolic rule that matches your graph. c. Write an equation for the graph shown. The y-intercept is d. Andrea used the calculator graph shown at the right to help her solve the equation x 4 x 3 27x x 54 = 0. She determined that the roots of the equation were x = 1 and x = 3. Yolanda, who was working in a group with Andrea, looked at the graph and said there had to be another root and found that 6 was also a solution. Verify that 1, 3, and 6 are all roots. How did Yolanda know there had to be another root? 390 UNIT 6 POLYNOMIAL AND RATIONAL FUNCTIONS
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# Linear Inequalities with One Variable
Here we talk about inequalities of the form ax>b or ax<b,ax>=b,ax<=b. If a>0 then inequality ax>b is equivalent to the inequality x>b/a, i.e. set of solutions of this inequality is interval (b/a,+oo). If a<0 then inequality ax>b is equivalent to the inequality x<b/a, therefore, set of solutions of this inequality is interval (-oo,b/a).
At last, if a=0 then inequality has form of 0*x>b, i.e. it doesn't have solution when b>=0 and is true for all x when b<0.
Example . Solve 2(x-3)+5(1-x)>=7(2x-5).
After a simplification we obtain:
2x-6+5-5x>=14x-35;
2x-5x-14x>=-35+6-5;
-17x>=-34.
Now, we divide inequality by -17 and change sign of inequality: x<=2.
Therefore, (-oo,2] is set of solutions of given equation. |
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# How to Find the Median
## Welcome, great people!
Have you ever wondered how to find the median? You're in the right place! In this article, we'll explore the concept of the median and learn how to calculate it. Whether you're a student studying statistics or just someone curious about numbers, understanding the median is essential. So let's dive right in!
## The Concept of the Median
Before we delve into the calculations, let's first understand what the median is. In statistics, the median is the middle value of a dataset when arranged in ascending or descending order. It divides the dataset into two equal halves. Unlike the mean, which can be heavily influenced by extreme values, the median provides a more robust measure of central tendency.
### Why is the Median Important?
The median is an essential statistic that helps us understand the distribution of data. It is particularly useful when dealing with skewed datasets or outliers. The median provides insight into the typical value or the central value of a set, making it a valuable tool in various fields such as economics, biology, and sociology.
### Calculating the Median
To calculate the median, follow these steps:
1. First, sort the dataset in ascending or descending order.
2. If the dataset has an odd number of values, the median is the middle value.
3. If the dataset has an even number of values, the median is the average of the two middle values.
Let's illustrate this with an example. Consider the following dataset: 3, 6, 9, 12, 15, 18. To find the median, first, sort the dataset in ascending order: 3, 6, 9, 12, 15, 18. Since the dataset has an even number of values, the median is the average of the two middle values, which in this case are 9 and 12. Thus, the median is (9 + 12) / 2 = 10.5.
### Advantages of Using the Median
1. Resistant to Outliers: The median is resistant to outliers, which means it is not affected by extreme values. This makes it a better measure of central tendency for skewed datasets or those with significant outliers.
👉 Outliers can significantly impact the mean but have less effect on the median, making it useful in such cases.
2. Easy to Understand: The median provides a straightforward and intuitive measure of central tendency. It represents the middle value of a dataset, making it easy to interpret for individuals with limited statistical knowledge.
👉 The median is commonly used in real-world scenarios to report the typical value, such as when determining the median household income in a country.
3. Works with Ordinal Data: The median is suitable for datasets with ordinal data, such as rankings or ratings. It allows us to identify the middle position or ranking within the dataset.
👉 For example, the median can be used to determine the middle position in a race or the median score in a test.
4. Robust Measure: As mentioned earlier, the median is a robust measure of central tendency. It is not influenced by extreme values or the exact values of the dataset, making it useful in situations where the mean may be misleading.
👉 For datasets with extreme values, such as average income in a country, the median provides a more accurate representation of the typical value.
5. Suitable for Skewed Distributions: The median is particularly useful when dealing with skewed distributions. Skewed datasets have a long tail in one direction, which can heavily influence the mean. In such cases, the median provides a more representative value.
👉 An example of a skewed distribution is the distribution of wealth, where a small percentage of the population holds a significant portion of the wealth.
6. Applicable to Non-Numeric Data: Unlike the mean, which only works with numeric data, the median can be applied to non-numeric data. It allows us to find the central value or position within a dataset, even if the data is not numerical.
👉 For example, the median can be used to determine the middle rank or position in a list of names.
7. Useful for Comparing Data: The median is useful for comparing data between different groups or categories. It provides a representative value for each group and allows for better comparisons.
👉 For example, the median salary can be used to compare income levels between different professions or industries.
### Disadvantages of Using the Median
1. Limited Descriptive Power: While the median is a robust measure of central tendency, it provides limited descriptive power compared to other measures such as the mean. It does not capture the full distribution of the dataset.
👉 Additional measures, such as the range or standard deviation, may be necessary to gain a better understanding of the data.
2. Ignores Relative Differences: The median treats all values equally and does not consider the relative differences between them. It only focuses on the middle value and disregards the magnitude of the values.
👉 For example, the median may not accurately reflect the difference between two groups if one group has significantly higher or lower values than the other.
3. Calculations Can Be Complex: While calculating the median is relatively straightforward for datasets with an odd number of values, it can be more complex for datasets with an even number of values. The need to sort the data and identify the middle values can be time-consuming.
👉 It's important to pay attention to the steps and ensure the dataset is correctly sorted to obtain an accurate median.
4. May Not Represent the "Typical" Value: In some cases, the median may not accurately represent the "typical" value of a dataset, especially if the distribution is heavily skewed or has multiple modes.
👉 Additional measures, such as the mode or quartiles, may be necessary to fully understand the characteristics of the dataset.
Now that we have explored the advantages and disadvantages of using the median, let's summarize the steps to find the median in a table:
Step Description
1 Sort the dataset
2 If the dataset has an odd number of values, the median is the middle value
3 If the dataset has an even number of values, the median is the average of the two middle values
## Frequently Asked Questions
### Q1: What is the main difference between the median and the mean?
A1: The main difference between the median and the mean is how they are influenced by extreme values. The median is resistant to outliers, while the mean can be heavily influenced by them.
### Q2: Can the median be applied to non-numeric data?
A2: Yes, the median can be applied to non-numeric data. It allows us to find the central value or position within a dataset, even if the data is not numerical.
### Q3: When should I use the median instead of the mean?
A3: You should use the median instead of the mean when dealing with skewed datasets or datasets with significant outliers. The median provides a more robust measure of central tendency in such cases.
### Q4: Does the median provide a complete picture of the dataset?
A4: No, the median provides a measure of central tendency but does not capture the full distribution of the dataset. Additional measures, such as the range or standard deviation, may be necessary to gain a better understanding of the data.
### Q5: How can I find the median in Excel?
A5: In Excel, you can use the MEDIAN function to find the median. Simply select the range of values and enter "=MEDIAN(range)" in a cell to obtain the median.
### Q6: Is the median affected by rounding errors?
A6: The median is not affected by rounding errors since it only considers the position of the values in the dataset, not their exact values.
### Q7: Are there any alternatives to the median?
A7: Yes, there are alternative measures of central tendency such as the mode and the trimmed mean, which exclude a certain percentage of extreme values. These measures may be more appropriate in specific situations.
## Conclusion
In conclusion, understanding how to find the median is essential in various fields where data analysis is involved. The median provides a robust measure of central tendency that is resistant to outliers and suitable for skewed datasets. It allows us to identify the middle value or position within a dataset, making it valuable for comparisons and understanding the distribution of data. While the median has its limitations, such as limited descriptive power, when used appropriately, it is a powerful tool for analyzing and interpreting data.
Now that you know how to find the median, why not apply this knowledge in your own data analysis? Take a look at the dataset you're working with and calculate the median to gain further insights. It's time to put your newfound skills into practice!
Disclaimer: The information provided in this article is for educational purposes only and should not be considered as professional advice. Always consult with a qualified statistician or data analyst for accurate and customized analysis. |
Home » BLOG » 10 Common Mistakes in SAT Probability [2023]
# 10 Common Mistakes in SAT Probability [2023]
Posted by on May 31, 2023
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10 Common Mistakes in :
Probability questions are a crucial part of the SAT Math section, testing your understanding of chance and likelihood. However, many students fall into common traps and make mistakes when tackling probability problems. By being aware of these pitfalls, you can avoid them and improve your chances of success on the SAT. Quickly, here are 10 of the most common mistakes to watch out for when dealing with probability:
## Misunderstanding the Question:
Understanding each probability question is compulsory. Take the time to carefully read and understand what the question is asking. Misinterpreting the problem can lead to incorrect calculations and answers.
## Ignoring Key Information:
Probability questions often provide important information that should not be overlooked. Pay attention to any given conditions, restrictions, or assumptions.
## Confusing Independent and Dependent Events:
Understand the difference between independent and dependent events. Independent events are not influenced by previous outcomes, while dependent events are affected by previous outcomes.
## Failing to Consider Sample Space:
The sample space represents all possible outcomes of an event. Forgetting to account for all possible outcomes can lead to incorrect probability calculations.
## Forgetting to Simplify Fractions:
Probability is typically expressed as a fraction. Always fractions to their lowest terms to ensure accuracy in your calculations.
## Using the Wrong Probability Formula:
Probability questions may involve calculating the probability of different scenarios. Make sure you select the appropriate formula based on the given situation.
## Neglecting to Multiply or Add Probabilities:
In some cases, you need to multiply or add probabilities to determine the overall probability of an event. Be mindful of these operations and apply them correctly. For example, ‘or' in probability calls for the addition of relevant fractions, while ‘and' calls for the multiplication of relevant fractions [A or B = A + B], [A and B = A * B].
RELATED => Find the probability that a number selected at random from 41...
## Assuming Equal Probability:
Not all probability scenarios involve equal likelihood. Be cautious of assuming equal probability without explicit information.
## Overlooking Complementary Events:
Complementary events are mutually exclusive outcomes that add up to a total probability of 1. Failing to consider complementary events can lead to incorrect calculations. For example, if the probability that an event will occur = 3/5, then the probably that the event will not occur = 1 – 3/5 = 2/5.
## Relying Solely on Intuition:
While intuition can be helpful, relying solely on it without proper calculations and reasoning may lead to incorrect answers. Always verify your intuition with proper probability techniques.
By being aware of these common mistakes, you can approach probability questions on the SAT with greater accuracy and confidence. Practice applying the correct techniques and double-check your calculations to avoid these errors and improve your overall performance. |
# HOW TO FIND THE FIRST THREE TERMS OF A GEOMETRIC SEQUENCE
How to Find the First Three Terms of a Geometric Sequence ?
A Geometric Progression is a sequence in which each term is obtained by multiplying a fixed non-zero number to the preceding term except the first term. The fixed number is called common ratio. The common ratio is usually denoted by r.
General form of geometric progression :
a, ar, ar2, ar3,...................
Here a = first term and r - t2/t1
## How to Find the First Three Terms of a Geometric Sequence - Questions
Question 1 :
Write the first three terms of the G.P. whose first term and the common ratio are given below.
(i) a = 6, r = 3
Solution :
First term (a) = 6
Second term = ar = 6(3) = 18
Third term = ar2 = 6(3)2 = 54
Hence the first three terms are 6, 18, 54.
(ii) a = 2, r = √2
Solution :
First term (a) = 2
Second term = ar = 2(2) = 2
Third term = ar2 = 2(2)2 = 2(2) = 22
Hence the first three terms are 2, 2, 22
(iii) a = 1000, r = 2/5
Solution :
First term (a) = 1000
Second term = ar = 1000(2/5) = 400
Third term = ar2 = 1000(2/5)2 = 1000(4/25) = 160
Hence the first three terms are 1000, 400, 160.
## How to Find the Indicated Term of a Geometric Sequence ?
nth term of a geometric sequence :
tn = ar n -1
Question 1 :
In a G.P. 729, 243, 81,… find t7 .
Solution :
tn = ar n -1
a = 729, r = 243/729 = 1/3 and n = 7
t7 = (729) (1/3)7 -1
= (729) (1/3)6
t7 = 1
Question 2 :
Find x so that x + 6, x + 12 and x + 15 are consecutive terms of a Geometric Progression.
Solution :
b = √ac
(x + 12) = √(x + 6) (x + 15)
Taking squares on both sides,
(x + 12)2 = (x + 6) (x + 15)
x2 + 122 + 2x(12) = x2 + 15x + 6x + 90
144 + 24x = 21x + 90
24x - 21x = 90 - 144
3x = -54
x = -18
Hence the value of x is -18.
Question 3 :
Find the number of terms in the following G.P.
(i) 4, 8, 16,…,8192 ?
Solution :
Let nth term be "8192"
tn = 8192
a = 4, r = 8/4 = 2
ar n -1 = 8192
4(2)n -1 = 8192
22(2)n -1 = 8192
2 n+1 = 213
n + 1 = 13
n = 12
Hence the 12th term of the above geometric sequence is 8192.
(ii) 1/3, 1/9, 1/27,................1/2187
Solution :
Let nth term be "1/2187"
tn = 1/2187
a = 1/3, r = (1/9)/(1/3) = 1/3
ar n -1 = 1/2187
(1/3)(1/3)n -1 = 1/2187
(1/3)1 + n-1 = 1/2187
(1/3)n = (1/3)7
n = 7
Hence the 7th term of the above sequence is 1/2187.
After having gone through the stuff given above, we hope that the students would have understood, "How to Find the First Three Terms of a Geometric Sequence".
Apart from the stuff given in this section "How to Find the First Three Terms of a Geometric Sequence"if you need any other stuff in math, please use our google custom search here.
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# Moment of Couple
By BYJU'S Exam Prep
Updated on: September 25th, 2023
The Moment of Couple, also known as torque, results from the forces forming the couple and the couple’s arm. Before going into specifics of the couple’s moment, let’s first understand the couple. A couple is a specific case of a moment. A couple is made up of two parallel forces that are equal in magnitude, opposite in a sense, and have no common line of action. It simply produces rotation, not translation. A couple’s resulting force is zero. However, the resultant of a couple is not zero; it is a pure moment.
In mechanics, a couple is defined as a pair of equal parallel forces pointing in opposite directions. The only effect a couple can have is on how a body turns. The magnitude of either force is multiplied by the angle between their action lines to get the couple’s turning effect or moment. Let’s now look further into the concepts of the moment of couple.
## What is a Moment of Couple?
The moment of couple is equal to any force multiplied by the perpendicular separation between the two forces. When two equal and opposite forces are applied simultaneously at different points on a body, their resultant force is zero. Still, these equal and opposite forces try to rotate the body in the same direction. These forces are called Couples.
### Moment of Couple Example
Such a force pair has a tendency to rotate that body. Consider A and B as the two points on a body on which two forces F of the equal result are applied in opposite directions parallel to each other. Due to the effect of this force, this body starts rotating in the anti-clockwise direction or tries to rotate about the axis passing through 0. Keep in mind that if the line of action of these two equal forces is the same or they act at the same point, they will cancel each other.
### Moment of Couple PDF
A moment of couple operates on a body as a result of a couple’s activity, causing the body to spin around a fixed point.
### SI Unit of Moment of Couple
A couple is a pair of forces with an equal magnitude but a different line of force action. The product of either force and the separation between the forces’ perpendicular axes yields the moment of the couple. Its SI unit of moment of couple is Nm.
## Characteristics of the Couples
A couple is a pair of forces equal in magnitude, oppositely directed, and separated by a perpendicular distance or moment. The most basic pair type consists of two equal and opposing forces with opposing lines of action. The following are the characteristics of the couple:
• The couple does not produce translational motion because the two forces that make up the couple are equal and in opposition.
• When it operates on a body, the net resulting force on the body is zero.
• Pure rotational motion in the body results from the algebraic sum of the moments of the two forces around any point in their plane is not zero.
• The moment of a pair around any point in its plane is fixed in size and orientation.
## Mathematical Representation of Moment of Couple
The moment of the couple is described mathematically as the product of the force and the perpendicular distance between the two forces’ lines of action. The arm of the Couple is another name for the distance that separates two forces’ directions of action perpendicularly in this context.
### Moment of Couple Formula
The applied force multiplied by the sum of the arms of any two forces gives the moment of force. Therefore, the moment of couple of forces is provided by,
Τ = F×d
The formula for the moment of a pair of forces clearly states that the moment of couple of forces will be more extraordinary if:
• The force is more significant in terms of magnitude.
• The distance between the lines of action of the two forces is greater because the arm of the pair of forces is longer.
## Practical Application of Moment of Couple
The resultant force of two equal and opposing forces acting simultaneously on distinct places of a body is zero. Still, these forces attempt to rotate the body in the same direction. These forces are referred to as Couple. A force pair of this type has a tendency to rotate that body. The following are practical applications for couples.
• We use our hands to force the handle when turning a moving bicycle.
• whenever we open and close the tap’s spout.
• A pair of forces is performed with the fingers to open and close the device’s lid.
• When the palms of both hands are run in the opposite direction while holding the pencil between them, the pencil begins to rotate.
• Another example of a pair is when the churn begins to rotate while holding the ends of the rope with one hand while moving the other in the opposite direction.
Important GATE Topics Fixed End Moment Gravity Of Earth Slope Deflection Equation Capacitors in Parallel Capacitors in Series Carnot Cycle Cement Test Clamping Circuit CMOS Converter Column Base
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Ratio
Learning outcome: After completing this module the student should know the meaning of percent, how to calculate percentages, how to divide the whole into the desired number of equal parts. Students should know what fractions are called simple and when to use them.
Introduction
One of the most common ways that fractions are used is money. A quarter is 1/4 of a euro, a dime is 1/100 of a euro. Knowing about fractions and how to compare them makes handling money much easier.
We also use fractions when we cook. Ingredients are often measured in fractions e.g. 1/4 kg instead of 250 g.
Even small children face the problem of sharing an apple equally. When they realize that none of them will get the whole apple, they split the apple in two parts.
###### 4.1 Fractions
The one (1) can be divided to smaller parts, e.g. into 2, 3, 4, 5 … and as many times as we need. What‘s more, we will keep those parts as new units and even count them:
One half 1/2, two halves 2/2, three halves 3/2; five halves 5/2, etc.
These are simple fractions.
Each equal part of a whole that is divided in two is called one-half. It is expressed as 1/2 and read as one over two or one upon two.
Source
We can divide the sheet further into four equal parts. Each equal part is called one-fourth or a quarter of the whole sheet. Thus, any whole can be divided into four equal parts and each part is one-fourth or a quarter of the whole. It is expressed as 1/4 and read as one over four or one upon four.
Each part of the fraction has its own name:
Divide the same line into 2, 3, 4 parts. Continue to divide. We are getting more and more lines. Their intervals are getting shorter though. Here are some examples of the number of segments we can get:
It is easy to add fractions with the same denominator (same bottom number):
Again: 6/8 can be simplified to 3/4
What do we do when the denominators (the bottom numbers) are not the same?
3/8 + 1/4 = ?
We need to make them the same before we can continue, because we cannot add them like that. We must somehow make the denominators the same.
In this case it is easy, because we know that 1/4 is the same as 2/8 :
3/8 + 2/8 = 5/8
However, if the bottom numbers (denominators) are different to make them identical we should multiply the nominator and denominator of the smallest fraction (to keep it easy) by 2. We have to multiply both the top and bottom, by the same number to keep the value of the fraction unchanged.
4 is the smallest denominator, if we multiply by 2, we get 8. This leaves the value of the fraction the same. Now that the fractions have the same bottom number (“8”), we can continue adding.
Add the top numbers and put them over the same denominator, as shown below: |
# What is graphing absolute value?
## What is graphing absolute value?
So basically it’s a graph that looks like a v pointing in the upward. Direction.
### What are two absolute values?
All you have to remember is thinking about if two numbers are equal and absolute value what does that mean it means that they sort of must have the same size.
#### What is the absolute value equation?
Recall that absolute value means distance from zero. The equation |x|=5 has two solutions. The statement asks for the values of x that are five units from zero. Both 5 and -5 are five units from zero. Therefore, x = 5 or x = -5.
Why do absolute value equations have 2 solutions?
And represents the distance between a and 0 on a number line. Has two solutions x = a and x = -a because both numbers are at the distance a from 0.
How do you solve an absolute value equation by graphing?
Solve absolute value equations by graphing – YouTube
## What is necessary to graph the absolute value function?
Recall that the absolute value of a number is its distance from 0 on the number line. To graph an absolute value function, choose several values of x and find some ordered pairs. Plot the points on a coordinate plane and connect them. Observe that the graph is V-shaped.
### What does two absolute value bars mean?
‖a‖ in general means the “norm” of a. Most commonly it means the Euclidean norm of the vector a. You could say “the geometric length of a” or “the magnitude of a” to refer to the same concept.
#### How do you solve for 2 absolute values?
Solving Equations with 2 Absolute Values – YouTube
What is absolute value example?
Absolute Value Examples and Equations
|6| = 6 means “the absolute value of 6 is 6.” |–6| = 6 means “the absolute value of –6 is 6.” |–2 – x| means “the absolute value of the expression –2 minus x.” –|x| means “the negative of the absolute value of x.”
Does absolute value equations have two solutions?
Summary. Absolute value equations are always solved with the same steps: isolate the absolute value term and then write equations based on the definition of the absolute value. There may end up being two solutions, one solution, or no solutions.
## How do you find two solutions with absolute value?
Solving an Absolute Value Equation with Two Solutions – YouTube
### How do you graph absolute value equations and inequalities?
How To Graph Absolute Value Functions – YouTube
#### What is an absolute value function in your own words?
The absolute value of a number is the number’s positive value. Formally, the absolute value of a number is its distance from zero on a number line. The absolute value function is f(x) = |x|. The output is the positive value of the input.
How do you solve double absolute value equations?
Double Absolute Value Equations.wmv – YouTube
What do the absolute value bars mean?
The absolute value of a number n is the distance of the number n from zero. The absolute value is denoted by vertical bars as | n |, and is read aloud as “the absolute value of enn”.
## What is the meaning of absolute number?
Absolute Values. The absolute value of a number refers to the distance of a number from the origin of a number line. It is represented as |a|, which defines the magnitude of any integer ‘a’. The absolute value of any integer, whether positive or negative, will be the real numbers, regardless of which sign it has.
### Why do we use absolute value?
When you see an absolute value in a problem or equation, it means that whatever is inside the absolute value is always positive. Absolute values are often used in problems involving distance and are sometimes used with inequalities.
#### How do you write an absolute value equation with two solutions?
Writing Absolute Value Equations Given Solutions – YouTube
How do you solve an absolute value equation step by step?
SOLVING EQUATIONS CONTAINING ABSOLUTE VALUE(S)
1. Step 1: Isolate the absolute value expression.
2. Step2: Set the quantity inside the absolute value notation equal to + and – the quantity on the other side of the equation.
3. Step 3: Solve for the unknown in both equations.
4. Step 4: Check your answer analytically or graphically.
How do you graph an absolute value equation?
Absolute Value Function (How to Graph) – YouTube
## What is a sentence for absolute value?
Market power is greatest when the absolute value of is greatest. The final two columns show the percentage of cases that are less than 0.01 and 0.02 in absolute value. The leading edge was detected using the absolute value of the time derivative d8/dt for discrimination. It contained among absolute value.
### What are the rules of absolute value?
The absolute value (or modulus) | x | of a real number x is the non-negative value of x without regard to its sign. For example, the absolute value of 5 is 5, and the absolute value of −5 is also 5.
#### What are the absolute value lines called?
bars
We use bars (vertical lines) on either side of a number to mean absolute value.
What is another name used for absolute value?
numerical value
Other names for absolute value include numerical value and magnitude.
What is meant by the absolute value of a variable?
Absolute value equations are equations where the variable is within an absolute value operator, like |x-5|=9. The challenge is that the absolute value of a number depends on the number’s sign: if it’s positive, it’s equal to the number: |9|=9. If the number is negative, then the absolute value is its opposite: |-9|=9.
### What are the main themes in Richard III?What are the main themes in Richard III?
What are the main themes in Richard III? Themes The Allure of Evil. When Richard claims that his deformity is the cause of his wicked ways, he seems to be
### What is WPA in networking?What is WPA in networking?
What is WPA in networking? Wi-Fi Protected Access (WPA) is a security standard for computing devices equipped with wireless internet connections. WPA was developed by the Wi-Fi Alliance to provide
### What is the Tom Hanks new movie?What is the Tom Hanks new movie?
What is the Tom Hanks new movie? A Man Called Otto is an American remake of the Oscar-nominated Swedish comedy movie based on Fredrik Backman’s best-selling novel A Man Called |
# Activity A1- Antiderivatives When the derivative, , of a function,
```ACOW INDEFINITE INTEGRALS MODULE
Updated 5/28/2016
Page 1 of 2
Activity A1- Antiderivatives
When the derivative, f ( x) , of a function, F ( x) , is known, we say that F ( x) is the
3
antiderivative of f ( x) . For example, given f ( x) x 2 2 x 4 , we can determine if
2
1 3
F ( x) x x 2 4 x 7 is the antiderivative of f ( x ) by finding if F '( x) f ( x) . Since
2
3
F '( x) x 2 2 x 4 f ( x) , we conclude f ( x ) is the antiderivative of F ( x) .
2
1. Which of the following is the antiderivative of f ( x) 3x 2 4 x 8 ?
(a)
F ( x) x 3 4 x 2 8
(b)
F ( x) 3x 3 2 x 2 8 x 7
(c)
F ( x) 3 x 3 4 x 2 8 x
(d)
F ( x) x 3 2 x 2 8 x 5
(e)
None of the above.
1 3
1
x x 2 4 x 7 , G ( x) x 3 x 2 4 x 10 , and
2
2
1
3
H ( x) x 3 x 2 4 x 4 are all antiderivatives of f ( x) x 2 2 x 4 . Why? Recall,
2
2
that the derivative of a constant is zero and thus we “lose” the constant in the derivative
process. Consequently, when we try to find the antiderivative of a function, we do not
know the actual value of the constant. Thus we say the most general antiderivative of
3
1
f ( x) x 2 2 x 4 is F ( x) x3 x 2 4 x C where C is any constant.
2
2
It is important to realize F ( x)
2. Which of the following is the most general antiderivative of f ( x) 4 x3 3x 2 7 ?
(a)
F ( x) x 4 x 3 7 x C
(b)
F ( x) x 4 x 3 7 x
(c)
F ( x) x 4 x 3 7 x 2
(d)
F ( x) x 4 x 3 7 x 7
When we are given a function and are asked to find it’s antiderivative, we must use the
Power Rule for Antiderivatives(RS2). Notice the Power Rule for Antiderivatives states
n 1 . This is necessary because if n 1 , then we would have the function
f ( x) x 1 . Trying to apply the power rule would give
ACOW INDEFINITE INTEGRALS MODULE
Updated 5/28/2016
Page 2 of 2
F ( x)
This is an undefined function because
1 0
x
0
1
is an undefined number.
0
Use the Power Rule for Antiderivatives to find the most general antiderivative of the
following functions.
3. f ( x) x5 .
4. f ( x) x .
5. f ( x)
1
x7
6. f ( x) 5 x3
``` |
# 1988 AHSME Problems/Problem 14
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Problem
For any real number a and positive integer k, define
${a \choose k} = \frac{a(a-1)(a-2)\cdots(a-(k-1))}{k(k-1)(k-2)\cdots(2)(1)}$
What is
${-\frac{1}{2} \choose 100} \div {\frac{1}{2} \choose 100}$?
$\textbf{(A)}\ -199\qquad \textbf{(B)}\ -197\qquad \textbf{(C)}\ -1\qquad \textbf{(D)}\ 197\qquad \textbf{(E)}\ 199$
## Solution
We expand both the numerator and the denominator.
\begin{align*} \binom{-\frac{1}{2}}{100}\div\binom{\frac{1}{2}}{100} &= \frac{ \dfrac{ (-\frac{1}{2}) (-\frac{1}{2} - 1) (-\frac{1}{2} - 2) \cdots (-\frac{1}{2} - (100 - 1)) }{\cancel{(100)(99)\cdots(1)}} }{ \dfrac{ (\frac{1}{2}) (\frac{1}{2} - 1) (\frac{1}{2} - 2) \cdots (\frac{1}{2} - (100 - 1)) }{\cancel{(100)(99)\cdots(1)}} } \\ &= \frac{ (-\frac{1}{2}) (-\frac{1}{2} - 1) (-\frac{1}{2} - 2) \cdots (-\frac{1}{2} - 99) }{ (\frac{1}{2}) (\frac{1}{2} - 1) (\frac{1}{2} - 2) \cdots (\frac{1}{2} - 99) } \end{align*}
Now, note that $-\frac{1}{2}-1=\frac{1}{2}-2$, $-\frac{1}{2}-2=\frac{1}{2}-3$, etc.; in essence, $-\frac{1}{2}-n=\frac{1}{2}-(n+1)$. We can then simplify the numerator and cancel like terms.
\begin{align*} \frac{ (-\frac{1}{2}) (-\frac{1}{2} - 1) (-\frac{1}{2} - 2) \cdots (-\frac{1}{2} - 99) }{ (\frac{1}{2}) (\frac{1}{2} - 1) (\frac{1}{2} - 2) \cdots (\frac{1}{2} - 99) } &= \frac{ \cancel{(\frac{1}{2} - 1)} \cancel{(\frac{1}{2} - 2)} \cancel{(\frac{1}{2} - 3)} \cdots (\frac{1}{2} - 100) }{ (\frac{1}{2}) \cancel{(\frac{1}{2} - 1)} \cancel{(\frac{1}{2} - 2)} \cdots \cancel{(\frac{1}{2} - 99)} } \\ &= \frac{\frac{1}{2}-100}{\frac{1}{2}} \\ &= \frac{-\frac{199}{2}}{\frac{1}{2}} \\ &= \boxed{\textbf{(A) } -199.} \end{align*} |
# DERIVATIVE OF a TO THE POWER x
To find derivative of a function in which you have variable in exponent, you have to use logarithmic derivative. The following steps would be useful to do logarithmic derivative.
Lett y = f(x) be a function in which let the variable be in exponent.
Step 1 :
Take logarithm on both sides.
Step 2 :
Apply the power rule of logarithm.
Step 3 :
Find the derivative and solve for ᵈʸ⁄d.
Derivative of ax with respect to x.
Let y = ax.
In y = axwe have variable x in exponent.
y = ax
Take logarithm on both sides.
ln(y) = ln(ax)
Apply the power rule of logarithm on the right side.
ln(y) = xln(a)
Find the derivative with respect to x.
(Since a is a constant, ln(a) is also a constant. When we find derivative xln(a), we keep the the constant ln(a) as it is and find the derivative of x with respect to x, that is 1)
Multiply both sides by y.
Substitute y = ax.
Therefore, the derivative ax is axln(a).
Note :
Based on the derivative of ax, that is axlna, we can get the derivative of any constant to the power x.
Examples :
Derivative 2x = 2xln(2)
Derivative 3x = 3xln(3)
Derivative 5x = 5xln(5)
## Solved Problems
Find ᵈʸ⁄d.in each of the following.
Example 1 :
y = ex
(x and y are variables and e is a constant)
Solution :
In y = ex, we have constant e in base and variable x in exponent.
y = ex
Take logarithm on both sides.
ln(y) = ln(ex)
Apply the power rule of logarithm on the right side.
ln(y) = xln(e)
(The base of a natural logarithm is e, lne is a natural logarithm and its base is e)
ln(y) = xlnee
In a logarithm, if the base and argument are same, its value is 1. In lnee, the base and argument are same, so its value is 1.
ln(y) = x(1)
ln(y) = x
Find the derivative with respect to x.
Multiply both sides by y.
Substitute y = ex.
Example 2 :
y = 72x
Solution :
In y = 72x, we have variable in exponent.
y = 72x
Take logarithm on both sides.
ln(y) = ln(72x)
Apply the power rule of logarithm on the right side.
ln(y) = 2xln(7)
Find the derivative with respect to x.
Multiply both sides by y.
Substitute y = 72x.
Example 3 :
y = 3x
Solution :
In y = 3x, we have variable in exponent.
y = 3x
Take logarithm on both sides.
ln(y) = ln(3x)
Apply the power rule of logarithm on the right side.
ln(y) = xln(3)
Find the derivative with respect to x.
Multiply both sides by y.
Substitute y = 3x.
Example 4 :
y = 2ln(x)
Solution :
In y = 2ln(x), we have variable in exponent.
y = 2ln(x)
Take logarithm on both sides.
ln(y) = ln(2ln(x))
Apply the power rule of logarithm on the right side.
ln(y) = ln(x) ⋅ ln(2)
Find the derivative with respect to x.
Multiply both sides by y.
Substitute y = 2ln(x).
Example 5 :
y = 5sinx
Solution :
In y = 5sinx, we have variable in exponent.
y = 5sinx
Take logarithm on both sides.
ln(y) = ln(5sinx)
Apply the power rule of logarithm on the right side.
ln(y) = sinx ⋅ ln(5)
Find the derivative with respect to x.
Multiply both sides by y.
Substitute y = 5sinx.
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# Converting Units
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To be able to convert from one type of unit to another, we need to know the equivalence between both units. We can convert between units as long as both use the same basic units because they must measure the same physical property.
## Consistency in converting units
Consistency is needed when converting units. We cannot convert time to length, but we can translate time to frequency because both use time as a base. We can also convert power units to watts as energy per second units, and so on. Lets look at these two examples in more detail.
We want to convert the oscillation of a pendulum in time to its frequency. The period (T ), expressed in seconds, is the time it takes to complete one cycle of an oscillation. The frequency (f ) is the number of occurrences of a repeating event per unit of time and is measured in hertz. The formula to convert from period to frequency is f = 1/T . The unit equivalence using this formula is:
The inverse of the value ‘x’ for T in seconds gives us the value Y in Hertz.
If the pendulum takes 3.2 seconds to come and go, we need to divide 1 by 3.2 seconds.
This gives us 0.3125 [Hertz].
Lets say we have a machine that consumes 60 watts of power each second. We want to convert power consumed to energy per second. The equation that links power, energy, and time is:
Here, P is the power in watts, E is the energy in joules, and t is the time it takes to consume or produce energy in seconds. If the machines consumption and energy production are measured each second, then 60 watts means 60 joules every second.
The relationship below expresses this better. Here, every watt unit is equivalent to a unit of Joules per second.
Replace the watts with 60:
Now lets say we have a machine that produces 100 joules watts each minute. We want to know how much power is produced every second. We have to divide the amount of energy in joules by the number of seconds it took the machine to produce 100 watts.
As we know, Joules per second is watts.
Figure 1. Hertz is widely used to measure frequency. Source: JCHaywire, Flickr (CC BY-SA 2.0).
## Converting larger units to smaller units
To convert larger units to small ones, we need to multiply by a factor. If we wish to convert from different scales and units, we need to multiply by two factors to scale the number and convert between derived units.
### Converting basic units from different scales
To convert between basic units from a smaller to a larger scale, we need to multiply by a factor. If A is ten times B, we need to multiply B by 10 to obtain A. Lets look at some examples.
We want to convert 1,234 tons to kilograms. We know that a ton is 1000 kilograms, so we can carry this out by multiplying 1,234 by 1000. This gives us 1234 kilograms.
We want to convert 0.3 metres to millimetres. We know that 1 millimetre is equal to 1 ⋅ 10 ^ -3 meters, so we need to divide 0.3 by 1 ⋅ 10 ^ -3, which gives us 300 millimetres.
We can also convert from metres to millimetres by multiplying by 1000, as 1 metre equals 1000 millimetres.
### Converting derived units from different scales
To convert between derived units and from a larger scale to a smaller one, we need to multiply by several factors. Consider the following example.
Convert 10km/h to m/s.
Our calculations are more complex here. First, we need to convert 10 kilometres to metres. To convert kilometres to metres, we use the factor of 1 ⋅ 10 ^ 3, giving us a velocity of 10000[m/h].
Now we need to convert from hours to seconds. This factor equals 3600, as 1 hour is equal to 60 minutes, and each minute to 60 seconds.
We must, therefore, divide 10000m by 3600s.
The result is 2.8m/s.
You can use a rule of thumb to calculate km/h to m/s just by dividing the number of km/h by 3.6.
If we do this at 10km/h, we obtain the same result:
## Converting smaller units to larger units
To convert smaller to larger units, we need to divide by a factor. As noted earlier, if we wish to combine conversion from different scales and units, we need to divide by two factors, one to scale the number and another to convert between derived units.
### Converting basic units from different scales
To convert between basic units from a smaller to a larger scale, we need to divide by a factor. If, for instance, A is ten times larger than B, we need to divide A by 10 to obtain B. See the following two examples:
We want to convert 23.4m to kilometres. As one kilometre is 1000 metres, we need to divide 23.4 by 1000, which gives us 0.023 kilometres.
We wish to convert 400 kelvin to megakelvin. The prefix mega means 1 ⋅ 10 ^ 6, so one megakelvin is one million kelvin. Dividing 400 kelvin by 1,000,000 gives us 0.0004 megakelvin.
### Converting derived units from different scales
To convert between derived units from small to large scales, we need to use several factors. A more complex conversion from watts to kilonewton-metres per second is required, for instance, in the example below.
We have a machine that consumes 1300 watts. The prefix kilo is equivalent to 1 ⋅ 10 ^ 3 in standard form. This means that we need to divide 1300 watts by 1 ⋅ 10 ^ 3 to get kilowatts.
In a second step, we need to convert kilowatts to newton-metres per second. As 1 watt is equivalent to 1 newton-metre per second, this is straightforward. 1.3 kilowatts are equal to 1.3 kilonewton-metres per second.
## Converting units from different systems
We might need to convert units from different systems, such as the imperial and SI systems. Converting temperature, volume, and length between imperial and SI units are three common operations. A simple way to convert between imperial and SI is by using weights. Multiplying the imperial or SI values by the correct weight gives us the value in the other unit system.
Figure 2. Gallons are a commonly used unit from the imperial system. Source: Matt Gibson, Flickr (CC BY 2.0).
The following table lists the conversion weights for converting between the imperial and SI systems.
Imperial to SI SI to imperial Imperial unit Conversion weight SI unit Conversion weight 1 gallon 3.7854 litres 1 litre 0.264172 gallons 1 mile 1.60934 kilometres 1 kilometre 0.621371 miles 1 foot 0.3048 metres 1 metre 3.28084 feet 1 pound 0.453592 kilograms 1 kilogram 2.20462 pounds
To convert between Fahrenheit and Celsius, we need to use the formulas below:
### Examples of converting between imperial and SI units
The conversion of units between systems is very common in everyday life, as the imperial system and the US Customary System of units (USCS) are still widely used. See the following examples.
The outside temperature is given as 32 degrees Fahrenheit. How much is that in Celsius?
Replacing with 32, we get:
It is a cold day!
You need to refuel your rental car during your holidays in the USA. The car is from Europe, and the tank has a capacity of 40 litres. The filling station sells the fuel in gallons, which cost $3.10 (USD). How much would it cost you to fill up the tank? First, you need to convert the 40 litres to gallons applying the unit factor weight in the table above, which tells you that 1 litre is equivalent to 0.26 gallons. Then, you need to multiply this by the price of$3.10.
10.4 ⋅ 3.1 [USD] = 32.24 [USD]
## Converting Units - Key takeaways
• Converting units allows us to translate values from one type of physical quantity to another as, for instance, when we need to translate the work done by a device to the amount of energy per second used by that device to perform the work.
• Converting units between different scales helps us understand the scale of the values with which we are working as, for instance, when we convert the speed of an aeroplane from km/h to m/s.
• Unit conversion is present in all fields of science and technology.
To convert one measurement into another one, we need to make sure that both measurements use the same basic units. We also need to know the equivalence between the units used in the two measurements.
One example is to convert a measurement done in cm to meters. Both measure the same physical quantity that is length, both use the same units (meters) and then you will only need to multiply the factor that converts cm to meters or 1[cm]=0.01[m].
No, to be able convert between derived units, they need to have the same base units. We cannot, for instance, convert hertz to kilograms as one measures the property of mass and the other the property of time.
## Final Converting Units Quiz
Question
Can we convert between two basic units like second and kilogram?
No, unit conversion between basic units is meaningless.
Show question
Question
Convert 3 hertz to a period in seconds.
T= ⅓ second.
Show question
Question
Do Watts and Joules per second represent the same physical quantity?
Yes, they both represent energy.
Show question
Question
Can we convert between different derived units?
Yes, provided they share the same basic quantities.
Show question
Question
Convert 25km/h to m/s.
6.94m/s.
Show question
Question
Convert 32m/s to km/h.
115.2km/h.
Show question
Question
Convert 0.6 kilowatts to Joules/second.
600 joules/second.
Show question
Question
Why can we not convert between two different basic units?
Because they are not measuring the same physical property.
Show question
Question
If you want to convert units from kilograms to grams, how many factors do you need?
One.
Show question
Question
Can you convert work to energy per time?
Yes, you can.
Show question
Question
Can you convert the energy per time to work units?
Yes, you can.
Show question
Question
If you want to convert a derived unit to basic units using a different scale, how many factors do you need?
You need more than one factor.
Show question
Question
Convert 6km/h to m/s.
1.6m/s.
Show question
Question
Convert 10m/s to km/h.
36km/h.
Show question
Question
Convert 3.4 kilokelvin to Celsius.
3126.85 Celsius.
Show question
Question
If a pendulum makes 100 oscillations per second, how many hertz is this?
100 hertz.
Show question
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Statistics: Standard Deviation Prob.
1. Jan 19, 2007
Mesmer
1. The problem statement, all variables and given/known data
This is my data: 42.4, 65.7, 29.8, 58.7, 52.1, 55.8, 57.0, 68.7, 67.3, 67.3, 54.3, 54.0 I need to find the standard deviation of this list of data.
2. Relevant equations standard deviation= $$: s^2=\frac{\sum_{i=1}^n(x_i -\overline{x})}{n-1}$$ Actually the standard deviation is the square root of this expression.
3. I can use a statistics package like minitab or excell to solve this but how would I do it my hand? where would I start?[/b]
Last edited: Jan 19, 2007
2. Jan 19, 2007
mattmns
Your formula is not correct, is that what messed you up? It should be:
$$s^2=\frac{\sum_{i=1}^n(x_i -\overline{x})^2}{n-1}$$
If that was not the problem then is there something you don't know how to do with the formula? Do you know what $x_i$ and $\overline{x}$ are?
$x_i$ represents the data (for example x_1 is 42.4 and x_4 is 58.8)
And $\overline{x}$ is the mean = $$\frac{x_1+x_2+...+x_n}{n}$$
Here is an example.
My data are the following: 14.3, 17.5, 15.6
So the mean, $$\overline{x} = \frac{14.3+17.5+15.6}{3} = 15.8$$
So then
$$(x_1 -\overline{x})^2 = (14.3 - 15.8)^2 = (-1.5)^2 = 2.25$$
$$(x_1 -\overline{x})^2 = (17.5 - 15.8)^2 = (1.7)^2 = 2.89$$
$$(x_1 -\overline{x})^2 = (15.6 - 15.8)^2 = (-.2)^2 = .04$$
Hence,
$$s^2 = \frac{2.25 + 2.89 + .04}{3-1} = \frac{5.18}{2} = 2.59$$
So our standard deviation s is then the square root of that which means s = 1.60935
Last edited: Jan 19, 2007
3. Jan 19, 2007
Mesmer
Thanks, I did latex my equation wrong, but you also cleared up my confusion about how the expression works. :) I see my mistake in understanding now.
4. Jan 20, 2007
2ltben
The equation is also for variance. If you want to find the actual standard deviation, you'll still need to take the square root of that.
5. Jan 20, 2007
HallsofIvy
Staff Emeritus
By the way since
$$\Sigma_{i=1}^n (x_i- \overline{x})^2= \Sigma_{i=1}^nx^2- \overline{x}\Sigma_{i=1}^n x_i+ \overline{x}^2\Sigma_{i=1}^n 1$$
$$= \Sigma_{i=1}^n x_i^2- 2n\overline{x}+ n\overline{x}= \Sigma_{i=1}^n x_i^2- n\overlne{x}$$
(because
$\Sigma_{i=1}^n x_i= n\overline{x}$
and
$\Sigma_{i=1}^n 1= n$
$$\sigma^2= \frac{1}{n-1}\Sigma_{i=1}^n x_i^2- \frac{n}{n-1}\overline{x}$$
That's often easier to calculate. |
Question Video: Deriving the Formula for Calculating the Area of a Triangle Using Sine | Nagwa Question Video: Deriving the Formula for Calculating the Area of a Triangle Using Sine | Nagwa
Question Video: Deriving the Formula for Calculating the Area of a Triangle Using Sine Mathematics • First Year of Secondary School
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In the given figure. Find an expression for the height β in terms of π, π, and π. Find an expression for the area of the triangle in terms of π, π, and π.
02:48
Video Transcript
In the given figure, find an expression for the height β in terms of π, π, and π. Find an expression for the area of the triangle in terms of π, π, and π.
Letβs begin by finding an expression for the height β in our triangle. If we look carefully, we can partition this triangle somewhat. And in doing so, we create a smaller triangle with an angle of π degrees and a length of π.
Now in fact, this length π represents the length of the hypotenuse of this right-angle triangle. And we know that itβs a right-angle triangle because angles on a straight line sum to 180 degrees. And since we have an angle of 90 degrees, we can find the angle on the other side of the dotted line by subtracting 90 from 180. That gives us 90 degrees. And therefore, we have a right-angle triangle. And since π is the longest side in the triangle, itβs the side opposite the right angle we know itβs a hypotenuse.
We can also say that the side in this triangle labelled as lowercase β is the opposite side of the triangle. Itβs the side opposite the included angle π degrees. And when we notice that we know the length of the opposite and the hypotenuse, we can see that we need to use the sine ratio, where sin of π is equal to opposite over hypotenuse.
Substituting what we know about our right-angle triangle into this formula, we see that sin π is equal to β, which is the opposite, over the hypotenuse, which is π. So sin π is equal to β over π. And since the aim of this question is to find an expression for the height β, we need to make β the subject. And we do so by multiplying both sides of our equation by π. β over π multiplied by π is simply β. And sin π multiplied by π is written π sin π.
And we see that in fact this question was a little bit misleading. An expression for the height β in terms of π, π, and π doesnβt actually include π. Itβs simply π sin π.
Next, weβre going to find an expression for the area of the triangle in terms of π, π, and π. This time, we recall the formula for area of a triangle. Itβs a half multiplied by its base multiplied by its perpendicular height. The base of our triangle is π units. And the perpendicular height is β. So we could say that the area is a half πβ.
But of course, weβre looking to find an expression for the area of the triangle in terms of π, π, and π. We just saw that β is equal to π sin π. So we can see that the area is a half π multiplied by π sin π. And if we fully simplify this expression, the area of our triangle is a half ππ sin π.
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# Calculator Strategies for the Revised GRE
For many students, the addition of the GRE’s onscreen calculator to the revised exam is a godsend. These students take solace in the notion that this new calculator will help them solve tons of questions. The truth of the matter is that almost all math questions on the revised GRE can be solved without a calculator. Furthermore, in many cases, it will actually take long to solve a question using a calculator that it will to use other techniques. Finally, the test-makers are taking questions that can be easily solved with a calculator and changing the numbers in order to render the calculator useless.
For example, a former GRE question would have asked you to evaluate – . The slow solution was to perform the actual (tedious) calculations. The fast solution was to recognize that this difference of squares can be factored as , which equals , which equals .
Since this question would be too easy to solve using the onscreen calculator, the test-makers will change the question to where and have too many digits for the calculator to handle. As such, you’ll have to solve this question using factoring techniques.
Aside: the onscreen calculator displays up to eight digits. If a computation results in a number greater than then an ERROR message is displayed. When you evaluate you get a 9-digit number.
Now, despite the test-makers’ attempts to remove the calculator from your arsenal, there are times when you can make a few adjustments to a question and then quickly answer it with the calculator.
Now, in my last post, we solved the following question using a variety of techniques and strategies:
Column A Column B
1. The quantity in Column A is greater
2. The quantity in Column B is greater
3. The two quantities are equal
4. The relationship cannot be determined from the information given
Notice that these numbers yield products that are too big for the calculator to handle. However, with a few adjustments we can use a new strategy with the calculator to answer the question.
One solution is to first divide each column by 1,000,000. When we do this, we get:
Column A Column B
From here, we can rewrite this as:
Column A Column B
And this is the same as:
Column A Column B
At this point, we can use the calculator enter all of these values, and each resulting product will have fewer than 8 digits.
So, with a small modification, we can answer this question using a calculator.
Now, can you think of another approach that allows you to use a calculator to solve the original question (without dividing by 1,000,000 or any other powers of 10)?
Here’s the original question:
Column A Column B
Another approach is to first divide both sides by 641,713 to get:
Column A Column B
Then, divide both sides by 897,189 to get:
Column A Column B
At this point, we can enter all of these values into the calculator and compare the columns.
Next, we’ll examine another strategy to thwart the revised GRE and use the onscreen calculator to solve questions that, at first glance, appear to render the calculator useless:
The square root of 2 billion is between
1. 2,000 and 5,000
2. 5,000 and 15,000
3. 15,000 and 30,000
4. 30,000 and 50,000
5. 50,000 and 90,000
Try to identify at least two ways to solve the above question.
Aside: Please notice that 2 billion is too large to fit in the onscreen calculator.
## Non-calculator strategy:
This approach uses the following rule:
First we need to recognize that:
From here, we can see that since and then must lie between 4 and 5. In other words, we can say that equals 4.something.
If equals 4.something, then must lie between 40,000 and 50,000.
As such the answer must be D.
## Calculator strategy:
With a slight modification, we can use the onscreen calculator to solve the question within seconds.
First recognize that:
From here we can use the calculator to evaluate both roots. When we do this, we get:
So, the answer must be D.
### More from Magoosh
By the way, students who use Magoosh GRE improve their scores by an average of 8 points on the new scale (150 points on the old scale.) Click here to learn more.
### 6 Responses to Calculator Strategies for the Revised GRE
1. Eric November 2, 2014 at 10:52 pm #
For the first one, I prefer the solution given in the Magoosh videos: if I use a calculator, 64×45=2880. 90×32=2880. Since the actual numbers of A are a bit bigger, and the actual numbers of B are a bit smaller (rounded up), then A must be bigger. Simple, fast. 😉
2. Bhavesh July 25, 2014 at 7:45 am #
This is one of the import skill to save critical time.
Thanks
• Chris Lele July 28, 2014 at 11:33 am #
Glad it was helpful! Anything that helps save time–without sacrificing accuracy–is a good thing 🙂
3. Vanan December 9, 2011 at 7:36 pm #
Hi Brent,
Your tips are really awesome and Magoosh is really doing a great job publishing excellent tips for the students to ace GRE.. Hats off Magoosh!!!
Regards,
Vanan.
4. Praveen December 9, 2011 at 10:19 am #
Hello Sir,
thanks for the simple and useful tips! they make u think.. “that was so obvious y didn’t i think that at the first place! “.
Magoosh is the only site which offers so many articles with invaluable tips for free! Keep up the good work 🙂
cheers!
• Brent December 9, 2011 at 12:06 pm #
Hi Praveen,
Thanks for the feedback!
Cheers,
Brent
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#### Provide solution for RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 37 Subquestion (vi) textbook solution.
Answer : $y=x^{2}+\cos x$
Give : $\frac{d y}{d x}+y \tan x=2 x+x^{2} \tan x$
Hint : Using $\int \frac{1}{1+x^{2}} d x$
Explanation : $\frac{d y}{d x}+y \tan x=2 x+x^{2} \tan x$
$=\frac{d x}{d y}+(\tan x) y=2 x+x^{2} \tan x$
This is a linear differential equation of the form
\begin{aligned} &\frac{d x}{d y}+P x=Q \\ &P=\tan x \text { and } Q=2 x+x^{2} \tan x \end{aligned}
The integrating factor $If$of this differential equation is
\begin{aligned} &I f=e^{\int P d x} \\ &=e^{\int \tan x d x} \\ &=e^{\log |\sec x|} \; \; \; \; \; \; \; \; \; \; \quad\left[\int \tan x d x=\sec x+C\right] \\ &=\sec x \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[e^{\log e^{x}}=x\right] \end{aligned}
Hence, the solution is
\begin{aligned} &y I f=\int Q I f d x+C \\ &=y \sec x=\int\left(2 x+x^{2} \tan x\right) \sec x d x+C \\ &=y \sec x=I+C \ldots(i) \\ &=I=\int\left(2 x+x^{2} \tan x\right) \sec x d x \end{aligned}
\begin{aligned} &=\int 2 x \sec x d x+\int x^{2} \tan x \sec x d x \\ &=2\left(\frac{x^{2}}{2} \sec x-\int \frac{x^{2}}{2}(\tan x \sec x) d x\right)+\int x^{2} \tan x \sec x d x+C \\ &=\frac{2 x^{2}}{2} \sec x-\int \frac{2 x^{2}}{2}(\tan x \sec x) d x+\int x^{2} \tan x \sec x d x+C \\ &=x^{2} \sec x-\int x^{2} \tan x \sec x d x+\int x^{2} \tan x \sec x d x+C \\ &=x^{2} \sec x+C \end{aligned}
Substituting in (i)
$=y \sec x=x^{2} \sec x+C$
Dividing by $\sec\; x$
\begin{aligned} &=y=x^{2}+\frac{C}{\sec x} \\ &=y=x^{2}+C \cos x \ldots(i i) \end{aligned}
Now
\begin{aligned} &y(0)=1 \text { when } x=0, y=1 \\ &\quad=1=0^{2}+C \cos (0) \\ &\quad=1=+C(1) \quad[\cos 0=1] \\ &\; \; \; \; =C=1 \end{aligned}
Substituting in (ii)
\begin{aligned} &=y=x^{2}+(1) \cos x \\ &=y=x^{2}+\cos x \end{aligned} |
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