Ask question

Ask question

All categories

3 years ago

14

The perimeter of a rectangular playground can be no greater than 120 meters. The width of the playground cannot exceed 22 meters

. What are the possie lengths of the playground?

1 answer:

Maslowich3 years ago

6 0

Answer:

There are an infinite number of values satisfying the requirements; every couple of numbers satisfying the following conditions are valid:

base = 60-w meters

width = w meters

0 < w <= 22

Step-by-step explanation:

Since the playground has a rectangular shape, let us us call b the base of the rectangle and w its width. In order for the rectangle to satisfy the condition of P = 120, we need for the following equation to satisfy:

2b + 2w = 120

Solving for b, we get that b = (120 - 2w)/2 = 60 - w .

Given a particular value (w) for the width, the base has to be: (60-w).

Therefore, the possible lengths of the playground are (60-w, w), where 60-w corresponds to the base of the rectangle and w to its width. And w can take any real value from 0 to 22.

You might be interested in

Which statements are true for triangle lnm? check all that apply. the side opposite ∠l is nm. the side opposite ∠n is ml. the hy

balu736 [363]

The statements that are true about ΔLNM are:

Hypotenuse = LN

Opposite side to ∠L = NM

Side opposite to ∠N = ML.

<h3>What is a Right Triangle?</h3>

A right triangle has right angle (angle 90) that is opposite to the longest side (hypotenuse).

From the image given, the right triangle has a right angle at angle M, which is oppposite to the hypotenuse, LN.

Therefore, the true statements about the triangle are:

Hypotenuse = LN
Opposite side to ∠L = NM
Side opposite to ∠N = ML.

Learn more about the right triangle on:

brainly.com/question/2437195

#SPJ4

6 0

2 years ago

Write an equivalent expression for each of the following

Anna11 [10]

1) 6x
2) 2x + 4
3) 5y + 11

hope this helps!

6 0

3 years ago

interpret r(t) as the position of a moving object at time t. Find the curvature of the path and determine thetangential and norm

Igoryamba

Answer:

The curvature is $\kappa=1$

The tangential component of acceleration is $a_{\boldsymbol{T}}=0$

The normal component of acceleration is $a_{\boldsymbol{N}}=1 (2)^2=4$

Step-by-step explanation:

To find the curvature of the path we are going to use this formula:

$\kappa=\frac{||d\boldsymbol{T}/dt||}{ds/dt}$

where

$\boldsymbol{T}}$ is the unit tangent vector.

$\frac{ds}{dt}=|| \boldsymbol{r}'(t)}||$ is the speed of the object

We need to find $\boldsymbol{r}'(t)$ , we know that $\boldsymbol{r}(t)=cos \:2t \:\boldsymbol{i}+sin \:2t \:\boldsymbol{j}+ \:\boldsymbol{k}$ so

$\boldsymbol{r}'(t)=\frac{d}{dt}\left(cos\left(2t\right)\right)\:\boldsymbol{i}+\frac{d}{dt}\left(sin\left(2t\right)\right)\:\boldsymbol{j}+\frac{d}{dt}\left(1)\right\:\boldsymbol{k}\\\boldsymbol{r}'(t)=-2\sin \left(2t\right)\boldsymbol{i}+2\cos \left(2t\right)\boldsymbol{j}$

Next , we find the magnitude of derivative of the position vector

$|| \boldsymbol{r}'(t)}||=\sqrt{(-2\sin \left(2t\right))^2+(2\cos \left(2t\right))^2} \\|| \boldsymbol{r}'(t)}||=\sqrt{2^2\sin ^2\left(2t\right)+2^2\cos ^2\left(2t\right)}\\|| \boldsymbol{r}'(t)}||=\sqrt{4\left(\sin ^2\left(2t\right)+\cos ^2\left(2t\right)\right)}\\|| \boldsymbol{r}'(t)}||=\sqrt{4}\sqrt{\sin ^2\left(2t\right)+\cos ^2\left(2t\right)}\\\\\mathrm{Use\:the\:following\:identity}:\quad \cos ^2\left(x\right)+\sin ^2\left(x\right)=1\\\\|| \boldsymbol{r}'(t)}||=2\sqrt{1}=2$

The unit tangent vector is defined by

$\boldsymbol{T}}=\frac{\boldsymbol{r}'(t)}{||\boldsymbol{r}'(t)||}$

$\boldsymbol{T}}=\frac{-2\sin \left(2t\right)\boldsymbol{i}+2\cos \left(2t\right)\boldsymbol{j}}{2} =\sin \left(2t\right)+\cos \left(2t\right)$

We need to find the derivative of unit tangent vector

$\boldsymbol{T}'=\frac{d}{dt}(\sin \left(2t\right)\boldsymbol{i}+\cos \left(2t\right)\boldsymbol{j}) \\\boldsymbol{T}'=-2\cdot(\sin \left(2t\right)\boldsymbol{i}+\cos \left(2t\right)\boldsymbol{j})$

And the magnitude of the derivative of unit tangent vector is

$||\boldsymbol{T}'||=2\sqrt{\cos ^2\left(x\right)+\sin ^2\left(x\right)} =2$

The curvature is

$\kappa=\frac{||d\boldsymbol{T}/dt||}{ds/dt}=\frac{2}{2} =1$

The tangential component of acceleration is given by the formula

$a_{\boldsymbol{T}}=\frac{d^2s}{dt^2}$

We know that $\frac{ds}{dt}=|| \boldsymbol{r}'(t)}||$ and $||\boldsymbol{r}'(t)}||=2$

$\frac{d}{dt}\left(2\right)\: = 0$ so

$a_{\boldsymbol{T}}=0$

The normal component of acceleration is given by the formula

$a_{\boldsymbol{N}}=\kappa (\frac{ds}{dt})^2$

We know that $\kappa=1$ and $\frac{ds}{dt}=2$ so

$a_{\boldsymbol{N}}=1 (2)^2=4$

3 0

3 years ago

Given f(x)=|x|. Explain the difference between the graphs of g(x)=|x+5| and h(x)=|x|+5.

ale4655 [162]

G(x) would have a translation of 5 to the left, while h(x) would have a translation of 5 up

4 0

3 years ago

A cleaners recommend 1 1/2 cup of cleaner to 12 cups of water what is the ratio of cleaner to water in simplest form

Evgen [1.6K]

Answer:

8 cups of water : 1 cup of cleaner

Step-by-step explanation:

To find the ratio, divide one value by the other. 12÷1½=8, so the ratio is 8 cups of water : 1 cup of cleaner.

6 0

3 years ago