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77julia77 [94]
3 years ago
7

Shenelle has 100 meters of fencing to build a rectangular garden. The garden's area (in square meters) as a function of the gard

en's width www (in meters) is modeled by: a(w)=-(w-25)^2+625. What is the maximum area possible?
Mathematics
2 answers:
Vesnalui [34]3 years ago
5 0

Answer:

625

Step-by-step explanation:

Check the question to make sure you have the right  question

It's not 25

ad-work [718]3 years ago
4 0

Answer:

625 metres

Step-by-step explanation:

Given the area function expressed as a(w)=-(w-25)^2+625.

The maximum area occurs at when da/dw = 0

da/dw = -2(w-25)

0 = -2(w-25)

-2(w-25) = 0

w - 25  = 0

w = 25

Substitute w = 25 into the modeled equation;

Recall a(w)=-(w-25)^2+625.

a(25)=-(25-25)^2+625

a(25) = 0+625

a(25) = 625

Hence the maximum area possible is 625 metres

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Alyssa plans to borrow $3,180 at 6.5% for 3.5 years . Find the amount of simple interest she should expect to pay.
irga5000 [103]

Answer:

784.16

Step-by-step explanation:

use the formula for simple interest =

P(1+r)^t

so

3180(1.065)^3.5 - 3180

3 0
2 years ago
URGENT PLEASE HELP
OleMash [197]

Answer:

6.32

Step-by-step explanation:

6^2 - 2^2

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=6.32455532

2d.p = 6.32

3 0
2 years ago
Find the area of the region that lies inside the first curve and outside the second curve.
marishachu [46]

Answer:

Step-by-step explanation:

From the given information:

r = 10 cos( θ)

r = 5

We are to find the  the area of the region that lies inside the first curve and outside the second curve.

The first thing we need to do is to determine the intersection of the points in these two curves.

To do that :

let equate the two parameters together

So;

10 cos( θ) = 5

cos( θ) = \dfrac{1}{2}

\theta = -\dfrac{\pi}{3}, \ \  \dfrac{\pi}{3}

Now, the area of the  region that lies inside the first curve and outside the second curve can be determined by finding the integral . i.e

A = \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} (10 \ cos \  \theta)^2 d \theta - \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \ \  5^2 d \theta

A = \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} 100 \ cos^2 \  \theta  d \theta - \dfrac{25}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \ \   d \theta

A = 50 \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \begin {pmatrix}  \dfrac{cos \ 2 \theta +1}{2}  \end {pmatrix} \ \ d \theta - \dfrac{25}{2}  \begin {bmatrix} \theta   \end {bmatrix}^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}}

A =\dfrac{ 50}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \begin {pmatrix}  {cos \ 2 \theta +1}  \end {pmatrix} \ \    d \theta - \dfrac{25}{2}  \begin {bmatrix}  \dfrac{\pi}{3} - (- \dfrac{\pi}{3} )\end {bmatrix}

A =25  \begin {bmatrix}  \dfrac{sin2 \theta }{2} + \theta \end {bmatrix}^{\dfrac{\pi}{3}}_{\dfrac{\pi}{3}}    \ \ - \dfrac{25}{2}  \begin {bmatrix}  \dfrac{2 \pi}{3} \end {bmatrix}

A =25  \begin {bmatrix}  \dfrac{sin (\dfrac{2 \pi}{3} )}{2}+\dfrac{\pi}{3} - \dfrac{ sin (\dfrac{-2\pi}{3}) }{2}-(-\dfrac{\pi}{3})  \end {bmatrix} - \dfrac{25 \pi}{3}

A = 25 \begin{bmatrix}   \dfrac{\dfrac{\sqrt{3}}{2} }{2} +\dfrac{\pi}{3} + \dfrac{\dfrac{\sqrt{3}}{2} }{2} +   \dfrac{\pi}{3}  \end {bmatrix}- \dfrac{ 25 \pi}{3}

A = 25 \begin{bmatrix}   \dfrac{\sqrt{3}}{2 } +\dfrac{2 \pi}{3}   \end {bmatrix}- \dfrac{ 25 \pi}{3}

A =    \dfrac{25 \sqrt{3}}{2 } +\dfrac{25 \pi}{3}

The diagrammatic expression showing the area of the region that lies inside the first curve and outside the second curve can be seen in the attached file below.

Download docx
7 0
3 years ago
6. The perimeter of a parallelogram is 100 meters. The width of the parallelogram is 4 meters less than its length. Find the len
Ierofanga [76]

Answer:

x = 23, and y = 27.

Step-by-step explanation:

Lets say,

x = width

y = length

this means,

x(2) + y(2) = 100

It also says that the width is 4 meters less than the length.

if we put that into equation form it would look like this,

y - 4 = x

using these 2 equations, we can use the subustusion method to find the x.

x(2) + y(2) = 100

turns in to this :

step 1.     (y - 4) (2) + y(2) = 100 - given

step 2.      2y -8 + 2y = 100 -  distribute

step 3.      4y -8 = 100 - Combine like terms

step 4.       4y  = 108 - Get rid of the 8

step 5        4y/4  = 108 /4 - divide

step 6 .          y = 27

So if y is 27, then

x(2)  +  27(2) = 100

x(2) + 54 = 100

100 - 54 = 46, so x(2) is 46, but to get x, we need to divide 46 by 2, since x multiplied by 2 equals 46.

46  / 2 = 23 = x

so x = 23, and y = 27.

and the rule that y - 4 = x still applies, since 27 - 4 = 23.

I hope this wasn't confusing! It did take a long time tho.

5 0
3 years ago
Write in index form 12 × 12 × 12 × 12 × 12 × 12 × 12 × 12
iris [78.8K]

Answer:

\sf{12^{8} }

Step-by-step explanation:

12 × 12 × 12 × 12 × 12 × 12 × 12 × 12 = \sf{12^{8} }

\sf{12^{8} } = 429,981,696

An easy way to think of this is that there are 12 is being multiplied by itselg 8 times (there are 8 12s). Similar to how squared numbers work...

Hope this helps! If you need further explanation, let me know!

- profparis

8 0
2 years ago
Read 2 more answers
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