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3 years ago

I did this problem and I was told I did it wrong how do I fix it??

Mathematics

1 answer:

Ksivusya [100]3 years ago

5 0

Answer:

(1/8)(cos(4x) -4cos(2x) +3)

Step-by-step explanation:

Your answer is correct as far as it goes. You now need to use a power-reducing identity on the cos(2x)² term in your answer. The appropriate one is ...

cos(x)² = (1/2)(1 +cos(2x))

In the context of this problem, using this formula gives you ...

sin(x)⁴ = (1/4)(1 -2cos(2x) +(1/2)(1 +cos(4x))

sin(x)⁴ = (1/8)(cos(4x) -4cos(2x) +3)

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4 years ago

Read 2 more answers

A homeowner plans to enclose a 200 square foot rectangular playground in his garden, with one side along the boundary of his pro

Anestetic [448]

Answer:

Therefore the length and width of the playground are 15.5 feet and 12.9 feet respectively.

Step-by-step explanation:

Given that, a homeowner plans to enclose a 200 square foot rectangle playground.

Let the width of the playground be y and the length of the playground be x which is the side along the boundary.

The perimeter of the playground is = 2(length +width)

=2(x+y) foot

The material costs $1 per foot.

Therefore total cost to give boundary of the play ground

=$[ 2(x+y)×1]

=$[2(x+y)]

But the neighbor will play one third of the side x foot.

So the neighbor will play $=\$(\frac13x)$

Now homeowner's total cost for the material is

$=\$[ 2(x+y)-\frac13x]$

$=\$[2x+2y-\frac13x]$

$=\$[2y+x+x-\frac13x]$

$=\$[2y+x+\frac{3x-x}{3}]$

$=\$[2y+x+\frac{2}{3}x]$

$=\$[2y+\frac53x]$

$\therefore C(x)=2y+\frac53x$

where C(x) is total cost of material in $.

Given that the area of the playground is 200.

We know that,

The area of a rectangle is =length×width

=xy square foot

∴xy=200

$\Rightarrow y=\frac{200}{x}$

Putting the value of y in C(x)

$\therefore C(x)=2(\frac{200}{x})+\frac53x$

The domain of C is $(0,\infty )$ .

$\therefore C(x)=2(\frac{200}{x})+\frac53x$

Differentiating with respect to x

$C'(x)= - \frac{400}{x^2}+\frac53$

Again differentiating with respect to x

$C''(x) = \frac{800}{x^3}$

To find the critical point set C'(x)=0

$\therefore 0= - \frac{400}{x^2}+\frac53$

$\Rightarrow \frac{400}{x^2}=\frac{5}{3}$

$\Rightarrow x^2 =\frac{400\times 3}{5}$

$\Rightarrow x=\sqrt{240}$

$\Rightarrow x=15.49 \approx15.5$

Therefore

$\left C''(x) \right|_{x=15.5}=\frac{800}{15.5^3}>0$

Therefore at x= 15.5 , C(x) is minimum.

Putting the value of x in $y=\frac{200}{x}$ we get

$\therefore y=\frac{200}{15.5}$

=12.9

Therefore the length and width of the playground are 15.5 feet and 12.9 feet respectively.

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3 years ago