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2 years ago

Evaluate the expression using an appropriate double angle formula.. cos^2(15) - sin^2(15)

Mathematics

1 answer:

riadik2000 [5.3K]2 years ago

7 0

Cos²(15°) - sin²(15°) = cos(15° * 2) = √(3/2 so this is the answer by using apropriate formula double angle identity

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Integration by Parts Evaluate e-2x cos(2x) dx.

kifflom [539]

Let

$I = \displaystyle \int e^{-2x} \cos(2x) \, dx[/]texIntegrate by parts:[tex]\displaystyle \int u \, dv = uv - \int v \, du$

with

$u = e^{-2x} \implies du = -2 e^{-2x} \, dx \\\\ dv = \cos(2x) \, dx \implies v = \dfrac12 \sin(2x)$

Then

$\displaystyle I = \frac12 e^{-2x} \sin(2x) + \int e^{-2x} \sin(2x) \, dx + C$

Integrate by parts again, this time with

$u = e^{-2x} \implies du = -2 e^{-2x} \, dx \\\\ dv = \sin(2x) \, dx \implies v = -\dfrac12 \cos(2x)$

so that

$\displaystyle I = \frac12 e^{-2x} \sin(2x) - \frac12 e^{-2x} \cos(2x) - \int e^{-2x} \cos(2x) \, dx + C\\\\ \implies I = \frac{\sin(2x)-\cos(2x)}{2e^{2x}} - I + C \\\\ \implies 2I = \frac{\sin(2x) - \cos(2x)}{2e^{2x}} + C \\\\ \implies I = \boxed{\frac{\sin(2x) - \cos(2x)}{4e^{2x}} + C}$

6 0

1 year ago

Like terms 4x + 2x + 7x

nasty-shy [4]

Since the 3 of them are like terms, we have to ADD all 3 of them.

4x+2x+7x= 13x

4 0

2 years ago

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I need help ASAP.I need help graphing this problem

Viktor [21]

Answer:

The answer to your question is below

Step-by-step explanation:

Each square values two units

I hope it helps you.

8 0

2 years ago

What is 9-(30) please help

Murrr4er [49]

21 and download math papa

6 0

2 years ago

If an open box has a square base and a volume of 115 in.3 and is constructed from a tin sheet, find the dimensions of the box, a

Karolina [17]

Let h = height of the box,
x = side length of the base.

Volume of the box is $V=x^{2} h = 115$ .
So $h = \frac{115}{ x^{2} }$

Surface area of a box is S = 2(Width • Length + Length • Height + Height • Width).
So surface area of the box is
$S = 2( x^{2} + hx + hx) \\ = 2 x^{2} + 4hx \\ = 2 x^{2} + 4( \frac{115}{ x^{2} } )x$
$= 2 x^{2} + \frac{460}{x}$
The surface are is supposed to be the minimum. So we'll need to find the first derivative of the surface area function and set it to zero.

$S' = 4x- \frac{460}{ x^{2} } = 0$
$4x = \frac{460}{ x^{2} } \\ 4x^{3} = 460 \\ x^{3} = 115 \\ x = \sqrt[3]{115} = 4.86$
Then $h= \frac{115}{4.86^{2}} = 4.87$
So the box is 4.86 in. wide and 4.87 in. high.

5 0

2 years ago

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