Calculate DE to the hundredths place.

Mathematics

1 answer:

kompoz [17]2 years ago

5 0

<h3>Answer: 9.03</h3>

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Explanation:

For now, focus solely on triangle HGF.

We'll need to find the measure of angle F.

Use the law of cosines

f^2 = g^2 + h^2 - 2*g*h*cos(F)

(4.25)^2 = 8^2 + 6^2 - 2*8*6*cos(F)

18.0625 = 100 - 96*cos(F)

18.0625-100 = -96*cos(F)

-81.9375 = -96*cos(F)

cos(F) = (-81.9375)/(-96)

cos(F) = 0.853515625

F = arccos(0.853515625)

F = 31.403868 degrees approximately

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Now we can move our attention to triangle DEF.

We'll use the angle F we just found to find the length of the opposite side DE, aka side f.

Once again, we use the law of cosines.

f^2 = d^2 + e^2 - 2*d*e*cos(F)

f^2 = (4.75+8)^2 + (11+6)^2 - 2*(4.75+8)*(11+6)*cos(31.403868)

f^2 = 81.563478

f = sqrt(81.563478)

f = 9.031250 approximately

Rounding to two decimal places means we get the final answer of DE = 9.03

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igor_vitrenko [27]

It looks like you want to compute the double integral

$\displaystyle \iint_D (x+y) \,\mathrm dx\,\mathrm dy$

over the region D with the unit circle x ² + y ² = 1 as its boundary.

Convert to polar coordinates, in which D is given by the set

D = {(r, θ) : 0 ≤ r ≤ 1 and 0 ≤ θ ≤ 2π}

and

x = r cos(θ)

y = r sin(θ)

dx dy = r dr dθ

Then the integral is

$\displaystyle \iint_D (x+y)\,\mathrm dx\,\mathrm dy = \iint_D r^2(\cos(\theta)+\sin(\theta))\,\mathrm dr\,\mathrm d\theta \\\\ = \int_0^{2\pi} \int_0^1 r^2(\cos(\theta)+\sin(\theta))\,\mathrm dr\,\mathrm d\theta \\\\ = \underbrace{\left( \int_0^{2\pi}(\cos(\theta)+\sin(\theta))\,\mathrm d\theta \right)}_{\int = 0} \left( \int_0^1 r^2\,\mathrm dr \right) = \boxed{0}$