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3 years ago

Please help,ive been stuck on it fir two days

Mathematics

2 answers:

____ [38]3 years ago

5 0

Step One
Find the radius of the cone.
d = 6 which is given.
r = d/2
r = 6/2
r = 3

Remark
That's one of the sides of a right triangle. The other side is the slant of 9 inches. This is the hypotenuse.

Step Two
Find the height
h = ??
r = 3
Slant = 9

Use the Pythagorean theorem.

h^2 + r^2 = s^2
h^2 + 3^2 = 9^2
h^2 + 9 = 81 subtract 9 from both sides.
h^2 = 81 - 9
h^2 = 72
h = sqrt(72)

Step three
Express the answer
h = sqrt(72)
h = 8.49 <<<<<< Answer, but see below.

If you need this as a square root, then factor 72
72 = 6*6*2

h = sqrt(6* 6 * 2) The rule for square roots is if you have 2 of them you can take one outside the square root sign and throw the other one away.

h = 6 * sqrt(2) <<<< Answer

san4es73 [151]3 years ago

3 0

The square root of 73. Since the theorem is a^2+b^2=c^2 and only works with right triangles. You take 2/2 the base so you have a right triangles with a hypotenuse if 9 and a side of 3. Then you just plug in to solve.

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Let

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For demonstration on how to tackle this sort of problem, I'll only work through the case where $n$ is odd. We can write

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For the remaining integral, we can integrate by parts, taking

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$\implies\displaystyle\int\sin^{n-2}ax\cos^2ax\,\mathrm dx=-\dfrac1{3a}\sin^{n-3}ax\cos^3ax+\dfrac{a(n-3)}{3a}\int\sin^{n-4}ax\cos^4ax\,\mathrm dx$

For this next integral, we rewrite the integrand

$\sin^{n-4}ax\cos^4ax=\sin^{n-4}ax(1-\sin^2ax)^2=\sin^{n-4}ax-2\sin^{n-2}ax+\sin^nax$
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So putting everything together, we found

$I(n,a)=I(n-2,a)-\displaystyle\int\sin^{n-2}ax\cos^2ax\,\mathrm dx$
$I(n,a)=I(n-2,a)-\left(-\dfrac1{3a}\sin^{n-3}ax\cos^3ax+\dfrac{n-3}3\displaystyle\int\sin^{n-4}ax\cos^4ax\,\mathrm dx\right)$
$I(n,a)=I(n-2,a)-\dfrac{n-3}3\bigg(I(n-4,a)-2I(n-2,a)+I(n,a)\bigg)+\dfrac1{3a}\sin^{n-3}ax\cos^3ax$
$\dfrac n3I(n,a)=\dfrac{2n-3}3I(n-2,a)-\dfrac{n-3}3I(n-4,a)+\dfrac1{3a}\sin^{n-3}ax\cos^3ax$

$\implies I(n,a)=\dfrac{2n-3}nI(n-2,a)-\dfrac{n-3}nI(n-4,a)+\dfrac1{na}\sin^{n-3}ax\cos^3ax$

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