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3 years ago

Calculate the volume of a ball having a radius of 8 cm

Mathematics

1 answer:

34kurt3 years ago

3 0

Answer:

267.9 $cm^{3}$

Step-by-step explanation:

volume of a circle = 4/3 * π * $r^{2}$

=4/3 *3.14 * $8^{2}$

=12.56/3 *64

=803.84/3

=267.94

=267.9 $cm^{3}$

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Solve for x in a right triangle (show work)

AnnZ [28]

Answer:

35.09°

Step-by-step explanation:

You can use cos theta to find the value of x.

Let us solve now.

They have already given the lengths of the hypotenuse and adjacent.

Hypotenuse = 33

Adjacent = 27

Let us solve now.

Cos x = Adjacent ÷ Hypotenuse

Cos x = 27 ÷ 33

Cos x = 0.8181

x = Cos⁻¹ 0.8181

x = 35.09°

Hope this helps you :-)

Let me know if you have any other questions :)

5 0

2 years ago

Using power series, solve the LDE: (2x^2 + 1) y" + 2xy' - 4x² y = 0 --- - -- -

sattari [20]

We're looking for a solution of the form

$y=\displaystyle\sum_{n\ge0}a_nx^n$

with derivatives

$y'=\displaystyle\sum_{n\ge0}(n+1)a_{n+1}x^n$

$y''=\displaystyle\sum_{n\ge0}(n+2)(n+1)a_{n+2}x^n$

Substituting these into the ODE gives

$\displaystyle\sum_{n\ge0}\left(\bigg(2(n+2)(n+1)a_{n+2}-4a_n\bigg)x^{n+2}+2(n+1)a_{n+1}x^{n+1}+(n+2)(n+1)a_{n+2}x^n\right)=0$

Shifting indices to get each term in the summand to start at the same power of $x$ and pulling the first few terms of the resulting shifted series as needed gives

$2a_2+(2a_1+6a_3)x+\displaystyle\sum_{n\ge2}\bigg((n+2)(n+1)a_{n+2}+2n^2a_n-4a_{n-2}\bigg)x^n=0$

Then the coefficients in the series solution are given according to the recurrence

$\begin{cases}a_0=y(0)\\\\a_1=y'(0)\\\\a_2=0\\\\2a_1+6a_3=0\implies a_3=-\dfrac{a_1}3\\\\a_n=\dfrac{-2(n-2)^2a_{n-2}+4a_{n-4}}{n(n-1)}&\text{for }n\ge4\end{cases}$

Given the complexity of this recursive definition, it's unlikely that you'll be able to find an exact solution to this recurrence. (You're welcome to try. I've learned this the hard way on scratch paper.) So instead of trying to do that, you can compute the first few coefficients to find an approximate solution. I got, assuming initial values of $y(0)=y'(0)=1$ , a degree-8 approximation of

$y(x)\approx1+x-\dfrac{x^3}3+\dfrac{x^4}3+\dfrac{x^5}2-\dfrac{16x^6}{45}-\dfrac{79x^7}{125}+\dfrac{101x^8}{210}$

Attached are plots of the exact (blue) and series (orange) solutions with increasing degree (3, 4, 5, and 65) and the aforementioned initial values to demonstrate that the series solution converges to the exact one (over whichever interval the series converges, that is).