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3 years ago

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Which expressions are equivalen] to x^3 + 9x^2 ?

1 answer:

Minchanka [31]3 years ago

4 0

Answer:3a. Stu ?

Step-by-step explanation:

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I have a math test coming up can people help me?!?!?!

kogti [31]

i can help if you want

5 0

3 years ago

Read 2 more answers

A cylindrical tank with radius 5 m is being filled with water at a rate of 3 cubic meters per minute. How fast is the height of

sasho [114]

Answer:

Step-by-step explanation:

Let's calculate the volume of the tank per each meter in height.

The volume of a cylinder is πr²h, where h is the height.

A height of 1 meter in a tank with a radius of 5 meters would hold a volume of:

Vol = (3.14)*(5 meters)^2 *(1 meter)

Vol (m^3) = 78.54 m^3 per 1 meter height.

If the tank were filled at a rate of 3 m^3/min, it would rise at at a rate of:

(78.54 m^3/meter)/(3 m^3/min) = 0.0382 meters/minute [38.2 cm/min

4 0

2 years ago

Find the surface area for a cube that has sides all 2yds.

Reika [66]

Answer:

24

Step-by-step explanation:

5 0

2 years ago

Solve for y where y(2)=2 and y'(2)=0 by representing y as a power series centered at x=a

Crank

I'll assume the ODE is actually

$y''+(x-2)y'+y=0$

Look for a series solution centered at $x=2$ , with

$y=\displaystyle\sum_{n\ge0}c_n(x-2)^n$

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

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

with $c_0=y(2)=2$ and $c_1=y'(2)=0$ .

Substituting the series into the ODE gives

$\displaystyle\sum_{n\ge0}(n+2)(n+1)c_{n+2}(x-2)^n+\sum_{n\ge0}(n+1)c_{n+1}(x-2)^{n+1}+\sum_{n\ge0}c_n(x-2)^n=0$

$\displaystyle\sum_{n\ge0}(n+2)(n+1)c_{n+2}(x-2)^n+\sum_{n\ge1}nc_n(x-2)^n+\sum_{n\ge0}c_n(x-2)^n=0$

$\displaystyle2c_2+c_0+\sum_{n\ge1}(n+2)(n+1)c_{n+2}(x-2)^n+\sum_{n\ge1}nc_n(x-2)^n+\sum_{n\ge1}c_n(x-2)^n=0$

$\displaystyle2c_2+c_0+\sum_{n\ge1}\bigg((n+2)(n+1)c_{n+2}+(n+1)c_n\bigg)(x-2)^n=0$

$\implies\begin{cases}c_0=2\\c_1=0\\(n+2)c_{n+2}+c_n=0&\text{for }n>0\end{cases}$

If $n=2k$ for integers $k\ge0$ , then

$k=0\implies n=0\implies c_0=c_0$

$k=1\implies n=2\implies c_2=-\dfrac{c_0}2=(-1)^1\dfrac{c_0}{2^1(1)}$

$k=2\implies n=4\implies c_4=-\dfrac{c_2}4=(-1)^2\dfrac{c_0}{2^2(2\cdot1)}$

$k=3\implies n=6\implies c_6=-\dfrac{c_4}6=(-1)^3\dfrac{c_0}{2^3(3\cdot2\cdot1)}$

and so on, with

$c_{2k}=(-1)^k\dfrac{c_0}{2^kk!}$

If $n=2k+1$ , we have $c_{2k+1}=0$ for all $k\ge0$ because $c_1=0$ causes every odd-indexed coefficient to vanish.

So we have

$y(x)=\displaystyle\sum_{k\ge0}c_{2k}(x-2)^{2k}=\sum_{k\ge0}(-1)^k\frac{(x-2)^{2k}}{2^{k-1}k!}$

Recall that

$e^x=\displaystyle\sum_{n\ge0}\frac{x^k}{k!}$

The solution we found can then be written as

$y(x)=\displaystyle2\sum_{k\ge0}\frac1{k!}\left(-\frac{(x-2)^2}2\right)^k$

$\implies\boxed{y(x)=2e^{-(x-2)^2/2}}$

6 0

3 years ago

Find the values of the six trigonometric functions for angle θ, when AC = 26 and BC = 24.

mart [117]

Answer:

See below

Step-by-step explanation:

The first step is to find the length of AB. By the Pythagorean Theorem:

$AB=\sqrt{26^2-24^2}=\sqrt{676-576}=\sqrt{100}=10$

Now, you can proceed:

$\sin \theta= \dfrac{24}{26}=\dfrac{12}{13} \\\\\\\cos \theta= \dfrac{10}{26}=\dfrac{5}{13} \\\\\\\tan \theta= \dfrac{24}{10}=\dfrac{12}{5} \\\\\\\csc \theta= \dfrac{13}{12} \\\\\\\sec \theta=\dfrac{13}{5} \\\\\\\cot \theta=\dfrac{5}{12}$

7 0

3 years ago