2 years ago

Pleaseeee help ASAP rocky!!!

1 answer:

7 0

Answer:E

Step-by-step explanation:

You might be interested in

marissa [1.9K]

P-5??? It's an expression

6 0

2 years ago

koban [17]

Set

$\begin{cases}x=r\cos\theta\\y=r\sin\theta\end{cases}\implies\mathrm dA=r\,\mathrm dr\,\mathrm d\theta$

The region $R$ is given in polar coordinates by the set

$R=\left\{(r,\theta)\mid2\le r\le3,0\le\theta\le\dfrac\pi2\right\}$

So we have

$\displaystyle\iint_R\sin(x^2+y^2)\,\mathrm dA=\int_0^{\pi/2}\int_2^3r\sin(r^2)\,\mathrm dr\,\mathrm d\theta=\boxed{\frac\pi4(\cos4-\cos9)}$

8 0

3 years ago

cluponka [151]

$\frac{1}{4} x+\frac{11}{4}$

we can rewrite it as

$\frac{1}{4}* x+\frac{1}{4}*11$

we can see that 1/4 is in multiple in both terms

so, we can factor it out

so, we get

$\frac{1}{4}( x+11)$ ..............Answer

5 0

3 years ago

Arturiano [62]

The answer is A, because the Y-intersect starts at -2

6 0

3 years ago

marusya05 [52]

The solution to this problem is 5/2a^2b^5

5 0

2 years ago