3 years ago

What is b - 14= a solve for b please

1 answer:

3 0

B= a+14 Would be your answer because you'd need to get B alone therefore You need to add 14 on both sides to give you B=A+14

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Evaluate the given integral by changing to polar coordinates.∫∫ x2y dA, where D is the top half of the disk with center the orig

amid [387]

Answer:

Required value is $\frac{1250}{3}$ .

Step-by-step explanation:

We have to find the value of,

$\int\int_{D} x^2y dA\hfill (1)$

where, $D=\{(5,\theta) : 0\leq r\leq 5, 0\leq \theta \leq \pi \}$

To convert it in polar form let $x=r\cos\theta, y=r\sin\theta$ and substitute in (1) we get,

$\int\int_{D} x^2y dA$

$=\int_{0}^{\pi}\int_{0}^{5}(r\cos\theta)^2(r\sin\theta)rdrd\theta$

Assuming,

$u=\cos \theta$ then,

$du=-\sin\theta d\theta\implies-du=\sin\thetad\theta[tex]For [tex]\theta=0, u=1$ and $\theta=\pi$ gives u=-1. Hence,

$\int\int_{D} x^2y dA$

$=\int_{0}^{\pi}\int_{0}^{5}(r\cos\theta)^2(r\sin\theta)rdrd\theta$

$=\int_{1}^{-1}u^2(-du)\int_{0}^{5}r^4dr$

$=\int_{-1}^{1}u^2du\int_{0}^{5}r^4dr$

$=\Big[\frac{u^3}{3}\Big]_{-1}^{1}\times\Big[\frac{r^5}{5}\Big]_{0}^{5}$

$=\frac{2\times 5^4}{3}=\frac{1250}{3}$

5 0

3 years ago

How do I solve the cube root of -(x-7)^4? PLS EXPLAIN

Shkiper50 [21]

Renmember that
$\sqrt[3]{ab}=( \sqrt[3]{a})( \sqrt[3]{b})$
and
$\sqrt[n]{x^m} =x^ \frac{m}{n}$

split it up

$\sqrt[3]{-1(x-7)^4} =( \sqrt[3]{-1} )( \sqrt[3]{(x-7)^4})$ =
$(-1)((x-7)^ \frac{4}{3})=(-1)(x-7)(x-7)^ \frac{1}{3}=-(x-7) \sqrt[3]{x-7}$ =
$(7-x) \sqrt[3]{x-7}=7\sqrt[3]{x-7}-x\sqrt[3]{x-7}$