I need help with this question.

Mathematics

1 answer:

Tamiku [17]3 years ago

3 0

Answer:

51x

Step-by-step explanation:

3x^3 - 7x + 3x^3 + 4x

27x - 7x + 27x + 4x

20x + 31x

51x

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Let divf = 6(5 − x) and 0 ≤ a, b, c ≤ 12. (a) find the flux of f out of the rectangular solid 0 ≤ x ≤ a, 0 ≤ y ≤ b, and 0 ≤ z ≤

dusya [7]

Continuing from the setup in the question linked above (and using the same symbols/variables), we have

$\displaystyle\iint_{\mathcal S}\mathbf f\cdot\mathrm d\mathbf S=\iiint_{\mathcal R}(\nabla\cdot f)\,\mathrm dV$
$=\displaystyle6\int_{z=0}^{z=c}\int_{y=0}^{y=b}\int_{x=0}^{x=a}(5-x)\,\mathrm dx\,\mathrm dy\,\mathrm dz$
$=\displaystyle6bc\int_0^a(5-x)\,\mathrm dx$
$=6bc\left(5a-\dfrac{a^2}2\right)=3abc(10-a)$

The next part of the question asks to maximize this result - our target function which we'll call $g(a,b,c)=3abc(10-a)$ - subject to $0\le a,b,c\le12$ .

We can see that $g$ is quadratic in $a$ , so let's complete the square.

$g(a,b,c)=-3bc(a^2-10a+25-25)=3bc(25-(a-5)^2)$

Since $b,c$ are non-negative, it stands to reason that the total product will be maximized if $a-5$ vanishes because $25-(a-5)^2$ is a parabola with its vertex (a maximum) at (5, 25). Setting $a=5$ , it's clear that the maximum of $g$ will then be attained when $b,c$ are largest, so the largest flux will be attained at $(a,b,c)=(5,12,12)$ , which gives a flux of 10,800.

7 0

3 years ago

Jordan wrote the following description: Three fewer than one fourth of x is 12. Write an

muminat

Answer:

x/4 - 3 = 12

Step-by-step explanation:

"one fourth of x" is x/4.

Then "three fewer than one fourth of x is 12" becomes, symbolically,

x/4 - 3 = 12

3 0

3 years ago

Law of cosine. Please help

Anna35 [415]

Answer:

B) 41.7

Step-by-step explanation:

Let $a=25$ , $b=32$ , and $c=37$ . Use the Law of Cosines equation to find angle A=x which is directly across from side $a$ :

$\displaystyle \cos(A)=\frac{b^2+c^2-a^2}{2bc}\\ \\\cos(A)=\frac{32^2+37^2-25^2}{2(32)(37)}\\ \\\cos(A)=\frac{1024+1369-625}{2(1184)}\\\\\cos(A)=\frac{2393-625}{2368}\\ \\\cos(A)=\frac{1768}{2368}\\\\A=\cos^{-1}\biggr(\frac{1768}{2368}\biggr)\\ \\A\approx41.7^\circ$