8b-4=-4+8b solve for x

Mathematics

2 answers:

guapka [62]3 years ago

8 0

Answer:

Step-by-step explanation

8b=8bs

cancel 8 on both sides

divide both sides by b

your left with 1=s

s=1

dolphi86 [110]3 years ago

6 0

Answer:

The answer would be 1.

x=1

Step-by-step explanation:

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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

Solve for n 2n+7=6n-5

rjkz [21]

Answer:

n=3

Step-by-step explanation:

you have 2n+7=6n-5

start with subtracting 2n from both sides giving you 7=4n-5

now take the -5 and add 5 to both sides, giving you 12=4n

now divide both sides by 4, now giving you 3=n

the variable has to be on the left every time you finish your work living with n=3

5 0

4 years ago

Read 2 more answers

Suppose that x=6 and y=8. If y is inversely related to x, what is y when x=3?

andrey2020 [161]

$\bf \qquad \qquad \textit{inverse proportional variation}\\\\
\textit{\underline{y} varies inversely with \underline{x}}\qquad \qquad y=\cfrac{k}{x}\impliedby 
\begin{array}{llll}
k=constant\ of\\
\qquad variation
\end{array}\\\\
-------------------------------\\\\
\textit{we also know that }
\begin{cases}
x=6\\
y=8
\end{cases}\implies 8=\cfrac{k}{6}\implies 48=k
\\\\\\
therefore\qquad \boxed{y=\cfrac{48}{x}}
\\\\\\
\textit{when x = 3, what is \underline{y}?}\qquad y=\cfrac{48}{3}$