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3 years ago

-6(2-3n)-6(-7-2n)=30 solve for n

Mathematics

2 answers:

g100num [7]3 years ago

7 0

-12 + 18N + 42 + 12N = 30
30N + 42 - 12 = 30
30N + 30 = 30
30N = 0
N = 0

hopefully that’s correct

Jet001 [13]3 years ago

6 0

Answer:

n=0

Step-by-step explanation:

See pictures

Hope this helps ^-^

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8≥x+4>5 i need answer

dexar [7]

Answer:

1 < x ≤ 4

Step-by-step explanation:

Given

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Pls help it due in a hour

Levart [38]

Answer:

The coordinates of the Midpoint of AB will be: (1/2, 5)

Step-by-step explanation:

Given the points

A(-2, 6)

B(3, 4)

Finding the midpoint between (-2, 6) and (3, 4)

$\mathrm{Midpoint\:of\:}\left(x_1,\:y_1\right),\:\left(x_2,\:y_2\right):\quad \left(\frac{x_2+x_1}{2},\:\:\frac{y_2+y_1}{2}\right)$

$\left(x_1,\:y_1\right)=\left(-2,\:6\right),\:\left(x_2,\:y_2\right)=\left(3,\:4\right)$

$M.P=\left(\frac{3-2}{2},\:\frac{4+6}{2}\right)$

$=\left(\frac{1}{2},\:5\right)$

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3 years ago

Find the extreme values of f subject to both constraints f(x,y) = 2x^2+3y^2-4x-5, x^2 + y^2 <=16

Softa [21]

$f(x,y)=2x^2+3y^2-4x-5$
$f_x=4x-4=0\implies x=1$
$f_y=6y=0\implies y=0$

$f(x,y)$ has only one critical point at $(x,y)=(1,0)$ . The function has Hessian

$\mathbf H(x,y)=\begin{bmatrix}f_{xx}&f_{xy}\\f_{yx}&f_{yy}\end{bmatrix}=\begin{bmatrix}4&0\\0&6\end{bmatrix}$

which is positive definite for all $(x,y)$ , which means $f(x,y)$ attains a minimum at the critical point with a value of $f(1,0)=-7$ .

To find the extrema (if any) along the boundary, parameterize it by $x=4\cos t$ and $y=4\sin t$ , with $0\le t$ . On the boundary, we have

$f(x(t),y(t))=F(t)=2(4\cos t)^2+3(4\sin t)^2-4(4\cos t)-5=32\cos^2t+48\sin^2t-16\cos t-5$
$F(t)=35-16\cos t-8\cos2t$

Find the critical points along the boundary:

$F'(t)=16\sin t+16\sin2t=16\sin t+32\sin t\cos t=16\sin t(1+2\cos t)=0$
$\implies t=0,\dfrac{2\pi}3,\pi,\dfrac{4\pi}3$

Respectively, plugging these values into $F(t)$ gives 11, 47, 43, and 47. We omit the first and third, as we can see the absolute extrema occur when $F(t)=47$ .

Now, solve for $x,y$ for both cases:

$t=\dfrac{2\pi}3\implies\begin{cases}x=4\cos t=-2\\y=4\sin t=2\sqrt3\end{cases}$

$t=\dfrac{4\pi}3\implies\begin{cases}x=4\cos t=-2\\y=4\sin t=-2\sqrt3\end{cases}$

so $f(x,y)$ has two absolute maxima at $(x,y)=(-2,\pm2\sqrt3)$ with the same value of 47.

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4 years ago

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-6(2-3n)-6(-7-2n)=30 solve for n​

-6(2-3n)-6(-7-2n)=30 solve for n