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tekilochka [14]

3 years ago

7

The ratio of the volume of sphere A to the volume of sphere B is

"TexFormula1" title="\frac{125}{64}" alt="\frac{125}{64}" align="absmiddle" class="latex-formula">
What is the ratio of the radius of sphere A to the radius of sphere B?

1 answer:

Assoli18 [71]3 years ago

7 0

Answer:

Step-by-step explanation:

let radius of shere A=R

radius of sphere B=r

$\frac{volume~of~sphere ~A}{volume~of~sphere~B} =\frac{\frac{4}{3} \pi R^3}{\frac{4}{3} r^3} =\frac{125}{64} \\(\frac{R}{r} )^3=(\frac{5}{4} )^3\\\frac{R}{r} =\frac{5}{4}$

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Solve by substitution 3x - 5y = -1 , x-y = -1

nevsk [136]

Answer:

(-2, -1)

Step-by-step explanation:

first you must reorganize the equations so that one of them has x or y equaling to something

x - y = -1 --> x = y - 1

then substitute that in for x

3(y - 1) - 5y = -1

and solve for y

3y - 3 - 5y = -1

-2y = 2

y is -1

then plug it back in to an equation to find x

x + 1 = -1

x is -2

6 0

2 years ago

Round 0.10947 to the nearest hundredths place.

tamaranim1 [39]

0.11 would be to the nearest hundredths

4 0

3 years ago

Read 2 more answers

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.

5 0

3 years ago

The Diamond Rule says two factors, m and n, must multiply to get to the top number and add to

tino4ka555 [31]

Answer: -3 and -6

Step-by-step explanation: I took the quiz

7 0

3 years ago

Read 2 more answers

Whats the answer to this?

mars1129 [50]

Answer:

15 degrees

X is an Acute Angle

Step-by-step explanation:

x+ 75 = 90

x =15

4 0

3 years ago