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3 years ago

Please help number 7 and number 8

Mathematics

1 answer:

ladessa [460]3 years ago

8 0

7. You are finding the area of the box. Not the cube, just a surface. Length times width of the surface. Eliminate other numbers and multiply

24.5×19.5

then

26×24.5

8. You are doing the same thing almost. Word problems are scary, but look for the numbers.

4inch ×8inch × 12inch

Find the volume
Multiply the volume by 3, that'd one answer
Devide the ORIGINAL volume by 2. don't divide the volume you multiplied by 3.

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Community Gym charges a $50 membership fee and a $70 monthly fee. Workout Gym charges a $200

Reil [10]

Answer:

community gym

50+(70×12)

=$890

workout gym

200+(60×12)

=$920

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3 years ago

Solve the formula t= 4u/e, for the variable e

Komok [63]

Answer:

$e=4u/t$

Step-by-step explanation:

we have

$t=\frac{4u}{e}$

Solve for the variable e

That means -----> Isolate the variable e

Multiply by e both sides

tex](e)t=\frac{4u}{e}(e)[/tex]

$et=4u$

Divide by t both sides

$e=4u/t$

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4 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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3 years ago

A share of ABC stock was worth $60 in 2000 and worth $1920 in 2005.

Sergio [31]

Answer:

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3 years ago

Analyzing a Function The population of bobcats in northern Arizona since 2008 can be modeled using the function b(t) = –0.32t2 +

bulgar [2K]

"b" represents the population of bobcats.

1) "t" represents the number of years after 2008.

2) The domain (t) starts at 0 (which is the year 2008) ⇒ t ≥ 0

3) The range (b) starts when t = 0. b =-0.32(0)² + 2.7(0) + 253 ⇒ b ≥ 253

4) The graph is discrete because the population must be in whole numbers.

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3 years ago

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