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Stells [14]
3 years ago
15

Which correctly describes a cross section of the right rectangular prism if the base is a rectangle measuring 15 inches by 8 inc

hes? Check all that apply.
-
A cross section parallel to the base is a rectangle measuring 15 inches by 8 inches.
A cross section parallel to the base is a rectangle measuring 15 inches by 6 inches.
A cross section perpendicular to the base through the midpoints of the 8-inch sides is a rectangle measuring 6 inches by 15 inches.
A cross section perpendicular to the base through the midpoints of the 8-inch sides is a rectangle measuring 4 inches by 15 inches.
A cross section not parallel to the base that passes through opposite 6-inch edges is a rectangle measuring 6 inches by greater than 15 inches.

Mathematics
2 answers:
grigory [225]3 years ago
5 0
<span>A cross-section parallel to the base is a rectangle measuring 15 inches by 8 inches.

</span><span>A cross-section perpendicular to the base through the midpoints of the 8-inch sides is a rectangle measuring 6 inches by 15 inches.
</span>
<span>A cross-section not parallel to the base that passes through opposite 6-inch edges is a rectangle measuring 6 inches by greater than 15 inches.

the cross sections that are parallel and perpendicular will have the same measurements as the non-intersected sides. the last one will be a diagonal so the intersected edge is 6 and it creates a right triangle so it must be larger than 15 inches.

</span><span>
</span>
irinina [24]3 years ago
4 0

Answer:

1, 3, 5

Step-by-step explanation:

E20

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bazaltina [42]

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1st plan: 24 movies 2nd plan: 10

Step-by-step explanation:

if the 2nd plan has to be the exact value of the 1st plan you would have to divide 6.25 from 62.50.

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A gardener grows sunflowers and records the heights, y , in centimeters, each day, x . The table shows the gardener’s data.
PolarNik [594]

Answer:

Explanation is in a file

Step-by-step explanation:

5 0
2 years ago
Which of the following represents the graph of f(x) = one−half to the power of x?
puteri [66]

f(x) = \frac{1}{2}^x

In this exponential function , the base is \frac{1}{2}

if base is less than 1 then it a decreasing function

We can verify it using a table

We plug in some x and values and find out f(x) to get point for the graph

x -----> f(x)

-1 -----> 2

0 ------> 1

1 -------> 0.5

The table is same as the points in first graph.

So first graph represents the graph of f(x) = \frac{1}{2}^x

4 0
3 years ago
Combine the following expressions<br> 2√18x^3 - 3√8x^3<br><br> A. -√x<br> B. 0<br> C.√x
rosijanka [135]

This is what I have but it doesn't match with your answers

8 0
2 years ago
Find a particular solution to <img src="https://tex.z-dn.net/?f=%20x%5E%7B2%7D%20%20%5Cfrac%7B%20d%5E%7B2%7Dy%20%7D%7Bd%20x%5E%7
Digiron [165]
y=x^r
\implies r(r-1)x^r+6rx^r+4x^r=0
\implies r^2+5r+4=(r+1)(r+4)=0
\implies r=-1,r=-4

so the characteristic solution is

y_c=\dfrac{C_1}x+\dfrac{C_2}{x^4}

As a guess for the particular solution, let's back up a bit. The reason the choice of y=x^r works for the characteristic solution is that, in the background, we're employing the substitution t=\ln x, so that y(x) is getting replaced with a new function z(t). Differentiating yields

\dfrac{\mathrm dy}{\mathrm dx}=\dfrac1x\dfrac{\mathrm dz}{\mathrm dt}
\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac1{x^2}\left(\dfrac{\mathrm d^2z}{\mathrm dt^2}-\dfrac{\mathrm dz}{\mathrm dt}\right)

Now the ODE in terms of t is linear with constant coefficients, since the coefficients x^2 and x will cancel, resulting in the ODE

\dfrac{\mathrm d^2z}{\mathrm dt^2}+5\dfrac{\mathrm dz}{\mathrm dt}+4z=e^{2t}\sin e^t

Of coursesin, the characteristic equation will be r^2+6r+4=0, which leads to solutions C_1e^{-t}+C_2e^{-4t}=C_1x^{-1}+C_2x^{-4}, as before.

Now that we have two linearly independent solutions, we can easily find more via variation of parameters. If z_1,z_2 are the solutions to the characteristic equation of the ODE in terms of z, then we can find another of the form z_p=u_1z_1+u_2z_2 where

u_1=-\displaystyle\int\frac{z_2e^{2t}\sin e^t}{W(z_1,z_2)}\,\mathrm dt
u_2=\displaystyle\int\frac{z_1e^{2t}\sin e^t}{W(z_1,z_2)}\,\mathrm dt

where W(z_1,z_2) is the Wronskian of the two characteristic solutions. We have

u_1=-\displaystyle\int\frac{e^{-2t}\sin e^t}{-3e^{-5t}}\,\mathrm dt
u_1=\dfrac23(1-2e^{2t})\cos e^t+\dfrac23e^t\sin e^t

u_2=\displaystyle\int\frac{e^t\sin e^t}{-3e^{-5t}}\,\mathrm dt
u_2=\dfrac13(120-20e^{2t}+e^{4t})e^t\cos e^t-\dfrac13(120-60e^{2t}+5e^{4t})\sin e^t

\implies z_p=u_1z_1+u_2z_2
\implies z_p=(40e^{-4t}-6)e^{-t}\cos e^t-(1-20e^{-2t}+40e^{-4t})\sin e^t

and recalling that t=\ln x\iff e^t=x, we have

\implies y_p=\left(\dfrac{40}{x^3}-\dfrac6x\right)\cos x-\left(1-\dfrac{20}{x^2}+\dfrac{40}{x^4}\right)\sin x
4 0
2 years ago
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