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Vinil7 [7]
3 years ago
12

2. The picture shows a feeding trough that is shaped like a right prism.

Mathematics
1 answer:
Nina [5.8K]3 years ago
8 0

Answer:

B

Step-by-step explanation:

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alexandr402 [8]

Answer:

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3 years ago
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Given that (x+2) is a factor of x^3+8, find the other quadratic factor.
Semmy [17]
If you know the formular a^3+b^3=(a+b)(a^2-ab+b^2), you can solve this problem.
8 is 2 cubed, so x^3+2^3=(x+2)(x^2-2x+4)
so the other quadratic factor is x^2-2x+4
3 0
3 years ago
Find the surface area
mote1985 [20]

Answer:

Step-by-step explanation:

l = 7 in

w = 11 in

h = 5 in

Surface area = 2*(lw +wh +hl)

        = 2*(7*11 + 11*5 + 5*7)

       = 2* (77+ 55 + 35)

     = 2 * 167

     = 334 square inches

3 0
3 years ago
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Applications 24. A rocket is launched into the air. Its height in feet, after x seconds, is given by the equation The starting h
SVEN [57.7K]

The given function is

h(x)=-16x^2+300x+20

According to this function, the starting height of the rocket is 20 feet because that's the initial condition of the problem stated by the independent term.

Additionally, we find the maximum height by calculating the vertex of the function V(h,k).

h=-\frac{b}{2a}

Where a = -16 and b = 300.

\begin{gathered} h=-\frac{300}{2(-16)}=\frac{300}{32}=\frac{150}{16}=\frac{75}{8} \\ h=9.375 \end{gathered}

Then, we find k by evaluating the function

\begin{gathered} k=-16(9.375)^2+300(9.375)+20 \\ k=-1406.25+2812.5+20=1426.25 \end{gathered}

Hence, the maximum height is 1426.25 feet.

At last, to know the time need to hit the ground, we just use h=9.375 and we multiply it by 2

t=2\cdot9.375=18.75

Hence, the rocket hits the ground after 18.75 seconds.

6 0
1 year ago
Find a particular solution to the nonhomogeneous differential equation y'' + 4 y = cos(2x) + sin(2x).
alukav5142 [94]
The characteristic solution follows from solving the characteristic equation,

r^2+4=0\implies r=\pm2i

so that

y_c=C_1\cos2x+C_2\sin2x

A guess for the particular solution may be a\cos2x+b\sin2x, but this is already contained within the characteristic solution. We require a set of linearly independent solutions, so we can look to

y_p=ax\cos2x+bx\sin2x

which has second derivative

{y_p}''=(-4ax+4b)\cos2x+(-4bx-4a)\sin2x

Substituting into the ODE, you have

y''+4y=\cos2x+\sin2x
\implies4b\cos2x-4a\sin2x=\cos2x+\sin2x
\implies\begin{cases}4b=1\\-4a=1\end{cases}\implies a=-\dfrac14,b=\dfrac14

Therefore the particular solution is

y_p=-\dfrac14x\cos2x+\dfrac14x\sin2x

Note that you could have made a more precise guess of

y_p=(a_1x+a_0)\cos2x+(b_1x+b_0)\sin2x

but, of course, any solution of the form a_0\cos2x+b_0\sin2x is already accounted for within y_c.
6 0
3 years ago
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