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2 years ago

The prism is completely filled with 1750 cubes that have edge length of 15 ft.

Mathematics

2 answers:

vladimir1956 [14]2 years ago

5 0

It is 14 i got it on my test

Firdavs [7]2 years ago

3 0

The answer would be 14. HOPE THIS HELPED U :)

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A window cleaning service charges under $75 to clean all of the windows of a house. Write an inequality to represent the amount

Alona [7]

Answer:

C < $75.

Step-by-step explanation:

Let C be the charge of a window cleaning service, that cleans all the windows of the house.

Since it charges less than $75. this means

C < $75.

So, this is the required inequality equation.

8 0

2 years ago

The Punic Wars began in 264 B.C. and ended in 146 B.C. How long did the Punic Wars last??

pishuonlain [190]

Answer:

The first two wars were long—23 years and 17 years, separated by an interval of 23 years. The third war lasted nearly three years. It started 52 years after the end of the second war. All three wars were won by Rome, which subsequently emerged as the greatest military power in the Mediterranean Sea.

Step-by-step explanation:

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3 years ago

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The diameter of a baseball is approximately 3 inches. What is its volume? Round your answer to the nearest whole number.

OLga [1]

Answer:

I am not sure, but i think it is A

Step-by-step explanation:

8 0

2 years ago

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a Harley Davidson motorcycle shop maintains an inventory of four times as many new bikes as used bikes. if there are 60 new bike

almond37 [142]

You would divide the new bikes by four to find out the amount of used bikes
so 15

6 0

2 years ago

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D^2(y)/(dx^2)-16*k*y=9.6e^(4x) + 30e^x

MA_775_DIABLO [31]

The solution depends on the value of $k$ . To make things simple, assume $k>0$ . The homogeneous part of the equation is

$\dfrac{\mathrm d^2y}{\mathrm dx^2}-16ky=0$

and has characteristic equation

$r^2-16k=0\implies r=\pm4\sqrt k$

which admits the characteristic solution $y_c=C_1e^{-4\sqrt kx}+C_2e^{4\sqrt kx}$ .

For the solution to the nonhomogeneous equation, a reasonable guess for the particular solution might be $y_p=ae^{4x}+be^x$ . Then

$\dfrac{\mathrm d^2y_p}{\mathrm dx^2}=16ae^{4x}+be^x$

So you have

$16ae^{4x}+be^x-16k(ae^{4x}+be^x)=9.6e^{4x}+30e^x$
$(16a-16ka)e^{4x}+(b-16kb)e^x=9.6e^{4x}+30e^x$

This means

$16a(1-k)=9.6\implies a=\dfrac3{5(1-k)}$
$b(1-16k)=30\implies b=\dfrac{30}{1-16k}$

and so the general solution would be

$y=C_1e^{-4\sqrt kx}+C_2e^{4\sqrt kx}+\dfrac3{5(1-k)}e^{4x}+\dfrac{30}{1-16k}e^x$

8 0

2 years ago