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3 years ago

A square pyramid has a base with a total area of 144m^2 and a volume of 384m^3, what is the slant height of the pyramid?

Mathematics

1 answer:

Nastasia [14]3 years ago

7 0

Answer:

Step-by-step explanation:

Volume of a square based pyramid = $\frac{1}{3}a^{2}h$ where;

a² is the base area of the pyramid since it is square shaped

h is the slant height.

Given Base area = 144m² and Volume = 384m³, on substituting this values given in the formula we have;

$384 = \frac{1}{3}*144h \\cross\ multiplying\ \\384*3=144h\\h = \frac{384*3}{144} \\h = \frac{1,152}{144}\\ h = 8m$

The slant height of the pyramid is 8m

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Sam inherits $250,000, which he invests at a rate of 9% compounded annually. How much will Sam's investment be worth in 25 years

Murljashka [212]

Answer:

the amount that should be worth in 25 years is $2,155,770

Step-by-step explanation:

The computation of the amount that should be worth in 25 years is shown below:

As we know that

Future value = Present value × (1 + rate of interest)^number of years

= $250,000 × (1 + 0.09)^25

= $250,000 × 1.09^25

= $2,155,770

Hence, the amount that should be worth in 25 years is $2,155,770

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3 years ago

PLEASE HELP ME????<br> What is the value of the expression below? A. 8 B. 4 C. 1 D. 2

shutvik [7]

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3 years ago

Consider the following. (A computer algebra system is recommended.) y'' + 3y' = 2t4 + t2e−3t + sin 3t (a) Determine a suitable f

drek231 [11]

First look for the fundamental solutions by solving the homogeneous version of the ODE:

$y''+3y'=0$

The characteristic equation is

$r^2+3r=r(r+3)=0$

with roots $r=0$ and $r=-3$ , giving the two solutions $C_1$ and $C_2e^{-3t}$ .

For the non-homogeneous version, you can exploit the superposition principle and consider one term from the right side at a time.

$y''+3y'=2t^4$

Assume the ansatz solution,

${y_p}=at^5+bt^4+ct^3+dt^2+et$

$\implies {y_p}'=5at^4+4bt^3+3ct^2+2dt+e$

$\implies {y_p}''=20at^3+12bt^2+6ct+2d$

(You could include a constant term <em>f</em> here, but it would get absorbed by the first solution $C_1$ anyway.)

Substitute these into the ODE:

$(20at^3+12bt^2+6ct+2d)+3(5at^4+4bt^3+3ct^2+2dt+e)=2t^4$

$15at^4+(20a+12b)t^3+(12b+9c)t^2+(6c+6d)t+(2d+e)=2t^4$

$\implies\begin{cases}15a=2\\20a+12b=0\\12b+9c=0\\6c+6d=0\\2d+e=0\end{cases}\implies a=\dfrac2{15},b=-\dfrac29,c=\dfrac8{27},d=-\dfrac8{27},e=\dfrac{16}{81}$

$y''+3y'=t^2e^{-3t}$

$e^{-3t}$ is already accounted for, so assume an ansatz of the form

$y_p=(at^3+bt^2+ct)e^{-3t}$

$\implies {y_p}'=(-3at^3+(3a-3b)t^2+(2b-3c)t+c)e^{-3t}$

$\implies {y_p}''=(9at^3+(9b-18a)t^2+(9c-12b+6a)t+2b-6c)e^{-3t}$

Substitute into the ODE:

$(9at^3+(9b-18a)t^2+(9c-12b+6a)t+2b-6c)e^{-3t}+3(-3at^3+(3a-3b)t^2+(2b-3c)t+c)e^{-3t}=t^2e^{-3t}$

$9at^3+(9b-18a)t^2+(9c-12b+6a)t+2b-6c-9at^3+(9a-9b)t^2+(6b-9c)t+3c=t^2$

$-9at^2+(6a-6b)t+2b-3c=t^2$

$\implies\begin{cases}-9a=1\\6a-6b=0\\2b-3c=0\end{cases}\implies a=-\dfrac19,b=-\dfrac19,c=-\dfrac2{27}$

$y''+3y'=\sin(3t)$

Assume an ansatz solution

$y_p=a\sin(3t)+b\cos(3t)$

$\implies {y_p}'=3a\cos(3t)-3b\sin(3t)$

$\implies {y_p}''=-9a\sin(3t)-9b\cos(3t)$

Substitute into the ODE:

$(-9a\sin(3t)-9b\cos(3t))+3(3a\cos(3t)-3b\sin(3t))=\sin(3t)$

$(-9a-9b)\sin(3t)+(9a-9b)\cos(3t)=\sin(3t)$

$\implies\begin{cases}-9a-9b=1\\9a-9b=0\end{cases}\implies a=-\dfrac1{18},b=-\dfrac1{18}$

So, the general solution of the original ODE is

$y(t)=\dfrac{54t^5 - 90t^4 + 120t^3 - 120t^2 + 80t}{405}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-\dfrac{3t^3+3t^2+2t}{27}e^{-3t}-\dfrac{\sin(3t)+\cos(3t)}{18}$

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4 years ago

Jason eats 10 ounces of candy in 5 days. a. How many pounds does he eat per day?

Nadusha1986 [10]

A. He eats 1/8 a pound of candy each day.
B. It would take him a total of 8 days to eat a pound of candy.

A. Since he ate 10 ounces in 5 days, you divide 10 by 5 to find he ate 2 ounces per day. Since there are 16 ounces in a pound that would be your denominator (bottom of fraction) and 2 would be the numerator (top of your fraction). You will get 2/16 which if you simplify (divide by 2) you get 1/8.

B. Knowing that there are 16 ounces per pound and he ate 1/8 pound ( 2/16 pound or 2 ounces) a day, you can either look at the simplified denominator to get the answer, or, divide 16 by 2 to get your answer.

I hope this helps!

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3 years ago

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Alja [10]

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4 years ago

Read 2 more answers