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2 years ago

Can’t figure this problem out need help

Mathematics

1 answer:

MatroZZZ [7]2 years ago

6 0

Answer: 6.5 ft³

Step-by-step explanation:

P =4l => l = 7.3/4 = 1.825 ≈ 1.8 ft

V = Ab×h/3 = l²×h/3

= (1.8)²×6/3

= 3.24×2

= 6.48 ft³

≈ 6.5 ft³

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The sum of 42 fish and x fish is 68 fish. What is the unknown number of fish?

lozanna [386]

Answer:

the unknown number of fish is 26

Step-by-step explanation:

I figured this out by subtracting 68 - 47 = x.

x = 26

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2 years ago

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A stick 43 inches long is broken into two pieces, so that one piece is twice as long as the other. How long are the two pieces?

allsm [11]

Answer:

One piece is $\frac{43}{3}$ inches and another piece is $\frac{86}{3}$ inches

Step-by-step explanation:

Let smaller piece length be $s$

Let larger piece length be $l$

Since total of two pieces is 43, we can write:

$s+l=43$

Also, one piece is twice as long as the other (so longer piece is definitely twice as the short piece), we can write:

$l=2s$

Substituting equation 2 into equation one, we can solve for s:

$s+l=43\\s+(2s)=43\\3s=43\\s=\frac{43}{3}$

Thus longer piece is:

$l=2s=2(\frac{43}{3})=\frac{86}{3}$

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3 years ago

Find the number of pairs (a,b) of integers such that a+2/a+5=b/3

Viktor [21]

Answer:

Step-by-step explanation:

The expression can be rearranged to ...

b = 3 -9/(a+5)

In order for b to be an integer, (a+5) must be an integer divisor of 9. There are exactly 6 of those: ±1, ±3, ±9.

The attached table shows the values (a, b) = (x₁, f(x₁)).

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3 years ago

Number 4 i don’t know how to do this one

Kruka [31]

Answer:

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Step-by-step explanation:

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2 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 f 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

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

3 0

3 years ago