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3 years ago

Linear Equations w/ Distribution 8(z+3)=32

Mathematics

1 answer:

Lemur [1.5K]3 years ago

8 0

Answer:

z=1

Step-by-step explanation:

8(z+3)=32

Multiply 8 into the parenthesis.

8z+24=32

Subtract 24 both sides.

8z=8

Divide 8 both sides.

z=1

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Which linear inequality is represented by the graph?

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Step-by-step explanation:

not 100% but if you go across 1 and up 2 you get to tje line

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3 years ago

9(5x + 1) ÷ 3y From the expression above, provide an example of each of the following: sum, term, product, factor, quotient, and

tatuchka [14]

5+1=6x9=54/3=18 use pemdas to solve it

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3 years ago

grace and her friends were to roll the two dice 100 times, how many of the 100 times could they expect to roll the same number o

LenaWriter [7]

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3 years ago

Part 1. Please show work. Please assist me with these math problems.

Vitek1552 [10]

Answer: 1) 2, 3, 4, 9, 32, 279, 8896, 2481705, 22077238784

2) 2,490,924

3) Neither

<u>Step-by-step explanation:</u>

$S_n=S_{n-2}\cdot (S_{n-1}-1)\quad \\\\S_1=2\quad given\\S_2=3\quad given\\S_3=S_1(S_2-1)\quad =2(3-1)\quad =4\\S_4=S_2(S_3-1)\quad =3(4-1)\quad =9\\S_5=S_3(S_4-1)\quad =4(9-1)\quad =32\\S_6=S_4(S_5-1)\quad =9(32-1)\quad =279\\S_7=S_5(S_6-1)\quad =32(279-1)\quad =8,896\\S_8=S_6(S_7-1)\quad =279(8896-1)\quad =2,481,705\\S_9=S_7(S_8-1)\quad =8896(2481705-1)\quad =22,077,238,784$

$\sum^8_{k=1}S_k\\\\=S_1+S_2+S_3+S_4+S_5+S_6+S_7+S_8\\\\= 2,490,924\\$

Neither

Not arithmetic because the difference between each of the terms is not the same.

S₂ - S₁ S₃ - S₂ S₄ - S₃

3 - 2 = 1 4 - 3 = 1 9 - 4 = 5

Not geometric because the ratios between each of the terms is not the same.

$\dfrac{S_2}{S_1}=\dfrac{S_3}{S_2}\qquad \rightarrow \dfrac{3}{2}=\dfrac{4}{3}\qquad \rightarrow \text{cross multiply to get}: 9\neq 8$

8 0

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

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

3 0

4 years ago