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3 years ago

Which expressions are equivalent to 7^8 x 7? Select all that apply.

Mathematics

1 answer:

nlexa [21]3 years ago

7 0

Answer:

Step-by-step explanation:

7^8*7=

5764801 * 7= 40353607

A. 7^3*7^3= 343*343= 117649 <--- not the answer

B. 7^18/7^9= 40353607 <---- right answer

C. (7^3)^3= 343^3= 40353607 <------ right answer

D. 7^5+7^4= 16807+2401= 19208 <---- not the answer

E. 7^5*7^4= 16807*2401= 40353607 <---- right answer

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I think its answer A

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4 years ago

What is the slope of the line shown below? (enter your answer as a decimal if necessary.)

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Answer: The slope is 2 -Please mark brainliest

1. Find the X and Y values of each point

X1: 1

X2: -1

Y1: 3

Y2: -1

2. Subtract X1 from X2 and Y1 from Y2

Y: -1-3=-4

X: -1-1=-2

3. Divide the X value by the Y value

-4/-2=2

The slope of the line shown on the coordinate plane is 2.

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3 years ago

Read 2 more answers

Find a particular solution to <img src="https://tex.z-dn.net/?f=%20x%5E%7B2%7D%20%20%5Cfrac%7B%20d%5E%7B2%7Dy%20%7D%7Bd%20x%5E%7

Digiron [165]

$y=x^r$
$\implies r(r-1)x^r+6rx^r+4x^r=0$
$\implies r^2+5r+4=(r+1)(r+4)=0$
$\implies r=-1,r=-4$

so the characteristic solution is

$y_c=\dfrac{C_1}x+\dfrac{C_2}{x^4}$

As a guess for the particular solution, let's back up a bit. The reason the choice of $y=x^r$ works for the characteristic solution is that, in the background, we're employing the substitution $t=\ln x$ , so that $y(x)$ is getting replaced with a new function $z(t)$ . Differentiating yields

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac1x\dfrac{\mathrm dz}{\mathrm dt}$
$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac1{x^2}\left(\dfrac{\mathrm d^2z}{\mathrm dt^2}-\dfrac{\mathrm dz}{\mathrm dt}\right)$

Now the ODE in terms of $t$ is linear with constant coefficients, since the coefficients $x^2$ and $x$ will cancel, resulting in the ODE

$\dfrac{\mathrm d^2z}{\mathrm dt^2}+5\dfrac{\mathrm dz}{\mathrm dt}+4z=e^{2t}\sin e^t$

Of coursesin, the characteristic equation will be $r^2+6r+4=0$ , which leads to solutions $C_1e^{-t}+C_2e^{-4t}=C_1x^{-1}+C_2x^{-4}$ , as before.

Now that we have two linearly independent solutions, we can easily find more via variation of parameters. If $z_1,z_2$ are the solutions to the characteristic equation of the ODE in terms of $z$ , then we can find another of the form $z_p=u_1z_1+u_2z_2$ where

$u_1=-\displaystyle\int\frac{z_2e^{2t}\sin e^t}{W(z_1,z_2)}\,\mathrm dt$
$u_2=\displaystyle\int\frac{z_1e^{2t}\sin e^t}{W(z_1,z_2)}\,\mathrm dt$

where $W(z_1,z_2)$ is the Wronskian of the two characteristic solutions. We have

$u_1=-\displaystyle\int\frac{e^{-2t}\sin e^t}{-3e^{-5t}}\,\mathrm dt$
$u_1=\dfrac23(1-2e^{2t})\cos e^t+\dfrac23e^t\sin e^t$

$u_2=\displaystyle\int\frac{e^t\sin e^t}{-3e^{-5t}}\,\mathrm dt$
$u_2=\dfrac13(120-20e^{2t}+e^{4t})e^t\cos e^t-\dfrac13(120-60e^{2t}+5e^{4t})\sin e^t$

$\implies z_p=u_1z_1+u_2z_2$
$\implies z_p=(40e^{-4t}-6)e^{-t}\cos e^t-(1-20e^{-2t}+40e^{-4t})\sin e^t$

and recalling that $t=\ln x\iff e^t=x$ , we have

$\implies y_p=\left(\dfrac{40}{x^3}-\dfrac6x\right)\cos x-\left(1-\dfrac{20}{x^2}+\dfrac{40}{x^4}\right)\sin x$

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3 years ago

Four students are sitting in the first row. In the second row, 4 students sit behind each of those students, and so on. After th

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<h3>✽ - - - - - - - - - - - - - - - ~<u>Hello There</u>!~ - - - - - - - - - - - - - - - ✽</h3>

➷ If 1 student leaves each row, there are now 3 students in each row.

2 x 3 = 6

There are 6 students in the fifth row

➶ Hope This Helps You!

➶ Good Luck (:

➶ Have A Great Day ^-^

↬ ʜᴀɴɴᴀʜ ♡

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3 years ago

Read 2 more answers

For the equation: 3x+1=2x+7 what are the 1 st and 2 nd opposite operations?

qaws [65]

Answer:

Step-by-step explanation:

3x +1 = 2x +7 subtract 1 from both sides ( first)

3x +1 -1 = 2x + 7 -1

3x = 2x + 6 subtract 2x brom both sides (second)

3x -2x = 2x -2x +6

5x = 6

4 0

3 years ago