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3 years ago

8. Solve the system using substitution. Show all work. (3 points)

1 answer:

3 0

Ok. first you have to pick x or y or whatever variable you have and the number multiplied with it has to be the same in each equation but one has to be positive and one negative.
lets pick x

y=-2x+6
3y=x-3/ *2

so the x are positive and negative we have to multiply the equation with 2

y=-2x+6
6y=2x-6

now you have to add the equations. you do that by adding each side of the equatins together like this:

y+6y=-2x+6+2x-6 the x will cancel out

7y=0
y=0

now just go back to one of the previous equations and put the y in it.

0=-2x+6
2x=6
x=3

here you go

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3 years ago

Integrate this two questions

tankabanditka [31]

Simplify the integrands by polynomial division.

$\dfrac{t^2}{1 - 3t} = -\dfrac19 \left(3t + 1 - \dfrac1{1 - 3t}\right)$

$\dfrac t{1 + 4t} = \dfrac14 \left(1 - \dfrac1{1 + 4t}\right)$

Now computing the integrals is trivial.

$\displaystyle \int \frac{t^2}{1 - 3t} \, dt = -\frac19 \int \left(3t + 1 - \frac1{1-3t}\right) \, dt \\\\ = -\frac19 \left(\frac32 t^2 + t + \frac13 \ln|1 - 3t|\right) + C \\\\ = \boxed{-\frac{t^2}6 - \frac t9 - \frac{\ln|1-3t|}{27} + C}$

where we use the power rule,

$\displaystyle \int x^n \, dx = \frac{x^{n+1}}{n+1} + C ~~~~ (n\neq-1)$

and a substitution to integrate the last term,

$\displaystyle \int \frac{dt}{1-3t} = -\frac13 \int \frac{du}u \\\\ = -\frac13 \ln|u| + C \\\\ = -\frac13 \ln|1-3t| + C ~~~ (u=1-3t \text{ and } du = -3\,dt)$

$\displaystyle \int \frac t{1+4t} \, dt = \frac14 \int \left(1 - \frac1{1+4t}\right) \, dt \\\\ = \frac14 \left(t - \frac14 \ln|1 + 4t|\right) + C \\\\ = \boxed{\frac t4 - \frac{\ln|1+4t|}{16} + C}$

using the same approach as above.

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2 years ago

What is 7 plus the sum of m and -17

olasank [31]

Answer:

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Step-by-step explanation:

7 plus = 7 +

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7 + (m + -17)

We can get rid of the parentheses

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Combine like terms

7 + -17 = -10

m + -10

Get rid of the addition sign

m - 10

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3 years ago