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3 years ago

Help me please Solve the system algebraically using substitution.

Mathematics

1 answer:

cluponka [151]3 years ago

3 0

Answer:

y = 5/3

x = 8/3

Steps:

y = x - 1

2x + y = 7

Substitute y = x-1

(2x + x - 1 = 7)

Simplify: 3x - 1 = 7

Isolate x: 3x - 1 = 7: x=8/3

Plug the valuse of x into the other equation (y = x- 1): y = 8/3 - 1

8/3 - 1 = 5/3

y = 5/3

y = 5/3, x = 8/3

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Determine whether the improper integral converges or diverges, and find the value of each that converges.

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Answer:

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Step-by-step explanation:

Assuming this integral:

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We can do this as the first step:

$5 \int_{-\infty}^0 e^{60x} dx$

Now we can solve the integral and we got:

$5 \frac{e^{60x}}{60} \Big|_{-\infty}^0$

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3 years ago

What are the solutions for x to the equation (x-3)(x+8)=0

hichkok12 [17]

For this case we have a factored expression, of a second degree equation:

( $(x-3) (x + 8) = 0$

To find the solutions:

$x-3 = 0$

we add 3 to both sides of the equation:

$x=3$

$x+8=0$

We subtract 8 on both sides of the equation:

$x = -8$

Thus, the solutions of the equation are:

$x_ {1} = 3\\x_ {2} = - 8$

Answer:

$x_ {1} = 3\\x_ {2} = - 8$

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