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1 year ago

I need help please Asap

Mathematics

1 answer:

s2008m [1.1K]1 year ago

4 0

Answer:

Step-by-step explanation:

Equation one:

x = -5, x = -1 (Both are real)

Equation two:

No real solutions

Equation three:

x = -3 (Real)

Equation four:

No real solutions

The easiest way to figure out if an equation has real solutions is to factor it. If it is factorable, then it has real solutions. If it isn't, then it doesn't have real solutions.

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Multiplying rational expresssions solve t^2-t-12/t+1 * t+1/t+3 Show answer with working

Andreyy89

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now you can rewrite the expression as one term; (t-4)(t+3)(t+1)/(t+1)(t+3)
cancel out (t+3) and (t+1) which leaves the answer as (t-4)</span>

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3 years ago

Simplify the expression. 8x^-10 y^'6 -2x^2y^-8 Write your answer without negative exponents.

sergij07 [2.7K]

Answer:

$8x^{-10}y^6 - 2x^2y^{-8} = \frac{8y^{14} - 2x^{12}}{x^{10}y^8}$

Step-by-step explanation:

Given

$8x^{-10}y^6 - 2x^2y^{-8}$

Required

Simplify

Rewrite as:

$8x^{-10}y^6 - 2x^2y^{-8} = \frac{8y^6}{x^{10}} - \frac{2x^2}{y^8}$

Take LCM

$8x^{-10}y^6 - 2x^2y^{-8} = \frac{8y^6*y^8 - 2x^2 * x^{10}}{x^{10}y^8}$

Apply law of indices

$8x^{-10}y^6 - 2x^2y^{-8} = \frac{8y^{14} - 2x^{12}}{x^{10}y^8}$

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3 years ago

What is the answer to this

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Answer:

5 f

Re-order terms so that constants are on the left

. 5

5 f

solution

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3 years ago

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Step-by-step explanation:

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