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3 years ago

-10=-14v+14v what is V

Mathematics

1 answer:

KatRina [158]3 years ago

7 0

-10 = -14v+14v

-10 = 0

Here, v can't have a definite value...so the question has no solution

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4-3x > 2x -1 what does it need to be to be greater or less than I guess

Gre4nikov [31]

There are probably mutliple solutions to this but heres one I found

x=0

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EXPLANTION

4-3(0) > 2(0)-1

4-0 > 0-1

4 > -1

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3 years ago

I will give brainlist for correct answer.

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Answer:

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Step-by-step explanation:

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Answer:

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2 years ago

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3 years ago

What is the difference? StartFraction x Over x squared minus 16 EndFraction minus StartFraction 3 Over x minus 4 EndFraction Sta

katovenus [111]

Answer:

The option "StartFraction negative 2 (x + 6) Over (x + 4) (x minus 4) EndFraction" is correct

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Therefore $\frac{x}{x^2-16}-\frac{3}{x-4}=\frac{-2(x+6)}{(x+4)(x-4)}$

Step-by-step explanation:

Given problem is StartFraction x Over x squared minus 16 EndFraction minus StartFraction 3 Over x minus 4 EndFraction

It can be written as below :

$\frac{x}{x^2-16}-\frac{3}{x-4}$

To solve the given expression

$\frac{x}{x^2-16}-\frac{3}{x-4}$

$=\frac{x}{x^2-4^2}-\frac{3}{x-4}$

$=\frac{x}{(x+4)(x-4)}-\frac{3}{x-4}$ ( using the property $a^2-b^2=(a+b)(a-b)$ )

$=\frac{x-3(x+4)}{(x+4)(x-4)}$

$=\frac{x-3x-12}{(x+4)(x-4)}$ ( by using distributive property )

$=\frac{-2x-12}{(x+4)(x-4)}$

$=\frac{-2(x+6)}{(x+4)(x-4)}$

$\frac{x}{x^2-16}-\frac{3}{x-4}=\frac{-2(x+6)}{(x+4)(x-4)}$

Therefore $\frac{x}{x^2-16}-\frac{3}{x-4}=\frac{-2(x+6)}{(x+4)(x-4)}$

Therefore the option "StartFraction negative 2 (x + 6) Over (x + 4) (x minus 4) EndFraction" is correct

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3 years ago

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