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2 years ago

PLEASE HELP ME!!!!!!!!!!!!!!!!!!

Mathematics

1 answer:

Varvara68 [4.7K]2 years ago

5 0

Answer:

Step-by-step explanation:

These are functions, the number in the parentheses shows you what number you will replace x with

If f(x) = x + 3, then f(-1) is -1 + 3

So, -1 + 3 = 2. g(2) = 2(2) + 1 = 4 + 1 = 5

Hope this helps friend.

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Is x2 greater then x

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Inverse equation t=120r/r+120 R=???

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r = 120t/(120-t)

Step-by-step explanation:

Multiply by the denominator, isolate r terms, then divide by the coefficient of r.

$t=\dfrac{120r}{r+120}\qquad\text{given}\\\\t(r+120)=120r\qquad\text{multiply by $(r+120)$}\\\\120t=120r-tr\qquad\text{subtract $tr$}\\\\\dfrac{120t}{120-t}=r\qquad\text{divide by the coefficient of $r$}\\\\\boxed{r=\dfrac{120t}{120-t}}$

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Fran is making a punch for her birthday party. She needs 4 1/5 liters of juice for the punch, but she only has 1 4/5 liters. How

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3 years ago

Read 2 more answers

For what values of x is f(x) = |x + 1| differentiable? I'm struggling my butt off for this course

pav-90 [236]

By definition of absolute value, you have

$f(x) = |x+1| = \begin{cases}x+1&\text{if }x+1\ge0 \\ -(x+1)&\text{if }x+1$

or more simply,

$f(x) = \begin{cases}x+1&\text{if }x\ge-1\\-x-1&\text{if }x$

On their own, each piece is differentiable over their respective domains, except at the point where they split off.

For x > -1, we have

(x + 1)' = 1

while for x < -1,

(-x - 1)' = -1

More concisely,

$f'(x) = \begin{cases}1&\text{if }x>-1\\-1&\text{if }x$

Note the strict inequalities in the definition of f '(x).

In order for f(x) to be differentiable at x = -1, the derivative f '(x) must be continuous at x = -1. But this is not the case, because the limits from either side of x = -1 for the derivative do not match:

$\displaystyle \lim_{x\to-1^-}f'(x) = \lim_{x\to-1}(-1) = -1$

$\displaystyle \lim_{x\to-1^+}f'(x) = \lim_{x\to-1}1 = 1$

All this to say that f(x) is differentiable everywhere on its domain, except at the point x = -1.

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3 years ago