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iogann1982 [59]
3 years ago
5

Please help Me. NO LINKS PLEASE. 8th Grade

Mathematics
1 answer:
dexar [7]3 years ago
6 0

Answer:

it ok               but i would love to answer anthor one for u

Step-by-step explanation:

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If y = 2 x , what happens to y as x becomes larger?
e-lub [12.9K]

Answer:

it doubles?

Step-by-step explanation:

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3 years ago
X2 + 3x + 2<br> What is the factor?
musickatia [10]
Answer: (x+1)(x+2)
step by step explanation:
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3 years ago
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 <em>x</em> > -1, we have

(<em>x</em> + 1)<em>'</em> = 1

while for <em>x</em> < -1,

(-<em>x</em> - 1)<em>'</em> = -1

More concisely,

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

Note the strict inequalities in the definition of <em>f '(x)</em>.

In order for <em>f(x)</em> to be differentiable at <em>x</em> = -1, the derivative <em>f '(x)</em> must be continuous at <em>x</em> = -1. But this is not the case, because the limits from either side of <em>x</em> = -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 <em>f(x)</em> is differentiable everywhere on its domain, <em>except</em> at the point <em>x</em> = -1.

4 0
2 years ago
What is 5/12 times 6 in multiply fractions with whole numbers
KiRa [710]
5/12*6  =5/12*6/1 crosscancle you get 5/2 the you get 2 1/2
3 0
3 years ago
Which of the following equations could represent the points in the table? Select all that apply.
schepotkina [342]

Answer:

Options (A) and (C)

Step-by-step explanation:

From table attached,

Let the equation of a linear equation that represents the input-output values in the table is,

y - y' = m(x - x')

Where (x', y') is a point lying on the graph.

And m = slope of the line passing through two points (x_1,y_1) and (x_2,y_2)

m = \frac{y_2-y_1}{x_2-x_1}

Slope of the line passing through (0, 11) and (1, 5) will be,

m = \frac{11-5}{0-1}

m = -6

Therefore, equation of the line passing through (0, 11) and slope = -6 will be,

y - 11 = -6(x - 0)

Equation of a line passing through (1, 5) and slope = -6 will be,

y - 5 = -6(x - 1)

Equation of a line passing through (2, -1) and slope = -6 will be,

y + 1 = -6(x - 2)

Equation of a line passing through (3, -7) and slope = -6 will be,

y + 7 = -6(x - 3)

Therefore, Options (A) and (C) are the correct options.

4 0
3 years ago
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