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Marta_Voda [28]
1 year ago
13

Graph Y=-2x-3 and y=-x+6

Mathematics
1 answer:
lesantik [10]1 year ago
7 0

Answer:

here ya go :)

the black line is y=-2x-3

the lavender line is y=-x+6

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Find the area and perimeter of the figure below, explain and show work pls
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This is my work and only part 1

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3 years ago
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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.

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3 years ago
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Answer:

The First one is the answer

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2 years ago
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Question in attachment​
Damm [24]

Answer:

a) scale factor = 1.5

b) d length is 7.5

c) Area scale factor is 2.25

Step-by-step explanation:

Given QR =6

And ST =9

Length Scalar factor = enlarged length /previous length

 =9/6

 =1.5

Length Scalar factor =1.5

b)we know PS enlarged length of PQ

Length scale factor = PS/PQ

1.5 =PS/5

PS =5×1.5

PS =d =7.5

d length is 7.5

c)Formula :

Area scale factor = (length scale factor)2

=(1.5)2

=2.25

Area scale factor is 2.25

5 0
2 years ago
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