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vlabodo [156]
3 years ago
12

The perimeter of a rectangle is 48 centimeters. The relationship between the length, the width, and the perimeter of the rectang

le can be described with the equation
`2\cdot\operatorname{length}+2\cdot width=48`



Find the length, in centimeters, if the width is `x` centimeters (answer as an expression)
Mathematics
1 answer:
Evgesh-ka [11]3 years ago
4 0

Answer:

The length in centimeters is (24-x) centimeters

Step-by-step explanation:

Mathematically, we are told that the equation linking the length, width and perimeter is as follows;

2(length) + 2(width) = 48

But in this case, we are told that the width is x centimeters

2(length) + 2(x) = 48

2(length) = 48-2x

length = (48-2x)/2

length = 24 - x

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2 years ago
Saturn is 8.867 × 108 miles away from the Sun. Uranus is 1.787 × 109 miles away from the Sun. Approximately how many times farth
GrogVix [38]

Answer:

0.2 times

Step-by-step explanation:

Saturn: 8.867 × 108 = 957.636

Uranus: 1.787 × 109 = 194.783

You take (Uranus: 194.783)

& you divide it by (Saturn: 957.636)

You should get a funky number like 0.203399830415732 but since they said approximately you only need to use the first 2 (don't forget to round)

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4 0
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.

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3 years ago
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pogonyaev

Answer:

the answer would be C

Step-by-step explanation:

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