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antiseptic1488 [7]
3 years ago
13

Maya is trying to break or tie the league record for home runs in her softball league. She solved the inequality mc019-1.jpg to

find the minimum number of home runs she must average per week over the five remaining weeks of the season to break or tie the record. What error did Maya make when she chose this inequality to model the situation?
She chose the incorrect relation and should have solved 5x<30

She modeled “average” incorrectly and should have solved x/5<_30

She chose the incorrect relation and should have solved 5x_>30

She modeled “average” incorrectly and should have solved x+5<_30
Mathematics
2 answers:
Lelechka [254]3 years ago
7 0
The correct answer would be letter C. She needs to either break or tie the record. The less than or equal to symbol in the original inequality would be her scoring less or tying the record. Using the greater than or equal to sign would represent her breaking the record or tying with it. Hope this helped.
Sladkaya [172]3 years ago
4 0

Answer:

c

Step-by-step explanation:

me smart birb

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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
PLEASE HELP THIS IS EASY BUT IM JUST BAD AT MATH
balandron [24]

Answer:

Its B

Step-by-step explanation:

Try and read throught it slowly and everytime you hit a variable go down the the answer and find where it is. That helped me a lot

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81 more than x
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