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zheka24 [161]
3 years ago
5

[10 Points, Algebra 2-simplifying complex fractions]

Mathematics
1 answer:
IrinaK [193]3 years ago
6 0
The restrictions are what the denominator can not be. The denominator cannot be zero. So if the denominator has x+1 in it. If you put -1 in place of x. You would get zero. -1 is a restriction
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The table below shows the cumulative amount of money that George saved over the course of several weeks.
noname [10]

Answer:

Step-by-step explanation:

8 0
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Write 35.2 in a word form
Brilliant_brown [7]
35.2 in word form is: Thirty Five and 2 Tenths.
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7 0
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Question pls help!!!!!!!!!
konstantin123 [22]

Hey!

--------------------------------------------

Solution:

72 tropical fish and 5/9 sold.

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3 0
4 years ago
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As a sales person at scrapbooks and treasures Alyssa receives a monthly base pay plus commission on all that she sells she sells
max2010maxim [7]
Let 
x =  base pay per month
y = commission per $1 of sales

Then the monthly pay for selling $100 of merchandise is
P = x + 100y

In one month, Alyssa earned $388 for selling $400 worth of merchandise.
Therefore
x + 400y = 388           (1)

In another month, she earned $520 for selling $1000 worth of merchandise.
Therefore
x + 1000y = 520        (2)

Subtract (1) from (2).
x + 1000y - (x + 400y) = 520 - 388
600y = 132
y = 0.22
From (1), obtain
x = 388 - 400y = 388 - 400*0.22 = 300

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3 0
4 years ago
PLEASE HELP! See attachment:
beks73 [17]
\bf f(x)=x^2e^{-2x}\\\\
-----------------------------\\\\
\cfrac{dy}{dx}=2xe^{-2x}+x^2\cdot -2e^{-2x}\implies \cfrac{dy}{dx}=xe^{-2x}(2-2x)
\\\\\\
\cfrac{dy}{dx}=\cfrac{x(2-2x)}{e^{2x}}\implies 0=\cfrac{x(2-2x)}{e^{2x}}
\\\\\\
0=x(2-2x)\implies 
\begin{cases}
0=x\\
0=2-2x\implies 2x=2\implies x=1
\end{cases}

now, bear in mind, that zeroing out the denominator, also gives critical points, usually asymptotic points, where the derivative is undefined, now, in this case, the denominator is never zero, so we don't get any from the denominator, just from the numerator, and are 0 and 1

now check the picture below

running a first-derivative test on it, those are the values on those regions

you get a negative, regardless of what it might be, what matters is the sign
you get a positive, and then a negative

so, f(x) goes down, then up then down

now, you can see, there's on relative minimum and a relative maximum

6 0
3 years ago
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