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dybincka [34]
3 years ago
14

Do I have to find a common denominator

Mathematics
2 answers:
Katen [24]3 years ago
6 0
Yea only for fractions man
antoniya [11.8K]3 years ago
5 0
When it comes to fractions yes you do. But to make it easier just multiply the denominators and that should give u a common denominator, they don't always have to be least common.
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Isabella had a week to read a book for a school assignment. She read an average of 36 pages per day for the first three days and
iogann1982 [59]
Ok, so what you want to do, is to start by adding the values together. So, she read an average of 36 pages for the first three days. So, you will multiply 36 by 3. Then, you will have 44 pages for the next three days, or 44 * 3. And finally, you have the last 10 pages. So, it is (36 * 3) + (44*3) + 10. So, 36 times 3 is 108, and 44 times 3 is 132. So, you now have 108 + 132 + 10. Add all of those values together, and your final answer is
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3 years ago
Scarlett is playing a video game she spends 900 minerals to create 18 workers each worker costs the same number of minerals writ
Rufina [12.5K]

Answer:

W=50M

Step-by-step explanation:

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There is your answer!

4 0
3 years ago
Read 2 more answers
A number n is 5 more than 12
zzz [600]
X + 12 = n
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6 0
3 years ago
Consider the following equation: f(x)=x^2+4\4x^2-4x-8 name the vertical asymptote(s)
zhenek [66]

ANSWER

The vertical asymptotes are


\Rightarrow x=2\:or\:x=-1

<u>EXPLANATION</u>

We have

f(x)=\frac{x^2+4}{4x^2-4x-8}


For vertical asymptotes we set the denominator to zero and solve the quadratic equation;

4x^2-4x-8=0


\Rightarrow x^2-x-2=0

We split the middle term to obtain,

x^2+x-2x-2=0

\Rightarrow x(x+1)-2(x+1)=0


\Rightarrow (x-2)(x+1)=0


\Rightarrow x=2\:or\:x=-1


Therefore the vertical asymptotes are


x=2\:or\:x=-1





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4 years ago
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The answer is 368.65.
6 0
3 years ago
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