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allsm [11]
3 years ago
8

Simplify. Evaluate the numerical bases. 3^2 b^2 c^-4 *3^-1 b^-5 c^7

Mathematics
1 answer:
photoshop1234 [79]3 years ago
8 0
Hope I can be of assistance!

3^2b^2c^{-4}\cdot \:3^{-1}b^{-5}c^7 \ \textgreater \  \mathrm{Apply\:exponent\:rule}:\ \:a^b\cdot \:a^c=a^{b+c}
3^{-1}\cdot \:3^2=\:3^{2-1}=\:3^1=\:3 \ \textgreater \  3b^{-5}b^2c^{-4}c^7

\mathrm{Apply\:exponent\:rule}: \:a^b\cdot \:a^c=a^{b+c} \ \textgreater \  b^{-5}b^2=\:b^{2-5}=\:b^{-3}
3b^{-3}c^{-4}c^7

\mathrm{Apply\:exponent\:rule}: \:a^b\cdot \:a^c=a^{b+c} \ \textgreater \  c^{-4}c^7=\:c^{-4+7}=\:c^3
3b^{-3}c^3

\mathrm{Apply\:exponent\:rule}: \:a^{-b}=\frac{1}{a^b} \ \textgreater \  b^{-3}=\frac{1}{b^3} \ \textgreater \  3\cdot \frac{1}{b^3}c^3

\mathrm{Multiply\:fractions}: \:a\cdot \frac{b}{c}=\frac{a\:\cdot \:b}{c} \ \textgreater \  \frac{1\cdot \:3c^3}{b^3}

Finally
\mathrm{Apply\:rule}\:1\cdot \:a=a \ \textgreater \  \frac{3c^3}{b^3}

Hope this helps!
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Given:

Angelica reads 15 pages of a book each day.

In 10 days, she reads 200 pages.

Assume the relationship is linear.

To find:

The rate of change and initial value.

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The slope intercept form of a linear function is:

y=mx+b            ...(i)

Where m is the slope and b is the y-intercept.

Angelica reads 15 pages of a book each day. So, the rate of change is 15 pages per day and m=15.

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Putting x=10,y=200,m=15 in (i), we get

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Answer:

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3. So the range of savings vary from 4180 to 5230$$.

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