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Mazyrski [523]
3 years ago
8

What is the product? (3a^2b^7) (5a^3b^8)

Mathematics
1 answer:
kumpel [21]3 years ago
6 0

Answer:

\boxed{\sf \ \ \ 15a^5b^{15} \ \ \ }

Step-by-step explanation:

hello,

(3a^2b^7) (5a^3b^8)=15a^{2+3}b^{7+8}=15a^5b^{15}

hope this helps

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PLEASE HELP I AM DESPERATE I WILL GIVE THANKS, 5 STARS, AND THE BRAINLIEST!!!
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Answer:

-8

Step-by-step explanation:

p(x) = - 2x - 3, x ≤ 0    You would use this equation for finding p(0), since 0 is equal to 0.

p(x) = - 2x - 3, x ≤ 0  ← You can ignore the x≤0 part.

p(0) = - 2(0) - 3              Input the value 0 as x.

p(0) = 0 - 3                    Simplify

p(0) = -3    

Next,

p(x) = -x² - 4, x > 0        You would use this equation for finding p(1), since 1 is greater than 0.    

p(x) = -x² - 4, x > 0    ←  You can ignore the x>0 part.  

p(1) = -(1)² - 4             Input the value 1 as x. 1² is equal to 1 times negative is -1.

p(1) = -1 - 4                 Simplify

p(1) = -5

Then,

p(x) = - 2x - 3, x ≤ 0    You would use this equation for finding p(-2), since -2 is less than 0.

p(x) = - 2x - 3, x ≤ 0  ← You can ignore the x ≤ 0 part.

p(-2) = - 2(-2) - 3          Input the value -2 as x. A negative times a negative is

p(-2) = 4 - 3                  a positive.

p(-2) = 1                        Simplify

Finally, you have to fine the value of \frac{p(0)+p(1)}{p(-2)}

\frac{p(0)+p(1)}{p(-2)}          

\frac{(-3) + (-5)}{(1)}            Input the values of p(0), p(1), and p(-2).

\frac{-8}{1}                      Simplify

-8

The answer is -8.

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Answer:

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