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swat32
3 years ago
9

What is the quotient of -8x^6/4x^-3 ? A. x^9/32 B. 4x^9 C. -12/x^2 D. -2x^9

Mathematics
2 answers:
NARA [144]3 years ago
8 0
\cfrac{-8x^{6}}{4x^{-3}} =-2x^{6-(-3)}=-2x^9
natka813 [3]3 years ago
5 0
I believe the answer to the question is D.
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What's a terminating decimal?
kykrilka [37]
Greetings!

"What's a terminating decimal?"...

A terminating decimal is a decimal that ends/has a finite number of digits.
This is opposite to a repeating decimal, which never ends.

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Hope this helps.
-Benjamin
6 0
3 years ago
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If an object is dropped from a height of 116 feet, the function h(t)=-16t^2+116 gives the height after t seconds. When will the
satela [25.4K]
Check the picture below.

\bf \stackrel{h(t)}{0}=-16t^2+116\implies 16t^2=116\implies t^2=\cfrac{116}{16}
\\\\\\
t^2=\cfrac{29}{4}
\implies 
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5 0
4 years ago
Randy sold corn for 12 ears for $2.How much would Randy make if he sold 600 ears of corn?
meriva

Answer:

$120

Step-by-step explanation:

12 ears = $2

600 ears = ?

to determine the cost of 600easrsof corn  we find the cost of 1 ears of corn first

1 ears = 2/12 =0.1666

= 0.2 approximately

so 600ears of corn = 0.2 × 600 =120

5 0
3 years ago
Please help. 1. The population of a town was 500 in 2010. The population grows at a rate of 9.6% annually. (a) Use the exponenti
pogonyaev

Answer:

Ok I did the math. t= 1248.

Step-by-step explanation:

Multiplying 500 by 9,6% is the start. Then, multiply ever number after that. Ex. 500 x 9.6%=548.  Then multiply 548 by 9.6% and so on. Hope this helps!!

3 0
3 years ago
the intensity, I, of light varies inversely as the square of the distance, D2 , from the light source. If I is 150 units when D
podryga [215]

\bf \textit{\underline{x} varies inversely with }\underline{z^5} ~\hspace{5.5em} \stackrel{\textit{constant of variation}}{x=\cfrac{\stackrel{\downarrow }{k}}{z^5}~\hfill } \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \stackrel{\textit{\underline{I} varies inversely with }\underline{D^2}}{I = \cfrac{k}{D^2}} \qquad \qquad \textit{we also know that } \begin{cases} I=150\\ D=6 \end{cases}

\bf 150=\cfrac{k}{6^2}\implies 150=\cfrac{k}{36}\implies 5400=k~\hfill \boxed{I=\cfrac{5400}{D^2}} \\\\\\ \textit{when I = 24, what is \underline{D}?}\qquad \qquad 24=\cfrac{5400}{D^2}\implies D^2=\cfrac{5400}{24} \\\\\\ D^2=225\implies D=\sqrt{225}\implies D=15

5 0
4 years ago
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