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blagie [28]
3 years ago
14

If we recreated the scene from Fast & Furious 7 and dropped a Challenger SRT® Hellcat Redeye Widebody from a C-130 aircraft

at 5,280 ft, how much horsepower would it take to drive past it before it hits the ground if you’re 1 mile away?
Mathematics
2 answers:
padilas [110]3 years ago
6 0

Answer:

7

Step-by-step explanation:

MissTica3 years ago
4 0

Answer:

After calculating all this, I came up with an answer around 474.34812 which is close to one of the offered answers of 475 HP, so I selected that.

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slava [35]

Answer:

2.52 lbs=40.32 oz

Step-by-step explanation:

I used the mass formula and I even searched it.

2.52 lbs=40.32 oz

Hope this helps! :)

7 0
2 years ago
A plant grew 3 1/4 inches over a 6 1/2 month period.what was the average monthly growth rate for the plant
Solnce55 [7]

Answer:

0.5 inches per a month

Step-by-step explanation:

you turn them to a decimal to make them easier and then divide them to get the answer.

5 0
3 years ago
Look at pic<br><br> Time in Weeks<br> Which panda was heavier when they were born?
arlik [135]
What the top guy said
7 0
3 years ago
For your summer babysitting jobs you have 10 weeks to work before your family vacation and before you return to school in the fa
seropon [69]
Uh ummm the question?
5 0
3 years ago
Find the right quotient. 36m 5 n 5 ÷ (12m 3)
polet [3.4K]
ANSWER

The right quotient is

3  {m}^{2} {n}^{5}

EXPLANATION


The given expression is :

\frac{36 {m}^{5} {n}^{5}  }{12 {m}^{3} }



\frac{36 {m}^{5} {n}^{5}  }{12 {m}^{3} }  =  \frac{36}{12}  \times  \frac{ {m}^{5} }{ {m}^{3}} \times  {n}^{5}


\frac{36 {m}^{5} {n}^{5}  }{12 {m}^{3} }  =  3\times  \frac{ {m}^{5} }{ {m}^{3}} \times  {n}^{5}


Recall that,


\frac{ {a}^{m} }{ {a}^{n} }  =  {a}^{m - n}
We apply this property to obtain;



\frac{36 {m}^{5} {n}^{5}  }{12 {m}^{3} }  =  3 \times   {m}^{5 - 3}  \times  {n}^{5}


\frac{36 {m}^{5} {n}^{5}  }{12 {m}^{3} }  =  3  {m}^{2} {n}^{5}

5 0
2 years ago
Read 2 more answers
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