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vovangra [49]
3 years ago
8

Find the distance between the two points. ​(-4​,-1​) and ​(-39​,-121​)

Mathematics
2 answers:
REY [17]3 years ago
7 0
The distance of the two points equals to

AVprozaik [17]3 years ago
5 0

Answer:

<h3>The answer is 125 units</h3>

Step-by-step explanation:

The distance between two points can be found by using the formula

d =  \sqrt{ ({x1 - x2})^{2} +  ({y1 - y2})^{2}  } \\

where

(x1 , y1) and (x2 , y2) are the points

From the question the points are

(-4,-1) and (-39,-121)

The distance between them is

d =  \sqrt{ ({  - 4 + 39})^{2} +  ({ - 1 + 121})^{2}  }  \\  =  \sqrt{ {35}^{2}  +  {120}^{2} } \:  \:  \:  \:   \\  =  \sqrt{1225 + 14400}  \\  =  \sqrt{15625}  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:   \\  = 125 \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:

We have the final answer as

<h3>125 units</h3>

Hope this helps you

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3 years ago
............................
prohojiy [21]
Ahh yes, negative exponents always give us a scare once and a while. All the negative means is to flip the position of its base. For instance, if x has a negative exponent and x in the denominator, all you would have to do is move x to the numerator with the same power (except it's no longer negative). Before we substitute x and all the other variables which the values given, let's eliminate the negatives first.

After flipping positions/eliminating the negative exponents it should look like this:

\frac{ 2^{2} x^{3} }{ y^{5} }
which simplifies to 
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now that everything is simplified, and all negative exponents are eliminated we can substitute x with 2, and y with (-4).
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Final Answer: - \frac{1}{32} [/tex]
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