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weqwewe [10]
2 years ago
10

Kay is running a 13.5 km race. If she runs at a rate of 8 2/5 km/he for 1.5 hours, how many kilometers will she have left to run

to finish the race?
Mathematics
1 answer:
Monica [59]2 years ago
7 0

The number of kilometers she will have left to run to finish the race is 0.9 km.

<u>Given the following data:</u>

  • Total distance = 13.5 km
  • Speed = 8\frac{2}{5} \;km/hr
  • Time = 1.5 hours

To determine the number of kilometers she will have left to run to finish the race:

First of all, we would calculate the distance she has covered by using this formula;

Distance = speed \times time\\\\Distance = 8\frac{2}{5} \times 1.5\\\\Distance = \frac{42}{5} \times \frac{3}{2} \\\\Distance = \frac{63}{5}

Distance = 12.6 meters.

<u>For the </u><u>distance left</u><u>:</u>

Distance\;left = Total\;distance - Distance\;covered\\\\Distance\;left =13.5 -12.6

Distance left = 0.9 km.

Learn more about distance here: brainly.com/question/10545161

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xeze [42]
<h3>Answer: 1</h3>

where x is nonzero

=======================================================

Explanation:

We'll use two rules here

  • (a^b)^c = a^(b*c) ... multiply exponents
  • a^b*a^c = a^(b+c) ... add exponents

------------------------------

The portion [ x^(a-b) ]^(a+b) would turn into x^[ (a-b)(a+b) ] after using the first rule shown above. That turns into x^(a^2 - b^2) after using the difference of squares rule.

Similarly, the second portion turns into x^(b^2-c^2) and the third part becomes x^(c^2-a^2)

-------------------------------

After applying rule 1 to each of the three pieces, we will have 3 bases of x with the exponents of (a^2-b^2),  (b^2-c^2) and (c^2-a^2)

Add up those exponents (using rule 2 above) and we get

(a^2-b^2)+(b^2-c^2)+(c^2-a^2)

a^2-b^2+b^2-c^2+c^2-a^2

(a^2-a^2) + (-b^2+b^2) + (-c^2+c^2)

0a^2 + 0b^2 + 0c^2

0+0+0

0

All three exponents add to 0. As long as x is nonzero, then x^0 = 1

4 0
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