Consider the two triangles.

Mathematics

1 answer:

Masja [62]4 years ago

3 0

For the answer to the question above, you just simply use the Pythagoras theorem to find the 3rd length, a^2+b^2=c^2.
E.g 15^2+20^2= length of hypotenuse^2
225+400=c^2
625=c^2
25=c
Which makes no. <span>1. AB = 25 and HG = 15 the answer. I hope this helps</span>

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1. |4n-15|=|n|<br> 2. |2c+8|=|10c| <br> can you please show the work for both

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Problem 1

<h3>Answers: n = 3 or n = 5</h3>

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Work Shown:

$|4n - 15| = |n|\\\\4n - 15 = n \ \text{ or } \ 4n-15 = -n\\\\-15 = n-4n \ \text{ or } \ -15 = -n-4n\\\\-15 = -3n \ \text{ or } \ -15 = -5n\\\\-3n = -15 \ \text{ or } \ -5n = -15\\\\n = \frac{-15}{-3} \ \text{ or } \ n = \frac{-15}{-5}\\\\n = 5 \ \text{ or } \ n = 3\\\\$

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Problem 2

<h3>Answers: c = 1 or c = -2/3</h3>

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Work Shown:

$|2c+8| = |10c|\\\\2c+8 = 10c \ \text{ or } \ 2c+8 = -10c\\\\8 = 10c-2c \ \text{ or } \ 8 = -10c-2c\\\\8 = 8c \ \text{ or } \ 8 = -12c\\\\8c = 8 \ \text{ or } \ -12c = 8\\\\c = \frac{8}{8} \ \text{ or } \ c = \frac{8}{-12}\\\\c = 1 \ \text{ or } \ c = -\frac{2}{3}$