#2**+3 **

S = 2r^{3}t

To solve this equation for t , first let's divide both sides by 2 .

\(\frac{S}{2}\) = r^{3}t

Next divide both sides by r^{3} .

\(\frac{S}{2r^3}\) = t

t = \(\frac{S}{2r^3}\)

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4x^{2} - 376 = 24

To solve this equation for x , first let's add 376 to both sides.

4x^{2} = 24 + 376

4x^{2} = 400

Then divide both sides by 4 .

x^{2} = 400/4

x^{2} = 100

Then take the ± square root of both sides.

x = ±√100

x = ± 10

This means there are two values of x that make the equation true,

and those two values are 10 and -10 .

hectictar Aug 17, 2018

#3**+3 **

For the third question...

∠A and ∠B must be acute angles because the third angle is 90°, and the sum of the angle measures in a triangle is 180°, and each angle has to have a measure greater than 0°. (If either ∠A or ∠B were 90° then two of the sides would be parallel, and they could never meet.)

First let's find the sine and cosine of angle A .

sin A = opposite / hypotenuse

sin A = 12 / 20

sin A = 3 / 5

cos A = adjacent / hypotenuse

cos A = 16 / 20

cos A = 4 / 5

Next let's find the sine and cosine of angle B .

sin B = opposite / hypotenuse

sin B = 16 / 20

sin B = 4 / 5

cos B = adjacent / hypotenuse

cos B = 12 / 20

cos B = 3 / 5

hectictar Aug 17, 2018