Given d = 1 2 at2, solve for a. a) a = t2 2d b) a = 2d t2 c) a = d - 1 2 t2 d) a = 1 2 t2 - d

Given expression: .

We need to solve it for a.

First we need to get rid 2 from denominator.

In order to get rid 2 from denominator, we need to multiply both sides by 2.

Multiplying both sides by 2, we get

Now, dividing both sides by t^2, we get

.

last option

step-by-step explanation:

given that δdef is right and isoceles,

assuming that side df is the hypotenuse means that de an ef are the other 2 sides that intersect at 90° and are of equal length.

because they are of equal length, we can say that they are congruent

because they intersect at 90°, their slopes must be opposite reciprocals.

only the last option satisfies these conditions.