The area of a trapezoid is one-half the sum of the lengths of the parallel sides multiplied by the height:

a = ½(b + d)(h)
a = ½(3 + 4)(3)
a = ½(7)(3)
a = ½(21) = \( \frac{21}{2} \)
a = 10\(\frac{1}{2}\)

To factor a quadratic expression, apply the FOIL method (First, Outside, Inside, Last) in reverse. First, find the two Last terms that will multiply to produce -45 as well and sum (Inside, Outside) to equal 4. For this problem, those two numbers are -5 and 9. Then, plug these into a set of binomials using the square root of the First variable (y²):

y² + 4y - 45
y² + (-5 + 9)y + (-5 x 9)
(y - 5)(y + 9)

The area of a triangle is equal to ½ base x height:

a = ½bh
a = ½ x 4 x 8 = \( \frac{32}{2} \) = 16

To solve this equation, repeatedly do the same thing to both sides of the equation until the variable is isolated on one side of the > sign and the answer on the other.

-5c + 1 > \( \frac{c}{6} \)
6 x (-5c + 1) > c
(6 x -5c) + (6 x 1) > c
-30c + 6 > c
-30c + 6 - c > 0
-30c - c > -6
-31c > -6
c > \( \frac{-6}{-31} \)
c > \(\frac{6}{31}\)

The circumference of a circle is the distance around its perimeter and equals π (approx. 3.14159) x diameter: c = π d. The area of a circle is π x (radius)² : a = π r².