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.

-3x + 5 > \( \frac{x}{2} \)
2 x (-3x + 5) > x
(2 x -3x) + (2 x 5) > x
-6x + 10 > x
-6x + 10 - x > 0
-6x - x > -10
-7x > -10
x > \( \frac{-10}{-7} \)
x > 1\(\frac{3}{7}\)

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.

-8b - 6 > 2 + 8b
-8b > 2 + 8b + 6
-8b - 8b > 2 + 6
-16b > 8
b > \( \frac{8}{-16} \)
b > -\(\frac{1}{2}\)

To find the diagonal we need to know the length of one of the square's sides. We know the area and the area of a square is the length of one side squared:

a = s²

so the length of one side of the square is:

s = \( \sqrt{a} \) = \( \sqrt{1} \) = 1

The Pythagorean theorem defines the square of the hypotenuse (diagonal) of a triangle with a right angle as the sum of the squares of the other two sides:

c² = a² + b²
c² = 1² + 1²
c² = 2
c = \( \sqrt{2} \)

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 -3 as well and sum (Inside, Outside) to equal -2. For this problem, those two numbers are -3 and 1. Then, plug these into a set of binomials using the square root of the First variable (y²):

y² - 2y - 3
y² + (-3 + 1)y + (-3 x 1)
(y - 3)(y + 1)

A cube is a rectangular solid box with a height (h), length (l), and width (w). The volume is h x l x w and the surface area is 2lw x 2wh + 2lh.