To solve this equation, isolate the variable for which you are solving (c) on one side of the equation and put everything else on the other side.

5c + 2z = 4c - 5z + 8
5c = 4c - 5z + 8 - 2z
5c - 4c = -5z + 8 - 2z
c = -7z + 8

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.

-8c + 2 < \( \frac{c}{-6} \)
-6 x (-8c + 2) < c
(-6 x -8c) + (-6 x 2) < c
48c - 12 < c
48c - 12 - c < 0
48c - c < 12
47c < 12
c < \( \frac{12}{47} \)
c < \(\frac{12}{47}\)

When solving an equation with two variables, replace the variables with the values given and then solve the now variable-free equation. (Remember order of operations, PEMDAS, Parentheses, Exponents, Multiplication/Division, Addition/Subtraction.)

A monomial contains one term, a binomial contains two terms, and a polynomial contains more than two terms.

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 = \( \sqrt{a} \) = \( \sqrt{16} \) = 4

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 = 4² + 4²
c² = 32
c = \( \sqrt{32} \) = \( \sqrt{16 x 2} \) = \( \sqrt{16} \) \( \sqrt{2} \)
c = 4\( \sqrt{2} \)