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.

2b - 9 > \( \frac{b}{7} \)
7 x (2b - 9) > b
(7 x 2b) + (7 x -9) > b
14b - 63 > b
14b - 63 - b > 0
14b - b > 63
13b > 63
b > \( \frac{63}{13} \)
b > 4\(\frac{11}{13}\)

The first step to solve a quadratic equation that's set to zero is to factor the quadratic equation:

y² - y - 20 = 0
(y + 4)(y - 5) = 0

For this expression to be true, the left side of the expression must equal zero. Therefore, either (y + 4) or (y - 5) must equal zero:

If (y + 4) = 0, y must equal -4
If (y - 5) = 0, y must equal 5

So the solution is that y = -4 or 5

Perimeter is equal to the sum of the four sides:

p = a + b + c + d
p = 8 + 3 + 6 + 6
p = 23

An exterior angle of a triangle is equal to the sum of the two interior angles that are opposite:

d° = b° + c°

To find angle c, remember that the sum of the interior angles of a triangle is 180°:

180° = a° + b° + c°
c° = 180° - a° - b°
c° = 180° - 53° - 29° = 98°

So, d° = 29° + 98° = 127°

A shortcut to get this answer is to remember that angles around a line add up to 180°:

a° + d° = 180°
d° = 180° - a°
d° = 180° - 53° = 127°

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 = \( \sqrt{a} \) = \( \sqrt{4} \) = 2

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 = 2² + 2²
c² = 8
c = \( \sqrt{8} \) = \( \sqrt{4 x 2} \) = \( \sqrt{4} \) \( \sqrt{2} \)
c = 2\( \sqrt{2} \)