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° - 35° - 47° = 98°

So, d° = 47° + 98° = 145°

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° - 35° = 145°

An equation is two expressions separated by an equal sign. The key to solving equations is to repeatedly do the same thing to both sides of the equation until the variable is isolated on one side of the equal sign and the answer on the other.

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 = \( \sqrt{a} \) = \( \sqrt{36} \) = 6

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 = 6² + 6²
c² = 72
c = \( \sqrt{72} \) = \( \sqrt{36 x 2} \) = \( \sqrt{36} \) \( \sqrt{2} \)
c = 6\( \sqrt{2} \)

You need to find the value of a so solve the first equation in terms of z:

9a + z = -6
z = -6 - 9a

then substitute the result (-6 - 9a) into the second equation:

2a + 9(-6 - 9a) = 1
2a + (9 x -6) + (9 x -9a) = 1
2a - 54 - 81a = 1
2a - 81a = 1 + 54
-79a = 55
a = \( \frac{55}{-79} \)
a = -\(\frac{55}{79}\)

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.

-6x - 9 > \( \frac{x}{-6} \)
-6 x (-6x - 9) > x
(-6 x -6x) + (-6 x -9) > x
36x + 54 > x
36x + 54 - x > 0
36x - x > -54
35x > -54
x > \( \frac{-54}{35} \)
x > -1\(\frac{19}{35}\)