I'd start by writing an equation for each of the right triangles. (Pythagorean theorem)
y² + 9² = z²
x² + z² = (4+9)²
4² + y² = x²
we want to find z so combine the equations by substituting the other variables x,y out.
substitute y² for (x² - 4²) in 1st equation.
(x² - 4²) + 9² = z²
now by rearranging the 2nd equation we can substitute x² for (13² - z²)
(13² - z²) - 4² + 9² = z²
169 - z² - 16 + 81 = z²
234 - z² = z²
234 = 2z²
234/2 = z²
117 = z²
√(117) = z
√(9*13) = z
3√(13) = z
13 goes in the box
Answer: B
Step-by-step explanation:
3/4 x 4/4 = 12/16
Answer:
144°
Step-by-step explanation:
First, find the area of the circle, with the formula A = 
r²
Plug in 10 as the radius, and solve
A = r²
A = (10²)
A = 100
Using this, create a proportion that relates the area of the sector to the degree measure of the arc.
Let x represent the degree measure of the arc of the sector:
= 
Cross multiply and solve for x:
100x = 14400
x = 144
So, the degree measure of the sector arc is 144°
S = (-5,0)
T = (2,1)
Step-by-step explanation:
Step 1 :
Given
Q = (3,6) and R = (-4,5). P = (-1,3)
Let S be (a,b) and T be (c,d)
The diagonals of a parallelogram bisect each other. so in order to ensure that QRST is a parallelogram, P must be the mid point of the diagonals QS and RT.
Step 2 :
P is the midpoint of QS
So we have (3+a) ÷ 2 = -1 and (6 + b) ÷ 2 = 3
=> 3 + a = -2 and 6 + b = 6
=> a = -5 and b =0
So S should be (-5,0)
Step 3 :
P is the midpoint of RT
So we have (-4+c) ÷ 2 = -1 and (5 + d) ÷ 2 = 3
=> -4+ c = -2 and 5 + d = 6
=> c = 2 and d =1
So T should be (2,1)
Step 4 :
Answer :
S = (-5,0)
T = (2,1)
T / 3/4 ; t = 9 3/4
9 3/4 / 3/4
9 3/4 = 39/4
39/4 / 3/4
39/4 x 4/3
39 x 4 = 156
4 x 3 = 12
156/12 / 4/4 = 39/3 = 13
