The answer is
domain-1
range-5
Answer:
m∠QPR = 35° m∠QPM =40° m∠PRS = 30°
Step-by-step explanation:
ΔPRQ is a right triangle with right angle at R. So m∠QPR = 90 - 55 = 35
ΔQPM is a right triangle with right angle at M. So m∠QPM = 90 - 40 = 40
Arc RQ = 2(35) = 70 and arc SR = 2(25) = 50.
So arc PS = 180 - (arc RQ + arc SR) = 180 - (70 + 50) = 180 - 120 = 60
Now arc PS is the intercepted arc for ∠PRS.
Therefore, m∠PRS = 60/2 = 30
I used the fact that an inscribed angle has a measure 1/2 the measure of the intercepted arc several times. Also, I used the fact that the acute angles of a right triangle are complementary. And, finally I used the fact that an inscribed angle in a semicircle is a right angle.
I hope this helped.
Answer:
P = 64 cm
Step-by-step explanation:
<C = 150° - 90° = 60°
AB = AC
∠B = ∠C = 60° } => ΔABC = EQUILATERAL TRIANGLE =>
=> BC = 10 cm
P = 10cm + 8cm + 6cm + 10cm + 6cm + 6cm + 8cm + 10cm
P = 3x10cm + 2x8cm + 3x6cm
P = 30cm + 16cm + 18cm
P = 64 cm
For this problem you are going to want to take 25 and find the square root. You get 5
This is the length of one side
Since it is a square and squares have 4 sides you will have to multiply 4 by 5
This gives you 20
Your answer is:
20m
