Answer:
Step-by-step explanation:
(qx^2-5qx-3)÷(x+2) gives remainder 14q-3
Equating remanders 14q-3=4
14q=7
q=1/2
a)
FH = 102, FG = 5x + 9. HG = 9x - 5 <em>Given</em>
FG + HG = FH <em>Segment addition postulate</em>
5x + 9 + 9x - 5 = 102 <em>Substitution</em>
14x + 4 = 102 <em>Simplify (added like terms)</em>
14x = 98 <em>Subtraction property of equality</em>
x = 7 <em>Division property of equality</em>
b) FG = 5x + 9 = 5(7) + 9 = 35 + 9 = 44
c) HG = 9x - 5 = 9(7) - 5 = 63 - 5 = 58
Answers: a = 7, b = 44, c = 58
Answer:
5333 i believe
Step-by-step explanation:
Answer:
Step-by-step explanation:
First find the first radius
c=88cm
=2*22/7*r=88cm
=r=14cm
22/7 (pie as common)(R1^2+R2^2=637cm)
14^2+R2^2=637cm
196+R2^2=637cm
R2^2=637-196=441
R2=√441
R2=21
first radius =14cm
second radius =21cm
22 because -9 and 9 squared is 81 subtract that from 103
