Answer:
3. reflective property
4.AE=CE
5.<AED=<CED 5.SSS congruence
Answer:
If a certain cone with a height of 9 inches has volume V = 3πx2 + 42πx + 147π, what is the cone’s radius r in terms of x?
Step-by-step explanation:
V = 3πx2 + 42πx + 147π
V=3π(x2 + 14x +49)
9.42(x2 + 14x +49)
9.42(x2 + 14x +14) -14 + 49= 0
9.42(x + 7)^2 + 35= 0
9.42(9.42(x + 7)^2 = - 35)9.42
(x + 7)^2 = - 35/9.42)
√(x + 7)^2=√- 35/9.42
x + 7 = - 1.927
x= - 1.927 - 7
x= - 8.927
V = 3π(- 8.927)^2 + 42π(- 8.927) + 147π
V=750.69 - 1177.29 + 461.58
<u>V=34.98</u>
h= 9 inches
V = 13πr2h
34.98 = 13(3.14) (r^2) (h)
34.98 = 40.82 (r^2) 9
34.98 = 367.38 r^2
34.98/ 367.38 = 367.38 r^2/ 367.38
0.095= r^2
We know that , in a circle radius perpendicular to chord will bisect the chord.
OM=18, so OQ=QM=18/2=9.
Given QU=8
from figure OQU is a right angled triangle , so OU^2=OQ^2 + QU^2
OU^2 = 9*9 + 8*8 = 81+72=153;
OU=sqrt(153) = 12.37 =13(approx);
From given statements of congruent NT and OU will also be congruent or identical. So, NT=OU=13
Answer: Adrian will need to fix 25 bikes to reach a break even point.
