4t=r
a=pir^2
sub 4t for r
a=pi(4t)^2
a=pi16t^2
a(t)=16pi(t^2)
A. a(t)=16pi(t^2)
B. sub 4 for t
a(4)=16pi4^2
a(4)=16pi16
a(4)=16*16*3.14
a(4)=803.84 square units
A. a(t)=16pi(t^2)
B. 803.84 square units
Answer:
5
Step-by-step explanation:
You can solve this by using the system of equations.
Jan - 4.95 = 2H + 3C
Wayne - 5.45 = 3H + 2C
Use elimination.
-3(2H + 3C = 4.95)
2(3H + 2C = 5.45)
Solve. And you'll get:
-6H + (-9C) = -14.85
6H + 4C = 10.9
Cross out -6H and 6H because they cancel out. And you're left with:
-9C = -14.85
4C = 10.9
Add -9C with 4C, and -14.85 with 10.9.
-5C = -3.95
Divide each side with -5.
C = $0.79
Now to figure out what H is, just substitute the C in one of the equations with 0.79.
5.45 = 3H + 2(0.79)
5.45 = 3H + 1.58
-1.58 -1.58
3.87 = 3H
3.87/3 = 3/3(H)
1.29 = H
Finished!
Not to good with math but the answer may be B
