1. find what percentage of the trip Keith drove, then his wife must have driven the remaining percentage
2.
1.5J = (1/2)K
3J = K
divide by K
3J/K = 1
divide by 3
J/K = 1/3
3. with radius r, area = πr^2
reduce radius by 10% ---> new radius = .9r
new area = π(.9r)^2 = .81π
decrease = .19π
.19π/π = .19 or 19%
1) If Keith drove 259 miles of a 500-mile trip and his wife drove the rest of the way, what percent of the trip did his wife drive?
a) 25.9%
b) 48.2%
c) 49.8%
d) 51.8%
e) 74.1%
2) If 150% of J is equal to half of K, and K is not equal to 0, then J/K =
a) 1/5
b) 1/3
c) 1/2
d) 2
e) 3
3) The length of a radius of a circle is decreased by 10%. This causes the area to be decreased by
a) 19%
b) 20%
c) 21%
d) 25%
2 answers
asdwa
2 answers