A and b are legs
a^2+b^2=c^2
lets say
a>b
so
a=13+3b
c=14+3b
a^2+b^2=c^2
(13+3b)^2+b^2=(14+3b)^2
9b^2+78b+169+b^2=9b^2+84b+196
10b^2+78b+169=9b^2+84b+196
minus 9b^2 both sides
b^2+78b+169=84b+196
minus 84b both sides
b^2-6b+169=196
minus 196 both sides
b^2-6b-27=0
factor
(b+3)(b-9)=0
set to zero
b+3=0
b=-3, false, dimentions cannot be negative
b-9=0
b=9
shorter leg is 9
a=13+3b
a=13+3(9)
a=13+27
a=40
c=14+3b
c=14+27
c=41
side legnths are
9in, 40in, 41in
The answer is b. Brent completed 3 more questions than alexa did
Alright, since there are 5 numbers, and the mean (or average) is (sum)/(amount of numbers), we have (sum)/5=14. Multiplying both sides by 5, we have the sum being 80. The median of 10 means that in a, b, c, d, e, 10 has to be c and the numbers have to be in ascending order. A and b must be 10 or lower, while d and e must be 10 or higher. Putting some random numbers in, we can have 1, 1, 10, 15, and e. We left e there because the sum needs to be 80, and since 1+1+10+15=27, 80-27=53=e. This, however, would not work if e was less than 10 and we therefore would have needed to make some numbers lower to compensate for this. Our answer is therefore 1, 1, 10, 15, 53
Looks like you got it right :) Each package contains 8 chocolates, so to find 8 containers, just multiply by 8 to get 64, which would be H.
