Ok so student in cleveland walks 2.5 miles away from cleveland every hour from 8
so the distance, x is x=2.5t
the riding one rides 4 hours later, so t-4
11 mph
y=11(t-4)
A. x=2.5t
y=11(t-4)
when do they meet is when the distance when they are equal or when x=y
2.5t=x=y=11(t-4)
solve for t
2.5t=11(t-4)
2.5t=11t-44
-8.5t=-44
divide both sides by -8.5
t=5.17647
answer is 5.18 hours
A.
x=2.5t
y=11(t-4)
B. 5.18 hours
Answer:
6 and 8
Step-by-step explanation:
All you have to do is find the difference between the numbers in each factor pair and make sure the factors combine to give 48.
2 and 4 . . . . . have a difference of 2, but 2·4 ≠ 48
4 and 6 . . . . . have a difference of 2, but 4·6 ≠ 48
6 and 8 . . . . . have a difference of 2, and 6·8 = 48
24 and 2 . . . . do not have a difference of 2
Answer:
$1820
Step-by-step explanation:
35 x 52 = $1820
