Answer:
yes she will have enough, they will have 17 dollars left
Step-by-step explanation:
84 times 5 is 420 so she will have enough
Step-by-step explanation:
Given that,
Amy jogs 1/3 of a mile in 1/15 of an hour and john takes 1/30 of an hour to jog 1/5 of a mile.
Speed of Amy,

Speed of John,

John will jog farther in one hour as he is moving with faster speed. The speed of John is 1 mph more than Amy.
Perimeter divided by base??
Answer:

Step-by-step explanation:
We are given the expression to be simplified:

Let us take common a term with a power of 5 from the numerator and the denominator of the given expression.
We know that:

Let us use it to solve the powers of 5 in the given expression.
we can write:


The given expression becomes:

Taking common
from the numerator and
Taking common
from the denominator

The answer is:
