Answer:
45 minutes
Step-by-step explanation:
At 30 mph for 1/4 hour, Peter has a 7.5 mile head start. After he leaves, Mitchell closes that gap at the rate of 40-30 = 10 miles per hour. It will take him ...
t = d/s
t = (7.5 mi)/(10 mi/h) = 0.75 h
to catch Peter.
Mitchell will catch Peter in 45 minutes.
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<em>Alternate Solution</em>
Another way to look at it is that Mitchell's 10 mph advantage is 1/3 of Peter's speed, so it will take 1/(1/3) = 3 times the period of Peter's head start:
3 × 15 minutes = 45 minutes . . . for Mitchell to catch Peter
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You can write equations involving time and distance and see where the distances traveled become the same. You need to be careful choosing the time reference, since you're concerned with Mitchell's travel time. I personally prefer to work "head start" problems by considering the differences in time and speed, as above. This is where you end up using the equations approach, anyway.
Answer:
idk this stuff yet
Step-by-step explanation:
The question gives the relationship to be
We will interpolate to solve the question.
First Question: 147 mi
Cross multiplying,
Answer:
C = $2.40
n = 6
Step-by-step explanation:
For Noelle,
Equation that represents the monthly cost 'C',
C = 0.2n + 1.20
Here, n = number of checks written in a month
For Micah,
Monthly cost for writing checks 'C' = $1.20
Number of checks 'n' = 3
Since, Cost of writing checks ∝ Number of checks written
C' ∝ n
C' = kn
k = 
Here, k = proportionality constant
For C' = 1.2 and n = 3
k = 
k = 0.4
Equation will be,
C' = 0.4n
For any month C = C'
Therefore, 0.2n + 1.20 = 0.4n
0.4n - 0.2n = 1.20
0.2n = 1.20
n = 6
Number of checks written by Noelle and Micah = 6
For n = 6,
C = 0.2(6) + 1.20
C = $2.40
Cost of writing checks = $2.40
