Jean has a car, Last year, cost of running her car £4200 and £0.16. for every kilmetre the car travelled.
1 answer:
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Answer:
7a. D)
7b. $78
Step-by-step explanation:
7a: We're given the information that there's a daily fee of $50 and $0.35 per mile driven, which is m. We can write "$0.35 per mile driven" as 0.35m. Then, the total amount of money it costs represented as an expression is 50 + 0.35m, thus, making it answer D).
7b: In this problem, we have to take our previous expression 50 + 0.35m, plug in the value of m, and simplify.
Thus, the answer is $78.
Have a lovely rest of your day/night, and good luck with your assignments! ♡
So
if you assume that the month has 30 days and that the library opens at midnight, then
24 hours in a day
5 pm=12+5=17 hours
on wednessday=17-2=15 hours
wednessday=1/7 of week
so we find 1/7 of 30 which is 30/7=4 and 2/7
then subtract that from 30
30-4 and 2/7=25 and 5/7
ok so then we have
25 and 5/7 days is 17 hours and
4 and 2/7 days is 15 hours
so just multipy them and add
25 and 5/7 times 17=437.143 hours
4 and 2/7 days times 15 =64.2857
add
437.143+64.2856=501.429
so aprox 501.429
the real equation is![[(\frac{1}{7})(n)(15)]+[ (\frac{6}{7}) (n)(17)]=hours](https://tex.z-dn.net/?f=%5B%28%5Cfrac%7B1%7D%7B7%7D%29%28n%29%2815%29%5D%2B%5B%20%28%5Cfrac%7B6%7D%7B7%7D%29%20%28n%29%2817%29%5D%3Dhours)
where n represents the number of days in the month
apros 501.429
if you assume that the month has 30 days and that the library opens at midnight, then
24 hours in a day
5 pm=12+5=17 hours
on wednessday=17-2=15 hours
wednessday=1/7 of week
so we find 1/7 of 30 which is 30/7=4 and 2/7
then subtract that from 30
30-4 and 2/7=25 and 5/7
ok so then we have
25 and 5/7 days is 17 hours and
4 and 2/7 days is 15 hours
so just multipy them and add
25 and 5/7 times 17=437.143 hours
4 and 2/7 days times 15 =64.2857
add
437.143+64.2856=501.429
so aprox 501.429
the real equation is
apros 501.429