PLEASE HELP, MEDAL AND FAN!!!!
ONE QUESTION!

- anonymous

PLEASE HELP, MEDAL AND FAN!!!!
ONE QUESTION!

- schrodinger

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- anonymous

A cab charges $1.75 for the first mile and $0.25 for each additional mile. Write and solve an inequality to determine how many miles Eddie can travel if he has $15 to spend.
$1.75 + $0.25x ≤ $15; x ≤ 53 miles
$1.75 + $0.25x ≥ $15; x ≥ 53 miles
$0.25 + $1.75x ≤ $15; x ≤ 8 miles
$0.25 + $1.75x ≥ $15; x ≥ 8 miles

- anonymous

halo

- anonymous

Yea

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## More answers

- anonymous

ok so does the initial charge of 1.75 depend on how many miles he drives?

- anonymous

no

- anonymous

well after the first mile, no it doesnt, so that will not change after the first mile, so you would not multiply the 1.75 by the number of miles right? the number of miles is x

- anonymous

so that gets rid of the last two options

- anonymous

you would multiply x by .25?

- anonymous

correct

- anonymous

so that leaves the first two optioins

- anonymous

so 1.75+.25x=?

- anonymous

ok so what does that represent there, what you just wrote it represents the total cost right?

- anonymous

yes?!

- anonymous

yes it does

- anonymous

so if Eddie has only 15 dollars to spend then the total cost cannot be greater than 15 correct?

- anonymous

otherwise he would not have enough money

- anonymous

yes

- anonymous

oh so its an inequality not an equation

- anonymous

correct, so the total cost must be less than or equal to 15, so which inequality is it?

- anonymous

B?

- anonymous

what does the less than or equal to sign look like?

- anonymous

<
----

- anonymous

correct, so it must be what then?

- anonymous

cost < or = to 15

- anonymous

but x, or the numbers of miles he drives, cant be more than 53, the miles it takes to spend all of his money

- anonymous

so a

- anonymous

correct, and when you solve the inequality for x i assume you will get x <= 53

- anonymous

could you help with another one?

- anonymous

close this one and post another one and i or someone else will help you

- anonymous

ok!

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