Ask your own question, for FREE!
Mathematics 14 Online
OpenStudy (thats_kyy):

HELP ME PLEASE I NEED HELP... A cab charges $1.50 for the flat fee and $0.50 for each mile. Write and solve an inequality to determine how many miles Sharon can travel if she has $25 to spend. $0.50 + $1.50x ≤ $25; x ≤ 16 miles $0.50 + $1.50x ≥ $25; x ≥ 16 miles $1.50 + $0.50x ≤ $25; x ≤ 47 miles $1.50 + $0.50x ≥ $25; x ≥ 47miles

OpenStudy (thats_kyy):

May someone please explain this to me to be honest @sammixboo @satellite73

OpenStudy (thats_kyy):

@Nnesha

OpenStudy (thats_kyy):

@Ashleyisakitty @Jaynator495

OpenStudy (thats_kyy):

@pooja195 @poopsiedoodle

OpenStudy (thats_kyy):

@satellite73 @Nnesha

OpenStudy (thats_kyy):

@Zale101

OpenStudy (thats_kyy):

@misty1212

OpenStudy (anonymous):

It would be the third one because 25 is the max she can spend so whatever the other side is, it has to be less than 25. $1.50 is the base price, so you have $1.50 + $0.50x, which represents the base price plus an additional $0.50 for every mile

OpenStudy (thats_kyy):

@misty1212 @misty1212

OpenStudy (thats_kyy):

@SolomonZelman

OpenStudy (anonymous):

it is usually C lets check

OpenStudy (thats_kyy):

I just need understanding to be honest..

OpenStudy (thats_kyy):

ok

OpenStudy (anonymous):

the flat rate is 1.50 so that is the constant out front

OpenStudy (anonymous):

50 cents a mile so for x miles it cost .5x

OpenStudy (thats_kyy):

okay..

OpenStudy (anonymous):

and she can spend at most $25 so \[1.50+.5x\leq 25\]

OpenStudy (thats_kyy):

wait why is 1.50 the constant rate

OpenStudy (anonymous):

it is not a rate, it is the flat fee

OpenStudy (anonymous):

oh i called it a "flat rate" sorry

OpenStudy (anonymous):

like a flat rate envelope, but nvm the rate is 50 cents the flat fee is 1.50

OpenStudy (thats_kyy):

ok

OpenStudy (anonymous):

C again imagine

OpenStudy (thats_kyy):

ok

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!