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Mathematics 17 Online
OpenStudy (anonymous):

?

OpenStudy (anonymous):

@Kainui

OpenStudy (anonymous):

@hazyskies @happyyanee4

OpenStudy (anonymous):

@HelpBlahBlahBlah @hhelpplzzzz @Squirrels @aaronq @kohai

OpenStudy (anonymous):

@One098 @Saeeddiscover

OpenStudy (anonymous):

@AuroraB

OpenStudy (anonymous):

Do you know how to do this @One098

OpenStudy (anonymous):

@kyley_bear @sunnycali @autuom @KenLJW

OpenStudy (anonymous):

@mathmath333

OpenStudy (anonymous):

@xo.minnie.xox

OpenStudy (one098):

A or C..

OpenStudy (anonymous):

can you explain it @One098 please

OpenStudy (one098):

Well i said A or C because the answer I got was 4.6, however I don't know how to get the other number.. Anyways, I got that by plugging it into the d. Then I subtracted 25t from one side to get it over to the other side. Now it reads as 130=15+25t. Then I subtracted 15 from both sides, then I divided both by 25.

OpenStudy (anonymous):

ok thank you

OpenStudy (anonymous):

does anyone know how to get the full answer

OpenStudy (mathmath333):

\(\large\tt \color{black}{15=|130-25t|}\) \(\large\tt \color{black}{15=\pm(130-25t)}\)

OpenStudy (mathmath333):

\(\large \tt \color{black}{\text{you will get two values of t}}\)

OpenStudy (anonymous):

how do you do the second problem @mathmath333

OpenStudy (mathmath333):

type it

OpenStudy (anonymous):

I don't know what the plus minus thing is @mathmath333

OpenStudy (one098):

it means it can be either

OpenStudy (anonymous):

so the answer is c @mathmath333

OpenStudy (mathmath333):

\(\large\tt \color{black}{15=\pm(130-25t)}\) \(\large \tt \color{black}{\text{there can be two possiblities}}\) \(\large\tt \color{black}{15=(130-25t) ~~and ~~15=-(130-25t) }\) \(\large\tt \color{black}{25t=130-15) ~~and~~ 15=-130+25t }\) \(\large\tt \color{black}{25t=115 ~~and~~ 25t=130+15 }\) \(\large\tt \color{black}{t=\dfrac{115}{25}~~ and~~ 25t=145}\) \(\large\tt \color{black}{t=\dfrac{115}{25} ~~and~~ t=\dfrac{145}{25}}\) \(\large\tt \color{black}{t=4.6 ~~and~~ t=5.8}\)

OpenStudy (anonymous):

ok that's c

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