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Mathematics 17 Online
OpenStudy (anonymous):

i will give medals for who ever helps me!!

OpenStudy (anonymous):

what do you need help with?

OpenStudy (anonymous):

Whats the question?

OpenStudy (anonymous):

this is the question Solve for t. d = rt

OpenStudy (anonymous):

d/r=t

OpenStudy (anonymous):

Since t is on the right-hand side of the equation, switch the sides so it is on the left-hand side of the equation. rt=d Divide each term in the equation by r. rt/r=d/r Simplify the equation. t=d/r

OpenStudy (anonymous):

okay what about another one?

OpenStudy (anonymous):

sure

OpenStudy (anonymous):

Solve for n. V=n/g

OpenStudy (anonymous):

Since n is on the right-hand side of the equation, switch the sides so it is on the left-hand side of the equation. \[\frac{ n }{ g }=v\] Multiply both sides of the equation by g. \[n=v \times(g)\] Multiply v by each term inside the parentheses. Multiply v by (g) to get v(g). \[n=v(g)\] Multiply v by the g inside the parentheses. \[n=v \times g\] Multiply v by g to get gv. \[n=gv\] ^^^ That's your answer!

OpenStudy (anonymous):

one more please(:

OpenStudy (anonymous):

Sure!

OpenStudy (anonymous):

Solve for t. c = rt + k

OpenStudy (anonymous):

Okay, give me a minute!

OpenStudy (anonymous):

okay and thank you very much

OpenStudy (anonymous):

Since t is on the right-hand side of the equation, switch the sides so it is on the left-hand side of the equation. \[rt+k=c\] Since k does not contain the variable to solve for, move it to the right-hand side of the equation by subtracting k from both sides. \[rt=-k+c\] Move all terms not containing t to the right-hand side of the equation. \[rt=c-k\] Divide each term in the equation by r. \[\frac{ rt }{ r }=\frac{ c }{ r }-\frac{ k }{ r }\] Simplify the equation. \[t=\frac{ c }{ r }-\frac{ k }{ r }\] That's it!

OpenStudy (anonymous):

thats no a answer

OpenStudy (anonymous):

What's your choices?

OpenStudy (anonymous):

A.t =c-k/r B.t = c+k/r C.t = c/r – k D.t = c-r/k

OpenStudy (anonymous):

It's A

OpenStudy (anonymous):

thanks

OpenStudy (anonymous):

They just combined. And you're very welcome! :)

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