If 5t = 45 and tk = 1, what is the value of k ?

RockerChick

New member
If 5t = 45 and tk = 1, what is the value of k ?
I know that t = 9 obviously, but i can't seem to pick which answer choice fits...can you help me?
I would think if it equals 1 that k should be 1....since anything times 1 = 1...help!

(a) 1/45

(b) 1/9

(c) 1/5

(d) 5

(e) 9

Subhotosh Khan

Super Moderator
Staff member
Re: 5t =...

RockerChick said:
If 5t = 45 and tk = 1, what is the value of k ?
I know that t = 9 obviously, but i can't seem to pick which answer choice fits...can you help me?
I would think if it equals 1 that k should be 1....since anything times 1 = 1...help!

(a) 1/45

(b) 1/9

(c) 1/5

(d) 5

(e) 9
Here we go again...

Please show us your work - so that we do nnot waste time repeating what you have already done (and not showing).

RockerChick

New member
Re: 5t =...

For this one this is all i've done, I know t = 9
But I cant figure out which answer choice fits for the other unknown

Subhotosh Khan

Super Moderator
Staff member
Re: 5t =...

RockerChick said:
For this one this is all i've done, I know t = 9
But I cant figure out which answer choice fits for the other unknown
The same way you found t=9 from 5t = 45 --- by dividing.

Now we have tk = 1 ---> 9k = 1 ---> k = ???

stapel

Super Moderator
Staff member
RockerChick said:
For this one this is all i've done, I know t = 9
But I cant figure out which answer choice fits for the other unknown
So you have "9k = 1", but you don't know where to go from there. That tells us that you don't know how to solve one-step linear equations; this is useful information.

Since we cannot provide the hours of instruction you need, please study some of the many great lessons available online:

. . . . .Google results for "solving linear equations"

Please study at least two lessons, and then attempt the last step of this exercise. If you still (even after plug-n-chug) cannot determine the answer, please reply showing all of your work and reasoning. Thank you!

Eliz.

Loren

Senior Member
Strange. I just posted an answer to this question from a person using the same/similar name on another site.