solve D=ABC for A A=
A=D/BC
thank you very much for your help
You're welcome, do you understand how to come to that answer?
no, but would I write it as, A= D over BC?
Yes, it is \[A= \frac{ D }{ BC }\] You get this because in order to isolate A, you have to divide D by BC. It'd be like if you had 25=5x, to get the value for X, you'd divide 25 by X, getting\[x=\frac{ 25 }{ 5 }\] which is equal to 5, therefore X=5. Hope this helps.
ty it did, Could i bother you for help on another question please?
Sure, I can try. What's the question?
a tugboat goes 140 miles upstream in 10 hours. the return trip downstream takes 5 hours. Find the speed of the tugboat without a current and the speed of the current. the speed of the tugboat is ___ mph, and the speed of the current is ____ mph.
simplify the answer please, if you can help
So I'm assuming that without a current means when the boat is going downstream, right?
yes correct
Ok. Well, when the boat is going upstream is has an average speed of 14 miles an hour. When it returns, it's going 140 miles in 5 hours, so 28 miles an hour. (To get those numbers, you simply divide the distance traveled by the time it took to get there) I'm GUESSING that the speed of the current would be the difference between the two, equaling a speed of 14 mph for the current as well.
how confident are you?
Actually that's incorrect, sorry. I just looked it up and I found a way to solve: http://answers.yahoo.com/question/index?qid=20120412090646AAlcO6o Let me know if you still don't understand and I'll explain it more than they did.
I am lost on how to do it
Ok. You need to set up an equation. t=tugboat speed c=current speed Since t+c=14 and t-c=28, you can combine those two to equal 2t=42 Divide that by 2 to solve for t and you get the tugboat's speed as 21 mph. If you put that with the first equation you'll get 28+c=14 Solve for c and you'll get 7mph.
so the speed of the tugboat is 21mph and the speed of the current is 7 mph?
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