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[deleted user]
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Sep 02, 2015 01:54PM
Algebra One
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I was invited to this right on time! I need help with a few problems. The question is: "Solve each equation for the indicated variable."
6. W = Fd, for d.
7. P = W/t, for W.
8. P = W/t, for t.
9. ak - r = on, for k.
6. W = Fd, for d.
7. P = W/t, for W.
8. P = W/t, for t.
9. ak - r = on, for k.
I need work also or I will get points taken off.
Elizabeth wrote: "6 would actually be D=W/F"
Right, sorry.
Right, sorry.
6 is just divide by F on both sides. The f's on the right cancel leaving you with D
♡ ƝιcσℓєAℓєχιѕ ♡ wrote: "I need work also or I will get points taken off."
Okay so the general idea is to isolate the variable. So for 6, you have to get D by itself and to do that, you need to divide both sides by F.
Okay so the general idea is to isolate the variable. So for 6, you have to get D by itself and to do that, you need to divide both sides by F.
7 multiply by T on both sides8. First multiply both by t giving you Pt=W, then divide by P. t=W/P
9.first add over r. ak=on+r then divide it all by a. k= (on+r)/a
I don't get how you did the work for 9?
Add the R to the ON, then you have AK=R+ON. Divide both sides by the A to get R by itself
Ok thanks! I think I get it now.
Ok, I actually need more help.
"In baseball, the equation E = 9R/I gives a pitcher's earned run average E, where R is the number of earned runs the player allowed and I is the number of innings pitched."
10b. Last season, a pitcher had an earned run average of 2.80 and allowed 70 earned runs. How many innings did the pitcher pitch last season?
"In baseball, the equation E = 9R/I gives a pitcher's earned run average E, where R is the number of earned runs the player allowed and I is the number of innings pitched."
10b. Last season, a pitcher had an earned run average of 2.80 and allowed 70 earned runs. How many innings did the pitcher pitch last season?
2.80=9(70)/I. Multiply by I to get 2.80I=630. I=225.
Oh wait I just got how you got 70.
It's the R, because the problem said he allowed 70 earned runs.
Yea, I don't know why I didn't know that. xD
We learned how to convert repeating decimals into fractions last year in pre-algebra, but I never understood what to do if not all the digits repeat.
As in, I know 0.56 (with 5 and 6 repeating) would really be 56/99, but what if the five terminated? :/
As in, I know 0.56 (with 5 and 6 repeating) would really be 56/99, but what if the five terminated? :/
If you mean 0.6 it would be 6/10 and you simplify it to 3/5.
If you mean 0.6 it would be 6/10 and you simplify it to 3/5.
The five would only terminate if the calculator rounded up or something like that, anything /99 is infinite.
Sorry, I meant if the five repeats and the six terminates, I was tired when I wrote that. :)
Oh, if it's .5555555555556, it's usually the calculator rounding up. :)
Thanks (I appreciate your help), but how do I convert that to a fraction? :/
Standard form would be in the y= mx+b form I'm pretty sure. In this case first subtract over the x then divide by -2.
I thought slope intercept was standard form.
I may be wrong, it's been awhile since I've taken algebra. I may be a math minor now but I don't really remember what the difference is.
Oh okay! Then yeah, you should be right. Sorry for the confusion.
