Algebra 1 question?! Please help
Favorites|Homepage
Subscriptions | sitemap
HOME > > Algebra 1 question?! Please help

Algebra 1 question?! Please help

[From: ] [author: ] [Date: 12-12-15] [Hit: ]
10 points for best answer! Thanks!:)-You haveF = (9/5)C + 32You want to solve for C; that just means get C on one side of the equation and everything else on the other.You can do whatever you want to one side of the equation as long as you do the exact same thing on the other.F = (9/5)C + 32F - 32 = (9/5)C + 32 - 32 (subtract 32 from both sides)F - 32 = (9/5)C (the +32 and -32 on the right cancel).(F - 32) * (5/9) = (9/5C) * 5/9 (multiply both sides by 5/9)(F - 32) * (5/9) = C (the *9/5 and *5/9 on the right multiply to 1)So when F is 5:(F - 32) * (5/9) = C(5 - 32) * (5/9) = C(-27) * (5/9) = C-135/9 = C-15 = CDone :)-So this means get C by itself.......
The formula for converting degrees Celsius (C) to degrees Fahrenheit (F) is F= 9/5C + 32. Solve the formula for degrees Celsius (C). Then find the temperature in degrees Celsius (C) when the temperature in degrees Fahrenheit (F) is 5.

10 points for best answer! Thanks!:)

-
You have
F = (9/5)C + 32

You want to "solve" for C; that just means get C on one side of the equation and everything else on the other. You can do whatever you want to one side of the equation as long as you do the exact same thing on the other.

F = (9/5)C + 32
F - 32 = (9/5)C + 32 - 32 (subtract 32 from both sides)
F - 32 = (9/5)C (the +32 and -32 on the right cancel).
(F - 32) * (5/9) = (9/5C) * 5/9 (multiply both sides by 5/9)
(F - 32) * (5/9) = C (the *9/5 and *5/9 on the right multiply to 1)

So when F is 5:
(F - 32) * (5/9) = C
(5 - 32) * (5/9) = C
(-27) * (5/9) = C
-135/9 = C
-15 = C

Done :)

-
So this means get C by itself. To get C by itself you have to get rid of all terms on the same side; since 32 is a seperate term, subtract 32 from both sides to get F-32=(9/5)C. Now you need to get rid of the (9/5) If you're not comfortable with algebra yet, do it in two parts by multiplying each term by 5 (or equivalently, multiplying each side by 5, but when you multiply the left-side by 5 you must distribute the 5 to each term) to get 5F-(5*32)=9C, both sides by 9 to get (5F-(5*32))/9=C; but you can break up the numerator of the left side to be (5F/9)-((5*32)/9)=C. The reason I wrote it the second way is so you could see that this can be written, then, as (5/9)F-(5/9)*32=C; this will be good evidence of the more direct way. Remember when we got F-32=(9/5)C? We could have gotten C by itself all in one step by dividing both sides by 9/5, or another way of saying this would be multiplying both sides by 5/9. If, for example, we have (1/2)X=1, we could divide both sides by 1/2 and get ((1/2)X)/(1/2)=(1/(1/2)), but (1/2)/(1/2)=1, so we get X=(1/(1/2)). But remember that dividing by a fraction means flipping the fraction in the denominator and then multiplying, so 1/(1/2)=1*(2/1)=2. So if we divided both sides by 1/2 in (1/2)X=1 we get X=2. Another way to think about this, though, is to multiply by the "reciprocal," which is just a fancy way of saying multiply by whatever the fraction is when it's flipped. So in the (1/2)X=1 case, the reciprocal of (1/2)=(2/1)=2, so 2*(1/2)*X=1*2 so X=2, the same thing.
12
keywords: help,Algebra,Please,question,Algebra 1 question?! Please help
New
Hot
© 2008-2010 http://www.science-mathematics.com . Program by zplan cms. Theme by wukong .