Help... Math is hard........?

Question

S=L(1-r), Solve for R

S=LW+WH+LH, Solve for W

Find 3 Consecutive numbers whose sum is 126.








I don't have anyone to help me... My parent's never got to this level of math, I'm the oldest kid, and none of my friends are answering their phones.... Any help at all would be appreciated.

Answer 1

To solve for R, you can manipulate both sides of the equation in the same way to isolate the R.

S=L(1 - R)...divide both sides by L
S/L = L(1-R)/L...cancel the L's on the right hand side
S/L = 1 - R...subtract ones from both sides
S/L - 1 = 1 - 1 - R...eliminate the 1s on the right hand side
S/L - 1 = -R...multiply both sides by -1 to make the R positive
-1 (S/L - 1) = R...simplify the left hand side
-S/L + 1 = R

For the second one, same process:
S = LW + WH + LH...subtract LH from both sides so all the parts with W are on one side of the 'equals', and all the parts without are on the other
S - LH = LW + WH...factor the W out of the terms on the right
S - LH = W(L + H)...divide both sides by L+H to isolate the W
(S - LH)/(L + H) = W

For the three consecutive numbers, think of it like this. Let A be the smallest of the numbers. The next one will be A+1, and the third will be A+1+1, or A+2, right?

So:
A + (A + 1) + (A + 2) = 126...add the left hand side together
3A + 3 = 126...subtract 3 from both sides
3A = 123...divide both sides by 3
A = 41. That means the numbers are 41, 42, and 43.

Answer 2

don't worry, it's not too bad. u just have to learn how to isolate a variable.

1) S =L(1-R)
S= L -LR
S-L = -LR
(S-L)/-L = R

2) S=LW+WH+LH, Solve for W
S-LH = LW + WH
S-LH = W(L+H)
(S-LH)/(L+H) = W


3) sum of 3 consec #'s that equal 126

let x be first #

x + (x+1) + (x+2) = 126
3x + 3 = 126
3x = 123
x = 41

check: 41 +42+43 = 126~~~! :)

Answer 3

the secret of math is to study the textbook examples and to do the homework on the right of the sections.The extra you do a particular concept in math the further you'll understand it.a good textbook will then prepare it to on a daily basis existence in a situation. once you prepare your self contained in the study branch, you'd be taken aback at how a lot you realize. i'm fifty 3 years previous and characteristic been faraway from college for a lengthy time period and am taking a school algebra classification the position I have an A ordinary. you should concentration!!!

Answer 4

well in order to solve for the variable R you have to get it by itself. the first step would be to use the distributive property and distribute the L to 1 and -r. after doing this it would look like
S=L-LR
the L doesn't have a one because it is understood as being one already. then you would want to subtract L from the right side and you would have
S-L=LR
to get R by its self simply divide by L and you have
S-L=R and of course the S-L is over L since you just divided it. IM me for more help, math is hard but it can also be fun and rewarding.

Answer 5

41+42+43=126

r=(-s/L) +1

Answer 6

let the4 no.s be x,x+1,x+2
x+x+1+x+2=126

find x then other no..s got it...

and s/l=1-r
r=1-s/l


and s-lh=w(l+h)

w=(s-lh)/l+h.


o.k.got it..tata