Ok.. Math problems. Algebra 2?

Comments

• nathan-khan_6096ebf63e8e9

  May 2021

  1. P= 2L +2W

  from the right hand side, we can see that 2 is common to both L and W, therefore, we'll bring it out like this:

  P=2(L+W)

  if this parenthesis are solved for, we'll get back to the original equation. Now didvide both side by 2 to get:

  P/2=2(L+W)/2

  on the right hand side (RHS), 2 will cancel out one another to get:

  P/2=L+W

  now, if W moves to the other side (Left), by crossing over the = sign, it changes it sign from + to -. As in:

  (P/2)-W=L . Therefore,

  L= (P/2)-W. Solved

  2. C=2pi r

  all we need to do here is to find a way to let r be alone and that is by dividing both sides by 2pi:

  C/2pi= 2pi r/2pi

  where 2pi will cancel one another on the RHS to get

  C/2pi= r. Therefore,

  r= C/2pi. Solved.

  3. P=3r+2s.

  All we need to do is to find a way of making s be alone on one side and the rest on the other side. First step is by making 3r cross to the other side and this will change its sign form + to -. As in:

  P-3r=2s

  s is not alone yet, to finalize this, we'll have to divide both sides by 2:

  (P-3r)/2=2s/2

  with 2cancelling 2 on the RHS, we'll be left with:

  (P-3r)/2=s. Therefore,

  s= (P-3r)/2. Solved

  4. A=1+prt

  again, we'll look for a way to make r be alone on one side. Firstly, let 1 cross to the other side where it will change to -1:

  A-1=prt

  now, divide both sides by pt:

  (A-1)/pt= prt/pt

  with pt cancelling out on the RHS, we'll have:

  (A-1)/pt=r. Therefore,

  r= (A-1)/pt. Solved.

  5. C=5(F-32)/9

  can be re written as:

  C/1=5(F-32)/9

  firstly, we'll cross multiply, 5(F-32) will be multiplied by 1 = C will be multiplied by 9.:

  5(F-32)=9C

  now, open the bracket:

  5F-160=9C

  with -160 moving to the other side, it changes sign form - to +:

  5F=9C+160

  divide both sides by 5:

  5F/5=(9C+160)/5

  F=(9C+160)/5. Solved

  or F= (9C/5) + (160/5)

  F= (9C/5) +32. Solved.

  6. A= h/2(a+b)

  firstly, cross multiply:

  h= 2A(a+b). Solved

  please try to understand, it makes me remember when i was coming up in maths . Just try, maths is very interesting.
• sonia-hughes_6096ebf71f08d

  May 2021

  Tip: see, all of these problems require you isolate the unknown, e.g. L, on one side via using math operations. I will do the 1st one for you in detail so you may hopefully learn it and do the rest on your own:

  P=2L+2W ; solve for L

  - add - 2W to both sides of the equation

  P - 2W = 2L + 2W - 2W => the 2Ws equal to zero

  - so now you have

  P - 2W = 2L

  - now divide both sides by 2 to get rid of the coefficient of L-the number preceding it

  (P - 2W) / 2 = (2/2) L => 2/2 = 1 in math, you don't write the coefficient of 1

  - now divide each term in the parenthesis on the left by 2

  P/2 - (2W/2) = L

  P/2 - W = L

  or

  L = P/2 - W
• anonymous

  May 2021

  1.P=2L+2W ; solve for L

  2L=P-2W

  L=P/2-W

  2.C=2 pi r; solve for r

  r=C/2pi

  3.P= 3r+2s; solve for s

  2s=P-3r

  S=(P-3r)/2

  4.A=1+prt; solve for r

  prt=A-1

  r=(A-1)/pt

  5.C= 5(F-32)/ 9 ; solve for F

  9C=5(F-32)

  9C=5F-160

  5F=9C+160

  F=(9C+160)/5

  F=1.8C+32

  6.A= h/2(a+b); solve for h

  a)if A= h/2*(a+b);

  then

  h=2A/(a+b)

  b) if A= h/ (2(a+b)); then

  h=A*2(a+b)
• man-of-steel

  May 2021

  you do the opposite for each side. opposite of addition is subtraction and opposite of multiplication is division.