Can someone please help me with these two Algebra problems?

Question

Here are the two if them

1. A = (1/2) (b+B)h ; solve for B

2. S = rT + (1/2)aT^2; solve for a

Please help me

Answer 1

Question 1
2A / h = b + B
B = (2 A / h) - b

Question 2
2 ( S - r T ) = a T ²
a = 2 ( S - r T ) / T ²

Answer 2

To solve these, move the factors to the other side of the equal sign by doing the opposite operation. Such as:

if added - then subtract
if subtracted - then add
if divided - then multiply
if multiplied - then divide

Last, to move fractions over to the other side, just simply flip them upside down 1/2 = 2/1, and multiply on the other side of the equal sign! :)

1. A = (1/2) (b+B)h ; solve for B

Multiply both sides by (2/1)
2A = (b+B)h

Divide both sides by h
(2A) / h = b+B

Subtract b on both sides
[(2A) / h] - b = B


2. S = rT + (1/2)aT²; solve for a

Subtract rT on both sides
S - rT = (1/2)aT²

Multiply by (2/1) on both sides
2(S - rT) = aT²

Divide by T² on both sides
2(S - rT) / T² = a

Answer 3

one million) sparkling up via Factoring 4z²-9=0 it is what's usually a distinction of squares when you consider that the two words are suited squares: (2z)² = 4z², 3² = 9. You factorize it like this: on the 1st ingredient: sq. root of the 1st term plus sq. root of the 2nd, on the 2nd ingredient: sq. root of the 1st minus sq. root of the 2nd. (2z + 3)(2z - 3) = 0 Then, to sparkling up for z, via fact the multiplication equals 0, each and each ingredient could be equivalent to 0 and sparkling up the equation. 2z + 3 = 0 2z = -3 z = -3/2 ? FIRST answer 2z - 3 = 0 2z = 3 z = 3/2 ? 2nd answer ___________________________ discover the gap between the two given factors making use of the pythagorean theorem. (5,8) (-7, -2) the concept says c² = a² + b², the place a and b are the strains that style the ninety° perspective, and c is the hypotenuse. in case you plot those factors in a graph, de distinction between ordenates are the legs. a = 5 - (-7) = 5 + 7 = 12 b = 8 - (-2) = 8 + 2 = 10 c² = 12² + 10² c² = a hundred and forty four + one hundred c² = 244 c = ?244 c = 15.sixty two ? answer

Answer 4

1. A = (1/2)(b+B)h
First: divide both sides by "h" (when you move a term to the opposite side, always use the opposite sign).

A/h = [(1/2)(b+B)h]/h

*Cross cancel "like" terms.

A/h = (1/2)(b+B)

Sec: eliminate the fraction - multiply 2 with both sides.

2(A/h) = 2[(1/2)(b+B)]
2(A/h) = 1(b+B)
2(A/h) = b + B

Third: solve for "b" by isolating it on one side - subtract "b" from both sides.

2(A/h) - b = b -b + B
2(A/h) - b = B, or (2A)/h -b = B

Answer 5

Okay..you just need to isolate B using algebra:
A = (1/2) (b+B)h
distributing: you get
A = (bh + Bh)/2
cross multiplying:
2A = bh + Bh
transposing:
2a-bh = Bh
then dividing by h:
2a - bh
---------- = B = ans!
h

next:
S = rT + (1/2)aT^2
transposing rT;
S - rT = aT^2/2
cross multiplying;
2(S-rT) = aT^2
then divide by T^2
2(S-rT)
---------- = a = ans!!
T^2

hope this clears it up!

Answer 6

1. Multiply both sides by 2. Divide both sides by h. Then subtract b from both sides.

2. Subtract rT from both sides. Multiply both sides by 2. Divide both sides by T^2.

Answer 7

Hi,

divide and conquer :-)

A = (1/2)(b+B)h
2A = (b+B)h
2A/h = b + B
2A/h - b = B

S = rt + (1/2)aT^2
S - rt = (1/2)aT^2
2(S - rt) = aT^2
(2(S - rt))/T^2 = a

REgards,
Chas.

Answer 8

1. divide both sides by (1/2) h
2A/h = b + B
now subtract b from both sides
(2A/h) - b

2. subtract rT from both sides
S - rT = (1/2)aT^2
divide both sides by (1/2)T^2
2(S-rT)/T^2 = a
or (2S - 2rT)/ T^2
or 2S/T^2 - 2r/T

Answer 9

B = (2A/h) - b

a = (2(S-rT)/T^2)

Answer 10

oh nvm just noticed everyone else said exactly what i just did