HELP with these math problems (square roots)?

Question

1 what is the square root of a^2+b^2???
2 If 50 =square root of 3x and y = 2 x, find y

Mult choice for #2
sqrt3/25
100sqrt3
100/sqrt2
(100sqrt3)/3
25sqrt3

the sqrt isn't over the x in #2

Answer 1

1. square root of (a^2 + b^2) = sqroot (a^2 + b^2) depends upon numerical values of "a" and "b".

2 If 50 = square root of 3x and y = 2 x, find y.

On squaring both sides: 50 = sqroot(3x) --> 2500 = 3x

--> x = 2500/3............(A)

Now; y = 2x = 2*2500/3 = 5000/3 ≈ 1666.66, using (A).

Answer 2

1. No answer. A common mistake is to think the square root and the squares on the a and b both cancel to give you, a + b, but that is an incorrect use of cancelling squares/square roots. Only if it were sqrt[(a + b)^2] would it equal a + b.

2. sqrt(3)x = 50 --- divide both sides by sqrt(3)
=> x = 50/sqrt(3) --- now substitute this into y = 2x
=> y = 2*[50/sqrt(3)] = 100/sqrt(3) --- now multiply the top and bottom by the sqrt(3)
=> y = [100/sqrt(3)]*[sqrt(3)/sqrt(3)] = 100sqrt(3)/3

Hope this helps.

Answer 3

c^2...also known as the Hypotenuse.

50*50=2500
2500=3x

y is 2/3's of 2500

Remember that a square root (of provided) is a number that when multilpied with itself results in the original amount.

You can complete the math from there. :D

Answer 4

2. 50 = sqrt(3x)
2500 = 3x (square both sides)
x=2500/3
y=2x
y=2(2500/3)=5000/3
y=5000/3

Answer 5

5000/3.
50^2=3X=2500
y=2*2500/3
=5000/3

Answer 6

well...
for the first one, you can't really simplify it...
for the second,
sqrt(3x)=50
3x=2500
2x=5000/3