Answer:
Numerator = 2(b^2+a^2) or equivalently 2b^2+2a^2
Denominator = (b+a)^2*(b-a), or equivalently b^3+ab^2-a^2b0-a^3
Step-by-step explanation:
Let
S = 2b/(b+a)^2 + 2a/(b^2-a^2) factor denominator
= 2b/(b+a)^2 + 2a/((b+a)(b-a)) factor denominators
= 1/(b+a) ( 2b/(b+a) + 2a/(b-a)) find common denominator
= 1/(b+a) ((2b*(b-a) + 2a*(b+a))/((b+a)(b-a)) expand
= 1/(b+a)(2b^2-2ab+2ab+2a^2)/((b+a)(b-a)) simplify & factor
= 2/(b+a)(b^2+a^2)/((b+a)(b-a)) simplify & rearrange
= 2(b^2+a^2)/((b+a)^2(b-a))
Numerator = 2(b^2+a^2) or equivalently 2b^2+2a^2
Denominator = (b+a)^2*(b-a), or equivalently b^3+ab^2-a^2b0-a^3
The closest one is 5 to the power of 1 over 3.
sq rt of 3 = 1.732
sq rt of 5 = 2.236
5^1/3 = 1.667
5^1/6 = 0.833
5^2/3 = 8.333
5^3/2 = 62.5
You basically plug in the x in the equation, for example if x = 4 and the equation is y = 2x + 5 you plug in 4 and the equation becomes y = 2(4) + 5. you then multiply and get y=8+5. from here u just add it and u get ur answer
It’s not exactly equal to one, but in many cases in math they ask for a rounded answer.
The distance between two points with the given coordinates in space is;
<u><em>Distance = 17 units</em></u>
We are given the coordinates;
(32, 12, 5)
(20, 3, 13)
- Formula for the distance between two points that have (x, y, z) coordinates is;
d = √((x₂ - x₁)² + (y₂ - y₁)² + (z₂ - z₁)²)
From the equation given to us, we can see that;
x₁ = 32
x₂ = 20
y₁ = 12
y₂ = 3
z₁ = 5
z₂ = 13
- Using the <em>formula for the distance</em>, we have;
d = √((20 - 32)² + (3 - 12)² + (13 - 5)²)
d = √(144 + 81 + 64)
d = √289
d = 17
Thus, the <em>distance</em> between the two points is 17 units
Read more at; brainly.com/question/20974053
