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
Answer:
<h2>

</h2>
Step-by-step explanation:
x - y² = -1 ⇒ - y² = - x - 1 ⇒ y² = x + 1 {x≥-1}
4x² + 9y² = 72
4x² + 9(x + 1)² = 72
4x² + 9(x² + 2x + 1) = 72
4x² + 9x² + 18x + 9 - 72 = 0
13x² + 18x - 63 = 0 ⇒ a = 13, b = 18, c = -63

Answer:
n(n+1)(n+5)/3
Step-by-step explanation:
there is no value, as we don't know n.
but we can create a summary formula/ function definition :
this is the sum for k = 1 to n of k×(k+3)
k×(k+3) = k² + 3k
so, the overall sum splits into the sum of k² for k=1 to n, and the sum of 3k for k=1 to n.
and the sum of 3k is 3 times the sum of k for k=1 to n.
Σk² for k=1 to n = [n(n+1)(2n+1)]/6
Σk for k=1 to n = n(n+1)/2
3×Σk for k=1 to n = 3×n(n+1)/2
so, we have a function formula
n(n+1)(2n+1)/6 + 3n(n+1)/2 = n(n+1)(2n+1)/6 + 9n(n+1)/6 =
= n(n+1)(2n+1+9)/6 = n(n+1)(2n+10)/6 = n(n+1)(n+5)/3
The bus stop method is just like division.
Ex
137÷5=
1.) 5 goes into one zero times so there is a zero on top.
2.) 5 goes into 13 two times, so after the zero there is a 2.
3.) 5 goes into 37 seven times so there is a 7 after the 2.
4.) There is a 2 left and five can't go into school therefore that is a remainder.
027R2
5/137
<u>- 10</u>
37
<u>-35</u>
2
Hoped I helped:D
Answer:
Step-by-step explanation:
Let
n ------> a number
we know that
The algebraic expression that represent the phrase"5 less than a number" is equal to subtract 5 from the number or a number minus 5
so