9514 1404 393
Answer:
Step-by-step explanation:
If we let x represent one of the numbers, then you want ...
x(58 -x) = 201.25
x² -58x +201.25 = 0 . . . . . put in standard form
(x² -58x +29²) +201.25 -29² = 0 . . . . complete the square
(x -29)² -639.75 = 0
x -29 = ±√639.75 . . . . . add 639.75 and take the square root
x = 29 ±√639.75 . . . . . . add 29
The two numbers of interest are ...
29 +√639.75 ≈ 54.2933
29 -√639.75 ≈ 3.7067
Ummmmnm...... the answer is 4a^8 b^3
I solved this like.....
12a^3 b^5 divide by 3 x a^5 b^-2
Calculate the quotient
4a^3 b^5 a^5 b^-2
Calculate the product
4a^8 b^3
And yea that’s how I got my answer :)
LHS ⇒ RHS:
Identities:
[1] cos(2A) = 2cos²(A) - 1 = 1 - 2sin²(A)
[2] sin(2A) = 2sin(A)cos(A)
[3] sin(A + B) = sin(A)cos(B) + cos(A)sin(B)
[4] cos(A + B) = cos(A)cos(B) - sin(A)sin(B)
cos(x) - cos(x + 2Θ)
= cos(x) - (cos(x)cos(2Θ) - sin(x)sin(2Θ)) [4]
= cos(x) - cos(x)(1 - 2sin²(Θ)) + sin(x)(2sin(Θ)cos(Θ)) [1] [2]
= cos(x) - cos(x) + 2sin²(Θ)cos(x) + 2sin(Θ)sin(x)cos(Θ)
= 2sin²(Θ)cos(x) + 2sin(Θ)sin(x)cos(Θ)
= 2sin(Θ)(sin(Θ)cos(x) + sin(x)cos(Θ))
= 2sin(Θ)sin(x + Θ)