The third term of the expansion is 6a^2b^2
<h3>How to determine the third term of the
expansion?</h3>
The binomial term is given as
(a - b)^4
The r-th term of the expansion is calculated using
r-th term = C(n, r - 1) * x^(n - r + 1) * y^(r - 1)
So, we have
3rd term = C(4, 3 - 1) * (a)^(4 - 3 + 1) * (-b)^(3-1)
Evaluate the sum and the difference
3rd term = C(4, 2) * (a)^2 * (-b)^2
Evaluate the exponents
3rd term = C(4, 2) * a^2b^2
Evaluate the combination expression
3rd term = 6 * a^2b^2
Evaluate the product
3rd term = 6a^2b^2
Hence, the third term of the expansion is 6a^2b^2
Read more about binomial expansion at
brainly.com/question/13602562
#SPJ1
1/40 is a fraction... did u make a typo?
Your answer is going to be 2304
The solutions appear to be
{π/2, 2π/3, 4π/3}.
_____
Replacing sin(2x) with 2sin(x)cos(x), you have
2sin(x)cos(x) +sin(x) -2cos(x) -1 = 0
sin(x)(2cos(x) +1) -(2cos(x) +1) = 0 . . . . factor by grouping
(sin(x) -1)(2cos(x) +1) = 0
This has solutions
sin(x) = 1
x = π/2and
2cos(x) = -1
cos(x) = -1/2
x = {2π/3, 4π/3}
4. vertical angles are congruent