A: (x + 5i)^2
= (x + 5i)(x + 5i)
= (x)(x) + (x)(5i) + (5i)(x) + (5i)(5i)
= x^2 + 5ix + 5ix + 25i^2
= 25i^2 + 10ix + x^2
B: (x - 5i)^2
= (x + - 5i)(x + - 5i)
= (x)(x) + (x)(- 5i) + (- 5i)(x) + (- 5i)(- 5i)
= x^2 - 5ix - 5ix + 25i^2
= 25i^2 - 10ix + x^2
C: (x - 5i)(x + 5i)
= (x + - 5i)(x + 5i)
= (x)(x) + (x)(5i) + (- 5i)(x) + (- 5i)(5i)
= x^2 + 5ix - 5ix - 25i^2
= 25i^2 + x^2
D: (x + 10i)(x - 15i)
= (x + 10i)(x + - 15i)
= (x)(x) + (x)(- 15i) + (10i)(x) + (10i)(- 15i)
= x^2 - 15ix + 10ix - 150i^2
= - 150i^2 + 5ix + x^2
Hope that helps!!!
I beleieve the answer is false
Answer:
Follow the rule of BODMAS (brackets, of, division, multiplication, addition, subtraction)
Ans: 58
Step-by-step explanation:
42 + 4(6 - 2) Process --> 6 - 2 = 4
= 42 + 4 * 4 Process --> 4 * 4 = 16
= 42 + 16 Process --> 42 + 16 = 58
= 58
Hope this helps :)
The parametric equations for x and y describe a circle of radius 10 m, so the length of the base of the fence is the length of the circumference of a circle of radius 10 m. The formula for that circumference (C) is ...
... C = 2πr
... C = 2π·(10 m) = 20π m
The height as a function of angle (t) is found by substituting for x and y.
... h(t) = h(x(t), y(t)) = 4 + 0.01·((10cos(t))²-)10sin(t))²) = 4+cos(2t)
The average value of this over the range 0 ≤ t ≤ 2π is 4, since the cosine function has two full cycles in that range, and its average value over a cycle is zero.
Thus, the area of one side of the fence is that of a rectangle 20π m long and 4 m wide. That will be
... (20π m)·(4 m) = 80π m²
The amount of paint required to cover both sides of the fence is
... 2×(80π m²)×(1 L)/(10 m²) = 16π L ≈ 50.3 L
_____
You can work out the integral for area as a function of t. When you do, you will find it gives this same result.
