A sphere. It is a perfectly circular and even shape, so all cross sections would be circles.
Write the left side of the given expression as N/D, where
N = sinA - sin3A + sin5A - sin7A
D = cosA - cos3A - cos5A + cos7A
Therefore we want to show that N/D = cot2A.
We shall use these identities:
sin x - sin y = 2cos((x+y)/2)*sin((x-y)/2)
cos x - cos y = -2sin((x+y)/2)*sin((x-y)2)
N = -(sin7A - sinA) + sin5A - sin3A
= -2cos4A*sin3A + 2cos4A*sinA
= 2cos4A(sinA - sin3A)
= 2cos4A*2cos(2A)sin(-A)
= -4cos4A*cos2A*sinA
D = cos7A + cosA - (cos5A + cos3A)
= 2cos4A*cos3A - 2cos4A*cosA
= 2cos4A(cos3A - cosA)
= 2cos4A*(-2)sin2A*sinA
= -4cos4A*sin2A*sinA
Therefore
N/D = [-4cos4A*cos2A*sinA]/[-4cos4A*sin2A*sinA]
= cos2A/sin2A
= cot2A
This verifies the identity.
Answer:
I used desmos and plotted all three circles. The point that hey all intersected was (-1, 3).
Step-by-step explanation:
Xy = -109i
We could find the value of i by substitute the algebraic form of x and y to the equation above
xy = -109i
(10 - 3i)(3 - 10i) = -109i
(10)(3) -3i(3) + 10(-10i) - 3i(-10i) = -109i
30 - 9i - 100i -30i² = -109i
multiply both side by -1
-30 + 9i + 100i + 30i² = 109i
30i² + 9i + 100i - 109i - 30 = 0
30i² - 30 = 0
30i² = 30
i² = 1
i = -1 or i = 1
Then find the value of x and y if i = -1
If i = -1, therefore
x = 10 - 3(-1)
x = 10 + 3
x = 13
y = (3 - 10i)
y = 3 - 10(-1)
y = 3 + 10
y = 13
x/y = 13/13 = 1
Then find the value of x and y if i = 1
x = 10 - 3(1)
x = 10 - 3
x = 7
y = (3 - 10i)
y = 3 - 10(1)
y = 3 - 10
y = -7
x/y = 7/-7 = -1
The value of x/y is either 1 or -1
The value of constant of variation "k" is 
<em><u>Solution:</u></em>
Given that the direct variation is:
y = kx ----- eqn 1
Where "k" is the constant of variation
Given that the point is (5, 8)
<em><u>To find the value of "k" , substitute (x, y) = (5, 8) in eqn 1</u></em>

Thus the value of constant of variation "k" is 