Answer:
a) 
b) 
c) 
Step-by-step explanation:
Given the curve
a) If the x-coordinate of P is 
, then the y-coordinate is 
so point P has coordinates 
If the x-coordinate of Q is 
, then the y-coordinate is 
so point Q has coordinates 
b) The gradient of the secant RQ is
c) If 
then the gradient 
<span>20.28% is the answer</span>
X(u, v) = (2(v - c) / (d - c) + 1)cos(pi * (u - a) / (2b - 2a))
y(u, v) = (2(v - c) / (d - c) + 1)sin(pi * (u - a) / (2b - 2a))
As
v ranges from c to d, 2(v - c) / (d - c) + 1 will range from 1 to 3,
which is the perfect range for the radius. As u ranges from a to b, pi *
(u - a) / (2b - 2a) will range from 0 to pi/2, which is the perfect
range for the angle. So, this maps the rectangle to R.
Four point six three zero.
If you meant:
(2ya)^4=16y^8 raise the left side to the power of four
16y^4*a^4=16y^8 divide both sides by 16y^4
a^4=y^4 take the quartic root of both sides
a=±y
