Hints:
(a)
find the radius r which corresponds to the distance between (-3,-2) and (0,-6). Then
(-3-a)^2+(-2-0)^2 = r^2
(b)
y=0 is horizontal (1)
y=2x (2) is a line through the origin sloping upwards
y=-0.5x+k (3) is a line sloping downwards with x-intercept: -0.5x+k=0, or x=2k.
You'll need to find the intercept between lines (2) and (3) and determine the height of the triangle.
(3)
This means that (x^2+1) divides P(x)-5x, or
P(x)-5x = (x^2+1)(x+3)=
=x^3+3*x^2+x+3
Therefore
P(x)=x^3+3*x^2+x+3 + 5x
=x^3+3*x^2+6x+3
=> A=-3
A circle of center (-3 , -2) passes through the points (0 , -6) and (a , 0). Find a.
The triangle bounded by the lines y = 0, y = 2x and y = -0.5x + k, with k positive, is equal to 80 square units. Find k.
When the polynomial P(x) = x3 + 3x2 -2Ax + 3, where A is a constant, is divided by x2 + 1 we get a remainder equal to -5x. Find A.
And would like to see work that can justify the answer. Thank you!
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