(-5, 15)
This is because y = 11 is a flat line that is 4 away from the point. Thus, the new point must also be 4 away from the line.
Given
consider substituting 
to get a proper quadratic equation,
Solve for 
; we can factorize to get
Solve for 
:
The first equation has no real solution, since 
for all non-zero . Proceeding with the second equation, we get
If we want to find all complex solutions, we take 
so that the first equation above would have led us to
Answer:
x = 10 cm, y = 5 cm gives a minimum area of 300 cm^2.
Step-by-step explanation:
V= x^2y = 500
Surface area A = x^2 + 4xy.
From the first equation y = 500/x^2
So substituting for y in the equation for the surface area:
A = x^2 + 4x * 500/x^2
A = x^2 + 2000/x
Finding the derivative:
dA/dx = 2x - 2000x^-2
dA/dx = 2x - 2000/x^2
This = 0 for a minimum/maximum value of A, so
2x - 2000/x^2 = 0
2x^3 - 2000 = 0
x^3 = 2000/ 2 = 1000
x = 10
Second derivative is 2 + 4000/x^3
when x = 10 this is positive so x = 10 gives a minimum value of A.
So y = 500/x^2
= 500/100
= 5.
1st term = n
2nd term = n+1
3rd term = n+2
4th term = n+3
The fifth term in the sequence would be n + 4.
For the fifth term in the sequence a = 3(n+4) - 2.
This should get you started.
