4x^2-3x+4y^2+4z^2=0
here we shall proceed as follows:
x=ρcosθsinφ
y=ρsinθsinφ
z=ρcosφ
thus
4x^2-3x+4y^2+4z^2=
4(ρcosθsinφ)^2-3(ρcosθsinφ)+4(ρsinθsinφ)^2+4(ρcosφ)
but
ρ=1/4cosθsinφ
hence we shall have:
4x^2-3x+4y^2+4z^2
=1/4cosθsinθ(cosθ(4-3sinφ))+4sin^2(φ)
Answer:
y= - 1/6x -1
Step-by-step explanation:
Given:
Perpendicular line has slope of reciprocal opposite to the given: - 1/6, so the equation:
Finding b using the intersected point:
- -2= -1/6*6 +b
- b= -2 +1= -1
So the equation is:
Answer:
So you can cancel out the 16. See, the -16 is negative, and if we add +16 to it, then it cancels out and is 0. That way, you have x=1.
Step-by-step explanation:
-16x+x=-15
+16x +16
0+x=1
x=1
There are i<span>Infinitely many solutions </span>
9(n+3)=7n-3
1) distribute 9 into (n+3)
9n+27=7n-3
2) subtract 7n on both sides
2n+27=-3
3) subtract 27 on both sides
2n=-30
4) divide by 2
n=-15
