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:
Yes
Step-by-step explanation:
6((3×5)-5)
6(15-5)
6(10)
60
Hope this helps!! :)
