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:
Step-by-step explanation:
you have a special triangle 30-60-90
the ration in this traingle is x, x
, 2x, where 2x is the hypothenuse
the hypothenuse is 12
x=12/2=6
the other leg is 6=10.39
Solution: We know that e is numerical constant and is equal to 2.71828.
Therefore log( 6 x 2.71828) = 1.21
Answer:
202m
Step-by-step explanation:
Please see attached picture for full solution.
Let x meters be the height of the building.
Answer:
=2w2+9w−35
Step-by-step explanation:
(2w+−5)(w+7)
(2w)(w)+(2w)(7)+(−5)(w)+(−5)(7)
2w2+14w−5w−35
2w2+9w−35
