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:
x = 4 ± 
Step-by-step explanation:
Given
(x - 12)(x + 4) = 9 ← expand the left side using FOIL
x² - 8x - 48 = 9 ( ad 48 to both sides )
x² - 8x = 57
To complete the square
add ( half the coefficient of the x- term )² to both sides
x² + 2(- 4)x + 16 = 57 + 16
(x - 4)² = 73 ( take the square root of both sides )
x - 4 = ± ( add 4 to both sides )
x = 4 ±
-23.2 I think this is the answer
Answer:
No they are not intergers
Step-by-step explanation:
Hope this helps
The general form of the given equation is 2x+y-6 = 0.
<u>Step-by-step explanation</u>:
- The given linear equation is 2x+y=6.
- The general form of the equation is AX+BY+C=0.
where,
- A is the co-efficient of x.
- B is the co-efficient of y.
- C is the constant term.
<u>From the given equation 2x+y=6, it can be determined that</u> :
The co-efficient of x is 2. It is in the form AX = 2x. Thus, no change is needed.
The co-efficient of y is 2. It is in the form BY = 1y. Thus, no change is needed.
The constant term 6 should be replaced to the left side of the equation, since the right side of the equation must be 0 always.
While moving the constant term form one side of the equation to other side, the sign changes from +ve to -ve.
Therefore, the general form is given as 2x+y-6 = 0.
