Radius, r = 3
The equation of a sphere entered at the origin in cartesian coordinates is
x^2 + y^2 + z^2 = r^2
That in spherical coordinates is:
x = rcos(theta)*sin(phi)
y= r sin(theta)*sin(phi)
z = rcos(phi)
where you can make u = r cos(phi) to obtain the parametrical equations
x = √[r^2 - u^2] cos(theta)
y = √[r^2 - u^2] sin (theta)
z = u
where theta goes from 0 to 2π and u goes from -r to r.
In our case r = 3, so the parametrical equations are:
Answer:
x = √[9 - u^2] cos(theta)
y = √[9 - u^2] sin (theta)
z = u
Answer:Hope This Helps ☺️
Step-by-step explanation:
She is not correct because she did not substitute the same number in both expressions in Step 1
Step-by-step explanation:
CASE 1: substitute 1 for x to both sides of the equations
L.H.S
-(4x-5)+2(x-3)
-(4 (1) - 5)+ 2(1-3) = - (-1) + 2(-2) = 1 - 4 = -3
R.H.S
-2x - 5
-2(1) - 5 = -2-5 = -7
Hence for x= 1
-(4x-5)+2(x-3) ≠ -2x -5
Because -3 ≠ -7
CASE 2: substitute -1 for x to both sides of the equations
L.H.S
-(4x-5)+2(x-3)
-(4 (-1) - 5)+ 2(-1-3) = - (-9) + 2(-4) = 9 - 8 = 1
R.H.S
-2x - 5
-2(-1) - 5 = 2-5 = -3
Hence for x= -1
-(4x-5)+2(x-3) ≠ -2x -5
Because 1 ≠ -3
Answer:
She is not correct because she did not substitute the same number in both expressions in Step 1
2b: If x is 2.1, then one side of the rectangle is 2.1, and another is 2.1*5=10.5. Thus, the perimeter is 2*(2.1+10.5)=2*12.6=25.2.
3: One side of the square with side-length three will not be on the outside, so we have 3*3=9 inches perimeter from the square of side-length 3. The square of side-length 6 has 3 from the top side missing from the outer perimeter, because it coincides with a side of the square of side-length three. This square contributes 6*4-3=33 inches. The total perimeter is 33+9=42 inches.
