3^2*3^2 (2x+3)=3^2+4x+6=3^4x+8
3^3*3^x-2=3^x+1
4x+8=x+1
3x=-7
x=-7/3
See attachment although I gather this is not your question. I did the graph(s) for the function,
Answer:
416 is your answer!!!
Step-by-step explanation:
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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:
- 3a + 6
Step-by-step explanation:
subtract and add like items together like
- 4a + a = -3a
b-b= 0
and
9-3 = 6
so the answer is
-3a +6
