C. 523.33
the formula is 4/3*3.14*r^3
R being radius (:
Missing information:
How fast is the temperature experienced by the particle changing in degrees Celsius per meter at the point
Answer:
Step-by-step explanation:
Given
Express the given point P as a unit tangent vector:
Next, find the gradient of P and T using: 
Where
So: the gradient becomes:
![\triangle T = [(sin \sqrt 3)i + (cos \sqrt 3)j] * [\frac{\sqrt 3}{2}i - \frac{1}{2}j]](https://tex.z-dn.net/?f=%5Ctriangle%20T%20%3D%20%5B%28sin%20%5Csqrt%203%29i%20%2B%20%28cos%20%5Csqrt%203%29j%5D%20%2A%20%20%5B%5Cfrac%7B%5Csqrt%203%7D%7B2%7Di%20-%20%5Cfrac%7B1%7D%7B2%7Dj%5D)
By vector multiplication, we have:
Hence, the rate is:
Answer:
-20c^3
Step-by-step explanation:
((−5c)(d−4))((2cd2)2)
=
−20c3d4
/d4
=-20c^3
One possible solution is
f(x) = x^4
g(x) = x-3
Since
f(x) = x^4
f(g(x)) = ( g(x) )^4
f(g(x)) = ( x-3 )^4
=================================================
Another possible solution could be
f(x) = x^2
g(x) = (x-3)^2
Because
f(x) = x^2
f(g(x)) = ( g(x) )^2
f(g(x)) = ( (x-3)^2 )^2
f(g(x)) = (x-3)^(2*2)
f(g(x)) = (x-3)^4
Answer:
y= 8(x-5)² + 6
Step-by-step explanation:
Parabolas have two equation forms, namely; the standard and vertex form.
In the vertex form, y = a(x - h)² + k, the variables h and k are the coordinates of the parabola's vertex.
In this case; a=8, h =5, and k=6
Therefore;
The vertex equation will be
y= 8(x-5)² + 6
