<span>f '(x) = 2(3x2<span> + 5) + 6x(2x - 1)</span></span><span>= 6x2<span> + 10 + 12x</span>2<span> - 6x</span></span><span>= 18x2<span> - 6x + 10</span></span>
Answer:
it's not 16/,13 ; )
Step-by-step explanation:
Answer:
y/x = 2.5
Step-by-step explanation:
y ∝ x
change ∝ to = by adding constant k or c
y = kx
Divide both sides by x
y/x = k, also k = y/x
Input the values
k = (7 1/2)/3. Change 7 1/2 to improper fraction to give 15/2
k = (15/2)/3
k =
÷ 3
Change the division sign to multiplication which will change 3 to its inverse(1/3)
k =
× 
k = 
Divide the numerator and denominator by 3 to get 5/2
k = 5/2
From y/x = k,input k into the eqn
y/x = 5/2
y/x = 2.5
Answer:
Let's suppose that one cubic foot of marble is x.
20x = 3400
But because the weight has a range of +-100, then 20x is between 3300 and 3500. Write it like this:
3300<20x<3500
Divide by 20:
165<x<175
The meaning of this is that one cubic foot of marble weight between 165 and 175 pounds
I assume the cone has equation
(i.e. the upper half of the infinite cone given by
). Take

The volume of the described region (call it
) is

The limits on
and
should be obvious. The lower limit on
is obtained by first determining the intersection of the cone and sphere lies in the cylinder
. The distance between the central axis of the cone and this intersection is 1. The sphere has radius
. Then
satisfies

(I've added a picture to better demonstrate this)
Computing the integral is trivial. We have
