Answer:
6:16 simplifies to 3:8
Step-by-step explanation:
7 + 6+ 1 + 2 = 16
Ratio of comedies to the total
6:16 simplifies to 3:8
Answer:
Maximum rate of change at the point (-1,2) = √17
Direction is the direction of the gradient
Step-by-step explanation:
The gradient of a function (scalar or vectorial ) is a vector in the direction of maximum rate of change then
f( x,y ) = x*2y + 2y
grad = δ/δx i + δ/δy j + δ/δz k
grad f(x,y) = [ δ/δx i , δ/δy] = [ 2y , x+2 ]
at the point ( -1 , 2 )
grad f(x,y) = [4 , 1]
| grad f(x,y) | = √ (4)² + (1)² = √17
Answer:
James is 37, Rob is 44, and Kate is 35
Step-by-step explanation:
Set up an equation where x is James' age:
Rob's age can be represented by x + 7, since he is 7 years older than James
Kate's age can be represented by x - 2, since she is 2 years younger than James
Add these together in the equation and set it equal to 116:
(x + 7) + (x - 2) + (x) = 116
Add like terms and solve for x:
(x + 7) + (x - 2) + (x) = 116
3x + 5 = 116
3x = 111
x = 37
So, James is 37.
Find Rob's age by adding 7 to this:
37 + 7
= 44
Find Kate's age by subtracting 2:
37 - 2
= 35
So, James is 37, Rob is 44, and Kate is 35
At the start, the tank contains
(0.25 lb/gal) * (100 gal) = 25 lb
of sugar. Let
be the amount of sugar in the tank at time
. Then
.
Sugar is added to the tank at a rate of <em>P</em> lb/min, and removed at a rate of

and so the amount of sugar in the tank changes at a net rate according to the separable differential equation,

Separate variables, integrate, and solve for <em>S</em>.







Use the initial value to solve for <em>C</em> :


The solution is being drained at a constant rate of 1 gal/min; there will be 5 gal of solution remaining after time

has passed. At this time, we want the tank to contain
(0.5 lb/gal) * (5 gal) = 2.5 lb
of sugar, so we pick <em>P</em> such that

At the museum cafe the answer is