I think multiple? try it..
We want to combine b(t) and t(h) to create a function that described the number of bacteria, b, as a function of time, h.
This function can be called b(t(h)), or b(h).
Consider the function b(t), or bacteria as a function of temperature.
b(t) = 20t² - 70t +300
This describes the number of bacteria, given the temperature.
t(h) described the temperature as a function of time, specifically hours after refrigeration:
t(h) = 2h + 3
Since the time h can tell us the temperature t, and the temperature t can tell us the # of bacteria b, we can create a function that tells us the number of bacteria, b, given hours following refrigeration, h.
To do this, we plug t(h) in for every t in the b(t) function:
b(t(h)) = 20 (2h+3)² - 70(2h+3) + 300
We can also call this function b(h), since we now can express b as a function of h.
Simplify the function:
b(h) = 20 (4h²+12h+9)-14h-210+300
b(h) = 80h² +100h +270
The Answer is B
Answer:
In this special case we know that 
and 
and for this case we can add all the values 12+18 +12+18=60 and that represent the perimeter
Step-by-step explanation:
For this case we know that the perimeter is given by:
In this special case we know that and and for this case we can add all the values 12+18 +12+18=60 and that represent the perimeter
I am assuming that you can only pick one answer per question.
Let's imagine there are two questions on the test. I would:
1) Consider the first question. How many possible ways could you answer it?
2) Consider the second question. How many ways can you answer that?
If you wrote out all the possibilities, how many combinations of answers would you get across the two questions?
<h2>9.</h2><h3>Given</h3>
<h3>Find</h3>
- linear approximation to the volume when the radius increases 0.4 cm
<h3>Solution</h3>
The equation for volume of a sphere is
... V = (4/3)π·r³
Differentiating gives
... dV = 4π·r²·dr
Filling in the given numbers gives
... change in volume ≈ 4π·(15 cm)²·(0.4 cm)
... = 360π cm³ ≈ 1130.97 cm³ . . . . . . volume of layer 4mm thick
<h2>11.</h2><h3>Given</h3>
- an x by x by 2x cuboid with surface area 129.6 cm²
- rate of change of x is 0.01 cm/s
<h3>Find</h3>
<h3>Solution</h3>
The area is that of two cubes of dimension x joined together. The area of each such cube is 6x², but the two joined faces don't count in the external surface area. Thus the area of the cuboid is 10x².
The volume of the cuboid is that of two cubes joined, so is 2x³. Then the rate of change of volume is
... dV/dt = (d/dt)(2x³) = 6x²·dx/dt
We know x² = A/10, where A is the area of the cuboid, so the rate of change of volume is ...
... dV/dt = (6/10)A·dx/dt = 0.6·(129.6 cm²)(0.01 cm/s)
... dV/dt = 0.7776 cm³/s
