Answer:

Step-by-step explanation:
So, the function, P(t), represents the number of cells after t hours.
This means that the derivative, P'(t), represents the instantaneous rate of change (in cells per hour) at a certain point t.
C)
So, we are given that the quadratic curve of the trend is the function:

To find the <em>instanteous</em> rate of growth at t=5 hours, we must first differentiate the function. So, differentiate with respect to t:
![\frac{d}{dt}[P(t)]=\frac{d}{dt}[6.10t^2-9.28t+16.43]](https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7Bdt%7D%5BP%28t%29%5D%3D%5Cfrac%7Bd%7D%7Bdt%7D%5B6.10t%5E2-9.28t%2B16.43%5D)
Expand:
![P'(t)=\frac{d}{dt}[6.10t^2]+\frac{d}{dt}[-9.28t]+\frac{d}{dt}[16.43]](https://tex.z-dn.net/?f=P%27%28t%29%3D%5Cfrac%7Bd%7D%7Bdt%7D%5B6.10t%5E2%5D%2B%5Cfrac%7Bd%7D%7Bdt%7D%5B-9.28t%5D%2B%5Cfrac%7Bd%7D%7Bdt%7D%5B16.43%5D)
Move the constant to the front using the constant multiple rule. The derivative of a constant is 0. So:
![P'(t)=6.10\frac{d}{dt}[t^2]-9.28\frac{d}{dt}[t]](https://tex.z-dn.net/?f=P%27%28t%29%3D6.10%5Cfrac%7Bd%7D%7Bdt%7D%5Bt%5E2%5D-9.28%5Cfrac%7Bd%7D%7Bdt%7D%5Bt%5D)
Differentiate. Use the power rule:

Simplify:

So, to find the instantaneous rate of growth at t=5, substitute 5 into our differentiated function:

Multiply:

Subtract:

This tells us that at <em>exactly</em> t=5, the rate of growth is 51.72 cells per hour.
And we're done!
For this case we have that by definition, the density is given by:

Where:
M: It is the mass
V: It is the volume
According to the data of the statement we have:

So, the mass is given by:

Thus, the mass of the brass ornament is
Answer:

I belive the answer is 1.
This is because out of 22 ribbon there are 10 which are bue so the probability of blue ribbons is 10/22 and probability of red ribbon is 12/22. So your question was the probability of choosing a red ribbon then a blu ribbon. Because of the word "then" , this means it is simply asking you to add their probablity so after adding 10/22 with 12/22 your answer will come to 22/22 which is equal to 1.
Like terms:
1.4c, -2c, -7.3c
11.4 has no like terms
Answer:
1. Josh's father: f Josh: f/2 - 2
2. f + (f/2 - 2) ≥ 49
3. 3/2 f - 2 ≥ 49
3/2 f ≥ 51
3f ≥ 102
f ≥ 34
4. 34