Answer:
9n-3
Step-by-step explanation:
since 9n has a variable, u can't add or subtract 9n with 3
but if you were multiplying: 9n(3) then it would be 27n
:)
Answer:
a=22.6549
Step-by-step explanation:
Calculated based on 2 given angles and 1 given side.
∠C = 180° - A - B = 0.99484 rad = 57°
a = c·sin(A)/sin(C) = 22.6549
b = c·sin(B)/sin(C) = 12.33874
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!
Answer The line two origins are 6 and 5.1
Step-by-step explanation:
Hope this helps ^^
Answer:
The solutions are linearly independent because the Wronskian is not equal to 0 for all x.
The value of the Wronskian is 
Step-by-step explanation:
We can calculate the Wronskian using the fundamental solutions that we are provided and their corresponding the derivatives, since the Wroskian is defined as the following determinant.

Thus replacing the functions of the exercise we get:

Working with the determinant we get

Thus we have found that the Wronskian is not 0, so the solutions are linearly independent.