You know that if you are multiplying a number by 5, and multiplying the same number by 6, 6 is going to be bigger
400x=22,000
22,000/400=x
55=x Hopefully this is right, and hopefully this helps
Answer:
(a) x = -2y
(c) 3x - 2y = 0
Step-by-step explanation:
You can tell if an equation is a direct variation equation if it can be written in the format y = kx.
Note that there is no addition and subtraction in this equation.
Let's put these equations in the form y = kx.
(a) x = -2y
- y = x/-2 → y = -1/2x
- This is equivalent to multiplying x by -1/2, so this is an example of direct variation.
(b) x + 2y = 12
- 2y = 12 - x
- y = 6 - 1/2x
- This is not in the form y = kx since we are adding 6 to -1/2x. Therefore, this is <u>NOT</u> an example of direct variation.
(c) 3x - 2y = 0
- -2y = -3x
- y = 3/2x
- This follows the format of y = kx, so it is an example of direct variation.
(d) 5x² + y = 0
- y = -5x²
- This is not in the form of y = kx, so it is <u>NOT</u> an example of direct variation.
(e) y = 0.3x + 1.6
- 1.6 is being added to 0.3x, so it is <u>NOT</u> an example of direct variation.
(f) y - 2 = x
- y = x + 2
- 2 is being added to x, so it is <u>NOT</u> an example of direct variation.
The following equations are examples of direct variation:
make the denominators the same so you cancompare
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!
