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:
-2, 8/3
Step-by-step explanation:
You can consider the area to be that of a trapezoid with parallel bases f(a) and f(4), and width (4-a). The area of that trapezoid is ...
A = (1/2)(f(a) +f(4))(4 -a)
= (1/2)((3a -1) +(3·4 -1))(4 -a)
= (1/2)(3a +10)(4 -a)
We want this area to be 12, so we can substitute that value for A and solve for "a".
12 = (1/2)(3a +10)(4 -a)
24 = (3a +10)(4 -a) = -3a² +2a +40
3a² -2a -16 = 0 . . . . . . subtract the right side
(3a -8)(a +2) = 0 . . . . . factor
Values of "a" that make these factors zero are ...
a = 8/3, a = -2
The values of "a" that make the area under the curve equal to 12 are -2 and 8/3.
_____
<em>Alternate solution</em>
The attachment shows a solution using the numerical integration function of a graphing calculator. The area under the curve of function f(x) on the interval [a, 4] is the integral of f(x) on that interval. Perhaps confusingly, we have called that area f(a). As we have seen above, the area is a quadratic function of "a". I find it convenient to use a calculator's functions to solve problems like this where possible.
Step by Step Solution
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STEP
1
:
Equation at the end of step 1
(((2y - 1)2) - 4 • (2y - 1)) + 4
STEP
2
:
Trying to factor by splitting the middle term
2.1 Factoring 4y2-12y+9
The first term is, 4y2 its coefficient is 4 .
The middle term is, -12y its coefficient is -12 .
The last term, "the constant", is +9
Step-1 : Multiply the coefficient of the first term by the constant 4 • 9 = 36
Step-2 : Find two factors of 36 whose sum equals the coefficient of the middle term, which is -12 .
-36 + -1 = -37
-18 + -2 = -20
-12 + -3 = -15
-9 + -4 = -13
-6 + -6 = -12 That's it
20 cans in kilograms = 336 grams * 20 cans / 1000 = 6.72 kg
Answer:
22% = 154
100%= ?
=100×154/22
= 700
Step-by-step explanation:
