Answer:
Step-by-step explanation:
<u>Exponential Growth
</u>
The natural growth of some magnitudes can be modeled by the equation:
Where P is the actual amount of the magnitude, Po is its initial amount, r is the growth rate and t is the time.
The initial number of bacteria is Po=40 and it doubles (P=2Po) at t=20 min. With that point we can find the value of r:
Simplifying:
Solving for 1+r:
![1+r=\sqrt[20]{2}](https://tex.z-dn.net/?f=1%2Br%3D%5Csqrt%5B20%5D%7B2%7D)
The exponential function that models the situation is:
a vertical axis, I assume it means a vertical axis of symmetry, thus it'd be a vertical parabola, like the one in the picture below.
![\bf ~~~~~~\textit{parabola vertex form} \\\\ \begin{array}{llll} y=a(x- h)^2+ k\qquad \qquad \leftarrow vertical\\\\ x=a(y- k)^2+ h \end{array} \qquad\qquad vertex~~(\stackrel{}{ h},\stackrel{}{ k}) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \begin{cases} h=0\\ k=0 \end{cases}\implies y=a(x-0)^2+0 \\\\\\ \textit{we also know that } \begin{cases} x=-2\\ y=3 \end{cases}\implies 3=a(-2-0)^2+0\implies 3=4a \\\\\\ \cfrac{3}{4}=a~\hspace{10em}y=\cfrac{3}{4}(x-0)^2+0\implies \boxed{y=\cfrac{3}{4}x^2}](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~%5Ctextit%7Bparabola%20vertex%20form%7D%20%5C%5C%5C%5C%20%5Cbegin%7Barray%7D%7Bllll%7D%20y%3Da%28x-%20h%29%5E2%2B%20k%5Cqquad%20%5Cqquad%20%5Cleftarrow%20vertical%5C%5C%5C%5C%20x%3Da%28y-%20k%29%5E2%2B%20h%20%5Cend%7Barray%7D%20%5Cqquad%5Cqquad%20vertex~~%28%5Cstackrel%7B%7D%7B%20h%7D%2C%5Cstackrel%7B%7D%7B%20k%7D%29%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20%5Cbegin%7Bcases%7D%20h%3D0%5C%5C%20k%3D0%20%5Cend%7Bcases%7D%5Cimplies%20y%3Da%28x-0%29%5E2%2B0%20%5C%5C%5C%5C%5C%5C%20%5Ctextit%7Bwe%20also%20know%20that%20%7D%20%5Cbegin%7Bcases%7D%20x%3D-2%5C%5C%20y%3D3%20%5Cend%7Bcases%7D%5Cimplies%203%3Da%28-2-0%29%5E2%2B0%5Cimplies%203%3D4a%20%5C%5C%5C%5C%5C%5C%20%5Ccfrac%7B3%7D%7B4%7D%3Da~%5Chspace%7B10em%7Dy%3D%5Ccfrac%7B3%7D%7B4%7D%28x-0%29%5E2%2B0%5Cimplies%20%5Cboxed%7By%3D%5Ccfrac%7B3%7D%7B4%7Dx%5E2%7D)
Part A:
Given a square with sides 6 and x + 4. Also, given a rectangle with sides 2 and 3x + 4
The perimeter of the square is given by 4(x + 4) = 4x + 16
The area of the rectangle is given by 2(2) + 2(3x + 4) = 4 + 6x + 8 = 6x + 12
For the perimeters to be the same
4x + 16 = 6x + 12
4x - 6x = 12 - 16
-2x = -4
x = -4 / -2 = 2
The value of x that makes the <span>perimeters of the quadrilaterals the same is 2.
Part B:
The area of the square is given by
The area of the rectangle is given by 2(3x + 4) = 6x + 8
For the areas to be the same
Thus, there is no real value of x for which the area of the quadrilaterals will be the same.
</span>
Answer:
6647
Step-by-step explanation:
The computation of the advertising revenue for 15 years is shown below:
Given that
r(t) = -18t^2 + 720t - 103
where
t = number of years = 15
now put the value of t in the above equation
= -18(15)^2 + 720(15) - 103
= -4050 + 10,800 - 103
= 6647
Answer:
4 |x+7| divided by 3.
Step-by-step explanation:
This cannot be divided any further.
