Answer:
Step-by-step explanation:
I'm sure you want your functions to appear as perfectly formed as possible so that others can help you. f(x) = 4(2)x should be written with the " ^ " sign to denote exponentation: f(x) = 4(2)^x
f(b) - f(a)
The formula for "average rate of change" is a.r.c. = --------------
b - a
change in function value
This is equivalent to ---------------------------------------
change in x value
For Section A: x changes from 1 to 2 and the function changes from 4(2)^1 to 4(2)^2: 8 to 16. Thus, "change in function value" is 8 for a 1-unit change in x from 1 to 2. Thus, in this Section, the a.r.c. is:
8
------ = 8 units (Section A)
1
Section B: x changes from 3 to 4, a net change of 1 unit: f(x) changes from
4(2)^3 to 4(2)^4, or 32 to 256, a net change of 224 units. Thus, the a.r.c. is
224 units
----------------- = 224 units (Section B)
1 unit
The a.r.c for Section B is 28 times greater than the a.r.c. for Section A.
This change in outcome is so great because the function f(x) is an exponential function; as x increases in unit steps, the function increases much faster (we say "exponentially").
so, let's keep in mind that
so let's make a quick table of those solutions, say A, B, C solutions with x,y,z liters of acid, with an acidity of 0.25, 0.40 and 0.60 respectively.
we know she's using "z" liters and those are 3 times as much as "y" liters, so z = 3y.
![\bf \begin{cases} x+y+3y=78\\ x+4y=78\\[-0.5em] \hrulefill\\ 0.25x+0.4y+0.6(3y)=35.1\\ 0.25x+0.4y=1.8y=35.1\\ 0.25x+2.2y=35.1 \end{cases}\implies \begin{cases} x+4y=78\\\\ 0.25x+2.2y=35.1 \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ x+4y=78\implies \boxed{x}=78-4y \\\\\\ \stackrel{\textit{using substitution on the 2nd equation}}{0.25\left( \boxed{78-4y} \right)+2.2y=35.1}\implies 19.5-y+2.2y=35.1](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Bcases%7D%20x%2By%2B3y%3D78%5C%5C%20x%2B4y%3D78%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%200.25x%2B0.4y%2B0.6%283y%29%3D35.1%5C%5C%200.25x%2B0.4y%3D1.8y%3D35.1%5C%5C%200.25x%2B2.2y%3D35.1%20%5Cend%7Bcases%7D%5Cimplies%20%5Cbegin%7Bcases%7D%20x%2B4y%3D78%5C%5C%5C%5C%200.25x%2B2.2y%3D35.1%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20x%2B4y%3D78%5Cimplies%20%5Cboxed%7Bx%7D%3D78-4y%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Busing%20substitution%20on%20the%202nd%20equation%7D%7D%7B0.25%5Cleft%28%20%5Cboxed%7B78-4y%7D%20%5Cright%29%2B2.2y%3D35.1%7D%5Cimplies%2019.5-y%2B2.2y%3D35.1)
![\bf 1.2y=15.6\implies y=\cfrac{15.6}{1.2}\implies \blacktriangleright y=13 \blacktriangleleft \\\\\\ x=78-4y\implies x=78-4(13)\implies \blacktriangleright x=26 \blacktriangleleft \\\\\\ z=3y\implies z=3(13)\implies \blacktriangleright z=39 \blacktriangleleft \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill \stackrel{25\%}{26}\qquad \stackrel{40\%}{13}\qquad \stackrel{60\%}{39}~\hfill](https://tex.z-dn.net/?f=%5Cbf%201.2y%3D15.6%5Cimplies%20y%3D%5Ccfrac%7B15.6%7D%7B1.2%7D%5Cimplies%20%5Cblacktriangleright%20y%3D13%20%5Cblacktriangleleft%20%5C%5C%5C%5C%5C%5C%20x%3D78-4y%5Cimplies%20x%3D78-4%2813%29%5Cimplies%20%5Cblacktriangleright%20x%3D26%20%5Cblacktriangleleft%20%5C%5C%5C%5C%5C%5C%20z%3D3y%5Cimplies%20z%3D3%2813%29%5Cimplies%20%5Cblacktriangleright%20z%3D39%20%5Cblacktriangleleft%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20~%5Chfill%20%5Cstackrel%7B25%5C%25%7D%7B26%7D%5Cqquad%20%5Cstackrel%7B40%5C%25%7D%7B13%7D%5Cqquad%20%5Cstackrel%7B60%5C%25%7D%7B39%7D~%5Chfill)
Answer: The distance is 15
Step-by-step explanation:
To find the distance between L and Y find the difference in the x and y coordinates then square them and add them to find their square root.
-7-5= -12
0-9 = -9
12^2 + 9^2 = d^2 where d is the distance
144 + 81 = d^2
225 = d^2
d= 
d= 15
Can you put more information into this question
Answer:
The graph should be stretched rather than become narrower.
Step-by-step explanation:
To figure this out, just create some example points.
At x = 0, your y-value will always be 0. But if you were to plug in the value 1, you would get a y-value of 1 in y=x^2, but a value of 0.5 in y=0.5x^2. If you were to plug in a value of 2, you would get a value of 4 in y=x^2, but a value of 2 in y=0.5x^2.
If you continue this pattern for a few more points, then plot them, you will see that adding a coefficient of 0.5 simply stretches the graph
