Answer:
<h2>2</h2>
Step-by-step explanation:
Given g(x)=5x-4 and g(x)=f^{-1}(x)-3
Substituting g(x) = 5x-4 into the second equation we have;
5x-4 = f^{-1}(x)-3
f^{-1}(x) = 5x-4+3
f^{-1}(x) = 5x-1
To get f(x), let us first make y to be equal f^{-1}(x)
y = 5x-1
expressing x in terms of y to get f(x), we have;
5x = y+1
x = y/5+1/5
replacing y with x, we will have;
y = x/5 + 1/5
F(x) = x/5 + 1/5
Comparing x/5 + 1/5 with ax+b, a = 1/5 and b = 1/5
5a + 5b = 5(1/5)+ 5(1/5)
5a+5b = 1+1
5a+5b = 2
Answer:
<u>So, the way a ratio works is </u><u>"for every 1 of this, we have 4 of this"</u><u> for example. (the ratio I just described would be 1:4.) </u>
<u />
I believe I've seen this question before. If the ratio of dolls to teddy bears is 9:3 then for every 9 dolls you have 3 teddy bears.
So, since we have 240 dolls and a 9:3 ratio to teddy bears, all we have to do is take how many dolls we have (240) and divide it by 3 to see how many teddy bears we have.
So:
240 ÷ 3 = 80
If your ratio is 9:3, then 80 is your answer.
<u>Hope this helps and have a nice day!</u>
<u></u>
Y - y1 = m(x - x1)
so equation
y + 9 = -2(x - 10)
answer is A
y + 9 = -2(x - 10)
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)
Here we will tell you what 329 is rounded to the nearest hundred and also show you what rules we used to get to the answer. First, 329 rounded to the nearest hundred is:
300
<span>Remember, we did not necessarily round up or down, but to the hundred that is nearest to 329.</span>