as you already know, to get the inverse of any expression, we start off by doing a quick switcheroo on the variables, and then solve for "y".
![\bf \textit{Logarithm Cancellation Rules} \\\\ log_a a^x = x\qquad \qquad \stackrel{\textit{we'll use this one}}{a^{log_a x}=x} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \stackrel{f(x)}{y}=\log_2(x+1)\implies \stackrel{\textit{quick switcheroo}}{\underline{x}=\log_2(\underline{y}+1)}\implies 2^x=2^{\log_2({y}+1)} \\\\\\ 2^x=y+1\implies 2^x-1=\stackrel{f^{-1}(x)}{y} \\\\[-0.35em] ~\dotfill\\\\ 2^2-1=f^{-1}(2)\implies 3=f^{-1}(2)](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7BLogarithm%20Cancellation%20Rules%7D%20%5C%5C%5C%5C%20log_a%20a%5Ex%20%3D%20x%5Cqquad%20%5Cqquad%20%5Cstackrel%7B%5Ctextit%7Bwe%27ll%20use%20this%20one%7D%7D%7Ba%5E%7Blog_a%20x%7D%3Dx%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20%5Cstackrel%7Bf%28x%29%7D%7By%7D%3D%5Clog_2%28x%2B1%29%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Bquick%20switcheroo%7D%7D%7B%5Cunderline%7Bx%7D%3D%5Clog_2%28%5Cunderline%7By%7D%2B1%29%7D%5Cimplies%202%5Ex%3D2%5E%7B%5Clog_2%28%7By%7D%2B1%29%7D%20%5C%5C%5C%5C%5C%5C%202%5Ex%3Dy%2B1%5Cimplies%202%5Ex-1%3D%5Cstackrel%7Bf%5E%7B-1%7D%28x%29%7D%7By%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%202%5E2-1%3Df%5E%7B-1%7D%282%29%5Cimplies%203%3Df%5E%7B-1%7D%282%29)
5n - 1 < 24 is the inequality in question. Add 1 to both sides, obtaining:
5n < 25. Divide both sides by 5, obtaining n < 5 (answer)
For this case we have:
10 scented markers: Column 1, row 3.
6 are permanent markers: Column 1, row 1
Half of the unscented markers are erasable: (1/2) * 30 = 15 Column 2, row 2.
Therefore, the selected table corresponds to the attached image.
Answer:
See attached image.
Answer:
v
Step-by-step explanation:
1. Any number above 13 works. Why? Because 20-7=13, and to be greater than 20, you must add a number larger than 13.
Examples: 14+7 > 20, 30+7 > 20, 100+7 > 20
2. Any number below 25/3 (which is also 8.3 with a repeating 3) works. Why? Because 25/3=8.3 with a repeating 3, and to remain less than 25, you must multiply by a number less than 8.3 with a repeating 3.
Examples: 3(8) < 25, 3(5) < 25, 3(0) < 25
3. 4 buses. 1 bus will hold 60 students, 2 will hold 120, 3 will hold 180, and 4 will hold 240. The question is trying to trick you into putting now 3.3333333333... buses because that's what 200/60 is, but there is no such thing as a third of a bus. So you need at least 4 buses. (There will be an extra 40 spaces for passengers on the 4th bus, but that is okay.)
To find this answer I did 200/60 and got 3.3 with a repeating 3. You must round to the higher whole number. Rounding down to 3 buses leaves you with 20 students without a bus.
4. 19 boxes. 18 boxes will only hold 288 candies. The question is trying to trick you into putting down 18.75 boxes because that's what 300/16 is, but there is no such thing as 75% of a box. So you need at least 19 boxes. (There will be an extra 4 spaces for candies in the 19th box, but that is okay.)
To find this answer I did 300/16 and got 18.75. You must round to the higher whole <span>number. Rounding down to 18 boxes leaves you with 12 candies without a box.</span>
