Answer:
the answer is x = 4
Step-by-step explanation:
............
Answer:
10 quarters = $2.50
10 nickels = $0.50
that leaves $0.20 for other coins (dimes / pennies)
Step-by-step explanation:
First, suppose she has only quarters and nickels and no other coins. Then if C is the identical number of coins of each type, then 5C + 25C = 320, so 30C = 320 and 3C = 32, but there is no integer solution to this. So she must have at least one other type of coin.
Assume she has only quarters, nickels, and dimes. Then if D is the number of dimes, 5C + 25C + 10D = 320, which means 30C + 10D = 320, or 3C + D = 32. The smallest D can be is 2, leaving 3C = 30 and thus C = 10. So in this scenario she would have 10 quarters, 10 nickels, and two dimes to make $2.50 + $0.50 + $0.20 = $3.20.
This has to be the highest number, because if she had 11 quarters and 11 nickels, that alone would add up to 11(0.25) + 11(0.05) = $3.30, which would already be too much.
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)
Answer:
Aziz's money was $30
Osman's money was $15
Step-by-step explanation:
Let
x-----> Aziz's money at first
y-----> Osman's money
we know that
------> equation A
-----> equation B
substitute equation A in equation B and solve for y
Find the value of x
therefore
At first
Aziz's money was $30
Osman's money was $15
The answer would be A i hope this helped
