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)
<u>the correct question is</u>
The denarius was a unit of currency in ancient rome. Suppose it costs the roman government 10 denarii per day to support 4 legionaries and 4 archers. It only costs 5 denarii per day to support 2 legionaries and 2 archers. Use a system of linear equations in two variables. Can we solve for a unique cost for each soldier?
Let
x-------> the cost to support a legionary per day
y-------> the cost to support an archer per day
we know that
4x+4y=10 ---------> equation 1
2x+2y=5 ---------> equation 2
If you multiply equation 1 by 2
2*(2x+2y)=2*5-----------> 4x+4y=10
so
equation 1 and equation 2 are the same
The system has infinite solutions-------> Is a consistent dependent system
therefore
<u>the answer is</u>
We cannot solve for a unique cost for each soldier, because there are infinite solutions.
Ok so we are using the Pythagorean theorem so were gonna do A^2 + B^2 = C^2
It don't say which leg we have so you can use either a or b.
Which would get us 14^2 + B^2 = 50^2
14*14 =196 , 50 *50 =2500
Then we have to subtract the two.
So 2500 - 196 = 2,304
Then you take the take the square root of that answer and you have your other leg.
The square root of 2,304 is 48
So the missing leg = 48
Answer:
The probability that a part was manufactured on machine A is 0.3
Step-by-step explanation:
Consider the provided information.
It is given that Half of a set of parts are manufactured by machine A and half by machine B.
P(A)=0.5
Let d represents the probability that part is defective.
Ten percent of all the parts are defective.
P(d) = 0.10
Six percent of the parts manufactured on machine A are defective.
P(d|A)=0.06
Now we need to find the probability that a part was manufactured on machine A, and given that the part is defective
:
Hence, the probability that a part was manufactured on machine A is 0.3
Answer:
z = 189/44
Step-by-step explanation:
The "varies jointly" relationship can be expressed by ...
y = kxz
We can find k from the given values.
40 = k(10)(9)
40/90 = k = 4/9 . . . divide by the coefficient of k
Now we want to find z for given values of x and y. That can be found from ...
y = (4/9)xz
9y/(4x) = z . . . . . multiply by 9/(4x)
Filling in the new numbers, we have ...
z = 9·105/(4·55)
z = 4 13/44 = 189/44 ≈ 4.2954...(repeating 54)
