Answer:
c.i^ 3
Step-by-step explanation:
i^7/i x i^3
i ^ 7 / i ^ 4
i^ 7 - 4
i^ 3
No.
This is because the y value of 1 repeats itself, which never happens in an exponential function. This function is likely quadratic, which forms a parabola.
Absolute value (|2|) makes all the numbers in it positive.
|5-9|= -4 with absolute value -> 4
|5|-9: only 5 is in the absolute value:
|5|-9 -> 5-9=-4
number 1: 4
number 2: -4
Step-by-step explanation:
As far as I understand, they can use all 26 letters (alphabet) and all 10 digits (0, 1, ..., 9), and the password looks like this:
letter-letter-letter-letter-letter-letter-number-number (6 letters, 2 numbers).
▪︎ They can reuse the letters, so they can choose the 1st letter out of 26 possible letters, then they still can choose the 2nd letter out of 26 (it's ok if they have both A's for example, they can reuse them) and so on. So each time they can choose one letter out of 26. Thus, it's 26*26*26*26*26*26.
▪︎ The same thing with numbers. Each time they can choose one number out of 10, so it's 10*10.
▪︎ Therefore, the answer is 26*26*26*26*26*26*10*10 = 30,891,577,600. That's almost 31 billion passwords.
Answer:
C. The graph of f(x) is shifted 11 units up.
Step-by-step explanation:
Given
![g(x) = f(x) + 11](https://tex.z-dn.net/?f=g%28x%29%20%3D%20f%28x%29%20%2B%2011)
Required
Interpret g(x)
When a function is shifted up in b units, the rule is:
![(x,y) \to (x,y+b)](https://tex.z-dn.net/?f=%28x%2Cy%29%20%5Cto%20%28x%2Cy%2Bb%29)
Replace b with 11
![(x,y) \to (x,y+11)](https://tex.z-dn.net/?f=%28x%2Cy%29%20%5Cto%20%28x%2Cy%2B11%29)
Replace y with f(x)
![(x,f(x)) \to (x,f(x)+11)](https://tex.z-dn.net/?f=%28x%2Cf%28x%29%29%20%5Cto%20%28x%2Cf%28x%29%2B11%29)
So, we have:
![g(x) = f(x) + 11](https://tex.z-dn.net/?f=g%28x%29%20%3D%20f%28x%29%20%2B%2011)
Hence; f(x) is shifted 11 units up.