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julsineya [31]
3 years ago
9

Annabelle's password is a 3-digit code that uses the numbers 0, 1, and 2. How many possible outcomes are there for Annabelle's p

assword if the digits can repeat?
27
9
21
3
Mathematics
2 answers:
jekas [21]3 years ago
7 0

Answer: The answer is 27

Step-by-step explanation:

Anit [1.1K]3 years ago
5 0

Answer:

You have 8 choices for the third digit (digits 0 - 9 excluding the numbers you chose for the first and second digit). So by the multiplication principle (MP), there are 9*9*8 ways to do this. So there are 648 numbers that are 3 digit numbers and have no repeated digits.

Step-by-step explanation:

You might be interested in
Lf f(x) = 5x, what is f^-1(x)?
Nonamiya [84]

Answer: f^{-1}(x)=\frac{x}{5}

Step-by-step explanation:

By definition the domain of an inverse function f^-1(x) is the range of f(x) and the range of the inverse function is equal to the domain of the principal function f(x).

If you have a function f(x)=5x, then to find the inverse function, follow these steps:

1. Make y=f(x)

 f(x)=y=5x

 y=5x

2. Solve for the variable "x":

x=\frac{y}{5}

3. Exchange the variable "x" with the variable "y":

y=\frac{x}{5}

4. Exchange "y" withf^{-1}(x). Then the inverse function is:

f^{-1}(x)=\frac{x}{5}

7 0
3 years ago
What are the pattern conbination between these number 18 48 88 128 178 38 68
valentina_108 [34]
18+30=48
48+40=88
88+40=128
128+50=178

Thats all i found so far in the pattern
30,40,40,50
then it drops
by 140 then adds by 30..

I dont know how else to explain it
7 0
3 years ago
Solve the absolute value equation and enter the solution in set notation.<br><br> |2x−3|−7=0
ziro4ka [17]

Answer:

The solutions are −5 and 1 .

4 0
3 years ago
Prove the theorem (AB )^T= B^T. A^T
Lisa [10]

Answer:

(AB)^T = B^T.A^T  (Proved)

Step-by-step explanation:

Given  (AB )^T= B^T. A^T;

To prove this expression, we need to apply multiplication law, power law and division law of indices respectively, as shown below.

(AB)^T = B^T.A^T\\\\Start, from \ Right \ hand \ side\\\\B^T.A^T = \frac{B^T.A^T}{A^T}.\frac{B^T.A^T}{B^T} (multiply \ through) \\\\                = \frac{A^{2T}.B^{2T}}{A^T.B^T} \\\\=\frac{(AB)^{2T}}{(AB)^T} \ \ (factor \ out \ the power)\\\\= (AB)^{2T-T}  \ (apply \ division \ law \ of \ indices; \ \frac{x^a}{x^b} = x^{a-b})\\\\= (AB)^T \ (Proved)

3 0
3 years ago
Plz awnser :) but explain i actually wanna know how to do this. long division, 1300÷15=?
chubhunter [2.5K]
(Divide) 1,300÷15=86.6666666667
6 0
4 years ago
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