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Hitman42 [59]
3 years ago
8

Abby must create a three-character password for her computer the is two different, upper-case letters followed by one digit. if

she only is willing to use the letters in her first name and a digit in the current year(2011), how many possible passwords can she create? one possible password to include s YB1
Mathematics
1 answer:
velikii [3]3 years ago
5 0
The number of different permutations  of 2 letters from the name Abby 
= 3P2 =  3! / 3-2! = 3*2*1 = 6
Number of different  digits in 2011 = 3

Number of possible passwords = 6*3 = 18 (answer).
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Plz help and thanks so much for it
kirza4 [7]
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          First, let's convert the variables to real numbers:   11+16+(16-11)+(11-16)
             Now, let's solve that equation.                                    11+16+5+(-5)
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7 0
4 years ago
Solve the system using elimination.<br> 2x + 18y = -9<br> 4x + 18y = -27
Elan Coil [88]

First, let's cancel out the x by multiplying 2x + 18y = -9 by -2.

-2 ( 2x + 18y = -9) = -4x -36y = 18

Then, we combine the two equations.

-4x + 4x = 0

18y - 36y = -18y

-27 + 18 = -9

Our new equation is -18y = -9.

Now, divide both sides by -18.

-18y / -18 = y

-9/ -18 = 1/2

y = 1/2

We can plug in a value for y since y = 1/2 now.

Let's use 2x + 18y = -9

Plug in y.

2x + 18(1/2) = -9

2x + 9 = -9

Then, subtract 9 from both sides.

2x = -18

Divide by 2.

2x/2 = x

-18/2 = -9

x = -9

Lastly, we can plug in both x and y values to see it works.

2(-9) + 18(1/2) = -9

-18 + 9 = -9

Therefore, the values of x and y does work.

x = -9

y = 1/2

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yKpoI14uk [10]

\\ \tt\leadsto tan\theta=\dfrac{Perpendicular}{Base}

\\ \tt\leadsto tan29=\dfrac{h}{400}

\\ \tt\leadsto h=400tan29

\\ \tt\leadsto h=221.7ft

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2 years ago
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artcher [175]

Answer:

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Step-by-step explanation:

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FinnZ [79.3K]

Answer:

x = 1

Step-by-step explanation:

12.3 + 2.3 = 14.6

If you multiplied 12.3 by 1 it would still be 12.3

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