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kotegsom [21]
3 years ago
10

You need to have a password with 4 letters followed by 3 even digits between 0 and 9, inclusive. If the characters and digits ca

nnot be used more than once, how many choices do you have for your password?
Mathematics
1 answer:
Gekata [30.6K]3 years ago
4 0

Answer: 21528000 choices

Step-by-step explanation:

Since there are 26 letters of the alphabet and no alphabet is to be used more than once,

then the first letter can be chosen in 26ways

Second letter can be chosen im 25ways

Third letter can be chosen in 24ways

Fourth letter can be chosen in 23ways.

Since we can only use even digits, then the even digits we have are "0,2,4,6,8"

Then the first digit can be chosen in 5ways

Second digit can be chosen in 4ways

Third digit can be chosen in 3ways.

Total number of choices for this pass word = 26×25×24×23×5×4×3 = 21528000 choices

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we need the graph/ lines,

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wherever 2 lines intersect, that is a solution

if they are paralell, no solutions

if they cross, 1 solution

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7 0
3 years ago
SYNTHETIC DIVISION: (2x^4-7x^3+14x^2-55x+21) divided by (2x-7)
Orlov [11]

Answer:

x3+7x-3

Step-by-step explanation:

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2 years ago
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Help please! anything will help thanks ya'll.!
zzz [600]
The cosine of A is (adjacent side)/(hypotenuse) = 55/73. Do the division, then find the angle whose cosine is that number. The whole thing takes maybe 6 or 7 seconds with your calculator. It's choice-B.
6 0
3 years ago
I don’t think I wanna do this
Vedmedyk [2.9K]
$89.10 is the best guess.
7 0
2 years ago
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How to solve this I don’t get this material
Nata [24]
6^2+(24/6+3^2)
First, solve the powers.
6^2=6*6=36
3^2=3*3=9
Now substitute.
36+(24/6+9)
Solve the parentheses.
36+(4+9)
Then, solve.
36+4+9=49


4 0
3 years ago
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