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gayaneshka [121]
3 years ago
15

1. The company has agreed on a 9- character password format, comprised of uppercase letters, lower case letters, and digits (0-9

). However, the rules specify that the first 3 characters have to be digits and the next 6 can be any upper case or lower case letter. How many possible passwords are there if: a. repetitions are allowed: b. repetitions are not allowed:
Mathematics
1 answer:
pochemuha3 years ago
5 0

Answer: a) 1977060966, b) 146581344.

Step-by-step explanation:

Since we have given that

There are 10 numbers,

There are 26 lower case letters,

There are 26 upper case letters,

Since the first 3 characters have to be digits and next 6 can be upper or lower.

So, a) repetitions are allowed:

10^3\times (26+26)^6\\\\=10^3\times 52^6\\\\=1977060966

b) repetitions are not allowed:

^{10}P_3\times ^{52}C_6\\\\=720\times 20358520\\\\=146581344

Hence, a) 1977060966, b) 146581344.

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12,909 rounded to the nearest ten is 12,910
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Mashutka [201]

Answer: -0=

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What is f(-4) if f(x)=|7x-3|
MrRissso [65]
Substitute the given into the function

f(-4) = |7(-4) - 3|
now solve
f(-4) = |-28 - 3|
f(-4) = | -31| 
the absolute value of |-31| is 31
so, 
f(-4) = 31

Hope this helps :)
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4 years ago
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For which triangle is the length of the hypotenuse an INTEGER?
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4 0
4 years ago
Solve the following 3 × 3 system. Enter the coordinates of the solution below.
love history [14]
The system is:

i)    <span>2x – 3y – 2z = 4
ii)    </span><span>x + 3y + 2z = –7
</span>iii)   <span>–4x – 4y – 2z = 10 

the last equation can be simplified, by dividing by -2, 

thus we have:

</span>i)    2x – 3y – 2z = 4
ii)    x + 3y + 2z = –7
iii)   2x +2y +z = -5 


The procedure to solve the system is as follows:

first use any pairs of 2 equations (for example i and ii, i and iii) and equalize them by using one of the variables:

i)    2x – 3y – 2z = 4   
iii)   2x +2y +z = -5 

2x can be written as 3y+2z+4 from the first equation, and -2y-z-5 from the third equation.

Equalize:  

3y+2z+4=-2y-z-5, group common terms:
5y+3z=-9   

similarly, using i and ii, eliminate x:

i)    2x – 3y – 2z = 4
ii)    x + 3y + 2z = –7

multiply the second equation by 2:


i)    2x – 3y – 2z = 4
ii)    2x + 6y + 4z = –14

thus 2x=3y+2z+4 from i and 2x=-6y-4z-14 from ii:

3y+2z+4=-6y-4z-14
9y+6z=-18

So we get 2 equations with variables y and z:

a)   5y+3z=-9 
b)   9y+6z=-18

now the aim of the method is clear: We eliminate one of the variables, creating a system of 2 linear equations with 2 variables, which we can solve by any of the standard methods.

Let's use elimination method, multiply the equation a by -2:

a)   -10y-6z=18 
b)   9y+6z=-18
------------------------    add the equations:

-10y+9y-6z+6z=18-18
-y=0
y=0,

thus :
9y+6z=-18 
0+6z=-18
z=-3

Finally to find x, use any of the equations i, ii or iii:

<span>2x – 3y – 2z = 4 
</span>
<span>2x – 3*0 – 2(-3) = 4

2x+6=4

2x=-2

x=-1

Solution: (x, y, z) = (-1, 0, -3 ) 


Remark: it is always a good attitude to check the answer, because often calculations mistakes can be made:

check by substituting x=-1, y=0, z=-3 in each of the 3 equations and see that for these numbers the equalities hold.</span>
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