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Anna007 [38]
3 years ago
9

. Quantas senhas com 4 algarismos diferentes podemos escrever com os algarismos 1, 2, 3, 4, 5, 6?

Mathematics
1 answer:
ruslelena [56]3 years ago
4 0

Answer:

We can write 360 distinct passwords using the numbers 1, 2, 3, 4, 5, and 6.

Step-by-step explanation:

We have to find how many passwords with 4 different digits can we write with the numbers 1, 2, 3, 4, 5, and 6.

Firstly, it must be known here that to calculate the above situation we have to use Permutation and not combination because here the order of the numbers in a password matter.

Since we are given six numbers (1, 2, 3, 4, 5, and 6) and have to make 4 different digits passwords.

  • Now, for first digit of the password, we have 6 possibilities (numbers from 1 to 6).
  • Similarly, for second digit of the password, we have 5 possibilities (because one number from 1 to 6 has been used above and it can't be repeated).
  • Similarly, for the third digit of the password, we have 4 possibilities (because two numbers from 1 to 6 have been used above and they can't be repeated).
  • Similarly, for the fourth digit of the password, we have 3 possibilities (because three numbers from 1 to 6 have been used above and they can't be repeated).

So, the number of passwords with 4 different digits we can write = 6 \times 5 \times 4 \times 3  = 360 possibilities.

Hence, we can write 360 distinct passwords using the numbers 1, 2, 3, 4, 5, and 6.

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A circle has the equation 2x²+12x+2y²−16y−150=0.
KonstantinChe [14]

Answer: B. The coordinates of the center are (-3,4), and the length of the radius is 10 units.

Step-by-step explanation:

The equation of a circle in the center-radius form is:

(x-h)^{2} +(y-k)^{2}=r^{2} (1)

Where (h,k) are the coordinates of the center and r is the radius.

Now, we are given the equation of this circle as follows:

2x^{2}+12x+2y^{2}-16y-150=0 (2)

And we have to write it in the format of equation (1). So, let's begin by applying common factor 2 in the left side of the equation:

2(x^{2}+6x+y^{2}-8y-75)=0 (3)

Rearranging the equation:

x^{2}+6x+y^{2}-8y=75 (4)

(x^{2}+6x)+(y^{2}-8y)=75 (5)

Now we have to complete the square in both parenthesis, in order to have a perfect square trinomial in the form of (a\pm b)^{2}=a^{2}\pm+2ab+b^{2}:

<u>For the first parenthesis:</u>

x^{2}+6x+b^{2}

We can rewrite this as:

x^{2}+2(3)x+b^{2}

Hence in this case b=3 and b^{2}=9:

x^{2}+2(3)x+3^{2}=x^{2}+6x+9=(x+3)^{2}

<u>For the second parenthesis:</u>

y^{2}-8y+b^{2}

We can rewrite this as:

y^{2}-2(4)y+b^{2}

Hence in this case b=-3 and b^{2}=9:

y^{2}-2(4)y+4^{2}=y^{2}-8y+16=(y-4)^{2}

Then, equation (5) is rewritten as follows:

(x^{2}+6x+9)+(y^{2}-8y+16)=75+9+16 (6)

<u>Note we are adding 9 and 16 in both sides of the equation in order to keep the equality.</u>

Rearranging:

(x-3)^{2}+(y-4)^{2}=100 (7)

At this point we have the circle equation in the center radius form (x-h)^{2} +(y-k)^{2}=r^{2}

Hence:

h=-3

k=4

r=\sqrt{100}=10

8 0
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