Your answer is 7^(- 2) or 
or 
A quadratic equation is written in four equivalent forms below.
1 answer:
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Answer:
The correct option is;
No, the distance from (0, 0) to (2, √6) is not 3 units
Step-by-step explanation:
The given parameters of the question are as follows;
Circle center (h, k) = (0, 0)
Point on the circle (x, y) = (3, 0)
We are required to verify whether point (2, √6) lie on the circle
We note that the radius of the circle is given by the equation of the circle as follows;
Distance² = (x - h)² + (y - k)² = r² which gives;
(3 - 0)² + (0 - 0)² = 3²
Hence r² = 3² and r = 3 units
We check the distance of the point (2, √6) from the center of the circle (0, 0) as follows;
Therefore;
(2 - 0)² + (√6 - 0)² = 2² + √6² = 4 + 6 = 10 = √10²
Which gives the distance of the point (2, √6) from the center of the circle (0, 0) = √10
Hence the distance from the circle center (0, 0) to (2, √6) is not √10 which s more than 3 units hence the point (2, √6), does not lie on the circle.
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 = = 360 possibilities.
Hence, we can write 360 distinct passwords using the numbers 1, 2, 3, 4, 5, and 6.