The smallest square number which can be divisible by 11 and 5
1 answer:
Answer:
3025
Step-by-step explanation:
let's look at it in this concept.
For a number to be divisible by 11 and 5. it must be a multiple of the LCM of 11 and 5.
LCM of 11 and 5=55
therefore the number is 55x, where x is a positive integer.
it is a said that the number is a perfect square
therefore the square root of 55x must be an integer.
the smallest value of x to make 55x a perfect square is....
Therefore the number is.... .
<em>sweet</em><em> </em><em>right</em>
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Answer:
Where 0 < x < 3
The location of the local minimum, is (2, 0)
The location of the local maximum is at (0, 16)
Step-by-step explanation:
The given function is f(x) = (x + 2)⁴
The range of the minimum = 0 < x < 3
At a local minimum/maximum values, we have;
∴ (-x + 2)³ = 0
x = 2
When x = 2, f''(2) = -12×(-2 + 2)² = 0 which gives a local minimum at x = 2
We have, f(2) = (-2 + 2)⁴ = 0
The location of the local minimum, is (2, 0)
Given that the minimum of the function is at x = 2, and the function is (-x + 2)⁴, the absolute local maximum will be at the maximum value of (-x + 2) for 0 < x < 3
When x = 0, -x + 2 = 0 + 2 = 2
Similarly, we have;
-x + 2 = 1, when x = 1
-x + 2 = 0, when x = 2
-x + 2 = -1, when x = 3
Therefore, the maximum value of -x + 2, is at x = 0 and the maximum value of the function where 0 < x < 3, is (0 + 2)⁴ = 16
The location of the local maximum is at (0, 16).