I am a perfect square that is 11 more than a perfect square and 13 less than a perfect square. whcih number am I?
The # is grater than 11:
4^2 = 16.
5^2 = 25.
6^2 = 36.
7^2 = 49.
36 is 11 greater than 25 and 13 less than 49. Therefore the # 1s 36.
hello a perfect square that is 11 more than a perfect square and 13 less than a perfect square. whcih number am I?
To solve this problem, let's break it down step by step:
Step 1: Let's assume that the perfect square you are is represented by "x^2."
Step 2: You are 11 more than a perfect square. This can be represented as "x^2 + 11."
Step 3: You are also 13 less than another perfect square. This can be represented as "x^2 - 13."
Step 4: Based on the given information, the equation becomes: x^2 = x^2 + 11 = x^2 - 13
Step 5: Simplifying the equation further, we get: x^2 - x^2 = x^2 + 11 - x^2 - 13
Step 6: Simplifying the equation even more, we get: 0 = 11 - 13
Step 7: Simplifying further, we get: 0 = -2
Step 8: Since this is an inconsistent statement (0 cannot be equal to -2), there is no value of x that satisfies the given conditions.
Therefore, there is no specific number that you are.