Three consecutive odd integers are such that the square of the third integer is 1515 greater thangreater than the sum of the squares of the first two. one solution is 11, 33, and 55. find three other consecutive odd integers that also satisfy the given conditions.

3 consecutive odd integersx, x + 2, and x + 4

(x + 4)^2 = (x^2 + (x + 2)^2) + 15
x^2 + 8x + 16 = x^2 + x^2 + 4x + 4 + 15
x^2 + 8x + 16 = 2x^2 + 4x + 19
2x^2 + 4x + 19 - x^2 - 8x - 16 = 0
x^2 - 4x + 3 = 0
(x - 3)(x - 1) = 0

x - 3 = 0x + 2 = 3 + 2 = 5x + 4 = 3 + 4 = 7
x = 3
so another one that satisfies this is : 3,5,7 <===

x - 1 = 0      x + 2 = 1 + 2 = 3x + 4 = 1 + 4 = 5
x = 1
and another one is : 1,3,5 <===

Period is the time it takes for one “cycle” to complete. in the parent sine function, period is 2*pi since sin(0) = 0, sin(pi/2) = 1, sin(pi) = 0, sin(3pi/2) = -1, and sin(2pi) = 0. at that point, the sequence repeats itself. i’m not sure if your graph is similar to the sine function since you didn’t provide a picture, but you can find the period by looking at a graph.