Prove algebraically that the square of any odd number is always 1 more than a multiple of 8. let n stand for any integer in your working

Explanation:

If n is "any integer", then 2n+1 is "any odd number."

The square of any odd number is then ...

  (2n+1)² = 4n² +4n +1 = 4n(n+1) +1

Since n is any integer, one of n and n+1 will be an even integer, so the product 4n(n+1) will be divisible by 8.

Then the sum 4n(n+1) +1 is one more than a number divisible by 8, hence ...

  the square of an odd number is 1 more than a multiple of 8.

x=-2

infinite number of lines, so there is an infinite number of equations.

step-by-step explanation:

there are an infinite number of line that passes through the point (-2,5)

x=-2   is one line that passes through the point

y = 5 is another line that passes through the point

(these two lines are perpendicular to each other).

since the number of   numbers are infinite and the slope is the ratio of two numbers,(change in y over change in x) we can have an infinite number of slopes.     our only condition is that it passes through the point (-2,5)

honestly bro aint no one got time to answer these dumb question just do it p***y

step-by-step explanation: