WILL GIVE 100 POINTS AND BRAINLIEST! Prove that the square of the second number out of three consecutive odd numbers is four greater than the product of the first and the third numbers.

Let's choose the three odd consecutive numbers 1, 3, and 5.

3^2=9

1•5=5

9-5=4

Let's try the same thing, but with the numbers 101, 103, and 105.

103^2=10609

101•105=10605

10609-10605=4

So, yes, the square of the second number out of three consecutive odd numbers is four greater than the product of the first and the third numbers.

Pick 3 consecutive odd numbers:

3, 5, 7

Square the 2nd number: 5 ^ 2 = 25

Multiply the first and 3rd:

3 x 7 = 21

25-21 = 4

This is true.

Try another set of numbers:

9, 11, 13

11^2 = 121

9 x 13 = 117

121-117 = 4

Again it’s true.

x = ± 6i

step-by-step explanation:

x^2 = -36

take the square root of each side

sqrt(x^2) = sqrt(-36)

we know sqrt(ab) = sqrt(a) sqrt(b)

x = ±sqrt(36) sqrt(-1)

we know that sqrt (-1) = i

x = ± 6i

step-by-step explanation:

6.48/4 4 goes into 6.48 1.62 times