Given that n is an interger and that n>1 prove algebraically that n^2-(n-2)^2-2 is always an even number

Answer and Step-by-step explanation:

If n is assumed to be an even number, then obviously the expression will be even because the square of an even number is even and subtracting 2 from an even number makes it remain an even number. Therefore, let's assume n is not an even number, then n = 2x + 1 where x > 0. We have:

(2x + 1)² – (2x + 1 – 2)² – 2 = (2x + 1)² – (2x – 1)² – 2 = 4x² + 4x + 1 – (4x² – 4x + 1) – 2 = 4x² + 4x + 1 – 4x² + 4x – 1 – 2 = 8x – 2 = 2(4x – 1). Since the product is a multiple of 2, then the expression is even if n is odd. Therefore the expression is always even.

x = 10.1

step-by-step explanation:

all of the angles of a tirangle add up to 180º

180 - 65 = 115 (since we already know the value of one angle)

(2x + 7) + (8x + 7) = 115

(combins like terms) 10x + 14 = 115

(isolate the variable subtract 14 from both sides): 10x = 101

(find the variable divide both sides by 10): x = 101/10

x = 10.1

no, the equation would be invalid since they are both not equal.

step-by-step explanation:

here you got

4y-6+4y 2-9

first, you would simplify the equation on the form.

4y-6+4y 2-9

then, move all the terms that contains a y to the left and all the other terms to the right.

+4y-4y+-9-6

finally, you will simplify the left and the right side of the equation.

y=-3