Which of the equations are true identities? A. N ( n − 2 ) ( n + 2 ) = n 3 − 4 n B. ( x + 1 ) 2 − 2 x + y 2 = x 2 + y 2 + 1

Both A and B are true identities

Step-by-step explanation:

A. N ( n − 2 ) ( n + 2 ) = n 3 − 4 n

We need to show that (left-hand-side)L.H.S = R.H.S (right-hand-side)

So,

 n ( n − 2 ) ( n + 2 ) = n(n² - 2²)     (difference of two squares)

                               = n³ - 2²n       (expanding the brackets)

                               = n³ - 4n         (simplifying)

So,                L.H.S  = R.H.S

B. ( x + 1 )² − 2x + y² = x² + y² + 1

We need to show that (left-hand-side)L.H.S = R.H.S (right-hand-side)

So,

( x + 1 )² − 2x + y² = x² + 2x + 1 - 2x + y²   (expanding the brackets)

                             = x² + 2x - 2x + 1  + y²   (collecting like terms)

                             = x² + 1 + y²        

                             = x² + y² + 1                    (re-arranging)

So,              L.H.S  = R.H.S

So, both A and B are true identities since we have been able to show that L.H.S  = R.H.S in both situations.

+25 = x

step-by-step explanation:

here x is the hypotenuse of a right triangle whose legs are 7 and 24.   by the pythagorean theorem, 7² + 24² must equal x²:

x² = 49 + 576 = 625.   thus, x = +√625 = +25 = x

(x, y)→(x − 8, y − 7)

step-by-step explanation:

enjoy!