Answer:
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.
Silver stars are worth 2 and gold stars are each worth 5
It's kinda blurry and I don't think that's math
24/5 ÷ 6/15
= 24/5 × 15/6
= 360/30
= 12
Answer:
The fourth and last answer is correct
Step-by-step explanation:
For the fourth one you have an exponent of an exponent, the base stays the same and the exponents get to be multiplied.
6^-1^16
For the last one both -8 x -2 and 8 x 2 equal 16.
