1st number = n
2nd number = n+1
3rd number = n+2
sum of the squares of 3 consecutive numbers is 116
n² + (n+1)² + (n+2)² = 116
n² + (n+1)(n+1) + (n+2)(n+2) = 116
n² + [n(n+1)+1(n+1)] + [n(n+2)+2(n+2)] = 116
n² + n² + n + n + 1 + n² + 2n + 2n + 4 = 116
n² + n² + n² + n + n + 2n + 2n + 1 + 4 = 116
3n² + 6n + 5 = 116 Last option.
Hey there!!
Given equation :
... -4 ( x + 10 ) - 6 = -3 ( x - 2 )
Using the distributive property :
... -4x - 40 - 6 = -3x + 6
Combining like terms :
... -4x - 46 = -3x + 6
Adding 46 on both sides :
... -4x = -3x + 52
Adding 3x on both sides :
... -x = 52
... x = -52
Hope helps!
Answer:
a b
Step-by-step explanation:
Answer:
-65
Step-by-step explanation:
-80 - -15
Two minuses become plus
= -80+15
Subtract and keep the sign of the higher number
= -(80-15)
= -65
Answer:
I think Its C
Step-by-step explanation:
I did it on unit test so idk if its right or wrong
