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.
Answer:
The staircase would extends 1.20 m if the changes were made.
Step-by-step explanation:
Acording tho the statement the total heigth is 5.04 m, that is 504 cm, each steps rises 21 cm. with this information we can calculte how many steps are in the whole staircase:

To calculate how much the staircase will extend, whe have to multiply the horizontal depth times the number of steps.
using the run of 24 cm that gives us a lengt of 576 cm and if the changes were made the new length will be 696 cm.
So the staircase would extend an amount of (696 - 576), that is 120 cm, or 1.20 m, if the changes were made.
There is no illustration, so it is impossible to answer this question. I apologize.
First isolate the group with h in it
minus 2pir^2 from both sides
s-2pir^2=2pirh
divide both sides by 2pir
(s-2pir^2)/(2pir)=h
Answer:
32º
Step-by-step explanation:
Fº-32 ×5/9
5/9Fº-160/9
5/9Fº =160/9
(5Fº)9=9(160)
45Fº=1440
divide both sides by 45
Fº=32