2007-12-11 17:28:49 · 6 answers · asked by Zach J 1 in Science & Mathematics ➔ Mathematics
946
2007-12-11 17:32:22 · answer #1 · answered by WhatWasThatNameAgain? 5 · 0⤊ 0⤋
It must be between 900 and 1000 because
900^2 = 810,000 < 894,916 < 1,000,000
(900 + a)^2 = 894,916
810,000 + 1,800a + a^2 = 894,916
a^2 + 1,800a = 84,916
let a be (10b + c), where b is a natural number between 0 and 9, and c<10
(10b + c)^2 + 1800*(10b + c) = 100b^2 + 20bc + c^2 +
+ 18,000b + 1,800c
100b^2 + 20bc + c^2 + 18,000b + 1,800c = 84,916
If b=5 18,000 * 5 = 90,000 > 84,916, thus b must be smaller than 5
Try b=4
100*4^2 + 20*4c + c^2 + 18,000 * 4 + 1,800c = 84,916
1,600 + 80c + c^2 + 72,000 + 1,800c = 84,916
c^2 + 1,880c + 73,600 = 84,916
c^2 + 1,880c = 11,316
11,316 / 1880 = 6.019 ...
Try c=6
6^2 + 1,880c = 36 + 11,280 = 11,316 -- Bingo!
900 + 10*b + c = 946
946^2 = 894,916
2007-12-12 01:53:24 · answer #2 · answered by Amit Y 5 · 0⤊ 0⤋
946
2007-12-12 01:34:18 · answer #3 · answered by someone else 7 · 0⤊ 0⤋
I have no calculator.
â894916 > 945 as 945^2 = (94*95)25 = 893025
â894916 < 950 as 950^2 = (9*10)2500 = 902500
=> â894916 = 946
2007-12-12 01:50:00 · answer #4 · answered by sv 7 · 0⤊ 0⤋
= â894,916
= 946
Answer: 946
2007-12-12 01:34:25 · answer #5 · answered by Jun Agruda 7 · 3⤊ 0⤋
i believe its 946
thank you google
2007-12-12 01:33:15 · answer #6 · answered by green_money73 1 · 0⤊ 0⤋