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yousifgorgees101
29.02.2020 •
Mathematics
The sum of the squares of three consecutive integer numbers is 1454. Find the numbers.
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Ответ:
The three consecutive numbers which have its sum of the square as 1454 is 21, 22 and 23
Let
the three consecutive numbers be
x
x x + 1
x x + 1x + 2
sum of their squares:
(x²) + (x + 1)² + (x + 2)² = 1454
x² + x² + 2x + 1 + x² + 4x + 4 = 1454
3x² + 6x + 5 = 1454
3x² + 6x + 5 - 1454 = 0
3x² + 6x - 1449 = 0
divide through by 3
x² + 2x - 483 = 0
x² + 2x = 483
add 1 to both sides
x² + 2x + 1 = 483 + 1
x² + 2x + 1 = 484
factorize
x(x + 1) + 1 (x + 1) = 484
(x + 1) (x + 1) = 484
(x + 1)² = 484
x + 1 = √484
x + 1 = 22
x = 22 - 1
x = 21
Therefore, the three consecutive numbers are
x = 21
x + 1
= 21 + 1
= 22
x + 2
= 21 + 2
= 23
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Ответ:
n = 21, 22, 23.
Step-by-step explanation:
let the 3 consequitive numbers be n, n +1 , n+2
fro the question,
let squares of three consecutive integer numbers be
n², (n+1)² , (n+2)²
let the sum of the squares of the three consecutive integer numbers be
n² + (n+1)² +(n+2)² =1454
by opening the bracket
we have;
n² + (n² + 2n +1) + (n²+ 2n + 4) =1454
n² + n² + 2n + 1 + n² +4n +4 = 1454
therefore;
3n² + 6n + 5 = 1454
3n² + 6n +5 - 1454 =0
3n² + 6n - 1449 = 0
divide through by 3
we have;
n² + 2n - 483 = 0
n² + 2n = 483
add 1 to both sides to make an equation
n² + 2n + 1 = 483 +1
n (n +1) + 1 ( n+1) = 484
since the equation in the bracket is thesame, therefore we pick 1
n +1 (n +1) = 484
(n+1)² = 484
Square on both sides
n + 1 = √484
n +1 = 22
n =22 - 1
n = 21
since its three consecutive integers
and n = 21
therefore n + 1 = 21 + 1 = 22
N +2 = 21 + 2 = 23
Therefore n = 21, 22, 23.
Ответ:
He needs 2.5 lbs of sugar and 2.5 lbs of water
since 2.5 if 50% of 5 then thats how much sugar he needs.