2007-04-26 16:29:38 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Seven...
2007-04-26 16:37:03 · answer #1 · answered by blktiger@pacbell.net 6 · 0⤊ 0⤋
Let the integer be x.
"twice the square" 2x^2
" is 35 more" 2x^2-35
" than nine time the integer" = 9x
Solve 2x^2-9x-35=0 for the positive root (its's 7)
2007-04-26 23:34:34 · answer #2 · answered by cattbarf 7 · 0⤊ 0⤋
I have no idea what an integer is, but 7 x7 x2 = 98
7 x 9 = 63
98-63 = 35
7?
Thank god my PC has a calculator.
What is trial and error x 15 Min's worth?
2007-04-26 23:45:48 · answer #3 · answered by Anonymous · 0⤊ 0⤋
2x^2 = 35 + 9x
2x^2 - 9x - 35 = 0
x = 7 or x = -2.5
Only 7 is an integer.
2007-04-26 23:34:25 · answer #4 · answered by novangelis 7 · 0⤊ 0⤋
do it this way:
2x^2 = 35+9x
=> 2x^2 - 9x - 35 = 0
From the quadratic formula you get x=7 and -2.5.
Which means 7 is the integer you're looking for.
2007-04-26 23:37:03 · answer #5 · answered by Anonymous · 0⤊ 0⤋
7
2 x 49=98
98-35=63
63=7 x 9
2007-04-26 23:33:36 · answer #6 · answered by Anonymous · 0⤊ 0⤋