A 15m ladder leans against a wall, its bottom is 9m from the wall. How much would the lower end have to be pulled away from the wall so that the top of the ladder would slide down the same amount?
a=9 b=? c=15 then.. b=12
a=9, b=12, c=15..
c^2= a^2 * b^2
15^2=(9+x)^2 + (12-x)^2
225=81+18x+x^2 + 144-24x+x^2
225=2x^2-6x+225
then? what's nexT??
[please show ur solution :D]
2007-03-22 00:35:20 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
From where you got to-:
225 = 2x^2 - 6x + 225
(Take 225 from each side)
0 = 2x^2 - 6x
2x^2 = 6x (now divide by x)
2x = 6
x = 3
2007-03-22 00:40:42 · answer #1 · answered by Doctor Q 6 · 0⤊ 0⤋
c^2 = a^2 + b^2 (but I guess you just typed that wrongly).
2x^2 - 6x = 0
x^2 - 3x = 0
x(x - 3) = 0
x = 0 or x = 3
It's a little better to factorise the equation and then give a reason for discarding the zero solution: that describes the initial position.
The final answer is therefore x = 3m.
2007-03-22 08:06:49 · answer #2 · answered by Anonymous · 0⤊ 0⤋
x=3?
2007-03-22 07:40:32 · answer #3 · answered by steven c 2 · 0⤊ 0⤋
How can you be so close to the answer bu not get it? Just solve for x!
0=2x'2-6x (I'm assuming your arithmatic is right.)
Plug that into the quadratic equation;
x=3
2007-03-22 07:43:28 · answer #4 · answered by tgypoi 5 · 0⤊ 0⤋
225=2x^2-6x+225
225 gets canel
0=2x^2-6x
x=-1/4
2007-03-22 07:45:48 · answer #5 · answered by amal 2 · 0⤊ 1⤋