0.05t^2-10t+100=0
Please show me the steps
2007-10-18 11:29:20 · 4 answers · asked by fijiprize 1 in Science & Mathematics ➔ Mathematics
This is a quadratic equation of the form ax^2 + bx + c =0
a = 0.05
b = -10
c = 100
The best way is to use the quadratic equation
t = + -(b^2-4ac)^.5 /(2a). You should find this equation in your book.
The + - in front means you will probably get two roots or stated differently, 2 places where this graph will cross the x axis.
2007-10-18 11:36:18 · answer #1 · answered by Jim M 3 · 0⤊ 0⤋
0.05t^2 - 10t + 100 = 0
You can use the quadratic formula to find the solutions
- (-10) +/- square root of (100 - 4(100)(0.05) all divided by 0.1
Gives you:
(- (-10) +/- (square root of 80)) / 0.1
So that's 40 * square root of 5 + 100, and 40 * square root of 5 - 100 for your solutions
Enjoy!
2007-10-18 11:41:21 · answer #2 · answered by zelljrc 2 · 0⤊ 0⤋
0.05t^2-10t+100=0
t = [10 +/- sqrt(10^2 -4(100)(.05))]/(2*.05)
t = [10 +/- sqrt(80)]/.1
t = [10 +/- 4sqrt(5)]/.1
t = 100 +/- 40sqrt(5)
2007-10-18 11:36:48 · answer #3 · answered by ironduke8159 7 · 0⤊ 0⤋
use the quadratic equation
2007-10-18 11:34:34 · answer #4 · answered by DWRead 7 · 0⤊ 0⤋