Let r = usual driving rate
let t = usual driving time
We need to figure out t
The distance she covers in her usual time at her usual rate is r*t
The distance she covers in her new time at her new rate is:
(1+t)*((2/3)r)
Set this equal to each other and solve for t.
rt = (2/3)r + (2/3)rt
(1/3)rt = (2/3)r
(1/3)t = (2/3)
t = 2
So her usual time is 2 hours. (There's probably a faster way to do this)
For Domain<span>, you just need to watch out for Square Roots and/or Fractions! 1) a POLYNOMIAL (no square roots or fractions), i.e. f(x) = x^2 + 3x + 1, </span>domain<span> is "all real numbers." 2) a FRACTION (w/ no square root), i.e. f(x) = (2x+1)/(x^2+5x+6). Set bottom "not equal" to zero.</span>
Well currently, we don't have sufficient evidence to prove that. we don't even know anything about j and l