Find f(3a) of $\displaystyle f(x)= -2x^2 + 3x - 4$
and

The Solutions to $\displaystyle 6m^2 - 7m + 1 = 0$
and

This is the equation used ot express the amount of items a business has at a given time. T = months.

$\displaystyle N(t) = 5t^3 - 70t^2 + 245t$
0<_ t <_ 10

Find when there are no items left..

2. Originally Posted by mibamars
Find f(3a) of $\displaystyle f(x)= -2x^2 + 3x - 4$
and
For questions like these, you just plug in whatever they put in brackets in the position that x is in.

$\displaystyle f(x) = -2x^2 + 3x - 4$

$\displaystyle \Rightarrow f(3a) = -2(3a)^2 + 3(3a) - 4$

And you just simplify that

The Solutions to $\displaystyle 6m^2 - 7m + 1 = 0$
This is a relatively simple quadratic equation to solve, just foil.

$\displaystyle 6m^2 - 7m + 1 = 0$

$\displaystyle \Rightarrow (6m - 1)(m - 1) = 0$

$\displaystyle \Rightarrow m = \frac {1}{6} \mbox { or } m = 1$

This is the equation used ot express the amount of items a business has at a given time. T = months.

$\displaystyle N(t) = 5t^3 - 70t^2 + 245t$
0<_ t <_ 10

Find when there are no items left..

$\displaystyle 0 = 5t^3 - 70t^2 + 245t$ ...........factor out the common 5t

$\displaystyle \Rightarrow 0 = 5t(t^2 - 14t + 49)$ ..............now foil the last quadratic

$\displaystyle \Rightarrow 0 = 5t (t - 7)(t - 7)$ ..............simplify a bit

$\displaystyle \Rightarrow 0 = 5t(t - 7)^2$

$\displaystyle \Rightarrow 5t = 0 \mbox { or } (t - 7)^2 = 0$

$\displaystyle \Rightarrow t = 0 \mbox { or } t = 7$

$\displaystyle t = 0$ is trivial, so we take $\displaystyle \boxed { t = 7}$