Rodrigo Underwood

2022-02-03

How do you evaluate ?

patriette65s

Explanation:
h(x) is just a way of writing "a function (call it h) in x", meaning whatever is on the other side of the equation. Given a particular value (-2) you just put it into the equation and solve it.
$h\left(x\right)=-\left(\frac{5}{3}\right)|x|+6$
$h\left(x\right)=-\left(\frac{10}{6}\right)+6;h\left(x\right)=-\left(\frac{10}{6}\right)+\frac{36}{6}$
$h\left(x\right)=\frac{26}{6};h\left(x\right)=\frac{13}{3}$ or 4.33

Do you have a similar question?