How do you evaluate $h\left(x\right)=-\frac{5}{3}\left|x\right|+6\text{for}\text{}x=-2$?

Answer & Explanation

patriette65s

Beginner2022-02-04Added 15 answers

$h\left(x\right)=\frac{13}{3}\text{at}\text{}x=-2$
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.
$h\left(x\right)=-\left(\frac{5}{3}\right)\left|x\right|+6$ $h\left(x\right)=-\left(\frac{10}{6}\right)+6;h\left(x\right)=-\left(\frac{10}{6}\right)+\frac{36}{6}$ $h\left(x\right)=\frac{26}{6};h\left(x\right)=\frac{13}{3}$ or 4.33