Question

given h(x)=e^-3x+9 when h(5)

Answer 1

\(\bf h({\color{red}{ x}})=e^{-3{\color{red}{ x}}}+9 \qquad \qquad h({\color{red}{ 5}})=e^{-3({\color{red}{ 5}})}+9\)

Answer 2

I get that, how do I evaluate and round to the nearest ten-thousandth? My book does not really explain the process clearly...

Answer 3

\(\large \begin{array}{llll}
1&0,000th\\
&\uparrow \\
&\textit{4 zeros}
\end{array}\implies
\begin{array}{ccllll}
12.4567{\color{red}{ 89}}\implies 12.456&7{\color{red}{ 8}}\\
&\downarrow \\\
&12.4568
\end{array}\)

Answer 4

10,,000th means round up to 4 decimals
so you use grab up to the 4th, then look to its right
if the next number is below 5, you keep the last digit as is

if the next number is 5 or above, then the last digit "bumps up"

Answer 5

Thank you!

Answer 6

yw