Graph the ellipse and locate the foci.
$$16x^2+9y^2=144$$

Step 1 ：Given the equation of the ellipse is $16x^2+9y^2=144$.

Step 2 ：The standard form of an ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ where a and b are the semi-major and semi-minor axes respectively.

Step 3 ：Rewrite the given equation in the standard form, we get $\frac{x^2}{9}+\frac{y^2}{16}=1$.

Step 4 ：From the standard form, we can identify the values of a and b. Here, $a=4$ and $b=3$.

Step 5 ：The foci of the ellipse are located at a distance of $\sqrt{a^2-b^2}$ from the center along the major axis.

Step 6 ：Calculate the distance to the foci, we get $\sqrt{4^2-3^2}=2.6457513110645907$.

Step 7 ：Final Answer: The foci of the ellipse are located at a distance of approximately $\boxed{{\displaystyle 2.6457513110645907}}$ units from the center along the major axis.