ADP7118

Very simple question.

On page 16 of the datasheet it is explained how to keep the noise of externally configured regulators low.

The theory makes sense. Reduce the gain of the circuit down to unity at high frequencies so as to not amplify the broadband noise by the DC circuit gain.

What I am not understanding is how AD arrive at a DC gain of 10, for the configuration shown, using the 5V, fixed, regulator.

The datasheet explains that all the fixed voltage devices work at a gain of unity. So to get the 5V to output 10V it needs to be configured for a gain of 2. Which is what you'd expect given the feedback resistor values with both being 100k.

What am I missing?
 
EDIT: I think somebody had too many drinks at lunch.

Links?
https://www.mouser.com/datasheet/2/609/adp7118-1503549.pdf

At DC the gain is set to 2, so 5V comes to 10V.

For AC the gain is knocked-down by the 10k resistor with the 1uFd cap. The 1uFd bypasses the 100k at about 1.7Hz. Just as the text says. The 1uFd is like 10k impedance by 17Hz, and then the gain is like 109k/100k or 1.09. And the noise reduction factor is about 2:1.
 

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