Write a polite but firm email to a landlord who has not fixed the heating for three weeks. I want the repair, not a fight.
NVIDIA Nemotron 3.5 Super VL
Subject: Urgent: Heating Repair Needed – [Your Name], Unit [Your Unit Number]
Dear [Landlord or Property Manager Name],
I hope you are doing well.
I am writing to follow up on the heating issue in my apartment. Unfortunately, the heat has not been working for three weeks now, and the unit is becoming uncomfortably cold.
I understand that repairs can sometimes take time to coordinate, but this issue is now affecting the livability of my home. I would like to resolve this amicably and get the heating fixed as soon as possible.
Could you please let me know when a technician will be scheduled to come out? I am available [mention your general availability, e.g., weekday afternoons or weekends] to provide access.
Thank you for your prompt attention to this matter. I look forward to hearing from you.
Best regards,
[Your Name] [Your Phone Number]
Asked Nov 1, 2026. Unedited.
Write a React hook that debounces a value and cancels on unmount. Include the TypeScript types.
NVIDIA Nemotron 3.5 Super VL
Here is a clean, TypeScript implementation. It tracks the previous value so it only resets the timer when the value actually changes, and it cleans up the timeout on unmount (or when the value/delay changes).
import { useEffect, useRef } from 'react';
/**
* Debounces a value. Returns the value after `delay` ms of inactivity.
* The timeout is cancelled automatically when the component unmounts,
* when the delay changes, or when the value changes.
*/
export function useDebounce<T>(value: T, delay: number): T {
const debouncedValueRef = useRef<T>(value);
const timeoutRef = useRef<ReturnType<typeof setTimeout> | null>(null);
// Update immediately when the delay changes so the new delay applies right away
useEffect(() => {
debouncedValueRef.current = value;
}, [value, delay]);
useEffect(() => {
// If the value hasn't actually changed, keep the existing debounced value
// and do not restart the timer.
if (debouncedValueRef.current === value) {
return undefined;
}
// Cancel any pending timeout
if (timeoutRef.current !== null) {
clearTimeout(timeoutRef.current);
}
// Set a new timeout to commit the debounced value
timeoutRef.current = setTimeout(() => {
debouncedValueRef.current = value;
}, delay);
// Cleanup: cancel on unmount or before re-running this effect
return () => {
if (timeoutRef.current !== null) {
clearTimeout(timeoutRef.current);
}
};
}, [value, delay]);
return debouncedValueRef.current;
}Example usage
import { useState } from 'react';
import { useDebounce } from './useDebounce';
export default function Search() {
const [query, setQuery] = useState('');
const debouncedQuery = useDebounce(query, 500);
// Only fetch after the user stops typing for 500ms
// useEffect(() => { ... fetchResults(debouncedQuery) }, [debouncedQuery]);
return (
<input
value={query}
onChange={(e) => setQuery(e.target.value)}
placeholder="Search..."
/>
);
}Key points
| Concern | How it's handled |
|---|---|
| Type safety | Generic <T> works for strings, numbers, objects, etc. |
| Cancel on unmount | The effect cleanup runs clearTimeout when the component unmounts. |
| Stale-value avoidance | A ref holds the latest committed debounced value; the effect only schedules a new timer when the input actually changes. |
| Delay changes | If delay changes, the previous timer is cleared and the new one is scheduled. |
If you need to execute a callback (rather than just a value) after a debounce, consider a useDebouncedCallback hook instead.
Asked Nov 1, 2026. Unedited.
A drug test is 99% accurate, and 0.5% of people use the drug. Someone tests positive. What is the probability they use it, and why does the answer surprise people?
NVIDIA Nemotron 3.5 Super VL
Assuming “99% accurate” means the test correctly identifies drug users 99% of the time (true positive rate = 99%) and correctly identifies non-users 99% of the time (true negative rate = 99%):
Use Bayes’ theorem
[ P(\text{user}\mid\text{positive}) = \frac{P(\text{positive}\mid\text{user}) \cdot P(\text{user})}{P(\text{positive})} ]
[ = \frac{0.99 \times 0.005}{(0.99 \times 0.005) + (0.01 \times 0.995)} ]
[ = \frac{0.00495}{0.00495 + 0.00995} = \frac{0.00495}{0.0149} \approx \boxed{0.332} ]
So the probability is about 33.2%, or roughly 1 in 3.
Why this surprises people
Intuitively, many hear “99% accurate” and assume a positive result means there is a 99% chance the person uses the drug. That ignores the base rate: only 0.5% of the population are users, so they are vastly outnumbered by non-users.
Even though the false-positive rate is just 1%, it is applied to the 99.5% of people who don’t use the drug. That produces a large number of false positives (about 9,950 per million people), which nearly doubles the number of true positives (about 4,950 per million). As a result, a positive test is only about twice as likely to come from a user as from a non-user, pulling the posterior probability down to roughly one-third.
This is the classic base-rate fallacy: people overweight the test’s accuracy and underweight how rare the condition is in the population.
Asked Nov 1, 2026. Unedited.
What it’s good at
Ranked against 42 models, from public leaderboards and our own blind tests where we have them. How we rank
- #26of 42 for math
- #28of 42 for coding
- #29of 42 for everyday chat
The facts
NVIDIA Nemotron 3.5 Super VL is a vision-language model in NVIDIA’s Nemotron family, with open weights. It reads images and video as well as text.
Scores from artificialanalysis.ai and lmarena.ai.
- Made by
- NVIDIA
- API price, per 1M tokens
- $0.20 in, $0.50 out
- Speed
- No public score
- Artificial Analysis Index
- No public score
- LMArena text
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- LMArena coding
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- Updated
- Oct 26, 2026