2026 HSC CHAT (4 Viewers)

Socialism

§øç¡ålîšm - SHE/HER please 💜
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Normally I see them everywhrre back when Jonathon was on here. I wonder where they've gone?
they are still around, a little. but yeah idk

honestly i dont remember enough from back then LOL half the time ur the one reminding me of things 😭
 

Socialism

§øç¡ålîšm - SHE/HER please 💜
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I thought you could under certain circumstances though (unlikely in this scenario)
Ngl I should have created a diff account too bc I'm way way too easy to find
no no u can

its just not like other websites where u can just change your handle or something
 

Study to success

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View attachment 47381
HELPP I CANT DO THESEE AHSHFHFJ
I think there supposed to be like a semicircle/ hyperbola thing. Maybe try graphing it on Desmos so u have a rough idea on how it’s supposed to look. I think maybe finding the vertex and intercepts of the equation in the denominator can help u find the domain and range
 

cheesynooby

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View attachment 47381
HELPP I CANT DO THESEE AHSHFHFJ
domain:
1. solve the quadratic and identify its concavity to find the x values where it is negative e.g. if it has x intercepts at x = -2 and 2, and is concave up the quadratic is negative between x = -2 and 2. since u cant square root a negative and cant divide by 0, that gives u the domain (x < -2, x > 2)

range:
2. find the range of the quadratic by finding the y coord of the vertex. e.g. vertex at (1, 3) and concave down = range: y ≤ 3
3. square root that (keeping in mind u can't square root a negative)-> 0 ≤ y ≤ sqrt3
4. find the reciprocal of that (a graph of y = 1/x could be useful - imagine the range of 1/x if you could only plug in values between 0 and sqrt3 (from step 3) -> y ≥ 1/sqrt3
 
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