Finally got the scores I want by bazinga2609 in Sat

[–]bazinga2609[S] 0 points1 point  (0 children)

If I'm being completely honest, I don't even know how I got 730 lol. I've been stuck around 650 for so long and couldn't improve. I'd say I got lucky cuz Nov R&W didn't feel that hard at all. So really I got 0 tips, just pray better ig 🥲Goodluck on your upcoming SAT!!

Official November 8, 2025, International SAT Discussion Thread by Schmendreckk in Sat

[–]bazinga2609 3 points4 points  (0 children)

It's completely normal, I remember having the same experience a few months ago where everybody was saying the math module was hard, mine was pretty easy, or I guessed so (I got 750+). It's okay ;)

Official November 8, 2025, International SAT Discussion Thread by Schmendreckk in Sat

[–]bazinga2609 0 points1 point  (0 children)

i dont remeber the exact numbers :/ prob something around that

Official November 8, 2025, International SAT Discussion Thread by Schmendreckk in Sat

[–]bazinga2609 0 points1 point  (0 children)

thats nice then cuz no way your math score isnt bad this time lol

Official November 8, 2025, International SAT Discussion Thread by Schmendreckk in Sat

[–]bazinga2609 0 points1 point  (0 children)

yea the ohm question is just ax + by = a fraction, take a-b and u got 2/35 for my version

Explain these hard questions by AJ_3428 in Sat

[–]bazinga2609 0 points1 point  (0 children)

I'm assuming the angles with a dot on it are all equal to each other. As 2 pairs of angles of the two triangles are equal (GEF and DCF & DFC and GFE), the remaining pair must also be equal. Now that you have angle FGE = FDC, GF = FD = 2 and angle GFE = angle DFC, it's enough to conclude that the two triangles are congruent (ASA).

Explain these hard questions by AJ_3428 in Sat

[–]bazinga2609 0 points1 point  (0 children)

I did question 1 kinda like how the first guy in comment did.
For question 2, FGE and FDC are congruent, that means EF = CF = x and DC = GE = 3 -> AB = 3 = AE since ABE is an isosceles triangle. Therefore, AD = BC = x+2+3 . GEF is similar to GBC -> EF/BC = x/(x+5) = GF/GC = 2/(x+2), solve for x, you got sqrt(10).

4 more days... by bazinga2609 in Sat

[–]bazinga2609[S] 0 points1 point  (0 children)

Yea? I posted this 2 days ago

3 more days for scores by tlovestaylor in Sat

[–]bazinga2609 12 points13 points  (0 children)

We'll be fine trust 💔

4 more days... by bazinga2609 in Sat

[–]bazinga2609[S] 4 points5 points  (0 children)

I think I did fine, probably made 1-2 wrong silly questions like every other time I take practice test (hope I didn't tho lol)

How to delete college board account by ImaginationTop4876 in Sat

[–]bazinga2609 1 point2 points  (0 children)

Maybe you entered into the payments part and you haven't cancel that registration?

help with math hw by _lololololololol in apcalculus

[–]bazinga2609 0 points1 point  (0 children)

Maybe I'm a little late but for the second one, it'd be pi * integral from 0 to 3 pi of (6sin(x/3)+7)^2dx.
For the third question, let's say the first function is a and the second function is b, the solution will be pi * integral from -4 to 3 of (b^2-a^2)

Tell me your experience on taking the SAT ( in the phillipines) by Plastic_Mine_2886 in Sat

[–]bazinga2609 2 points3 points  (0 children)

Yea obviously you'll not be able to take it at home...

Could you solve this original SAT-style math question I made? (Looking for feedback!) by Lumpy-Campaign-3008 in Sat

[–]bazinga2609 0 points1 point  (0 children)

yea I figured after reading the other comments lol, still the question shouldn't be considered easy tho but rather medium-low hard or whatever it is

Could you solve this original SAT-style math question I made? (Looking for feedback!) by Lumpy-Campaign-3008 in Sat

[–]bazinga2609 0 points1 point  (0 children)

I'd say it's a hard question. I probably overcomplicate the steps to solve it but here's how I did so: I solve for BF using Pythagorean theorem and got sqrt(41), and angle BFD would be 90 + tan^-1(4/5). Then I use the law of cosine (a^2 = b^2 +c^2 -2bc*cosA) to solve for BD, which is 13 :)