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jimmyiswrong
Приєднався 23 лип 2022
Doing Complex Number Precalculus on New years day
Doing Complex Number Precalculus on New years day
Переглядів: 2
Відео
Blooket Ice Monster Box Opening with Cheats!(I got a chroma)
Переглядів 27921 день тому
Blooket Ice Monster Box Opening with Cheats!(I got a chroma)
i'm doing AIME I 2024 today ima come back here later to say what score I got. also I made AMC 14 2025 winter get rekt nerd.
I got 4 on it.
@@christophoilet crazy
WE NOT MAKIN IT TO USAJMO WITH THIS ONE
@chicken_rice0123 nah the answers to AIME 2025 will all be 000 and we get 15/15.
@@christophoilet yes trust me bro
bruh my dad wants to qualify for USAJMO
ok
reckless driving trial
@chicken_rice0123 mine was a stolen car trial
note: I had the yeti before
jimmy has to cook for the AUGSAOM problem 3 because aint no way he can solve problem 10: \[ \left\{ \begin{array}{lll} \min\limits_{x \in \mathbb{R}^{24}} & \frac{\displaystyle \int_{0}^{\infty} \int_{0}^{x_1} \int_{0}^{x_2} \int_{0}^{x_3} \dots \int_{0}^{x_{24}} \prod_{i=1}^{24} \sin(x_i) \exp(x_i^2) \, dx_{24} \cdots dx_1}{\displaystyle \prod_{i=1}^{24} (1 + x_i^2) + \sum_{i=1}^{24} \Gamma(x_i) + \zeta(3) + \operatorname{erf}(x_{24})} \\[10pt] \hspace{0.2cm} \text{s.t.} & abla f(x) = \begin{bmatrix} \frac{\partial}{\partial x_1} \left( \prod_{i=1}^{24} \cos(x_i) ight) \\ \vdots \\ \frac{\partial}{\partial x_{24}} \left( \sum_{i=1}^{24} \exp(x_i) \sin(x_i) ight) \end{bmatrix}, \\[10pt] & \mathbf{M}x = \begin{bmatrix} \Gamma(x_1) \\ \sin(x_2 + x_3) \\ \vdots \\ \zeta(x_{24}) \end{bmatrix}, \text{ where } \mathbf{M} \in \mathbb{R}^{24 \times 24}, \\[10pt] & \displaystyle \sum_{i=1}^{24} x_i^2 - \prod_{i=1}^{24} x_i \leq 1, \\[10pt] & \displaystyle \int_{0}^{\pi/2} \prod_{i=1}^{24} \sin(x_i t) dt \leq \exp\left(\sum_{i=1}^{24} x_i ight), \\[10pt] & x_i \geq 0, \, i = 1, 2, \dots, 24. \end{array} ight. \]
hell nah i'm not doing that
@chicken_rice0123 that's nothing compared to the equivalent AGIMO problem: \[ \left\{ \begin{array}{lll} \min\limits_{\mathbf{x} \in \mathbb{H}^{1024}} & \frac{\displaystyle \int_{0}^{\infty} \cdots \int_{0}^{x_1} \prod_{i=1}^{1024} \sin\left(x_i^2 ight) \exp\left(x_i^3 ight) \prod_{j=1}^{1024} \cos\left(\sum_{k=1}^j x_k ight) dx_{1024} \cdots dx_1} {\displaystyle \prod_{i=1}^{1024} \left(1 + x_i^4 ight) + \sum_{i=1}^{1024} \Gamma(x_i) + \zeta(4) + \operatorname{erf}\left(\sum_{i=1}^{1024} x_i^2 ight)} \\[20pt] \text{s.t.} & abla f(\mathbf{x}) = \begin{bmatrix} \frac{\partial}{\partial x_1} \left( \prod_{i=1}^{1024} \cos(x_i^3 + x_i) + \sum_{j=1}^{1024} \zeta(x_j) ight) \\ \frac{\partial}{\partial x_2} \left( \prod_{k=1}^{1024} \Gamma(x_k) \sin\left(x_k^2 ight) ight) \\ \vdots \\ \frac{\partial}{\partial x_{1024}} \left( \sum_{m=1}^{1024} x_m^5 \cos\left(x_m^2 ight) + \prod_{j=1}^{1024} \exp\left(x_j ight) ight) \end{bmatrix} \\[20pt] & \mathbf{M}_1 \mathbf{x} + \mathbf{M}_2 \mathbf{x}^2 + \mathbf{M}_3 \mathbf{x}^3 = \begin{bmatrix} \Gamma(x_1) + \prod_{j=1}^{1024} \sin(x_j^2) \\ \sum_{k=1}^{1024} \cos(x_k^3) + \zeta(x_2) \\ \vdots \\ \prod_{p=1}^{1024} \left(\Gamma(x_p) + x_p^4 ight) \end{bmatrix}, \quad \mathbf{M}_1, \mathbf{M}_2, \mathbf{M}_3 \in \mathbb{R}^{1024 \times 1024}, \\[20pt] & \sum_{i=1}^{1024} x_i^3 - \prod_{i=1}^{1024} x_i^2 + \sum_{j=1}^{1024} \prod_{k=1}^j \sin(x_k^3) \leq 1, \\[20pt] & \int_{0}^{\pi/2} \prod_{i=1}^{1024} \sin(x_i t) dt \leq \exp\left(\sum_{i=1}^{1024} x_i + \sum_{j=1}^{1024} x_j^2 ight), \\[20pt] & \sum_{i=1}^{1024} \left(x_i^2 + \left|x_i ight|_{\mathbb{H}} ight) \leq 1024, \\[20pt] & \mathbf{A} \mathbf{x} \mathbf{B} + \mathbf{C} \mathbf{x}^2 \mathbf{D} = \mathbf{x}, \quad \mathbf{A}, \mathbf{B}, \mathbf{C}, \mathbf{D} \in \mathbb{C}^{1024 \times 1024}, \\[20pt] & \prod_{j=1}^{1024} \sum_{k=1}^j \sin\left(x_j^3 + x_k^2 + \zeta(x_k) ight) = \zeta\left(\prod_{i=1}^{1024} x_i ight) + \Gamma\left(\sum_{i=1}^{1024} x_i^5 ight), \\[20pt] & \prod_{i=1}^{1024} \cos\left(\sum_{j=1}^i \prod_{k=1}^j x_k^2 ight) = \int_{0}^\infty \prod_{i=1}^{1024} \sin(x_i^3 t) dt, \\[20pt] & x_i \geq 0, \quad i = 1, 2, \dots, 1024, \quad x_i \in \mathbb{H}. \end{array} ight. \]
@chicken_rice0123 my dad got 11/25 on the AMC 8Z
@chicken_rice0123 my dad got 11/25 on the AMC 8Z.
@chicken_rice0123 my dad got 11/25 on the AMC 8Z
are you going to do ACIME or MATRIX: docs.google.com/document/d/18Opri_3eWj0TzDYu4821E38fpBv0esxCz2rlLTqjOQo/edit?tab=t.0
guess that means you're taking MATRIX
me commenting on this from home during school hours be like
what are you doing bruh
why didn't you come
also we're playing kinetic energy in band
also in math we did a participation quiz and there's a paper homework so haha
the teacher didn't post the homework so yeah you can't get the hw
m
by the way the other way to solve it be like: we need to find the points where: \[ abla f = \lambda abla g \] - \( f(x, y) = 3x + 4y \) - \( g(x, y) = x^2 + y^2 - 10 = 0 \) - \( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} ight) = (3, 4) \) - \( abla g = \left( \frac{\partial g}{\partial x}, \frac{\partial g}{\partial y} ight) = (2x, 2y) \) \[ abla f = \lambda abla g \] \[ 3 = 2\lambda x \quad \text{(1)} \] \[ 4 = 2\lambda y \quad \text{(2)} \] \[ x^2 + y^2 = 10 \quad \text{(3)} \] from equations (1) and (2), we can solve for \( \lambda \): \[ \lambda = \frac{3}{2x} = \frac{4}{2y} \] \[ \frac{3}{x} = \frac{4}{y} \implies 3y = 4x \implies y = \frac{4}{3}x \quad \text{(4)} \] \[ x^2 + \left( \frac{4}{3}x ight)^2 = 10 \] \[ x^2 + \frac{16}{9}x^2 = 10 \] \[ \frac{25}{9}x^2 = 10 \] \[ x^2 = \frac{90}{25} = \frac{18}{5} \] \[ x = \pm \sqrt{\frac{18}{5}} = \pm \frac{3\sqrt{10}}{5} \] Using \( y = \frac{4}{3}x \): \[ y = \frac{4}{3} \cdot \frac{3\sqrt{10}}{5} = \frac{4\sqrt{10}}{5} \quad \text{or} \quad y = -\frac{4\sqrt{10}}{5} \] The points are \( \left( \frac{3\sqrt{10}}{5}, \frac{4\sqrt{10}}{5} ight) \) and \( \left( -\frac{3\sqrt{10}}{5}, -\frac{4\sqrt{10}}{5} ight) \). \[ 3x + 4y = 3 \cdot \frac{3\sqrt{10}}{5} + 4 \cdot \frac{4\sqrt{10}}{5} = \frac{9\sqrt{10}}{5} + \frac{16\sqrt{10}}{5} = \frac{25\sqrt{10}}{5} = 5\sqrt{10} \] \[ 3x + 4y = 3 \cdot -\frac{3\sqrt{10}}{5} + 4 \cdot -\frac{4\sqrt{10}}{5} = -\frac{9\sqrt{10}}{5} - \frac{16\sqrt{10}}{5} = -5\sqrt{10} \] the maximum value is \( 5\sqrt{10} \), and it occurs at the point: \[ \left( \frac{3\sqrt{10}}{5}, \frac{4\sqrt{10}}{5} ight) \]
bro took like 4 months to learn how to solve max of 3x + 4y under the constraint x^2 + y^2 = 10. also i have 200 subs again wow i wonder how that happened
about 1 - 1 + 1 - 1 + 1 . . . , that equals 1/2 in a cesaro summation, but it does not converge in a traditional sense. also, a/(1 - r) only works if it does not diverge(i.e. when |r| > or = 1, it does not work)
trump winning 247-214
he got 6 battleground states
shoooooooooooot
@@chicken_rice0123 trump only needs 23
trump might be second double president after Grover Cleveland in 1893
Hey chicken rice what aops class do u take which intermediate algebra ima join ur class
3938
my username is NinJaPro
brilliant way to go *****!
how was trick of treating
@@chicken_rice0123 did you see someone get run over by a car and become not alive
when
@@chicken_rice0123 some time in the 21st century
@@christophoilet ?
and why did bro come to my house during Halloween
ahhhhhhh howd you know
@@chicken_rice0123 wow I wonder how I knew
also i was doing amc 10a 2023, i got 141 so yay
about the problem did: 1. I wrote the question wrong, it's $x^3 - 22x^2 + 80x - 67$ instead of $x^3 - 22x^2 + 40x - 67$. 2.\frac1A + \frac1B + \frac1C = p^2 + q^2 + r^2 - (pq + qr + pr), not \frac1A + \frac1B + \frac1C = p^2 + q^2 + r^2 - 2(pq -+qr + pr)
brilliant
cool man
cool man
cool man
cool man
cool man
cool man
cool man
cool man
Cool man
peak content
omg
LOL the way u said haha magnus😂
Yay😊
cool😃
Tan have 1 point
Great video! Very inspiring! Please show how to use the labelpath command 🙏🙏🙏
We need a video covering USAMO 2024 Problem 6.
Nah we need a video covering: Three pains of twin women are married to twin men such that each woman's twin sister is married to her husband's twin brother. If we need to separate them into 6 mixed pairs, how many ways are there such that no one is paired with their spouse or their twin's spouse?
@@chicken_rice0123 Nah we need a video covering: Let n ≥ 2 be an integer. Let x_1 ≥ x_2 ≥ . . . ≥ x_n and y_1 ≥ y_2 ≥ . . . ≥ y_n be 2n real numbers such that \begin{align*} 0 &= x_1 + x_2 + \cdots + x_n = y_1 + y_2 + \cdots + y_n\\ \text{and }1 &= x_1^2+x_2^2+\cdots+x_n^2=y_1^2+y_2^2+\cdots+y_n^2. \end{align*} Prove that \sum_{i=1}^n(x_iy_i-x_iy_{n+1-i})\ge\frac{2}{\sqrt{n-1}}
nice
jimmy chess strategy : "oh no my (piece)"
hey yo find the max of (x+y): xy-(x+y)=gcd(x,y)+lcm(x,y)
@@chicken_rice0123 in the space C[a, b] of complex continuous functions on the interval [a, b], a standard inner product <. , . > is : 2. the maximum value of $f(x) = \frac{3}{4x^2+4x+3}$ 3. we have this latex code over here: \[ f(x) = \left\{\begin{array}{rl} -x & \text{if}\ x < 0 \\ 0 & \text{if}\ x = 0 \\ x & \text{if}\ x > 0 \end{array} ight. \] what does it represent. 4. the euclidean inner product on F^n, where F is either real or complex is defined as :
yay
furst
skibidi toilet 74 is coming out. not like you care or anything, just saying
yeah, i don't really care. the lore has become boring
also mrbeast is 1 mil away from beating t-series
not like you care tho
ua-cam.com/video/0bhFagWP_W0/v-deo.html
@@chicken_rice0123 not like I asked
Gj today tho
Bro dont hang pieces in hopes of stalemate 😂
also for problem 5 you can just use pythagorean identity.
also what is dat second method bro, like what da hail
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@@chicken_rice0123 $ \ban {chicken_rice0123} Reason : spamming$
also on saturday me, lucas, tianyi, jason, ryan, and anshul went to Westview high school to do math tournament. We got first place and we beat ptms
@@chicken_rice0123 alright but graph this wave : 12cos(pi/8 x + pi/24) + 12
@@christophoilet bro you do realize that learning calculus without foundations like algebra and geometry is trying to run without knowing how to walk
sin of 4π /3 rad, which is 240 degrees, is not hard to find bruh even I can find it. also why is there only introduction to geometry.
how did bro let this happen 💀
yo
Its funny how the seconds are on the right
Second violins
@@dingpanda oh you know music theory?
do you play piano
@@chicken_rice0123 not really I’m self taught
I play violin
hey chris create your own intro
captain america march ua-cam.com/video/cU4HbLvqufQ/v-deo.html
this felt very familiar and then i realized this is the same background music as lucas
ye
Wowow
bru it says translate to english, and when I press it, it changes from wowow to wow
@@christophoilet hehe
also what is up with the actually footage being much smaller.