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ama im in math class

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#1
melon_fan

no one is responding to this aight

#2
Powercvrd
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🗿

#3
LyCan52
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Unga BunGa

#4
QlQ
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jst 0:02

#5
AstroGalaxy
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tell ure teacher to give u some money

#6
CookieSir
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what math class?

#7
Piter
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What is integral of cos(x)

#8
An_dH
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Help me solve my electric circuits homework using Gauss methods bro

#9
denwhy
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give your math test to me, i can solve it for u

#10
Valgod
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Whats 12+9-7

#11
starxito_sike
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The coefficient of x^3 in the expansion of ( p + x/p )^4 is 144. Find the possible values of the constant p.

#12
Warlordwibz
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What is 6/2(1+2) = ?

#13
Meyourlightmare
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(k + 2) (1- [wz+h+j-q]²- [(gk+2g+k+1)(h+j) + h −z]²- [16(k+1)³(k + 2)(n+1)² + 1 - ƒ²² - [2n+p+q+z-e]²- [e³ (e+2)(a+1)2+1-0²]²- [(a² - 1)y² + 1 − x²]² – [16ry (a²-1)+1— u²² — [n+1+v - y]²- [(a² - 1)²+1-m²]² – [ai + k + 1 - l - i]²-[((a+u² (u² - a))² - 1)(n+4dy)² + 1 − (x + cu)²]² — [p+l(a− n − 1)+b(2an + 2a - n² - 2n − 2) — m]² - [q+y(a-p-1)+s(2ap + 2a - p² - 2p — 2) — x]² —[z+pl(ap)+t(2ap-p2-1)-pm]²)>0

#14
Liem
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Meyourlightmare [#13]

(k + 2) (1- [wz+h+j-q]²- [(gk+2g+k+1)(h+j) + h −z]²- [16(k+1)³(k + 2)(n+1)² + 1 - ƒ²² - [2n+p+q+z-e]²- [e³ (e+2)(a+1)2+1-0²]²- [(a² - 1)y² + 1 − x²]² – [16ry (a²-1)+1— u²² — [n+1+v - y]²- [(a² - 1)²+1-m²]² – [ai + k + 1 - l - i]²-[((a+u² (u² - a))² - 1)(n+4dy)² + 1 − (x + cu)²]² — [p+l(a− n − 1)+b(2an + 2a - n² - 2n − 2) — m]² - [q+y(a-p-1)+s(2ap + 2a - p² - 2p — 2) — x]² —[z+pl(ap)+t(2ap-p2-1)-pm]²)>0

i wholeheartedly agree

#15
s0ber
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starxito_sike [#11]

The coefficient of x^3 in the expansion of ( p + x/p )^4 is 144. Find the possible values of the constant p.

p^2 = 4/144
p = 2/12 = 1/6

#16
s0ber
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An_dH [#8]

Help me solve my electric circuits homework using Gauss methods bro

wasnt gauss law for electric field

#17
craftydroid
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Piter [#7]

What is integral of cos(x)

sin(x) + c

#18
ShanksFan69
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An_dH [#8]

Help me solve my electric circuits homework using Gauss methods bro

Bro don't take electrical i regret it to this day

#19
ShanksFan69
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Why is zero factorial 1

#20
propulsion
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s0ber [#16]

wasnt gauss law for electric field

for closed objects right?

#21
meyournightmare
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Meyourlightmare [#13]

(k + 2) (1- [wz+h+j-q]²- [(gk+2g+k+1)(h+j) + h −z]²- [16(k+1)³(k + 2)(n+1)² + 1 - ƒ²² - [2n+p+q+z-e]²- [e³ (e+2)(a+1)2+1-0²]²- [(a² - 1)y² + 1 − x²]² – [16ry (a²-1)+1— u²² — [n+1+v - y]²- [(a² - 1)²+1-m²]² – [ai + k + 1 - l - i]²-[((a+u² (u² - a))² - 1)(n+4dy)² + 1 − (x + cu)²]² — [p+l(a− n − 1)+b(2an + 2a - n² - 2n − 2) — m]² - [q+y(a-p-1)+s(2ap + 2a - p² - 2p — 2) — x]² —[z+pl(ap)+t(2ap-p2-1)-pm]²)>0

as a doc can confirm

#22
An_dH
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s0ber [#16]

wasnt gauss law for electric field

Gauss elimination

#23
UBClears
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The square root of 69 is 8 something

#24
uwukitten
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Meyourlightmare [#13]

(k + 2) (1- [wz+h+j-q]²- [(gk+2g+k+1)(h+j) + h −z]²- [16(k+1)³(k + 2)(n+1)² + 1 - ƒ²² - [2n+p+q+z-e]²- [e³ (e+2)(a+1)2+1-0²]²- [(a² - 1)y² + 1 − x²]² – [16ry (a²-1)+1— u²² — [n+1+v - y]²- [(a² - 1)²+1-m²]² – [ai + k + 1 - l - i]²-[((a+u² (u² - a))² - 1)(n+4dy)² + 1 − (x + cu)²]² — [p+l(a− n − 1)+b(2an + 2a - n² - 2n − 2) — m]² - [q+y(a-p-1)+s(2ap + 2a - p² - 2p — 2) — x]² —[z+pl(ap)+t(2ap-p2-1)-pm]²)>0

the opposite of night is day not light 😭

#25
Manas_o_0
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LyCan52 [#3]

Unga BunGa

Punga?

#26
melon_fan
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CookieSir [#6]

what math class?

math 2 I’m a freshman

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