#help-0
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Would it make sense to say the range of e^x would be all yER such that y>0 or would it only be y>0 and why?
imo, both are fine since its implied that y is a real number
but if you want to be extra clear, the first option is good
and I can write it like {yER:y>0} right
yes
Alright thanks
yw
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Closed by @wheat isle
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✅
if you can't type the set membership symbol ∈, it's better to replace it with in rather than E
{y in R : y > 0}
just a style thing
@wheat isle Has your question been resolved?
yER visually more appealing than y in I stand my point

mic drop
or how y’all mathematicians call it

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i dont quite understand where the red 1/x comes from, id like a bit of help ty
heres the function
je ne comprends pas trop d'où viens le rouge
mais ensuite le bleu?
$$\pdv{}{y}\left(\frac{1}{x}yh'_t(t)\right) = \frac{1}{x}\pdv{}{y}\left(yh'_t(t)\right)$$
_aplatypus
tu refais ta chain rule
un peu comme à la 1ère étape
t'as $$\pdv{}{y}\left(yh'_t(t)\right) = \pdv{}{y}\left(y\right) h'_t(t) + y \pdv{}{y}\left(h'_t(t)\right)$$
euh
il me manque un truc là
_aplatypus
ah voila (ça c'est juste une product rule pas trop de problème là dessus normalement)
la dérivée partielle tout à droite c'est une fonction composée
t dépend de x et y ici
_aplatypus
d/dt (h' (t)) c'est h''(t)
et dt/dy = 1/x
là c'est la fonction composée que j'utilise
@lyric geyser
yus
même en plusieurs variables, les dérivées ça reste très mécanique
c'est juste un peu plus le bordel à gérer
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assume there is an equilateral triange ABC with lenghts 10, and there is a circle that touches sife AB and AC and touches the outer triangle circle, and the points where the circle touched AB and AC arecalled f and g. what is the area of triangle AFG?
currently i have no idea on what to do
all i know is that AFG is an ewuilateral triangle i think
2 5 7 10 what is next number ?
read #❓how-to-get-help ,not here
also it is a very basic aritmathic series
So Say Answer
not an arithmetic series
we don't do that here
i just scanned trough the wuestion and thought it was a diffrence of 3 series
NO ANY ONE CAN ANSWER MY QEUSTION
I KNOW THE ANSWER BUT I WANT KNOW HOW MUCH YOU KNOW
then tell the answer kid
FUCK OFF
OK YOU KNOW I ASK YOU QUESTION
guys how much your precdicting hes underaged
got a diagram for this?
uhh
THE ANSWER IS 15
if you haven't then now's as good a time as ever to make one
IS TERM TO TERM ADD 2 THEN ADD 3
the question is in text so i have to make one myself
yes you do
PROVE
10+2=12
why
if you want genuine help, claim your own channel, don't invade others
and don't test helpers
do you guys think he is 10 or 11
and don't talk in all caps
or maybe 9
@heady panther stop.
WHY
@raw jetty do not encourage channel intrusion.
I COME TO USE APP
you're breaking several rules.
think this is more than worthy of a <@&268886789983436800>
yeah ok
BUT YOU NOT RESPECT TO ME
do you reccomend any website/app to draw those geometric stuff on phone?
draw on paper
GOOD BYE THAT NOT MY PLACE
i dont have a pen
i have paper but no writing tools
do you have a stylus for your phone?
i think you're going to be better of coming back to this problem when you can properly write down the work
sorry if the drawing sucks
the last time i tried to approach it i set the point where the smaller circle touches the outer circle to be like N or smth, then i tried to find it ysing the basis that AFG=FGN but i have no idea if that was even correct or not
@raw jetty Has your question been resolved?
@raw jetty Has your question been resolved?
<@&286206848099549185>
draw the radii from the centre of the small circle to the points of contact
and to A as well
Okay so here is what I did
Find radius of bigger circle and find height of triangle
Now set three variables GC=y
OC=r
OD=x
where O is centre of smaller circle and D is the bisector of BC
Now you have OG is perpendicular to AC
So you got two right triangles OAG and ODC
you'll get two equations in r,y,x
Observe the length DE is just diameter of larger circle minus height of triangle
where E is the point of contact of two circles
So you can get another relation between r and x
3 equations and 3 variables
solvable
@raw jetty
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All the time I invested...
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Find at positive x the minimum value x^4+3/x
$\frac{x^{4}+3}{x}$
This?
beard420
or $\frac{3}{x}+{x^{4}}$
sherif_player
So did you study calculus?
No
Oh
So solving this problem would be mush easier if you have studied it.
But in what way do you want it
I need to solve it.
How would you even do this without derivatives
Yeah that is what I wanted to ask
$\frac{-3}{x^2} + {4x^3}$
sherif_player
You said that you don't know calculus.
How did you know that it is the derivative?
Yeah then substitute this in the original expression which is this
To know the minimum value
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x^2+(a-b)x+(1-a-b)=0 find the condition for a for which roots are imaginary ∀ b∈R
I have been taught the method of taking D>0 and then making another quadratic in b and then taking its D>0 for the other part (which was real and unequal roots) but i am not able to understand what to use in this
First D<0
did that...and i get b^2+(4-2a)b +(4a-4+a^2)
And when you get the quadratic in b. It's Discriminant must be greater than 0 since b is a real number
Questions has said for all b belongs to R
yes i did it and i get a<1 but the answer is no solution
Wait there is no solution, right
yes
What condition do we have?
This is less than 0
yeah <0
What condition must a quadratic follow to be always negative
umm....d<0
And b<0
ohh yes
ummm lemme see what u wrote and think again lmao
isnt it a<0?
ohh so can i write it like this? b^2+(4-2a)b +(4a-4+a^2)<0 but for a quad to be <0 a<0 but it has >a so there is no sol? @wet nest
but if i make D<0 then i get a>1
Leave a
alright
got it
can u solve another question related to range of quad??
or i have to create another channel?
Send the question
y=(x^2-3x-2)/(x^2-2x-5) find range
instead of x there was tan but as tan has its range as all real values so i took it as x
What is the answer?
dont know
umm
ok so
first perform long division
or
lets do it without division
we see limits at both inf and -inf are 1
now we need all critical points
they are
uhh i havent learn till that much
zeroes of f'(x) and places where its undefined
you dont know diffrentiation yet?
we are taught a method of taking the denomintor to left and making a quad in x and solving by suing D formula
just a little bit to solve physics questions.....in which i am not good
diffrentiation is simpler method
you dont happen to be in 11th grade in india do you
i am lmao
knew it
lmao
ohh in maths? or in physics
the discriminant method is annoying af
from youtube i meant
lack of diffrentiation makes this hella annoying
ok so
take denominator to other side
yeah it does but i gotta do it by this method
lol
write it as quadratic in x in standard form
<@&268886789983436800>
done it and got roots as (40 ±root(-32))/48
i meant solving the quad in y gives that which we got by D of z
x
wait so
just close this channel and go to a new one
yeah..i'll ping ya there
jeez
..close
im sorry, taken care of
.close
Closed by @autumn vigil
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oh....i closed
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Hey, I am currently working on my terrible math, and seem to not be understanding this rather basic principle.
What exactly is the reason that in this exercise y isn't 2/(p+q) but 8/(p+q+4), why can't you simplify?
basic fraction simplification involves
multiplying/dividing the whole of the numerator and denominator by the same value,
(effectively multiplying by 1)
and/or cancelling a common factor
you can't divide the numerator by 2
while only dividing only a single term in the denominator by 2
in your case it's not quite that,
but either way, not valid
which changes the ratio
Makes complete sense, just so I am sure I understand correctly, does this imply that (even though obviously undesirable) this would be correct?
mhm
don't overthink the value of 4/4
$y = \frac{2}{\frac{p}{4} + \frac{q}{4} + \frac{4}{4}}$
Hayley
yeah, so adding a +1 would be correct in this case at the end presumably
or am i misunderstanding still?
+1 in the denominator, in the place of the 4/4,
$y = \frac{2}{\frac{p}{4} + \frac{q}{4} + \cancel\frac{4}{4}1}$
Hayley
Compile Error! Click the
reaction for more information.
(You may edit your message to recompile.)
oh god
lol
$y = \frac{2}{\frac{p}{4} + \frac{q}{4} + \cancelto{1}{\frac{4}{4}}}$
ℝamonov
good enough
okay thanks for the answers & help guys, this makes sense now; even though latex doesnt seem to enjoy this very much :p
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For e I got this
How do I put this as an inequality
Is it x bigger than or x smaller than 1?
"increasing" means the derivative is nonnegative
Meaning?
what is unclear
Yeah Ik that but how does that help
have you seen these symbols before
< , >
Yeah inequalities which is supposed to be the final answer
< is "less than"
e.g. 2 < 3 is a true statement
and > is "greater than"
e.g. 1 > 0 is also a true statement
Ik that
But here I got 2 answers for x that equal the same
Idk what to do with that
!show
Show your work, and if possible, explain where you are stuck.
This…..
where are your 2 values?
what do you mean "equal the same"
(X-1)(x-1)
If I put ‘‘em in a number line
The answer could be positive if it’s less than or bigger than 1
do you know (x-1)(x-1) = (x-1)^2 ?
Yeah
are squares of numbers ever negative?
No
or equivalently, does the graph of (x-1)^2 ever go below the x-axis
Do I just write all values of x are increasing?
just use words
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Can you state the commutative property?
What
commutative property in math means you can switch the order of numbers being added or multiplied without changing the result.
oops
@proper heron Has your question been resolved?
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v3ryberry
It is
your work all looks fine, but you didn't actually show the question
$\frac{1}{4} \left(x + 3\right)^1 = \frac{1}{4}x + \frac{3}{4}$
dldh06
I'm assuming the first image is the answer key
yeah the first answer is the answer key
ahh i see thx
I was just wondering if my answer was wrong
no wonder it looks correct haha
i have no idea why they felt the need to put the 1 exponent
It's not clear what your answer as I don't see a y = answer
I see all the math, but what's the final answer you got?
is that the equation in the upper right corner?
best if you can write stuff in order and maybe put a box around the answer 😁
No you guys answered it already
I didn't understand why my answer was different from the answer key
but if both the answers mean the same thing then it's ok
Answer key was just in a different format, that was all
Because don't forget x/4 = (1/4)x
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Hey guys I am currently in class 11th And on 7/7/23 I have my maths exam can Anyone Give me overview or idea about what exactly linear inequalities is ?
thank you Bot you owe me
What?
@harsh pewter Has your question been resolved?
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Help plz


HAHAHA
95% of people can’t solve!
More than 95

!elliptic curve meme
🍎 = 154476802108746166441951315019919837485664325669565431700026634898253202035277999
🍌 = 36875131794129999827197811565225474825492979968971970996283137471637224634055579
🍍 = 4373612677928697257861252602371390152816537558161613618621437993378423467772036
!solved
Ok at least explain why it can’t be solved then
In number theory, Fermat's Last Theorem (sometimes called Fermat's conjecture, especially in older texts) states that no three positive integers a, b, and c satisfy the equation an + bn = cn for any integer value of n greater than 2. The cases n = 1 and n = 2 have been known since antiquity to have infinitely many solutions.The proposition was f...
I wrote the proof in the margin of my book but i cant find it
Sorry
I would be able to if i knew where it was tho
No issue, shouldn’t be too hard to remember then
Somehow its slipping my mind
btw srry if this isn’t math related but you haven’t used your boosts ok me out srry once again 
@woeful current Has your question been resolved?
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There are two parking garages in Beacon Falls. Garage A charges $7.00 to park for the first 2 hours, and each additional hour costs $3.00. Garage B charges $3.25 per hour to park. When a person parks for at least 2 hours, write equations to model the cost of parking for a total of x hours in Garage A and Garage B.
help?
try writing out a table of numbers for 1,2,3,4,5 hours
i need an equation first
no you don't
Garage A charges $7.00 to park for the first 2 hours, and each additional hour costs $3.00
When a person parks for at least 2 hours, write equations
but you don't need an equation to do this
so 14 for the first 2 hours
right?
nope
how?
Garage A charges $7.00 to park for the first 2 hours
so 14 for those 2 hours
again, no
how?
$7 for two hours
OOHHHHHHHHHHHHHHHHHHHH
doesn't matter if you park for 1 minute
omg
do this for each garage
garage A
7, 7, 10, 13, 16,
Garage B
3.25, 6.50, 9.75, 12, 15.25
A looks right. B does not
are you done?
but 1 miniute is not an hour
oh no you need equations
im confused with the equation
try plotting
have you seen equations of lines before
if i park for 1 minute, is that still 7 dollars?
y = mx + b maybe
yes
how tho?
garage A is harder than garage B. i suggest working on B first
just english
3.25x
great you're done with B
but 1 minute is not 2 hours
indeed it isn't
so its not 7
no. it is
how tho?
ya but we havent even made an equation for first graph
you specifically said you didn't want to graph
you don't need to graph to find the equation for garage A
sorry we havent made an equation for garage a
this is a type of equation
watch this lesson
https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:absolute-value-piecewise-functions/x2f8bb11595b61c86:piecewise-functions/v/piecewise-function-example
try making a piecewise one
huh?
but dont i need an equation?
.
.
let me watch it
that picture is itself a piecewise defined equation
use your numbers from here to similarly define one
7, 7, 10, 13, 16,
how tho?
like if i use 7
and do it from 1-5?
what's 5 here
the hours?
7 is the price for the first two hours
at hour 3, the price is $10
and at hour 5, the price is $16
so 7 wouldn't be the price for the first 5 hours
bro so we use all the numbers from price 7-16?
and graph it from 1-5?
can we just write and equation from the table?
1:7
2:7
3:10
4:13
5:16
try it yourself
it wont work
like i cant find anything
and the calc cant either
hello?
Your piecewise function should be one function for x <=2 and another function for x> 2
Like the -5 here
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Status 1
Do you remember how to complete the square @digital moon?
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Try to simplify $\frac{a^2+a(a+1)+(a+1)^2}{a^3(a+1)^3}$ and see what you get
Cain
Try to get telescoping series
It's actually 1/a^3 - 1/(a+1)^3
I got to go now
Good luck
@alpine sable Has your question been resolved?
,w expand a^2 + a(a+1) + (a+1)^2
huh, I guess so
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Oop
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1+1=?
3
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read the given distances carefully
i probably shouldnt say this but have u tried googling the question lol
there could be a step by step solution thingy online that u can use
fair point
considering how lazy and stupid my teacher is she probably hasnt changed the hw in years
considering she doesnt even have the time to respond to emails
dang 💀
o;-;
I can't tell what you did wrong
i mean you did something wrong
but I can't tell what it was, how did you get those values? like the angle
try gautmath
triangle calculator
mine comes out near to 243 ;-;
that's what I got as well
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I’m unsure how to approach part B and C
For b, you just have to substitute t = 0 into the equation since that's what it means
For c, it is very similar
So N = 450(3)?
For part B
Actually
It would just be 450 right
450(3^0) = 450(1)
Yes
And part C i is 450(3^3)?
Assuming you sub in 3 for t
Actually
450(3^(0.9*3))
Cheers
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Are the steps i used here correct ?
I got the answer but im kinda skeptical about the simplification
Like where i cancelled 2 and 12
Is it right to do that step ?
near the beginning you went from $\frac{r}{1/2}$ to $\frac{r}{2}$ and that doesn't seem right
Hayley
cancelling 2 and 12 is fine there
Oeh okay
Nono its r/2 i made a mistake
I assume you want to find the distance of the point (x1,y1) intersection of both lines from (3,4)
Oh what about the x substitution in the beggining
Yeye
this one? I assume you have an equation somewhere that says 10x + 5y + 10 = 0
One sec
It's 12
Yeh
i was a math major so numbers bigger than 10 are out of scope for me
oh! yeah the solving each of the equations for x and y is fine, once I figured out that your $r$ wasn't a $\pi$
Hayley

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this is the problem i am working on
i was able to solve the differential equation but i am stuck on how to find the domain
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do you know what y(1) = 0 means?
I’m not entirely sure could you clarify for me
is it when x is 1 y = 0?
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can someone pls explain to me why the inequality reverses
how is it negative?
np
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need some help here
Show your work, and if possible, explain where you are stuck.
Mr. Bot, not possible cause I have no clue where to start. That's why I need help
<@&286206848099549185>
I finished that problem, now I need help with a new one
no
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When I set this up i get m = c1e^x + c2e^-x
does it matter which order i do this in
c1 e^-x + c2 e^x is the same
okay i just want to verify, my solution manuel does it differently so i wanted to make sure there wasnt some rule i was missing
thank you 🙂
okay another question
Im getting down to the point where im getting the determinate of the wonskian, and doing the thing thats similar to cramers rule.
im getting 2, e^xcoshx and e^-xcoshx
so i simplifly the 2 expressions on the right, divide them by 2 and then take the integral?
IE e^xcoshx = 1/2(e^2x +1)
then divide that by 2 and take the integral right?
i don't remember enough about odes to help you there, sry
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1shero1
well if you combine all these fractions again you will be left with (some polynomial)/q(x)
and that (some polynomial) is p(x) if you choose the A_i correctly
there are proofs
but none that I know that give more intuition
than just "you can do it, here is even an algorithm for how to"
Idk, but I guess you could reason that any constant A_I times (x-a_i) will also be a polynomial as long as A_I is not 0
so when you do add these up
it becomes this
to be more formal, (sum of A_i)x - (sum of a_i)(sum of A_i)/q(x)
but you dont want that because?
writing this down formally is just an exercise in using sum and product notation
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Need help pls
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help
What step are you on?
1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above
methisalwaysright
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the only nontrivial factors of a quadratic are linear
You have to specify over what field it is not factorisable
usually Q in the high school sense
Likely that it has nor real roots
But it lacks context as Zac said
Or Sac or EAC idk how to write it sry
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not sure about irrational roots
i think we regard irrational roots as not irreducible
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do you know how percentages work
well you either do or you don't
why urgent?
are you in a test?
we don't help on tests.
it's against the rules and even bannable.
well that sorted itself out relatively quickly
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Could anyone help me with trigonometry and finding side lengths using trig ratios and multiplication
Question pic on the way?
one sec lemme pull up the powerpoint
so the angle was 32 degrees and the side length was 4.6cm
Use sine ratio
draw a diagram and label all these and send a pic
no
it is the ratio of opposite side and hypotenuse in a right triangle
oh like SohCahToa?
Yes
im sorry but I have absolutely zero clue what that means
do you think you could walk me through what the question is asking?
[sin(32) = \frac{4.6}{x}]
Akira
@primal flume solve for x
Start with cross multiplication first
Yes
k
@primal flume also did you understand why I did it like this
not really
yep
You need always remember sohcahtoa definitions
yep i have them in my notebook
That's good
did i do that right?
isnt it sin32 times x?
oh yeah
,calc sin (32)
Result:
0.55142668124169
8.341997506616519?
Akira
yeah sorry just trying to process what i just did
It should be like this
....
ye
Isn't the question asking to find AC? Then you need find the Hypotenuse
That's why we used the sine ratio
Yeah
Yep
and you use x because its an unknown value
Yeah
but why use cross mulitplication?
Akira
@primal flume
By cross multiplying these two fractions, you can easily solve for the variable at hand thanks to the reduction to only one equation
?
Yes
ah yes
nearly
Okay
Angle CB?
the opposite of a right angle triangle is the side opposite to the non right angle
isnt it?
It's the hypotenuse
what?
oh ok
so we figured out the adjacent right?
or sorry i measured the adjacent instead of the opposite
They already given you the opposite
Not sure if you need to find Hypotenuse or adjacent
Let's find adjacent
Tan or cos
But your question asking to find AC
Which is Hypotenuse
It seems like someone already said it
^^
is this the adjacent side?
Use cos
It's fine
Oh okay
Is this a different one?
so i was supposed to measure the opposite but i measured the adjacent
I see
sin(32)x = 2.8?
oh
Have you heard of wolfram like do you know how to use it?
no
,w sin(32)x = 2.8?
k thanks
np
do you have to go?
@ebon sparrow you probably wont see this but thanks for all your help :)
mods ill close it soon i just have to take notes
.
.
I just opened my pc
you can solve the cos ratio just like this
you can watch khan academy for this if you wish
still stuck?
no im just figuring out the other questions
this one right?
oh so you're moving to another question
yep
then you're all set
you can still open a new channel at any time
ok
i might just keep this one open because im nearly at the next question
which i dont know
can i see it
okay so you need to use cosine ratio this time
what do you need find for A or H?
what are the numbers given
i gotta measure them one sec
okay
,w tan(21)x = 3.6?
oh that was another question
oh lol
no thats opposite side
