help-43

[old mica]

for example, if you stretch f(x) vertically by a factor of two, you're going to end up with 2sqrt(x + 1) - 4, not -2 as you might think

[jovial widget]

im really confused with the sqr (x+1) part uhh the d value

since its in a bracket i would have to transform it to the right since its a negative

but because of relativity would that work like im moving to the right 5 time or 4 times or 3 times 😓

[old mica]

adding a negative number under the square root shifts the graph to the right by that amount c:

in f(x), the graph terminates at x = -1 (because for x < -1, we'd have to take the sqrt of a negative)

in g(x), it terminates at x = 4

so we need to shift the graph how many units, and in which direction?

we just wanna transport -1 to 4

[jovial widget]

so +5

?

[old mica]

the graph gets shifted +5 units yeah

that's 5 units to the right

[jovial widget]

TYSMMM!! i think i understand a buncha the gaps now tyy

[old mica]

happy to help, and also: welcome to the mathcord!

[jovial widget]

oh one more question im so sorry

for this one the 1 in the f(x) parent function would represent the k value and not the a value ?

for reference this is like the letters we use because i know its different sometimes

im like 90% sure its the k value but just want to verify 😅

[old mica]

the 1 is the d value

or rather, -1 is

for a rational function the a, k, d, c looks like

,, \frac{a}{k(x - d)} + c

[boreal girder]

higher!

[jovial widget]

ohh so it would be a

[old mica]

oh, you meant the 1 in the numerator

yeah, that's a

[jovial widget]

tyyy again!!!!! tysm

[jovial widget]

i have another question

😅

i dont think the functions shown are the either of the parents of the transformed function im lookign for unless im missing something

[scarlet saffron]

i think it's asking you to graph y = f(x+2) - 1 for both f(x) = sqrt(x) and f(x) = 1/x, i.e. you do y = sqrt(x+2) - 1 and y = 1/(x+2) - 1 separately

[jovial widget]

OHHH so i sub in the function ok thats quite obviious i shouldve got that

this iswhy i dont do math at 1am . THANK YOU

what would be the domain and range of a reciprocal like the one pictured in blue? would it be just xer and yer?

[sterile crag]

-1 isn’t in the range

and -2 isn’t in the domain

[compact pewter]

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[chrome hemlock]

help me anybody, i dont want a long explanation i legit have been stuck on this forever and just want an answer 😭

[compact pewter]

Please don't occupy multiple help channels.

[cerulean steeple]

!noans

[compact pewter]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[cerulean steeple]

find the area of the segment

[chrome hemlock]

i dont know how😭i would if i could

[cerulean steeple]

the area of the segmnt is just the sector minus the triangle

[unkempt sail]

Do you know area of sector?

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[woeful zodiac]

Hi! Does anybody have a link to a nice proof of the property that if a/b and c/d are farey neighbours then |ad-bc|=1

[compact pewter]

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[cerulean steeple]

found this online, maybe itll help

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[woeful zodiac]

can you share the enture proof?

[woeful zodiac]

.eopen

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[hushed magnet]

its just on wikipedia

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[livid laurel]

Suppose 2^k - 2 | 2^(2^k) - 2
Is it always hold that k | 2^k

[inland nacelle]

If k were to | 2^k, then k is a power of 2

[turbid trench]

$$Suppose 2^k - 2 | 2^(2^k) - 2
Is it always hold that k | 2^k$$

[inland nacelle]

OR k = 1

[livid laurel]

Wht

[turbid trench]

$$ Suppose (2^k - 2) |[( 2^(2^k)) - 2]
Is it always hold that k | 2^k $$

[inland nacelle]

aiyo

[livid laurel]

So its true or no

[inland nacelle]

!noans

[compact pewter]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[livid laurel]

I read a sollution and says that it is trivial / wellknown

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[tall hare]

How can I solve this question?

[compact pewter]

@tall hare Has your question been resolved?

[unkempt sail]

isn't this a bernoulli equation

take 5y to LHS

and multiply by integrating factor

@tall hare

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[unkempt sail]

yes

this

is what you have to do

[tall hare]

but how do I integrate the e^5x?

[unkempt sail]

thats simple

once you take 5y to LHS

[tall hare]

i meant (6x^2+10x+2)e^5x

[unkempt sail]

notice that IF is e^-5x

[tall hare]

since that's Q(x) right?

[unkempt sail]

which you can take from the e^5x in RHS

[tall hare]

I've only seen examples where there is dy/dx alone

what do I do if there is something like this in this case if we take e^5x to the left hand side?

[unkempt sail]

Follow me

When we take 5y to LHS

The equation becomes dy/dx - 5y = blah blah
Blah blah being the term involving x

Here try to identify what P(x) and Q(x) are @tall hare

[compact pewter]

@tall hare Has your question been resolved?

[tall hare]

P(x) = -5

and Q(x) = (6x^2+10x+2)e^5x

@unkempt sail

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[still sundial]

Hello! I'm using conic equations to model an aircraft fuselage cross section. I would like to use this shape - However, I would like the min/max y values to be 0 and 2, whilst the min/max x-values to be -0.75 and +0.75. How could I achieve this whilst keeping the shape as similar as possible? Thanks

[still sundial]

<@&286206848099549185> assemble, thanks!

[pallid rune]

!15min

[compact pewter]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[pallid rune]

but anyway,

!status

[compact pewter]

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 was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[humble mortar]

use coordinate transformation

[tired bear]

thankfully, desmos already has the edge point coordinates ready
maybe you can use that and manipulate the equation you have

[humble mortar]

x -> x/a will stretch the plot a in the x-axis
x -> x - b will translate the plot right by b units

same for y

[tired bear]

yeah

[opal blade]

2.53125*(x / 1.1516)^2 + 2.2*((y - 0.94944) / 1.2203)^4 - 0.3 * ((y - 0.94944) / 1.2203) = 1

My humble attempt. (very messy)

[still sundial]

Was it not 15 minutes?

xx34 to xx49

[still sundial]

Thank you, this looks great! I'll try to make final adjustments from here

[opal blade]

Glad to be of help :D

[still sundial]

I managed to adjust the equation slightly to 2.529*(x/1.1516)^{2}+2.2*((y-0.94)/1.2203)^{4}-0.3*((y-0.94944)/1.2203)=1, which essentially is correct to 2 decimal places each way (which works for me, as I'm working to the nearest 0.01m/1.0mm

Thanks

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[jovial widget]

i understand why because it makes the value =0 but i dont understand how you get that

[sterile crag]

start at the bottom of the y axis and scan upwards and whenever there is like a gap somewhere like when y = -1 the range doesn’t include the gap

there is no x such that f(x) = -1. maybe it’s hard to see, but the curve on the right is a little above -1 and the left it’s a little below -1

[jovial widget]

so for this one x =/= 4?

[sterile crag]

do you mean y?

[jovial widget]

oops yes

[sterile crag]

yea

[jovial widget]

and x =/= 0?

i think

[sterile crag]

yea

[compact pewter]

@jovial widget Has your question been resolved?

[jovial widget]

im struggling to find the k value for this function

this is what i have so far

[compact pewter]

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[jovial widget]

.reopen

[compact pewter]

✅

[placid knot]

is this transformations

[jovial widget]

yes

[jovial widget]

this is wrong i just realized i have the numbers off it should be x2 what i have

[placid knot]

and k is like a variable that indicates one of the transformations

[placid knot]

ooh okay thanks my bad haha

[jovial widget]

and i think thats right im just struggling to find how much its stretched by

i know its a value in between 1 and 0 because it stretches

[placid knot]

wait yeah i see what you mean

its really weird cos you can see its stretched but there's no clear x coordinate other than 0

[jagged blade]

If you move 2 units in x from the vertex, how much does the normal curve change in the y direction?

[placid knot]

oh wait yeah we have the vertex

[jagged blade]

@jovial widget

[placid knot]

we have the vertex to which you can typically denote the function as y = a(x+h)^2 + k
which is almost exactly what you have, just without your vertical stretch 'a'
you can solve for 'a' here though by using another given point

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[rich yacht]

Shouldnt there be a plus C on the right hands side?

[quiet sentinel]

C is just a made up constant, like you could argue that the +c of the integration is actually -c, making it positive on the other side

It doesnt matter where it is as long as a constant is taken into account

[rich yacht]

oh ok

thanks

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[pure orchid]

Let f: \mathbb{R} \to \mathbb{R} be an odd function, i.e.,
f(-x) = -f(x) \quad \text{for all } x \in \mathbb{R}.
Suppose that for all x \in \mathbb{R}, the following inequality holds:
(|x| + 2) f(x) \leq |x + 1| - |x - 1|.
Find the expression (formula) for f(x).

[pure orchid]

How

Help please

[open plaza]

isolate f(x)

then draw a rough graph of the function on the right hand side

[compact pewter]

@pure orchid Has your question been resolved?

[pure orchid]

close

!close

[open plaza]

.close

btw did you solve it yet

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[sand summit]

I'm confused, as I thought that the electric field inside of a closed shape was zero at static equilibrium

I tried that for my first answer, and it was wrong as well

[kind crane]

where does it say it's at static equilib

use gauss' law

[sand summit]

that's a really good idea

We've just been learning it 🤦‍♂️

@kind crane it's charge density over Epsilon0, right?

Is the number supposed to be so high?

Nevermind, ignore that

Is this the right process?

1. Determine charge density with total charge and sphere volume
2. Calculate total charge for a sphere with half the radius
3. divide total charge by epsilon0

[compact pewter]

@sand summit Has your question been resolved?

[sage geyser]

New here

[sand summit]

do you know electrostatics?

[sand summit]

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[vale peak]

I watch a youtube video and still did not understand how they got the solutiuon

[vale peak]

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[keen granite]

I assume you got it? 😅

[vale peak]

can i hear your thoughts on how to solve it?

[keen granite]

.reopen

[compact pewter]

✅

[keen granite]

we don't only just care about whether the the function is increasing or decreasing but by how much it is increasing or decreasing

notice that for option C the before it switches that graph would have us show that its increasing faster than when it starts decresing

but this is not true when we look at our red graph

[vale peak]

why would B also be wrong?

[keen granite]

there it's decreasing then increasing

[glass turret]

because the graph is increasing first then decreasing

[vale peak]

What about for this problem

[devout latch]

Where is the slope of the red curve 0?

[vale peak]

no idea, this is how i got the question

[devout latch]

the decreasing lines give you different possibilities for local rate of change of the function, whether it goes up or down and how much so around each point

[vale peak]

yea the slope is 0

[devout latch]

for example at x=-3 this slope/rate of change is >0 for all the lines proposed and the red curve is indeed going up at x = -3

[devout latch]

not for all x values, if the slope was always 0 you wouldn't get a red curve like this, just a flat constant line.

[vale peak]

so the answer would be orange?

[devout latch]

yeah but why?

[vale peak]

becasue the function value is at 0

at x= -1

[devout latch]

yep and that's the only "slope curve" proposed where the slope = 0 at x=-1 where the red curve's slope is 0

[vale peak]

thanks

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[pallid crypt]

can someone explain the third line

[pallid crypt]

no clue how that integrated to 1/2 sec^2(x)

[worn loom]

$$ \frac{\dd{ }}{\dd{x}} \sec(x) = \sec(x)\tan(x) $$

[boreal girder]

StrangeQuarkAL

[vale star]

See if this helps

Find the integral

If sec²x now

Some what like this,complete it in this way

[worn loom]

$$ \frac{1}{2} \tan^2 (x) $$
is also a valid antiderivative

[boreal girder]

StrangeQuarkAL

[pallid crypt]

interesting

my first thought was

split it into (sec)(sec.tan)

and then do parts again

but i have no idea what sec integrates to haha

[worn loom]

it's not that nice

[vale star]

Sec²x integrates to tanx
And u were already given that

So y bother splitting

[pallid crypt]

yeah thats why i scrapped it

[worn loom]

$$\int sec(x) \dd{x} = ln(|\sec(x) + \tan(x)|) + c $$

[boreal girder]

StrangeQuarkAL

[pallid crypt]

yikes thats long

[worn loom]

but yeah, you don't need that

[pallid crypt]

alr thanks

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[slim lodge]

Can anyone help me with this dumb matlab thing

[slim lodge]

btw i have no idea what kinda problem this is i never took physics or whaytever

im js a little confused on where im suppose to start

[rotund wave]

do you understand the problem

[slim lodge]

no

[rotund wave]

which part

do u understand V

[slim lodge]

i just know thats the equation for voltage at a specific electric potential

they give me some variables

[slim lodge]

i lowk just dont understand this

i know it goesgives me the y valueto put the y-axis on

and x

i think

idk what im suppose to do with the points

i know meshgrid takes in 3 arguements

x y and z

uhhh

yea

[rotund wave]

find V at those points

V is a function of (x,y)

q1, q2 are fixed numbers (charges)

r1, r2 depend on position (x,y)
they are the distance from (x,y) to the charges

[slim lodge]

okay

[slim lodge]

how am i supopose to get the values for r1 and r2

?

[rotund wave]

where is the first charge?

imagine it's an electron or smth

[slim lodge]

uh first charge is q1 = 2*10^-10? at (0,3,0)

idk

💀

[rotund wave]

what is the location of the first electron

[slim lodge]

im not really sure, i see that r1 and r2 are the distances of the charges from the pooint in question

but im not sure what that tells me

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[gritty rain]

So im kinda confused on how to start this problem? Am i supposed to find all the ways that numbers could add up to 10?

[open mason]

what math level is this

[kind viper]

yes

[hearty lake]

Yes.

[quartz yoke]

you can do shorter

[kind viper]

note the minimum sum is 7 so there actually aren't a lot of ways to do it

[rotund sphinx]

It’s an old AMC 10 question

[gritty rain]

Why is the minimum sum 7?

oh

nvm

[quartz yoke]

1 at least

[kind viper]

somewhat above basic probability

[quartz yoke]

7 times

[gritty rain]

mb

[open mason]

damn bro idk this

i thought i could help

[gritty rain]

So once i get those possibilities what do i do with them?

[hearty lake]

Get a count of them.

[quartz yoke]

soit x1...x7 7 v.i.i., la loi conjointe S = sum_1^7 x_i a P(S=10)...

is faster when you notice a pattern in the law of probability. which would also give you answer for 11, 12...

[gritty rain]

ah

do you just know what place this is from lol

[quartz yoke]

sorry i don't know the english terms

wdym

[kind viper]

that's your numerator lol

[gritty rain]

ah ok

thanks

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[sudden raven]

i got 25 with p=14, q=11 i want to know if its right

[hushed magnet]

that would be easier if you showed your solution

few people are willing to just solve the problem on their own just to compare their result to yours

[sudden raven]

sure sec

i will write it again because the sol is written in another language

[compact pewter]

@sudden raven Has your question been resolved?

[sudden raven]

my solution

[compact pewter]

@sudden raven Has your question been resolved?

[compact pewter]

@sudden raven Has your question been resolved?

[compact pewter]

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[signal valley]

help

[signal valley]

can anyone explain to me

why the +2 means you move two units to the left

instead of two units to the right

and vice versa

seems counterintuitive

[humble mortar]

because x+2 means "move the grid (not the graph) 2 units right"

likewise, 2x means "scale the x-axis (not the graph) by 2 units"

[signal valley]

well that doesnt make any sense

oh nvm

i see thx

which case do i use

i dont understand what that symbol means

the one that looks like a 3 but facingthe other way

is it the bottom one

(2, infinity)

[bleak dock]

that means $\in$

[boreal girder]

south

[signal valley]

yea

what is that symbol

[bleak dock]

yes, because 7 is in the range (2, infinity)

[signal valley]

yea but what that symbol mean tho

[bleak dock]

it means 'in'
properly, it means that x is a member of the set....

[humble mortar]

That means
\begin{enumerate}
\item if $-\infty < x < -7$, then $g(x) = x^2 - 5$
\item if $-7 \le x \le 2$, then $g(x) = 9x - 17$
\item if $2 < x < \infty$, then $g(x) = (x+1)(x-5)$
\end{enumerate}

[boreal girder]

Shuba

[signal valley]

oh

so it just means if

like a conditional

ok thx

[humble mortar]

(2,5) is an interval meaning every number between 2 to 5, not including 2, and not including 5.
(2,5] is an interval meaning every number between 2 to 5, not including 2, and including 5

[bleak dock]

no, the $\in$ symbol does not mean if

[boreal girder]

south

[bleak dock]

the 'if' part is given by the big case bracket after "g(x) ="

[humble mortar]

$x \in (2,5]$ means "x is in the range 2 (not inclusive) to 5 (inclusive)"

[boreal girder]

Shuba

[signal valley]

ahh ok thx

[bleak dock]

yeah that's the thing you wanted to know right

no worries!

if you're done type .close pls

[signal valley]

.close

[compact pewter]

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[compact pewter]

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[gleaming fox]

Hi so I am confused on this predicate logic question, specifically 4.10b. I am just having trouble understanding the Edge(x, z) ^ Path(z, y) portion of the solution, and how it acts recursively. Could someone kinda provide their own perspective so I can possibly understand the solution further? Thanks

[compact pewter]

@gleaming fox Has your question been resolved?

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@gleaming fox Has your question been resolved?

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[hasty quiver]

I need help here

[thorny urchin]

what have you tried

[hasty quiver]

I reached, 6sin(x)^2 +2sin(2x)-1 >=0

[glass turret]

hm

[hasty quiver]

then i tried to break the sin2x to cosxsinx

but that wont get me anywhere

[thorny urchin]

then i tried to break the sin2x to cosxsinx
do that

but keep the cos^ and sin^2 as they are

[hasty quiver]

i did

mhm

lemme try

brb

[thorny urchin]

and try factorising that

[remote flower]

ig u can make a quadratic

yea

[hasty quiver]

got it

ty lads

[unkempt sail]

oh yeah instead of converting into quadratic
you could maybe convert the square terms into cos 2x

[remote flower]

yeah but sin^2x will remain ig

[unkempt sail]

no

sin^2(x) = (1 - cos2x)/2

cos^2(x) = (1 +cos 2x)/2

use this

[remote flower]

oh u mean it like that

[unkempt sail]

yes

[hasty quiver]

doesnt work

[unkempt sail]

you sure?

[hasty quiver]

Yes

but quadric way works

[compact pewter]

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[signal valley]

help

[signal valley]

Are there any exponent laws I’m missing

[kind crane]

,tex .exp rules

[boreal girder]

riemann

[kind viper]

1/5^2 is not 0.5

[signal valley]

And are they all accurate

[kind viper]

so #2 is incorrect here

[signal valley]

Oh

[kind viper]

#1, #3 and #4 are correct but the 5/3 is a bit distracting if you want just a pure example of the law

[signal valley]

I meant you could multiply it by 5

[kind viper]

mmm

[signal valley]

It’s not a decimal point

Like they’re equivalent

They are all correct right

[kind viper]

what is this supposed to mean

[signal valley]

Remember how you helped me earlier

You were like dividing by 1/10 is the same as multiplying it by 10

Or something like that

That’s what the *5 means

Like I’m showing it’s equivalent

[kind viper]

yeah but uh

[young raft]

yes but here you are dividing by 5^2

not by 1/5^2

or 1/5

[kind viper]

you're trying to show an equivalence between a number on its own and an operation...

you're kinda off the mark there

i don't think there's even a way to salvage this

like $5^{-2} = \frac{1}{5^2}$ is correct but then the ${} \cdot 5$ bit is... idk what to do with it personally

[boreal girder]

Ann

[signal valley]

So you’re saying it’s incorrect

[kind viper]

more than that, im not even seeing a correct idea buried under it :P

[boreal dawn]

Never seen that notation in my life...

[signal valley]

ok

listen

ill tell u what im thinking ok

so do u remember this problem

how you helped me

you explained to me that dividing by 1 / y^10

is the exact equivalence of multiplying it by 10

you will get the same answer

[kind viper]

y^10 not just 10

anyway this is because you actually had a thing being divided by 1/y^10

while in your 5^-2 example there is nothing like that

my thing was never meant to be put into notation like that

[signal valley]

ok

what im just trying to say is

[scarlet saffron]

if you're dividing by 1/[something] that's the same as multiplying by [something]

[signal valley]

you can re write 1 / y^10 as multiplied by 10

[kind viper]

you're still mispronouncing it. and rather severely at that

[scarlet saffron]

but in your example you're dividing by 5^2, which isn't in the form 1/[something]

[signal valley]

i see

so i must be misunderstanding

[kind viper]

you are misunderstanding or overthinking or maybe a bit of both

[signal valley]

So is the right hand statement equivalent

[scarlet saffron]

yes

[kind viper]

dutch ahh angle

[signal valley]

So would this also be equivalent

The one at the bottom

[scarlet saffron]

yes

[signal valley]

So that’s why I put the *5 there

[scarlet saffron]

that only works if you have a fraction within a fraction though

[signal valley]

Like this

[scarlet saffron]

which you don't have in the case of (\frac{1}{5^2})

[boreal girder]

calculus gaming

[signal valley]

Right here the idea is connected

Ohh

I see

So you’re saying it’s wrong because

It must be in a fraction

Like fraction inception

Fraction in a fraction

[scarlet saffron]

yeah

because (\frac{1}{5^2}) is just (\frac{1}{25}) which you can't do much with

[boreal girder]

calculus gaming

[signal valley]

Ok I’ll write that down

ok

but is there any others im missing?

everything else is correct though

i think..

[scarlet saffron]

refer to this

[kind viper]

id say write down three columns

name, general form, example

[signal valley]

ok

wait

so as long as i like understand

exponent rules and log rules

and like how to manipulate

and trig and geometry i should be fine right?

is there anything im missing idk

[scarlet saffron]

depends on what you're preparing for

[kind viper]

this understanding comes only with a good volume of practice

so yes take these notes but also dont expect that to be the end of it

[signal valley]

im built different

😬

[kind viper]

i have known nobody who went straight from "struggle with applying exponent laws" to "fluent in algebra" without spending a good deal of time in between

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@signal valley Has your question been resolved?

[compact pewter]

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[rancid stag]

What do you call doing this:
4 - (4/5)^5 * 4
= 4 (1 - (4/5)^5)

I need to learn how to do this, but I don't have a name for what's being done here so I don't know what keywords to use to search for more information about this

[subtle helm]

factoring?

[rancid stag]

Thank you!

.close

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[sand summit]

I know that I need to find the kinetic energy, and from there I can calculate velocity.

However, I'm not exactly sure how to calculate kinetic energy. Is it just equal in magnitude to the change in potential energy, but opposite sign? I don't think this is it, though, because wouldn't that make work = 0?

Is it double the magnitude of the change in potential energy, with an opposite sign?

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[sand summit]

.close

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[signal valley]

help

[signal valley]

can someone explain to me

why when you square something

say for instance 144

its equal to +-12

so it can either be -12 or 12

[stark marsh]

note (12)^2 , (-12)^2 and not -12^2 cuz that equals -144

[molten coral]

this is not true

but you're maybe saying something else

x^2 is always nonnegative

its always equal to x * x, and its never + or -

[scarlet saffron]

if you have something like x^2 = 144, then x can be either -12 or +12 because a negative squared becomes a positive
(i.e. (12)(12) = 144; (-12)(-12) = 144)

note however that if you have something like sqrt(144) that typically only refers to the positive value 12 (this is really just a convention in order to make sure that square root is a function, since functions can (typically) only have 1 output)

[molten coral]

the issue is that if you take y=x^2 and invert it by swapping y and x, its no longer a function x

[upbeat gorge]

• When you square 12, you get (12^2) = 144.
• When you square -12, you get (-12)(-12) = 144 too.
• When you take √144, this is only defined for positive numbers, so you will only get 12. √144 does not equal -12
• When solving x^2 = 144, there are two solutions, because you can square two numbers to get 144: -12 and 12

[upbeat gorge]

i actually disagree with saying it's a "convention." If we define square roots for x < 0, then we assume the properties of square roots hold for x < 0. But then you can use those properties to do some very weird things:

sorry for being pedantic, it will happen again unfortunately 😭

[compact pewter]

@signal valley Has your question been resolved?

[scarlet saffron]

i mean we can define square root to be a multivalued function

https://en.wikipedia.org/wiki/Multivalued_function

also we can define sqrt to be the negative value

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[deft tangle]

Limit x tends to 0

√(1-cosx^2)/(1-cosx)

[craggy sluice]

I mean first of all, tell me what 1-cos^2 is

[deft tangle]

okay so here i felt that

Let me write my work first

Guys

[subtle helm]

[ \lim_{x \to 0} \frac{\sqrt{1 - \cos (x^2)}}{1-\cos x}]
or
[ \lim_{x \to 0} \sqrt{\frac{1-\cos (x^2)}{1-\cos x}}]

[deft tangle]

Yes

[subtle helm]

which

top or bottom

[deft tangle]

Both are wrong

[native shard]

lol

[deft tangle]

√(1-cos(x^2))/(1-cosx)

[subtle helm]

oh

[boreal girder]

k

[subtle helm]

@deft tangle top or bottom

[deft tangle]

Limit ( x ) tends to 0 ( \frac{\sqrt{1 - \cos(x^2)}}{1 - \cos x} ).

[boreal girder]

Andy

[deft tangle]

Top

[subtle helm]

ok

[deft tangle]

So my work is change into sin

√2 |sin(x^2/2)|/2sin^2(x/2)

so here if I see that x>0+,0-

x^2 will make it positive

so both upper and lower part will be positive

So limit exists and by sinx/x thing i will get √2

[craggy sluice]

you n=know L' Hopital right?

[deft tangle]

My answer is √2

Is it correct?

[subtle helm]

👍

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[quasi python]

any way to convert 10^x into something like

10^x = 10^a + 10^b
or
10^x = a + 10^b
or
10^x = a + e^b
or
10^x = e^a + e^b
?

[compact pewter]

Please don't occupy multiple help channels.

[quasi python]

or something similar like that?

[brazen quiver]

There aren't nice ways.

What are you actually trying to do?

(like, if we want to make 10^x = a + 10^b we can select some a, then compute b by simply solving for it to get b = log_10(10^x - a) but no further simplification is possible; there are no log rules for addition.)

@quasi python

[quasi python]

I'm trying to calculate high precision of 10^x

the problem with doing it the
10^x = 10^a * 10^b
is that there is a bigger precision issue when using multiplication on limited Mantissa precision

with addition there is no precision issue

so I want to do it with addition

[compact pewter]

@quasi python Has your question been resolved?

[compact pewter]

@quasi python Has your question been resolved?

[compact pewter]

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[quartz yoke]

Integrate mod (x^3-x)

[quartz yoke]

Ul 2 and ll -1

[grim nimbus]

oh my god

ok you got it peakachu

[quartz yoke]

Wrong channel buddy

[grim nimbus]

i was typing as you were typing

it was a free channel

anyways moving on

[quartz yoke]

Oh ok

[azure vault]

!status

[compact pewter]

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 was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[quartz yoke]

1

I split the mod x in 2 terms, 1 with -ve and the other with +ve

But they split into 3 terms

[azure vault]

wdym? can you show work?

[quartz yoke]

-(x3-x) and x3-x

Then used the property of definite integrals

1 min

[azure vault]

I mean there are three different intervals to consider if that's what you're saying

[quartz yoke]

How

X3-x=0

Then x2=o

Then x =0

[azure vault]

huh?

how does one go from x^3 - x = 0 to x^2 = 0

[quartz yoke]

Send x to rhs, then cancel out 1

[azure vault]

?????????

x^3 = x means x^2 = 0??? is that what you're saying

[quartz yoke]

Umm ye 💀😭 ig

[azure vault]

ooooof

[quartz yoke]

Seems like im wrong

[azure vault]

do something more rigorous than "cancelling" first

what do you mean by cancelling, mathematically

[quartz yoke]

I dont understand

[azure vault]

do you add something on both sides? subtract?

[quartz yoke]

Same terms on both sides

[azure vault]

etc...

[quartz yoke]

What

[azure vault]

what operation are you doing

cancelling isn't the correct term, whatever you're trying to say

[quartz yoke]

Oh wait 💀

I shud take x as common then j get x=0 1 -1 right?

[azure vault]

yes

[quartz yoke]

Then i get two +ve terms and 1 negative

[azure vault]

you're not cancelling rigorously speaking, you're dividing

and you're dividing by x

so first of all make sure x isn't 0

so x = 0 OR ... and then you divide by x

then you get x^2 = x/x = 1

not 0

[quartz yoke]

Oh

Ye i forgot 😭

Sorry mb

But then i have another doubt

Since the limits are -1 to 2

We split 3 terms as -1 to 0, 0 to 1 and 1 to 2

But which of 1 these terms are -ve?

-1 to 0 right?

Ok nvm i got it

[azure vault]

it's 0 to 1 that's negative

[quartz yoke]

Its 0 to 1

[azure vault]

it goes + - +

[quartz yoke]

Ye

Ye thanks

.close

[compact pewter]

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[quartz yoke]

.reopen

[compact pewter]

✅

[quartz yoke]

Integrate (x sinx)/(1+cos^2 x)

Ul pi ll 0

.close

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[ornate sable]

Can someone explain why there is a collection of sets in the plane?

[kind viper]

wdym why

it just is

what's confusing about it?

it's like the collection of all possible* plane shapes

[ornate sable]

So is he saying there is a collection of sets because each set represents the area of one shape in the plane but if there multiple we group all those sets in another set?

[hushed magnet]

a shape in the plane is a set

[ornate sable]

In the sense that we represent as a set for our area function? Or that by creating a shape in the plane you create a corresponding set

[hushed magnet]

what do you think a shape is if not a set

the set of all the points which belong to that shape

[kind viper]

maybe she thinks of shapes as sui-generis objects

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[compact pewter]

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[compact pewter]

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[signal valley]

Help

[signal valley]

I don’t know what to do

How do you deal with negative square roots

Idk if I made a mistake

[red basin]

Idk if your workings are correct, I haven't looked at them properly

But all that would mean is there aren't any real solutions

Not necesarily a mistake

As a very simple example, think about x^2 = -1

You get x = sqrt(-1)

I didn't go wrong, there just aren't any solutions

[molten badger]

$3x^2-6x+11=0$

$3(x-1)^2 +8=0$

It's just no real solution

[boreal girder]

Alexis_Fx

[signal valley]

this is the question

this is my work

so far i am correct i think

i just dont know how to deal with

the sqrt -96 / -6

like dealing with negative square roots

thats what i need help on

[red basin]

Have you been taught about complex numbers yet?

[signal valley]

no

[red basin]

Okay

In that case, if you see a negative root, the answer is "there are no solutions"

Because you can't take the square root of a negative number, it doesn't exist

[slate sand]

Have you not been just doing things relating to sqrt(-a) for positive a

You literally have asked questions about this like 5 times in the last 2 hours

[signal valley]

sorry

im just trying to cram for my math placement test

i havent done math in a long time

thats why ive been on this server for like 30+ hours

[slate sand]

I’m not berating you

(Sorry if it came off that way)

[signal valley]

oh

i havent done math in like 3 years or something like that

[slate sand]

What I meant was, look at the connection between what you’ve just asked moments ago

[signal valley]

ohhhhhh

[slate sand]

You just did something about sqrt(-a)

[signal valley]

yea i did

but i dont really understand it

ig

[slate sand]

Well 96 = 12 * 8?

So like 3 * 2⁵

[signal valley]

yes

ok

[slate sand]

So you can pull out a 2⁴

[signal valley]

makes sense

[slate sand]

So sqrt(-96) = what

[signal valley]

where did you get 2⁴

[slate sand]

From here

[signal valley]

from i thought it was 2⁵

[slate sand]

Yeah but 5 is not even

[signal valley]

why does it have to be even?

[slate sand]

You can’t square root the 2⁵ but you can for 2⁴

Cos square roots sorta divides the power by 2

[signal valley]

oh

so youre trying to look for a perfect square

such as 4

[slate sand]

Yes

That’s what I meant by “pull out”

[signal valley]

oh

[slate sand]

I’m looking for a perfect square to pull out of the square root

[slate sand]

So what’s after here

[forest token]

(no real solns)

[compact pewter]

@signal valley Has your question been resolved?

[signal valley]

it would be

srry i had to go to the abhtroom

4sqrt6i

[red basin]

Kind of a minor point but I've never really liked that phrase. It implies that there are secret solutions that we're just not gonna talk about. But as an equation over the real numbers, there truly are no solutions

But yes

[signal valley]

ahh i see

thx for thje help frost

[compact pewter]

Channel closed

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Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[wary mulch]

can someone walk me through this challenge question

[molten badger]

!status

[compact pewter]

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 was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[wary mulch]

I rubbed it out tho

because it was only like the first step i got stuck on

so what i did was get the derivative of the function

and evaluuate it at the stationary points

thats where i got stuck

i tried the completing the square method

[molten badger]

Okay ,have you found the differential of the cubic functions?

[wary mulch]

yes

first derivative?

[molten badger]

Yup

[wary mulch]

yep

[molten badger]

So it should be $y\prime = 3ax^2 + 2bx +c$

[boreal girder]

Alexis_Fx

[wary mulch]

i got different

[molten badger]

Oh, what did you get then?

[wary mulch]

i got f'(x)= 3x^2+2x+c

[molten badger]

You need to keep a,b,c,d since this is for general cubic function

[wary mulch]

how do u know?

[molten badger]

The question doesn't state any specific function, just cubic functions so we have to write it as $ax^3 + b^2 + cx + d = y$

[boreal girder]

Alexis_Fx

[wary mulch]

im still confused

i dont get it

[molten badger]

Okay, for you what is a cubic function?

[wary mulch]

power of

3

[molten badger]

Yes, so $3x^3 + 4x^2 + x +1=y$ and $x^3 + x^2 + x +1 =y$ both are cubic function right?

[boreal girder]

Alexis_Fx

[wary mulch]

agree

[molten badger]

So every function has the form of $ax^3 + bx^2 + cx + d=y$ with a not equal to 0 is a cubic function agree?

[wary mulch]

agree

[boreal girder]

Alexis_Fx

[molten badger]

Okay, that's what you have to work with. Not $x^3 + x^2 + x + 1$

[boreal girder]

Alexis_Fx

[wary mulch]

👍

but x^3 +x^2 + x + 1

is also a cubic

oh u said

a

not equal to 0?

[molten badger]

Yes

[wary mulch]

but

i wrote

3x^2+2x+c

a is not equal to 0

[molten badger]

It's differential of x^3 + x^2 + x isn't it?

[wary mulch]

yea

[molten badger]

So a=1 in your function your wrote

[wary mulch]

true

[molten badger]

x^3 + x^2 + x +1 is a function but it doesn't represent all the cubic functions possible

[wary mulch]

this q is hard

should we just

do this q at the end, i have another question

[molten badger]

Just try to use $ax^3 + bx^2 + cx + d$

[boreal girder]

Alexis_Fx

[molten badger]

Okay

[wary mulch]

am i on the right track?

q 7. a

[molten badger]

What's the goal here

[molten badger]

Aren't we doing f(x+h)-f(x)

[wary mulch]

it asked us to find

f(x+h) AND f(x)

[molten badger]

The first the lines seem correct. Idk what are you doing after that

It say show f(x+h)-f(x) not f(x+h) + f(x)

[wary mulch]

what does this mean?

[molten badger]

You have done it in your first 3 lines

[wary mulch]

okk i understand now, thanks

.close

[compact pewter]

Channel closed

Closed by @wary mulch

Use .reopen if this was a mistake.

[compact pewter]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[deft tangle]

How to find a×b×c=30 total integers solution?

[deft tangle]

Positive integer,negative integer, non negative integer, non positive integer

[pine osprey]

writing 30 as a product of 3 integers might help

does a,b,c has to be different ?

[soft bobcat]

Prime factorisation consists of only 3 primes so is nice

[serene junco]

the only solution for a, b, c > 1 is 2×3×5

from there, you can flip two of the signs to negative

which would give 3 more solutions

and then there's the solutions where one of the numbers is 1

that would give you 1×5×6, 1×2×15 and 1×3×10

[deft tangle]

Nope

[serene junco]

then there's also the solution where two of the numbers are 1

1×1×30

[deft tangle]

2×3×5=30

[soft bobcat]

Remember to give negative to 2 of each product to account for all ints

[serene junco]

accounting for all of those, you get 5 solutions where everything is positive, and 15 more solutions where two of them are negative and cancel out

but we counted -1×1×-30 twice, so there's 19 solutions total

[deft tangle]

You are too fast but i am not understanding your steps

First tell me about positive integers

And non negative integers will have same solution?

[pine osprey]

there isn't really steps except dealing with differents a,b,c first and then do the cases where they can be the same, adding all the signs changing

[kind viper]

none of a, b and c can be 0 anyway so yes

[serene junco]

wait before that, i just want to clarify
does the question count 2×3×5 and 3×2×5 as different solutions?

[soft bobcat]

Ordered pair

Would be so many more

[deft tangle]

Maybe yes?

[serene junco]

some of the probability questions i've solved have done that and it's sort of annoying when it isn't specified

[deft tangle]

Let me check the answers for better help

[soft bobcat]

Actually nm it must be ordered pair

[deft tangle]

27

For positive

[soft bobcat]

If it's given a,b,c specifically

[serene junco]

oh ok then they are ordered pairs

alright so

[pine osprey]

should have been worded as (a,b,c) \in Z^3 in some sense

[serene junco]

if none of a, b, c are equal to 1
then the only solution is 2×3×5 and all of the ways you can rearrange that (6 solutions)

if one of them is equal to 1
the solutions are 1×2×15, 1×3×10, and 1×5×6 and all of the ways you can rearrange those (6 + 6 + 6 solutions)

if two of them are equal to 1
the only solution is 1×1×30 and all the ways you can rearrange that (3 solutions)

[civic quartz]

(ordered triplet)

[serene junco]

oop

yeah

[serene junco]

in total, that's 6 + 6 + 6 + 6 + 3 = 27 solutions

[deft tangle]

Do you understand Hindi?@executor.1011

@soft bobcat

[soft bobcat]

What

Yes

[deft tangle]

Fine.

[serene junco]

oh that's convenient

[deft tangle]

Did you understand what they discussed?

[serene junco]

i do too

[deft tangle]

Ohh that's good for me

[soft bobcat]

Yh talk to them

[deft tangle]

Toh bhai kya kaisa ho gya

Sab blunder ho gya

Can we do it again?

[serene junco]

hm

uhhhhh which step did i lose you at

[deft tangle]

I don't know properly where to start

You started with negative integers too

pehle ham sirf positive k liye karte hain

[serene junco]

these solutions are only positive

[deft tangle]

abc>1 toh one choice is 2×3×5

[serene junco]

correct

[deft tangle]

Now i will include 1

i can give one to anyone

1 (2,15)

1(15,2)

Total 6 cases

[soft bobcat]

? That product is not 30

[deft tangle]

1( 3 10) again 6 cases

[serene junco]

oh oops i misread

yeah it should be 2 3 5

[deft tangle]

Sorry it is 5

1 ( 5 6) again 6 cases

[serene junco]

yeah that's correct

[deft tangle]

6+6+6+1 case?

[serene junco]

this is also 6 cases

[deft tangle]

Ohh wait

1 1 30

So 3 more

6+6+6+3+1?

Ohh 27?

[serene junco]

yes

[deft tangle]

2 3 5 hmm

[serene junco]

6+6+6 + 3 + 6

[deft tangle]

Very very nice

Ab bat karte hain apn negative walo ki

Non negative m bhi 27 aayega because 0 ko lena hi ni

[serene junco]

yep

[deft tangle]

Ab nagative me hum 2 ka sign - le sakte hain

[serene junco]

let's take 2 3 5 as an example
how many ways are there to add negative signs

[deft tangle]

-2 -3 5

-2 3 -5

2 -3 -5

[serene junco]

yeah

[deft tangle]

2 signs so 3C2

[serene junco]

that means for every positive solution, there are 3 negative solutions

except for 1 1 30, 30 1 1 and 1 30 1

for those there are only 2 options

oh no wait

we're considering ordered triplets so it's still 3

[deft tangle]

-1 -1 30

1 -1 30

-1 1 -30

Yes 3

[serene junco]

so this applies to everything

which means you can just say the number of negative solutions are 27 * 3

[deft tangle]

What??

We just need only negative solution

So it is 27 no?

[serene junco]

27 is the number of positive solutions

[deft tangle]

Yes