help-28

[sharp osprey]

or maybe it is r^2

and it is approaching 0 so it kills the cos and sin terms

is it possible for you to elaborate on why this is the case? how can I convince myself of this?

or if the cute cat reaction was a mistake again then nvm my question!

like if i had chosen pi = theta

[steel nest]

it was not a missclick this time, r is positive because if we allowed r to be negative we could assign 2 coordinate to the same point radially opposite

r is in 0,infinity for the same reason we impose theta to be between 0 and 2pi when describing point in polar coords

[sharp osprey]

right but radially opposite isnt interesting to us since we are approaching origo

and radially opposite of that is away from origo

[steel nest]

at least that my reason

[steel nest]

yeah but you can get that ray with another theta keeping r positive

[sharp osprey]

and this is kind of showing r approaching 0+?

[steel nest]

for two different angle phi an theta

yeah

[sharp osprey]

would this be if r is approaching 0?

or somethign

or is that just nonsense

[steel nest]

no the arrows point to (0,0)

[sharp osprey]

i thought that was only if r approaches 0+

what would r approaching 0 look like

i assumed it would look different

than 0+

[steel nest]

you mean from left and right if we let r be negative (which we do not)?

[sharp osprey]

yes so, to do this question correctly, only r approaching 0+ is acceptable, but now im wondering, indepedent of that, what would r approaching 0 from both 0+ and 0- look like visually

^ i am asking in order to better convince myself that r has to approach 0+ and not 0 or 0-

or is that the whole point of saying they are rays

that they only move in one direction

but in that case, why would 0- be wrong

since that would also be one direction

is it just defined that 0- means we are moving away from origo?

[steel nest]

but we don’t do r tends to 0 from the left usually

[sharp osprey]

hmm

is it a coincidence that you put the point of r approaches 0- in the middle of the line and not near origo?

[steel nest]

yeah it’s a coincidence

[sharp osprey]

sorry i mean point as it pointy-bit

ah alright

but okay so

does that mean i could have

just as easily said r approaches 0-

instead of 0+

but it is convention

to do it from 0+

[steel nest]

it’s a bit more than convention if we could let r tends to 0 from the left it would mean r could be negative, and then the polar coord system would allow to label the same point with two coords and that’s annoying to work with tbh

[sharp osprey]

i dont understand why it would need two new coords

or maybe more precisely i dont understand why r would have to be negative

for 0-

cant i still call r a positive length

like for the infinite many rays

they will end up in the negative quadrants

[sharp osprey]

even for 0+

depending on the angle

[steel nest]

for instance this is a way to describe the same point two different way if we allowed negative r

[sharp osprey]

i dont understand why r would be negative

it is only moving closer to 0 from a different direction

okay so

it is more than convention

in the sense that it is less annoying

but 0- would get me the right result?

as well?

but it is just tedious?

[steel nest]

because when you are saying 0-, i assume you mean r tends to 0 from the left which by definition means taking the limit with r<0 getting arbitrarily close to 0

[steel nest]

it’s unconventional and it mess up with what the polar coords are, a different way of assigning uniquely two number to a point in the plane to describe it

you mess up the uniquely bit

[sharp osprey]

is a length undefined on the negative horizontal axis?

i thought the whole point of a length was that it was always positive no matter where i draw it

[steel nest]

it’s positive

[sharp osprey]

or okay not the whole point

but a point

so if i draw the length there

i dont see how it would change anything?

the length is still positive

[steel nest]

this is fine length 2

[sharp osprey]

ok

and yes right

and if i wanted to shrink that length

0-

it would move closer to 0

but still be positive

or rather

i guess it really wouldnt be moved closer to 0

it is fixed at 0

but it would shrink toward 0, the length would get shorter in that direction

but why would that give me a differental result it doesnt make intuitive sense to me

isnt this exactly what we are doing

with the length of 2 being drawn

on the other side

[steel nest]

in this picture r is approaching 0 from the right too 0+ but with theta = pi

[sharp osprey]

yes

this is what im picturing i think unless the height changing is important

i view it as having the same height of 0

[steel nest]

no it’s not important i did it this way because if i did not we would get hella overlap

[sharp osprey]

okay good!

[sharp osprey]

but yeah so, if we mirrored this image

isn't that the "right way", r approaching 0+?

the numbers get mirroed but yeah lol

[sharp osprey]

this is when the length r = 2 is approaching 0?

okay that doesnt make sense

[steel nest]

yes but theta = 0

[sharp osprey]

like what i wrote

[steel nest]

and here theta is also fixed and equals 180 degrees

[sharp osprey]

oh

and yeah i mean 0 is defined

where it is

so that is why i get two different points

with 180 degrees

even if the length is still positive?

it just has to do with where the angle 0 is defined?

but how are we able to describe the other rays

arent these rays a problem

[steel nest]

you could describe them as the set of points {(rcos(theta),rsin(theta)) r > 0} and picking the theta appropriately

if you want to describe them explicitly with a parametrisation

what i wanted you to appreciate is how the inputs move toward (0,0) when we fix theta and make r tends to 0+

so this limit notation with the r+ would hopefully make more sense

imo we strayed a lot from your original problem though

[sharp osprey]

that is okay my intention wasn't really specifically to solve the problem, i want to understand it better

sorry if it feels like you wasted your time but i truly appreciated the help

and i think i understand it a bit better now, but the 0+ is a bit strange to me still

[steel nest]

no if you have a better geometric feel for this my time was not wasted

[sharp osprey]

just because im thinking of all the other rays

i have an improved geometric feel for it now but im not enitrely satisified but i understand everyone has limited time to engage with me XD

[steel nest]

yeah it do be getting late where i am

[sharp osprey]

i will for sure ask my tutor on Wednesday more about this, whether or not you have more time to engae or not

yeah ok i wont hold you any longer then

thanks for the help

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[abstract hawk]

why do they do this

it would make all of them equal???

i dont get it

[foggy vapor]

Context?

[lime ether]

body is probably at rest

but yes show context

[abstract hawk]

they have to be of opposite directions

3 forces are going clockwise

1 force is counter clockwise, Force 3

[sharp osprey]

that's weird

[abstract hawk]

so one of them is force 3

so any combination of two forces will create a net zero torque

so any two forces with one of them being force 3 will work

i looked at the solution and they do this

when comparing any possible two forces

they want to do this

like why

they are pretty much making every force equal to each other

even though it states this first

the whole question if it helps

to achieve a net zero torque with two forces, the two forces need to be going opposite directions

force 3 is the only force that goes counter clockwise

so force 3 is one of the forces

[rocky rapids]

I suppose yeah

[rocky rapids]

Also couldve just sent this from the start

[abstract hawk]

yeah i realized that so i sent it

mb

[rocky rapids]

Do you know what the formula for torque is?

[abstract hawk]

fr

Fr

[rocky rapids]

yep

[abstract hawk]

but you cant just substitute

the forces are equal to weird amounts compared to each other

[rocky rapids]

We can bruteforce

[abstract hawk]

what im trying to ask here

[rocky rapids]

Since F3 is counterclockwise, we can assume T3 is negative

[abstract hawk]

they did this and solved it

and it works

[rocky rapids]

Yep

[abstract hawk]

but i just dont understand why they do this

all they did was switch the forces with 2F to have only F and vice versa

like how does that even help

[rocky rapids]

What are you not understanding from this part?

F3 = 2F?

[abstract hawk]

the whole step

how does that help solve the problem

what is that doing

[rocky rapids]

It's just a way to represent it simpler

[abstract hawk]

but its not?

without doing it you get the wrong answer

but when you do it you get the right answer

because if i just apply the torque formula and substitute

i dont ever find the rotational equilibrium

but once they change these force values when comparing, it magically finds this rotational equilibrium

[rocky rapids]

It makes the value more concrete

[abstract hawk]

so here torque three has 2F and torque 4 has just F

[rocky rapids]

yes

[abstract hawk]

but when you revert them back to original values where torque 3 will just have F and torque 4 will have 2F

you get the wrong answer

why did they change the force values

<@&286206848099549185>

[rocky rapids]

I just thought about this

What they meant "F" was F is a number, not a variable

[abstract hawk]

yes

[rocky rapids]

But F2 and F4 are variables

[abstract hawk]

its a constant

[rocky rapids]

Yes

That's why they switched it to F

[abstract hawk]

?

what

wdym

[rocky rapids]

If you just say F2 and F4 it's not concrete enough to work with

Using F2 = F4 = F makes it easier to calculate stuff briefly

[abstract hawk]

that doesnt seem like the case though

[rocky rapids]

F1 = F3 = 2F

[abstract hawk]

i dont get it

[rocky rapids]

Let's say F is 10 N

[abstract hawk]

ok

[rocky rapids]

Then F1 = F3 = 2(10 N) which F1 = F3 = 20 N

Instead of writing it as 20 N = 20 N = 2x10 N = 2x10 N we can use that

[abstract hawk]

bro but how does that justify it

it has some relationship with trigonometry i think

[rocky rapids]

It does

[abstract hawk]

what would it be then

what is this relationship

that im missing

[rocky rapids]

tau = F x R x sin(theta), correct?

[abstract hawk]

ye

[brave ruin]

`T3 = F3*(sin30)2R
= F3R

T4 = F4*(sin90)2R
= F42R
= 2F4*R

Given 2F4=F3 so
T4 = F3*R = T3 `

So it still works out. It gives the same result if you don't assume the F part.

[abstract hawk]

oh wow

how did i not see that

but like how does it work

when they

convert the F

what is happening

[brave ruin]

Now it's more readable lol

Ok i ask you if

[abstract hawk]

that would suck

i just dont get why

[brave ruin]

Nvm I didn't give good example

[glossy valve]

Shubham1029

$$\documentclass{article}
\usepackage{amsmath}
\begin{document}

\textbf{Q.} If 
\[
\frac{F_1 - 3}{2} = \frac{F_2 - 4}{3} = \frac{F_3 - 7}{5},
\]
find F_1 + F_2 + F_3.

\textbf{Solution:}

Let 
\[
\frac{F_1 - 3}{2} = \frac{F_2 - 4}{3} = \frac{F_3 - 7}{5} = F.
\]

Then, express each in terms of F:
\[
F_1 = 2F + 3, \quad F_2 = 3F + 4, \quad F_3 = 5F + 7.
\]

Now, add them:
\[
F_1 + F_2 + F_3 = (2F + 3) + (3F + 4) + (5F + 7) = 10F + 14.
\]

\textbf{Answer: } F_1 + F_2 + F_3 = 10F + 14

\end{document}
$$
```Compilation error:```! LaTeX Error: Can be used only in preamble.

See the LaTeX manual or LaTeX Companion for explanation.
Type  H <return>  for immediate help.
 ...                                              
                                                  
l.49 $$\documentclass
                     {article}
Your command was ignored.
Type  I <command> <return>  to replace it with another command,
or  <return>  to continue without it.```

[brave ruin]

Only if I could delete this shi

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[raven mauve]

can i get help please

i know a =1

[junior oyster]

so uhh, did u try the binomial expansion?

[raven mauve]

no sure how to do it

does it have something to do with this?

[junior oyster]

ah right, see x and y here arent really functions. x can be (1) and y can be (-2x)

[raven mauve]

so like the first thing in the bracket is x

followed by y?

[junior oyster]

yeee

[raven mauve]

ok

so what next

[junior oyster]

and then u distribute the r to y (do the calculations). Then see for what value of r = 1 and r = 2

because ur trying to find the coefficients of x^1 and x^2

[raven mauve]

so what is n and r in this case

[junior oyster]

n is 6 (total power) and r goes from 0 to 6

[raven mauve]

so n = the power
r = the x power we are looking for?

[junior oyster]

this is a summation, as seen by a+bx+cx^2+dx^4+ex^5+fx^6

[junior oyster]

yes

and once u find the number, use the rest of the calculation to find the coefficient

[raven mauve]

sorry i’m being dumb rn

but is this correct so far

[junior oyster]

since r = 1 for b, coefficient should be 6C3*1^(6-3) times (-2)^3

oh wait not 3, 1

[junior oyster]

hmm if one i raised to the power of 6-1, why is (-2) raised to the power of 6?

[raven mauve]

no idea

i’m just following the formula

[junior oyster]

but if 1^(n-r) and n=6 and r = 1, then shouldnt (-2)^r be (-2)^1

[raven mauve]

oh yeah

right

oops

so it simplifies to 6x1-2

said 6 choose 1 is 6

oh i forgot the times symbol in front of -2

[brave ruin]

(1+a)ⁿ has a really simple binomial expansion.

n choose 0 multiplied by a⁰
+
n choose 1 multiplied by a¹
+
n choose 2 multiplied by a²
+
...
n choose n multiplied by aⁿ

.

And your question only asks for first 3 terms so you only need to do the first 3 terms.

Values of 6 choose 0, 1, 2 is already given in the Pascal's triangle.

[raven mauve]

b = -12

c=60

?

[brave ruin]

Yes

[raven mauve]

ty

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[pallid blaze]

The answer is 1/6 and I've understood my mistake, i.e using the Taylor series in composite function wrongly
But still I can't the correct method

[onyx glen]

(1-cos(x))/x^4 is troublesome

and the stuff you did afterward is also sus

cause this thing is actually divergent at 0

[grave elm]

I'd add doing +1-1 in numerator maybe

[onyx glen]

i would expand sin^2(x) and sin^4(x) down to the x^4 term in each

[pallid blaze]

The composite - cos is not divergent tho?

[onyx glen]

wdym

i'm saying the fraction $\frac{1-\cos(x)}{x^4}$ diverges.

[glossy valve]

Ann

[pallid blaze]

I mean to say, that the fn is contious for all x

[onyx glen]

you are going off track...

[pallid blaze]

I can't seem to make it work?

[pallid blaze]

Sry

[onyx glen]

please do not call me "bro"!

[pallid blaze]

Sorry bro

🙏

[onyx glen]

could you edit that message where you did it

ok

can you show me what you have so far with my instructions

like what exactly do you mean with "can't seem to make it work"

[pallid blaze]

When I take x^4 inside the 2 and 4 degree bracket, it becomes x^8 and x^16 respectively and I will get infinites inside which are of different magnitude

Am I missing something??

🥲

@onyx glen

[onyx glen]

you seem to be missing a lot but idk if i can say much right now specifically.

[pallid blaze]

If you can't help, then don't be a prick about it

[onyx glen]

i tried to be polite about it

i am just physically tired right now

i dont appreciate being accused of prickery when there was none intended on my part

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[leaden ermine]

@pallid blaze

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[thick ivy]

help me prove or disprove section c

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[thick ivy]

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[raven mesa]

If I have a parallelipiped

[raven mesa]

to find the volume I need scalar triple product

but in my notes it says to take the modulus of the scalar triple product, how can I do that when the end result is a numerical value resulting from dot product

[lime ether]

by taking the absolute value

[raven mesa]

oh

say the scalar triple product resulted in the following expression:

-2m²+9m-11 (m is an unknown in two of the vertices)

how do you turn that into absolute value

[lime ether]

|-2m^2 + 9m - 11|

[raven mesa]

what I don't understand is that in my teacher's notes she goes further, and does as seen in the image

but I don't see how that works, when usually you do that with the components of a vertex

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[wispy kayak]

can some1 plz help

[woeful pasture]

how do u define the period of a sinusoidal function

[wispy kayak]

2pi/k right?

[woeful pasture]

like

definition

[wispy kayak]

1 cycle of a sin function?

[woeful pasture]

yes

one cycle

is moving from one max point to another (adjacent) completing a cycle

[wispy kayak]

oh ok i get that i think

[woeful pasture]

so

whats the period

[wispy kayak]

one cycle of this function but im not sure how to calculate that

[woeful pasture]

ok

its the distance between ${-5\pi/7}$ and ${-3\pi/7}$, right?

[glossy valve]

k

[wispy kayak]

yes

[woeful pasture]

so whats the distance

[wispy kayak]

i would usually convert to degrees b/c for me its easier to understand and then subtract -3pi/7 from -5pi/7 but im not sure if thats right

[woeful pasture]

thats correct

just subtract!

so

what's ur period

[wispy kayak]

oh ok thank you, im doing the calculations and my calulators acting weird, might take a min im sorry

i got -51.43pi/180 or -51.43 degrees

[woeful pasture]

,calc 2pi/7

[glossy valve]

Result:

0.89759790102566

[woeful pasture]

${-\frac{5\pi}{7} - (-\frac{3\pi}{7}) = -\frac{5\pi}{7} + \frac{3\pi}{7} = -\frac{2\pi}{7}}$. Since distance cannot be negative, period = ${\frac{2\pi}{7}}$.

[glossy valve]

k

[wispy kayak]

oh ok i didnt know you could do that

thats way easier than what i did lol

[woeful pasture]

ye💀

[wispy kayak]

ok thank you so much this was really helpful

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[half oyster]

trying to learn graphing

[half oyster]

seems like range is a little tricky

would I just use all this stuff to somehow figure out range?

like use end behavior, min/max values

first/second deriv test

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[dusk jay]

<@&286206848099549185>

[hot herald]

In the future, please only ping helpers after 15m of asking your question (and have not received an adequate response)
anyway

what have you tried?

[dusk jay]

i wanna do my sparxmath homework quickly

so hard btw im in year 9

[hot herald]

did you watch the video guide?
do you know your parallel line theorems?

[dusk jay]

yes

still hard

[hot herald]

what did the video say? did it show a worked solution for a similar question?
what parallel line theorems do you know?

[mellow berry]

you know they are parallel lines right?

[dusk jay]

yes

I GOT 2 INCORRECT 1 MORE AND QUESTION CHANGES

[mellow berry]

do you know the sum of co-interior angles?

[timid goblet]

make a linear eqn in 2 variables

[dusk jay]

how the hell do i do that

i know the sum is 360

[mellow berry]

in parallel lines

the sum of co interior angles is 180

[timid goblet]

yea so x + y + 67 + 108 = 360 (assumption: euclidean geometry)

[dusk jay]

why yall make it so complictated just gimme the answer 😢

yes ik that

idk the x and y

[hot herald]

ignore them, what they're saying doesn't help you

[mellow berry]

TUV=180-108

[dusk jay]

72?

[mellow berry]

yes

[dusk jay]

and the second one?

[mellow berry]

solve for that in the similar way

[dusk jay]

180-87?

[mellow berry]

87?

67

[dusk jay]

i mean 67 mb

[mellow berry]

right

all done

[dusk jay]

YYAYYAYAYAYYA

I GTG EAT

bye

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[untold palm]

total surface area of lateral surface area?

its 10

[woeful pasture]

Laterally increase by 4? Circles increase by 16?

It’s 2pirh + 2pir^2

So if I plug in 4r instead of r

[untold palm]

no its 2pirh + 2pir^2

[woeful pasture]

Ye mb

[untold palm]

yeye

[woeful pasture]

8pirh + 32pir^2

[untold palm]

just plug in 1 and 4 as values

[woeful pasture]

The ratio ain’t the same for side and top?

[hushed barn]

I think it matters that we know what the height is in terms of the radius. Or is that not given?

[untold palm]

the answer varies with the value of h

[hushed barn]

Can you state what the original question was exactly?

[inner parcel]

Do you mean that the height and the radius both are quadrupled?

[untold palm]

it would make sense for the question to ask lateral surface area as it is not specified

if so its 4

the answer varies with h then

[hushed barn]

Is there any context to the question? Because if not, the question seems weird

I think you can bring this up with your teacher and argue that the height needs to be mentioned in terms of the radius for there to even exist an answer

Maybe they weren’t looking for an integer answer, but still that would mean that you would need to define some variables

Total surface area of a cylinder is 2pir*(h+r), right? Take h=2r or h=3r, then there’s a difference when you say that you quadruple the radius

Definitely not. Height is different

[untold palm]

try to do it urself once, you'll get it

[inner parcel]

So i just put random numbers for the radius(2) und height(4) and it increases by 10
Then I put radius(2) and height(6) and it inreases by 10 again

[hushed barn]

Because we specified two different heights from the beginning, one being 2r and the other being 3r

2r only equals to 3r when r is 0, which is impossible

No, I was trying to provide an example

It’s not in the question

What was the answer that the teacher was looking for?

What shapes are you referring to?

Yes it would be 16 if the height was also quadrupled. Maybe that’s the answer that they were looking for

Okay

You cannot really apply that logic here. Just because only one out of the to things are scaled doesn’t mean that that it’s a half of 16.

I think you should just bring this up to your teacher and tell that the question is ambiguous. You should tell that the height needs to be specified in terms of the radius otherwise there’s more than one answer

Still ambiguous. The height always needs to be expressed in terms of the radius for there to exist a scale factor to begin with. Or that the height is increased by the same scale factor as the radius

[full forum]

@oak rover Has your question been resolved?

[brave ruin]

Curved surface area increased by 4 times while top and bottom's area increased by 16 times

I think it needs to be answered seperately like this

Since we can't answer in terms of total surface area

[full forum]

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[torn jolt]

cubic equations

[torn jolt]

inspection method

[torn jolt]

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

[glossy valve]

James

[torn jolt]

what i did

i got first bracket

(x-1)

i got first term of 2nd on

$x^2$

[glossy valve]

James

[torn jolt]

then

now want middle term

i tried

$-3 x x^2 = -3x^2$

[glossy valve]

James

[torn jolt]

but i donno what to do from here

bc that aint middle term

but i gotta do smth w it

[torn jolt]

so yeah

[gritty flax]

how so ?

1 is a root ?

[torn jolt]

well i

[gritty flax]

indeed it is

nvm

[torn jolt]

k

[gritty flax]

so (x-1)P(x) = cubic

find P(x) with the coefficient a,b,c of quadratic

[brittle parcel]

100% valid way to do it
you can also use polynomial long division
you can also use synthetic division

[gritty flax]

yeah poly division is cool

[torn jolt]

well im working on inspection method

[brittle parcel]

let me google what that means

[gritty flax]

inspection = identification i bet

[torn jolt]

uh

[gritty flax]

i ended up on taxes rules of my country

[torn jolt]

$ax^3 +bx^2 + cx + d = 0$

[glossy valve]

James

[brittle parcel]

inspection means this way apparently

[steel solar]

Try to factor by grouping

[torn jolt]

but like

i dont wanna change my mehod

bc

[steel solar]

Just for verification

[torn jolt]

it specifies

[brittle parcel]

doesn't seem to be a standard term but from the sources that DO call it inspection it's what msb db said

[torn jolt]

OOOO

wait

sorry

i see

give me a min

i found middle term i believe

$(x-1)(x^2 +4x + 3)$

[glossy valve]

James

[torn jolt]

can someone confirm

that it = to tht

[brittle parcel]

,w equation to send a query to wolfram alpha

,w 1+1=sqrt(4)

[glossy valve]

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Requested by gfauxpas

[steel solar]

That's not what I got Yeah that's good

Wait

[torn jolt]

😄

thanks for leading me to my

moment of realisation

❤️

.close

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[limber sable]

How would I go about simplifying this?

[abstract hawk]

i forgot how this works

like why does change in time end up on the bottom

and not the top

dividing something is like multiplying the reciprocal?

change in time would go on the top

but here it doesnt

[limber sable]

This is the answer according to the book

[foggy vapor]

This channel is occupied. Open your own and delete the messages you sent here if you could

[umbral dome]

you would want to get the two fractions onto a common denominator

[limber sable]

What would I do from here? Would I need to develop all that or something? Is there a simpler less complicated way?

[keen vector]

is ur goal just to get the answer in the book

just factorise the numerator

[limber sable]

I’d like to ideally understand how the book got to its answer

[keen vector]

its answer is probably the same as urs

like the 2 expressions are equal

[foggy vapor]

Your numerator isn't too bad. Just factor out (2sqrt(x) + 7)^2 and expand the rest

[limber sable]

I’ve already confirmed that my original one was correct, I just want to know how to simplify it so that it matches the one in the book

[keen vector]

to show they're equal it's algebra not calculus anymore

[limber sable]

(I’m kinda trash at both lol)

[keen vector]

also factor out 3

[limber sable]

Did I do it right? Would I just need to develop and solve the rest from here?

[limber sable]

Sorry, I don’t quite see how I could factor 3 out

[foggy vapor]

Yes seems legit

[foggy vapor]

Yoink it from 3x^2 - 6x, on the right term

[limber sable]

Oh I missed that, thanks

[foggy vapor]

The rest isn't too bad. You have one FOIL and the rest are just like terms you need to Group up

[keen vector]

this looks like quotient rule

[limber sable]

What’s FOIL mean?

[limber sable]

I’m sorry, my brain is fried rn, but what’s the quotient rule?

[keen vector]

the rule u used to differentiate

[foggy vapor]

When you expand an expression of the form (a+b)(c+d)

[keen vector]

i'm guessing

[limber sable]

Oh

Ohhhh

[abstract hawk]

@limber sable thanks for taking over my channel

@foggy vapor you stated this channel is occupied and then proceeded to help them

[foggy vapor]

Yes, you were the one who was intruding. Mae was the one who claimed this first

[rough tundra]

calm down dude you can open another one easily 😭

[abstract hawk]

oh wtf

we sent our messages at the same time

it was open when i sent it

mb

[foggy vapor]

It's alright

[limber sable]

Did I expand this right?

I don’t think so, but I also can’t spot an error

Nevermind, I’ve found my mistake

Or one of them at least

I’ve figured it out! Thank you for the help,

.close

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[deft glacier]

Q13. Suppose that the number of terms in an A.P. is 2k, k € N. If the sum of all odd terms of the A.P. is 40, the sum of all even terms is 55 and the last term of the A.P. exceeds the first term by 27, then k is equal to :
(1) 6
(2) 5
(3) 8
(4) 4

[deft glacier]

my answer is comming as 4 or 8

[sturdy valve]

show ur work please

[deft glacier]

i dont got phone to show my work

neither do i know AP

i just know the basics

[sturdy valve]

can u type it then? how do we know where u went wrong?

[deft glacier]

ah so

N1=Ndash-27

[thorn marten]

The sum of all the terms = sum of the even terms + sum of the odd terms

[deft glacier]

yeah

27 can only come in AP

if N is 3

so

3=27-3

nvm

i am just hit and trialing at this point

[sturdy valve]

never do this lol

u are assuming common difference has to be an integer

[deft glacier]

yea

i never studied AP

i just know that same number adding is ap

[sturdy valve]

maybe study theory of AP

and come back to this question

[deft glacier]

ah 😦

[sturdy valve]

its not very hard

will take maximum 30 minutes of ur time

[deft glacier]

i know that sum of odd number is n sqaure

and even number is n(n+1)

cant we take n as 2k

and in put

[sturdy valve]

sure but theres a reason they have given u it as 2k

they want u to split it into even terms (k number of terms) and odd terms (k number of terms)

[deft glacier]

what

[sturdy valve]

if u take n=2k
then ull have to take n/2 terms for each and it will get complicated

[deft glacier]

oh

i will study theory

tmr

its about 12pm

12 am

how do u even split add and even terms

[deft glacier]

.

[sturdy valve]

^^

[deft glacier]

yes then

95+55+40

[sturdy valve]

what is formula for sum of AP

[deft glacier]

idk

[sturdy valve]

and i think u mean 95 = 55+40

[sturdy valve]

please learn it and come

[deft glacier]

yess

[deft glacier]

ok :>

which country u are in?

[sturdy valve]

india

[deft glacier]

yeah so

this question can be done with 10th or 11th knowledge?

[sturdy valve]

yes i learnt it in 10th

[deft glacier]

oh

that question is of jee mains

[sturdy valve]

oh ok lol

[knotty grail]

can you see that there are k odd terms and k even terms

[deft glacier]

no

[knotty grail]

odd terms meaning

[deft glacier]

45 is not equal to 50

[knotty grail]

terms at odd positions

[deft glacier]

how do we even know that

n is odd or even

[knotty grail]

like 1st term of AP, 3rd term of AP, 5th term of AP and so on

2k is an even number

there are 2k terms

read the question

[deft glacier]

yeah

yes

so

number of odd terms=number of even terms?

[knotty grail]

yeah

[deft glacier]

45/n=50/n?

[knotty grail]

what are 45 and 50

[deft glacier]

sum of odd and even terms

[knotty grail]

it's given 40 and 55

[deft glacier]

oh.

[knotty grail]

and i don't understand what you're trying to do

[deft glacier]

like the number of odd terms*the first term=40 or 55

and same goes for number of even terms

then if we divide it by n

bro yk what

[knotty grail]

all the odd terms are not equal to the first term

[deft glacier]

i will just read the thoery

[knotty grail]

yeah

so you didn't know what an AP is..

[deft glacier]

i knew

the number getting added

just don tknow the rest

[knotty grail]

it's something like 1, 5, 9, 13, 17..

[deft glacier]

yeah

ik

that

ONLY that ik :>

[knotty grail]

hm

you could get through it by intuition

[deft glacier]

trying that

[knotty grail]

but I suggest you read theory anyway

[deft glacier]

lets si

[knotty grail]

it's difficult to always get intuitive answers

and there are other problems where you are forced to use the formula for the sum of an AP

[deft glacier]

yes

[full forum]

@deft glacier Has your question been resolved?

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[thorn marten]

Solve the equation (x-2)(x-3)(x-4) + 5 = 0

[loud haven]

so whats the issue?

[thick hedge]

cubic I guess

[loud haven]

i guess, but its not like there arent formula for that

or trial and error if all fails

[thick hedge]

,w (x-2)(x-3)(x-4)

[glossy valve]

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[onyx glen]

thing has no easily guessable real roots so you're SOL

[thick hedge]

so we have x^3-9x^2+26x-19=0

use the cubic formula here I guess

[analog shale]

origin of the problem? this doesn't have a nice form solution

[onyx glen]

$10 says it's an ex recto problem

[strange basalt]

Cardono????? Wdym cubic formula

[analog shale]

yeah i think thats what they meant by cubic formula, just cardano

[loud haven]

either cardano or the generalized version

[onyx glen]

@thorn marten

why did you post your own pfp here

anyway let's keep quiet until @thorn marten decides to grace us with his presence once again

that was weird and out of nowhere.

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@thorn marten Has your question been resolved?

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[jovial meadow]

yo guys pppllllllssssssss dm me help me to understand powers must talk frensh but its ok if its english

[full forum]

Please don't occupy multiple help channels.

[steel solar]

.close

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[full forum]

@thorn wagon Has your question been resolved?

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@thorn wagon Has your question been resolved?

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[plucky remnant]

Could I express P as
P=2E_{k}/v

[plucky remnant]

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[sudden condor]

context ?

[plucky remnant]

Ah 🥲

[sudden condor]

the timing lol

[plucky remnant]

That was like the same second xd

[sudden condor]

[sudden condor]

it should work cuz simplifying would yield mv

[plucky remnant]

Yeah

But for light it's just
P=E/v

v=c

[sudden condor]

.reopen

[full forum]

✅

[sudden condor]

well E here shouldn't be related to mass but rather frequency if I remember correctly

[plucky remnant]

Do you remember the name of that E?

[sudden condor]

I remember the equation
E=hf

[plucky remnant]

What's h

Planck?

[sudden condor]

yeah plancks constant

[plucky remnant]

Okay I see

But what kind of energy would it be?

[sudden condor]

for the name I'm not really sure

[plucky remnant]

Okay

[sudden condor]

and the equation can be simplified further using wavelength

p=E/c
p=hf/c
[f/c=1/lambda]
p=h/lambda and also lambda =h/p for de brole's wave (not sure I wrote the name correctly)

[plucky remnant]

Aah I see

Looks like it is called electromagnetic energu

Which makes sense

[sudden condor]

hope this clears it up

[plucky remnant]

Well thank you for the explenation

[plucky remnant]

Yuo

I completwly forgot about E=hf

[sudden condor]

[plucky remnant]

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[soft quiver]

Have you read about the basic proportionality theorem?

[toxic ermine]

no but is that a required thing to know

[soft quiver]

It'll be easier if you know it

[toxic ermine]

well for x i got 1.25

does that work

[soft quiver]

I got x=5 using basic proportionality theorem

[toxic ermine]

so the whole thing is 7.5?

if thats the case, shouldn't the ratio be 1.5*2.5 ?

[soft quiver]

Ratio of what?

[toxic ermine]

of 3 and 2

[soft quiver]

Yes

Ratio of 3 and 2 is 1.5 regardless the question

[toxic ermine]

can you write out how you did it

[placid oak]

you've got two similar triangles there

[toxic ermine]

yea

[toxic ermine]

can i do 3/2 = x/2.5

[placid oak]

can you figure out the angle in the bottom left corner, once using each similar triangle - they should be equal

[soft quiver]

No

[placid oak]

so, the entire bottom length is 2.5, not just the extra length?

[toxic ermine]

oh wait no 3/2 = x+2.5 /2.5

[soft quiver]

Yes

[placid oak]

no, this isn't correct

draw out the two similar triangles separately

[toxic ermine]

should look like that

[placid oak]

no

[toxic ermine]

whats wrong with it

well x = 5

from that

[placid oak]

this 🙂

[toxic ermine]

yea that

[placid oak]

this one is correct 🙂

[toxic ermine]

so its 5?

[placid oak]

yes

[toxic ermine]

oh

i get it

thanks

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[steel solar]

Do you know the definition of complementary?

[signal pecan]

Yeah

It’s something that adds up to 90°

Right?

[tender nacelle]

be more specific

[signal pecan]

That’s all I know im sorry

[tender nacelle]

A pair of complementary angles are two angles that add up to 90 degrees

So not a single "something", it's two angles

[signal pecan]

Oh okay

So how would you do it with letters?

[steel solar]

?

[signal pecan]

Like how do I know it adds up to 90°

[limpid kiln]

Angle Addition Postulate

If you have a 90 degree angle and it is split into two angles

By AAP you know they are complementary

[steel solar]

(If there is a right angle contained by 2 angles)

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[cedar minnow]

Need some help thinking through this coset proof.

[cedar minnow]

I'm kinda stuck. Not sure where to go next.

[full forum]

@cedar minnow Has your question been resolved?

[analog shale]

maybe start with the easy case that a=b

hint: do a contradiction, if both a,a^2 are in aH what happens (assuming you are doing left cosets for the definition of [G:H])

[cedar minnow]

What is aH? is that the other coset?

[analog shale]

$aH$ denotes the coset $aH:={ah: h\in H}$, since your index is $2$, and $a\not\in H$, we know it is the other coset.

[glossy valve]

qwertytrewq

[analog shale]

I'm assuming some background: that cosets are disjoint, and, in some sense, equinumerous.

[haughty saddle]

Gurt: yo

[cedar minnow]

Yeah they would be disjoint.

I feel understanding how you get aH is the piece I'm missing. Since I don't understand how that is constructed. So just reading it and making it make sense in my head

[analog shale]

by construction a is in the other coset, and we wish to show that a^2 is in H (this is the case where a=b).

so in particular, if a^2 is not in H it must be in the other coset, but we have previouse established that the other coset is aH (since a is in the other coset).

So we can try and prove by contradiction that somthing will go wrong if both a, a^2 are in aH, where aH is the other coset.

[cedar minnow]

Yeah that approach makes sense. Thanks! I'm going to give this a try.

[cedar minnow]

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[young night]

here is the equation to the oringal one $f\left(x\right)=\left{2\le x\le5:\ -2x+7,\ 0\le x\le2:\ x+1,\ -4\le x\le0:\ 1,\ -5\le x\le-4:\ 3x+13\right}$

[glossy valve]

ppq#7826

[limpid scroll]

do you know graph shifting?

[young night]

yes

[limpid scroll]

then where are you having trouble

[young night]

doing it all to this set

would this be right

@limpid scroll

[limpid scroll]

ok wait lemme check

[limpid scroll]

ye all of them are correct

[limpid scroll]

i am sorry i cant understand this one

[young night]

but would the overlapping of colors be fine or not

[young night]

i added the oringal to this to show the transfomrations

[limpid scroll]

why did some of them change shape

[young night]

idk but the first one is fine

[limpid scroll]

ye the first one is correct

[young night]

t=alright thanks

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[covert goblet]

Uhh how do we know f is continuous

A surjective map between two metric spaces isn't necessarily continuous is it?

[silver topaz]

it isnt, but it is needed for this problem

what they want you to prove is that the continuous image of a connected space is connected

[covert goblet]

but the question never stated that f was continuous

[silver topaz]

that is true 😄

[covert goblet]

did they just leave it out by mistak

lmao

[silver topaz]

well, it could be that by "map" they mean a continuous function

but this is purely speculation on my part

if you take "map" to just mean any old function, then obviously this question is wrong 😄

[covert goblet]

yea, I think it's just wrong

There were theorems that went "Let f be a map. f is continuous iff..."

None of which apply here afaik, but it does show they don't think a map is automatically contunious

[silver topaz]

but do note: if f:X->Y is a continuous surjective map and X is connected, then Y is connected is a true statement and the proof of this is presented here

[covert goblet]

thanks

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[silver topaz]

glad to help 😄

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[covert goblet]

Greetings again, need some help checking my intuition for the second part here

[covert goblet]

I claim that $f$ is not differentiable at nonzero because $f$ is not even continuous. Suppose $f$ is continuous at some nonzero rational $x$. Then by the sequential characterisation of continuity, we must have that for every sequence $(r_n)$ that converges to $x$, the sequence $(f(r_n))$ converges to $f(x) = x^2$. But consider a sequence $(r_n)$ that converges to $x$ where all the $(r_n)$ is irrational (we can find this because of the density of the irrationals in the reals). It must be that $\lim (r_n) = 0$ since every $r_n = 0$, Then we have $0 = x^2$ for nonzero $x$, a contradiction

[glossy valve]

astral

[onyx glen]

yes you're spot on

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@covert goblet Has your question been resolved?

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[full cobalt]

Need help can't slove this

[misty violet]

I can think of a way but doesn't know if it's the easiest

[full cobalt]

I mean it's for my friend so as long as it works it's good

[misty violet]

Ok taking AD=h

Do u know the formula for area of trapezium?

[full cobalt]

I forget everything i studied last year

[misty violet]

Ok

[weary shadow]

is that gcse?

[misty violet]

The area is like A(ABD)+A(BDC)

Which is 1/2h4+1/2h9

[full cobalt]

It's a math i .q test

For university

My friend is applying to

[misty violet]

U get this? @full cobalt

[full cobalt]

Where could I get it

Kinda didn't

[misty violet]

Taking AD=h

A(ABCD) = A(ABD) + A(BDC)

=1/2 h 4 + 1/2 h 9

[devout nova]

if you observe carefully all the right triangles are actually similar

so you can just use triangles BAD and ADC

[full cobalt]

.closs

,closs

[misty violet]

Hh .close

[full cobalt]

.close

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[random lichen]

how do i do this

[random lichen]

idk how to do b

[grave elm]

you know how to express 7 - 4x - x^2 in the form p - (x + q)^2

can you also express 7 - 4(y+3) - (y+3)^2 in the form p - ((y+3) + q)^2?

[random lichen]

uhh

wait

[grave elm]

a hint is that it's basically the same thing

you just replace x with y + 3

nothing more

[random lichen]

oh

then just solve?

[indigo frost]

yep

[random lichen]

ok

so does x replace the two brackets there

[woeful pasture]

U can interpret that as just moving the function

[random lichen]

or just the first one

[grave elm]

I think it will be better to replace x in p-(x+q)^2 with y + 3 instead

[indigo frost]

x=y+3

[grave elm]

so you'd get p - (y+3+q)^2 = 0

[random lichen]

ohh

[indigo frost]

you can i.e solve the first equation and then change the solutions you get with y+3

,, x=5 \text{ becomes } y+3=5

[glossy valve]

<rajel />

[random lichen]

ya i got -2 +- sqr root 11

but im like

still unclear about replacing it

[grave elm]

what p and q did you get btw?

just for sake of illustration

[indigo frost]

hmm , that doesnt seem correct tho

[grave elm]

,w expand 11 - (x + 2)^2

oh i see what you did there

[flint venture]

hello

linear algebra chanel?

[indigo frost]

its not 11

oh it is

my bad

[grave elm]

Do you remember the algebra you did to get from 7 - 4x - x^2 to 11 - (x + 2)^2?