help-23

[marsh walrus]

you need to write a convincing argument for why your 'answer' is true

the definitions are there to help you

you just state what you want to show, cite the definition, and do the little work you need to do to tie it together

[covert berry]

i tried to be more "formal"
can you check this one?

[safe radish]

@covert berry Has your question been resolved?

[safe radish]

@covert berry Has your question been resolved?

[clear blade]

you mean x^4 not in P2 right?

then it’s correct

[safe radish]

@covert berry Has your question been resolved?

[covert berry]

.close

[safe radish]

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[karmic pelican]

I am having serious trouble problem solving when it comes to permutations and even understanding them in the first place.

So, I know the addition principle and the multiplication principle and how to use them.

BUT, I can't use permutations and I can't understand it because... It's so weird. I aksi need a concrete explanation of what the word "Ways" mean in math, because it's getting mixed up. For example: How many ways can you sort a team of people?

[safe radish]

@karmic pelican Has your question been resolved?

[safe radish]

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[safe radish]

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[minor barn]

This might be a silly question but do I have to substitute every x with the limit or can I choose which x's I want to substitute?

[minor barn]

for example, lim(sin(x) - cos(x))^tan(x-pi/4), with x approaching pi/4

could I only substitute the x in the tan and leave the others?

what do you mean?

[flat frigate]

QuasiStar 超新星

[minor barn]

what makes this case stand out?

[tawdry ether]

isn't it just because as lim x->pi/4 that tan(x-pi/4) -> 0 so no matter what the base is the answer will be 1

[dusty crescent]

Actually not

[minor barn]

iirc if the base is 0 and the exponent is 0 then its undefined

or something

[dusty crescent]

Exponents with negative bases, for example (-1)^pi, are not generally defined.

So the limit does not exist

[minor barn]

wtf

why

[dusty crescent]

This is a principal branch

[minor barn]

apparently you can't have a negative base with ^ being a fraction number

that's interesting

[tawdry ether]

well you can but that'll just be in complex numbers

[minor barn]

also I just checked this and it also results in the answer being 1

[minor barn]

yeah but why though

oh

squared

i mean

square root

lol

[dusty crescent]

It's not that simple. Many properties of exponents only work for positive bases.

[minor barn]

@dusty crescent hey I was wondering if I could selectively substitute the x's on all limits without any problems

[dusty crescent]

If both limits converge, and the base is positive, then yes

[minor barn]

wdym both limits converge?

[minor barn]

also I did it on this excersise and it gave me the same result as wolfram did

i substituted the x in tan

= 1

[dusty crescent]

Like if f(c) approaches a and g(x) approaches b then g(x)^f(x) approaches b^a assuming b≥0

[minor barn]

could you perhaps give me an example?

[dusty crescent]

Your example was great. It would in that context be correct to substitute in pi/4 directly.

Another example would be lim (e+x)^(20x) when x-> 0.

This turns into e^0=1

[minor barn]

which example would make it not possible to substitute? it if in my excersise the base was sin(x) - 2cos(x), then b would be negative, so therefore I can't do this trick right?

[dusty crescent]

Yes and if you consider lim x^sin (1/x) when x tends to 0 from the right, then the exponent does not converge so you have to find a different appoach.

In this case it diverges

[minor barn]

it doesn't converge because the exponent would be infinite while the base would be 0 right

[dusty crescent]

That is 1) not true and 2) not a valid proof

[minor barn]

damn

how do you know it doesn't converge then?

[dusty crescent]

The exponent stays in the range (-1, 1)

So it is not infinity.

And even if it tended to infinity, it could have done so slowly enough for the limit to exists.

It's not all about what the different part converge to. It also matters how fast they converge.

When x tends to 0 in this case, x^sin(1/x) wobbles rapidly between 0 and 1. Here, desmos is useful for intuition.

However, I should sleep now

[minor barn]

Aw man

thanks for your help anyways

goodnight bro

.close

[safe radish]

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[broken tide]

Here’s the question and my working out, I’m not sure why I got it wrong

[broken tide]

To be fair I’m not 100% confident that that’s the way I’m supposed to work it out .but it should work I think?

[safe radish]

@broken tide Has your question been resolved?

[thin bridge]

I don't understand your steps

or what you're trying to do

[minor barn]

@broken tide

<@&286206848099549185>

[thin bridge]

@minor barn the channel is now occupied by someone else. if you need further assistance claim a new channel

[sweet warren]

Can't see the image clearly

[minor barn]

i am trying to help him by getting more helpers

helping a brother out

[thin bridge]

don't ping for others

[minor barn]

why not

its been over 15 mins

[thin bridge]

especially if someone is already trying to assist

[broken tide]

Oh shoot . I forgot it only works on right angled triangles lmaoo

I’m not rlly sure how to go about solving it then ?

[thin bridge]

try following the instructions in the question

[broken tide]

I’m not really sure how ..

Because I thought h and k were one point

[thin bridge]

can you create an equation with h,k knowing
(h,k) lies on y = x + 1

[broken tide]

OOOH TYTY

Hang on but how would I know that it was 3 and 4

[sweet warren]

,tex \sqrt{(h-x)^2+(k-2)^2} = 5

[flat frigate]

Lia 👅
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[thin bridge]

!nosols

[safe radish]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[broken tide]

Well that’s not really the solution

It’s just a part of the figuring out

[sweet warren]

You can make it quadratic equation

[thin bridge]

could tell them what to apply and have them try applying it

instead of doing it for them

[broken tide]

. Uh how does that apply

Sorry I’m not super good at maths that’s why I’m here

[sweet warren]

Squaring both sides of the equation

[thin bridge]

can you create an equation with h,k knowing
(h,k) lies on y = x + 1

[broken tide]

Isn’t that Pythagoras

[thin bridge]

distance formula is an application of Pythagoras

[sweet warren]

,tex so we have the equation $k = h + 1$

[flat frigate]

Lia 👅

[broken tide]

Like I understand that the sides are 3 and 4, but how would I figure that out?

[sweet warren]

,tex Substituting $k = h + 1$ into the distance formula equation

[flat frigate]

Lia 👅

[thin bridge]

could tell them what to apply and have them try applying it
instead of doing it for them

[sweet warren]

I am telling them to apply it

not doing it 🗿

[thin bridge]

you've essentially completely done the first two parts of the question for them

[broken tide]

I’m super lost sorry 😭

[sweet warren]

🗿

my bad

[broken tide]

I still don’t

Uuh

So I can’t really see how this helps

[thin bridge]

don't worry about that image

[broken tide]

Oh .

So uh what

[thin bridge]

can you create an equation with h,k knowing
(h,k) lies on y = x + 1

[broken tide]

Ya so k = h+1 I get that

[sweet warren]

So look substitute k=h+1 in to distance formula

[broken tide]

OOH WAIT

[thin bridge]

can you apply the distance formula to create an equation for
distance between (h,k) and (0,2) is 5

[broken tide]

YEAH DISTANCE

[sweet warren]

Yes

[broken tide]

Sorry yeah that makes sense lmaoo I just added an extra step for no reason

[sweet warren]

When you have substituted tell us what you have

[broken tide]

That would be like this right?

[sweet warren]

No

[broken tide]

Ah ?

[thin bridge]

yes

[broken tide]

[sweet warren]

How 🗿

Oh wait nvm

[thin bridge]

that's just from distance formula

[broken tide]

But I’m not 100% sure how to ..uh

[sweet warren]

you didn't substitute the k = h+1

[thin bridge]

you didn't square the 5 on the right side here

[sweet warren]

yes

[broken tide]

Wait was I supposed to uh

I think I’m lost again

[sweet warren]

Yes to cancel the square root

[thin bridge]

yes, do same stuff you both sides

[broken tide]

OOH EYAH

YEAH

Okok

[thin bridge]

you cant just square one side to cancel the root and leave the other untouched

[broken tide]

Mbmb

[sweet warren]

You still confused?

[broken tide]

I think I can get it now

[sweet warren]

Okay

[broken tide]

I gotta get them to be 4 and 3 squared

So it = 25

[sweet warren]

Yes

[thin bridge]

solving will give you the solutions

[broken tide]

I got ..a bunch of answers

[sweet warren]

is that your final answers?

[broken tide]

Since I need to get them to = 16 + 9 or 9 + 16

[thin bridge]

that's unreliable

[broken tide]

There’s a bunch of ways to do that

[broken tide]

I think that’s all the possible values ?

At least whole numbers

[sweet warren]

Something is wrong

[broken tide]

Ya idk what

[thin bridge]

16 and 9 isn't the only way to get 25,
values don't necessarily need to be perfect squares, so this way isn't really idesl

[broken tide]

Yea so there’s really an infinite number of answers

Which isn’t right

[slate quail]

My god

[thin bridge]

you've yet to use k = h + 1

[slate quail]

This server is clutch

[sweet warren]

yeah

[thin bridge]

ideally you should substitute that into the quadratic equation

[broken tide]

I’m not sure where I need to use that ?

[thin bridge]

which will result in a quadratic equation with one variable

[sweet warren]

Substituting $k = h + 1$ into the distance formula equation

[flat frigate]

Lia 👅

[thin bridge]

you posted in an occupied channel

[broken tide]

idk how.. what

how do i do that

[thin bridge]

replace the k in your equation with (h+1)

[slate quail]

Sorry

[thin bridge]

literal definition of substitution

[broken tide]

which equation im so confused

really sorry i feel like im being difficult

[thin bridge]

replace the k in
(0-h)^2 + (2-k)^2 = 25
with (h+1)

[broken tide]

OOH

YA I CAN DO THAT

[thin bridge]

() included to maintain order of operations

[broken tide]

but just to clarify, why is this wrong?

because it seems like it would work

[thin bridge]

some of the pairs you listed don't result in a sum of 25

[broken tide]

huh?

but like if h = -4, then the first part will be 16. then if k = -1, then the second part will be 9. 16 + 9 is 25?

[thin bridge]

and of the pairs you manage to find, you'd need to identify ones where k = h + 1

[broken tide]

OOH YEA

i forgot about that

sorry sorry that makes sense thank you

I GOT IT RIGHT :DDD

tysm sorry i took a while to catch on

[thin bridge]

even funding the nice pairs is inefficient / tedious
(easy to miss, so it's better to apply the restriction and substitute earlier)

[broken tide]

.close

[safe radish]

Channel closed

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[safe radish]

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[hybrid wedge]

AD = 14 ; ER=25 SE=??

@knotty turret is this the qstn?

[knotty turret]

Yes

[thin bridge]

!status

[safe radish]

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

[torpid compass]

@knotty turret join EA with imaginary line

[knotty turret]

So its 25

[torpid compass]

yes

[knotty turret]

Ea

Ok

[torpid compass]

ES is perpendicular bisector of AD

[knotty turret]

Sorry.. what? hehe

It's also 14?

[torpid compass]

no

wait sry gtg

[knotty turret]

Oh ok

[torpid compass]

but s is mid point of AD

[knotty turret]

Ok ok

Hi @hybrid wedge hahaahha

[hybrid wedge]

Do you know what perp bisector means?

[knotty turret]

Line in the center which divides the chord to 2 equal parts?

Im so lost this is embarrassing

[hybrid wedge]

The perp bisector is the line perp to the given line and bisector(divides line into 2 equal parts) of the given line

Btw which grade are you in?

[knotty turret]

10 😭

I was absent that day

[hybrid wedge]

Did you get this?

[knotty turret]

Yes

[hybrid wedge]

Line joining centre of circle and midpoint of chord is the perp bisector of the chord (property of circle)

[hybrid wedge]

Which country?

[knotty turret]

Sorry cant give those type of info (mom would get mad)

[hybrid wedge]

So what is the value of AS now?

[hybrid wedge]

Np, i can understand

[knotty turret]

7?

[hybrid wedge]

Yes

[knotty turret]

So whats the next step?

[hybrid wedge]

ASE is a right angled Δ and you know two sides

[knotty turret]

Ohhh so ill use pythagorean again?

[hybrid wedge]

Yes

[torpid compass]

yes

[knotty turret]

Thank youuu

Omg next is angles

Answer is 24

[safe radish]

Channel closed

Channel closed due to the original message being deleted.
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• Show your work, and if possible, explain where you are stuck.
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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[knotty turret]

X is 92?

[knotty turret]

Dont mind the drawing lol

[twilit hare]

yes

the sum of angles in a cyclic quad is equal to 180 - by this property

[knotty turret]

How about this tho

[torpid compass]

360?

[knotty turret]

Eyy the circle improved hahahaha

[torpid compass]

half of angle of 70

this is a property

[knotty turret]

35?

[junior smelt]

[opposite angles, @twilit hare probably meant]

[torpid compass]

ig so

[knotty turret]

Is this right?

[torpid compass]

yes

[knotty turret]

Ooo nice thanks again

[twilit hare]

sorry typo- the sum of opp angles*

[twilit hare]

yeah thanks

[knotty turret]

(Thats 75) so x = 105?

Hi guys hahahahaha

[torpid compass]

yea

but you know the approach?

[knotty turret]

No hehe

Just did the same as the last one

How do i do this haha 35 is the only given

(Btw the middle is y)

Need to find the minor arc

[torpid compass]

length of minor arc?

[knotty turret]

Yes

PT

[torpid compass]

OE=PT?

[knotty turret]

?

Wait ill do another drawing haha

Wait what hehe

[torpid compass]

see take YOE as a triangle

[knotty turret]

Ok

[torpid compass]

YO = YE = radius

[knotty turret]

Oh ok ok

[torpid compass]

when 2 sides of a tringle are equal then its called isosceles triangle

[rose holly]

Angle yeo = you cpct

[knotty turret]

Wheres u

[torpid compass]

not everyone can indentify similar triangles

whers u

btw

[torpid compass]

ok u saw tht isoceles triangle right?

[knotty turret]

Yes

[torpid compass]

The two angles opposite to the equal sides are equal

got it?

[knotty turret]

Yes

[torpid compass]

now angle YEO = angle YOE

[knotty turret]

Oh ok ok

[torpid compass]

whats angle YOE?

[knotty turret]

YEO? hahhaha (sorryy in advance)

[torpid compass]

yes = 35 degrees

now use angle sum property in triangle YOE

[knotty turret]

Y is 110
O and E is 35?

[torpid compass]

yes

you know formula for length of arc ?

[knotty turret]

Is it

[torpid compass]

yes

you know theta value

[knotty turret]

35?

[torpid compass]

110

from the center

[knotty turret]

Oh ok ok

[knotty turret]

How about r tho

[torpid compass]

they dint mention r

[knotty turret]

This said 2pier tho

[torpid compass]

so weare finding in variable r

[knotty turret]

No the measure of the arc pt

[torpid compass]

ya we r finding the length in terms of formula cause they dint mention the value of r

[knotty turret]

Ooo ok

[torpid compass]

in the formula substitute the value of theta as 110

yea you can simplify it if you want

btw OE = PT because length of chords are equal because distance from center is also equal

we found OE because we know OE = PT

[knotty turret]

Oh ok so P and T are also 35

[torpid compass]

ya if you join YP and YT

[knotty turret]

Whats next?

[torpid compass]

this is the answer

i just explained why we found OE instead of PT

[knotty turret]

Oh really ok ok

If simplified

11rpi/18

Ok thank youu

[torpid compass]

yw

[knotty turret]

Ill just guess the remaining questions cause i cant do it anymore 🥳

[torpid compass]

exam?

[knotty turret]

No just a activity

That is not graded but req to answer

[torpid compass]

ok

[knotty turret]

Oh the last one is 18

But i think its wrong tho

All of them hahaha

[torpid compass]

1st one is correct

[knotty turret]

Nice one hahaha

[torpid compass]

2nd is also correct

last one join OL

use pythogrss theorem

[knotty turret]

Oh that again 😭

So 4?

[torpid compass]

yea

[torpid compass]

x = 8

[knotty turret]

Ok ok

The last 3 questions makes me want to throw the computer

[torpid compass]

😂 .

[knotty turret]

Like huhhhh

[torpid compass]

find x?

[knotty turret]

Is he trying to make a new logo or something

[knotty turret]

Yea

[torpid compass]

lol

[torpid compass]

use the property

angle p will be half of 100

[knotty turret]

50?

[torpid compass]

if the 100 degrees is center

[torpid compass]

yes

now x+50 = 180 (its a line)

[knotty turret]

130?

[torpid compass]

wait a minn

i think i did wrong

sry

[knotty turret]

Oh hahaha its ok

[torpid compass]

ok x = 50

property for this

[knotty turret]

💥

Mind blown

[torpid compass]

idk how to define this property

[knotty turret]

So 96 is just a display?

(Im out of words hahahaha)

[torpid compass]

idk wht is 96 degrees doing over there

[knotty turret]

Ok lets goo last 2 questions hahahaha

Both are identifying the shaded region 💥

[torpid compass]

umm sry but gtg

bye

[knotty turret]

Ok byeee

[safe radish]

@knotty turret Has your question been resolved?

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[lean otter]

Hi,

[lean otter]

im actually dumb but help pls

James jogged 3000m from 5:45 to 6:15 on a sat morning. Find the speed he jogged that morning.

A

And this,
An eagle flew a distance of 36km in 40m/min.
Find the speed of the eagle in m/min

Pls help

<@&286206848099549185>

Hi

pls

<@&286206848099549185>

[vale robin]

assuming your speed is constant it will be distance/time

[lean otter]

for which question

the first or sec

[vale robin]

for both

[paper jewel]

5:45 to 6:15 (assuming the difference is only 30 mins) - 3000 m in 30 mins = 3000/30 = 100 m/min

[lean otter]

ohh ok

Wut abt the next one

[paper jewel]

What is 40m/min?

[lean otter]

speed

40meters per minute

[real wasp]

36 / 40 = 0.9km
40 / 40 = 1min

So speed is 0.9km/min

[lean otter]

ohh

why 40/40?

[paper jewel]

36km = 36*1000 m = 36000 m so 36000/40 m/min = 900 m/min

[real wasp]

Well you want to find th km/min right

[lean otter]

yeah

[real wasp]

Its 36km every 40 minutes

40mins / 40mins = 1 min

[lean otter]

ohh ok ty

i got it alr byee! ^^

[paper jewel]

Try to understand the questions and solve them by own. Don't ask your school homework questions though

[ruby bolt]

@lean otter

[lean otter]

naur its my mistakes on the test

[ruby bolt]

How old are you

[lean otter]

13

grd 8

[ruby bolt]

ok

[lean otter]

byee cya

.close

[safe radish]

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[surreal mauve]

Am I allowed to do this?

[surreal mauve]

Hello?

<@&286206848099549185>

[safe radish]

@surreal mauve Has your question been resolved?

[surreal mauve]

No

[safe radish]

@surreal mauve Has your question been resolved?

[safe radish]

Channel closed

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[safe radish]

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Remember:
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• Show your work, and if possible, explain where you are stuck.
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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[lean otter]

Are you guys able to read my question?

[desert pasture]

yes, if no one helps ping helpers in 15 minutes

[lean otter]

thanks, who are the helpers?

because it's been 30minutes

[quasi bison]

@Helpers

[lean otter]

I started a new channel

<@&286206848099549185>

[quasi bison]

yes this is legible

!status

[safe radish]

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

[lean otter]

1

I don't know how the apply the lagrange mean value theorem

@quasi bison, can you solve it?

[quasi bison]

!noans

[safe radish]

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.

[quasi bison]

can you state the thm itself

[lean otter]

f(b)-f(a)/b-a

I understand the theorem, I just don't know how to apply it to that exercise

it means that there exists a point at which the rate of change (derivate) equals the rate of change from a to b entirely (so if you take a line through a and b, you get the same rico as a certain point c on the function).

I understand the theorem 😭

[quasi bison]

no,

that is far from a complete statement of the theorem

[quasi bison]

there exists a point c such that (f(b)-f(a))/(b-a) = f'(c)

or to put it another way there exists c such that f(b)-f(a)=f'(c)(b-a)

do you know what the derivative of cos(x) is? @lean otter

[lean otter]

-sinx

I'm not gonna lie you probably have no clue how to solve it yourself @quasi bison

you're asking me what the derivative of cos x is ...

[gusty crest]

don't underestimate ann

[quasi bison]

yeah, to see if you know

but like idk, if you want to throw punches at me i can leave you hanging here

[quasi bison]

i was now going to tell you to express this with f(x) = cos(x)

[lean otter]

okay so -sin(c)(b-a)=cos(b)-cos(a)

[gusty crest]

are you wondering what to do next @lean otter or no

[lean otter]

yes

[gusty crest]

you are close

you are after a relation between cosb - cosa and b - a

idk how to hint this without giving away the answer

you want |cosb - cosa|

can you manipulate your current equation to find some equation involving |cosb - cosa|?

[lean otter]

got it!

😛

thanks rasp

[gusty crest]

won't lie i only got it bc of ann's hints too

but awesome and np 🎉

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[surreal mauve]

Am I doing this right?

[surreal mauve]

<@&286206848099549185>

[formal swallow]

Can't see the pictures clearly

[surreal mauve]

Ok wait

[surreal mauve]

Now?

[lean otter]

do you know the formula for the thrid column

[surreal mauve]

Yes

How else would I have done it?

I'm just asking if the third step is right

[safe radish]

@surreal mauve Has your question been resolved?

[surreal mauve]

No

<@&286206848099549185>

HELP PLZZZZ

Argh I'm done

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[silk token]

Imagine we've the next exercise: find if the matrix A is diagonalizable. I would follow the next steps Find the autovalues (d) with its charasteristic polinomial |A-d*I|=0, then, I would have to find the numbers of autovectors for each autovalue so i can know if the geometrical multiplicty is equal to the algebraical multiplicity of the autovalues. But I want to know if there's other way to know it, so I've noticed that if rg(A-d*I|0)=2 we would have an only autovector because in the undeterminated system we would have to put an only parameter to solve it, and if rg(A-d*I|0)=1 I would have to put 2 parameters to solve it, so it generates 2 autovectors, which means that in that cases (which are the most of the cases that I've seen) if rg(A-d*I)=2 which at the same time is rg(A-d*I|0) I would have an only base, but if that's equal to 1 I would have two bases. So I would be able to find out the geometrical multiplicity without knowing the autovectors (the ones that we need, I mean the bases of the autospaces), just calculating the dimension of our vectorial space minus the rank of the matrix, but if I want to use that method that suposition must always work, otherwise said. I want to know if: for a space R^n the geometrical multiplicity (G) matches to the next formula: G=n-k being k the rank of the rank of the matrix (A-d*I)

[safe radish]

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[severe moon]

can someone help me with this question pls

[severe moon]

i've already solved for the cost of one apple, one banana and one orange using system of equations

but idk what to do next

can someone help me pls

[safe radish]

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[severe moon]

.reopen

[safe radish]

✅

[severe moon]

<@&286206848099549185>

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[eternal carbon]

it doesn't look like this question has a unique solution

[safe radish]

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[severe moon]

WDYM?

[eternal carbon]

to begin you'd divide 3.12 by 1.2 because his 3.12 is 120% of the price of the bulk purchases

[severe moon]

it should have a solution

[severe moon]

i don't get it

[eternal carbon]

like

[severe moon]

especially ". Individual fruit bought costs 20% more than the corresponding price in any of the special
offer baskets."

[eternal carbon]

you calculate the cost of a banana to be 0.4 right

[severe moon]

yes

[eternal carbon]

that means that if he bought the banana individually, he would pay 0.48

[severe moon]

how?

[eternal carbon]

that's what this means

[severe moon]

ye but idk what it means

[eternal carbon]

buy the banana as part of a basket = pay 0.4
buy the banana alone = pay 0.48

that's why i'm telling you

[severe moon]

but how it is 0.48

cuz i don't understand that line

[eternal carbon]

do you know how percents work?

what's 20% more than 100?

[severe moon]

first u need to find 20% of 100

then add that to 100

[eternal carbon]

yes

[mellow lodge]

well technically you can just multiply by 1.2

[severe moon]

ye

[eternal carbon]

so you have that 20% more of 0.4 is 0.48

[severe moon]

so ur just saying that this line means if u buy an individual fruit, it costs 20% more than the cost of the same fruit inside a special offer basket

[eternal carbon]

yes

this also means that the 3.12 he paid at the end is 20% more than the cost of the basket price of those fruits

[severe moon]

oh ye

so would it be 0.48, 0.36 and 0.24

@eternal carbon

<@&286206848099549185>

[eternal carbon]

well you could do this but i wouldn't

[severe moon]

?

[eternal carbon]

it's easier to just use the basket prices and divide 3.12 by 1.2

[severe moon]

you get 2.6

[eternal carbon]

yes

[severe moon]

so you divide 2.6 by 2 bananas, 2 oranges and 3apples

right?

@eternal carbon

[eternal carbon]

yes

at least 2 bananas, at least 2 oranges, at least 3 apples... so you can just subtract the sum of those prices from 2.6

[safe radish]

@severe moon Has your question been resolved?

[eternal carbon]

uh

have you tried this @severe moon

[severe moon]

3.12-2.4

=0.72

0.48+0.24=0.72

so 4a, 3b and 2o

[eternal carbon]

the reason i suggested to work with the basket prices was because 0.4, 0.3, and 0.2 are nicer numbers than 0.48, 0.46, and 0.24

[eternal carbon]

but yes this is a potential solution

there is another one though

[severe moon]

3 apples, 2 bananas, 4 oranges

wait

but If it says "I dont fancy any of the mixtures of the baskets" then maybe cou can disregard the 4 apples, 3 bananas, 2 oranges solution since it is one of the baskets

[eternal carbon]

ah that's what it was

[safe radish]

@severe moon Has your question been resolved?

[safe radish]

@severe moon Has your question been resolved?

[severe moon]

but this combination is also in the basket

so what is the answer

?

<@&286206848099549185>

[eternal carbon]

there's a third one 😅

[severe moon]

wtf

what tf is the answer then

[eternal carbon]

2 bananas 2 oranges 6 apples

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[river hinge]

How do u solve the one in the middle

Last number is 9i

[round shore]

you expand and get two equations, one real, and one imaginary

then you have a system of two linear equations

[river hinge]

How do you do it

?

[round shore]

expand using distributive

[river hinge]

But what then

[round shore]

you get 8x + 4xi + y - 4yi + 7 = 3x + 3xi - 12i + 6 + 9i

simplifying

5x + xi + y - 4yi + 1 + 3i = 0

taking the real, we get 5x + y + 1 = 0

taking the imaginary, we get xi - 4yi + 3i = 0

we can divide the second equation by i, x - 4y + 3 = 0

and now its just a system of two linear equatiosn

@river hinge

do you understand?

[safe radish]

@river hinge Has your question been resolved?

[river hinge]

And what’s the answe overall

[round shore]

x = -1/3 , y = 2/3

[river hinge]

What how

[round shore]

Its a system of linear equations

5x + y + 1 = 0, and x - 4y + 3 = 0

You get it by taking the real part and the imaginary part of the equation and writing them as two separate equations

[river hinge]

Yes but how you solve this part

[round shore]

The linear equations?

[river hinge]

Yes

[round shore]

Multiplying the second equation by 5, and subtracting them, we get

5x + y + 1 = 0

5x - 20y + 15 = 0

21y - 14 = 0

or y = 2/3

plugging in our value for y in the first equation,

5x + 2/3 + 1 = 0

5x = -5/3

x = -1/3

so x = -1/3, y = 2/3

[river hinge]

Ah ok

Thank you very much

I really appreciate your patience 🫶

[safe radish]

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[lean otter]

How do I solve this

[quiet plume]

,rccw

[quiet plume]

B is solved by using the linearity of the integral, i.e. the integral of the sum of to things is the sum of their integrals.

C uses substitution.

A as well

The point is you want to express these integrals in terms of integrals you do know, and you know the integral of f(x) from a to b.

In A, you have to substitute something so that you have just f(u), not f(x-5) (the variable doesn't matter)

[safe radish]

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[safe radish]

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[remote sail]

Could someone quickly do question and give me the answer for C

I got the answer but my answer is different from the answers given and I feel like they might be wrong

[plucky elk]

!show

[safe radish]

Show your work, and if possible, explain where you are stuck.

[remote sail]

I calulated the area underneath the velocity graph to give the distance

I used both integration and just using triangle area

and both times I got 1.6

integration was

[plucky elk]

what's the given answer

[remote sail]

32/15

[plucky elk]

,calc 32/15

[flat frigate]

Result:

2.1333333333333

[plucky elk]

yea book doesn't seem right

should be 1.6

[remote sail]

alr thanks

.close

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[sharp sparrow]

Does anybody know how to solve number 2?

Im genuinely so lost

This is in calculus

[safe radish]

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[safe radish]

@sharp sparrow Has your question been resolved?

[eternal carbon]

do you know the formula for area of an equilateral triangle given its length?

[copper pewter]

It has a formula???

[eternal carbon]

yes

[sharp sparrow]

I have it at home. Sorry I just had to leave. Can I get back to you in the morning?

[eternal carbon]

sure but i'll probably be gone

[sharp sparrow]

That’s okay. The assignment is due Friday

[safe radish]

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[azure charm]

if we have this abelian group F(A) on the arbitrary set A (defined as on pic), can we say that there's only one function in F(A) that maps points {n_1, n_2, ...} to {k_1, k_2, ...} and points not from set {n_1, n_2, ...} to 0, is unique?

[lean otter]

yes, points in the set {n_1, n_2, ...} to corresponding values in the set {k_1, k_2, ...}.
All other points in A (not in {n_1, n_2, ...}) to 0, and with the properties of abelian group it means we can suppose that there were another function g in F(A) that also maps {n_1, n_2, ...} to {k_1, k_2, ...} and all other points to 0, then continue your work you'll find that the function f is indeed unique within the group F(A).

sorry for bad english ig

[safe radish]

@azure charm Has your question been resolved?

[azure charm]

so, i can just say that function if uniquely defined by points x -> y?

[lean otter]

i dont understand?

[peak estuary]

a function is uniquely determined by where it sends stuff to

[azure charm]

yes, that's what i mean

thank you both

.close

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[south sparrow]

What open/research questions connect to this Theorem?
\begin{equation}
P(A+B)=P(A)+P(B)-P(A \cdot B)
\end{equation}

[south sparrow]

Is this a Theorem?

[flat frigate]

kytsu1

[plucky elk]

What's A and B

And what's dot

[south sparrow]

A and B are arbitrary events, A•B is A⋂B of event algebra.

The likelihood of zeeing a zebra or an elephant, or both, is the sum of the likelihoods of seeing a zebra, of seeing an elephant, and of seeing the two at the same time.

P(🦓 v 🐘) = P(🦓 ) + P(🐘 ) - P(🦓 ^ 🐘 )

[plucky elk]

https://youtu.be/jAfNg3ylZAI?si=eUDbgcU9I7iRNs_q

[south sparrow]

I guess one meaningful research question would be to how to apply this to playing the guitar.

The ultimate low floor, high ceiling task.

[halcyon carbon]

A guitar??

[toxic stratus]

[safe radish]

@south sparrow Has your question been resolved?

[south sparrow]

https://arxiv.org/search/?query="inclusion-exclusion"&searchtype=all&abstracts=show&order=-announced_date_first&size=50

These are papers related to the inclusion-exclusion principle, or P(🦓 v 🐘) = P(🦓 ) + P(🐘 ) - P(🦓 ^ 🐘 ). I just found an activity that seems useful: paste the entire paper to a computer agent, and discuss it in terms of lighthearted analogies. In the context of pebbles, dry leaves, ocean waves, wrinkles on elephants or stripes on zebras.

Mathematics being inaccessible seems to be an open research problem as well.

[safe radish]

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[safe radish]

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[supple shore]

is this function continuous even if the lim when x is goes to zero by the left doesn't exist ?

[spice grove]

If you're saying something like continuous for x $\in (0,1)$ then sure.

[flat frigate]

! What the hell am I doing here?

[supple shore]

yes, why for something else it's not true

[spice grove]

If you want to say is it continuous at x = 0, then it is not.

[supple shore]

but if x is in [O,1] and it's continuous so for x=0 it's continuous no ?

[spice grove]

If it's that, then you don't need lhl.

You just need to check the right hand limit.

Because you can't approach this function from the left if anything less than 0 is out of bounds.

[supple shore]

so it's continuous at x=0

[spice grove]

It is, if that's the interval.

If it were (-1,1) then no.

I hope you get it.

[supple shore]

ok thanks a lot 🙂

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[grave moss]

hello. i have to discover center and radius of circunference given by this equation. im having trouble with the y on the right side. can someone walk me through?

[spice grove]

You try to express this as:
(x-a)^2 + (y-b)^2 = r^2

And for that, you're going to have tl subtract 8y from both sides.

[safe radish]

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[gray flint]

If I have the function f:(0,inf)->R, f(x)=e^x+lnx, then , since this function is monotonic, it has at most one solution for a value c. Is this correct?

[solar hazel]

yes f(x) = c has at most one solution (exactly one, even) for any c

[gray flint]

Is this also because the degree is less than 2?

[solar hazel]

wdym degree

[gray flint]

Grade

[solar hazel]

what

[gray flint]

The highest power of x

[solar hazel]

[gray flint]

for a polynomial x^2+x+1 , what would you call the power 2?

Not grade?

[solar hazel]

i would say degree but that's not important here

f isn't a polynomial

[gray flint]

Ik but if it had x to power 2 or more for example, then it would have more than one solution

[solar hazel]

you do not seem to know

[solar hazel]

this made no sense

[gray flint]

what do i not know

[solar hazel]

you said "the degree is less than 2" which makes no sense

[gray flint]

Does it not have a degree since it contains an exponential and logarithmic function?

[solar hazel]

sure

[peak estuary]

and also sidenote, some polynomials like x^3 still only have one solution for each c

[gray flint]

thank you for your help

.close

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[mystic bear]

how do i evaluate this following expression as a radical expression

[lean otter]

[
81 = 3^{???}
]

[lean otter]

what is ???

[torpid compass]

write 81 as 3^

[mystic bear]

3^4

[thin bridge]

as a radical expression
you shouldn't be bringing radicals into this

[lean otter]

great

now

[
\p{3^4}^4
]

[lean otter]

how does this simplify

[mystic bear]

it's my homework sooo.....

[thin bridge]

does it explicitly mention using radicals for this specific question?

[mystic bear]

it says to evaluate each expression as a radical

[mystic bear]

(81)^4

[thin bridge]

then you should ignore the part about radicals
and simplify applying the basic exponent laws

[mystic bear]

i have an example

@thin bridge

this is what i'm supposed to do

[thin bridge]

that has major formatting issues

[mystic bear]

yea ik

[shell epoch]

wtf is that

[thin bridge]

did you type that out yourself

[mystic bear]

i typed it

its supposed to be 3 squared

but thats one example

[thin bridge]

$8^{-\frac23} = \frac{1}{8^{\frac 23}} = \frac{1}{\cbrt{8^2}} = \frac{1}{\cbrt{64}} = \frac 14$

[flat frigate]

ℝαμΩℕωⅤ

[mystic bear]

yeah that

[mystic bear]

can i do the same with this

[thin bridge]

fractional exponents are related to radicals,
you original expression doesn't have fractional exponents

did you format the expression correctly?

[mystic bear]

i forgot if the 81 was outside or inside

[thin bridge]

did you intend to have something like
$$81^{\frac 34}$$

[mystic bear]

yea that

my bad

[flat frigate]

ℝαμΩℕωⅤ

[shell epoch]

okay so to solve the other one you would 1. Express with a positive exponent 2. write in exponential form 3. simplify 4. evaluate then you should have your solution

I think

[shell epoch]

wait wat

bruh

[mystic bear]

teach me

[shell epoch]

das just 3^3

[mystic bear]

is this it

[thin bridge]

yes, that would be equivalent

[shell epoch]

ya

[mystic bear]

okay

oh it’s that simple

tysm!

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[glossy obsidian]

Hello I would like to know why my domain is wrong.

[quasi bison]

,rccw

[glossy obsidian]

It's supposed to be [-pi,0] and I don't understand why

[quasi bison]

what does this say?

f(11) = 2 ?????(sqrt(x)) - pi

is how i read it rn

[glossy obsidian]

It's the function we are supposed to obtain the inverse of. It's arcsin

Sorry for the writing lpl

[quasi bison]

what about the 11 st the start?

also i cannot even fathom how that squiggle could decipher as arcsin

[lime dust]

I think it’s an x not an 11 lol

[glossy obsidian]

Wym?

[quasi bison]

f(11) = ...

[glossy obsidian]

It's an x...

[quasi bison]

i read those two vertical lines as 1s...

you really gotta make your handwriting less illegible

[glossy obsidian]

Yeah my writing is not so great

[quasi bison]

anyway

the domain of f is [0,1] yeah

what is its range...

f(0) is -pi

f(1) is... 0?

and f is strictly increasing.

[glossy obsidian]

Oh wait

That is wrong

[quasi bison]

what is

[glossy obsidian]

The CD is range

The range of the inverse function is [0,1]

That part I know is correct

I just don't understand why is the domain supposed to be [-pi,0] instead of the [-2pi,0] (the result that I got)

[quasi bison]

-pi, not pi.

-pi/2 <= (y+pi)/2 <= pi/2 is off

[glossy obsidian]

Ye ye I forgot in the message

Just corrected it

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[blissful sleet]

Task:

Given n - number of verticies in a tournament graph (2 <= n <= 500), with help of at most 2000 requests find any vertex with out-degree <= 1 or determine that there's no such one.
The request is 2 vertices and the answer on it is direction of edge between them.

I'm stuck at the very beginning, because it's a new kind of task for me and i can determine only like 1.6% percent of edges (in the worst case), so it'd probably be better if you only give a hint, not the solution fully.

[quasi bison]

what is a tournament graph?

do you have a defn handy?

[blissful sleet]

Every pair of vertices have only one directed edge between them

[quasi bison]

ah

so like, a graph that would become K_n if we were to strip the direction from all the edges

[blissful sleet]

yes

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@blissful sleet Has your question been resolved?

[zinc hornet]

My instinct is that you should first compare an arbitrary vertex pair (1,2). Now you have make another comparison. That part is easy, the question is what should your second comparison be?

A reasonable choice is to select which one of the previously two vertices, {1,2} and another vertex which is not in the set {1,2}. If you keep choosing the "winning" vertex, you'll likely get nowhere because there is a non-zero probability it could require 498 more comparisons to confirm its outdegree. So you have no choice but to select the "losing" vertex to have any hope of getting more information efficiently.

[safe radish]

@blissful sleet Has your question been resolved?

[blissful sleet]

Sounds kind of working

Thanks

.close

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[charred yew]

Hi could someone explain the solution steps I highlighted

[wind shoal]

2^1/2+1/2

2^(2/2)

2^1= 2

[charred yew]

Do you not times out there brackets ?

[wind shoal]

well

you distribute

yes times

like you said

so square root of 2 times square root of 2

theres a rule of exponents

where you can add it up i forgot the formula

!expo

!exp

[charred yew]

How has it went from this to that

[wind shoal]

well they just brought out the 2

to simplify as much as possible from the denominator as well

@charred yew

product rule

[charred yew]

Ah right 👍

Cheers

,close

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[charred yew]

Yes

,close

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[quasi bison]

well what do you think

[olive sorrel]

no

for both

[quasi bison]

why

can you show a matrix A with A^9 * A != A * A^9?

and what about A * A^-1 != A^-1 * A?

[olive sorrel]

hm let me try

okay, i think they are both true actually

I couldn't find any counterexample and I get why A*A^9 commutes, but not sure why A * A^-1 commutes

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[rugged surge]

is the answer for this just 2i + 0j + 0k?

[safe radish]

Please don't occupy multiple help channels.

@rugged surge Has your question been resolved?

[plucky elk]

Do you know the definition of the laplacian

[rugged surge]

Yeaaa

Grad^2v

[plucky elk]

Expand that

Write it out in terms of partials with respect to x, y, and z

[rugged surge]

So del^2/delx Ta = 2

Then we do the same for y and z?

[plucky elk]

Yes

[rugged surge]

So we just get 2i +0j+0k right? As

[plucky elk]

No

You got the definition wrong

[rugged surge]

Oh wait it will just be 2 since it's scaler?

[plucky elk]

Yes

You add them, not put them in a vector

[rugged surge]

Got it. Thankyouu

.close

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[untold mist]

Hello, just trying to test the series convergence, maybe someone can help.\
(\sum (\sqrt[3]{n^3+2} - \sqrt{n^2+7}) = \sum(( \sqrt[3]{n^3+2} - n )+ (n- \sqrt{n^2+7} ))= \sum (\frac{2}{(n^3+2)^{2/3} + (n\sqrt[3]{n^3+2}) + n^2} - \frac{7}{\sqrt{n^2+7}+n}))

[flat frigate]

Clippy

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@untold mist Has your question been resolved?

[untold mist]

No, it hasn't :). Any ideas💡 <@&286206848099549185>?

[safe radish]

@untold mist Has your question been resolved?

[safe radish]

@untold mist Has your question been resolved?

[safe radish]

@untold mist Has your question been resolved?

[drowsy karma]

$\text{use such formula appropriately:}\\a^{6}-b^{6}=\left( a^{3} \right)^{2}-\left( b^{3} \right)^{2}=\left( a-b \right)\left( a^{2}+ab+b^{2} \right)\left( a^{3}+b^{3} \right)$

[flat frigate]

Joanna Angel

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[steep magnet]

How do I find the distance if theyre parallel

I cant set them equal to eachother to see the coordinates of where they will meet cause they never meet

Whats the first step to solve this

[plucky elk]

draw a line perpendicular between them and find the distance along that line

[steep magnet]

So I get the inverse of the slope (-2x), but what do I do with that

[plucky elk]

once you have the slope, you find a point on either line and use point slope to get the equation of that line

,tex .point slope

[flat frigate]

riemann

[steep magnet]

Ah oke, I think I know what to do now, 1 moment

Got it thanks!

❤️

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