help-13

[modest dove]

so this??

[amber ridge]

which is (x)^2 + 2(x)(1/x) + (1/x)^2

[modest dove]

$(x)^2 + 2(x)(1/x) + (1/x)^2$

[wraith dagger]

Jshy <3

[amber ridge]

yeah

[modest dove]

Ohhh right

ok, i just couldnt interpret that in a straight line

okay yes, that makes sense

where has the -2 come from then?

[amber ridge]

what do you get when you expand out everything and simplify

[modest dove]

From this? @amber ridge

[amber ridge]

yeah

[modest dove]

$\frac{x^4+2x^2+1}{x^2}$

[wraith dagger]

Jshy <3

[modest dove]

Have I done that correctly?

Not too sure where this is going

[amber ridge]

idk what you’re doing either haha

it’s pretty simple, no need to overthink it

it’s just applying (a+b)^2 = a^2 + 2ab + b^2

[modest dove]

im now really confused, could you guide me through this?

Im quite lost at this stage

I understand this

im just confused where the -2 comes from at the end

[modest dove]

How can you get from

$x^2+\frac{1}{x^2}$

[wraith dagger]

Jshy <3

[modest dove]

to the next equation

What you showed me above was getting from the final answer to the first answer

How would you go the opposite way?

[amber ridge]

well it’s not totally obvious how you’d do it

you could probably derive it from completing the square

[modest dove]

right okay

I will need to work a bit more on this question

thanks for your help

.close

[cedar kiln]

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[austere iron]

Hey there

[austere iron]

Question is: There are two points; A(2,1) and B(-1,-3). Will the perpendicular bisector of line segment AB pass through the point (2,-2)

No clue where to start here.

I do know the basic formulas like m = y2-y1/x2-x1 , M= (x1+x2, y1+y2) etc.

[digital cliff]

what is the midpoint?

[austere iron]

Let me find it

@digital cliff It's (-0.5, -1)

[digital cliff]

are you sure about the x value?

[austere iron]

Pretty sure, yeah 2+(-1) divided by 2

[digital cliff]

what is 2+(-1)?

[austere iron]

-1

ait

1

And then its 1 over 2

Which is 0.5

[digital cliff]

yeah, so it should be +0.5

[austere iron]

yeah my bad

[digital cliff]

no worries, next you need the gradient/slope of the line joining the points

[austere iron]

So m= y2-y1 over x2-x1?

[digital cliff]

yeah

[austere iron]

And I put in the values of A and B right

[digital cliff]

yeah

[austere iron]

m= -4 over 3

over -3*

[digital cliff]

you can simplify that

[austere iron]

? I can?

to -1.333333

[digital cliff]

no

what is (-4)/(-3)

they have a common factor

[austere iron]

12?

[digital cliff]

its division, not multiplication

[austere iron]

I'm not sure.. 4 over 3?

[digital cliff]

yup, the '-' cancel leaving you with 4/3

okay, so now, do you know what a perpendicular bisector actually is?

[austere iron]

Kind of

I imagine it as a dotted line cutting through AB or whatever

90 degree angle

[digital cliff]

cutting through where?

[austere iron]

the midpoint

[digital cliff]

exactly right

its a line perpendicular to AB that passes through its midpoint

[austere iron]

so if it's running through the midpoint, the midpoint is (0.5, -1) and not (2,-2)

[digital cliff]

the line does pass through (0.5,-1) but we cant check if it passes through (2,-2) yet, we need to do a bit more

[austere iron]

ah

[digital cliff]

knowing the gradient/slope of AB is 4/3, do you know what the gradient of a line perpendicular to it would be?

[austere iron]

I probably should know, but I don't

[digital cliff]

no worries, if we have 2 gradients $m_1,m_2$, then $m_1$ and $m_2$ are perpendicular if $m_1\cdot m_2=-1$, they are negative reciprocals of eachother

[wraith dagger]

AℤØ

[austere iron]

oh right right I remember something about m1 and m2

and how m2 basically

is like for example 3 over 4

but flips and becomes -4 over 3

[digital cliff]

exactly

[austere iron]

so is this the case we're in?

[digital cliff]

the gradient of out perpendicular bisector is indeed -4/3

wait no, sorry, its -3/4

the original was 4/3

but your logic is right

[austere iron]

I've had problems somewhat similar to these without using the negative reciprocal, how do I know when it's one of the steps?

[digital cliff]

do you have an example?

[austere iron]

I don't have an example, but like; why are we using the negative reciprocal in this problem?

[digital cliff]

basically, our goal here is to find the equation of the perpendicular bisector

to do this we need to know a point it passes through: the midpoint of AB

and its gradient: the negative reciprocal of the gradient of AB

[austere iron]

is gradient another word for slope?

[digital cliff]

yeah

[austere iron]

ah okay

Okay sorry for trailing off a bit

[digital cliff]

no problem

[austere iron]

So we have M= 0.5, -1 and m= -3 over 4

[digital cliff]

we do, are you familiar with this equation:
$y-y_1=m(x-x_1)$

[wraith dagger]

AℤØ

[austere iron]

Actually, I'm not

[digital cliff]

it essentially is just this except y2 and x2 are arbitrary - then multiply both sides by x2-x1

since m is constant for a linear line, it allows you to get an equation for the line if you know its slope and a point it passes through

[austere iron]

like y = mx + b

[digital cliff]

yeah

[austere iron]

So I plug what we have into that?

[digital cliff]

yup plug your midpoint into y1 and x1 and your slope into m

[austere iron]

What about the other y and x

[digital cliff]

they are arbitrary - its like how in y=mx+b, y and x dont have a defined value yet as they are variables

[digital cliff]

doing this basically gives you the y=mx+b equation

[austere iron]

y-(-1)=m(x-0.5)

[digital cliff]

yup, just plug in your m value too

[austere iron]

y-(-1)=-3/4(x-0.5)

(3 over 4 not divided by 4)

[digital cliff]

perfect, now, you could rearrange this into y=mx+b but you dont really have to for this question

since we are checking if it passes through (2,-2) if you plug in y=-2 and x=2, then the left hand side of the = will be the same as the right hand side, if the line passes through it

[austere iron]

-2-(-1)=-3/4(2-0.5)?

The sides aren't the same?

[digital cliff]

if they arent equivalent then the perpendicular bisector doesnt pass through (2,-2)

[austere iron]

I don't think they're equivalent

[digital cliff]

have you evaluated it?

[austere iron]

-3/4(2-0.5) is 9 over 8

and -2 - (-1) is just

-1

[digital cliff]

(-9/8)

[austere iron]

my bad

[digital cliff]

but yes, they arent equal

as you can see it indeed doesnt pass through (2,-2)

[austere iron]

Thats desmos, right?

[digital cliff]

yeah

[austere iron]

I tried plotting it on desmos but I couldnt get the lines to run through

[digital cliff]

its because youre plotting points, not lines

[austere iron]

Oh

[digital cliff]

you needed to find the gradient and stuff if you want to plot it out

[austere iron]

ah

Okay well I appreciate your help, I have another question; if you don't have the time to answer it I understand

like a separate problem

[digital cliff]

i can have a look

[austere iron]

Okay well

Question is:
A line has a y-intercept of 4 and an x-intercept of -5

What is the shortest distance from the origin to this line?

Only thing I'm sure of is that the origin is 0,0

[digital cliff]

do you have any idea when a point is at a shortest distance from a line?

its when the line from the point to the line is perpendicular to the line itself

[austere iron]

I don't fully understand that

[digital cliff]

its not a very good diagram, but the shortest distance will be the second line from the point to the white line

im not sure how i can explain it properly, however the shortest distance between the point and that line, is a distance joining them that is perpendicular to the line

[austere iron]

So this is whats relevant?

[digital cliff]

yeah, the bits that go beyond the white line dont matter

[austere iron]

Right then so the white line is just like, the graph

and the origin is in the middle of that

[digital cliff]

the origin would be that white dot

[austere iron]

Oh

So the origin isn't 0,0?

[digital cliff]

it is (0,0) i havent drawn any axes, my diagram is more generic

[austere iron]

Oh okay

[digital cliff]

https://www.youtube.com/watch?v=aA15QO4YD6U&ab_channel=corbettmaths

this might give you a better idea about the perpendicular part

[austere iron]

That line is y=x + 1

is my line put into that same equation?

or is it just

(-5,4)

[digital cliff]

your line isnt the same equation

you know that the y intercept is 4 and the x-intercept is -5

if we have a generic line y=mx+b

when x=0, y=4

and when y=0, x=-5

can you find m and b with those?

[austere iron]

No?

[digital cliff]

why not?

[austere iron]

Oh wait I think you can

You can find m with x and y

[digital cliff]

in what way?

[austere iron]

m = (x1+x2/2, y1+y2/2)

[digital cliff]

that would provide the midpoint of 2 points, which yes you could do, but you dont need that

you could also use (y2-y1)/(x2-x1) however there is a simpler way

[austere iron]

Ok

[digital cliff]

we have y=mx+b, when x=0,y=4 so 4=b

similar process for finding m, when y=0, x=-5
0=-5m+4

[austere iron]

why isn't x -5

and vice versa

[digital cliff]

because x=-5 at the y intercept, we have 2 points basically (0,4) and (-5,0)

we are just subbing them into y=mx+b

[austere iron]

alright

So what can I do to start?

[digital cliff]

have a go at determining m and b with the provided info, since we need the equation of that line

[austere iron]

I don't know how to do this

I just wrote "4=m(-5)b"

Which I don't think I was supposed to do, but I don't what else to do

[digital cliff]

not quite, no

we have y=mx+b right? and we have (0,4) and (-5,0)
you should sub those points into y=mx+b separately

so for (0,4) we get 4=m(0)+b

so we know b is 4

[austere iron]

Is that it? b = 4 ?

[digital cliff]

that simple

now you can use the other point to find m

[austere iron]

How?

[digital cliff]

0=-5m+4

[austere iron]

Ooh my lord I'm naive

I forgot that one of the values is supposed to be 0

[digital cliff]

no worries😅

[austere iron]

so I subtract 4 from both sides?

[digital cliff]

i mean, you could, i would just add 5m to both sides but it doesnt really matter

you just need to make m the subject

[austere iron]

m = 4 over 5

[digital cliff]

yup

so what is the equation of the line

[austere iron]

0=5(4/5)+4

[digital cliff]

that isnt a line equation, youve also written that 8=0 which isnt true

[austere iron]

oh wait

[austere iron]

is this what you mean?

[digital cliff]

you dont need to use that

[austere iron]

Oh.. I'm lost then

[digital cliff]

remember, we had y=mx+b, you now know what m and b are, so just replace them with their values

[austere iron]

I thought I did that

[digital cliff]

not quite

i can see what you tried to do, but it wasnt what you should have done since it doesnt make much sense

[austere iron]

b= 4
y= 0
x= -5

[digital cliff]

b, yes, the other 2, no

[austere iron]

and m = 4/5

oh

[digital cliff]

y and x are variables in our line equation, not constants

they dont have a strict value

[austere iron]

What's the proper equation

[digital cliff]

$y=\frac{4}{5}x+4$

[wraith dagger]

AℤØ

[austere iron]

oh

interesting

[digital cliff]

how so?

[austere iron]

I pretty much forgot what variables and constants are so I automatically changed the values of y and x

[digital cliff]

ah i see

[austere iron]

Yeah I haven't had to do such weird problems before

[digital cliff]

yeah, try not to confuse points on a line and the line itself

however thats most of the work out the way really

[digital cliff]

now you need a line that passes through the origin and is perpendicular to this

hint: it just has the form y=mx

[austere iron]

is not just y=4 over 5x?

[digital cliff]

that would be parallel to the line, not perpendicular

[austere iron]

I don't know how to do this, I'm not sure how to draw it and/or visualize it

[digital cliff]

this is what you have done

you just need to remember what the slope of a perpendicular line is

[austere iron]

needs to run through like that right

[digital cliff]

thats a perpendicular line yeah, but it should pass through the origin

[austere iron]

I just don't know what values to sub in to get that

[digital cliff]

remember when we did it on the last question?

[digital cliff]

this

[digital cliff]

you need to find the m that would be perpendicular to this

[austere iron]

negative reciprocal?

[digital cliff]

yup

[austere iron]

y=-5/4

[digital cliff]

exactly

since it goes through the origin, in y=mx+b, the b=0 so we just have y=-5/4 x

S is our shortest distance

so theres just one more thing we need to find before we can get the value of S

any idea?

[austere iron]

Not really but I'll likely facepalm myself upon learning/realizing what it is

[digital cliff]

the equation for the distance between two points is $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ we have one point so far (0,0) so we need the second point

[wraith dagger]

AℤØ

[austere iron]

oh right! I know that equation

[digital cliff]

that second point is where these 2 lines intersect

[austere iron]

wait but what the heck are the values I sub in

[digital cliff]

you cant use this just yet

[digital cliff]

you still need to find this

[austere iron]

With what formula?

m=y2-y1/x2-x1 wouldnt make much sense would it

[digital cliff]

do you know how to find where 2 lines intersect?

[digital cliff]

not quite, no

[austere iron]

uuh well we have y=-5/4x and

y=4/5x?

[digital cliff]

not quite no

we have y=4/5 x+4 and y=-5/4 x

when we have 2 lines: y=mx+b and y=px+q
we can find where they intersect by setting them equal to eachother:
mx+b=px+q, then solve for x

[austere iron]

like solving by substitution?

[digital cliff]

basically

[austere iron]

Okay I presumed that and got (0,4)

[digital cliff]

that wouldnt make much sense

[digital cliff]

our second line doesnt pass through that point

[austere iron]

Oh I don't know then

[digital cliff]

you need to solve $-\frac{5}{4} x=\frac{4}{5} x+4$

[wraith dagger]

AℤØ

[austere iron]

But that just gives -80/41

I dont know how to solve that

otherwise

[digital cliff]

thats the x value where they intersect

you now need the y

[austere iron]

Do I get y the same way

[digital cliff]

technically you could, but theres no need, you can just sub x=-80/41 into either equation

[austere iron]

So y=100/41

[digital cliff]

yup, so they intersect at (-80/41 , 100/41)

[austere iron]

Such ugly numbers

[digital cliff]

finally, you are able to now use this

[digital cliff]

ik

[austere iron]

which is x/y 1 though?

does it matter?

[digital cliff]

wdym?

[austere iron]

like x= -5
y = 4

and then

the other set is x = -80/41 and y= 100/41

[digital cliff]

oh i see, no it doesnt really matter

just mind your signs

[austere iron]

ok I got

d roughly equals 7.73

eee

[digital cliff]

not quite

what did you do?

[austere iron]

I did d= (-80/41-(-5))to the power of 2 + (100/41-4)to the power of 2

I subbed it into the formula

[digital cliff]

im not sure where you pulled -5 and 4 from

[austere iron]

oh

[digital cliff]

our point is the origin (0,0) remember

$\sqrt{(-\frac{80}{41}-0)^2+(\frac{100}{41}-0)^2}$

[wraith dagger]

AℤØ

[austere iron]

waits so it's 3.12?

[digital cliff]

yup

thats the shortest distance

[austere iron]

That's.. the entire thing?

[digital cliff]

indeed

[austere iron]

holy shit that was mind grueling I could've never done that myself

I didn't know the steps

[digital cliff]

😅 its alright once you figure out what you need, you can work backwards

[austere iron]

and I got the -5 and 4 from the very beginning e.e

since just subbing in 0,0 I don't know.. seemed wrong to me

I didn't think logically

I always forget the vital stuff like that

[digital cliff]

ah, yeah just remember these are the x and y intercepts of the line

[austere iron]

Doing longer equations

I've got two more questions like these but I'll have to figure them out in the morning

two more long(ish?) questions

[digital cliff]

if you follow the general stuff from here you should be alright

itll become natural the more you do it

[austere iron]

Thank you for the 1.5 hours of help though

[digital cliff]

no worries

[austere iron]

is there a way to archive this?

or something

[digital cliff]

hm, i dont think so, but if you see that search bar in the top right

type: from:Cwuka

and it should show you a list of your past messages so you can jump to here

[austere iron]

ah yeah

Thank you for your time I'll have to see if any helpers can help me in the morning

[digital cliff]

no problem, have a good night

[austere iron]

You too, or whatever time of day it may be for you

[digital cliff]

2am atm😅

[austere iron]

ah..

👋

[cedar kiln]

@austere iron Has your question been resolved?

[cedar kiln]

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[cedar kiln]

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[cedar kiln]

@opaque mantle Has your question been resolved?

[cedar kiln]

@opaque mantle Has your question been resolved?

[cedar kiln]

@opaque mantle Has your question been resolved?

[crimson sedge]

• How did you get x_i=1+(1/2)I?
• How did you get that f(1+(1/2)i)=(1+(1/2)i)-1?

Also good freakin' patience lol

First start by finding x_i

[crimson sedge]

As you can see from my message, I am perplexed by the answer you gave in the picture

Hmmm...

You could

But you didn't use the proper a and width

Weirdly enough you wrote that a is -10 above but then used 1 for some reason

Also, width is just Delta x

-10

Notice you had cancelled out the minus signs above

It was -1-(-10)

Math be like

Indeed

Yep

I suggest you simplify it to -10+1i-1=-11+i before rewriting it into the sum

Wait actually something feels wrong

I just rewrote the -1 you had put in but I'm starting to question why you even put it there

Oooh I'm dumb

Sorry I forgot to mention the second problem

So x_i is indeed -10+i

Actually let's just write it i-10

But f(i-10) would not be (i-10)-1

Notice when you did it the first time you forgot the square

It would be (i-10)^2-1

Ah fudge did I miss an off-by-one error again

I was thinking about that too

I guess it would be easier to change the x_i formula

Oh sure that works

I was gonna ask if i was 0-indexed or 1-indexed

If it is 1 indexed then everything works fine already

Yes

0-indexed would be a left Riemann sum

But if it's a mid Riemann sum then... :/

Another method which would always work is to add a shift to i in the formula, which I'll call s:
x_i=a+(i+s)(Delta x)
left: s=0
mid: s=1/2
right: s=1

Honestly I have no idea if there's a standard way to do it

I just made the shift method up

Did you write i=x under the sigma?

I can't quite make out what that symbol is

In that case, you can .close this channel

[cedar kiln]

Channel closed

Closed by @opaque mantle

Use .reopen if this was a mistake.

[cedar kiln]

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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.
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[fluid sedge]

may someone help me im dying

[fluid sedge]

i really need help im dying 💀

[crimson sedge]

Are they similar triangles?

[fluid sedge]

yes

[zenith sail]

Can you post the entire question? Something is missing here

[crimson sedge]

Yeah lol

@fluid sedge

[fluid sedge]

alrr

thats all

this is the other 2

[crimson sedge]

yeah that's supposed to be congruent not similar

[fluid sedge]

oh mb

[crimson sedge]

if two figures are congruent then you can just equate both the sides

[fluid sedge]

oh alr alr

thxx

[crimson sedge]

equate the length of RY and BT

[fluid sedge]

aite

[crimson sedge]

rotate the RUY triangle in the direction with the green arrow and then rotate it so that it matches the orientation of the BTE

[fluid sedge]

Ahhh

[crimson sedge]

which is how i figured RY and BT lie on top of each other and yeah you just equate the sides

Anyway that's it, if you don't have questions you can probably close the channel

[fluid sedge]

alrr tjx youu!

.close

[cedar kiln]

Channel closed

Closed by @fluid sedge

Use .reopen if this was a mistake.

[cedar kiln]

Available help channel!

Send your question here to claim the channel.

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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[limber ether]

Probability of a person having O- blood in the world: 3/30
30 ppl in a room
One person who gives their blood is o negative, whats the probability of the next person being o-negative

[soft tinsel]

2/29?

Idk it's a guess

[limber ether]

Dont think so

Cause then if 3 ppl have o- blood then the probability of the 4th person having it is 0

But thats not true

[soft tinsel]

Hmmm i see

[buoyant latch]

it should still be 3/30

[limber ether]

Thats what i wrote

[buoyant latch]

(assuming the selection of participants are random)

[limber ether]

But all my friends r telling me its 2/29

[limber ether]

Whats your reasoning behind this?

[buoyant latch]

well it's like flipping a coin

line up 30 coins

first one flips head

what is the chance you get heads on the 2nd flip

50%

[limber ether]

Hmm yes

[buoyant latch]

because each of the coinflips are independent events

[buoyant latch]

hence why i said this

[limber ether]

But what if u gathered every single person on earth

[buoyant latch]

doesn't matter

[limber ether]

And then u run out of o- ppl

[buoyant latch]

what if you gather all the coins in the world

[soft tinsel]

Then why it isn't like 1 over all the no. of blood groups in this world?

[buoyant latch]

that's why it matters whether they are independent or not

if there is only 1 O- person in the world

and the first one was O-

then obviously the conditional probability makes the 2nd impossible

[limber ether]

Over here the 1st one is O-

[buoyant latch]

but, here since the "population of earth" is a very big number

[limber ether]

So whats the probability of the 2nd person

Being O-

[buoyant latch]

revealing 1 to be O- doesn't mean much in the grand scheme

[limber ether]

Yeh my reasoning here was that 1 person being O- doesnt change the probability much

Since 0.1 of 8 billion is still 800 million

[buoyant latch]

that's why i said the group of people needs to be random

[limber ether]

What does that mean

[buoyant latch]

well

[limber ether]

The question just said theres 30 ppl in a room

[buoyant latch]

if you look at a family of people

a family of 30 people

since genetics play a big part in your blood type

[limber ether]

Then the probability obviously differs

[buoyant latch]

the 30% would not be accurate for your sample

hence "random" is needed

[limber ether]

Some ppl said u needed to put 2.9/29

[buoyant latch]

no

[limber ether]

Instead of 3/30

But its the same right?

Both r 1/10

[buoyant latch]

ok

30 people

50 people

8 billion people

it doesn't matter how many are in the room

P(first is O-) = 30/100
P(second is O-) = 29/99

P(2nd O- | first is O-) = P(2nd O- AND first is O-)/P(first is O-)
= (3/100 * 29/99)/(30/100) = 29.29%

so it's a little less right?

[limber ether]

Yes

[buoyant latch]

this is when there's only 100 people in the world

,calc (3000/10000 * 2999/9999)/(3000/10000)

[wraith dagger]

Result:

0.2999299929993

[buoyant latch]

this is with 10k people

[limber ether]

Damn

So its basically still the same

3/30

[buoyant latch]

,calc (2310000000/7700000000 * 2309999999/7699999999)/(2310000000/7700000000)

[wraith dagger]

Result:

0.29999999990909

[buoyant latch]

nice

[limber ether]

I see

[buoyant latch]

that's with 7.7 billion people

[limber ether]

It was basically in the form of a tree diagram

[buoyant latch]

like yes it does change the probability

but it's not like there's only 30 people in the world

[limber ether]

So i had to fill out the branch which said 2nd person is 0- after 1st person is also O-

[buoyant latch]

it's "negligible"

it's pretty much 3/30

[limber ether]

Do you think theyd cut marks if i put 3/30

Instead of the big decimal

Since it was a small box

[buoyant latch]

no

this is correct

[limber ether]

Alright thx

[cedar kiln]

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[delicate atlas]

Statistics, this is a correction redo but it’s for something months ago so this has already gone out my head. Not sure what to fix or do here

[cedar kiln]

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[cedar kiln]

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[honest glacier]

<@&286206848099549185>

<@&286206848099549185>

<@&268886789983436800>

<@&286206848099549185>

[crimson sedge]

Dont spam u

[livid hound]

do NOT abuse helper pings
do NOT ping mods for help

[crimson sedge]

Ahh man this guy

[cerulean sail]

!volunteers

[cedar kiln]

Helpers are just people volunteering their time to help you. Be polite.

[crimson sedge]

It's so ez find it

Answer is in the question

[cerulean sail]

No helper is obligated to help you, and do so out of the kindness of their hearts

[woven herald]

come back after reading the rules

[pliant ledge]

uh can you look at what i just wrote

i suggested some idead tho

ideas

[crimson sedge]

Read this u man

[pliant ledge]

i don't see the customization thing

i did read the rules

but i can't find the helper thing

[crimson sedge]

Not u bruh

[pliant ledge]

oh k

[fiery sundial]

May be he need the Euclid's Elements

[pliant ledge]

yo bro

[cedar kiln]

@honest glacier Has your question been resolved?

[cedar kiln]

@honest glacier Has your question been resolved?

[cedar kiln]

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[vocal raptor]

The distance between cities A and B is 350 km. A tourist traveled from city A to B and 3 hours later the second tourist who arrived in City B 1 hour later than the first. Find the speed of each tourist if their ratio is 5:7

[karmic patio]

The ratio is of speed or time?

[vocal raptor]

speed

can u help

[karmic patio]

trying to understand the question becuase it is paraphrased a bit incorrectly

[vocal raptor]

yeah translated from diff language

[karmic patio]

which language?

[vocal raptor]

Georgian

tell me what u find wrong so I tell it correct way

[karmic patio]

Oh I cant read georgian ,

[vocal raptor]

So the second started moving 3 hour after first started moving

[karmic patio]

yeah so correct me if i am wrong: So the distance is 350km. A reaches 1 hour before B and their speeds are in ratio 5:7?

[vocal raptor]

yes

[karmic patio]

ok lemme try

[vocal raptor]

b started moving 3 hr after a started tho

[karmic patio]

is the speed of first 10km/hr?

[vocal raptor]

50 km/h

[karmic patio]

oh damn

[vocal raptor]

maybe u got x=10 and speed of the first is 5x

[karmic patio]

yea

one sec

Yoo @vocal raptor I got it

Finally

[vocal raptor]

what did u do

can u tell quickly

[karmic patio]

the speed of second tourist is 43.75?

[vocal raptor]

no bro its literally more than first

since their ratio is 5:7

[karmic patio]

wait sorry

[karmic patio]

70 right?

[vocal raptor]

yes

[karmic patio]

70 kmph

ok I will send a picture

[vocal raptor]

oki thanks

[karmic patio]

Here tell me if you don't understand

[vocal raptor]

thank u very much mate

.close

[cedar kiln]

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[karmic patio]

no problem

[cedar kiln]

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[tame seal]

4a^2+4a+1 by 2a-1

[tame seal]

How to do thag

That

Hello?

.close

[cedar kiln]

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[limber bluff]

yeah

@tame seal still need help?

[cedar kiln]

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[keen lichen]

I am doing a dct problem, but I am confused by the answer in my book

[keen lichen]

Doesn't c equal the first instance in a series?

if that's the case, then how is c ≠ 1/2?

[tropic oxide]

well yeah it should be 1/2 and not 1

[keen lichen]

This is confusing. Maybe there is something regarding the larger context of the problem

[tropic oxide]

nah its just a slight (inconsequential) instance of inattention to detail

for convergence purposes it doesnt even matter what the first term is

what matters is that the common ratio is less than 1 in magnitude, which it is

and is stated as such

[keen lichen]

So because b_n converges, then a_n converges by the DCT?

[tropic oxide]

nothing is given such names here

we have two positive series, one greater than the other termwise
the greater one converges, therefore so does the smaller

[keen lichen]

Okay, thank you!

.close

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[novel stone]

Comparition rule for series
How do I apply the comparison rule for this series?

[novel stone]

I tried to compare it with n/n^2, which diverges

oh, so if the greater one is divergent

then it doesn't mean that the smaller one is too

I need to compare it with smth else

<@&286206848099549185>

[cedar kiln]

@novel stone Has your question been resolved?

[dire geode]

do you know $\lim_{n \to \y} \frac{\log(n)}{\sqrt{n}}=$?

[wraith dagger]

riemann

[novel stone]

0?

[dire geode]

yes

[cedar kiln]

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[dire geode]

.reopen

[cedar kiln]

✅

[dire geode]

modify n/n^2 so that it's convergent

[dire geode]

and using this fact

[novel stone]

n/n^3 is convergent

But I need something bigger than ln(n)/sqrt (n)

For convergence

Idk what the thought process is in this situation

[dire geode]

right so n/n^3 is the wrong one to compare to

[novel stone]

Maybe something bigger than ln(n)

But smaller than n

[dire geode]

.

[novel stone]

Oh wait lol

So the limit is 0

Than means

Sqrt(n) is bigger

So sqrt(n)/n

Hmm

Still not convergent

Oh bruh

I'm tired and it shows

[novel stone]

I need to prove that this is convergent

Should I use integrals rule?

[cerulean sail]

I've got a hint for you, what's 2/2?

(no that isn't a troll question before anyone asks and 's me )

[novel stone]

Triangle

[cerulean sail]

[cerulean sail]

Do you remember log rules?

[novel stone]

I do

[cerulean sail]

Think about why I'm asking that question

[novel stone]

1s

Hmm

[cerulean sail]

I’ll make a comment that it seems like you used that ln(x) < x, that is useful…

[novel stone]

Sorry

I think I need a break

I can't think about it rn

[cerulean sail]

To be fair that hint was (deliberately) vague what I will say is that the aim is to do this here

[novel stone]

So I should aim to find another convergent series

[dire geode]

fair. i wanted mike to suffer a little bit

[novel stone]

[dire geode]

this is one of those problems where i can't think of a hint that doesn't give it away completely

but maybe that's okay since mike's suffered enough already

[novel stone]

I need to start drinking coffee ngl

[dire geode]

multiply the top and bottom by something that gets your sequence of terms

$\frac{\log(n)}{\sqrt{n}}\cdot ? =\frac{\log(n)}{n^2}$

[wraith dagger]

riemann

[novel stone]

n^3/2???

[dire geode]

close!

yes. now that should tell you what to compare your sequence to

[novel stone]

So n^3/2 * log(n)/n^2?

[cedar kiln]

@novel stone Has your question been resolved?

[velvet mortar]

x-1/2=-2 solve for x.

[cedar kiln]

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[crimson sedge]

my question is that it says write the value of underroot 4900, so square root of 49 is 7 times 7 therefore can you say its 7 and 2 zeroes, the answer is incorrect

[crystal raptor]

Why 2 zeroes?

[crimson sedge]

that is what i did not understand

square root of 49 is 7, they have written underroot of 4900?

[crystal raptor]

Are you meaning to type "underroot"? What does that mean?

[wraith dagger]

MathIsAlwaysRight

[crystal raptor]

But why do you think you have to add two zeroes to the end of 7?

[crimson sedge]

i never came across a question where in an underroot there would be two zeroes mentioned

either they were underroot 49 or like underroot 100 and underroot 25

they wrote here underroot 4900

[short blade]

you should seek to understand what the square root operation is doing, not just base your answer on previous problems

the square root of a number x, is a number y so that y^2 = x

for example the square root of 4 is 2, since 2^2 = 4

as you’ve identified, 7^2 = 49

[crimson sedge]

why would they add two zeroes?

[short blade]

they didn’t “add” anything really

[vocal sparrow]

who is "they"

[short blade]

they’re just asking you to compute the square root of a specific quantity

[crimson sedge]

examiners

[vocal sparrow]

are they not allowed to ask for the square root of 4900 ?

[crimson sedge]

we like know about 49, 4900 was something that was not tested anytime before in past paper exams

and so i got entirely confused about zeroes

[lyric narwhal]

memorising square roots is not gonna get you anywhere.

[lyric narwhal]

the two "added zeroes" mean that the number inside has been multiplied by 100

[vocal sparrow]

so you should fall back on the definition of a square root and what it means to be the square root of something

[crimson sedge]

there were certain squareroots that were memorized

well thanks

i will try to solve the kind of questions

[maiden stirrup]

You’re on the right track but there is a simple logical problem with this. A 3 digit number multiplied by another 3 digit number cannot equal a number with less than 5 digits. You can see this by multiplying the smallest 3 digit number by itself; 100x100; which is equal to 10000. 4900 is less than 10000 which means that the square root of 4900 cannot be a number with more than 2 digits.

[crimson sedge]

thankyou so much, appreciate it!

[maiden stirrup]

Since you know that the square root of 49 is 7, you need to simplify the parent number into one that isolates the 49, you can do this by dividing 4900 by 100. Now you’re left with the square root of 49x100.

Now you can see that you have two numbers with simple square roots, 49 and 100. You can square root both of these to get 7x10, which will lead you to your answer: 70

[crimson sedge]

yes the answer is 70

.close

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[wraith otter]

I’ve tried a few answers but I’m really lost

[granite knoll]

!show

[cedar kiln]

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

[dusk finch]

to find roots we sometimes convert it to factored form

what kind of quadratic factored form has just one root?

[narrow badge]

to have one root, the delta=b^2-4ac should be equal to 0

[dusk finch]

||this actually has 2 solutions||

[dusk finch]

use this one

and for the other solution try to think about what happens when n=0

[cedar kiln]

@wraith otter Has your question been resolved?

[wraith otter]

How could I factor it with no a value

[wraith otter]

I already tried -1/4

[granite knoll]

What are a, b, and c in the equation?

[narrow badge]

yes, should be -1/4 and the other solution should be n=0

[granite knoll]

I was asking ASR

[wraith otter]

Ohh ok

Thanks Tracy

.close

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[ashen kite]

Hello! I’m trying to learn how to fill out a VIRP table for a physics project I’m working on. I created a circuit shown in the diagram, but am unsure how to fill out a VIRP table with it.

[ashen kite]

V = Voltage

I = Current

R = Resistance

P = Power

[crimson sedge]

(there's a physics server in #old-network)

[ashen kite]

Thanks!

[cedar kiln]

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[jovial eagle]

do i apply lhopital's rule here in order to solve?

[jovial eagle]

.close

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[acoustic slate]

Hello, i'm reading linear algebra done right:
here, sequences are just lists with infinite elements, right?

[short blade]

what do you mean by “list”

[short blade]

then yes

formally, a sequence is a function from the naturals to some set

[velvet mortar]

What you think of the book so far?

[acoustic slate]

and here, functions from S to F is equal to functions with domain S (these functions need to take every element of S) ?

For example, the functions in this set (R^[0,1]) have all domain [0,1], right?

[short blade]

yes

[acoustic slate]

thanks!

[acoustic slate]

I am just beginning to read it so no opinion so far.

.close

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[slender jungle]

Hello

[slender jungle]

Could someone help

[cunning adder]

find area of base

multiply by height

[slender jungle]

Thanks 😊

[cunning adder]

if u need help ping me

[cedar kiln]

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[cedar kiln]

Please don't occupy multiple help channels.

[rancid wren]

oops

.close

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[real smelt]

Suppose that a 200 pound cable is 100 ft long and hangs vertically from the top of a building that is very tall. How much work is required to lift the cable to the top of the building? Clearly set up your integral and provide the final answer with the appropriate units.

a. Explain how we know that the cable weighs 2 pounds per foot. Be very clear with your response.

b. Suppose that a 250 pound cable is 100 ft long and hangs vertically from the top of a building that is 200 feet tall. How much work is required to lift the cable to the top of the building? Clearly set up your integral and provide the final answer with the appropriate units.

[cunning adder]

!status

[cedar kiln]

What step are you on?
1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

@real smelt Has your question been resolved?

[cedar kiln]

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[blazing blade]

Hello, I am having trouble approaching this problem: https://gyazo.com/e1f423624fd20b4bc0a092c6520624de

[blazing blade]

Any guidance on how I can approach it?

[lethal jackal]

I assume you just calculate it the normal way, except take into account the saturation?

[cedar kiln]

@blazing blade Has your question been resolved?

[cedar kiln]

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[tired widget]

How can I do this one?

[cedar kiln]

@tired widget Has your question been resolved?

[lethal jackal]

what does ABC similar to ACD tell you?

oh wait this is a hot mess

[cedar kiln]

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[wise dove]

me pls

[wise dove]

whats the answer and explanation

[zealous compass]

do you know how to expand binomials

[wise dove]

can u gimme the answer i beg u sir

i genuinely need the answer

[lethal jackal]

lowest effort ever smh

[wise dove]

no joke my life is on the line

i will ask for explanation later

[lethal jackal]

if your life is on the line, then your Google / ChatGPT skills must be severely lacking

[wise dove]

pls

man

man pls

i dont wanna die

[zealous compass]

lmao

[lethal jackal]

quite literally puts meaning into "couldn't search Google to save your life"

[wise dove]

man pls

pls

[wraith dagger]

ed

[wise dove]

whats the answer

god damn

[radiant topaz]

the answer is expansion

[lethal jackal]

should've stayed in school and studied lmfao

[wise dove]

im handicapped help me out

yall show sum respect

[lethal jackal]

I will say that knowledge of complex numbers and the polar representation of them

very useful

[wise dove]

man pls

j the answer

i dont wanna die this yung

[radiant topaz]

no one does

[lethal jackal]

I'm sorry :/

[wise dove]

okay man

.close

[cedar kiln]

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[pseudo merlin]

hu

[pseudo merlin]

hi

how did i do this wrong

[eager kiln]

wrong brackets

[pseudo merlin]

ohhhhdhehehe

[eager kiln]

a^2 + b^2 = c^2

so

x^2 + x^2 =420^2

that's $2 x^2 = 420^2$

[wraith dagger]

sho

[pseudo merlin]

my answer is stil lwrong

[eager kiln]

hypotenuse is 420 not 210

[pseudo merlin]

ohhhhhhhh

[cedar kiln]

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[chrome meteor]

hi yalls! i need help for this question

[upbeat dune]

The green triangle is equilateral

Bruh why won't it send the pic

The 2 long looking sides are
15-5
And the short looking side is 5 + 5

[chrome meteor]

wow its because of that stupid not drawn to scale photo

thanks my dude

[cedar kiln]

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[cedar kiln]

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[crimson sedge]

1. Make a graph of function and fill in its property

[crimson sedge]

I got this what should the brackets be? the open or closed?

I guess it will be in <> other than infinite

<@&286206848099549185>

[cedar kiln]

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[cedar kiln]

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[crimson sedge]

smh

[crimson sedge]

<@&286206848099549185>

[velvet mortar]

=

[radiant kayak]

how do you calculate the angle of repose of the pyramid?

[crimson sedge]

Bruh

.close

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[cedar kiln]

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[obtuse frigate]

hi

[gritty galleon]

hi

[gritty galleon]

!status

[cedar kiln]

What step are you on?
1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[obtuse frigate]

Im lost on where to start

[gritty galleon]

I see

can u start by simplifying?

[obtuse frigate]

how? Ive expanded

[gritty galleon]

taking lcm of yz,zx,xy, and then write the the entire numertor as a product of a bunch of expressions?

[gritty galleon]

. @obtuse frigate

[obtuse frigate]

uhhh let me try

[gritty galleon]

@obtuse frigate u found an answer?

[obtuse frigate]

nope

[obtuse frigate]

im lost

starting from scratch 💀

.close

[cedar kiln]

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[lyric dove]

I have to use the central limit theorem to conduct a test on my own hypothesis. I was given a large dataset about the quality in pumpkins across 31 states over 2019 and 2020. The null and alternative hypothesis are in the photo. I've got up to the standard deviation of the proportion of good quality pumpkins out of the total. Does this look right where I've got to and where should I go next?

[cedar kiln]

@lyric dove Has your question been resolved?

[cedar kiln]

@lyric dove Has your question been resolved?

[cedar kiln]

@lyric dove Has your question been resolved?

[lyric dove]

.close

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[winged jewel]

Ive been trying part iii

and I set the particular integral form to be Pcos(10t) + Qsin(10t)

i then differentiated twice to get an equation for dq/dt and d^2q/dt^2

but in the markscheme the answer is -0.1cos(100t)

I have no idea where the 100t came from when the particular integral form uses 10t

[royal loom]

can I see the solution that you got?

,w x''+100x

[winged jewel]

ok

one sec

I apologize if my working is hard to see

[wraith dagger]

AustinU

[winged jewel]

If you'd like me to rewrite any of it more nearly then let me know

0.99sin10t

[royal loom]

okay

[wraith dagger]

AustinU

[winged jewel]

Yes

7iii.

I've looked through my working and I can't see any mistakes

There may be some but then I don't know how the ms got that answer

[royal loom]

hmm I am taking a look

okay @winged jewel

I solved it two ways

using undetermined coefficients

and using ERF

both give the same answer (which is different from the one in the answer key)

However, it is also different than your answer

Also wolfram and symbolab both give the same answer that I got

So it seems like both you and the answer key are wrong

I recommend using ERF for this one, much nicer than undetermined coefficients btw

[cedar kiln]

@winged jewel Has your question been resolved?

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[winged jewel]

.reopen

[cedar kiln]

✅

[royal loom]

Hello

[winged jewel]

thank you!!

ill redo the question

i had to go eat

whats erf?

[royal loom]

okay no worries, I don't know why the textbook solution seems off

exponential response formula

[winged jewel]

i dont know what that is

ive only learned the auxiliary equation method

unfortunatelyt

[lyric narwhal]

WHAT

i thought it was the error function

my mind is blown

oh

wait

you're talking about something else

nvm

[wraith dagger]

AustinU

[royal loom]

in different contexts it is both