help-0

[ashen copper]

i hate myself for doing this but i tried out every number to get to 80000 and i absolutely hate myself for it

i could go even closer

if that is what you want me to do ;_;

[spare vessel]

😭😭 Tysm for your effort😭😭

Bruh I started to hate my math teacher now

it's is not the teacher job to teach first😭😭😭

[ashen copper]

btw im just joking btw

dont think that im that insane

wha

[spare vessel]

But fr tysm 😁

[ashen copper]

the formula is

(1+r)^10 = FV/PV

since our future value (FV) is 80000 and our present value (PV) is 5000

we just insert them into the correct spots

which is

(1+r)^10 = 80000/5000

(1+r)^10 = 16

now we get rid of the ^10

1+r = 16^1/10

1+r = 1.319507911

now we subtract 1 from each side so r is alone

r = 1.319507911 - 1

r= 0.319507911

[ashen copper]

this is the base formula which should work most of the times if the future and present value are given

[spare vessel]

sheeshhhh Tysm😭😭😭

[ashen copper]

np

[spare vessel]

wait how it become 1.319...?

[ashen copper]

please correct me again if there is any issue with the formula

[ashen copper]

16^1/10 is 1.319507911

they should be around the same value

to go even precise the value of 16^1/10 is 1.3195079108

omg

yes

😎

[spare vessel]

oh it's not the results of calculator but the value of 16^1/10?

[ashen copper]

yes

[spare vessel]

I thought my calculator is broken HAHAHHA

[ashen copper]

btw

if you use 0.3195079108 for r (which should be correct) you get 80000.00002

which is even closer to 80000

[spare vessel]

so Im gonna change it to 0.319

Ok ok

[ashen copper]

better to change it to 0.3195

[spare vessel]

it is the round off?

[ashen copper]

because if you teacher asks what the percentage is

you can say

31.95%

[spare vessel]

Ok ok

Bruh right now your my teacher

[ashen copper]

the round off should be 0.32 if you want 2 numbers behind but that is futher away from 80000

[spare vessel]

I'm gonna change my teacher now

[ashen copper]

so the best version is 0.3195

💀

[spare vessel]

umm this here is this a round off? of 1.319507911?

[ashen copper]

1.319507911 is a round off from 1.319507908

[spare vessel]

Oh yeah sorry 😭

[ashen copper]

the only difference is the 11 and the 08

which makes no sense as i have just noticed

[spare vessel]

oh is it 1.319 it's not 1.39 right? HAHAHA

[ashen copper]

so

again

16^1/10 is 1.3195

BUT

the value for r which you are supposed to use in the formula

is 0.3195

[ashen copper]

i meant the two ends here

[spare vessel]

Yes but I just wanna ask about this, is this a typo it is suppose to be 1.319 and not 1.395?

[ashen copper]

OMG TY

yes it was a typo

thank you for correcting me

my fault

[spare vessel]

Ohhh My mind is clear as crystal clear now

Tysmmm Boritooo❤️❤️❤️❤️

[ashen copper]

ty for correcting my mistakes

we basically just traded knowledge :p

[spare vessel]

[ashen copper]

if there are no further questions you can close this channel :J

[spare vessel]

Yes tysm😊😊

imma screenshot all this first

[ashen copper]

îf you want to i can make a completely new summary from all this

so there are no messages inbetween

[spare vessel]

u sure?

[ashen copper]

yes

its no problem since i already know what to do

[spare vessel]

Tysm😭

[ashen copper]

(1+r)^10 = FV/PV -> Base formula FV = future value, PV = present value, r = annual interest

(1+r)^10 = 80000/5000

(1+r)^10 = 16 | take the 10th root of both sides or 10√

1+r = 16^1/10 -> 1+r = 1.3195079108 | now subtract 1 from both sides or -1

r = 0.3195079108 -> r = 0.3195 or 31.95%

[spare vessel]

Arigathxsomuchmuch Boritooo😭😭❤️❤️

[ashen copper]

npnp

so again if there are no further questions you can close this channel :o

[spare vessel]

Ty😭 I'm gonna close it now

Byeee

[ashen copper]

byez

[spare vessel]

close

wait how

[ashen copper]

.close*

i think

[spare vessel]

close*

[ashen copper]

😭

[spare vessel]

.close

[lone heart]

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[spare vessel]

.close

[ashen copper]

you just need to wait

[spare vessel]

Oh HAHAHAH ok byee tysm

[ashen copper]

bye and np ^^

[lone heart]

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[cinder sundial]

“comparisons where relevant,” it seems ambiguous IMO

Can someone explain

[zinc haven]

this isnt math this is ielts writing task 1

[cinder sundial]

I know

It is statistics

I suppose it wants me to compare where the similarities involve?

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[ripe mulch]

how would i express these in uh

index form

would (g) be like

10^2/5 * t^2/5

[daring rover]

10 is not squared

(10t)² ≠ 10t²

[ripe mulch]

oh

so

is it just 10t^2/5

[cyan drum]

10 is still under the 5th root

[ripe mulch]

10^1/5?

[cyan drum]

yes

[ripe mulch]

10^1/5 * t^2/5

[cyan drum]

correct

[ripe mulch]

then

(h) 8^1/8 * m^1/2

[cyan drum]

yes

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[serene steppe]

Hi, can someone point me in the right direction? It's been a while since I touched linear algebra, and a lot of concepts are rusty.

[serene steppe]

So the system is already in echelon form, and I'm trying to find the solutions

[mortal trellis]

introduce free variables until you cant anymore

then solve in terms of the free variables

then plug all the available options in (each free variable can be 0 or 1)

[serene steppe]

could you provide an example of this?

the basic variables are x1,x2,x3,x4 and the free variables are x5,x6,x7

[mortal trellis]

yes

and then you can solve for x1,x2,x3,x4 in terms of x5,x6,x7

[serene steppe]

i'm lost

[mortal trellis]

from the last equation you get that x4 = 1-x5-x6-x7

you can plug that into the equation above

and get that x3=x7-1

and then x2 = (something only with x5,x6,x7)

and then same for x1

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[serene steppe]

I just tried it, and I I got the solutions as shown in the answer key. Thank you @mortal trellis

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[north perch]

pls check if i solved it correctly

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[north perch]

<@&286206848099549185>

[alpine sable]

<@&286206848099549185>

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[dusk ibex]

There are 5 possible essay formats. 3 will be asked and you must answer any 2 of those asked. What is the minimum amount of formats you should study to ensure you never get asked something you haven't studied?

I know the answer is four by using logic, but how do I prove this mathematically?

[tiny minnow]

We take the worst possible scenario

the worst possible case the 3 question asked, one from 3 question you didn't study but the other else u studied u ended up with 2 correct still

[dusk ibex]

Ok so like an iteration on combination formula?

[dusk ibex]

https://tenor.com/view/huh-dog-huh-dog-gif-9454444978823724629

[tiny minnow]

for the worst case we only need 4 formats

[dusk ibex]

Yes but how do i prove it mathematically

[tiny minnow]

They aren't Mathematically (by doing some calculation) ways to prove it

The only thing u need is logic

Use Worst-case scenario

[dusk ibex]

Are you sure? It feels like there should be

[tiny minnow]

I am assure if u dont trust just wait others helpers:)

[exotic belfry]

Define what you mean with "how do i prove it mathematically".

you can only skip one format as you have to answer two of the 3 asked. so you have to learn all but one -> answer = 4.

[tiny minnow]

Mathematically is buy doing calculation i think

[exotic belfry]

mathematical proofs are much more than just doing some calculations. but maybe your example asks for a special type of proof.

[lone heart]

@dusk ibex Has your question been resolved?

[dusk ibex]

It is a question I made up, and I want to use a mathematical formula to find the exact minimum value

If there is none, maybe an algorithm?

[exotic belfry]

define x the number of answers you can skip.
define y the number of possible questions.
define z the number of questions you have to learn to ensure that you can answer enough questions.
z = y - x.
calulation enough?

x = 1, y = 5 -> z = 4.

[dusk ibex]

Thanks, but how do you calculate x without z?

[exotic belfry]

x is given. "3 will be asked and you must answer any 2 of those asked." -> you can skip 1.

[dusk ibex]

Oh dang I missed that thanks 💀

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[subtle delta]

CSMC 2022 Part B 2c)
prove n < √(c + √c) < n+1 the number of positive integers c is even

My solution:
Square all of inequality and put back together to get:
n^2 < c + √c < n^2 + 2n + 1 (*)

Property of number of solutions to an inequality of form a < b < c:

Number of solutions = c - a - 1

Apply to (*)

n^2 + 2n + 1 - n^2 - 1 = 2n

Thus the number of answers that satisfy this inequality is 2n, which is even.

[subtle delta]

is there anything wrong with this proof

because in the solutions they have a 3 page proof

https://www.cemc.uwaterloo.ca/contests/past_contests/2022/2022CSMCSolution.pdf

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@subtle delta Has your question been resolved?

[ember oak]

what is 1+1?

[lone heart]

@subtle delta Has your question been resolved?

[olive ice]

@subtle delta just asking out of curiousity, what is this property of number of solutions to the given inequality there?

if you have the time, would you mind explaining it to me?

[subtle delta]

Lets say you have an interval 2<a<9 right

the possible answers for a if a is an integer is

3,4,5,6,7,8

6 possible answers

which is 9-2-1

it is just using logic

its in between this interval

so 9-2

but we exclude the starting number so we subtract 1

[olive ice]

oh wow, that seems simple!

i wish we were taught this sort of thing at school, though i did pirate a logic book, i didn't see this there

do you have any logic books you would recommend?

[subtle delta]

sorry i dont really use logic books or anything of the sort

i realy just practice from computer science and math contest practice

[olive ice]

ah i see

[subtle delta]

math contests pretty much help ur overall creative thinking

if u havent started them

[olive ice]

are there any online that you know of?

[subtle delta]

are you from canada

[olive ice]

nope

[subtle delta]

waterloo has some very good contests

waterloo is the top canadian university

for math

i can send u a link to their past contests

also the AMC contest

[olive ice]

ok. So at least ill have the questions and answers, right?

[subtle delta]

yes

[olive ice]

that sounds good, i would like that

[subtle delta]

https://www.cemc.uwaterloo.ca/contests/past_contests.html

https://artofproblemsolving.com/wiki/index.php/AMC_Problems_and_Solutions

here are the two links

AoPS has some good books for math

if you would like to look at it

on the amc page there are resources as well

[olive ice]

wow!

this is all amazing.. didnt know all of this existed

thank you, this is nice of you

[subtle delta]

no problem

[olive ice]

i'll look at some of the questions, thanks man. Good luck with your studies

[subtle delta]

u too

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[subtle delta]

.close

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[olive cypress]

can someone help me understand how to approach this

I know I have to use the x^2+y^2+z^2 somewhere for the distance but I’m lost after that

[tall topaz]

It tells you to use Lagrange multiplier. Have you done this?

[olive cypress]

I don’t know how I should set it up first

what equation should I be starting from

and there’s an inequality as well so I’m even more confused

<@&286206848099549185>

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[heady harness]

I am projecting a vector onto a translated span and the textbook does it a bit differently. Am I wrong?

project [1,0] onto span{[2,1]} + {[4,0]}

we want to find t e R that minimizes |[1,0] - [4,0] + t[2,1]| which is the distance between the vector and the span

expanding and doing some calculus I arrive at t = 1.5 and so the answer is [7,1.5]

the textbook however finds t to be -6/5

[heady harness]

they set up the equation as |[1,0] - [4,0] - t[2,1]|^2

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[heady harness]

<@&286206848099549185>

[cobalt canopy]

Ye

May e the text book wrong and you can point this out to your professor

*maybe

[heady harness]

well, is my setup correct?

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@heady harness Has your question been resolved?

[heady harness]

No

[lone heart]

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[unkempt cipher]

Suppose that
∞

n = 0
anxn
converges to a function y such that
y'' − 2y' + y = 0
where
y(0) = 0
and
y'(0) = 1.
Find a formula that relates
an + 2,

an + 1,
and
an.

[unkempt cipher]

.close

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[plush geyser]

i’ve been struggling trying to find the slope intercept form for the line of fit

[wild umbra]

best u can do is form the equation of line with 2 point form and then convert it to slope intercept form

[plush geyser]

i did that but finding the B is my struggle

[wild umbra]

B ?

intercept?

u mean'

[plush geyser]

yes

[wild umbra]

!show

[lone heart]

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[plush geyser]

i chose the coordinates (12,47) and (25,50) and used them to calculate the slope of 3/13

i’m not sure if substituting one of the coordinate pairs to find the b intercept would be the right thing to do

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[olive cypress]

https://media.discordapp.net/attachments/823200439322935296/1173094143904448562/image0.jpg

[grave eagle]

lets say i have a(b)^y/12

oops

[olive cypress]

I don’t know how to start with this

[sonic ridge]

you've worked with lagrange multipliers before right? start by identifying the constraint

[olive cypress]

x^2+y^2<=1?

[sonic ridge]

yup,

[olive cypress]

but then there’s the distance

and idk how to deal with the inequality

[sonic ridge]

the constraint would be your distance

if you set g(x,y)=x^2+y^2

with the constraint that g <= 1

[olive cypress]

we were told to deal with the inside of inequalities separately using critical points and the boundary using Lagrange multipliers

what would the equation be that we use to solve the system

it might help me understand

[sonic ridge]

hold on clearly my lagrange multiplier knowledge is floating away

[olive cypress]

please dm me if you figure it out later

[gleaming granite]

are you still working on this?

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[olive cypress]

.reopen

[lone heart]

✅

[olive cypress]

yeah im so confused

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[final ibex]

Help plz Physics

[lone heart]

Please don't occupy multiple help channels.

[wet cipher]

uh oh thats alot wrong

maybe i can hel

which oen u need help

[final ibex]

yes

i had another post

srry it took time to respond

4. part b and question 5

if u can do part b question 4 that would be great

@wet cipher

<@&286206848099549185>

[vagrant heath]

Hey

[final ibex]

hi

[vagrant heath]

Part b question 4?

[final ibex]

yes

well i do need help on 5 but idk how much time or how much work u want to help me with

[vagrant heath]

I have time don't worry

Let's start with 4

So first do you understand why the diagram doesn't take an applied force?

[final ibex]

hm

[vagrant heath]

You don't?

[final ibex]

cuz its initial speed

but thats a guess

[vagrant heath]

Yeah, because like for a force to be applied the hockey stick has to still be in contact with the Puck. Once there isn't something pushing the Puck, the Puck can still be moving I that direction but there's no more force

Got that?

[final ibex]

ye

so the same would be if a baseball bat hits a ball

lets say the ball was still

[vagrant heath]

Yes, in that case we only have air resistance (in x) and gravity (in y)

[vagrant heath]

Why would the ball be still?

[final ibex]

no reason XD

[vagrant heath]

Ok now back to the question then

Ok so now, the only force in x is the Ff and the two forces in y are mg and Fn

Got that?

[final ibex]

ye

[vagrant heath]

Ok so to find the value of Ff how would you go about doing that?

[final ibex]

isnt if given

[vagrant heath]

Also just as a guess, are you in grade 11?

[final ibex]

ye

[vagrant heath]

No it's not

The kineticcoefficient of friction u is given

And yk the equation Ff = Fn • u

Yk that equation right?

[final ibex]

oh ye

[vagrant heath]

Ok so we need the value of Fn for that

What's the value of Fn?

[final ibex]

mxg?

[vagrant heath]

Yeah

Which is 9.8m

[final ibex]

we do 9.8 x 0.11

?

to find Ff

?

[vagrant heath]

And so in total Ff=(0.11)(9.8m) = 1.078m

[vagrant heath]

Remember to include the mass

[final ibex]

wym

mxg

do we gotta find m or

[vagrant heath]

You just had Ff=(9.8)(0.11) but it has to be Ff=9.8x0.11xm

[final ibex]

oh

in the end its just gonna be in m

since its displacement

[vagrant heath]

Listen, if the question doesn't give m that means you don't need it bc it's gonna cancel at some point in your equations

[final ibex]

oh ok

thats it for part b then

[vagrant heath]

Ok so we have Ff=1.078m , which is the only force in x and so when we're doing the equation f=ma we're gonna use 1.078m=ma

[vagrant heath]

No

[vagrant heath]

And so 1.078m=ma and the acceleration is a=1.078

[final ibex]

but we only lf displacement

[vagrant heath]

And then we have a=1.078, vf=0 and vi=4.4 and then we need to find d

[final ibex]

ye

it is

[vagrant heath]

Vi is initial velocity and vf is final velocity

[final ibex]

thats the formula i use

[final ibex]

thats what i meant i typo mb

lemme look at formula sheet

[vagrant heath]

Yes you had the right formula

[final ibex]

sure but we dont got time

[vagrant heath]

Yeah actually we should use a different formula

Tell me which formula we should use

That doesn't have t

[final ibex]

srry i hadda fidn it

vf2=v02 + 2a displacement x

[vagrant heath]

Yeah

Use it and find the d for me

[final ibex]

21.52 = d

its going left so negative

[vagrant heath]

No it's not negative

It's a that's negative

Show me your work

[final ibex]

physics teacher told me if its left or down it is

[vagrant heath]

Yeah but it's not going left in the first place

[final ibex]

all i did was add vo^2 and 2a

oh

im goofy

the ff is left

so positive

[vagrant heath]

Do the work and show me

[final ibex]

im doing it on paper and my phone dead

all i did was add those 2 values when i fixed its exponents and that was displacement

[vagrant heath]

Ah OK so you can't take a picture of it

Look,

0²=4.4²+2(-1.078)(d)
0=19.36-2.156d
2.156d=19.36
d=19.36÷2.156
d=9

[final ibex]

ahh i see

i had my a positive

so i just added and moved over

[vagrant heath]

I see

Well now yk your mistake

And that's it for 4b

Now we can work on 5

[final ibex]

ok

thanks

[vagrant heath]

No problem

I add more friends on discord than the amount of people I chat with

[final ibex]

ah ok

[vagrant heath]

Ok so the question is

I want you to draw for me the diagram of forces

[final ibex]

ok

[vagrant heath]

BRB in like 5 minutes

[final ibex]

i try

[vagrant heath]

I see

Can we actually work on this tomorrow inshaAllah?

In dm

[final ibex]

Sure but I got school

Ill just dm u

[vagrant heath]

Ok

[final ibex]

if u wanna do the problem and you dm me the answer and the steps then we dont have to

i can learn now if u just do steps and show answer

or whenever u wanna do that XD

[lone heart]

@final ibex Has your question been resolved?

[lone heart]

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[dry halo]

Hello

I'm trying to find the equation for i2, but I'm so stumped lol

I thought it was easy but I've been on this all day

Ah I solved it aside from the sign error

The correct solution is that but without the negatives ha

[lone heart]

@dry halo Has your question been resolved?

[lone heart]

@dry halo Has your question been resolved?

[lone heart]

@dry halo Has your question been resolved?

[lone heart]

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[sweet igloo]

HI could I please have some help with understanding the first and second step. I understand the base change formula as loga(x) = logb(x)/logb(a) but I am not sure how this is actually applied.

[alpine sable]

$\log_a b = \frac{1}{\log_b a}$

[ocean seal]

notamy

[sweet igloo]

is that it?

thank you

.close

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[cunning tulip]

anyone good with python?

[cunning tulip]

my question has something to do with math

and its in python

[chilly crag]

Sure just ask

[cunning tulip]

so my teacher is asking to write a code that asks the user a month and a day ( like 12, 31)
then it prints which day of the year it is

and i cant find a solution

[chilly crag]

You could write an array that represents the months. First entry would be the days in january etc

[cunning tulip]

i did that

[chilly crag]

Then if month is 3, then add the first 3 entries

And add the day in your month

Actually no

[cunning tulip]

wait

that works

i think

[chilly crag]

If month is n, add the first n-1 entries

[cunning tulip]

ill try it quickly qnd get back to you

wait

i put this days = [31,28,30,31,...]

do i have to put each month seperate

[chilly crag]

That looks good

If we are in month 4, that means we are through months 1,2,3 already but not month 4

So we could do for i in range(months-1) and add them

[lone heart]

@cunning tulip Has your question been resolved?

[lone heart]

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[wide kestrel]

find b using a and c

[wild umbra]

!xy

[lone heart]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[wide kestrel]

this is all I was given

i think I need to find B (the area) in relation to a and c

[full shard]

damn thats hard

[real parrot]

Its just use of some basic trigonometry

[real parrot]

maybe you need to find the length b not the area

[wide kestrel]

nah the area

[real parrot]

ah

I thought you have to find the length cos we have this same question in Vectors

[wide kestrel]

this is a grade 7 question though

[real parrot]

hmm

[wide kestrel]

im in grade 9, my grade 7 friends sent me this question

[real parrot]

just do area of bigger triangle - area of smaller triangle

[wide kestrel]

i used my knowledge to get this bs but i doubt it's right (theta is angle AC)

[wide kestrel]

yeah but how do I find the little length between the As

[real parrot]

trig

wait

use pythagoras theorem

x = $\frac{-2a + \sqrt{4c^2 - 4d^2 - 2a^2}}{2}$

[ocean seal]

Wither

[real parrot]

@wide kestrel

[wide kestrel]

what's D?

[real parrot]

the perp

[wide kestrel]

the little length continuing a?

[real parrot]

no thats x

d is the perpendicul

perpendicular

[wide kestrel]

of what?

[real parrot]

this

[wide kestrel]

that's A

[real parrot]

huh

[wide kestrel]

i'm bad at writing

but that's a

mb

[real parrot]

capital?

x = $\frac{-2a + \sqrt{4c^2 - 4A^2 - 2a^2}}{2}$

[ocean seal]

Wither

[real parrot]

here

@wide kestrel

[wide kestrel]

no capital

sry

im really stupid mb

[real parrot]

both of the lengths are a's then?

[wide kestrel]

yeah

somehow

[real parrot]

oh

then it'll be easy

[wide kestrel]

alright

[real parrot]

then

x = $\sqrt{b^2 - a^2}$

[ocean seal]

Wither

[wide kestrel]

b is the area of the small triangle though right?

[real parrot]

ah

im so dumb omg

[wide kestrel]

if b was that length the law of cosines would instantly deal with it if you just assumed angle ac as theta

[real parrot]

true but

[wide kestrel]

which is actually what i did

[real parrot]

trig at grade 7

[wide kestrel]

yeah i know

im confused

[real parrot]

not even the case in India

ok

x = $a + \sqrt{c^2 - a^2}$

[ocean seal]

Wither

[real parrot]

now ig u can solve it

@wide kestrel

[wide kestrel]

yeah

can you explain how you got that?

[real parrot]

is the answer $\frac{a^2}{2}$

[ocean seal]

Wither

[real parrot]

sure but is it the answer?

[wide kestrel]

not sure

[real parrot]

ok

but im pretty sure about that

ok listen

[wide kestrel]

wait im stupid

i forgot something

[real parrot]

?

[wide kestrel]

in this case the area of the small triangle there is just 70m2 right?

[real parrot]

wait a minute

yes

it is

[wide kestrel]

frick im stupid

i forgot how triangles worked

thought it was the full base

[real parrot]

and thats how we get a^2/2

[wide kestrel]

im so sorry

[real parrot]

no probs

[wide kestrel]

thanks for the help i got confused there

[real parrot]

even I got confused aha

[wide kestrel]

thanks man

.close

[lone heart]

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[rare grotto]

how to solve this using integration

[rare grotto]

sending work I did in a sec

The solution has a -2n-5

How??

help plz

[lone heart]

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@rare grotto Has your question been resolved?

[lone heart]

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[heady void]

Help finding range for (f/g)(x)

[north perch]

!status

[lone heart]

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

[heady void]

4

[north perch]

whats your answer

[heady void]

(f/g)(x)=(x+2)^2, x->R

cuz idk how to type that E thingy

[north perch]

wouldnt the range be (0,infinity)

since (x+2)^2cant be negative and also cant zero since x not equal to -2

[heady void]

oh wait

oh yeah

ok

thx

.close

[lone heart]

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[rapid hemlock]

can some one help me with q7 and 8

[lone heart]

@rapid hemlock Has your question been resolved?

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@rapid hemlock Has your question been resolved?

[lone heart]

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[clever depot]

What is de morgans law???

[prime badge]

negation of (A or B) is (not A and not B)

negation of (A and B) is (not A or not B)

they are both de morgan's law

[lone heart]

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[dawn quail]

which formula would you recommend to solve?

this one? (u . v) / |v|

[lone heart]

@dawn quail Has your question been resolved?

[dawn quail]

.close

[lone heart]

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[tame nova]

please i have no idea what to do

[tame nova]

We have $A=[x=\sqrt{(n²+1)-n}/n∈N]$

[ocean seal]

Swedish enjoyer

[tame nova]

$1) show that A⊂]0,1]$

[ocean seal]

Swedish enjoyer

[tame nova]

$2) Solve in N the equation \sqrt{n²+1}-1=1/2$

[ocean seal]

Swedish enjoyer

[tame nova]

$do we have A=]0,1]$

[ocean seal]

Swedish enjoyer

[tame nova]

<@&286206848099549185> im losing hope

[lone heart]

@tame nova Has your question been resolved?

[lone heart]

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[late roost]

i need help please

[tall topaz]

assuming n is a natural number show =1 is not prime and n=2 is prime

Then for n>2

Assume by contradiction there exists a prime number >2 that is even

Write n = 2k

Can you show this is not prime

[late roost]

thank u very much

[tall topaz]

[tame nova]

i literally asked for help just now why no one answered

and even pinged helpers

[late roost]

go and ask on math help (available)

[tall topaz]

Be patient someone will help eventually

[tame nova]

I DID

[tame nova]

.

[late roost]

euh how i delate my name and make this available agaiN?

[tame nova]

yes please

[late roost]

.close

[lone heart]

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[tame nova]

We have A={x=√(n²+1)-n and n∈ N}
1)show that A⊂]0,1]
2) Solve in N the equation √(n²+1)-1=1/2

[tame nova]

<@&286206848099549185>

please?

[tacit arch]

do rational simplification using algebra and conjugation

[tame nova]

what about the second one

[tacit arch]

what about it

[tame nova]

..

[tacit arch]

yes?

[tame nova]

what can i do to solve it

[tacit arch]

algebra

[tame nova]

i promise you if I only needed two or three words as a hint i wouldnt be here asking for help...respectfully

the whole exercice is algebra themed

[tacit arch]

then ask more specific questions

or show some work and why you're stuck

[tacit arch]

like it says here

[lone heart]

@tame nova Has your question been resolved?

[lone heart]

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[mossy pine]

Find the point on the line 3x+y+2=0 which is closest to the point (1,−2)

[mossy pine]

These are two different questions but the first one i got no clue

and the second one i got A and stcuk on B

I just need answers i dont need explainations 😄

[limpid spade]

No

[mossy pine]

What

[limpid spade]

We don't give snswers

[mossy pine]

🤯

[limpid spade]

Yeah

[mossy pine]

https://tenor.com/view/saturday-gif-9724160898038358982

[ebon sparrow]

It's against the rules

[limpid spade]

It's monday

[ebon sparrow]

Today is Monday

[mossy pine]

Dm answer?

[limpid spade]

Stop spreading misinformation

[mossy pine]

Math is rigged

It too hard

[ebon sparrow]

Not that either

[mossy pine]

But it will be secret no one find out!

[limpid spade]

Stop!

[mossy pine]

Your blowing my cover

[limpid spade]

Either you're willing to get helped or close this channel!

[ebon sparrow]

[mossy pine]

Ok show all the steps and secretly include the answer at the bottom so I can skip the steps and see answer

Im loophole system!

[worn fox]

!noans

[lone heart]

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.

@mossy pine Has your question been resolved?

[lone heart]

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[kind peak]

The lengths a<b<c<d are such that triangle with sides a, b, c is similar to the triangle with sides b, c, d. If for all possible a, b, c, d it is true d/a<e, what is the smallest that e can be?

[lone heart]

Please don't occupy multiple help channels.

[kind peak]

help please

[vapid shuttle]

@kind peak I closed your other help channel. Stick to one channel please.

[kind peak]

could anyone help me even where to get started, i understand b/a is equal to c/b etc and u can equate them to a common ratio eg k

that gets you to find d/a in terms of k, but doesn't help answer

[lone heart]

@kind peak Has your question been resolved?

[prime badge]

must be about triangle inequality

some common ratio more than x makes c too long

[lone heart]

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[forest garden]

What is the point (let's say, x = k) between two distinct real numbers n and m, where f'(k) = (f(n) - f(m))/(n-m) ? Can we even find k?
(to better explain, for example, if f(x) = x^2, (f(n) - f(m)) / (n - m) = f'((n + m)/2 . but let's say, f(x) = x^3. how would we find such point?)

[golden canyon]

This is called the mean value theorem (at least the existence of such a point), you can probably find the coordinates of it by setting up some some equations

[forest garden]

Oh hmmm... so it is possible to find it?

[golden canyon]

yeah

You'd get this $$\frac{n^{3}-m^{3}}{n-m}=3k^{2}$$

[ocean seal]

Jelle

[golden canyon]

n^3 - m^3 can be factored though

It might not be possible to find the exact value (or values in some cases) for some functions, but solutions will always exist

[forest garden]

Oh right

that's cool

isn't there like

a general equation for all f(x) = x ^ n (n is natural)?

[golden canyon]

Yeah $$a^{n}-b^{n}=\left(a-b\right)\left(a^{n-1}+a^{n-2}b+...+ab^{n-2}+b^{n-1}\right)$$

[ocean seal]

Jelle

[forest garden]

oh yeah, so n*k^(n-1) = a^(n-1) + b . a^(n-2) +....+ b^(n-1)?

[golden canyon]

Yes, you can divide by n and take the (n-1)th root

You can also show that the value for k actually lies between a and b

[forest garden]

daaaamn

that's so cool

alrighty

I appreciate your help

thanks

.close

[lone heart]

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[tall garden]

whats the best approach to do this

the answer is infinity

[carmine reef]

probably lhopitals

[tall garden]

now what

somehow it has to be inf/1

[carmine reef]

Simplify

[tall garden]

it ends up being undefined/2 no?

[carmine reef]

!show and simplify

[lone heart]

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

[carmine reef]

And no

Not really

[tall garden]

do we know its infinity because of the direction it gives

so we just need to know what the graph looks like or what

[carmine reef]

if we knew what the graph looked like we wouldn't have to do any of this

try moving the sec^2(x) to the denominator

[tall garden]

im confused

[tacit arch]

about what

[tacit arch]

their hint is sec(x) = 1/cos(x)

[tall garden]

right but how is that gonna work

where is infinity gonna come from

like this for example

makes sense

because of domination

they both go to infinity, but the cuberoot outgrows lnx

do we just treat an asymptote in my original problem similarly?

could I get some kind of explanation?

[lone heart]

@tall garden Has your question been resolved?

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[cursive bridge]

Hi, I have this problem: Let U={1, 3, 5, 7, 9,...,99} . In the set of functions from U to itself, we consider the relation R defined as follows: f R g if and only if the composite of f and g is an identity function in U. Is R an equivalence relation?

[prime badge]

no way

you can see it's not reflexive

[cursive bridge]

I have an exam in discrete math tomorrow Im so stressed😭

Im just not sure how to disprove this

or prove

do I literally take a function as an example

or is there something more general

[prime badge]

that's what i did

[cursive bridge]

I have another similar question but instead of the composite of f and g being an identity it has to be bijective

[lone heart]

@cursive bridge Has your question been resolved?

[cursive bridge]

so can someone help?

[cursive bridge]

<@&286206848099549185>

[lone heart]

@cursive bridge Has your question been resolved?

[lone heart]

@cursive bridge Has your question been resolved?

[cursive bridge]

welp

[lone heart]

@cursive bridge Has your question been resolved?

[lone heart]

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[scenic wing]

𝐹(𝑥)=𝑥+tan(𝑥)+𝐶

[scenic wing]

find the anti derivative

the answer says x+tanx +c

[tacit arch]

Wot

[scenic wing]

𝑓(𝑥)=sec2(𝑥)+1

this is the question

[tacit arch]

Oh

[scenic wing]

where did the + x come from

[tacit arch]

That was backwards

What's the derivative of x?

[scenic wing]

1

💀

[tacit arch]

Did that answer your question

[scenic wing]

yes

[lone heart]

@scenic wing Has your question been resolved?

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[fair glade]

Do I just ask a question here?

[torpid dirge]

yes

[fair glade]

I have absolutely no idea how to factor decimals. I think I have to use the quadratic formula but my teacher hasn't taught it yet and I'm hoping for another way.

😭

[torpid dirge]

you could just factor out the decimal

[fair glade]

I'm dumb, how exactly?

[torpid dirge]

so for some quadratic equation that starts with -0.1 a^2+...= -0.1(a^2...)

[fair glade]

how would I do it for this question?

<@&286206848099549185>

[opaque kelp]

Do you know how to complete the square?

[fair glade]

My teacher didn't teach it but I learned it myself 5 minutes ago

[opaque kelp]

Yep try completeing the square with this eqn

[fair glade]

would it be d=-1/10(a^2-85a+400)

and then I complete the square?

[opaque kelp]

Yes

[fair glade]

you are a savior omg

[opaque kelp]

From the completed square you know that in he bracket (x-h)^2+k x and k are the Turning points of the graph/highest lowest points

[fair glade]

(imagine my teacher takes off 1 mark bc we didnt learn it in class and therefore I shouldnt use it 🙄 )

[opaque kelp]

Nah your teacher can't do that

[fair glade]

(he did, twice)

[opaque kelp]

It's still a method u will learn further off

[fair glade]

true

[opaque kelp]

☠️

My teacher is fine with that as long as I don't mess up the qn

[fair glade]

lol

ok, nvm im just a dumb dumb

didnt even need to complete the sq

.close

[lone heart]

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[tall ivy]

how to find the quadratic function equation

pls explain slowly and patiently

i already know the 2nd difference is -4

also i know the standard form of quadratic functions

and bam thats all i know

[paper compass]

ok

have you learned vertex form?

[tall ivy]

yes

and x-intercept form too

[paper compass]

ok

how would you write this equation in vertex form?

[tall ivy]

y=a(x)^2

wait

[paper compass]

perfect

[tall ivy]

oh

yay

[paper compass]

so do you know how you would go about finding the a value?

[tall ivy]

put in a point for y and x?

[paper compass]

yep

so choose any point

[tall ivy]

1, -2

[paper compass]

sure

so by plugging those values in, what do you get for a?

[tall ivy]

-2

[paper compass]

yep

[tall ivy]

yey

wait is that it tho

[paper compass]

so if the a=-2 we have the equation y=-2(x)^2

and that is our equation

[tall ivy]

oooooooooo okokokok

wait thats in standard form too right
idk my brain isn't working rn

[paper compass]

yep

[tall ivy]

hoorayyy

tysmm

also when a problem says "clears the net"

does it mean

goes into the net? or goes over the net?

[paper compass]

can you show me where it says "clears the net"

[tall ivy]

oh wait wait goes over the net mb

the context is tennis ball so its over the net

[paper compass]

oh

ok, i think u can do the rest, u got this

[tall ivy]

I got stuck again 😭

now i have to think of functions that fit these rules

💀

is there a fast way to do this

[paper compass]

ok

give me a moment

[tall ivy]

kk

[paper compass]

go research what it means for a function to be even, odd, or neither

then based on that you can determine values for a, b, and c

but i gtg

sry

:(

[tall ivy]

even: f(-x) = f(x), it is symmetrical about the y axis
odd: f(-x) = -f(x), symmetrical in about the origin
Neither: neither of that, it's a linear thing

[paper compass]

basically

[tall ivy]

oh its okkk

if its urgent u can go :)

ima friend request though 😭

<@&286206848099549185>

[lone heart]

@tall ivy Has your question been resolved?

[tall ivy]

somebody help pls

<@&286206848099549185>

[lone heart]

@tall ivy Has your question been resolved?

[tall ivy]

no but yes

[lone heart]

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[lone heart]

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[amber idol]

<@&286206848099549185>

Guide me to solve this please

[split mantle]

Is there something to simplify?

[amber idol]

Solve to find value of x

Simplify for x

[zinc haven]

for our convenience, let $m=\log{x}$

[ocean seal]

FungusDesu

[fallow quiver]

let the log x = m and solve it

[zinc haven]

and dont forget $a^{\log_{a}{b}} = b$

[ocean seal]

FungusDesu

[amber idol]

Oh yes

[zinc haven]

$\sqrt{x}$ can be rewritten as $x^{\frac{1}{2}}$

[amber idol]

Yes

[ocean seal]

FungusDesu

[zinc haven]

this information should be sufficient for you to solve for x

if you have further question feel free to ask

[amber idol]

Yes

[amber idol]

Hey can you explain how is this possible

Im comfortable with exponents, you can use them for explaining

[zinc haven]

hope this explanation is good

[amber idol]

I get it

What is log^2 x ?

@zinc haven

The 3rd term

Is it log10^2 x

[zinc haven]

it would simply be m^2

[amber idol]

Oh ok

[zinc haven]

if you did what i have instructed and simplify everything, which shouldnt be too hard, you should have a polynomial

which you can solve to find m, then find x

[amber idol]

Ok

I got a polynomial

[zinc haven]

you got a polynomial indeed, you should be able to solve it with a calculator

[amber idol]

I did it without calculator

Thank you @zinc haven

😄

[zinc haven]

glad to help

[amber idol]

.close

[lone heart]

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