help-13

[sharp lotus]

perimeter is just the sum of all the sides right?

[crimson sedge]

yeahh

[sharp lotus]

ok so think of just one pair of sides in each triangle, let's think about red and green ok?

[crimson sedge]

ok

[sharp lotus]

the ratio RED to GREEN is the same in each triangle, thats the key idea

so you can write that as fractions:

[crimson sedge]

ok..

[sharp lotus]

a ratio is just a fraction right?

[crimson sedge]

yes

[sharp lotus]

$\frac{28}{35}=\frac{x}{15}$

[crimson sedge]

28/35

[wraith dagger]

Soosh

[crimson sedge]

i see

[sharp lotus]

so you see how the ratio of the big triangle is on the left, and the small triangle is on the right, where we wrote x for the unknown side?

with both sides of the equation using the red\green sides of the triangles?

[crimson sedge]

yes

[sharp lotus]

so now you have an equation you can solve for x

[crimson sedge]

so its x = 12

?

[sharp lotus]

exactly

ok, now see if you can setup a similar equation using the same idea to solve for y

(the blue side of the big triangle)

ill paste this again so we dont keep scrolling up

[crimson sedge]

so its 8/15 = y/35?

[sharp lotus]

yes, exactly

note: you could have picked also blue\red sides since you know x now, or what you picked which was blue\green sides, either way you will get same answer

[crimson sedge]

so, y = 18.66?

[sharp lotus]

seems right

so now you can add up the perimeters of both triangles, since the problem was asking for the difference in perimeters

[crimson sedge]

the perimeter of large triangle: 28 + 35 + 18.67 = 81.67
perimeter of the small triangle: 12+8+15 = 35

wiat

u said the problem was asking for the diff in perimeters

how did you know that?

[sharp lotus]

subtraction 😄

subtract small one from big one and ta-da

[crimson sedge]

yeah but like where does it say in the question that its looking for the difference in perimeters?

[sharp lotus]

"the perimeter of the larger triangle is k centimeters longer than smaller, what is k"?

[crimson sedge]

OHHHH

i get it

[sharp lotus]

i was just paraphrasing it

[crimson sedge]

thank you smm

[sharp lotus]

i didnt remember exact wording, but same thing

[crimson sedge]

yeahh i just realized

[sharp lotus]

when dealing with similar triangles, just remember this idea of comparing PAIRS of their sides as ratios

and make sure you match up the right ones even if they are rotated in weird ways etc., this idea will come up a lot in future maths : )

any other q?

[crimson sedge]

hmm not yet so far

butt

[sharp lotus]

what was final answer?

[crimson sedge]

final answer was 46.7

[sharp lotus]

?

[crimson sedge]

dyk statistics? im also having trouble with that

[sharp lotus]

not really to be honest 😄

but you should just .close this channel then open a new one and ask your question, im sure someone will come along and help

gl!

!done

[cedar kiln]

If you are done with this channel, please mark your problem as solved by typing .close

[crimson sedge]

okay. thank you!

.close

[cedar kiln]

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[crimson sedge]

please help

[cedar kiln]

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[cedar kiln]

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[little wave]

very dumb question but the range on this graph should be -1480, infinity, right?

[little wave]

hang on i gotta send pic

[rare vault]

seems possible

probably want the () [] though

[little wave]

what do u mean?

[rare vault]

does it include or exclude -1480

[little wave]

im just trying to find the range of the graph

i know the domain is -infinity, infinity

[rare vault]

yes, does the range include or exclude -1480

does the graph ever actually get the -1480

[little wave]

yeah

[rare vault]

(-infinity, infinity)

okay then the range is [-1480, infinity)

[little wave]

oh yeah i remember now

thank you!

.close

[cedar kiln]

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[cedar kiln]

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[warm flint]

The ratio of products produced by two workers is 0.95. How many products does each person make, knowing that this person makes 10 more products than the other?

[warm flint]

help

[gilded mesa]

Hey

So ratio is 0.95

We can write it as 95/100

So one makes products in multiple of 100 so second makes with same multiple of 95 concept of ratio

Like if first worker makes 100a products second worker makes 95a products

Now there deference is 10

So 100a-95a=10

a=2

Puting a back

We get first make 200,and second make 190

[warm flint]

i think it have a formula for that

y=a.x

[gilded mesa]

Formula??

[warm flint]

form

[gilded mesa]

I am not understanding you

[warm flint]

a form

[gilded mesa]

My answer is right don't worry

[warm flint]

thank

[cedar kiln]

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[reef sedge]

a.)

[reef sedge]

10gm/s*

[buoyant latch]

What have you tried

[reef sedge]

draw the diagram

i think i might have an idea now

do i just use the position vector

but, what im confused about is the 50m above ground.. why is that info nessescay

[buoyant latch]

It just means after 0s you’re 50m above the ground

It just shifts the entire system up

[reef sedge]

i think i got it

[reef sedge]

the j component right?

[buoyant latch]

j?

[reef sedge]

j component?

for vectors...

i take them and then turn them into paramtric equations..

hold up...

how do they still have a left over g at the bottom for the denominator

[idle tusk]

that's from the second term

the (g(x/...)^2) / 2

$5g \frac{x}{5g \sqrt{3}} = \frac{\sqrt{3}}{3}x$

[wraith dagger]

artemetra

[reef sedge]

i got this...

im talking about this g

[idle tusk]

yes ik

[reef sedge]

hmm...?

[idle tusk]

this is where it comes from

they reordered the terms

[reef sedge]

hmmm?...

lets do it

so it would be

x^2/75g

[idle tusk]

75?

how

[reef sedge]

$=\frac{x^2}{75g}$

[wraith dagger]

LW

[reef sedge]

5^2=25

*3

=75

sqrt(75)

[idle tusk]

and times 2

[reef sedge]

wait hold up

is it sqrt(75)

[idle tusk]

no sqrt

no

[reef sedge]

HUHH

WAIY WHAT

[idle tusk]

???

where do you see a square root

[reef sedge]

$5\sqrt{3}$

[idle tusk]

here

[wraith dagger]

LW

[reef sedge]

=75???

[idle tusk]

yes but you also have /2

[reef sedge]

yes

[idle tusk]

so 75*2 = 150

[reef sedge]

how is it 75 tho

im really tired right now

its just

sqrt(25*3)

which is sqrt 75...?

[idle tusk]

it is squared

[reef sedge]

bugger

i thought so

ok

so we have

$\frac{-g(\frac{x}{75g}){2}$

wtf

[wraith dagger]

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

[reef sedge]

my latex skills

WTF

[idle tusk]

something unmatched

$\frac{-g\frac{x}{75}}{2}$

[wraith dagger]

artemetra

[idle tusk]

eh

[reef sedge]

$\frac{-g(\frac{x}{5\sqrt{3}g})^{2}{2}$

[wraith dagger]

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

[reef sedge]

BRUH

[idle tusk]

write it by hand

and do it slow

[idle tusk]

like step-by-step

[reef sedge]

i am doing it by hand

so we have

$\frac{-g}{2}$

[wraith dagger]

LW

[reef sedge]

now:

[idle tusk]

g is squared

[reef sedge]

ohhhhhhhhh

that means

im working with

$y=\frac{x}{\sqrt{3}}-\frac{gx^2}{75g^2}\times\frac{1}{2}+50$

[idle tusk]

yes

[wraith dagger]

LW

[reef sedge]

then i do

cancel the g at top with bottom

to get

$y=\frac{x}{\sqrt{3}}-\frac{x^2}{2\times75g}+50$

[wraith dagger]

LW

[idle tusk]

mhm

[reef sedge]

so then

$y=\frac{x}{\sqrt{3}}-\frac{x^2}{150g}+50$

[idle tusk]

yes

[wraith dagger]

LW

[reef sedge]

bro my latex so slow today

[idle tusk]

it's okay lol

[reef sedge]

thanks for pointing out the mistake!

[idle tusk]

no problem

[reef sedge]

now!

lemme try b

i just set y=25...

wait

ok thats what the solution is

setting y=25

do u have any idea why u set it =25

why is y=25?

@idle tusk any chance u could explain?

[idle tusk]

y is height

[reef sedge]

why

[idle tusk]

"after t seconds, the particle is at a height of y"

so y is the height

[reef sedge]

hmmm

where

[idle tusk]

think of y as a function of height with respect to how much horizontal distance has already travelled

[idle tusk]

wdym where

[reef sedge]

uh

then why would i need to apply

+50

if its no longer 50 above the ground

[idle tusk]

because that's the initial height

[idle tusk]

why not

[reef sedge]

,w 25sqrt(3)*(sqrt(g^2+2g)+g), g=9.81

its not

its 25m now

[idle tusk]

bruh

lol

[idle tusk]

think of someone throwing a ball off a cliff

[wraith dagger]

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[reef sedge]

yes

[idle tusk]

the ball starts off at 50

goes up a bit

and falls down below 50

not sure what the issue is here

[reef sedge]

hmmm

i guess

im supposed to just set y=25

[idle tusk]

ye

and then it's just a quadratic wrt x

[reef sedge]

yes

thanks!

.close

[cedar kiln]

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[wraith dagger]

konxmok

[cedar kiln]

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[hexed wraith]

It says determine the total area of the following shapes

[blazing zephyr]

for number a does it mean volume?

[hexed wraith]

I don’t know

i don’t think so

[idle tusk]

i think it's surface area

[blazing zephyr]

!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 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

[hexed wraith]

1

I’ve read that the formula is 1/2 * perimeter of the base * height

[blazing zephyr]

isnt that for the triangles only?

[idle tusk]

i mean

all faces are triangles

[idle tusk]

yes that's for a single triangle

[blazing zephyr]

how about the bottom face

[hexed wraith]

are you guys on the first or second shape

[idle tusk]

oh yeah that's a square

oops

[blazing zephyr]

which are you having problem?

[hexed wraith]

both

the second one it gives two sides of the bottom triangle

[blazing zephyr]

you could use this for the triangles, then since the pyramid consists of 4 triangles (accounted by the alg) and a square (bottom face), you could use the formula then add the area of the base

[hexed wraith]

ok

what about the second shape

.close

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[near trout]

can anyone help me with 1b?

[digital cliff]

do you have any ideas

[grizzled plank]

so

[near trout]

nope so far 😹

[civic coral]

ayyy

[grizzled plank]

you can set the equations equal to each other

[civic coral]

ayyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyy

[grizzled plank]

gotta go

but just for idea:

[civic coral]

ayyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyy

[grizzled plank]

set them equal and try to factor

[near trout]

i got that for x already

[grizzled plank]

oh

[near trout]

just need to integrate

[grizzled plank]

soz

gtg

[near trout]

aight np

[civic coral]

shit

[near trout]

whadap 😹

[civic coral]

have you find out the answer for a

[near trout]

yes

0, -1, 1

[civic coral]

Then I suppose you do not know how to integrate it to find the shade area?

[near trout]

ye

[civic coral]

im so mad

tell me, in the range of -1 to 0

which of the two function has a greater outputs than the other.

[near trout]

-1 mate

0 will just result in 0 haha

[civic coral]

wdym

[near trout]

wont it be an absolute value

[civic coral]

in the range of x=-1 to x=0

which function has a greater output

than the other.

It is a key to solve the question.

can you distinguish the graph between f(x) and g(x)

[near trout]

nvm mate i got it

im just gonna make y = x+3

so i can just integrate the area under the curve and subtract with the area of the trapezoid

[civic coral]

well, if that works, everyone has his own way

[cedar kiln]

@near trout Has your question been resolved?

[cedar kiln]

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[crimson sedge]

In the second last equation (probability of passing at least two subjects), why have we added the probability of passing beta, gamma, delta at the end too?

[carmine whale]

Coz passing at atleast 2 subjects includes passing in 3 subjects

which was excluded in all the previous cases

So we have to add that seperately

[crimson sedge]

I see, that does make sense, I have to practice more to understand it better

Thank you!

[cedar kiln]

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[astral pagoda]

can someone help with these two exercises

[cedar kiln]

@astral pagoda Has your question been resolved?

[astral pagoda]

<@&286206848099549185>

im guessing its some elementary probability rules but honestly not sure 💀

ri[

rip

my qs always drowned sad

[cedar kiln]

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[bold pond]

what is this asking? im not sure how to do these, my notes say something about approaching infinity but i dont get what they mean

[digital cliff]

its essentially asking what the value of P(x) approaches as x approaches infinity or -infinity, ie as x just gets very very large in the + direction or the negative direction

[mighty shuttle]

find the limit as x approachs infty and -infty

[obtuse vapor]

help?

[bold pond]

please go to an unoccupied room

[obtuse vapor]

mb

[digital cliff]

with a polynomial, when infinities are involved, you can focus on the highest x power term only, since the contribution of the other terms will be much smaller in comparison

[bold pond]

but what am i supposed to do to solve i dont get that part i guess

ohh do ineed to find the zeros and multpulicties?

[digital cliff]

no no

[bold pond]

o k wrong thing

[digital cliff]

lets think about the +infinity limit

we only need to look at the 7x^5 really

what happens to 7x^5 as x becomes an increasingly large positive number

[bold pond]

it goes to infinity

[digital cliff]

theres your answer

[bold pond]

yea

ok

[digital cliff]

-infinity works in the same way

[bold pond]

ok so that is a theoretical question

i guess it was too obvious didnt think it was the answer for some reason

[bold pond]

i have a graph written down in my notes that has 2 curved lines in quad 1 and 3 and they just go to infinity on both axis

is that refferring to this?

[digital cliff]

like this?

,w graph x^5

[wraith dagger]

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[digital cliff]

your graph probably is of the whole function though rather than just the leading term, wont change the shape though

[bold pond]

I know this is getting confusing but this is what I mean.. I guess it's for vertical asymptotes?

[digital cliff]

ah different function i see

youre studying limits really atm
when you have a function with some fractional component
if the denominator can be 0 at some point there'll be a VA, usually, there can be holes instead
you can use limits to see which infinity itll go to from which side and whatnot

which is what youre doing there

[bold pond]

ok, i see, my next few HW quesations are about V.A and hortzontal As too

[digital cliff]

snazzy stuff

[bold pond]

haha

[cedar kiln]

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[scarlet fox]

Do I have to differentiate a function before integrating it?

[past oriole]

it depends on what you are doing

If you are calculating antiderivatives or the area under the curve then no

[raven shard]

in short, fundamental theorem of calculus

[scarlet fox]

so should I learn that before learning integration?

[raven shard]

uh, probably?

[scarlet fox]

Im a bit lost on this

[green torrent]

There is a YouTube creator called 3 blue 1 brown. I'd recommend his series on Calculus - just search 3 blue 1 brown calculus

That'll cover the basics of what derivatives and integrals are

[scarlet fox]

.close

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[wet fossil]

How do I prove this for n = k assuming it works for all n = k-1, k-2, k-3,...,2,1?

[wet fossil]

$\frac{1}{1^{2}} + \frac{1}{2^{2}} + ... + \frac{1}{(k-1)^{2}} + \frac{1}{k^{2}} \leq 2 - \frac{1}{k-1} + \frac{1}{k^{2}}$

[wraith dagger]

Normed

[wet fossil]

Need to prove : $\$
$\frac{1}{1^{2}} + \frac{1}{2^{2}} + ... + \frac{1}{(k-1)^{2}} + \frac{1}{k^{2}} \leq 2 - \frac{1}{k}$

[wraith dagger]

Normed

[warped drum]

$\frac{1}{1^{2}} + \frac{1}{2^{2}} + ... + \frac{1}{(k-1)^{2}} + \frac{1}{k^{2}} \leq \left( 2 - \frac{1}{k-1} \right) + \frac{1}{k^2}$

[wraith dagger]

LF

[wet fossil]

,w simplify 1/k^2 - 1/(k-1)

[wraith dagger]

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[wet fossil]

,w simplify 1/k^2 - 1/(k-1) + (1/(k^2(k-1)))

[wraith dagger]

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[wet fossil]

Haha

.close

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[dark pasture]

This is the problem I'm working on

[dark pasture]

I solved it a different way and got the answer 2/x³

What I can't figure out is how

Oh sorry that's sideways

[crystal raptor]

,rccw

[dark pasture]

Thank you!

So I guess the question is how to factor multivariable polynomials that lack concrete common factors

[cedar kiln]

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[cedar kiln]

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[primal sparrow]

I need help with #3 and #4. This is for an Acoustics class. So it's basically audio physics

[crimson sedge]

.close

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[nocturne iris]

Just need some help

[nocturne iris]

.close

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[subtle summit]

I need help with this

[crimson sedge]

!show

[cedar kiln]

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

[subtle summit]

x^2+4x-48=(x-a)(x-a)+b.

x^2+4x-48=x^2-2ax+a+b

[crimson sedge]

i mean i guess you can do it like this. But have you heard of completing the square?

thats a more general, pragmatic approach

[subtle summit]

this is a completing the square question?

[crimson sedge]

yes

it is haha

also

your foiling was wrong

its not x^2+4x-48

[subtle summit]

oh I think I did a miscalculation :/

[crimson sedge]

ye

do you want to go thru on how to FOIL it + completing the square?

[subtle summit]

no, I know how to foil

[crimson sedge]

okay then do that

[subtle summit]

but how do I tackle this problem exactly?

I complete the square for the other side?

[crimson sedge]

no

ignore the right side completely

just
1- FOIL your two binomials
2- complete the square

after that, compare with the right side

[subtle summit]

for the foil bit I got x^2-4x-32

[crimson sedge]

ok thats correct now

g'job

now complete the square

[subtle summit]

then for the second part I got (x-2)^2-36

[crimson sedge]

,w expand (x-2)^2 - 36

[wraith dagger]

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Requested by qylooo

[crimson sedge]

okay

yes thats correct

you are done pretty much

now just compare that with the right side

what is b and what is a

[subtle summit]

ohhh that's it-

[crimson sedge]

yeppie

[subtle summit]

wait I think I did a mistake somewhere cuz 'a' turned out to be 2 and not -2

[crimson sedge]

no u didnt

its because, notice, that it says (x**-**a)^2 not (x+a)^2

like

the negative is already there

they just want the number

if it was (x+a)^2 then -2 would be correct

[subtle summit]

oh I realise my mistake now-

.close

[cedar kiln]

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[crimson sedge]

you didnt really do-

oh okay bye

[cedar kiln]

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[eternal crescent]

Hey could someone check if my answer is good?

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[silk escarp]

Hello there! How would you do this?

[brisk prairie]

...

[silk escarp]

Sorry, I guess you were about to post your question but I got It first :v

[raven shard]

are you allowed to use calculus?

[silk escarp]

If you mean derivates, and those things, no

I got this problem from a book which is "pre- calculus" so I shouldn't use It

[raven shard]

ok cool

[silk escarp]

I translated It

[raven shard]

what's special about cotx and tanx?

[silk escarp]

What do you mean?

[raven shard]

what's the relation between those two functions

[silk escarp]

are reciprocal

[raven shard]

yes

so you can begin with that fact

(and honestly if you're not looking for a rigorous solution... once you realize that, this problem becomes very simple)

[silk escarp]

Should I transform cot x to 1/tanx?

[raven shard]

or better yet, use cotx = u

[silk escarp]

Something like this?

[raven shard]

uh.. huh?

what did you do

[silk escarp]

I tought It was going that way :'v

[raven shard]

if cotx = u,

then tanx = 1/u

so you shouldn't have any tan terms in your transformed expression

[raven shard]

yes

[silk escarp]

Should I operate It?

[raven shard]

are you looking for a rigorous solution?

once you have the terms in this form, you can reasonably guess the answer

and this path doesn't require any proof

[silk escarp]

6?

[raven shard]

yes

[silk escarp]

But, how can I demostrate It?

[raven shard]

are you looking for a proof, then?

[silk escarp]

Sure

[raven shard]

?

[silk escarp]

Yes

[cedar kiln]

@silk escarp Has your question been resolved?

[raven shard]

i'm actually not sure how to easily prove y = 6 is a global minimum without calculus

but proving that y > 2 (so putting this in context of the answer choices) is pretty easy

because the answer choices have minimum 2, 6, 8

[silk escarp]

Yep, I know. But I'm still wondering for a solution

I guess thinking in doing some inequations but It does not seem to work

Also, from the first condition. I can realize that from any value of "n", being n an integer, cot and tan are positive

Oh

Both go higher while the value of x increment

Maybe I can graph

[cedar kiln]

@silk escarp Has your question been resolved?

[cedar kiln]

@silk escarp Has your question been resolved?

[silk escarp]

Reposting if someone wanna help me

<@&286206848099549185> 😔 🤙

I even graphed and nothing

[gilded mesa]

You tried putting value of n

Tag me when you tried

[silk escarp]

Is this what you mean?

[gilded mesa]

Try n=0

[gilded mesa]

It makes it easy

n=0

Pi to 3pi/2

[silk escarp]

I know, tan and cot are positive there

[gilded mesa]

Ya that's right interval

[silk escarp]

Also replacing n<0, I get the same

[gilded mesa]

What is 1+tan^2

[silk escarp]

sec´2?

[gilded mesa]

Ya

[gilded mesa]

Yeah

Now what is sec when goes pi to 3pi/2

[silk escarp]

Ich habe keine Ahnung

I mean, It's the inverse of cos so

[gilded mesa]

Yeah

[silk escarp]

from 0 to 1?

or -1 to 0

oh

[-1 ; 0> ?

[gilded mesa]

Cos 3pi/2 is 0

It's inverse of cos

So it goes to infinity

And cospi =-1

[silk escarp]

Oh

[gilded mesa]

So sec^2 (pi) =1

[silk escarp]

Then from minus infinite to -1

[gilded mesa]

It goes [1, ♾️)

[silk escarp]

https://tenor.com/view/not-everyone-has-an-iphone-gif-7021079305409267183

[gilded mesa]

Think again

[silk escarp]

Oh

To the power of two (square)

[gilded mesa]

Yeah

[silk escarp]

Alright

[gilded mesa]

Now for cotx

[silk escarp]

You mean from 1 to ♾️

[gilded mesa]

Oh sorry you are right

It goes 1 to ♾️

[silk escarp]

:v

[gilded mesa]

Now for cot x

What is cot pi

[silk escarp]

From 0 to ♾️

[gilded mesa]

From 1 to ♾️

Sec^2(pi) is 1

[silk escarp]

Isn't cotanget of 3pi/2, zero?

[gilded mesa]

Ok now we are on cot

Cot (pi) first

[silk escarp]

Is is not defined on pi

[gilded mesa]

Yeah

[silk escarp]

It has no limit there

[gilded mesa]

What is cot (3pi/2)

[silk escarp]

0

[gilded mesa]

Your question seems wrong to me

[silk escarp]

Somewhere has to be a minimum I guess

[gilded mesa]

Oh yeah

[silk escarp]

I graphed (to graphic but in past I guess :v)

[gilded mesa]

We have to derivative it for minimum

[silk escarp]

The cot and tan function and yeah, while one increases the other one decreases

[gilded mesa]

Now see
Cot is going ♾️ to 0
And sec is going 1 to ♾️

We have to find a minimum point

Derivative it and put equal to 0

[silk escarp]

Alright but unfortunaly I don't know how to derivate xD

[gilded mesa]

Ok let me do it

[silk escarp]

Also, the book where I found the exercise is "pre-calculus" so It shouldn't be solved by calculus

Well you can do It but that's not the point

[gilded mesa]

Pi +pi/4

5pi/4 is maximum point

Put it in equation

Answer D is correct

[silk escarp]

Thank you but I'm searching a non-calculus solution 😅

Or at least may you share me your solution?

[gilded mesa]

Calculs and math are like never ending story

They are love birds

How you can separate them

[silk escarp]

🗣️ 🗣️ 🔥

You speak Spanish, right?

[gilded mesa]

No

[silk escarp]

I saw the Messi photo and I tought It

[gilded mesa]

Oh but messi is from argentina

How you think Spanish

[silk escarp]

Yes I thought you were Argentinian

[gilded mesa]

Guess what

[silk escarp]

Are you?

[gilded mesa]

I am not from argantina

[silk escarp]

Bangladesh?

[gilded mesa]

😆

No close

You are close same continent

[silk escarp]

Brasil?

Oh, Asia

India?

[gilded mesa]

Yeah india

Where are you from

[silk escarp]

It's an stereotype but Indians and mostly all asians are so good in Math and Science

[silk escarp]

Peru

[gilded mesa]

Ya genetics 🫣

Matters

[silk escarp]

Peru is basically India but without so much population and our president kills us

[gilded mesa]

I don't know much about Peru

Why your president is dictator

[silk escarp]

Is a joke haha

[gilded mesa]

?

Oh

You have constitution??

[silk escarp]

And I don't wanna explain It because is so complicate to understand

[silk escarp]

Sure

[gilded mesa]

But not big as ours 😜

[silk escarp]

Probably yes

Since all of our presidents since 1990 have been charged with corruption

[gilded mesa]

Oh my god

[silk escarp]

Anyways, and nice to meet you

May you share me your solution?

[gilded mesa]

Nice to meet you too

[silk escarp]

It's almost 12 AM and I must sleep :'v

[gilded mesa]

I will share you in morning??

[silk escarp]

Oh, sure

[gilded mesa]

DM me

[silk escarp]

I've already sent you a friend request

[gilded mesa]

Ok i will send good bye

[silk escarp]

Thanks and sweet dreams

I'll wait for It

.close

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[upper laurel]

since x(µ) = µ, why would d0 be 0 instead of µ

[dire geode]

Because x(mu) = 0 + mu

[upper laurel]

the last equation has x(mu) = 0
the top equation would also = 0 when mu is plugged in

[cedar kiln]

@upper laurel Has your question been resolved?

[cedar kiln]

@upper laurel Has your question been resolved?

[upper laurel]

I dont understand the paper Im reading but it appears this is a typo
the next time this equation appears, theres a µ +

nowhere else between this image and the previous image does x(z) appear

Ill see if this works out

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[vital rose]

Part 2) i was thinking getting A of tri ABD - A of tri ABH but idk how to get ABH. i thought tri ABI might be the same as ABH but the answer i got was wrong

[late bay]

its a

[vital rose]

yooo abh

[vital rose]

a?

[royal loom]

!nosols

[cedar kiln]

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.

[royal loom]

!show

[cedar kiln]

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

[vital rose]

my working's all over the place but i got area of ABD as 27cm^2, line segment AD as sqrt(117)

For area of tri ABI

hmm lemme do it again

216/13 i think

<@&286206848099549185>

i think i should be finding AH or HG but idk how to find that

then id be able to get area of AHB/GDH

[cedar kiln]

@vital rose Has your question been resolved?

[vital rose]

rip ;-;

[cedar kiln]

@vital rose Has your question been resolved?

[cedar kiln]

@vital rose Has your question been resolved?

[vital rose]

Nvm i got it:)

.close

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[past oriole]

Can I write a function that will give me the distance of a random point to a plane in $\mathbb{R}^3$?

[wraith dagger]

Gapi

[livid dust]

What do you mean by "random"?

[past oriole]

For any point

[crimson delta]

you can write the plane in hesse normal form

[livid dust]

so an arbitrary point? when you say random it kinda implies you're picking from some probability distribution

[crimson delta]

then you can just plug in

[hollow trail]

https://www.khanacademy.org/math/linear-algebra/vectors-and-spaces/dot-cross-products/v/point-distance-to-plane

[past oriole]

OK thannk you

.close

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[waxen kestrel]

Here's how I think: (1/365)^4*349C5 = 2,362 quintuples with shared birthdays

is it correct? I found a similar exercise where the formula were (1/365)^(k-1)*(1/365)*nCk

[cedar kiln]

@waxen kestrel Has your question been resolved?

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@waxen kestrel Has your question been resolved?

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@waxen kestrel Has your question been resolved?

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[trim jetty]

is $\lim_{x\to\infty}f(g(x)) = lim_{g(x)\to\infty}f(g(x))$ where $g(x)=2x$

[wraith dagger]

shrod1nger

[dusty hazel]

yeah

[trim jetty]

Wht is the intuition for that

[slate lintel]

x and 2x are essentially the same here, ones just a little more zoomed out than the other

but infinity stays at the same place

[still merlin]

infiinity timies anything postive is still infinity

[trim jetty]

basically, f(g(x)) is f(x) but x axis is stretched

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[pseudo grail]

I can't solve this integral. Does anyone have a tip or a clue?

[mental trail]

substitute nicely

[pseudo grail]

what do you mean by substitute ?

[mental trail]

u-substitution

u = ...

[pseudo grail]

like u = ln^2 ?

[mental trail]

close but not quite

[pseudo grail]

hmmm

[mental trail]

you want to make the 1/x disappear

but you know that this is the derivative of...

[pseudo grail]

lnx

[mental trail]

exactly

[pseudo grail]

wait wait wait

[pseudo grail]

found it !

thank you

.close

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[crimson sedge]

I need help

[crimson sedge]

My algebra 1 teacher doesnt teach shi

Just gives us worksheets and grades them no notes

Im so lost 😞

<@&286206848099549185>

[peak minnow]

is 32x^7 the area and is it a square?

[crimson sedge]

Yes

[peak minnow]

formula for area of square is
area = l^2
and for perimeter is
perimeter = 4l

taking sqrt on both sides of area u get
sqrt(area) = l
and using this in the perimeter formula u get
perimeter = 4(sqrt(area))

using area as 32x^7 u get perimeter = 4(sqrt(32x^7)) and just simplify

[crimson sedge]

Okay lemme do this math

I got this

16x^3sqrt2x

[peak minnow]

seems good

[crimson sedge]

Alright thanks

.close

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[cedar kiln]

@cold rivet Has your question been resolved?

[cold rivet]

<@&286206848099549185>

[brazen sky]

basicly that is how you write a number in base 10

and n is the number of digits - 1

[cold rivet]

Can you give me a solution?

[brazen sky]

use logarithm

[cold rivet]

Tried, but not able to simplify

[brazen sky]

?

use this rule

[cold rivet]

Took log base 10 on both sides, simplified it, but then it comes out to be
an.an-1...a0*10^n(n+1)/2=2^2010

[brazen sky]

ok look

and here b = n+1

so basicly n+1 = [1+log_10(2^2010)]

,w 2010log_10(2)

[wraith dagger]

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[brazen sky]

so n = 605

[cold rivet]

Thanks

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[rugged heron]

i'm not sure where to go from here on how to get x

[sinful bay]

on the right you already found x though

didnt you

[rugged heron]

mb that was the wrong letter lol

that was how I got the length of AC

[sinful bay]

oh you want AE

[rugged heron]

yeah

[sinful bay]

just use similar triangles

[rugged heron]

are they similar though

iit's not an altitude

[sinful bay]

ED is literally perpendicular to AB

hint: use AA similarity

[rugged heron]

that means they're similar?

oooh

oh my god

oh I see it now

okay thanks I think I know how to do the rest

.close

[cedar kiln]

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[sick fossil]

how do I start?

for a I got all real numbers since any number can fit under the square root

0 can fit

wut

[crimson sedge]

yea it can

which one you want to know

[sick fossil]

how do I prove that the domain of g is all real numbers tho?

can I just say that the square root and square cancels out?

[crimson sedge]

for 2nd one, substitute x with -x, if its still the same func then its even func

[crimson sedge]

start with log domain

i.e x>0

[crimson sedge]

its actually natural number

[sick fossil]

natural log domain?

[cosmic steppe]

Prove that x + sqrt(x^2 + 1) > 0 for all x in R

Which ain't too bad tbh

since sqrt(|-x|^2 + 1) > |-x| for all x in R

So you know that x + sqrt(x^2 + 1) > 0

[sick fossil]

so do I solve for x here?

[cosmic steppe]

No need to

[cosmic steppe]

This is sufficient proof imo

Just showing that no matter what x is

It'll always be positive

[sick fossil]

I'm trying to understand it better sry

since we know ln(x) has a domain of x > 0

we have to figure out the "x" in ln(x) right?

but since anything inside will always be positive

the domain is (-infinity, infinity)?

@cosmic steppe

[white latch]

domain (0, infinity) for ln(x)

[sick fossil]

and find any domain in the middle?

[white latch]

?

[sick fossil]

find the domain for x in ln(x)?

[white latch]

x > 0

Do you understand why it can’t be negative?

[sick fossil]

cuz natural log restricts it by x>0?

[white latch]

here is why

[sick fossil]

won't it not matter anyway?

since anything I put under the square root

will always come out to be positive?

so the domain will be all real numbers?

[white latch]

Lets graph log_(-5)_(x) where -5 is our base

If we try to plot the function some points will be fine, but when we deal with fractions as the powers thats where the problem arises. Ex. log_-5_(x) = 1/2
x = (-5)^(1/2) We can’t have a negative in a square root. This would be true for a bunch of other values as well, so we just negate the negative all together.

[sick fossil]

there would never be a negative under the square root tho

since there is a square

[white latch]

exactly

so its impossible

is my point

[sick fossil]

so all real numbers

[white latch]

No.

[sick fossil]

no?

[white latch]

We don’t consider negative x values or 0

x > 0 any real number that is greater than 0

[sick fossil]

I'm solving for the domain tho

even if I put in 0

[white latch]

that is the domain bro.

what is the question you are dealing with?

[sick fossil]

it still puts out a positive value for ln(x)

2a

[white latch]

post it here

[white latch]

ln(0) will not ever compute.

[sick fossil]

but there can never be a ln(0)

[white latch]

so what we have to ensure is that we solve for the restrictions in the domain such that its > 0

[sick fossil]

since if I put 0 in

[white latch]

thats my point

there can never be a 0 inside a log

Why are you overcomplicating this idea?

[sick fossil]

x can be 0 wdym

ln(0) doesn't exist sure

but if I input 0

it spews out ln(sqrt(2))

if it's a negative x

[white latch]

In the instance of your homework problem sure

[sick fossil]

it's still a positive value

[white latch]

but in general

ln(x) x cannot equal 0

[sick fossil]

I'm only solving for my homework

[white latch]

okay you gotta specify that.

[sick fossil]

when did I ever say I was solving for ln(x)

bruh read first

[white latch]

nah

U were saying ln(x) leading me to all this confusion before. Perhaps be more specific and you’ll get the help u want.

okay so what have u tried

to find the domain of g? We have to find the restriction what was ur first step?

[sick fossil]

I was trying to get a mathematical way to prove that the domain was all real numbers instead of just saying it with words

[white latch]

Very good. We always want to do that.

[sick fossil]

is it just as simple as saying sqrt(x^2 +1) is bigger than 0?

for x in real numbers?

[white latch]

so first lets analyze the restriction in the square root. x^2 = -1. Seems to have no restrictions.

Now note sqrt(x^2) = | x^(2/2) |

[sick fossil]

add the square root to that?

that's just |x|

[white latch]

Just thinking of different ways as to how we can analytically solve for the domain cause its clear the general methods are not working.

[sick fossil]

I know it's very easy visually to solve it

just trying to use math to prove it

[white latch]

well if we say ln(x + sqrt(x^2 + 1)) and we plug in a negative value or positive value the end result would be

[sick fossil]

it would still be ln(x), x > 0

[white latch]

x + | x^(2/2) | and any value we plug in if its negative well this would just cancel to 0 and we would just have to then consider adding the 1

[cosmic steppe]

Sorry I was in class

There's another way

If you know your hyperbolic

Because that's just arcsinh(x)

Just show that sinh(x) is one to one

[sick fossil]

huh?

when was trig functions a part of this?

[cosmic steppe]

ln(x + sqrt(x^2 + 1)) = arcsinh(x) by definition

[white latch]

huh?

have not seen this before. He needs to know a way within what he is already learning.

Aeon, so I did verify that the domain exists for all real numbers as I had believed before.

[sick fossil]

bruh u were spewing about x > 0 before u read the question

[sick fossil]

I'll prob use this then

[white latch]

oh shutup u said ln(x), so the whole time i thought u were saying ln(x) when u shoulda correct me. If u were attentive would’ve been able to do that in no time. Sure I made a silly mistake my bad, but u shoulda corrected me. Mb bruh.

[white latch]

thats good.

[raw gulch]

that is all about the domain , in my opinion

[white latch]

sqrt(x^2 + 1) > -x

ooh thats interesting it just proves that sqrt(x^2 + 1) will always be greater than x which was my thought process as well but unsure how to prove. Noice I learnt something from this.

[cedar kiln]

@sick fossil Has your question been resolved?

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[plush wharf]

What did I do wrong

[small dust]

where did u get that 12 from

[plush wharf]

10+2

[small dust]

notice the 10 is seperate from the 2^x term

the power of x is on the 2

not on the (10+2)

[plush wharf]

Yeah I thought u move the x forward so it become 10+2?

[small dust]

wym move it forward?

to deal with the x in the exponent, u should first deal with the 10 and then get 2^x on it's own

[plush wharf]

So I was supposed to divide 50 by 10 first?

Then 2^x=5?

[small dust]

not dividing

subtraction

10 is being added to 2^x

not multiplied

you get 2^x = 40

[plush wharf]

Then x=ln40/ln2?