help-13

[crimson sedge]

3?

[gritty viper]

3*cos(3) ?

[crimson sedge]

3/4

[crimson sedge]

Y=3cos3/4x

?

[gritty viper]

I have no idea where you got that from but it's correct

[crimson sedge]

Just wanna make sure did I put it in right

[gritty viper]

looks right

parentheses probably wouldn't hurt

[cedar kiln]

@crimson sedge Has your question been resolved?

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[cedar kiln]

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[nova snow]

How should I start this?

[nova snow]

I was thinking of dividing 15.8 by 2 to get the length of adjacent

and then using 57 degrees and 29 minutes to get the length of opposite?

but im not sure how i would get to AC

[surreal cave]

divide the given side by two and you have three components of each triangle then you could work out each once to find the length that makes up AC

because one diagonal bisects another in a kite, that is a property of a kite, yes?

[nova snow]

yeah

[surreal cave]

so I'll call point E the intersection between the two diagonals, what is the length of BE?

[storm mural]

we got it

[nova snow]

7.9cm if you are talking about that

[livid hound]

but im not sure how i would get to AC
after you have the lengths of those two opposite sides, segment addition

[storm mural]

same for the triangle below that

you can find AE and EC

[nova snow]

so i know this side is 12.39cm

[surreal cave]

yea now do that for the triangle below it

then add the results

to get the desired AC

[nova snow]

like this?

[surreal cave]

yh

[nova snow]

then I add 24.1 + 12.39?

so i get 36.49cm

[surreal cave]

looks right

uh oh Ramonov is typing im about to get yelled at

[livid hound]

wrong angle

for the second one

[surreal cave]

oh yea

[livid hound]

where's 71.85 coming from

sci-calcs are also capable of accepting dms inputs

so no need to convert dms to just degrees

[tropic oxide]

you dont get it, if theres no unit conversion its not real math

[nova snow]

wut

oh

its 72.85 degrees

not 71.85

[nova snow]

well thats how the textbook wants me to do it and plus its kinda less confusing for me if i convert it first anyway

[livid hound]

can you show the textbook explicitly saying thjat

[nova snow]

is it really that important

[livid hound]

people here commonly assume they have to do something when they don't

converting takes more work and if you're rounding as well, lowers precision

and makes you more prone to mistakes as evident above

[nova snow]

i see

well in class im primarily taught to round up things

rarely ever do we not round up values

[livid hound]

ideally round at the end

[nova snow]

hm

alright

How would I do this?

Would I just half 60m to make one side of the triangle 30m

and then use tan inverse

[storm mural]

hypotenuse is 30
opposite is 18

[nova snow]

hypotenuse??

[storm mural]

of the triangle

[nova snow]

dont you mean adjacent?

[storm mural]

no I don't

30 is longer than the adjacent

[nova snow]

but how do you know 30 is the adjacent

i mean

the hypotenuse*

how do you know 30 is the hypotenuse when the 60m refers to the base of the triangle

[storm mural]

because it's a bar of a bridge that rotates

it doesn't change length

[nova snow]

oh i think i get it now

when the hypotenuse side fully lowers it meets at half of 60m

[storm mural]

[cedar kiln]

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[tired latch]

(conditional probability product rule)

im not sure how to get a) and b)

[tropic oxide]

were those incorrect answers your doing?

[tired latch]

yes lol

i wanted to get the new randomization after 3 tries

to see the answer, then work my way backwards

[tropic oxide]

weh

so... wait

do you have any work for these

bc its probably better to look at that and see what mistakes you made

than to start anew not knowing where exactly you had gone wrong

[tired latch]

i used mental math

for a)

[tropic oxide]

💀

but can you reproduce the calculation that you did for it

[tired latch]

i added probability of getting a red ball (4/10+3/10) to get 7/10 which is wrong..

[tropic oxide]

how did 11/151 happen?

thats what im most interested in.

[tired latch]

11/15 real answer

added a 1 to get the answer after 3 tries

[tropic oxide]

oh so this is just random shit.

[tired latch]

same with b

yes!

💀

[tropic oxide]

couldve been a bit clearer about that, bc i didnt get that from your messages.

"disregard the stuff in the answer boxes -- this is random shit i put in so the system would give me the answer"

[tired latch]

i see

[tropic oxide]

anyway ok so like

lets get to the actual problem now

[tired latch]

anyways the title of the hmwk is (Conditional Probability Product Rule)

[tropic oxide]

A box contains 4 red balls, 3 white balls and 3 black balls. 2 balls are drawn at random from the box, with replacement of the first before the second is drawn. What's the probability of getting a red ball on the first draw and a white ball on the second?

you want the first draw to be a red ball -- what is the probability of that happening?

[tired latch]

4/10

[tropic oxide]

correct

we put the first drawn ball back in the box before the second draw. does this mean the contents of the box are the same or different, and if they are different, how will they differ?

[tired latch]

different bc we know what the first drawn ball is

[tropic oxide]

you take a ball out and you put it back in, and your box now has different contents than it started with?

[tired latch]

same

[tropic oxide]

exactly, the contents are the same.

you want the second draw to be a white ball -- what is the probability of that happening?

[tired latch]

3/10

[tropic oxide]

(formally, i should be asking you for the probability of it happening given that a red ball was drawn first, but i hope you can see that the color of the first draw has no bearing on the probabilities here)

yes, correct

[tired latch]

i then added both values thinking it was the answer

[tropic oxide]

P(first draw red) = 4/10, and P(second draw white) = P(second draw white | first draw red) = 3/10

no, you were wrong to add them.

what if there were different numbers of white and red balls in the box and the numbers just happened to add up to more than 1? you'd see instantly that adding can't be right here.

[tired latch]

😦

[tropic oxide]

do you know the definition of conditional probability?

[tired latch]

one probability depends on another

? 🙂

[tropic oxide]

no

too vague

let me be more direct in my question:

do you know the meaning of the notation P(A|B)

[tired latch]

Conditional Probability is a measure of the probability of an event occurring, given that another event (by assumption, presumption, assertion or evidence) has already occurred.

[tropic oxide]

bruh

[tired latch]

🙂

[tropic oxide]

don't copy wikipedia.

if you want to be able to compute things with conditional probability, then you have to know the formula that defines it.

do you or do you not know the definition of P(A|B)?

[tired latch]

probability of a given b

[tropic oxide]

that's how you read the notation.

[tired latch]

still not sure how they got 3/25 though

division probably

[tropic oxide]

okay, so the gist of it is that you don't know, and won't admit it to me.

$P(A|B) = \frac{P(A \cap B)}{P(B)}$

[wraith dagger]

Ann

[tropic oxide]

this look familiar?

[tired latch]

sure

[tropic oxide]

wdym "sure"

you should've seen this in class, or in your lecture notes, or in your friends' lecture notes, etc.

i say this without exaggeration: if you dont know this you dont know how to do anything with conditional probability

[tired latch]

event a was the probability of getting a red ball on the first draw (4/10)

[tropic oxide]

why'd you say "sure" instead of "yes", i wonder?

[tired latch]

and the white ball was (3/10)

i feel like we plug in the values into the equation

[tropic oxide]

a corollary of this formula is that $P(A \cap B) = P(A|B) P(B)$. in our case, for $B$ we have the event \textit{the first draw is red}, and for $A$ we have the event \textit{the second draw is white}.

[wraith dagger]

Ann

[tired latch]

interesting

[tropic oxide]

do you now understand why you were supposed to multiply these individual-draw probabilities and not add them

[tired latch]

(4/10 x 3/10)?

[tropic oxide]

don't use the letter x as a multiplication symbol...

[tired latch]

oh wowzers

[tropic oxide]

but yes, that's what i meant.

[tired latch]

that did not occur to me

[tropic oxide]

what didn't?

[tired latch]

= 3/25

when multiplied

hmm how about part b)

where the first ball is not replaced

[tropic oxide]

we will still be calculating the required probability as P(second white | first red) * P(first red)

but P(second white | first red) will be different, because the contents of the box are now different

[tired latch]

ohh so the red ball is now taken out

leaving it to be 4/9 red probability

(4/9 * 3/10) = 2/15 there we go

would you care to do more hmwk problems like this ?

i have 5 more problems to finish this set

[tropic oxide]

i very much would not care to go through the same shit 5 more times.

[tired latch]

this one's political so it will be fun.

got it!

.close

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[nova snow]

I'm not quite sure where to start for this one

[muted bear]

which angle ratio should you use?

[nova snow]

well im not sure what to start with first

[muted bear]

start with a

[nova snow]

it would still work with angle ABC even if its not a right angle?

[muted bear]

what do you mean

[nova snow]

its not a right angle triangle

so ratios wouldnt work on it?

[muted bear]

use the right angle triangle you do have

[tropic oxide]

triangle ABC is not a right triangle and so trig ratios do not apply directly as they do for right ones.

recommend giving that unnamed right-angle vertex a name.

[nova snow]

x

[tropic oxide]

point names are typically uppercase letters.

[nova snow]

are we talking about the whole right angle triangle

[livid hound]

use the the triangle with the 39° that actually has the w you're trying to find (and that is a right triangle)

[nova snow]

or the smaller one

[muted bear]

ABX

[nova snow]

okay so using ABX i should start with finding side w?

[livid hound]

finding w is what part a) is asking for

[tropic oxide]

,calc 15.9 * tan(39 deg)

[wraith dagger]

Result:

12.875566127801

[tropic oxide]

checks out.

[nova snow]

okay

and now I have to find the missing interior angle

still unsure of how to go about that

[tropic oxide]

name the angle which you want to find.

[nova snow]

alpha

[tropic oxide]

there's no alpha in this diagram.

[nova snow]

you said name it

[tropic oxide]

i said name it as in "say its name" not "give it a name"

perhaps that is my fault for not being unambiguous

[nova snow]

oh well its not given a name

its just referred to as an angle

[tropic oxide]

do you know how to name angles using three points?

[nova snow]

i dont think so

are you taking about that angle symbol with three letters next to it

[tropic oxide]

you name a point on one of the angle's sides, then the vertex of the angle, then a point on its other side.

this angle here would be named angle BCX.

[nova snow]

oh right

[tropic oxide]

this angle is the angle you wish to find.

[nova snow]

yes

so what steps should i take to find it

[tropic oxide]

a good start would be to find one of its trig ratios.

which one it'll be depends on which side lengths in triangle CXB you have on hand.

[nova snow]

CBX or BCX?

[tropic oxide]

are you asking which of "CXB", "CBX" or "BCX" is the correct way to refer to the triangle of which i speak?

[nova snow]

yes

[tropic oxide]

they all are.

[nova snow]

i remember you saying that the middle letter refers to the vertex awhile ago

[tropic oxide]

that's for angles.

not for triangles.

[nova snow]

oh

[tropic oxide]

on paper you might use the symbols $\angle$ and $\triangle$ to disambiguate what kind of object you're talking about

[wraith dagger]

Ann

[nova snow]

so is there any difference between CBX and BCX in regards to triangles then

[tropic oxide]

not in this case.

(it can make a difference when you are stating two triangles are similar -- in that case, the ordering of the vertices in each one's name has to reflect which vertices correspond to which)

[nova snow]

I see

well to find a trig ratio of the triangle BCX wouldnt it require to find at least 2 of the sides

[tropic oxide]

indeed, and how many sides of it are known to you at this point?

[nova snow]

only side w

[tropic oxide]

that's right, you know side BX.

there's another side you can find, and for that you need no trig knowledge.

[nova snow]

hm i assume you're referring to pythagoras?

Although Im not sure how we can find the adjacent side from this

because halving 15.9 would not find side CX

[tropic oxide]

no

[tropic oxide]

why halve it?

isn't it clear that CX = AX - AC?

[nova snow]

ohhh yeahh

that did cross my mind

[inland ocean]

Why don't you use the sine formula for the width

[tropic oxide]

sine formula?

[inland ocean]

$\frac{sinA}{a} = \frac{sinB}{b}$

[wraith dagger]

ColdTee

[tropic oxide]

law of sines...?

excessive.

[inland ocean]

Yes yes

Law of sines

[tropic oxide]

overkill

[nova snow]

8.5 = CX and then BX = 12.88

now we can use tan inverse

[tropic oxide]

yes now we can write down tan(BCX) = 12.88/8.5 and so angle BCX = arctan(12.88/8.5)

[nova snow]

whats arctan

[tropic oxide]

a better name for tan^-1

[nova snow]

oh

so is there like arc cos and arc sin as well

angle CBX = 56.58 degrees

[tropic oxide]

arccos and arcsin exist yes

no space

,calc atan(12.88/8.5)*180/pi

[wraith dagger]

Result:

56.577667812722

[tropic oxide]

ok yeah

[nova snow]

ive heard of things like hypsin as well what is that?

[tropic oxide]

hyperbolic sine...?

typically called sinh or sh in my experience.

[nova snow]

thats a thing?

hyperbolic sin

[tropic oxide]

these have to do with exponential functions

and come up in hyperbolic geometry sometimes

[nova snow]

oh alright

.close

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[ornate grove]

can i have help

i di soomething

completly wrong idk what

oops

[cedar kiln]

@ornate grove Has your question been resolved?

[sly creek]

sure

what ques

[sly creek]

Please provide the curves that bound the region and the axis of rotation, and we will be happy to help you with the solution and sketch.

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@ornate grove Has your question been resolved?

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[obtuse frigate]

hi

[obtuse frigate]

basically i did a vectors question but got a different answer from the mark scheme

this is my answer for question 1

this is what the answer is suppose to be

where did i go wrong?

<@&286206848099549185>

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@obtuse frigate Has your question been resolved?

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[fathom carbon]

Hi, I'm having trouble answering this question "How many five digit numbers can be formed, containing exactly two non-zero digits (not necessarily distinct), and such that the sum of the digits is a square number?" My guess is 36? Like first since 1 times 1 is a square, I did 11000, 10100, 10010, 10001. If you do the same thing until 9 (99000, 90900 etc) you would get 36 ways to form the 5 digit number.

[fair geyser]

well you're not supposed to be multiplying 1 and 1

it asks for sum

[cedar kiln]

@fathom carbon Has your question been resolved?

[fathom carbon]

Oh right. But then how do you do it

[crimson sedge]

Well, do you think 11000 satisfies the conditions you mentioned?

[fathom carbon]

yes... I believe so. 1 plus 1 equals two, which is a square number

[livid hound]

2 isn't a square number

[fathom carbon]

oh my god sorry you7re right

22000?

[crimson sedge]

The statement above me makes Ramonov a god XD

yep

that would be a valid one

but not 11000

13000 will also work

See, the minimum sum will be 2

and maximum will be 18

how many square numbers are inbetween?

[fathom carbon]

3

so three possibilities?

[crimson sedge]

4, 9 and 16

yeah ig

how you have to find a way to find numbers

which sum up to those

[fathom carbon]

22000, 13000, 63000, 36000, (do i just list it all?)

[crimson sedge]

1,3
2,2

1, 8
2, 7
3, 6
4, 5

7, 9
8, 8

[fathom carbon]

okok that makes sense

[crimson sedge]

Well see, umm

we can rearrange this as well

[fathom carbon]

right

[crimson sedge]

1 -> 3, 8
2 -> 2, 7
3 -> 6
4 -> 5
7 -> 9
8 -> 8

[fathom carbon]

so 13, 31, 22, 18, 81, 27, 72, 36, 63, 45, 54, 79, 97, 88

[crimson sedge]

well

10300 will also be valid

10003 will be as well

[fathom carbon]

ugh

but i get it now

i'm just gonna have to do all that

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@fathom carbon Has your question been resolved?

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[cursive pumice]

.

How many cycles per second in 10 HZ

[cursive pumice]

How many cycles per seconds in 10Hz

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[meager jungle]

find the slope of ac

[cedar kiln]

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[vestal jewel]

like $\int\frac{1}{x},dx$?

[wraith dagger]

FamilyFriendly

[vestal jewel]

we define $\ln(x) = \int_{1}^{x}\frac{1}{t},dt$

[wraith dagger]

FamilyFriendly

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[solar flare]

multiplying decimals with whole numbers ie 0.3x 10

[crimson sedge]

0.3 be represented as a fraction

[solar flare]

how do i change it to a fraction?

[crimson sedge]

Rewrite it first as 0.3/1, and multiply the fraction by 10^n, where n is the number of digits after the decimal point

Here, we got only 1 digit, so 10¹, or just 10

[solar flare]

so instead of 0.3 it would be 0.3/1 x 10/1?

and multiply them as a fraction?

[crimson sedge]

Yes

0.3/1 × 10/10 I mean.

Multiply both numerator and denominator

It's because we don't want to change the number itself, 10/10 is basically 1, but we multiply like that for a simplified version of 0.3

It's incorrect to multiply by whatever we want without good reason.

[solar flare]

ahhh okay

ill keep practicing. thanks!

.close

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[olive heart]

how do you draw this

i have the plug drawn out but dont know where to go next

[crimson sedge]

theres an elec engineering server in #old-network

[olive heart]

but this is easy

and its physics

.close

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[jade compass]

i do not have to rearrange the set right because it is given and the answer would be x?

[granite knoll]

what do you mean "it is given"

[whole pond]

i think he means ||the set is already put in ascending order||

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@jade compass Has your question been resolved?

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[stoic sluice]

Can we say that if we strech an object in 2D by the same factor on both x and y axis, it becomes the same thing as applying a homothety by that factor?

[thick cipher]

Yes

Because matricially a homothety would be $A = \begin{pmatrix} \lambda &0 \ 0& \lambda \end{pmatrix}$

[wraith dagger]

black_couscous

[stoic sluice]

yes, and applying a strech is either one or the other lambdas right, so if you apply both, it becomes this exact matrix which you posted

[thick cipher]

Yup

[stoic sluice]

ok, thanks for confirming. Btw, what does "black couscous" mean by curiosity

Are you from North africa?

[thick cipher]

Couscous is a traditional meal from north africa

My family is from North Africa

[stoic sluice]

ah ok, do you speak french?

[thick cipher]

I am french

[stoic sluice]

ohh

ok, je parle aussi francais

J'étais juste curieux de savoir

[thick cipher]

Aucun souci

[stoic sluice]

Merci encore et bonne journée

[thick cipher]

Bonne journée

[stoic sluice]

man hwo do you close threads on here again

.close

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[cedar kiln]

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[vague rapids]

,rccw

[faint jackal]

I'm like dead I cannot remember how to do this and it's due tomorrow

[cedar kiln]

@faint jackal Has your question been resolved?

[faint jackal]

<@&286206848099549185>

[vague rapids]

all 3 parts of 1?

[faint jackal]

Yeah

[vague rapids]

total number of pets is

3+4

[faint jackal]

Mhm

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[vague rapids]

use contradiction

[bronze briar]

so I said that let a/b = rad2 + rad3. Then I multiplied both sides by the conjugate and got a(rad2 - rad3)/b = 1, and thus b = a(rad2- rad3)

[bronze briar]

that's what im trying to do but i feel like i hit a roadblock

[vague rapids]

let it be equal to a/b

[bronze briar]

that's what i said

[bronze briar]

over here

[vague rapids]

whole square both side

send all rational terms to one side(either lhs or rhs)

[bronze briar]

does it matter at all that (rad3 + rad2)^2 is irrational?

[vague rapids]

yes it does

$\left(\sqrt{2}+\sqrt{3}\right)^{2}=2+3+2\sqrt{6}=5+2\sqrt{6}$

[bronze briar]

I agot 5 + 2(rad3)(rad2)

[wraith dagger]

B-eard

[vague rapids]

yeah

[bronze briar]

then after some algebraic manipulation, i got (a^2 - b^2)/(2b^2) = (rad3)(rad2)

[vague rapids]

$5+2\sqrt{6}=\frac{a^{2}}{b^{2}}\ \ \ ,\ \ \ \ 2\sqrt{6}=\frac{a^{2}}{b^{2}}-5\ \ ,\ \ \ \ \sqrt{6}=\frac{\left(\frac{a^{2}}{b^{2}}-5\right)}{2}$

[wraith dagger]

B-eard

[vague rapids]

But irrational≠rational

so this is a contradiction

[bronze briar]

oh right

thank you

.close

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[drifting palm]

Hello, I’m doing restricted domain of quadratic functions, and I don’t understand how to get “for f(x) restricted to”.

[cedar kiln]

@drifting palm Has your question been resolved?

[cedar kiln]

@drifting palm Has your question been resolved?

[cedar kiln]

@drifting palm Has your question been resolved?

[cedar kiln]

@drifting palm Has your question been resolved?

[drifting palm]

.cloae

.close

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[fleet temple]

for 25 im confused with the limit of integraiton

[fleet temple]

its saying that its from 0 to 2

but wouldnt it be 1 to 2 ? bc its bounded at y = 0

[cedar kiln]

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[cedar kiln]

@fleet temple Has your question been resolved?

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[crimson sedge]

need help figuring out the expression of this graph, from then on I've got it

And tbh that's it

Also a small walkthrough on how you did it so I can do the next one alone

Will return after if I still can't do it lmfao

this is practice so don't worry about a time limit or anything

just trying to figure out this calculus before my second attempt at the test

<@&286206848099549185>

[cedar kiln]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[crimson sedge]

ok

[cedar kiln]

@crimson sedge Has your question been resolved?

[crimson sedge]

no

For fucks sake no

[slow thicket]

@crimson sedgefind where the slope is 0

thats either a min or max

[crimson sedge]

what

[slow thicket]

what

[crimson sedge]

don't even worry about it

Resolve or whatever

[slow thicket]

.solve

.closed

oml

.close

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[spare pier]

Its easy I think

I just cant see it

for some reason

Because if we do this pattern

it will be log 9

and thats not equal to log 24

[spare pier]

Also no solutions in the book

[buoyant latch]

uh

the sum means log 1 + log 2 + log 3 + log 4

[spare pier]

yeah

[buoyant latch]

what's log 1

[spare pier]

uh log 1

[buoyant latch]

log 1 = 0 for all bases of log

[spare pier]

I need a calculator

[buoyant latch]

no

[spare pier]

Oh

[buoyant latch]

what's log 1

[spare pier]

i did not know that

[buoyant latch]

well

[spare pier]

what does that mean

[buoyant latch]

it is the equivalent statement as "Anything to the power of 0 equals 1"

[silver frost]

yes but u just plus the log 1

[spare pier]

in exponents right?

[buoyant latch]

yeah

[spare pier]

ok

[buoyant latch]

knowing this

what is log 1

[spare pier]

log 1 = any number to the power 0

right?

[buoyant latch]

log 1 = 0

that's the answer im looking for

[spare pier]

oh yeah

my bad

thats what I meant

[buoyant latch]

ok what's log 1 + log 1

[spare pier]

0 + 0

[silver frost]

0

[buoyant latch]

ok

[spare pier]

Oh...

so its just 0 = 0

[buoyant latch]

so we can see log a + log a ≠ log 2a

[spare pier]

if u expand it

or am I making stuff up

[silver frost]

it's loga^2

[silver frost]

.

[spare pier]

0^2

[buoyant latch]

my question now is how did you get log 9

[spare pier]

= 0

log 1 + log 2 + log 3 + log 4

my bad log 10

[buoyant latch]

.

[spare pier]

my math isnt mathing correctly

[buoyant latch]

you CANNOT just add logs together

as we showed

with log 1 + log 1 = 0 + 0 = 0

[spare pier]

Ok yeah

[buoyant latch]

if log 1 + log 1 = log 2 then log 2 is not equal to 0

[silver frost]

yes

[tropic oxide]

[buoyant latch]

but we know log 2 is not 0

[spare pier]

wait what

[tropic oxide]

(a lot of them)

[spare pier]

isnt log 1 = 0

so log 1 + log 1 would = 0

[buoyant latch]

it is

[spare pier]

so log 2 = 0

WAIT

u cant add them

[buoyant latch]

but what you did here suggests that we can add the arguments together

[spare pier]

yeah

[silver frost]

..

[buoyant latch]

which if we could, log 2 would equal 0

[spare pier]

I think I errored

[buoyant latch]

which is not true

[silver frost]

log1 + log1 = log 1* 1

[spare pier]

OH YEAH

I remember this

so it would be

6 * 4

[silver frost]

the law of log

[spare pier]

WHICH is 24

bruh

[buoyant latch]

yep

[spare pier]

im not this stupid irl

[silver frost]

4! = 24

[spare pier]

I swear

[spare pier]

thats the cooler way

[buoyant latch]

indeed

[silver frost]

[spare pier]

Thanks so much

https://tenor.com/view/kiss-gif-26557594

thank u for going out of ur way to help people

I find it inspiring

I will do the same when Im in college

[silver frost]

you're welcome

[spare pier]

to return the favor

[silver frost]

haha

[spare pier]

Endless help

Mwhaahaha

.close

[cedar kiln]

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[silver frost]

nice person

[spare pier]

:)

u too

[cedar kiln]

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[fading cipher]

hi, could anybody possibly tell me why my inequality sign is the wrong way round?

[vague rapids]

Yeah you can not transport x from lhs to rhs thay way

Better bring 16 to lhs

And then take common denominator

[vague rapids]

Because if x were to be negative the sign would have reversed

[fading cipher]

sorry, i should have made it clearer; it's a word problem where x is guaranteed to be a positive value (the amt of cups of a liquid).

[vague rapids]

Oh

Yeah then you can do that

Your answer is wrong

Because you technically took reciprocal on both sides

You just took reciprocal and transported 100(10) to rhs

So the sign must be reversed

Since the value of x is positive*

[fading cipher]

one sec, i'm trying to figure this out haha

[vague rapids]

Off topic but are you aware of wavy curve method?

[fading cipher]

definitely not 😭

[vague rapids]

Ahh okay

It would make inequalities clearer

But anyways, you don't need it to solve this

See what is better ro do is subtract 16 on both sides

Then make common denominator

[fading cipher]

right

[vague rapids]

After that since x is positive, the denominator is alos positive and we can eliminiate it since rhs is zero

[fading cipher]

okay, that makes much more sense now

thank you so much for the help, i rlly appreciate it :)) i'll be sure to check out the wavy curve method as well!

[vague rapids]

Yeah good

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[nova snow]

This is what I have so far

Not sure what the next step is

[tropic oxide]

why is theta wrapped in individual parentheses

also you might want to switch to degree mode instead of radian mode

,calc 11.9*sin( (15+49/60) deg)/5.4

[wraith dagger]

Result:

0.60064177542357

[tropic oxide]

this is not even theta itself this is sin(θ)

also kind of strange you decided to close the channel after what seemed to be 5 minutes of me not replying

[cedar kiln]

@nova snow Has your question been resolved?

[nova snow]

it was more than 5 minutes and i thought you went offline

[nova snow]

my calculator is in degrees

its not in radians

[tropic oxide]

could have pinged me lmao

[tropic oxide]

yes i realized that wasn't the issue. the issue was that you passed off the value of sin(θ) as if it were θ itself.

[nova snow]

youre way too scary to ping

[nova snow]

so for this result

what do we do with this?

[tropic oxide]

that is sin(θ) and we want θ itself

i remember telling you that the obtuse value of θ is 180-arcsin(that)

you maybe want to have me confirm once again that this is what i want you to do

[nova snow]

sin^-1(0.6006..)?

[nova snow]

@tropic oxide 🤔

[tropic oxide]

180-arcsin(0.6006)

[cedar kiln]

@nova snow Has your question been resolved?

[cedar kiln]

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[tulip river]

Suppose the weight of individuals in a given population follows a normal law with mean $70 \mathrm{~kg}$ and variance $20 \mathrm{~kg}^2 $.
Taking a random individual from the population, calculate the probability that it weighs at least $75 \mathrm{~kg}$

[wraith dagger]

Lyuka

[tulip river]

I found the answer by using this formula: $\mathbb{P}{x \geqslant 75}$ = $\Phi\left(\frac{75-70}{\sqrt{20}}\right)$ = 0,1314

[wraith dagger]

Lyuka

[tulip river]

Yet my teacher, on the official exam used this one: $$
\begin{aligned}
\mathbb{R}{X \geqslant 75}= & 1-F(75)=1-\Phi\left(\frac{75-70}{\sqrt{20}}\right)=1-\Phi\left(\frac{5}{\sqrt{20}}\right) \
& =1-\Phi\left(\frac{\sqrt{5}}{2}\right)=1-\Phi(1,118) \simeq 1-0,8686=0,1314
\end{aligned}
$$

[wraith dagger]

Lyuka

[tulip river]

I don't understand which one is wrong and which one is correct

(even if the answer is the same)

[cedar kiln]

@tulip river Has your question been resolved?

[tulip river]

nvm i understood whats the problem

.close

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[glacial flax]

Hello

[glacial flax]

I’m confused about this and don’t know how to do it

[copper plinth]

Put n=1 means 1 st term , n=2 means 2nd term and so on

[glacial flax]

I don’t understand

[copper plinth]

You know about sequence?

[glacial flax]

I kinda understand d

I don’t rly get this

[copper plinth]

What do you understand about sequence?

[glacial flax]

Nothing

[copper plinth]

Do you know natural numbers?

[glacial flax]

Nah

We just started the subject

In school

But I don’t really understand

[copper plinth]

It's 1,2,3,4,5,... etc

Natural number are 1,2,3,4 and it goes on and on.

Now you understand about natural number?

[glacial flax]

Yeah

So it just keeps going

[copper plinth]

Basically natural number is counting number.

[copper plinth]

Yeah it never stop. But we can count by that

[glacial flax]

But how do I do the question

With the brackets

[copper plinth]

So sequence is nothing but instead of 1,2,3... We can assume whatever we want instead of that natural

But every natural number have to assign one element

It's like a function from natural numbers to something else

Like 1,1,1,... And goes on and on is also sequence

[glacial flax]

Oooo

Ok I understand what a sequence is

Like if it’s 2

2,4,6,8

[copper plinth]

It's constant sequence usually we denote it by $(1)^n$

[wraith dagger]

kiddie_mano

[copper plinth]

Yeah. Can you tell me how can we represent it?

[glacial flax]

Wdym

[copper plinth]

I represent constant sequence 1,1,1... Using (1)^n . Do you understand you?

[copper plinth]

Where n is natural number

Do you understand why $(1)^n$ represent the sequence 1,1,1,...

[wraith dagger]

kiddie_mano

[glacial flax]

Like

It means

1x1

Is 1

It keeps going

Squared

[copper plinth]

Yeah for n=1, $(1)^1=1$ and n=2,$(1)^2=1$ and so on

[wraith dagger]

kiddie_mano

[glacial flax]

O

I get it

[copper plinth]

So tell me how to represent the sequence (2,4,6,8,10...) in that form?

[glacial flax]

Ok

N=2 (2)2

It that correct

[copper plinth]

It's $2n$

[wraith dagger]

kiddie_mano

[copper plinth]

So when n =1 you get 2×1=2

n=2 you get 2×2=4

And so on

In your question...the sequence is 8(n+4)

[glacial flax]

I get it

[copper plinth]

Tell me what the sequence look like?

[glacial flax]

That question is complicated

[glacial flax]

But I get this

[copper plinth]

Why so? You just have to put the values for n=1,2,3,4...

Then you get the sequence, right?

[glacial flax]

Yeah

But how do you do the times

With the 8(n+4)

[copper plinth]

Its 8×(n+4)

Find n+4 then multiply by 8

[glacial flax]

O

So first term would be 1+4=5

And then

Times by 8

So 40

[copper plinth]

Yep

[glacial flax]

Let’s go

Thank you

[copper plinth]

No mention

[cedar kiln]

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[wild estuary]

Im quite confused regarding quadratic inequalities

[wild estuary]

wait lemme get the pics

1 of the questions is does it want the y value? or is it looking at the y

[crimson sedge]

hmm?

[wild estuary]

The variable is y since the question says so but I think it’s still x

It’s gonna be really hard conveying the idea of what I mean since I have no clue what I’m doing but is it looking at x or y?

Since there’s kind of no x in the middle.

[inland roost]

you have y=0 and y=4 it’s just the region between it

[wild estuary]

hmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm

alright fine

.close

[cedar kiln]

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[sly hemlock]

i don't think i'm going in the right direction

how do i set this up?

[tropic oxide]

line 1: $\int 4v(u^2+9)^2\dd{u}$?

[sly hemlock]

all v

[wraith dagger]

Ann

[tropic oxide]

4v(v^2+9)^2 is not equal to (4v^3+36v)^2.

[sly hemlock]

and then find the integral

[inland roost]

yeah that’s right

but have you learned u sub?

[sly hemlock]

no it's next section

but that would make it easier

this just doesn't seem to go in the direction of the other examples i've seen

seems off

[granite knoll]

where does the "45" come from in the first term?

[sly hemlock]

distributing and expanding from the first step

[tropic oxide]

how does 4v^5 integrate to 45v^6/6

[sly hemlock]

i miswrote it

let me correct that

2/3vˆ6

is that correct

never seen four terms in any previous examples

[granite knoll]

take the derivative of your answer to verify

[sly hemlock]

ok

true

forgot

not right

forgot to take the constant out in the first step

i'll take a look and come back to it

thanks guys

.close

[cedar kiln]

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[young urchin]

<@&286206848099549185>

Can anyone explain me question

[half mantle]

represent 5-y in terms of x.

[young urchin]

X=10-y

[half mantle]

so what about 5-y?

[young urchin]

5-y=0

Y=5

[half mantle]

try to think about.

[young urchin]

About what?

X=10-5=5

X=5,y=5

[half mantle]

5 - y = 5 - (10 - x) = 5 - 10 + x = x - 5.

[young urchin]

X+y=10
X=10-y
Y=10-x

Min{4-x,5-y}

4-x=4-(10-y)=-6+y

5-y=5-(10-x)=-5+x

[half mantle]

then min{4-x, 5-y} = min{4-x, x-5}.

[young urchin]

How to conclude that which one is min

[half mantle]

draw two lines.

pick the lower one.

[young urchin]

Equation of line?

@half mantle ?

[half mantle]

y = 4-x, y = x-5.

and the y-coordinate of intersection is what you want.

[copper plinth]

I think answer is none of these

Do you know the formula for min(x,y)?

[young urchin]

No

What is the formula

[copper plinth]

$\frac{x+y-|x-y|}{2}$

[wraith dagger]

kiddie_mano

[young urchin]

What value of x and y

[copper plinth]

it's for min{x,y}

like if min{2,3} then x=2, y=3 ..

i think this formula may lead to some conclusion

But i'm not sure.

[young urchin]

But in question it's variable

{4-x,5-y}

[copper plinth]

yeah. But try it. I get unusual conclusion. Maybe that will help

[young urchin]

Any other method

[copper plinth]

I'm thinking

[copper plinth]

sorry my mistake. That's why i told you the answer is none of these

But that formula will help you to find the answer

[young urchin]

How

[copper plinth]

put x= 4-x and y= 5-y in that formula

then try to solve

[copper plinth]

why didn't you understand this method?

that method is far better

[young urchin]

I was asking that how to conclude that which one is min

Min{4-x,5-y}

[copper plinth]

we don't know unless we put the value

the question is we have to find (x,y) which gives largest min{4-x,5-y}. x is variable so we can't able to tell which one is going to be minimum .

[young urchin]

That's what I'm confused

[copper plinth]

do you get it now?

[young urchin]

If we can't tell which one is min and also we can't say which one is max

It is asking largest min ?

[copper plinth]

It's not true

[copper plinth]

yep

you have to find every possible min value and find out which pair give the largest value

[young urchin]

Do you mean every value of x and y

[copper plinth]

If these is no 'none of these' in option then it is easy to answer the question

[copper plinth]

yep. But y is depend on x right? so just x .

It's absurd but you have to find all the min values for x, that is you have find for all real number, which is uncountable

But you can find out by that formula

[copper plinth]

put the value for y

[copper plinth]

y is depend on x right? so why don't you make it for x only

[young urchin]

Ok

Y=10-x
Right?

[copper plinth]

yep

[young urchin]

What can you say

[copper plinth]

you made careless mistake

[young urchin]

How?

[copper plinth]

this step is correct

[copper plinth]

this is wrong. check

[young urchin]

You mean in this i must substitute for y

[copper plinth]

you substitute for x?

[young urchin]

Ok

[copper plinth]

substitute y=10-x

[copper plinth]

where that 9 goes?

if you do correctly you get $\frac{-1-|9-2x|}{2}$

[wraith dagger]

kiddie_mano

[young urchin]

We get x again

[copper plinth]

Do you get this step?

[young urchin]

Wait trying

[copper plinth]

ok

[young urchin]

I get something else

[copper plinth]

min{4-x,5-y}=$\frac{4-x+5-y-|4-x-5+y|}{2}$

[wraith dagger]

kiddie_mano

[copper plinth]

you confuse the formula with answer

[copper plinth]

here only you put the value of y

[young urchin]

Oh that's incorrect way i did

[copper plinth]

min(x,y) is just formula like min(a,b)=$\frac{a+b-|a-b|}{2}$

[wraith dagger]

kiddie_mano

[copper plinth]

yeah. 1 st of all understand the content. If you can't ask me

[young urchin]

Once again I will try

[copper plinth]

at the end it not '-y' it's '+y'

[young urchin]

Oh yes

[copper plinth]

grow up

[young urchin]

Didn't get this format

[copper plinth]

now here put y=10-x

[young urchin]

Oh so now I have to substitute

[copper plinth]

yep

[copper plinth]

now can you able to see whatever value you give for 'x' you will get negative value right?

[young urchin]

Ye

[copper plinth]

you when 9-2x=0 you get the largest value

right?

[young urchin]

Ye

[copper plinth]

so find x?

[young urchin]

4.5

[copper plinth]

yep that means 9/2

then find y

[young urchin]

Y=5.5

[copper plinth]

yeah.. 11/2

[young urchin]

Option a correct

[copper plinth]

yep.. Seems that way

[young urchin]

Thank you

You help me a lot

[copper plinth]

no prblm. I also think you can find the using another method too

try to understand from this

But I'm going.

[young urchin]

Ok

[copper plinth]

is there anything else?

[young urchin]

No thanks 👍

You belong to this channel

[copper plinth]

what does that mean?

[young urchin]

How can I get you next time

[copper plinth]

I don't know. Even if I'm not available someone will help you

[young urchin]

Ok

[copper plinth]

.close

[young urchin]

.close

[cedar kiln]

Channel closed

Closed by @young urchin

Use .reopen if this was a mistake.

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[marsh pond]

any issues with my algebra here?

[opal basin]

and what is this needed for again?

[marsh pond]

just curious lol

[lyric narwhal]

it's correct

[marsh pond]

ty

[lyric narwhal]

but the method is pretty

well

inefficient

[marsh pond]

how would you do it?