help-10

[dapper robin]

im dumb

wait then

[finite turret]

Makes sense right ?

[royal basin]

your teacher told you "you MUST only solve THE LEFT SIDE, and ONLY THE LEFT SIDE, and NOT THE RIGHT SIDE"?

[finite turret]

Ill@try

[whole dock]

Do this

[dapper robin]

+yea yea orry bro

i think i got it

[finite turret]

Both works i mean have to solve one side only to get the answer of the other side

[royal basin]

@dapper robin

augh

hit enter too soon

[dapper robin]

i got cos xsec X

[royal basin]

meant to ask you not to call me bro

[dapper robin]

so

im getting

sec X + tan X

we can write cos x as

90 - xec

and then there is some minus law

i gorgot it

[dapper robin]

god

forgot*

[whole dock]

That is correct

[finite turret]

These r the ones to use

[finite turret]

After

[whole dock]

Now multiply (1 + sinx) to numerator and denominator

[dapper robin]

which grade are u bro

[finite turret]

10

[dapper robin]

lemme open mine 1 second

[finite turret]

Eerm them we have to chnage the left one too ryt ?

[whole dock]

Not necessary

1 + sinx divided by 1+ sinx is 1

So it doesnt make any difference

[finite turret]

Im sorry i dont get ypu

[whole dock]

$\frac{1 - sinx}{cosx}*\frac{1+sinx}{1+sinx}$

[warm shale]

ColdTee

[whole dock]

$\frac{1+sinx}{1+sinx} = 1$

[finite turret]

Ohh

[warm shale]

ColdTee

[finite turret]

Thank youu

[whole dock]

Np

[finite turret]

Well then i have a question

[whole dock]

Ask

[finite turret]

Can i get the upside down answer by multiplying it by the denominator

[dapper robin]

but didnt u want rhs

proved

[finite turret]

Yeah im just curious to know

[whole dock]

Im not sure what you mean to say

[whole dock]

This is what you had to prove right

[dapper robin]

damn that is confusing

[whole dock]

Yes, this is what you did right

[dapper robin]

bro write it down is lhs = rhs form

[finite turret]

Is this how u do it when u need the inverse answer

[whole dock]

Like you mean a general formula or sth

[finite turret]

Same concept applies for this too ryt ?

[whole dock]

Yeah

[finite turret]

Okay thanks have a good day

.close

[obtuse pebble]

Channel closed

Closed by @finite turret

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[fathom kindle]

Help

[fathom kindle]

what do i do now to solve for x

48 = -2x^2 + 20x

[teal turret]

!status

[obtuse pebble]

What step are you on?

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

[fathom kindle]

um

what do I do now to solve for x

48 = -2x^2 + 20x

[teal turret]

Uh Aight so there’s a couple different methods we could use

[fathom kindle]

mhm

[wooden cipher]

what have you tried

[fathom kindle]

um idk

i just dont know how to do it lol

[teal turret]

We could use quadratic formula, completing the square

[wooden cipher]

ok move everything to one side

thats a good starting step for quadratics

[fathom kindle]

ok

no wait

my equation started like this

P = -2x^2 + 20x - 42

P = 6

6 = -2x^2 + 20x - 42

48 = -2x^2 + 20x

[teal turret]

Ok yea so just move the 48 over for now

[fathom kindle]

bruh what

isnt there a simplier way

to continue moving everything

until i end with x

[teal turret]

Not sure what u mean

We could do completing the square or quadratic formula, but with both methods, we should move all terms to one side before proceeding

[fathom kindle]

um ok wait

ok so

0 = -2x^2 + 20x - 48

[teal turret]

Nice, now which one of the methods would u like to proceed with

[fathom kindle]

ok just factor it

[teal turret]

Factoring, sure

So first, we wanna get a coefficient of “1” on the x^2 term, can u do that?

[fathom kindle]

ya waiy

wait

um

ok

im confused

i got

-2(x-6)(x-4)

[teal turret]

Nono don’t factor yet

Just get our x^2 term to have a coefficient of 1

How can we do so

[fathom kindle]

divide

everything

by

-2

[teal turret]

Good, and what do we get now

[fathom kindle]

um

x^2 -10 +24

[warm shale]

Stephen

[fathom kindle]

yeah

[teal turret]

-10x

[fathom kindle]

ya ya

ya sorry ik

[teal turret]

Now factor

[fathom kindle]

ok

ya

u get like

(x-6)(x-4)?

[teal turret]

Yep

[fathom kindle]

ok

[teal turret]

=0

[fathom kindle]

so what does that mean

i need the vertex tho

eywuijk

[teal turret]

Bruh

[fathom kindle]

LOL

[teal turret]

Why didn’t u say that at first

[fathom kindle]

GIRL IDK

but its ok i need both anyways

um

so back to my original question

[fathom kindle]

!

[teal turret]

So now u want to find vertex from this?

[fathom kindle]

yes

look

omg

[teal turret]

Right, now we need to get it into vertex form

[fathom kindle]

wait

look

P = -2x^2 + 20x - 42

this was my original equation

(x,p)

p is 6

so i plug in p for 6

so whats x

[teal turret]

Well we alrdy solved that

[warm shale]

Stephen

[teal turret]

What’s x?

Also, could u share a pic/ss of the original problem that’s asking for the vertex

Tag me if u come back

[fathom kindle]

um

[fathom kindle]

but thats two x intercepst

i literally just need one x

at what number will be 6

im just gonna reask my questio because im confused ill make a new channel lol

.clos

.close

[obtuse pebble]

Channel closed

Closed by @fathom kindle

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[narrow dew]

while x=0 I need to examine the differentiability

and continuity as well

I don't if that term is the right one

[fickle turret]

What have you tried?

[narrow dew]

I do not know how to approach this type of question

[fickle turret]

Do you need to show continuity only at x= 0 or at the full domain?

[narrow dew]

I know the product rule but it says x is not 0

only at x=0

[fickle turret]

OK, then note that -1 =< sin x =< 1

[narrow dew]

noted

[fickle turret]

Do you know how to show continuity then?

[narrow dew]

nope

[fickle turret]

Calculate the lim as x goes to 0

[fickle turret]

Using this

And apply sandwich theorem

https://tenor.com/view/scooby-doo-sandwich-dogs-dog-lunch-gif-16290238

[narrow dew]

I do not understand something

in this part

x cannot be 0

doesn't that mean lim x->0 is null

[fickle turret]

No

You can go infinitly close to 0 with the upper function

So you need to know the limit of x^2*sin(1/x)

[narrow dew]

so that means in this point f(x) isn't continuiable

[fickle turret]

Why?

[narrow dew]

in the second rule "The function must be defined at this point"

[shell kelp]

I think you mean f' right

[narrow dew]

Isn't the graph will look something like this

don't worry about this ups and downs

[fickle turret]

Sorry, can we see the lower half of the piecewise function?

[narrow dew]

thats the whole

this is the full image

[fickle turret]

And the hole is filled in by f(0)= 0

[narrow dew]

OH

by this part

[fickle turret]

Yes

[narrow dew]

I understand

I was just focused on the sin one sorry

[shell kelp]

Is it also asking for turning point

[fickle turret]

Can you follow this approach now?

[narrow dew]

nope

[shell kelp]

Tureklene bisey daha soruyor

Ah tmm

[narrow dew]

can you show me how I can do it

[fickle turret]

-1 =< sin x =< 1

-x^2 =< x^2*sin x =< x^2

You can calculate the lim of -x^2 and x^2 with ez- both are 0

[narrow dew]

yes

[fickle turret]

So following this the original lim needs to be 0, too

[obtuse pebble]

@narrow dew Has your question been resolved?

Channel closed

Closed by @narrow dew

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[unique kettle]

WTS: $sup(A \cup B) = max(supA,supB)$

[warm shale]

clip

[unique kettle]

Kind of stuck here, I have that for any $a \in A$ it follows that $a \leq sup(A)$ and that for any $b \in B$ it follows that $b \leq sup(B)$

[warm shale]

clip

[unique kettle]

From this it follows that any $a \in A \cup B$ will have that $a \leq max(supA,supB)$ and the same thing for b, ($b \in A \cup B$ will have that $b \leq max(supA,supB)$)

[warm shale]

clip

[unique kettle]

Therefore $sup(A \cup B) \leq max(supA,supB)$

[warm shale]

clip

[unique kettle]

I dont know how to go about showing equality here though

Im thinking I should just show that the max is the least upper bound in this scenario and thus force them to be equal

So do something like For any $\epsilon > 0,\ max(supA,supB)-\epsilon \leq sup(A \cup B)$ and then prove the two cases, one being supA larger sup B and the other being vice versa?

[warm shale]

clip

[unique kettle]

This would make the max the supremum here and imply that sup(A union B) = max(supA,supB), ???

<@&286206848099549185>

[obtuse pebble]

@unique kettle Has your question been resolved?

[unique kettle]

<@&286206848099549185>

[obtuse pebble]

@unique kettle Has your question been resolved?

[dark stirrup]

I've encountered this before

Slow replies, but I'll look this over, @unique kettle (if you're still around)

[unique kettle]

Yes I am thanks

[uncut galleon]

Step 1 : show that sup(A U B) >= max(sup A, sup B) (easy)
Step 2 : show that some real number x is below sup(A U B) if and only if it is below sup A or sup B

From step 2 you can deduce that sup(A U B) <= max(sup A, sup B)

@unique kettle

[unique kettle]

I already showed that, would not just then showing that max(…) is the l.u.b. be enough

[uncut galleon]

Well, if you have sup(A U B) <= max(sup A, sup B) and sup(A U B) => max(sup A, sup B) then what's the problem ?

Also, I'm not english native, plz don't use acronyms like l.u.b. 🙂

oh ok lowest upper bound

[unique kettle]

Sorry least upper bound

Yep

[uncut galleon]

You proved that sup(A U B) <= max(sup A, sup B) right ?

[unique kettle]

Yes

[uncut galleon]

Ok. You know that sup A <= sup(A U B) right ?

[unique kettle]

Yeahsame with supB

[uncut galleon]

yup

therefore max(sup A, sup B) <= sup(A U B) right ?

[unique kettle]

Yeah I know that works I am just wondering if the way I did it works because I already have that written down and I don’t see why it wouldn’t

[uncut galleon]

Send what you write

Assuming it's in English or in a language I understand :x

[unique kettle]

.

I’m not at home right now but it’s along the lines of what that says

[uncut galleon]

hmm

you only have max(sup A, sup B) <= sup(A U B) with that

not max(sup A, sup B) >= sup(A U B)

May I write it my way ?

[unique kettle]

But that’s with the implication that max is an upper bound and doing the makes it the least upper bound which implies that they are equal right?

[unique kettle]

Sure

[uncut galleon]

Ok, here I go

1. You know that $sup A \le sup A\cup B$ and $sup B\le sup A\cup B$. From that, you know directly that $max(sup A, sup B)\le sup A\cup B$\\
2. Let $x\in \bR$. $x< sup A\cup B \Rightarrow \exists a\in A\cup B$ such that $a>x \Rightarrow$ we have either $x\le sup A$ or $x\le sup B$. Therefore, $max(sup A, sup B)$ is an upper bound of all these $x$'s. Therefore, it is higher than their lowest upper bound. In other words :
   $$max(sup A, sup B)\ge sup A\cup B$$
   Finally, by antisymmetry, we know that $max(sup A, sup B) = sup A\cup B$

[warm shale]

Silfer

[unique kettle]

Ok I understand, would you mind answering a more generalized question that i have

[uncut galleon]

ok go ahead

[unique kettle]

Ok so if you know some supT is an upper bound for supS then supT>=supS

[uncut galleon]

You mean supT is an upper bound for S I guess ? But yes

[unique kettle]

So if you were to show that supT is the least upper bound for supS is it still supT>=supS or is it now supT=supS

[unique kettle]

Yes that

[uncut galleon]

Well, you would need to prove that supT = supS because supS is S's lowest upper bound by definition

[unique kettle]

Ok thanks for the help

[uncut galleon]

np

[unique kettle]

.close

[obtuse pebble]

Channel closed

Closed by @unique kettle

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[next lantern]

how would i calculate this

i thought i would do this but that was wrong

.close

[obtuse pebble]

Channel closed

Closed by @next lantern

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[wintry shoal]

Here is my work and the original question along with the needed resources, I cant seem to figure out where Im going wrong but I know its the wrong answer (sorry for bad handwriting)

[obtuse pebble]

@wintry shoal Has your question been resolved?

[wintry shoal]

<@&286206848099549185>

[wintry shoal]

<@&286206848099549185>

.close

[obtuse pebble]

Channel closed

Closed by @wintry shoal

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[obtuse pebble]

@worthy wagon Has your question been resolved?

[worthy wagon]

<@&286206848099549185>

[dusk ruin]

!status

[obtuse pebble]

What step are you on?

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

[dusk ruin]

@worthy wagon

[dusk ruin]

oh ok lol

do you know what the complement of an interval is?

[worthy wagon]

All the real numbers of an interval?

[dusk ruin]

isn't that the interval itself?

[worthy wagon]

No. Thats incorrect. I just tried it

[dusk ruin]

to my understanding, I thought the complement of an interval B over R would be the set of all other numbers in R not in the interval B

so the complement of (2,∞) would be (-∞,2] to my understanding, no?

[high lily]

No. Thats incorrect. I just tried it
what exactly did you try

[worthy wagon]

All i tried so far is what he told me. I tried to use the interval itself but it didint work

[dusk ruin]

oh whoops you misunderstood me, I apologise

[high lily]

ofc that won't work

[dusk ruin]

here, I was attempting to clarify that you thought the complement of the interval is "All the real numbers of an interval?", which I was asking you to confirm you thought the complement of an interval was itself

apologies for any confusion

[worthy wagon]

thats alright

[dusk ruin]

well, regardless

it appears your problem is that you do not know what the complement of an interval is.

correct?

[worthy wagon]

Yes

[dusk ruin]

excellent

so, the complement of a closed interval over a space, is, effectively, everything outside that interval

so let's say you have the interval (1,2] on R.

your interval is all numbers greater than 1 up to and including 2

this means your complement is all numbers less than 1, including 1, and all numbers greater than 2, not including 2

there is no overlap between intervals, but together they span all of R, all real numbers

@high lily is this correct?

[high lily]

seems alright

[dusk ruin]

hooray! that was a wild guess :P (based off what other complements I know are)

anyways, @worthy wagon do you think, with this explanation of what a complement is, you can complete your problem?

[worthy wagon]

I think soooooo

[dusk ruin]

I gtg shower, brb

[icy spire]

hi

i am new

[dark stirrup]

Introductions go in #discussion

But welcome

[worthy wagon]

I need help with another one

[worthy wagon]

<@&286206848099549185>

[dusk ruin]

are you not able to solve it the same way as the last one? @worthy wagon

[worthy wagon]

I think its confusing me because theres no infinity

[dusk ruin]

you're right

do you know about unions of intervals?

[worthy wagon]

sounds familer

[dusk ruin]

so for instance

(1,2]U(3,8]

is the set of numbers greater than but not equal to 1, but less than or equal to 2, as well. as the numbers grater than but not equal to 3, but less than or equal to 8

it's just a way of merging two intervals

does this help solve this problem? @worthy wagon

[worthy wagon]

im still having trouble

[dusk ruin]

with which part, specifically?

[worthy wagon]

How i'm able to transform that interval

[worthy wagon]

I still cant get this

[dusk ruin]

hmmm ok so

in this question, they want the complement of (1,6] right?

[worthy wagon]

yes

[dusk ruin]

so your interval, initially, is all the numbers greater than but not equal to 1, up to and including 6

so your complement is all other real numbers.

that means, your complement holds all numbers 1 and less, and all numbers greater than but not equal to 6.
together, they combine without overlap to cover all real numbers.

with me so far?

[worthy wagon]

So its basically the opposite?

[dusk ruin]

I guess, as much as there is such a thing as an opposite interval

you can't really have an interval full of anti-numbers that combined with this cancel each other out

the complement is close enough, as it is everything but the interval

well, anyways

that means, your complement holds all numbers 1 and less, and all numbers greater than but not equal to 6.

how would you turn these words into an interval of your own?

[worthy wagon]

1 >x>6?

[dusk ruin]

well, ok, let's break it down farther (since that is very wrong, though surprisingly close)

let's do half of this interval at a time

all numbers 1 and less

how would you turn this into an interval?

[worthy wagon]

1<x?

[dusk ruin]

including 1 (1 and less)

[worthy wagon]

I don't know how to do that on my keyboard

[dusk ruin]

≤ ≥

for me it's hold the option key

no worries, you can use <= or >= if you want instead

or, even better

use interval notation

that's the next step anyways

[dusk ruin]

what is this in interval notation?

[worthy wagon]

1≤x?

[dusk ruin]

that's x is all numbers greater than or equal to 1

you want all numbers 1 and less

use the form they want on the computer, ie (2,6] or [-3,∞), it'll make things easier later

[worthy wagon]

[-1,0)

[dusk ruin]

that would be all numbers greater than or equal to -1, but less than 0 (not including 0)

[worthy wagon]

(-1,0)

[dusk ruin]

do you understand how this notation works?

[worthy wagon]

This is confusing me

[dusk ruin]

I'm very sorry :/

well ok let's start from the top

https://media.discordapp.net/attachments/903430332324405288/1066557866632290304/image.png?width=1080&height=304

you have the interval (1,6]

you need its complement.

tell me, what is interval such that all numbers before (1,6] are contained in it?

the answer is (-∞,1]

(-∞,1] contains all numbers less than or equal to 1

[worthy wagon]

Oh, because the infinty contains all the previous numbers

[dusk ruin]

after that, the numbers belong to (1,6], or are bigger than those in (1,6]

[dusk ruin]

yes, an infinity encompasses everything when set as the edge (x<∞ really means x is anything)

well anyways, so we know

[dusk ruin]

well, that's just half the answer

(-∞,1] is all the numbers BEFORE the interval (1,6]

what about after?

[worthy wagon]

hmm

[dusk ruin]

the interval (1,6] stops at the number 6

6 is in that interval.

which means all numbers AFTER the interval (1,6] is all numbers greater but not equal to 6

do you think you know what those numbers are?

[worthy wagon]

[6,infinity)?

[dusk ruin]

almost

[ is inclusive

we don't want 6 in our range, since it's in the original interval. we don't want any overlap in our complement

[worthy wagon]

(6,infinity)

[dusk ruin]

exactly!

so we have all numbers before: (-∞,1], and all numbers after: (6,∞)

now we just have to combine the two

[worthy wagon]

(-∞,1] U (6,∞)?

[dusk ruin]

yup

make sure to copy paste the union symbol the website uses

don't use a capital U, I doubt it'll work

[worthy wagon]

It works.
Now I just need help with like one or two more

[dusk ruin]

P: sure thing

let me know what part you need help with

[worthy wagon]

would I solve for T first?

[dusk ruin]

yes

you would find the values of t where the ball hits the 6 foot height threshold

[worthy wagon]

This is what I got for T

@dusk ruin

[spice chasm]

Um

all you need to do is set it to 28t-16t² ≥ 6

[dusk ruin]

don't just give people answers, they need to understand how to get them

[dusk ruin]

so in this case you want when H=6

[spice chasm]

Mb

[dusk ruin]

it's ok, it's not an awful thing to do, just not very helpful for learning

[worthy wagon]

I got it right

[dusk ruin]

that's good!

did you use Jash's answer, or work it out on your own??

[spice chasm]

wait did you complete the square to get this

[worthy wagon]

Yes

One more

Ive had this problem before when it was a point and a slope, and between two points, but never between three points

[spice chasm]

Find the line through (1,4) and (2,5)

that way you can find the slope

[spice chasm]

and use the last point (1,6) to get the point and slope you’re familiar with

[worthy wagon]

So the slope I got for this is 1

<@&286206848099549185>

[wise talon]

!15min

[obtuse pebble]

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.

[wise talon]

but to answer your question

think about the derivative

[bronze pumice]

What’s a constant function?

[worthy wagon]

a constant function is a function whose value is the same for every input value, right?

[bronze pumice]

Alright, now, with that information, what shall be the rate of change on every interval?

[worthy wagon]

I mean, theres alot of numbers

thats what confuses me

oh wait

0

[bronze pumice]

Why? Consider a constant function, $y=4$

[warm shale]

Heisenbug

[bronze pumice]

What’s the rate of change? (Plot if you don’t understand)

[bronze pumice]

Correct.

[bronze pumice]

$ROC(x) = \frac{f(b) - f(a)}{b-a}$

Are you familiar with this formula?

[worthy wagon]

not really

[warm shale]

Heisenbug

[bronze pumice]

This is the formula for rate of change, if given in interval, i.e (a, b), where you know a,b. Do you know how to continue from here?

[worthy wagon]

What confuses me is that im given 2 x's and not a full point

[bronze pumice]

What confuses you? Consider b=8, a=3.

[spice chasm]

you can plug in x into f(x) to get a point

[timid silo]

0

[worthy wagon]

I got that

[timid silo]

10

[timid silo]

6

[worthy wagon]

This was has different options so i'll post multiple screen shots of it

Actually nvm, I got this one

Last one. This one is confusing because I have never been given points like this

<@&286206848099549185>

[worthy wagon]

<@&286206848099549185>

[worthy wagon]

<@&286206848099549185>

[opal ore]

ok hold on there bucko that's enough pinging

[opal ore]

number of flue cases

[opal ore]

day

[worthy wagon]

No, I already got that problem.

I'm talking about the last one I posted

[opal ore]

recall the distance formula: $d=\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

[warm shale]

messyinterval

[worthy wagon]

The points that I have to use arent clear to me

[opal ore]

$x_1 = x+ 4y , x_2 = x-4y$

[warm shale]

messyinterval

[opal ore]

$y_1=4y, y_2 = 4y $

[worthy wagon]

Oooh, I get it

.close

[obtuse pebble]

Channel closed

Closed by @worthy wagon

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[fallow zinc]

last page

[fallow zinc]

I need helo

help*
plz

plz

[zenith raft]

people usually prefer screenshots or text over pdfs 🙂

[fallow zinc]

Okay i will sc

[fallow zinc]

So

[obtuse pebble]

@fallow zinc Has your question been resolved?

[fallow zinc]

.close

[obtuse pebble]

Channel closed

Closed by @fallow zinc

Use .reopen if this was a mistake.

[fallow zinc]

.reopen

[obtuse pebble]

✅

[grizzled shore]

Have you tried anything

[dense imp]

!status

[obtuse pebble]

What step are you on?

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

[fallow zinc]

1

[trim portal]

Do you know what AA similarity test means? @fallow zinc

If 2 triangles have at least 2 angles the same, then they're similar. If you find pair of triangles with 2 angles in common, you can conclude that they're similar.

Another similarity is SSS. Side-side-side. If sides of 1 triangle are 3, 4 and 5 and sides of another triangle are 6, 8 and 10, you can conclude that these 2 triangles are similar, because 3/6=4/8=5/10. For this similarity test, I recommend you to order the side lengths in ascending order so you can compare them.

This is how you do SSS test

[fallow zinc]

OH OKAY

THANK YOU SO MUCH

FOR UR HELP

I REALLY APPECIATE IT

WE CAN ALSO USE ANGLE SYMBOL

ANYWAY

THANK YOU SO MUCH FOR YOUR HELP

[obtuse pebble]

@fallow zinc Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[urban ether]

(please pin)

[flint tartan]

bro.

discord.gg/physics

[urban ether]

ok thanks

.close

[obtuse pebble]

Channel closed

Closed by @urban ether

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[hard hill]

hello, I am having troubles soulving a trignometric equation, I have arrived to this point, is this true?

[trim portal]

Hello, is the 30 and 60 in degrees or in radians?

[hard hill]

in degrees

[elfin belfry]

What is the point of the square?

[trim portal]

And do you know the sin(a+b)=cos(a)sin(b)+sin(a)cos(b) formula?

[hard hill]

yes

[trim portal]

Oh you replaced cosines with sines, right?

[hard hill]

it is because I have used the sin^2(x) + cos^2(x) = 1 and I substituted the cos(x) = sqrt(1-sin^2(x))

[trim portal]

you can use the fact that cos(30)=1/2 and sin(30)=sqrt(3)/2 to directly substitute those values

[hard hill]

okey, thanks, do you see any other error?

[elfin belfry]

But why?

[trim portal]

it's not an error, you are just making it harder than it is. I would just keep the cosines there, and replace the sin(30) and cos(30)

[elfin belfry]

If you really want to do it that way, you need square roots

Here it won't cause any issues, but when the angles go higher than 90 degrees, it will cause issues with signs

[hard hill]

okey, so much thanks to both 😀

.close

[obtuse pebble]

Channel closed

Closed by @hard hill

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[onyx monolith]

How to solve this

[trim portal]

Small hint: SA+SB+SC is some constant, you will have to find out why

[obtuse pebble]

@onyx monolith Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[rapid quartz]

https://i.imgur.com/uArll0i.png

[kind hawk]

it seems like it's using the logic version of de-morgans law to prove the set version

not sure how much you want to go into logic but proving de-morgans law from nothing is probably quite a bit harder

[rapid quartz]

Seem to me that proofs should be built from the ground up. All resources (that I can find) show it the same way. Got any other suggestions or resources to check out? Or it is that time of "accept it and move on".

[kind hawk]

while in principle I agree that proofs should be built from the ground up, not in this particular case

just take the basic laws of logic and accept them

proving them from the ground up is hard

logic in general is very hard

[wintry swift]

what is the definition $x \in A \cap B$?

[warm shale]

ThM

[kind hawk]

x in A and x in B

[rapid quartz]

Yeah.

Thanks for the help Denascite. ❤️

[kind hawk]

youre welcome

[rapid quartz]

.close

[obtuse pebble]

Channel closed

Closed by @rapid quartz

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[acoustic pawn]

Is a linear map from R^2 to R^2 given by (x,y) --> (x,0) a linear transformation or just a map?

[acoustic pawn]

It maps to R^2 so its a transformation but it feels weird how the map maps all of R^2 onto R^1 so the space isnt rly transformed but changed

[zenith raft]

have you tried verifying the properties in the def of linear transformation?

[acoustic pawn]

Well according to the definition its a transformation since it maps to itself

[acoustic pawn]

Still confused and looking for some intuition on why that definition makes sense given this example

[zenith raft]

i said linear transformation tho

wait but you don't even know if it's a linear transformation yet, do you?

[acoustic pawn]

Yeah its just given to be a map

But the domain and codomain are the same

[zenith raft]

sooooo this haha

[acoustic pawn]

The properties being that its a linear map which maps onto itself or?

Bcs thats what i learned a linear transformation is

[zenith raft]

wait what, linear map and linear transformation are not the same thing for you?

[acoustic pawn]

No

[zenith raft]

[acoustic pawn]

A linear map is a map between vector spaces satisfying certain properties (linearity)

A linear transformation is a linear map from a vector space to itself

Like how a linear form is a linear map from a vector space to its field of scalars

[kind hawk]

map, function, transformation etc mean essentially the same thing

dw about it too much

[zenith raft]

apparently not right now

[acoustic pawn]

Oh we learned that a transformation is on a vector space while a map is between vector spaces

[acoustic pawn]

So a map between two different vector spaces would count as a transformation?

[kind hawk]

I guess apparently not in your course

[zenith raft]

pretty atypical

[acoustic pawn]

Oh but i ment according to you guys?

[kind hawk]

yes

[acoustic pawn]

Ohh okey

Thats crazy

[kind hawk]

I don't make any distinction between them at all

[zenith raft]

same loll

[acoustic pawn]

It sounds usefull tho since you can speak of eigenvectors of a linear transformation but not of a linear map

And linear transformations always have square matrices

As far as i know

[zenith raft]

linear transformations aren't just matrices when you work with vector spaces in general

[acoustic pawn]

I ment the corresponding matrix representation is always square

[zenith raft]

well yes me too

but

linear transformations (or maps whatever you call them lmaoo) don't necessarily have a "matrix representation" with other vector spaces

that aren't like, real number vector spaces

[kind hawk]

well they do if you pick a basis

that's like half the point of linear algebra. that every linear map on finite dim spaces is essentially a matrix

[acoustic pawn]

I thought all scalar fields were representable with matrices

Meaning a matrix can consist of scalars from any scalar field so a transform/map between spaces can always be represented by a matrix

[acoustic pawn]

Girl

[timid silo]

Can someone help me with this

[acoustic pawn]

#❓how-to-get-help

Anyway ill just interpret it as a linear map since thats how my.bo9k defined it and ig it doesnr rly matter to much if its a "good" definition if its atypical

[zenith raft]

loll

[kind hawk]

I mean it's still from R^2 to R^2

[obtuse pebble]

@acoustic pawn Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[timid silo]

how would i work this out? also, any pointers in being able to solve this quickly, like test conditions?

[timid silo]

also if i have another question do i post it here or?

are you familiar with the area ratio of similar shapes?

knowing the side ratio?

[sage geode]

Recall that the ratio of similar figures = the square of the ratio of side lengths

[timid silo]

no

[timid silo]

welp it's in the message above

[trim portal]

By AA similarity, the triangles are similar. Therefore they've same ratio's of heights and sides. Area of triangle is base*height. In the second triangle, base is 1.2*base in first triangle, and height must be 1.2 times as large, as height in first triangle

[timid silo]

yep

is it 30?

[trim portal]

Nope

Both base, and height are 1.2 times as large.

[timid silo]

36?

[trim portal]

Correct

[timid silo]

thanks

[trim portal]

Oh and the factor was derived by 12/10

[timid silo]

so i have another question, but do i post it here?

im new to the server

[trim portal]

You probably can post it here

[timid silo]

no

im turning 14 this year is it bad that i dont know that?

[trim portal]

So let's mark Item 1 as A, item 2 as B, item 3 as C, and item 4 as D.

[timid silo]

yep

[trim portal]

It's actually really easy

[timid silo]

easy as in “wow how tf do u not know that” or “it’s fine you can learn the concept quickly”

[astral dirge]

Can someone help me out on this question

[trim portal]

We can do this 4 equations, right?

[timid silo]

#❓how-to-get-help

[timid silo]

yep\

it's alr if you are 14 years old and struggling, everyone learns at different speeds.

a is 77

[trim portal]

Correct

[timid silo]

tbh i remember when i was 14 years old, a lot of my classmates struggled with math even easier than this lol

and they are completely fine

[trim portal]

What else can you do?

[timid silo]

137-88?

bc a is 77 and b is 11?

[trim portal]

Yes! What will 137-88 be equal to?

[timid silo]

49

[trim portal]

Okay, and what combination of letters is 49?

[timid silo]

c and d?

[trim portal]

Yep, so we have this. What can you do now?

[timid silo]

uhhh

hold on

divide it by 7?

[trim portal]

Correct. By replacing C in first equation with 6*D, you get new equation

[timid silo]

is c 42 or am i dumb

[trim portal]

And by dividing it by 7 you get D

[timid silo]

yep

[trim portal]

So you had 4 equation at start, and by replacing the letters with numbers or different letters you got the solution. And that's how you solve system of equations. I told you it's easy :)

[timid silo]

tysm

[trim portal]

You are welcome :)

[timid silo]

how do i close this

💀

[trim portal]

.close

[timid silo]

.close

[obtuse pebble]

Channel closed

Closed by @acoustic topaz

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[merry vault]

Hello

[merry vault]

To calculate sign(sigma), do you use a particular formula?

Or do you just go and figure out all the "errors" (?) and then do sign(sigma) = (-1)^m where m is the number of "errors"?

An "error" is defined as an index-pair (i,j) with 1 <= i < j <= n and sigma(i) > sigma(j)

[elfin burrow]

yes

you can count the number of transpositions

alternatively, you can decompose the permutation into disjoint cycles and count the number of even length cycles

[merry vault]

"error"=transposition ok

that seems to be the easiest way

ty

.close

[obtuse pebble]

Channel closed

Closed by @merry vault

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[green magnet]

Im dum with math pls help (இдஇ; ). Circular measure

[green magnet]

I cant seem to know how to find the angle

[trim portal]

What's your progress, have you determined any lengths or something?

[green magnet]

I only know tht half of the radius is 7🥲😞

[trim portal]

If half the radius is 7, what wil |OP| be?

[green magnet]

Tried to do sin cos tan but tht wont find the angle

[green magnet]

7

[trim portal]

Great, now you can find |PR| using pythagoran's theorem

[green magnet]

Whats PR for ಥ_ಥ

Yes

[trim portal]

Sorry, you don't need it. I misread the question

[green magnet]

Okayಥ_ಥ

[trim portal]

you know that cos(QOR)=7/14, right?

Try to think about what I just wrote, does it make sense to you?

7 is the adjacent side, 14 is hypotenuse. Cos is adjacent/hypotenuse

[green magnet]

Oooooohhh

Then use Cos to find the anglee

[trim portal]

You need to use inverse cosine. Or acos

it's the same

You know the cos(QOR)=1/2, so the angle is InverseCos(1/2)

gtg, if you will need more help ping Helpers

sorry, bye

[green magnet]

Issok i got it now!!

Thanks a lottt

.close

[obtuse pebble]

Channel closed

Closed by @green magnet

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[neat pewter]

I need help with these 3

[royal basin]

@neat pewter are those answers yours?

[neat pewter]

Yes

[royal basin]

do you have work to show how you got them?

maybe you were on the right track but screwed up somewhere

[neat pewter]

Yes

For the first one

I multiplied 13 12 and bunch of numbers to get

154440

For the second one I meant

[royal basin]

"13 12 and a bunch of numbers" is not very precise.

[neat pewter]

Let me look at my notes

13 12 11 10 9

[royal basin]

okay, so you calculated 13 * 12 * 11 * 10 * 9

for what purpose?

what does this product represent?

[neat pewter]

The probability

I’m really bad at matj so please bare with me

[royal basin]

13 * 12 * 11 * 10 * 9 cannot be the probability of anything

it is not a number between 0 and 1

[neat pewter]

So what steps do I do

First

[royal basin]

and even if it was, i'd then ask you why you are taking its reciprocal...

anyway

for this problem, i'd consider the following:

it does not actually matter how many cd's you have in your library.

it only matters that 5 cd's end up on the rack.

[neat pewter]

Yes

[royal basin]

and among all possible permutations of the 5 cd's on the rack, in how many cases are they in alphabetical order?

[neat pewter]

1

[royal basin]

that's right

and how many ways are there to place 5 objects in a row?

5 distinct objects i must add

[neat pewter]

I’m not sure

25?

[royal basin]

how and why did you get 25?

[neat pewter]

5 times 5

[royal basin]

ok, why 5 * 5?

[neat pewter]

I guessed I’m not sure

[royal basin]

well, your guess was way off the mark.

have you heard of such a thing as factorials?

[neat pewter]

I think so

[royal basin]

when calculating how many ways exist to put 5 cd's on a rack,

there are 5 options for what CD will go in the first slot,
4 options for what CD will go in the second slot (do you understand why it isn't 5 again?),
3 options for what will go in the third slot,
2 options for the fourth and finally 1 option for the fifth.

for a total of 5 * 4 * 3 * 2 * 1 options.

you have a lot of combinatorics review to do, that's for sure :P

[neat pewter]

Yes

So 120?

[royal basin]

yes, there are 120 ways to place 5 CD's on a rack.

[neat pewter]

So 1/120

[royal basin]

so the answer to problem #2 is 1/120, yes.

in general you should never just blurt out numbers.

[neat pewter]

Ok

Can u help with 1 and 3

Please

[royal basin]

5 cards are drawn randomly from a standard deck of 52 cards. Determine the probability that exactly 3 of these cards are aces.

okay, so...

how did you get the answer that you got

[neat pewter]

I guessed

I was really confused with this one

[royal basin]

so your 0.0004 was nothing but a guess...?

not even any calculations going into it?

[neat pewter]

I think I searched it

[royal basin]

you searched what and where

[neat pewter]

Online

And there was a similar problem and I tried the steps

And got 0.0004

[royal basin]

so you tried some steps.

[neat pewter]

Yea but in my head I don’t have it on paper

[royal basin]

you chose not to write it down on paper, then.

[tall tusk]

Bonjour Ann

[royal basin]

that's bad. you should write stuff down on paper more.

anyway, we're gonna need to do some more combinatorics.

[neat pewter]

Ok

[royal basin]

we will need to find the following counts:

• the number of ways to deal a hand of 5 cards out of 52
• the number of ways to pick 3 aces out of __ (how many are in a standard deck?)
• the number of ways to pick 2 non-aces out of __ (how many are in a standard deck?)

[neat pewter]

Ok

[royal basin]

do you know how to find these counts

[neat pewter]

No

[royal basin]

okay then look up "permutations and combinations"

come back here when you have done so and attempted the problem

[obtuse pebble]

@neat pewter Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[daring fjord]

I'm in gr 10 so this problem may seem basic to some, but how do I go about solving this?

[drifting loom]

its

simple

first factorize

the quadratic equation

11x^2-14w+3

and then add the factors

@daring fjord

[daring fjord]

thank you

.close

[obtuse pebble]

Channel closed

Closed by @daring fjord

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[queen osprey]

anyone have any tips on how to start this?

i tried Chinese Remainder theorem but im having trouble because of the left having multiple terms

[viral blade]

it is technically factorable

if that would help

otherwise you could do case by case mod 5 and then case by case mod 7

since the entire thing is a function of n^2 you could do case by case on n^2 in which case 0,1,4 are the only possible values

mod 5

(and something similar mod 7)

[queen osprey]

ohh i see. i would have to go through 12 cases

[viral blade]

yeah

[faint sphinx]

suppose n^7=1
15n^8==15n [7] fermat

and -14n^6 == -14 [7] (fermat again)

[viral blade]

hm

[ancient jacinth]

would substituting u for n^2 help

[viral blade]

isn't Fermat a^(p-1) is equivalent to 1

[faint sphinx]

multiply by a

[viral blade]

so 15n^8 would be 15n^2

mod 7

[faint sphinx]

yep that's what i meant

then you get that 15n^2-n^2=14*n^2 equivalent to 0

do the same with 5

[queen osprey]

ill try this way

[faint sphinx]

and there you go

[viral blade]

ye that's fastest

[faint sphinx]

you have to separate two cases tho !!

if p doesn't divides n

you have n^(p-1)==1 [7]

else n==0

[viral blade]

==0?

[viral blade]

only when p doesn't divide n

[faint sphinx]

yes, sorry i'm drunk

[viral blade]

dw lol

[faint sphinx]

pretty sure there is a billion other methods

i hate arithmetics

[queen osprey]

arent 2,3 also possible values?

[viral blade]

n^2 == 2 mod 5 isn't satisfyable

same for n^2 == 3 mod 5

[queen osprey]

ah i see

im very new to this, i find it incredible that u guys seem to be comfortable with the world of congruences

[queen osprey]

and u figure this out by checking n from 0 to 4 right?

because mod 5 is small enough

[viral blade]

yeah

after 5 they'll start to repeat

important facts about mod theory is that

adding and multiplying residues is "well defined"

in the sense that

a+b mod n will always be determined by a mod n and b mod n

[queen osprey]

yeah i get u

[viral blade]

(a+b) mod n = (a mod n + b mod n) mod n

same thing for multiplication

ab mod n is always the same as long as a mod n and b mod n are the same

[queen osprey]

is that by property of addition?

[viral blade]

idk

[queen osprey]

i knew it worked but i couldnt figure out a proof for it

[viral blade]

you can prove it like

a = pn + q

b = rn + s

a+b = (p+r)n + q + s

[queen osprey]

oh right thats nice

[viral blade]

a+b == q+s (mod n)

something like that

[queen osprey]

yep

[viral blade]

multiplication is similar

it's worth noting for example that

a == b mod n does not mean

2^a == 2^b mod n

so it's specifically addition and multiplication

[queen osprey]

alright got it. ive yet to encounter this case but ill keep it in mind now

ty for the help btw. i really appreicate it

[viral blade]

np

:)

[obtuse pebble]

@queen osprey Has your question been resolved?

[queen osprey]

to close this question, i said:
every interger n satisfies
n== 0 mod 5
n== 0 mod 7
which implies n == 0 mod 35 (by chinese remainder theorem)
so every n satisfies the congruence

@viral blade @faint sphinx ty <3 (i hope the last conclusion is right)

[obtuse pebble]

Channel closed

Closed by @queen osprey

Use .reopen if this was a mistake.

[viral blade]

yeah it does

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[indigo inlet]

what am i doing wrong in this picture

[warm canopy]

what happens to an inequality if you times or divide by a negative?

[indigo inlet]

you flip the sign

OHHH

[high lily]

also missed a - sign when indicating intention to divide

[indigo inlet]

thank you 🥲

so how would i know which choice to pick

[high lily]

well what do you have after fixing your mistake

[indigo inlet]

i think x<2

[high lily]

yes,
so overall you have
x<2 or x>=2

trying simplifying that and/or consider what values satisfy that

[indigo inlet]

so all values are solutions??

[high lily]

yes

[indigo inlet]

but why

i am confused on why it isn't there are no solutions

[high lily]

because its OR and not AND

[indigo inlet]

oh okay thank you very much

[high lily]

as long as a value satisfies either of those inequalities, its a solution

[indigo inlet]

that makes so much more sense

thank youu

.close

[obtuse pebble]

Channel closed

Closed by @indigo inlet

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[stable jacinth]

I need the parent function for each of these if possible

[fierce lagoon]

Google can easily tell you

x^2
x^3
sqrt(x)
|x|
1/x
e^x

[autumn adder]

Uhm the rules pls?

[fierce lagoon]

The parent functions are basically just the very, very basic form of each of them

But google can easily tell you all of them

[autumn adder]

Ig if we relied on Google only then the very whole motive of helping would be no more there

[high lily]

didn't you ask this yesterday?

[stable jacinth]

I'm trying to clarify, I used what was said yesterday (1/x) and didn't get the right values

[autumn adder]

Oh yeah he did

[autumn adder]

Which one?

Reciprocal function?

[fierce lagoon]

Maybe it wants x^(-1)

[stable jacinth]

yeah

[fierce lagoon]

Which is stupid

[stable jacinth]

btw

should I always use -3,-2,-1,0,1,2,3

as my x value

or are there ones where this isnt convinent

in a table of vlaues

[autumn adder]

Table of values for?

[stable jacinth]

finding the parent graph

[autumn adder]

Yeah you could use any.

Its just that, it'd be better to keep values small, just to ease your calculations

[autumn adder]

These are like optimal values, so that you dont miss on any parts (hopefully) whilst graphing the curve

Negative, zero and positive - make sure to include all of them, as the behavior of the function changes in different places / intervals

[stable jacinth]

yeah

[autumn adder]

Yup

[obtuse pebble]

@stable jacinth Has your question been resolved?

[stable jacinth]

.close

[obtuse pebble]

Channel closed

Closed by @stable jacinth

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[rich trout]

I dont know where to start

[winter marten]

Cos72=Cos(2•36)

Use double angle identities