help-10

[river kindle]

are the system of equations you solve

all = 0 except f'(x)=1

[light furnace]

that makes me wanna 🔫

[obtuse pebble]

@river kindle Has your question been resolved?

[obtuse pebble]

@river kindle Has your question been resolved?

[obtuse pebble]

@river kindle Has your question been resolved?

[obtuse pebble]

@river kindle Has your question been resolved?

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[steady ledge]

idk how to do this

[daring sorrel]

h(5)=g(f(5)). Work from this inside out. What is f(5)?

[steady ledge]

uhm

25?

:(

[daring sorrel]

How did you get that

[steady ledge]

cause 5 times 5 is 25 :(

this isnt my work its my friends i just need the answer

sorry :(

[daring sorrel]

Dw. The value is given by the table so you can get it by looking there

[steady ledge]

so the equation

to find the answer is in the table?

is it possible if u can just list out the steps w the answer please im just tryna impress a guy here 🙏 SORRY

[daring sorrel]

To find f(5) we look at the table. To find g(f(5)) we plug whatever we get f(5) equals to into the g(x) equation

Lol unfortunately that is against the rules

[steady ledge]

aw man

[steady ledge]

so if i tell him that

he'll get it??

🥺

[daring sorrel]

Probably. Just hope he doesnt ask a follow up question

[steady ledge]

okay THANK UU SO MUCH!!

[obtuse pebble]

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[robust garnet]

discrete /decision math problem pls help

[kind hawk]

first you need to figure out which numbers you want to include in the list, then you need to figure out in how many orders you can put them in a list

[robust garnet]

1st part is cool

the how many orders is the hard part for me

[gilded needle]

the possibility of a duplicate 1 makes that part slightly tricky

[robust garnet]

oh yeh

[drifting wraith]

only slightly tricky, you just separately count lists with no 1 and with one one, and with two ones

oh actually it's still tricky

and also there's infinite lists because you can write 6 as many times as you want?

you could pretend it says zero numbers that don't belong

[gilded needle]

it's poorly worded for sure

like are 4 and 4 "two square numbers" if they appear twice in the list?

[drifting wraith]

yeah it matters a lot if you're allowed to have 4,4

[robust garnet]

i got no clue lads

.close

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[ripe hazel]

this problem makes 0 sense to me

[ripe hazel]

because the "to verify" part asks for ratios that dont make sense at all

i dont get how E and G can have the same x value

these x coords are driving me fkn crazy

right know i know xy must be 4p, de 6p fg 7p

[viral blade]

X should be 4q,4r

for example

because it is halfway between L (0,0) and M (8q,8r)

[ripe hazel]

yea

[viral blade]

the rest can be solved similarly

[ripe hazel]

solving x d and f doesnt help solve y e and g

[viral blade]

Y is the midpoint of M and N, which have given coordinates

[ripe hazel]

so y is sqrt[(8p-8q)^2 + (8r)^2] ?

[viral blade]

That looks more like a distance formula

You need midpoint formula

[ripe hazel]

oops im dumb

its 8q+8p/2 right

[viral blade]

Yeah

With parentheses on the left

So 4q+4p

[ripe hazel]

ok i figured the rest out

.close

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[grim cape]

Is the integral F(x) domain [0,x] or R?

[obtuse pebble]

@grim cape Has your question been resolved?

[fathom flicker]

Big Fs domain is R

It says F:R->R

Right because you can make the upper bound of your integral, “x”, any real number

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[somber vale]

Any solution on how to solve this? The question is find the upper bounds and the lower bounds, the sepremum, the infermum ect

All i know is i need to find the set for the inequality

And i think its [? ; 2]

I dont have any idea on how to find the second number

[warm canopy]

This is equivalent to solving the inequality -2 <= x²-2x-3 <= 2

The solution will be the intersection of the solutions of both individual inequalities

[somber vale]

Can you tell me how to solve one of them? I suck at inequalities lol

So i can know the idea

[warm canopy]

Make one side equal to zero and then think about the graph of the quad on the other side and when it is positive or negative

You will need to find the roots along the way

[somber vale]

Okay thank you, when doing that i should remove the absolute value or keep it and remove it along the way?

[warm canopy]

There is no absolute value in either of these inequalities so you're fine

I just unpacked what |a| <= b means

[somber vale]

Okay thank you so much for help

[obtuse pebble]

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[lethal blaze]

Is this correct??

[rigid shard]

no

[worthy rivet]

step 3 is wrong

[rigid shard]

you did a mistake isolating the secx+cosecx and not dividing it out

the ans / simplification should be secx.cosecx

[lethal blaze]

wait how??

[worthy rivet]

,, secx+cscx=1/cosx + 1/sinx

[warm shale]

Lobter

[lethal blaze]

did you transform the numerator??

[worthy rivet]

$\frac{secx+cscx}{cosx+sinx}$

[warm shale]

Lobter

[worthy rivet]

=

$\frac{secx+cscx}{1/cscx+1/secx}$

[warm shale]

Lobter

[worthy rivet]

=

$\frac{secx+cscx}{(secx+cscx)/cscxsecx}$

[warm shale]

Lobter

[worthy rivet]

=

,, cscxsecx

[warm shale]

Lobter

[lethal blaze]

ohh you added the 1/cscx+1/secx

waiitt should not it be 1/cscxsecx?

[rigid shard]

this expression is in denominator

[worthy rivet]

Sorry I'm not really good with latex

[rigid shard]

so when you have something inverted in denominator you can take it up as numerator

[worthy rivet]

yeh

a/(b/c)=ac/b

[lethal blaze]

or i can just trasform it to sinxcosx right?

[worthy rivet]

no

[rigid shard]

you may

if you want to

the ans in sin cos would be 1 over sinx.cosx

[worthy rivet]

I think they meant something else

[lethal blaze]

wait im getting confused

[rigid shard]

hold up im gonna send a pic now 😭

[lethal blaze]

$\frac {1}{secxcscx}$

[warm shale]

akio

[lethal blaze]

why do I need to reciprocal it?

if I can just transform it to its reciprocal identities

which is

,, sinxcosx

[warm shale]

akio

[rigid shard]

yeah you may but the ans would be 1 over sinxcosx

[lethal blaze]

why 1 over

[worthy rivet]

@lethal blaze

[worthy rivet]

Do you understand this

[lethal blaze]

yess

[worthy rivet]

yes

[lethal blaze]

ohh wait

Is this it?

[worthy rivet]

yes

[lethal blaze]

Okey I got it now thank you

[worthy rivet]

Very well.

[rigid shard]

W

[lethal blaze]

Wait I still have some more

But im gonna try it first

[worthy rivet]

Sure 🙂

[rigid shard]

😭

[obtuse pebble]

@lethal blaze Has your question been resolved?

[worthy rivet]

Any problems, @lethal blaze ?

[lethal blaze]

yess I don't know how to procced in this

wait let me send a pic

this is a new equation

[worthy rivet]

Sure

[lethal blaze]

i tried simplifying the right side first

i got

$\frac{cos^2x}{sinx}$

[warm shale]

akio

[worthy rivet]

huh

[lethal blaze]

for the right side??

[worthy rivet]

is it csc(1+cos^2)?

[lethal blaze]

yess

[worthy rivet]

csc of that

?

or csc times that

[lethal blaze]

csc times (1+cos^2)

[worthy rivet]

oh

then yeah

proceed

wait no

[worthy rivet]

This is wrong

[lethal blaze]

but i don't know to make it happen in the left side

wait why?

[worthy rivet]

csc = 1/sin

[lethal blaze]

is it 2cos^2x/sinx?

[worthy rivet]

you forgot to divide the 1 by sin

[lethal blaze]

ohh yeah i forgot the 1+

[worthy rivet]

and for the LHS

Try combining fractions

[lethal blaze]

ahh okey wait

[lethal blaze]

wait you cross multiply?

[worthy rivet]

Well yes, how else would I combine the fractions

[lethal blaze]

by using the lcd

why is there over sin^2?

[worthy rivet]

the denominator is (1-cos)(1+cos)

so

it is 1-cos^2

which is sin^2

[lethal blaze]

ohh yeah

[worthy rivet]

yes

[lethal blaze]

okey now i got

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

[warm shale]

akio

[lethal blaze]

if I divide it by

[worthy rivet]

Why divide?

[lethal blaze]

$\frac{sin}{sin}$

[warm shale]

akio

[worthy rivet]

I mean yeah, you can put it back to the csc(1+cos^2) form

[worthy rivet]

but we have already shown that ^ is equal to RHS, and now we have shown that the LHS is also equal to this

[lethal blaze]

I separate the two then simplify it

Is this wrong?

[worthy rivet]

We have already solved the problem

[lethal blaze]

So this is correct

[worthy rivet]

You are back to this

[lethal blaze]

Yeah so the right and left is equal

[worthy rivet]

yeah

[lethal blaze]

Okey last

Is this the final or can I simplify it again

[worthy rivet]

You can simplify further

[lethal blaze]

Is this it?

[worthy rivet]

Sorry, I was having lunch

[lethal blaze]

its okey

[worthy rivet]

I suggest not expanding tanx

just use that cot=1//tan

[lethal blaze]

ahh ok

[worthy rivet]

I think you can continue from here

[lethal blaze]

Am i still in the right direction??

is the answer cos^4x??

[obtuse pebble]

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[balmy sand]

i want critique on the way im writing this proof. my teacher wants us to write induction in a certain way, which is basically the outline. but for the divisiblity ones, i think i've messed up the wording a bit?

Please ping me

[obtuse pebble]

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[reef rain]

pls help me with question 2 and 4, just started limits and trying to understand the whole concept of proving stuff

[reef rain]

i already watched this one video

https://youtu.be/29jRJO_QSW4?si=OSjoDQ2tDpPYBbrZ

there was a playlist for it and i watched a few but question 2 i dont know how to solve i just cant understand for some reason

pls help

[steel parcel]

do you know the first principles formula

[reef rain]

uhh

i dont think so 😭

im doing it how the video showed me

man ive never proved anything in math before this is the first time im doing this stuff

uni really fucked me over with this one

[steel parcel]

ok no worries

ok have you seen this before?

[reef rain]

i remember doing that in integration or something

like turning a function into some form or the other

ive seen it before i think but id appreciate if you could explain it out for me

[steel parcel]

okay

so when you first learn about differentiation

you will be introduced to first principles

[reef rain]

OOOH

ok now i remember

💀

[steel parcel]

this formula essentially allows you to find the derivative at a point when there’s a very small change (denoted by h)

the lim as x approaches 0 reinforced the very small change

[reef rain]

i never saw that lim thing in highschool but i remember i was daught first principle without that lim

[reef rain]

okok

[steel parcel]

haha okay

given this formula

we can know start doing your question

for question 2, our f(x) is x^2 and h approaches 2

[reef rain]

yes

[steel parcel]

we can do lim as x->2 ( lim h -> 0 (( (x+h)^2 - x^2)/h))

[reef rain]

ok

[steel parcel]

wait lemme think for a sec

[reef rain]

alr

[steel parcel]

ok yes let’s keep moving forward

expanding this we get

(x^2 + 2xh + h^2 - x^2)/h

(2xh + h^2)/h

(2x + h)

does it make sense so far?

[reef rain]

x^2 + 2xh + h^2 right

2xh or just 2h

[steel parcel]

oh yes 2xh

[reef rain]

ok got it

2x+1

?

[steel parcel]

now we apply the limit h -> 0

we get 2x

but the question asks for x -> 2

we can apply this limit now

2(2)

[reef rain]

wait one sec good sir

[steel parcel]

= 4

okay

[reef rain]

nvm im being stupid

got it

[reef rain]

okok

so we sub in values of h

[steel parcel]

sub h -> 0 then x -> whatever the question says

[reef rain]

OOO

[steel parcel]

yeah that’s it

[reef rain]

minor issue

[reef rain]

u say h -> 2 here

[reef rain]

but then h -> 0 here

no comprende

[steel parcel]

sorry i just make it a bit clear

[reef rain]

ooh okok

@steel parcel tysm

i appreciate you man thanks

[steel parcel]

no worries :)

[obtuse pebble]

@reef rain Has your question been resolved?

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[obtuse pebble]

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[timid silo]

If (-1)^n does not converge, then why does the rule "if the absoulte value converges" the sequence converges work

[past sand]

Where did you see that rule?

[timid silo]

https://youtu.be/FoNLQvf4NUs?si=ipHVLHzm2m7HVtil

He talks about it earlier in the video and around 2 hours and 25 minutes

[past sand]

He probably meant "if the absolute value converges to zero"

[timid silo]

Ohhh

So the absoute value of (-1)^n does not converge to 0, correct?

[past sand]

No it's just 1

[timid silo]

But because the absolute value of (-1/2)^n converges to 0, the whole sequence converges to 0?

[past sand]

Yes

[timid silo]

Alr thank you!

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[graceful marten]

Hia bae's

[graceful marten]

Part a, is giving me a headache

All ive got so far is

OP= OA + AP
a + x(-a+c)

so a(-1 + x) +xC

and thats about it but its def wrong as an answer but correct

I guess what I have to do is make MQ = AP and somehow factor it and see what ratio P is from one of them

.close

[obtuse pebble]

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[graceful marten]

.reopen

[obtuse pebble]

✅

[graceful marten]

HIAaaaaaaaaaa bae

Question a

Im unsure what to do but I did
MQ = AC

1/4a +c : -a+c

No clue what to do from here

:>

[obtuse pebble]

@graceful marten Has your question been resolved?

[obtuse pebble]

@graceful marten Has your question been resolved?

[graceful marten]

<@&286206848099549185>

[graceful marten]

.close

[obtuse pebble]

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[edgy tree]

Hey. Been stuck on this question for quite some time now.
Tried solving it by doing y2-y1/x2-x1 and tried other various methods. didn't work out.

The line of a graph in the xy- plane has slope
and contains the point (6,-1) . The graph of
a second line passes through the points (0,1)
and (1,0). If the two lines intersect at the point
(p,q), what is the value of p.q?
A) -6
B) -3
C) 6
D) 3

[latent walrus]

has slope (...?)

[edgy tree]

Slope is 1/3

[latent walrus]

what did you get for each line equation?

[edgy tree]

Can you elaborate further, I do not quite get it.

[latent walrus]

there are two lines:
the one with slope 1/3 passing through (6,-1)
and the one passing through (0,1) and (1,0)

what did you get for their respective equations

[edgy tree]

This?

That's what I did so far. but I have no idea on how to continue further

[latent walrus]

those are just points

do you know how to get line equations

y=...

[edgy tree]

oh.

you meant as in

y=mx+b?

The first one will be y= 1/3x-1

If I'm not wrong

I do not know about the second one

[latent walrus]

cant be

[edgy tree]

Hmm.

Quite clueless.

[latent walrus]

if x is 6, youd get y=1, (6,1) wasnt the point you were given

do you know point-slope form?

[edgy tree]

It's

y-y1

I do not remebmer

The forumla

For it

[edgy tree]

Yeah that

So I have to use that?

[latent walrus]

its the easiest way really

[edgy tree]

Okay well. then it should be

if we do it that way

(y-1) = 1/3(x-6)?

or

[latent walrus]

no

[edgy tree]

And it should be

Let me rewritte it.

[latent walrus]

also no

[edgy tree]

(y-(-1)) = 1/3(x-6)

How come?

[latent walrus]

this one is fine

[edgy tree]

Ah so

Just solve?

y+1 =1/3x - 2

Should be

[latent walrus]

sure

[edgy tree]

What would the next step be

[latent walrus]

make y the subject
then get the other line's equation

[edgy tree]

Y the subject

as in like

y = 1/3x -3?

[latent walrus]

mhm

[edgy tree]

Okay we got that

What did you mean by "other lines equation"

[latent walrus]

the other line

get its equation

[edgy tree]

the, (0,1) (1,0)?

[latent walrus]

the one that goes through (0,1) and (1,0)

[edgy tree]

Same thing? slope intercept?

[latent walrus]

yeah

just get the slope first

[edgy tree]

So

0-1/1-0

= 1/-1

1 up and 1 left

[latent walrus]

uh, sure

[edgy tree]

Afterwards we use slope intercept

which is

(y-1)= 1/-1(x-0)

[latent walrus]

why leave it as 1/(-1), thats rather strange practice

its just -1

[edgy tree]

Well

Sure

[latent walrus]

this is fine though

[edgy tree]

y = -1x+0 or y = -1x

[latent walrus]

neither

[edgy tree]

How come

[latent walrus]

y-1=-x

y=-x+1

[edgy tree]

Right right. we do not put the coeffecient of 1

But it's 0 though

for the y-intercept

[latent walrus]

no, it isnt

[edgy tree]

Alright hold on let me resolve

[latent walrus]

you wrote this yourself

y-1=-1(x-0)

y-1=-x

y=-x+1

[latent walrus]

just from your plotting you should know the y-int is not 0

[latent walrus]

you think -1(0)=-1?

[edgy tree]

My brain is fried

😂

Your correct

Okay

Fixed

[latent walrus]

👌

[edgy tree]

Alright we now have

2 slopes

[ancient tangle]

I Need HELO

Help guys

[latent walrus]

#❓how-to-get-help

[ancient tangle]

ITS NOT WORKING

GOSH

[edgy tree]

Alright so. we have y = 1/3x -3
and y = -x+1

[latent walrus]

indeed you do

[edgy tree]

The question states.

if the two lines intersect at a the point (p,q) then what is the value when you multiply p and q

[latent walrus]

it does

[edgy tree]

So how would I know the p and q

after finding the slopes

[latent walrus]

find the intersection

set the line equations equal to eachother and solve it

[edgy tree]

1/3x-3 = -x+1?

[latent walrus]

mhm

[edgy tree]

6

2/3x = 4

divide by 2/3

you get 6

Which is A?

Oh my bad, C*

So it counts as multiplying when we did it like y = y?

[latent walrus]

?

how so

[latent walrus]

1/3x +x is not 2/3 x

[edgy tree]

no but.

1/3x + x

we know coefficent of x is 1

so 2/3x

[latent walrus]

no?

[edgy tree]

yeah?

[latent walrus]

1/3 + 1 is not 2/3

[edgy tree]

oh.

4/3

[latent walrus]

yeah

[edgy tree]

I shouldn't do mental math

Alright hold on

3

Okayy

Seems easy when you understand what your doing

Thing is. I know how to do all these steps

The question is confusing

Okay well. I am going to solve a question and see if I have any issues

Which of the following numbers is NOT a
solution to the inequality 2-1/2x ≤ 2x+7?
A) -2.5
B) -0.5
C) 0.5
D) 2.5

Wouldn't this be substitution?

[latent walrus]

you could do that if you want

or you could solve the inequality

[edgy tree]

How would you solve the inequality

[latent walrus]

pretty much the same as how youd solve an equation, get x on one side

just note if you divide/multiply by a negative, then flip the inequality sign

[edgy tree]

Well I'm using substituation

And I got A as the correct answer

[latent walrus]

sure

[edgy tree]

x-y = 1/2y
If (x, y) is a solution to the equation above and
y is not equal to 0, what is the ratio x/y?
A) -3/2
B) -2/3
C) 2/3
D) 3/2

😅

Any idea on what to do?

[vital imp]

its D i think

cuz 3-2= 1/2x2

[edgy tree]

Well I don't have any questions for now.

Ill be back when I do so

.close

[obtuse pebble]

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[tired nacelle]

Hi
I have
f(x) = x/2E(x/3)
E is the integer part function
And also we have x!=0
And f(0)=0.5

He say proof that
|f(x) - 1/3| <= |x|\2

[tired nacelle]

I try to make table include x and E values and f(x) values and compare it to |x|/2 values since i don't know how to prove that

But even that i find that this Inequality is true for x>=3

So how i can prove something like this

[obtuse pebble]

@tired nacelle Has your question been resolved?

[tired nacelle]

<@&286206848099549185>

Anyone 👀

.close

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[pine umbra]

according to solution sheet, it is tan60= ?/15, 15xtan60=?, ?=25.98

i do not understand how to get to opposite and adjacent

[pine sail]

tan(60°) is NOT ?/15

So the solution is wrong.

[signal ledge]

uh?

no?

[pine sail]

Talking to me?

[signal ledge]

What are you saying

yeah

[pine umbra]

i dont understand

[signal ledge]

tan 30 = 1/√3

So 1/√3 is the ratio of your "?" And 15

[pine sail]

Kavin, might wanna recheck before re addressing me?

[signal ledge]

Which means 1/√3 = x/15

Sorry I goofed up, I meant to reply to d3vil

I mistook you to be the original poster

My baddd

[signal ledge]

x = 15/√3 = 5√3

[pine umbra]

so the solution is tan30= x/15 right?

[signal ledge]

Yes

[pine umbra]

urm

please help to explain how to tell which side is opposite and which side is adjacent

[signal ledge]

The side opposite to the angle in concern is the opposite side

The side connected to the angle in concern which also is connected to the right angle is the adjacent

[pine umbra]

is it possible for the ? side to be adjacent and 15 is opposite?

[signal ledge]

Ofcourse, but in that case your angle in concern is 60°

[pine umbra]

ahhhh, i understand now, thanks a lot man i was stuck on this for so long

.close

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[timid silo]

I need to find the x in degrees

[timid silo]

KLMN - rectangle

[thin star]

!15m

[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.

[timid silo]

<@&286206848099549185>

[little jewel]

Do you know trigonometry and similarity of triangles?

[timid silo]

Yes sir

[timid silo]

WAIT

Hmm

How does it help me @little jewel

[little jewel]

Maybe those 2 triangles are similar

[timid silo]

Maybe?I need to prove then

[little jewel]

Yeah

[timid silo]

Wait

I thibk

[little jewel]

Or there's also a corollary which you can use

[timid silo]

Angle L = angle M?

[little jewel]

It's || LP * PN = PM^2 ||

[timid silo]

How so

[little jewel]

Can you elaborate both angles?

[timid silo]

Not really just guessed

[little jewel]

It is a result that you get when you prove MPL is similar to NPM

[timid silo]

Hm

[little jewel]

Can you write down the values of the following angles: MLN, LMP, PMN, MNL in terms of x

Remember that KLMN is a rectangle

[timid silo]

Mhm

[timid silo]

x = 90° - MLN :D?

Wait

The minute

x = LMP - 90°

[little jewel]

So MLN = 90 - x

[timid silo]

Yes

[little jewel]

Check it again

[timid silo]

Wym it's correct

[little jewel]

LMP is < 90

[timid silo]

A

[little jewel]

So LMP - 90 is negative

[timid silo]

Hm

[little jewel]

But x cant be negative

[timid silo]

Ofc

[little jewel]

How did you get this?

[timid silo]

It's wrongsl

Wrong*

[little jewel]

Oki

[timid silo]

Just realized

[little jewel]

Try to find the remaining 3 angles in terms of x. It's easy

[timid silo]

Oke

[little jewel]

Then you will yourself be able to see how those 2 triangles are similar

You already have MLN = 90 - x

[timid silo]

Soo

angleKLN = angleMNL

Which is x

Can't I make similiar triangles

triangleLNK and triangleMPN

[little jewel]

How

[little jewel]

But we need to get info out of it

[timid silo]

Owh wait wtf

[little jewel]

Oki correct

[timid silo]

So we can make it similiar triangles

Cuz we got two angles

[little jewel]

Which two triangles

[timid silo]

LKN and MPN

[little jewel]

You are right they are similar

But it doesnt really help us alot

[timid silo]

But they don't help

Yeh..

[little jewel]

What help us with the given info is

If we could prove MPL similar to NPM

[timid silo]

Nothjng

WAIT

I got an idea

[little jewel]

You said MNL is x. You should be able to find NMP then

[timid silo]

Yehu

I got two angles

The right side I calculated 2 angles

Which are equal

Soo

I can make it similiar triangles

So

Hm

[little jewel]

Oki yes

[timid silo]

It doesnt help us tho

We need to find k

k

Or does it

[little jewel]

MPL is similar to NPM right?

[timid silo]

Yes

[little jewel]

Using the property of similar triangles, you can deduce PM^2 = PN * PL

[timid silo]

How

[little jewel]

.

[timid silo]

Never heard of it

[little jewel]

If you rearrange it, it is saying you have to deduce PM/PN = PL/PM

[timid silo]

Ye

So

Wait

Yeye

I see

[little jewel]

Your job is to use the fact that MPL is similar to NPM and establish this expression PM/PN = PL/PM

[timid silo]

So

Pm

Is sqrt(3)

[little jewel]

Yep

Well not exactly

[timid silo]

Why not

[little jewel]

How did you get that

[timid silo]

Owhh

well

sqrt(PL*PN)

Which is = sqrt(3) b

B would be like

the ratio

[little jewel]

PM = sqrt(PL*PN)

[timid silo]

Yeye

[little jewel]

=sqrt(3PN*PN)

[timid silo]

Yup

I make PN = b

So it's eazier

To use it

[little jewel]

So it means PM = bsqrt(3)

[timid silo]

Yup

Eoo

Soo

We can get angle

[little jewel]

which is PN*sqrt(3)

[timid silo]

MLP

Yeh

[little jewel]

Now using trigo, you got this

[timid silo]

Yes

Thx u alot

Asian 100% u are

[little jewel]

Np

LOL

[timid silo]

What?

I guessed?

[little jewel]

Yeah fair enough

[timid silo]

I knew it everybody goo d with math asians

[little jewel]

People outside asia can be great with maths too but I get where you're coming from

[timid silo]

Idk about that

Anyway thx alot we will see again some time 😉

.close

[obtuse pebble]

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[fervent jay]

Right, in line 3

They divide it into 2 seconds, cos(x)=0 and the longer one

How are they simply able to divide that

Holy shit I'm a moron

😭 DAWG IM SO STUPID

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[obtuse pebble]

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[final thunder]

Long day

[violet sentinel]

You're not stupid, you're learning, there's a bit difference

[obtuse pebble]

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[vivid wraith]

@violet sentinel is this place open?

[violet sentinel]

yep you're good to go!

[raw escarp]

so i have this math problem to which idk if i went wrong or my professor and I can't even understand final result is this the place

[vivid wraith]

ok so basically im unsure how I would solve this percentage uncertainty question.

[pine sail]

Significant figures.

[vivid wraith]

i looked at the mark scheme

[pine sail]

And error.

[vivid wraith]

but still have absolutely no clue

[pine sail]

Heard of those?

[vivid wraith]

cus they just put the answer

ik what sig figs are

[pine sail]

Least count?

[vivid wraith]

least count?

[pine sail]

Do you know what least count is?

[vivid wraith]

no

[pine sail]

okay alright

[vivid wraith]

this was the answer from the mark scheme

[pine sail]

If the reading were:
25.1
What would be the % uncertainty?

[raw escarp]

I'm assuming this is the wrong place

[vivid wraith]

would it be the .1?

[violet sentinel]

hello, if you have a math questions, feel free to go to #❓how-to-get-help . You can open a new thread there. This one is occupied.

[pine sail]

That's not % though is it?

[vivid wraith]

.1/100

[pine sail]

No.

[vivid wraith]

wait

no

• 100

[pine sail]

No.

0.1/25.1 * 100

For the % uncertainty

[vivid wraith]

oh

[pine sail]

yeah.

[vivid wraith]

so in this one

its

1/3.5 x 10^3

which gives

[pine sail]

NO

[vivid wraith]

oh

[pine sail]

Well, that's where the least count comes into play

That's like

The smallest measurement you can make

Which is the likely error.

When you measure something like 3.5

It means the least count is 0.1

As the reading could have been something like 3.47 but you had no means to measure it that way and just said 3.5 for simplicity

It could be 3.5 +- 0.1

So the % error is 0.1/3.5 * 100 in this case

[vivid wraith]

Why not 0.5 for least count

[pine sail]

Get it, somewhat?

[pine sail]

The entire 0.5?

[vivid wraith]

Oh wait rounding

I can be rounded up to 4

Down to 3

[pine sail]

yeah.

Can you proceed now?

I'll have a look at your work.

[vivid wraith]

So 3.5 uncertainty would be 0.1

So 0.1/3.5 * 100

[pine sail]

Correct.

What

• 100

Was essential

Alright good.

[vivid wraith]

2.88%?

[pine sail]

Now before you're off, I'll tell you.
In this case the uncertainty wasn't really 0.1
But 0.1 * 10^3
But when you find the % it'd be 0.1 * 10^3 / (3.5 * 10^3)

Keep that in mind.

[vivid wraith]

Ohh 2.9

[pine sail]

Yeah whatever that is.

Also keep your answer in the same number of significant figures as the reading.

So like, you had 2 originally, so 2.9

yeah.

[vivid wraith]

So what I should have done was to make em both in the units they gave me

So if I had 3.55

Would the uncertainty be 0.01?

[pine sail]

precisely.

[vivid wraith]

Ahhh

Okkk

Thx much

[obtuse pebble]

@vivid wraith Has your question been resolved?

[obtuse pebble]

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[keen notch]

@pine sail

Heyo, need a bit more help here.

Im not quite sure how to visualize whats being said here.

I can easily imagine a plate with a given thickness.

However once they start talking about Area of the Element I get lost.

[pine sail]

Well the element could be anything really.
It's the same as weight, you broke the body into infinite smaller weights and called them dW.
Similarly now there's infinite smaller areas.

Earlier we were concerned with the weight of the elementary particle.

Now its area.

[keen notch]

Ah, so instead of being given a discrete elemental weight,

we have to infer the weight given the size of the area, thickness of the plate and the density of the material, ?

[pine sail]

Yes.
Weight = density * area * thickness

But since the density is meant to be constant.

It'll eventually cancel.

[keen notch]

And im assuming its only this simple because we were told the plate was of uniform thickness**

[pine sail]

Precisely.

Later on, you'll do variable thickness, and even variable density too.

You'll have to account the density there.

Because after all it's the weight you need.

You'll be given density as f(x) and so on, so that you'll still be able to integrate.

[keen notch]

ah... great.. that sounds so fun

lol

Give me one more minute if you don't mind. I likely need help understanding first moments.

[pine sail]

You better hurry, I have got to go soon.

Haha.

[keen notch]

well actually this I understand.

however this next part seems insane.

but if you gotta go man, don't worry about it

[pine sail]

yeah, I wasn't expecting paragraphs worth this time lol.

Well I'm pretty sure there'll be someone soon enough.

Have fun.

[keen notch]

tyty

have a good one.

.close

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[weary heart]

How does dot product represent a scalar and at the same time it can be negative

[brazen gorge]

scalars can be negative

🧐

[dark stirrup]

One characteristic that dot product measures is how "parallel" two vectors can be to each other

If you dot two unit vectors and the result is -1, then they are anti parallel

[weary heart]

Doesn't the negative sign show that it's in the opposite direction?

[brazen gorge]

well, yes

but a scalar is (most likely in your case) any real number

including negative ones

[dark stirrup]

Dot product also can describe an abstract quality known as flux: how much one thing flows through something else

A loose analogy is putting a net under a steam of water to try to catch fish. If the net is head on to the flow of water, then you maximize the amount of fish you catch (positive flux), if you turn the net sideways, no fish can swim into the opening (zero flux), and if you turn it all the way around, then all the fish you caught will escape (negative flux).

[weary heart]

But for example I can also represent work by the dot product of force and distance

How can work be negative

[dark stirrup]

You put energy into the system

Aka potential energy

As an object falls to earth, it does work traveling through the gravitational force field. If you lift that same object up, it is experiencing negative work.

[weary heart]

So in genral a quantity can be a scalar and have a negative sign and it shows that it works in the opposite direction

[dark stirrup]

In general, no. Dot product is an abstract concept and the significance of negative depends on your application.

[weary heart]

But in all situation it follows the same law

A.B.cos($\theta$)

[warm shale]

Hamdy Hisham

[dark stirrup]

Yes. But work is not the only application to dot product

[weary heart]

Aha

Thnx

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[zealous oak]

if something has 1.6% chance of happening whats the chance of it happening once in 31 tries?

[empty cypress]

exactly once or at least once?

[zealous oak]

how do I solve the problem for myself?

[empty cypress]

... well it depends on what the problem actually is

theres 2 different methods depending on if you want exactly or at least

[zealous oak]

I want to know the exact number

[viral blade]

I don't think you know what they're asking

[empty cypress]

Do you want to know the chance of the event happening exactly once out of 31 tries, or the event happening one or more times out of 31 tries?

"happening once" is an ambigious phrase

[zealous oak]

the chance of it happening one or more times in 31 tries

sorry I misunderstood

[empty cypress]

The method for this is
$$1-P(\text{x not happening})^n$$

[empty cypress]

so in your case

$$1 - (1-0.016)^{31}$$

[zealous oak]

so x is the chance and n is the number of tries?

[empty cypress]

yes

the "classic" example of this is the question "What is the chance of rolling at least one 6 if I roll 6 dice?"

The answer is not 6 * 1/6, which is what a lot of people fall for, its $1-(1-\frac16)^6 \approx 0.6651$

[zealous oak]

thank you for explaining

[empty cypress]

Of course - if you look up the classic example I gave there are lots of explanations on how the formula is derived if you are interested

[zealous oak]

.close

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[molten ether]

how do i solve for L i'm not sure where to start

[obtuse pebble]

@molten ether Has your question been resolved?

[obtuse pebble]

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[fathom bridge]

did i implicitly differentiate this properly?

[final thunder]

No

[fathom bridge]

god damnit

how wrong is it

[final thunder]

Your second line is wrong

😭

[fathom bridge]

oh shit

[final thunder]

You need to use product rule on x^2y

And as a rule of thumb every time you differentiate y times it by dy/dx

[fathom bridge]

???????????????? dawg where u get that

x^2y is not in that

at all

[final thunder]

What’s $\frac{d}{dx} x^2y$

[fathom bridge]

oh i saw power rule

[final thunder]

Well

[fathom bridge]

lawl

[warm shale]

$Pure$

[fathom bridge]

why cant i pull y as the coefficient

[final thunder]

y is not a constant

y is a function of x

[fathom bridge]

truuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuue

[final thunder]

so you have to use chain rule

[fathom bridge]

prduct rule

chain is nil

[final thunder]

$\frac{d}{dx} x^2y = y\frac{d}{dx} x^2 + x^2\frac{dy}{dy}\frac{dy}{dx}$

[warm shale]

$Pure$

[final thunder]

Where the second term I used chain rule

[fathom bridge]

this looks better so far?

[final thunder]

Yeah

Now just solve for y‘

[final thunder]

Erm wait the RHS is -dy/dx right

[fathom bridge]

oh yeah

-10

[final thunder]

It shouldn’t be a number

But expression

One sec I’ll send u

This should be right

[fathom bridge]

sorry r u talking abt my awful -4

[final thunder]

What do you mean?

[final thunder]

When I read this I don’t really know what’s going on

[final thunder]

Oh

Lmao

It’s 4 😭😭😭

Well

One sec

After second line all you have to do is put everything with dy/dx on one side

And the other terms on the other side

[fathom bridge]

how the fuck, i need to start on a new paper

[obtuse pebble]

@fathom bridge Has your question been resolved?

[obtuse pebble]

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[fringe trellis]

Find the symmetry
F(x) = x /x-1

[fringe trellis]

Even:
F(x) = f(-x)

F(x) = -x / -x - 1
F(x) = -x / -(x+1)

F(x) = x /x + 1

[runic void]

only plug into x

[fringe trellis]

F(x) =/ f(-x)

[fringe trellis]

Yeah

[runic void]

$F(-x) = \frac{(-x)}{(-x)-1}$

[fringe trellis]

Yeah

But u bring out the negative in the bottom

It's x/x+1

[warm shale]

.doc

[fringe trellis]

Which means it's not even

And it doesn't match up with it being odd

Which is

-- (x) / (x-1)

[runic void]

you got f(-x)=x/x+1?

[fringe trellis]

Yeah

[runic void]

great

[fringe trellis]

As for odd

Is it that

[fringe trellis]

^

[obtuse pebble]

@fringe trellis Has your question been resolved?