help-17

[cedar linden]

and stressful

but i learnt >D

thanks my friend

[silk osprey]

you’re welcome

[cedar linden]

btw

you know why wolfram

doesnt give me the answer?

,w \int_{1}^{2} \frac{x}{(x - 2) (x - 4)}

[twin meteor]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by pablogarcia11

[silk osprey]

it says integral does not converge

it does give you the answer

infinity

[cedar linden]

uhhhhhhhhh

ajsdjadf

thanks

being incorrect

is correct

hahhaa

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[sonic spoke]

i'm confused at two different math problems being solved differently, specifically im lost as to the why behind the process

[sonic spoke]

this is the problem on the quiz:

[uncut cedar]

what's the question

[sonic spoke]

(one moment)

so on that step with the arrows -11/5r = -11, they want you to multiply each side by -5r/11 but in the equation with variables on both sides: fractions video, they do this example from the lesson:

step 1:

then they multiply both fractions with the common denominator:

i dont understand why, on the test, you multiply by -5/11 when on the lesson video they multiply each side by the common denominator

would you not want to first get like denominators by doubling 3/4x to 6/8x, and then multiply each side by -3/8x?

[strange crater]

both work

[subtle helm]

there's no "why," both methods are valid and lead to the same answer

[subtle helm]

you can do it however you prefer

[sonic spoke]

thank you both very much

cheers

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[median quiver]

Can someone help with part c and d?

[median quiver]

I have this for my work so far

[weary inlet]

what I would do is find the zeroes of v(t) by factoring

as they are when the particle changes direction

then just make a sign chart and plug in values for v(t)

^part c

[median quiver]

Ohhh okk

[weary inlet]

when the sign chart is negative for v(t) that interval (closed interval btw since its velocity) it is moving to the left, and when it is positive it is moving to the right

then for part d

you find the total distance by taking the integral of the absolute value of v(t)

from 0 to 2

by taking the integral of the absolute value it makes all of the times when it switches directions all become positive, so it gives the total distance

so basically for part d

[median quiver]

Is there a different way to get the answer other than integrating?

[weary inlet]

well you could use the position function

and just take into account when there are direction changes

but it would be simpler to just integrate

$\int_{0}^{2} |v(t)| \mathrm{d}t$

[twin meteor]

Potatomonke

[weary inlet]

honestly I dont think there are any changes in direction

during this interval

so you could just plug in x=2

to s(t)

[median quiver]

Mmmm okk thank youu

[weary inlet]

if you use position

make sure to take absolute value though

[median quiver]

Will do

[weary inlet]

because during that interval I think its negative

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[rough hound]

Is there a thing such as differential inequality like how there are differential equations?

[pallid zenith]

i have a feeling that this is just a calculus of variations thing, you should keep the channel open tho, im curious what ppl say dont let it close

[vast shale]

F(x) can be solved

This is a D.E with inequalities

If this is what you meant

[rough hound]

what if the inequality was strict?

[vast shale]

Hmm

[rough hound]

wait, I'm just learning about differential equations, and I just thought if there was a thing like differential inequality, I barely know how to solve DEs.

[vast shale]

Middle school?

[vast shale]

BTW you can solve this D.E normally

As if there is no inequality

[vast shale]

Try asking it in Calc

[rough hound]

No, high school

[vast shale]

It is going to be simple,no inequalities

[rough hound]

wouldn't this be possible? y' + y/x > 3x then y > x^2

nvm

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[undone aurora]

Kinda confused at what exactly happened here

are we just adding p_1 + p_6 for all of those because A = {1, 6}?

if so then how did they get 1/3 in the first line

oh im acc slow wow.... 1/6 + 1/6 = 2/6 = 1/3

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[abstract palm]

how does this equal?
[\begin{pmatrix}
1&0\0&1
\end{pmatrix}\left(1-\frac{1}{2!}+\frac{1}{4!}-\frac{1}{6!}+\cdots\right)+\begin{pmatrix}
0&i\i&0
\end{pmatrix}\left(1-\frac{1}{3!}+\frac{1}{5!}-\frac{1}{7!}+\cdots\right)=][=\begin{pmatrix}
\cos 1&i\sin 1\ i\sin 1 &\cos 1
\end{pmatrix}]

[twin meteor]

Slowaq

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[bitter ridge]

How would I go about this?

[unique elm]

!status

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1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
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[bitter ridge]

I started with f(1) into the limit but it will get me 0/0,

I did f'(1) but that should just be f'(x) but still division by 0

dont know where to go

[unique elm]

did you do L'H?

[bitter ridge]

isnt that what I did? I may have misunderstood, but i dont where to go from this begining point

wait

did reserach?

does it mean f'(x)/g'(x)

so 2 + 6/ 1

[unique elm]

using L'H will get you $\lim_{x->0}\frac{2f'(1+2x)+ 6f'(1+6x)}{1}$

[bitter ridge]

so 12 + 36 / 1

48

[twin meteor]

caspar

[unique elm]

yup!

[bitter ridge]

thank you

This is a different question but related to calc, but how does the P(1) mean

can I assume t = 1

I see a problem with my understanding of some the basic notation in calculus

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<@&286206848099549185>

please advice how I should start this question

above

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[ashen knot]

hey, I’m in research methods in psych right now. I need help finding and omitting the potential outliers in a paired data set (x & y values).

[ashen knot]

these are the statistics I calculated using vassar stats

the x values are minutes of exercise from participants last week, and the y values are their Beck Anxiety Inventory scores

any recommendation on how to find these outliers in a reliable method, in the easiest way possible based on the information I have already collected?

[strange crater]

is this real data or just a homework problem

[ashen knot]

real data.

from mostly colleges students

problem?

[strange crater]

then I wouldn't remove any data points unless there was measurement error

for example, I'd be willing to be that the 1800 minutes of exercise is an error (that'd be 4 hr 17 min exercise every day for 7 days)

[ashen knot]

here are the instructions by my prof:

After you have removed the under 18 people you need to remove any outliers.
Make sure you state what threshold was used (to determine the outliers), give the number of people removed, then give the new N.
State that analysis was done of this new N.

[ashen knot]

tbh that’s not too unrealistic, and it is obviously an outliers

[strange crater]

unless they're (professional) athletes, I'd argue it is very unrealistic

anyway, it seems your professor might want you to remove all data > 3 standard deviations from the mean or something like that

[ashen knot]

sorta like this?

idk why the SD was so big but it seems like I can only have two of them before the sample runs out?

[strange crater]

yeah

[autumn stratus]

wouldn't you want to remove outliers based on their residuals?

how strong is the relationship

[ashen knot]

I was watching videos that talked about that but I can’t seem to find a calculator anywhere online that will save me the work.

[ashen knot]

the negative r value suggests that there is a weak negative correlation

[autumn stratus]

ehh i mean your r is -0.1165 so its kind of meaningless

but if you want to do it

[ashen knot]

I have to for a good grade

[autumn stratus]

you just do the 1.5IQR rule on the residuals

set it up on excel or smthn

[ashen knot]

will do.

will I know how many data pairs I need to remove currently?

is it 100?

based on the mean plus 2 SDs equaling n=714?

[strange crater]

hopefully not, that's 12% of your data

[ashen knot]

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[strange helm]

what is the expansion of lnx

[mellow oyster]

like the taylor series expansion?

[strange helm]

yeah

[mellow oyster]

you can't expand it around 0 since it's undefined at 0, so maybe it would be better to expand ln(x-1)?

im pretty sure that's what's standard

[strange helm]

this works though

[mellow oyster]

hm it's not in the form of a taylor series since you don't just have a_nx^n in the sum

like you have (x-1)/(x+1) raised to some power

i'm not saying that you can't define a series expansion that behaves similarly near 0

[strange helm]

ah ok

[mellow oyster]

i'm just saying that other expansions like sinx and cosx and e^x pick a = 0 in their taylor series when computing, and you can't do that here since ln(x) isn't defined there, so you could expand around a = 1

you'll get an infinite series that's identical to lnx

it's just that the terms will look like a_n(x-1)^n

[strange helm]

yeye got it

thanzk

[mellow oyster]

haopy to help

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[clever hare]

pls helpppppp. I have to prove the identity

[winter thistle]

!hw

[bright escarp]

?

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As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[winter thistle]

thx

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[velvet cove]

Money is split between sara, Cathy, and Ruby. The ratio is 8:7:2. Ruby has $60 less than cathy. How much does Sara have

[velvet cove]

what am i doing wrong

i keep getting a negative value for cathy's money

[rough hound]

show your work

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

[velvet cove]

let s be sara's money
c = cathy's money
r = ruby's money
r=c-60

$8ks=7kc=2kr$

[twin meteor]

Jash

[velvet cove]

7c=2r
7c=2(c-60)

u get 5c=-120

c=-24

[mellow oyster]

i don't think the setup of the ratios is right

[somber holly]

c - r = 60 since ruby has less than cathy

[mellow oyster]

there might be a faster way to get the setup but i like thinking of ratios as fractions, then the total amount is (s + c + r), and you know that s = 8/17 (s + c + r), c = 7/17 (s + c + r), and r = 2/17 (s + c + r)

this is the system you'd solve

along with r = c - 60

[strange crater]

easier to just work with r and c alone IMO

[mellow oyster]

oh wait yeah you don't even need to involve s

[strange crater]

c / r = 7 / 2

[mellow oyster]

you have the ratios for c and r

ez

[velvet cove]

wait how do u get this

oh i see

i think

like sara has 8/17 of the whole thing

ye

[mellow oyster]

yeah

though it was pointed out that my solution isn't too efficient

though you should still get the right answer

[velvet cove]

what was wrong about this

[strange crater]

your ratio is wrong

it is 2c = 7r

[rough hound]

Actually 8k = s, 7k = c, and 2k =r, saying 8ks = 7kc = 2kr, is saying s^2=c^2=r^2, which is wrong. The compound ratio 8:7:2 is where the gcd(8,7,2)=1, which means it's relatively prime, so we can multiply them by a constant which equals s, c and r, now it's given 2k=7k-60 or k = 12, to get sara, just plug the value s =8k = 8.12 = 96

[wraith python]

The ratio is compared to some arbitrary unit A; S:C:R = 8A:7A:2A

You know that Ruby, 2A, has $60 less than Cathy, 7A.

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[glacial sparrow]

i got the general formula of the height as tn=1/2*n^2+4n+8, but how do i answer 3.2 with that info?

[strange crater]

what is "n" in your formula?

and what is "tn"

[glacial sparrow]

so Tn is just the general term where 1 is the first number of the heights

so you'd sub in 1 into that formula to get 12,5; then sub in 2 to get 18 and so on

[narrow saffron]

isn't that just the answer

[strange crater]

ok. I guess I'd let h = at^2 + bt + c

plug in 3 of the pairs that you have, and solve for a b and c

I'm not sure if there's a simpler way to do it

I'm also assuming it is quadratic. it's certainly not linear

[glacial sparrow]

yes it is

[glacial sparrow]

ill try solve htat

ok wait im not sure I understand

[rough hound]

n = t - 1

plug the value of n in your formula

[glacial sparrow]

so i just sub that in everywhere n is ?

[rough hound]

yes

but do you understand how I got n = t -1 ?

[glacial sparrow]

wait lemme try to see if it works

cuz i see how it is that

[strange crater]

oh yeah that's much better

[glacial sparrow]

yeahh i see

my problem was how i denoted it
i didnt think to write t as it is, t
i used n in both cases and it made it wrong

thanks @rough hound

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[ebon shadow]

find the equation for this parabola (vertex or standard form) that models the path of a ball given:

y intercept = (0,1.93)
x intercept = (1.2, 0)
initial height of launch = 1.93
highest point in graph = 5.15

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[formal obsidian]

can someone explain the margin notes on the bottom left side?

[elder scaffold]

think of it as 1 divided by a large number

like 1/1000000

which is really small

[formal obsidian]

alright obviously its very smol

but like how does it relate to rational functions and asymptotes

actually im kinda getting it

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[craggy ravine]

Can someone explain why 0 and numbers 2,6,8 are in their own cases. And for case 2, even though we used 3 options 2,6,8, why the first number is 5 options

[west nebula]

0 cannot be a leading digit, so it must be treated separately

for case 2, the first digit can be any of 1,3,5,6,8 (5 options, excluding 0 and the last digit chosen)

[craggy ravine]

Wait, but there is 6 options aside from 0, so the first digit should be like 6-(1 zero and 3 even numbers) which would be 4

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[sharp frigate]

6x + 9y = 70
x + y = 5
Do I subtract 9y and y and make it dissapear? my friends notes say so
(the notes are a different solution cuz this is what our math teacher does)

[wintry vine]

There are many approaches you can take, one of which is rearranging both equations for x, which will yield:

x = (70 - 9y)/6
x = 5 - y

and then solve for y:

(70 - 9y)/6 = 5 - y
70 - 9y = 30 - 6y
40 = 3y
therefore y = 40/3
and x = 5 - 40/3 = -25/3

[marsh folio]

i mean easiest way to solve this would be 6x + 6y = 30, 3y = 40, y = 13,33, x = -8,33

[sharp frigate]

but how do you subtract it instead? which is the one in my friends notes... I'm asking for that specifically because I feel like the other methods would cost me time-

[wintry vine]

like Seid said

you can multiply the second equation to get

6x + 6y = 30

and the first will still be

6x + 9y = 70

You can subtract from either way:

(6x + 6y) - (6x + 9y) = 30 - 70

6x + 6y - 6x - 9y = -40

You notice the x's cancel, therefore

6y - 9y = -40,

and thus y = 4-/3

Just be careful of the signs

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[sharp frigate]

t h a n k you o m g

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[vast shale]

Can someone explain the différence between geometrical sequence and arithmetic sequence

[dark kiln]

they aren't similar enough that you would confuse them

[wintry vine]

arithmetic sequence

a sub1, a sub2, a sub3, ... a subn where asub k + 1 - asub k = constant

geometric sequence

a sub 1, a sub 2.. a sub n where a subk + 1 / a sub k = constant

example

[vast shale]

Than why why the sum from 1 to n of 4/3 is geometrical while the sum of 3 from 1 to n for example is arithmetic

[wintry vine]

arithmetic

1, 2, 3, 4 because common difference = 1

geometric

1, 2, 4, 8 because common quotient = 2

[outer warren]

as written both are arithmetic

[wintry vine]

i dont understand what you mean

are you talking about infinite sum convergence

[vast shale]

Well let me write you down the problem

[outer warren]

both $\sum_{i=1}^{n} \frac 43, \sum_{i=1}^{n} 3$ are arithmetic series

[twin meteor]

ℝαμΩℕωⅤ

[wintry vine]

well they are, with common difference 0

[outer warren]

they could also be viewed as geometric with common ratio of 1

[wintry vine]

oh yes thats true as well

[outer warren]

sequences,series where all the terms are the same have this property

otherwise they'll be arithmetic,geometric or neither

[wintry vine]

cool, um this irrelevant to this problem, but can you take a look at my channel, i got stuck doing some mathematical physics

[vast shale]

@outer warren @wintry vine

[outer warren]

do you have abetter pic? its illegible

[vast shale]

Good enough?

@outer warren still there?

[outer warren]

still pretty hard to read

what's the original problem

[vast shale]

The original problem is in French that's why

[vocal sleet]

@vast shale Has your question been resolved?

[vocal sleet]

@vast shale Has your question been resolved?

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[vast shale]

wtf to do

[wintry vine]

what have you tried

this is a nice question

[vast shale]

Idk

[wintry vine]

graph transformation is my hint for you

[finite swallow]

it's a simple transformation, think a bit of what you can replace x in f(x+1) with so that you get f(x)

[peak matrix]

hmm i got a different approach here, both would work tho and you can choose which one to use. Consider f((x+1)-1)

[finite swallow]

x -> x-1

[vocal sleet]

@vast shale Has your question been resolved?

[vast shale]

uh idk how still

[vast shale]

.close

[vocal sleet]

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[vocal sleet]

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Please don't occupy multiple help channels.

[ocean scaffold]

is that area

[vocal sleet]

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[distant dagger]

need help with finding the area in the shaded region

[distant dagger]

im trying to find the inverse of the function x^3 - 4x but ofc its not easy

am i shading the wrong region or something

[peak matrix]

no you are not

[peak matrix]

do you know calculus?

[distant dagger]

trying to figure it out yup

[peak matrix]

antiderivative of x^3 - 4x shouldnt be an issue then

[distant dagger]

yea but dont i have to find the area between that curve and the y axis like dont i have to find its inverse?

[peak matrix]

ah so thats the issue

you could

but it would be much much much more complicated

[distant dagger]

i dont think its possible to find the area otherwise unless im tripping

[distant dagger]

i mean is there another way

[peak matrix]

yes there is

what shapes do you see now?

[distant dagger]

ohhh

yea that makes it a little easier

but what about that little part we left out

how do we find its area

[peak matrix]

which one?

this one?

[distant dagger]

yes yes

OHHHH

i got it

i just subtract the curve from the line

[peak matrix]

ye

[distant dagger]

aight thanks man i get it now

[peak matrix]

yw

[distant dagger]

.close

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[vocal sleet]

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[vast shale]

its easy ik

[vast shale]

<@&286206848099549185>

[edgy gulch]

well do you notice that the y's have the same coefficient and different signs

what can you do with that

[vast shale]

ummm

combine?

+-=+

[edgy gulch]

you mean add both of the equations together?

[vast shale]

umm well u asked what can i do with that :c

but im confused with the x/2 and x/4

[edgy gulch]

dw about it for now

by combining you mean add right?

[vast shale]

yes

[edgy gulch]

ok great

+y - y is?

[vast shale]

😃

[vast shale]

+y

[edgy gulch]

no..

+y - y is y - y

y - y is?

[vast shale]

+y - y

[edgy gulch]

yeah what is it equal to?

[vast shale]

umm

=

[edgy gulch]

what happens if you subtract two values that are the same?

something like 1-1, 2-2

[vast shale]

0

thats 0

[edgy gulch]

so whats y - y?

[vast shale]

0

:o

[edgy gulch]

ok good

so that y is just gone now

[vast shale]

okay

[edgy gulch]

now what is x/2 + x/4

[vast shale]

:c

division and addition

[edgy gulch]

if you dont know

first factor the x out

so you have x(1/2 + 1/4)

whats 1/2 + 1/4?

[vast shale]

u mean what can i do?

2x + 4x?ç

[edgy gulch]

no

you know how to add fractions?

[vast shale]

yes

2/6

or 1/6

[edgy gulch]

no?

[vast shale]

i forgot

[edgy gulch]

adding fractions require you to have the same denominator

which is the number on the bottom

[vast shale]

yes

[edgy gulch]

so 1/2 + 1/4

how to make it have the same number on the bottom

[vast shale]

change

[edgy gulch]

change what

[vast shale]

1

with 2

[edgy gulch]

ok thats the top

what about the bottom

[vast shale]

or multiply smt

[edgy gulch]

yes thats it

[vast shale]

4x2?

[edgy gulch]

no need

[vast shale]

4x1

[edgy gulch]

you can multiply the 1/2 with 2 top and bottom

so you get 2/4

and hey you have the same number on the bottom

so you can just add it with 1/4

2/4 + 1/4?

[vast shale]

how 0,5 x 2 is 2/4

[solar walrus]

whats the question

[edgy gulch]

its not multiplied by 2

its multiplied by 2/2

[vast shale]

oh man im confused

[edgy gulch]

ok here

you have 1/2

you want the bottom number to become 4

so you need to multiply by something to make it 4

what can you multiply it with

[vast shale]

2?

[edgy gulch]

ok good

so you multiply that on the bottom

but you also have to multiply it on the top

[vast shale]

so its 1/4?

[edgy gulch]

you havent multiply 2 on the top

[vast shale]

2/4

[edgy gulch]

ok good

2/4 + 1/4?

[vast shale]

3/4

[edgy gulch]

ok good

and this was multiplied with x

so 3x/4 yes?

[vast shale]

ok

[edgy gulch]

ok last thing

whats 5 + 1

[vast shale]

6

[edgy gulch]

now solve it

[vast shale]

3/4 + 6?

[edgy gulch]

no

the 6 is on the other side of the equal sign

so you have 3x/4 = 6

[vast shale]

aaaaaaçç+

[edgy gulch]

can you solve this?

[vast shale]

4=6/3

4=2

[edgy gulch]

what?

[vast shale]

bc the 3x is multiplying?

[edgy gulch]

uh huh

[vast shale]

no

?

[edgy gulch]

3x/4 = 6

solve for x

how would you do this

[vast shale]

AAAAAAAAAAAAAAAAAAAAa

3x=4x6

3x=24

24/3

8

[edgy gulch]

ok good

now put that into any one of the original equations

to find y

and you're done

[vast shale]

tysm :D

[vocal sleet]

@vast shale Has your question been resolved?

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[low stream]

$d\left(x,:y\right)=min\left(\left|x-y\right|,:1\right)$

[twin meteor]

Dani ²

[low stream]

Hello, I was trying to prove that this metric satisfies the triangular inequality property
but according to the guide, I must analyze by case: when |x-y| is less than 1 and when it is greater than or equal to, I honestly have no idea how this proves the property. If anyone has another method, I would appreciate an explanation.

[vocal sleet]

@low stream Has your question been resolved?

[vocal sleet]

@low stream Has your question been resolved?

[vocal sleet]

@low stream Has your question been resolved?

[vocal sleet]

@low stream Has your question been resolved?

[low stream]

.close

[vocal sleet]

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[slim maple]

is the point at infinity equal to the mythical complex infinity?

[slim maple]

some links for this topic:
https://en.wikipedia.org/wiki/Point_at_infinity
https://en.wikipedia.org/wiki/Riemann_sphere

[vast shale]

yes it is

[slim maple]

pog

thx

.close

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[glacial meadow]

how does my work look?

did i get the problem correct?

the prompt was - solve for Sn

[half imp]

the mistake is in this step

20 is not 19-1

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@glacial meadow Has your question been resolved?

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[marble tinsel]

how would i correct portray f/g

[marble tinsel]

i cant use the 6th root of (3-x)

[vocal sleet]

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[glacial meadow]

.close

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[mighty quiver]

Hello, I have a homework question on sin/cos and the teacher hasn't posted the answer sheet can someone let me know if my solution is correct?

Q: "Write the equation of a periodic function that has an amplitude of 6, vertical shift of 2. phase shift of 2pi/3 and period of 8pi."
my answer: "6sin1/4(x-2pi/3) + 2"

[hushed pewter]

Looks good

[mighty quiver]

Where do you reccomend I can brushup on graphing them lol

[hushed pewter]

https://www.mathsisfun.com/algebra/amplitude-period-frequency-phase-shift.html

[mighty quiver]

I need some help graphing y = -2sin(x-pi/4)

Period: 2pi, Amplitude: 2, phase shift pi/4, vertical shift = 0, correct?

[vocal sleet]

@mighty quiver Has your question been resolved?

[vocal sleet]

@mighty quiver Has your question been resolved?

[red spruce]

if you want a tool to graph them, you can use a graphing calculator you have installed (Calculator for Windows for example) or a graphing calculator online (Desmos for example)

or an irl graphing calculator

[vocal sleet]

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[valid bison]

any way to solve this without L hopital?

[viral copper]

series

[orchid wren]

Hm

[valid bison]

huh

[viral copper]

seems to be the only way

[valid bison]

what series

[viral copper]

taylor polynomials

for arcsin and arctan

[valid bison]

is it the same as sin x

and tan x

[orchid wren]

No

[valid bison]

the x-x^3/3!

[valid bison]

oof

[viral copper]

each function has a different one

[orchid wren]

Its kinda difficult to memorize

[valid bison]

so L hopital is my best option for this question

[orchid wren]

Ofc

[viral copper]

rhats why you derive with an integral and the binomial theorem every time

[viral copper]

sure if you want to do it three times lmao

[valid bison]

two times right?

[viral copper]

theres an x^3 in the denominator it's probably 3 times

try it out though

[valid bison]

okay

[orchid wren]

Hey

There may be a alternate way im not sure

But you can write sin^-1 in tan ^-1 form and use the tan^-1 a - tan^-1 b formula

[viral copper]

yeah that could work nicely as well

[orchid wren]

If it doesnt still multiply numerator and deno with 2

And use 2tan^-1 x formula

[valid bison]

it was twice yeah x got canceled at last

[valid bison]

lemme try it

i like other ways rather than L hopital

[vocal sleet]

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[vast shale]

what pattern these cloucks followin

[vocal sleet]

Please don't occupy multiple help channels.

[vast shale]

mb

pls

help me brothars

and sistars

[silk osprey]

you have a channel open for the same question already

close one of them

[orchid wren]

1:50 > 2:00 > 2:20 >2:50 > 3:30

see

• 10 then +20then +30

So its next will be +40 and last will be +50

@vast shale ^

3:30 + 50 mins = 4:20

[vast shale]

fr 1 THANKS bro!

uhh htheres another question

[hushed pewter]

!nosols

[vocal sleet]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[orchid wren]

Wth bro ?

I havent given answer there !!

[hushed pewter]

How is this whole thing not the solution?

[orchid wren]

:'''')

[hushed pewter]

Seriously tho. Do not post solutions. Have OP work through the problems themself and just be a guide.

[orchid wren]

See bro

the sum of all angles of a quadrilateral is 360 °

[vast shale]

right... so...

wait

[orchid wren]

So use that

Add all angles and equate to 360°

You will find x

[vast shale]

2x - 5 + x + 35 + x + 90 = 360?

[orchid wren]

Yeahh

[vast shale]

hmmm lemme show working

[orchid wren]

Yeah

Ofc

[vast shale]

2x - 5 + x + 35 + x + 90 = 360
4x + 120 = 360
4x = 360 – 120
4x = 240
X = 240/4
X= 60

[orchid wren]

Correct 😉

[vast shale]

nice

but whats this

[orchid wren]

Now without explaining the solution i think

Ita difficult to answer

Assume the shapes as variables

Let circle = x , diamond = y , rectangle = z, a line like shape = k

x + y = y+z + k

Again z = 2k

[orchid wren]

From this x = z+k

Again x = 2k + k [ z=2k]

Or x = 3k

Got this ?

[vast shale]

yea

thank you guy

i dont think im cooked on this

thanks bro!!1

[orchid wren]

Welc

[vocal sleet]

@vast shale Has your question been resolved?

[vocal sleet]

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[wind python]

how do i start with this question?

i think i can sub 180-44

[sullen shoal]

direction angle is angle between vector and x axis right

[wind python]

yes

[sullen shoal]

wht is bearing?

[wind python]

i dont know

[sullen shoal]

yes

[wind python]

it may involve sub 360 or 180

but i cant find the pattern or indicator on which one i should subtract it with

[sullen shoal]

ohh idk abt bearing

sry cannt help with tht

[wind python]

thats okay

[wind python]

i just learned it

so bearing is like how far away it is from 180

180 starts at the y axis up to down and goes clock wise

[sullen shoal]

ohh

[wind python]

so if i want to find the bearing of the red vector, we look at the graph, remember that 180 is on top and starts on top but also is the bottom

until you reach 26, since 26 is NOT past 180, we can add 180+26

314

[sullen shoal]

direction of red vector is 90-26

[wind python]

idk the way i explain stuff is weird

[wind python]

why do you say that?

[sullen shoal]

extend the vector

angle with +ve y axis is 26

so +ve x axis is 90-26

and angle of blue vector is 180-44

angle is always measured anticlockwise starting from x axis to line

[wind python]

this is wrong

you do 270-26

and then for red, you do 180-44

omg bearing is so weird

like i said, vertically, both sides are 180 degrees

horizontally, it is 90 degrees

so see that the blue arrow comes from 90 degrees?

subtract 90-22

now look at the red arrow

red started from 180

so subtract 180-34

[south flare]

why you said like that

[south flare]

.

[wind python]

o

idk i watched a video and to my understanding, the guy said that 180 starts at y on the top

and goes clockwise

[south flare]

what is the definiton of bearing angle

[wind python]

im like oh okay

oh a bearing is an angle less than 90

[south flare]

interesting

[south flare]

but it says 146 here

[south flare]

.

[wind python]

where?

oh yes

idk the way the guy described it was weird to me

i did 180-34

[south flare]

well

we need to start from the top of the y-axis

northern part

and we need to move clockwise to the vector

and the angle between them is bearing angle of the vector

[south flare]

okay its definiton is like that

so it's like starting at exactly 12 on an analog clock

[wind python]

yeah something like that

man this sucks

i found out 360 is a thing on bearing

[south flare]

or you can say positive y axis

[south flare]

isn't it 0 though

lol

maybe 360 is another alternative

and the direction angle is

the classic way

counterclockwise from the positive x-axis

[wind python]

well

it says here

i didnt know

[south flare]

clockwise

not counter

for the red's bearing

[wind python]

so is that arrow going right or down?

[south flare]

like this

[wind python]

ohhh

318???

OHH

[south flare]

yes

360-42

[wind python]

wait so at the start, since y is 180, it is also 360?

is it 360 because the line is touching the y axis?

[south flare]

i don't know what you're talking about

[wind python]

okay so

you see that line i circled

[south flare]

yes

not the red one right?

[wind python]

so to me, it looks like that red arrow started at the y axis and is now going counterclockwise

[wind python]

not the red one

[south flare]

yes i realised

[south flare]

ok i understand

[wind python]

so to me, in my head, i thought okay i just need to subtract 42 from 180 since i seen in the video that 180 is the starter point

AND that the line i circled goes back to 180 y

BUT

when i put that as my answer 180-42, it was wrong

until i got the answer revealed and it gave me this

204

im like wth why is it 204?

and then 360 was used to get 204

this is so complicated omg im so sorry

so does the line i circled touching the y axis mean its 360?

[south flare]

it's no problem

[wind python]

cuz you told me and i saw the video of the bearing starting clockwise

and at 180

[south flare]

it's just there to represent the angle in between

nothing much

[wind python]

i wish i could get more practice questions to understand

oh

[south flare]

so as i said, the all thing you must do is starting from positive y axis and move clockwise until you come to the vector

to find bearing angle

or 360-42 is the practical way to do this

[wind python]

So 360- everything?

[south flare]

no

From the number 360, you subtract the part you do not cross as you go clockwise

[wind python]

the part?

[south flare]

like we don't care about this part i painted

so we take it out

[wind python]

right

the thing is

[south flare]

the thing is this

[wind python]

i understand that

[south flare]

we exclude the 42° angle

[wind python]

the thing is

what if we had something like this

and the angle was like 34

[south flare]

you will start from the same place again

from the above

[wind python]

what do you mean?

[south flare]

y axis

[wind python]

oh

180-34?

[south flare]

180°+34° = 214°

is the bearing angle in this situation

[wind python]

yes

[south flare]

it doesnt work for this

you must do 180+34 here

but

[wind python]

why?

[south flare]

avoid memorizing such things

trust me it will be better

lol

[south flare]

because

when you go clockwise from positive y axis

until you come down

[south flare]

like this

[wind python]

youre right, i tend to overthink alot of simple stuff

[south flare]

you did a 180° angle

[wind python]

yes

[south flare]

you'll be okay

[south flare]

and you said in the example

the angle is 34

like between y axis and the vector

we will have to go over the angle in between because we need to reach the vector

and that angle is 34°

so it is just 180°+34° = 214° here

[wind python]

hmm

okay i have a plan

i explain how i solve this

you tell me if i made an error or not

[south flare]

okay

sure

[wind python]

okok

so to my understanding if i go past 180 vector, i have to add?

[south flare]

i think there are 8 different situations here

so you can't do this kind of thing every time

i won't work all the time because the places where the angles arise are different

so just do the general rule for this

[south flare]

actually you did it

here

[south flare]

start from positive y axis

[wind python]

wait no i didnt, that green marker is my prof

[south flare]

oh okay

[south flare]

go clockwise (for bearing angle)

[wind python]

red bearing is 164

[south flare]

yes of course

[wind python]

blue bearing is 221?

[south flare]

no it's the direction

[wind python]

🫠

[south flare]

your prof wrote the direction

there

[wind python]

the bearing was also 221

[south flare]

S.P is starting point okay

for the bearing

[wind python]

yes

[south flare]

i painted

put your hand there..

and

follow the lines

that's it

[south flare]

my drawings

like green and orange one

when you do this

for green, your hand will have drawn an angle of 164°

for orange, it will be 229°

After understanding what angles the question is talking about, you just need to know how to find these angles

so it's important that you even see that the red vector's bearing agle is where i colored it with green

even if you don't know it's 164°

I'm just trying to show that that's the angle

because to find out 164

is more simplier

or 229

what i mean is, you need to find the "angle" before you find the angle "measure"

[wind python]

esat

[south flare]

yes

[wind python]

can you explain this on a call

[south flare]

i'm not available rn

[wind python]

thats okay

okay so 164

okay so the starting point was blue

[south flare]

yes

[wind python]

and now we need to find the red arrow angle

[south flare]

of course

[wind python]

okay question

so since this red arrow is NOT past the middle point of Y, that automatically means i can subtract 180-16??

[south flare]

what is the middle point of y

[wind python]

this

thats the negative side of y, i just call it middle point of y

[south flare]

oh okay

still didn't understand it..

[wind python]

oh you dont understand?

[south flare]

yes

[wind python]

oh im sorry

[south flare]

no you're fine

[wind python]

okay uhh give me a sec

okay

so side right is 1

side left is 2

[south flare]

sure

oh it is negative side of x actually

not y

are you talking about 2?

[wind python]

nono

hold up

okay better

so we know the start point is the Y axis, top

and it goes clockwise

NOW

if it goes past the pink color and enters the yellow color, would that mean i would have to add the angle number with 180?

[south flare]

yes

[wind python]

okay wait wait

[south flare]

but there is a condition

[wind python]

oh?

[south flare]

the angle we're talking about must come out of the negative part of the y-axis

[wind python]

how do we know its coming out of the negative part of the y axis

like this?

that says 28 btw

[south flare]

yes as you can see

it touches y axis

[wind python]

okay wait how do we know it touches the y axis?

is it because of the line here?

[south flare]

when you draw the curve indicating the angle

[south flare]

yeah

[wind python]

okay great

[south flare]

i'm talking about this little curve

for example, if it gave you 72 degree part

[south flare]

like here

sorry i saw it like 18

xD

[wind python]

lol its okay

[south flare]

that's why i said 72 but doesnt matter i guess

[wind python]

72 is touching the x axis but its negative

[south flare]

imagine there is no stated 18 degree angle

there is just 72

[south flare]

yes but you can't do 180+angle

like this

you need to find the part that touches the y-axis

with doing 90-72

[wind python]

oh yes

yeah i just have 90 implanted in my head

omg

[south flare]

like think of the green thing as an angle

you should do 180+green angle's measure

which is 90-72

180+90-72 = 180+18 = 198

[wind python]

wait waaa

[south flare]

to find green part we do this

[wind python]

180+90-72?

[south flare]

yes

but think of it like

180+(90-72)

[wind python]

can i do this for all bearing angles?

wait

no i cant

[south flare]

yes it is not the 90 always

if it is on the x axis

things are different

well, can you draw any arrow and find its bearing angle?

draw an arrow, especially one that you're not sure how to find

[wind python]

this is how i see it