help-42

[nimble parrot]

IT GOES UNDER 7

Wth

Q should be 10.5 but I got 8

Help 🙏

[remote mural]

HOWWWWWWWWWWW

[nimble parrot]

What did I do wrong 😭 😭

[remote mural]

WHY ARE WE USING PI/6

[nimble parrot]

So it isn’t a coincidence that the y intercept and minimum are both 7

Look

[remote mural]

@inner zealot bro come help this guy i beg you😭

[nimble parrot]

😭

Wya kookiemon 😀

Time for a different problem. Kookiemon if u come back help 🙏

@remote mural hello 👋

@remote mural so apparently u plug in -pi/3 instead of x=0

[calm coral]

@nimble parrot Has your question been resolved?

[calm coral]

@nimble parrot Has your question been resolved?

[inner zealot]

Yes?

[nimble parrot]

I did it with pi/6 instead and i got it wrong

5sin(x-pi/6) + 1.5 + q
5sin(0-pi/6) + 1.5 + q
5(-1/2) + 1.5 + q > 7
q > 8

Then when I plug in 8 the equation goes below the minimum (7)

[inner zealot]

One moment.

[inner zealot]

First determine what q should equal. You have already done that. You found it was 10.5.

The horizontal shift will not affect the maximum or minimum values of the function.

[nimble parrot]

But I somehow determined q was 8 when I used pi/6 😭

[inner zealot]

The ony thing affected, as it relates to your question, will be where the sine wave intercepts the y-axis.

The max and min of cosine and sine functions are +1 and -1 respectively.

[nimble parrot]

I wanted to do a problem that wasn’t a “coincidence” that the y intercept and minimum were the same so I switched the horizontal shift.

[inner zealot]

So you know the function

( g(x) = 5 \* (-1) + 1.5 +q ) > 7

[nimble parrot]

Ohh I got 5 * -1/2 when I used pi/6

Is it cus I don’t plug in 0 for x and have to plug in something else

[inner zealot]

One of the constraints given by the problem is that g(x) >= 7.

Because the entire function must be greater than or equal to 7, you have to consider the minimum possible value of 5 * sin(x - H).

Regardless of the horizontal shift, the minimum value of sin(x - H) will always be -1. Likewise, its maximum possible value will always be +1.

So you can infer that the minimum possible value of 5 * sin(x - H) will be -5.

[nimble parrot]

Oh ok

[inner zealot]

Substitute that into your original function and you get

g(x) = 5 \* sin(x - pi/6) + 1.5 + q

g(x) = 5 \* (-1) + 1.5 + q

[nimble parrot]

Ok I think u would plug in x as -pi/3

Based off the graphing calculator I think

Ok this probably not on the test 😭

[inner zealot]

https://www.geogebra.org/classic/ggnp2vt5

Well, it's good to at least understand this anyways. Take a look at that Geogebra file. The horizontal shift is set to increment at pi/3 units.

You can see how the y-intercept changes.

[nimble parrot]

Ooo

[nimble parrot]

Thank u so much

[inner zealot]

yw

[calm coral]

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[brisk pelican]

do you know law of cosines

[remote mural]

Let us assign the variable $x$ to the length of BA = length of CA\par
\text{}\par
$$8^2 = x^2 + x^2 - 2x^2(\cos{30^\circ})$$
$$64 = 2x^2 (1 - \cos{30^\circ})$$
$$64 = 2x^2 (1 - \frac{\sqrt{3}}{2})$$
$$\frac{64}{1 - \frac{\sqrt{3}}{2}} = 2x^2$$
$$x = \sqrt{\frac{32}{1 - \frac{\sqrt{3}}{2}}}$$

[potent lotus]

Emploice Muswashans

[remote mural]

Now that we have the value of x, we can substitute it in the second triangle when the value of $BC = 4$ and $\angle BAC = \alpha$

[potent lotus]

Emploice Muswashans

[remote mural]

$$4 = (\sqrt{\frac{32}{1 - \frac{\sqrt{3}}{2}}})^2 + (\sqrt{\frac{32}{1 - \frac{\sqrt{3}}{2}}})^2 - 2(\sqrt{\frac{32}{1 - \frac{\sqrt{3}}{2}}})(\sqrt{\frac{32}{1 - \frac{\sqrt{3}}{2}}})(\cos{\alpha})$$

[potent lotus]

Emploice Muswashans

[remote mural]

$$2 = {\frac{32}{1 - \frac{\sqrt{3}}{2}}} - 2({\frac{32}{1 - \frac{\sqrt{3}}{2}}})\cos{\alpha}$$

[potent lotus]

Emploice Muswashans

[calm coral]

@limpid mulch Has your question been resolved?

[limpid mulch]

.close

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[split escarp]

Hello, I'm stuck on a problem here. Basically, in this one example I was given in notes we're simplifying trig problems, and part of it goes from (3tan(theta))^2 and that simplifies to 1 + tan^2(theta) im really confused how they got there, I thought maybe they just expanded it out but I cant seem to get that to work

[split escarp]

can someone sit with me here and help me figure this out?

nevermind, i figuredi t out

.close

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[calm coral]

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[remote mural]

Can anyone create a simple velocity displacement word problem in which velocity function is cosx and displacement is sinx

[remote mural]

here displacement is 0 right?

[gusty steeple]

correct

[remote mural]

So it means while travelling at velocity at (0,1) to decreasing velocity till ( π/2, 0) I covered +a displacement and then further with negative velocity from (π/2, 0) to (π, -1)covered -a displacement
This making displacement 0

And both +a and -a covered in same amount of time

@gusty steeple let me know if I am right with this

[gusty steeple]

yeah

[remote mural]

Yeah as in I am right or you will let me know

[gusty steeple]

yeah ur right

[remote mural]

+a and-a covered in same time no?

[gusty steeple]

yeah you covered +a in pi/2 seconds and -a in pi/2 seconds aswell

[remote mural]

Oh thanks man
You saved my day my life
Love u

[gusty steeple]

all g

[remote mural]

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[calm coral]

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[exotic falcon]

I'm really not sure what the relationship is here, or what my sir is trying to point out to me. Does anyone have any ideas?

This is theorem 4.8 for context

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@exotic falcon Has your question been resolved?

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[wheat flax]

Hey, so I'm trying to understand rotation matrices in linear algebra

[wheat flax]

I have this graph to understand how the matrix is formed

We are trying to determine how to rotate vector v by angle $\theta$ relative to unit vector n. We use vector $v\perp$ and n as basis. A circle is also drawn on the upper left vertex of the right triangle that intercepts the upper red vector. Firstly, I don't understand why $\alpha$ is introduced in order to calculate a rotation by $\theta$. Secondly, In the lecture it is stated that $Rn(v\perp) = cos (\theta) v\perp + sin(\theta) (n\times v)$ (Rn = Rotation) I don't understand this part, is it the polar coordinates of the upper red vector?

[potent lotus]

froud

[wheat flax]

<@&286206848099549185>

[snow spire]

I couldn't figure out this yet, I'm still doing prealgebra 😉 , but this animation seems to relate to that circle: https://blogs.mathworks.com/cleve/files/compass.gif

[wheat flax]

Ah thank you, Indeed the matrix in the animation corresponds to a rotation matrix, however I'd still like to understand my graph 😄

[snow spire]

It's from this article: https://blogs.mathworks.com/cleve/2022/05/18/rotation-matrices/

[wheat flax]

Pretty interesting article, the thing is it doesn't explain how the rotation matrix is formed and that's what my lecture tries to show

<@&286206848099549185>

[snow spire]

There is good explanation here also, judging from the first part of it: https://en.wikipedia.org/wiki/Rotation_matrix

" One way to understand this is to say we have a vector at an angle 30° from the x axis, and we wish to rotate that angle by a further 45°. We simply need to compute the vector endpoint coordinates at 75°. "

[wheat flax]

Thank you, but this still does not help me concerning my understanding of the graph above.

.close

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[snow spire]

.reopen

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[snow spire]

.

I'm trying to understand this problem

[snow spire]

This

@wheat flax I think α is introduced because vector v is at an α angle to vector n, so α is the angle being rotated. The rotation is done by changing coordinates, or trigonometric values of the angle of the vector. α has certain (x, y) coordinates, or (cos α, sin α) on a unit circle, and to rotate it, we change these coordinates. I guess we use a matrix, because we care about the coordinates according to some n base axis, and not the x axis.

The formula stated in the lecture seems to be this formula for rotating angles:

"If x and y are the endpoint coordinates of a vector, where x is cosine and y is sine, then the above equations become the trigonometric summation angle formulae. Indeed, a rotation matrix can be seen as the trigonometric summation angle formulae in matrix form. One way to understand this is to say we have a vector at an angle 30° from the x axis, and we wish to rotate that angle by a further 45°. We simply need to compute the vector endpoint coordinates at 75°. "

I guess the orthogonal v, the base vector, and the circle are there so we can use the trigonometric measurements.

As if the right angle part was a unit circle

Something related that seems very interesting, is that trigonometry gives a circular perspective to thinking. I don't yet know how, but this is apparently something fascinating about trigonometry.

https://tenor.com/view/cat-space-infinity-cosmic-mobile-game-gif-24837603

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[lament cloud]

can someone help ?

[unkempt drift]

Split the triangle into two smaller congruent triangles

[regal bronze]

you know sin and cos?

[unkempt drift]

https://external-content.duckduckgo.com/iu/?u=https%3A%2F%2Fd138zd1ktt9iqe.cloudfront.net%2Fmedia%2Fseo_landing_files%2Farea-of-isosceles-triangle-formula-image-2-1612625132.png&f=1&nofb=1&ipt=e9f760f4972d508d04f438298f4a84451c186d7937a0f6e1e7ea5fb4042dc858&ipo=images

Now use Pythagoras to find the perpendicular height of O above the line PQ

[lament cloud]

we're not suppose to use that here though

[novel herald]

i think

u can use the chord length to find the angle

i might be wrong tho

and then that way you can find the area of the segment and sector and substract the sector from the segment

[lament cloud]

why do we need the angle ?

[novel herald]

so that you can find the area of the sector and segment

[rare stream]

can use the Pythagorean theorem to find the height of the triangle hence the area i think

[novel herald]

oh yes can do that too

bisect PQ

[remote mural]

||its 12cm²||

[calm coral]

@lament cloud Has your question been resolved?

[lament cloud]

like this?

and this for area?

[remote mural]

why is the 9 in a square root?

[lament cloud]

25-16?

[remote mural]

but yeah

[remote mural]

then its 9?

not square root of nine?

[lament cloud]

noo 3 right

[remote mural]

okay so

its p²=9
p=sqrt9

[lament cloud]

yeah?

[remote mural]

it is 3

the way you wrote it is wrong

if you put 9 in a square root you also put what its equal to in a square root

[lament cloud]

ohhhh

[remote mural]

yep

[remote mural]

so what you weote here is that p²=3

which is wrong

[remote mural]

dis correct yeh

[lament cloud]

alright thankss

i have another though u got time?

[remote mural]

sure

[remote mural]

ahh this is easy alright

ok so do you know what the lenghts in a rhombus are always uh what is the word again

perpendicular?

90°

[lament cloud]

right angle? huh

[remote mural]

yes!

[remote mural]

you can use that to do good ol uh

[lament cloud]

this part?

[remote mural]

yep

its right triangle

you can do uh

pythaoreum

pyhagoreom?

[lament cloud]

wait

[remote mural]

i fogor

[lament cloud]

which triangle

[remote mural]

one in the rhombus

[remote mural]

the rhombus

[lament cloud]

now what 😭

[calm coral]

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[remote mural]

do pythagoreom theorem

oh reopen

[lament cloud]

4 is hypotnuse?

[calm coral]

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[snow hare]

how can i prove $\sqrt{2} + \sqrt{3}$ is a irrational number

[potent lotus]

milanesa de pollo

[calm coral]

@snow hare Has your question been resolved?

[proper marlin]

setting it to be a rational number and squaring both sides, this problem is equivalent to proving sqrt6 is irrational

[snow hare]

how to prove sqrt 6 then

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[snow hare]

how to calculate the image of this sucka

[eternal shard]

Consider horizontal asymptotes, meaning x -> oo and x -> -oo

[snow hare]

what about them

can you elaborate

[calm coral]

@snow hare Has your question been resolved?

[obsidian elm]

asymptotes are tangents which touch a given curve at infinity

just find the limit of the given func. with x-->infinity

[snow hare]

3

[obsidian elm]

wait whose image are we required to find

what's the point

[obsidian elm]

which point is it? the point whose image has to be found?

[snow hare]

IDK thats just the homework I got assigned bro

I am not a teacher, sorry

[obsidian elm]

hmm ofc not

you could mark it as ,"not clearly explained"

[calm coral]

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[surreal patio]

By rotation of axis, eliminate xy term. Identify the conic and its elements

[surreal patio]

the answer key tells me that this should be a equation of hyperbola , but i cant seem to get rid of X or Y terms

where am i wrong?

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[sweet tartan]

Lost on how to solve this one, the blue paper is my working, can anyone help please?

[modern peak]

use rationalization

i think

fairly certain

[remote mural]

multiply the denominqtor with sqrt(20)+sqrt(12)

[modern peak]

yep, rationalization

no

wrong

[quick karma]

multiply and divide with the conjugate of the denominator

yeah

[modern peak]

multiply the entire expression

[remote mural]

wuh

9h forgor

yeh

[modern peak]

yea u forgot num

[remote mural]

i fogor

[potent lotus]

ren

[modern peak]

@sweet tartan

[sweet tartan]

And then what does that do?

Rationalization

[potent lotus]

ren

[sweet tartan]

ohhhh ok

[modern peak]

here, it simplifies the denom

[remote mural]

[sweet tartan]

Ok got the answer

ty for the help!

[remote mural]

[sweet tartan]

.close

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[snow hare]

how do i calculate this limit

.close

[calm coral]

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[polar frost]

Weird

Subs x = 2 in doesnt cause anything

[snow hare]

yeah

it’s straightforward

[calm coral]

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[vivid cypress]

can someone help me with this? the topic is all about the ungrounded data

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[noble cedar]

? not sure why this text is red

[pure kayak]

?

wdym

[pure kayak]

oh it means a message didnt sent

[noble cedar]

oh

https://tenor.com/view/flash-light-mode-funny-gif-26499136

sorry for dark theme discord users

[stoic oyster]

my eyes are blown

rip

[noble cedar]

https://tenor.com/view/tenset-10set-fox-cup-gif-4636644335299830446

i know that from PDP^-1 the D matrix will get that sweet sweet ^n to it

but I'm not sure how to word this answer, because there's 3 matrices involved

[ancient helm]

n-1 copies of P^-1 and P will cancel

[noble cedar]

we already went over that

so it will be P(D^n)P^-1

[ancient helm]

wdym then

[stoic oyster]

yeah

[noble cedar]

but in terms of how to word it

for WebWork

[ancient helm]

like what the 4 entries should be for M^n?

[noble cedar]

yeah

[ancient helm]

find the conjugation

then D^n is easy to evaluate

and do the multiplications

[noble cedar]

https://tenor.com/view/i-dont-know-what-that-means-i-have-no-clue-what-youre-talking-about-clueless-huh-what-gif-17107312

[ancient helm]

decompose M as PDP^-1

[stoic oyster]

yeah we already did that

you just have to compute the actual product P D^n P^(-1) now

[noble cedar]

OK

I'll try with ^n

-2(3)^n

does this equal -6^n?

no it does not

looks like i'll have to keep them seperate

@stoic oyster @ancient helm can I simplify this any further?

or this should be fine, for the final answer

[ancient helm]

you can drop the 1 coefficients

but assuming that's the right decomposition, looks fine

[stoic oyster]

time to pray webwork isn't an ass

[noble cedar]

😄

[stoic oyster]

yay

[noble cedar]

100% on all WebWork

WebWork is worth only 10% of the final grade

what a shit ton of work that was

exams are worth too damn much

many days and hours of work spent in WebWork, but it can all go down the toilet in a few hours depending on how well you do for the midterm and final exam

is this the way math classes should be? or do math students feel math courses are graded a bit unfairly? there is no curve and over half the class always fails, which makes it extra brutal

[stoic oyster]

had a class a bit like that once, there wasn't supposed to be a curve, but they added one in the end anyway cause the pass rate would have been completely abysmal otherwise

was a first year class, I think that explains most of it

the rest went much more smooth

anyway I gtg

[noble cedar]

thanks again 🙂

final exam is next week, i'll keep practicing

[stoic oyster]

maybe you could continue such a thread in #discussion or #math-discussion if you want

cya

[noble cedar]

later!

.close

[calm coral]

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[strange lichen]

avidwunna numba won

[calm coral]

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[warm hill]

hey can someone explain why to find the maximum angle,

$cos^2\theta = sin^2\theta$

[potent lotus]

tim

[warm hill]

i get spherical polar coordinates in 3 variables

so i understand to parametrise the angle between the z axis and the radius is the angle of the cone

just dont know where that part came from

[calm coral]

@warm hill Has your question been resolved?

[warm hill]

@helper

<@&286206848099549185>

[calm coral]

@warm hill Has your question been resolved?

[calm coral]

@warm hill Has your question been resolved?

[young jolt]

Can you show what equation is (1)?

[warm hill]

Sorry im in bed atm but itll be the rsinthetacosphi + rsinthetasinphi+costheta if im correct

Could be mixing up phi and theta though

[young jolt]

Okok

[warm hill]

Sorry are u gonna help cuz im thinking of sleeping in a min unless u do

[young jolt]

Do you know the geometric interpretation of $\theta$?

[potent lotus]

bondalton

[warm hill]

Yes its the angle between the z axis and the radius

So itd be the angle of the cone in the parameterisation

But im just confused on where the equation comes from

As in $cos^2\theta = sin^2\theta$

[potent lotus]

tim

[young jolt]

Just subtitute the values of $x,y$ and $z$ in spherical coordinates into the equation $z^2=x^2+y^2$

[potent lotus]

bondalton

[young jolt]

Use the fact that $\sin^2 x+\cos^2 x=1$

[warm hill]

Oh

[potent lotus]

bondalton

[warm hill]

Ohh right and the max would be pi/4

Right lol

Thanks

[young jolt]

Exactly!

[warm hill]

I can now sleep peacefully

[young jolt]

No problem : D

[warm hill]

.close

[calm coral]

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[unkempt drift]

Cause the biker who finished doesn't need to wait for the other biker

The race time stops only when the second person crosses the line

So it will be quicker if they finish together

Right so you mean the person who isn't on the bike is walking right?

And also the 25km here should be 35km right

So you mean that the two people switch at 14km and at 35km

Alright, so if we look at the part that is from 35km to the finish line

(Cause the two people need to be in the same position at 35km to switch who is on the bike)

Then the finishing time is given by the slower of the person who walks and the person who bikes

Wait so both of them can be biking at the same time

So what are the 14 and 35 then?

You mean 14 km/h and 35 km/h right

km is a distance not a speed

Ohhhh

It's very important

Interesting

Let's actually try solving the question then
So let the time the 1st person walks be a and the time the 1st person walks be b
And let the time the 2nd person walks be c and the time the 2nd person bikes be d

So then we have:
10a + 35b = 42
14c + 25d = 42

No, one walks at 10 km/h and the other one walks at 14 km/h

I know so I didn't say they didn't start at the same time

[unkempt drift]

I think it comes down to this

Because the person walking is always slower than the person biking

So no matter how fast the person is on the bike, they still have to wait for the person walking to finish

They do

The race time stops when the second person crosses the line

If you call the average speed of the first person A and the average speed of the second person B

Then the time the race finishes is going to come from the smaller of A and B

Like the worst case scenario is when A = 10 and B = 25

Or min(10, 25) = 10

We can have A = 35 and B = 14, so min(35, 14) = 14

That's not much better

Where are you stuck?

I've explained the same thing so many times to you

So one person walks at 10 km/h and the other person walks at 35 km/h

I'm saying if person A walks at 10 km/h for the entire race

And person B walks at 35 km/h the entire race

[calm coral]

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[regal glen]

Hello. Can you please help me find the coordinates of points A,B,C in an equilateral triangle. The origin of the coordinates is in the center of the triangle

[regal glen]

The lengths of the sides were not given, just "a"

[calm coral]

@regal glen Has your question been resolved?

[surreal patio]

can u just make 3 equations from the given info and solve respectively for the coordinates?

let A be (x1,y1) B be (x2 , y2) C be (x3 , y3)

actually nvm that would be very lengthy

maybe use midpoint formula for all 3 sides

[regal glen]

It's not very difficult to find the coordinates here, I think.

[surreal patio]

and as we know origin is the centre , OM = ON = OP

M , N , P being mid points of AB , AC , BC

[surreal patio]

is this from vectors or ?

also what are x and y

[regal glen]

Yes, using vectors

[regal glen]

In what sense?

[surreal patio]

like why have u drawn them?

the red lines

[regal glen]

Just arbitrary axes (I need to arrange them that way)

[surreal patio]

oh ok

hm

[regal glen]

I've drawn it, but I don't know what to do yet

[surreal patio]

maybe try find vector OA , OB and OC

by head to tail rule OA + OB should be equal to AB which is a

nvm

[regal glen]

Here is the task condition (Translated using google) if you want. It needs to be solved using vectors and with exactly such axes

[calm coral]

@regal glen Has your question been resolved?

[winter elbow]

@regal glen Actually, there is a very simple solution to this question

However, it requires some techniques regarding polar or matrix

Do you have either of them?

[regal glen]

Yes, I know a little bit about matrix.

[winter elbow]

Great

Have you found the coordinates of A?

@regal glen

Ping me when you’re back

[winter elbow]

It’s much faster to rotate the vector than combining vectors

[surreal patio]

hmm , havent learnt that or if i have then i forgot

[regal glen]

Sorry. No, I didn't find

[winter elbow]

It’s fine 🙂

[winter elbow]

What’s the length of OA?

State it with a

[regal glen]

I don't know. I only know what AB=BC=AC=a

[winter elbow]

Do you know cosine law?

[regal glen]

Yes, but how will he help me?

[winter elbow]

Do you know the angle of green?

[regal glen]

No, but why are you asking this? I can't answer the questions you're asking.

[winter elbow]

Those are crucial information for solving this question

[regal glen]

Here is the task condition

[winter elbow]

Do you know OA does divide angle BAC into two equal angles?

[winter elbow]

Yeah, I’m solving this question

[regal glen]

Then the angle BOA=120?

[winter elbow]

Yes

Cosine law

[regal glen]

Thanks, what's next? I don't understand yet how this will help to find the coordinates.

Oh, I see

[winter elbow]

After so, you’ll be able to get the coordinates of A

That’s the first part of the question

50% process remaining

I’ll be back in a few hours

Time to study

[regal glen]

Okay, thanks a lot.

[calm coral]

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[calm coral]

@indigo dawn Has your question been resolved?

[calm coral]

@indigo dawn Has your question been resolved?

[calm coral]

@indigo dawn Has your question been resolved?

[calm coral]

@indigo dawn Has your question been resolved?

[winged shadow]

Can u type the question

[calm coral]

@indigo dawn Has your question been resolved?

[calm coral]

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[hexed glen]

In figure ABCD, angles EFB and FEB are the same also IAEI=ICFI.
Jestify that...

a) triangles BAF and BCE are congruent.
b) figure DEBF has two pairs of the same sides.

[calm coral]

@hexed glen Has your question been resolved?

[hexed glen]

<@&286206848099549185>

[pale blade]

!status

[calm coral]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[calm coral]

@hexed glen Has your question been resolved?

[calm coral]

@hexed glen Has your question been resolved?

[calm coral]

@hexed glen Has your question been resolved?

[hexed glen]

No

[lunar ridge]

ok since the two angles are same

the triangle BEF is isosceles

so BE = BF

try solving further

[calm coral]

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[magic light]

Help with probobility question

[cloud nexus]

How many total permutations are there and how many of them have all in a line

[magic light]

12p12

[cloud nexus]

Otherwise known as 12!

But yeah

[magic light]

ok, now what?

[cloud nexus]

How many of those have them in a line

[magic light]

4?

[cloud nexus]

Not quite

[magic light]

wait wdym How many of those have them in a line

[cloud nexus]

How many permutations

Have them in a line

Like

All behind same type

Like what the question is asking for

[magic light]

3?

[cloud nexus]

How did you get that

[magic light]

im a bit confused on what you are asking soz

[cloud nexus]

How many permutations

Have the trucks of same type together

Literally what the question is asking for

[magic light]

im so stuck 😭

[calm coral]

@magic light Has your question been resolved?

[calm coral]

@magic light Has your question been resolved?

[calm coral]

@magic light Has your question been resolved?

[novel herald]

no way it’s <30 atar gang

I might have it

Do u have the answer I js want to check first before I share

[minor falcon]

I think the 4! is to encompass all the different types of materials

then each group of trucks with same materials together can be arranged in 3! ways

[magic light]

Why do we divide by 12p12?

[minor falcon]

because its the probability, so you have the restricted ways/total ways

12p12 is the total ways to arrange the trucks

[calm coral]

@magic light Has your question been resolved?

[calm coral]

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[boreal flume]

In the triangle ABC, the angle ∠BAC = 70◦
On side BC, a point D is chosen and on
side AC, a point E is chosen, so that |AB| = |AD| = |AE|. It turns out that |DE| = |EC|. Decide
all angles of triangle ABC.

[lethal dawn]

Is A=70 C=35 and B=70

[boreal flume]

How do you calculate those?

[lethal dawn]

A E D B the inner degrees total is 360

Gave aed and ade x adb and abd z

[boreal flume]

So is it basically just an equation?

[lethal dawn]

Sure? Dont understand what you mean by equation

[boreal flume]

Like you can somehow from knowing abe or smthg like that can make it into an equation?

Ima try doing it from 360

[lethal dawn]

Ight

[boreal flume]

How do I know B, D when I don't know C?

[lethal dawn]

Did you do 70+2x+2y=360?

[lethal dawn]

You will be able to find c after finding x+z

[boreal flume]

I don't even know how to start, which is x and y

[lethal dawn]

Oh

Sec kinda suck at explaining

[lethal dawn]

A

[boreal flume]

okay

2x + 2y + 70 = 360 = 2x + 2y =290?

[lethal dawn]

Yes

[lethal dawn]

Continue simplify it a bit more

[boreal flume]

x + y = 145

[lethal dawn]

Now that you know x+y

Look at EDB

[boreal flume]

yeah

[lethal dawn]

Would be way better if i wrote but meh

Do you know what was x and what was y

I mean where were they

[boreal flume]

ABD is x? and ADE = y?

BD , E D

[lethal dawn]

Yes

Now look at edb again

[boreal flume]

Yes, D must be 180 - C and B??

no

EDB, I don't know where to go from there

I just know that D = x + something

or minus

[lethal dawn]

If you have a piece of paper writing the angles should show you

E D is x Other d and B y and we know x+y=145

[lethal dawn]

Oh i think i explained it wrongly

Is the x seperate

And y

E and left D should be x since EA=DA

Right D and B should be both Y since AD=AB

[boreal flume]

So E and A should be the same as B and A?

and they're y

[lethal dawn]

Was it isocopes in english

[boreal flume]

I haven't learnt these stuff so I get cnonfused

[lethal dawn]

2 sides equal triangle

[boreal flume]

B =70

D right = 70?

[lethal dawn]

This is what i meant by x and y

Since they are isocopes

[boreal flume]

Okay hmm

Idk what do like do I start finding out B first or

Or it doesnt matter

[lethal dawn]

We know x+y is 145 so far

[boreal flume]

The D = x + y

Right?

[lethal dawn]

Yeah

[boreal flume]

So C + x + y =180

[lethal dawn]

Yes

[boreal flume]

C must be 25?

[lethal dawn]

C is 35 since 145+35=180

[boreal flume]

Oh lol

Yea

[lethal dawn]

What do you do from there

[boreal flume]

E =110?

[lethal dawn]

Yes

Then what?

[boreal flume]

X=70

[lethal dawn]

Yes

And the rest can be found with x=70

Oh made a mistake lol

[lethal dawn]

B should be 75

[boreal flume]

Oh yeah damnnn alr

I solved it

[lethal dawn]

Nice

[boreal flume]

The trick thing was i didnt know what x and y was

[lethal dawn]

Yeh it woulda been way better if i wrote it lol

[boreal flume]

Yeah ty best explaitikn i undderstamd

🫡king

.close

[calm coral]

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[supple rock]

21

[supple rock]

How do I solve the 21 questions I don't understand how to simplify the terms till n times

[polar frost]

U dont

Apply chain rule

[supple rock]

How

Can you solve and show

[polar frost]

I will now provide the solution

But consider the red part as inside function and blue as outside function

[supple rock]

So here 3/2 is outside derivative and after doing derivative of that I do derivatie of nx+1?

I don't really understand when it's written in f(x) etc if possible can you write in a simpler manner 😅😅

[polar frost]

U need to apply product rule for the inside part

The purpose of this question is to test ur understanding on product and chain rule

[supple rock]

So I will do product rule of which terms [(n-1){x}+1][nx+1]?

.close

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[hollow fractal]

i cant do this limit

(i only can use transformations )

[ancient helm]

does that say 5em(2x)?

[odd moon]

what's 5em

[hollow fractal]

no

its sen(2x)

@ancient helm @odd moon

[odd moon]

sine?

[hollow fractal]

sin

[hollow fractal]

yes

[calm coral]

@hollow fractal Has your question been resolved?

[calm coral]

@hollow fractal Has your question been resolved?

[calm coral]

@hollow fractal Has your question been resolved?

[calm coral]

@hollow fractal Has your question been resolved?

[calm coral]

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[vagrant crypt]

Hello

[vagrant crypt]

Merp

Uhhh

Can prime number be negative?

Or

Can negative number be a prime?

And also, if composite?

Please ping

[old falcon]

whats your definition of prime?

just start there

,w what is a prime number

[potent lotus]

Results provided by WolframAlpha

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

Requested by janniku

[old falcon]

so maybe another question is like...what useful things do we know that result from having this definition of primes and what happens if we relax them?

like, can we just say that primes are only integers having only themselves and 1 as a factor

[vagrant crypt]

Uhhhh integer do have negative...

But can a prime be it or vice versa?

Uhhh

[brazen elbow]

primes were defined in positive integers. usually if you extend this definition beyond positive integers youd have 2 outcomes: revolutionary discovery or disaster

newton finding a much faster way to compute pi is an example of the first outcome, 1+2+3+...=-1/12 is an example of the second outcome

[vagrant crypt]

Uhhh

[vagrant crypt]

I Google and it comes with two schemes

One being under positive integer

Other being under real number

[calm coral]

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[calm coral]

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[normal wasp]

Sin2a=2/|a| and a is not 0. What value can a take A)12. B)0.5 C)-1/3. D)1/pi. E) 2/pi

[jolly hill]

have you tried playing around with the equation yet

[normal wasp]

Not sure what to do with it

[jolly hill]

try rearranging it

[normal wasp]

In what way

[jolly hill]

just have a go at rearranging stuff

[normal wasp]

Sin2a is 2sinacosa

Right?

[jolly hill]

i wouldnt say this is a half angle question

also with the brackets, is it

|a|*sin(2a) = 2

[normal wasp]

Mhh ok

Soi have this whats next?

@jolly hill

[jolly hill]

do you know what amplitude means

[normal wasp]

Like of sin cos?

Like max and min values of the function

[jolly hill]

are we missing an x here

like is the question |a|*sin(2ax) = 2

if so then its very trivial

if not then you actually have to solve an extremely weird transcendental equation

[normal wasp]

The question is what value can it take out of thees

[jolly hill]

okay but did you miss typing an x

[normal wasp]

No no x

[jolly hill]

okay this is really weird then

because since theres no x

|a|*sin(2a) = 2

sin(2a) = 2/|a|

thats solving a hyperbola equals to sine

which is transcendental

[normal wasp]

The problems should be simple since its a 1 point one

Idk what you’re saying

[jolly hill]

none of the points in the answers they provided are given

are you able to screenshot the problem ?

[normal wasp]

So none are the answer?

I don’t think its ment to be done like that

[jolly hill]

me neither, which is why i asked whether theres an x

[normal wasp]

The question ask what value for the function is possible

And a is not 0

[jolly hill]

okay

[normal wasp]

So what do i do?

[jolly hill]

ping the helpers because either theres an x or either none of the answers are right

[normal wasp]

<@&286206848099549185>

.close

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[rare bane]

how to approach this family of equations questions..?

[lime saddle]

Try substituting $r^2 = x^2 + y^2$

[potent lotus]

Leander

[rare bane]

then.. it gives r = 5..

[lime saddle]

Yes, so the other circles have their origin on the edge of a circle whose radius is 5 and origin is (0,0). So if each circle has a radius of 3, what is the minimum distance from (0,0) and what is the maximum possible distance from (0,0) for any point in the set?

@rare bane

[rare bane]

2 and 8?

[lime saddle]

Correct, and since 2 <= r <= 8, what are the boundaries for r^2?

[rare bane]

between 4 and 64..

cool

[lime saddle]

That’s all I can think of unless they defined some property for the locus

[rare bane]

this seems an appropriate ans.. thank you sir..\

shall i close this channel with your permissionb..

[lime saddle]

Go right ahead

[rare bane]

.close

[calm coral]

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[rare bane]

how they confirm that the roots will be equal?

[rare bane]

@sly geyser

how did they find the locus (in the last step) with the help of

$k^2 = 2ah - 4a^2$

[potent lotus]

Mirthun1012

[calm coral]

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[topaz jasper]

can anyone help me out with this problem?

[calm coral]

@topaz jasper Has your question been resolved?

[topaz jasper]

<@&286206848099549185> help me out

[dusky fjord]

What do you need help with

[topaz jasper]

can you help me solve that one?

[topaz jasper]

what is linear interpolation?

@dusky fjord sorry for the ping please help me out

[void frost]

You can find the equation of the line through two points - also called linear interpolant, which you can use to e.g. find points between these two points

[topaz jasper]

i already got an answer but

can you check it for me if it's right?

[void frost]

Just post it here along with the original problem

[topaz jasper]

is that right?

[void frost]

What does Q_1, D_2, ... denote?

[topaz jasper]

what do you mean?

[topaz jasper]

Quartiles, Deciles, and Percentiles

[void frost]

Ah

[topaz jasper]

haven't heard of this topic before?

[void frost]

I don't know many of the English definitions of stuff not in the usual math major, I'm from Germany. Haven't had this either yet, so maybe wait for someone who had this

[topaz jasper]

oh but can you help me answer it?

@void frost please help me out 😭

<@&286206848099549185> can anyone help me please

[void frost]

You can probably check with a calculator

5 8 3 1 2 9 4 10 15 12 13 8 9 12 15 4 12 11 8 7 6 12 14 15 6 11 3 5 3 9 6 7 10 9 10 9 13 14 14 13

[topaz jasper]

i tried xd all of does calculator are different answers

[topaz jasper]

i already did arrange it

[void frost]

Arranged, it's 1, 2, 3, 3, 3, 4, 4, 5, 5, 6, 6, 6, 7, 7, 8, 8, 8, 9, 9, 9, 9, 9, 10, 10, 10, 11, 11, 12, 12, 12, 12, 13, 13, 13, 14, 14, 14, 15, 15, 15

[topaz jasper]

yeah i did that

but im still not sure about my answer

[topaz jasper]

look at here

.close

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[void frost]

How are you getting 6.25 for the first quartile

You have 6 + 0.25*0

[topaz jasper]

.reopen

[calm coral]

✅

[topaz jasper]

yeah

idk

[void frost]

That's 6 + 0

= 6

[topaz jasper]

is it wrong?

[void frost]

Then it also agrees with the calculator's first quartile

[topaz jasper]

but 6 + .25

[void frost]

6 + 0.25 *** 0**, no?

11th is 6, 10th is 6

[topaz jasper]

dang it is actually 6

[void frost]

So we have 10th + 0.25 * (6 - 6) = 10th + 0.25 * 0 = 10th

[topaz jasper]

is there another no. that is wrong?

[void frost]

Wait

[topaz jasper]

okey

[void frost]

The calculator says 5 for the second decile, let's check if there is a mistake in yours..

[topaz jasper]

really but its 5+.2(1)?

so the final answer in no.1 is 6?

[void frost]

For no. 1, yes

[void frost]

The 9th entry is 5, no?

[topaz jasper]

i think?

[void frost]

Your table is wrong

[void frost]

It's the 9th here

And then we again have + 0.2 * 0

Which is 0

So 5 + 0 = 5

Agreeing with the calculator

[topaz jasper]

there 5 9

[void frost]

?

[topaz jasper]

yeah i missed up in there

[void frost]

You messed up here

There are 3 3s

[topaz jasper]

i mean the 9 has 5 of does

yeah is there another no. that i messed up?

[void frost]

Well I think because of your table, you messed up a few more..

Because of the mistake in it, you were reading off some wrong values

E.g. the third quartile should be 12

[topaz jasper]

how about now?

[void frost]

2nd decile is 5

It should be 5 - 5 in the brackets

And so 5 + 0.2 * (5-5) = 5 + 0 = 5

[topaz jasper]

oh yeahmb

i think it was the table problem

is there another no. that i messed up?

[void frost]

The calculator says the 20th percentile is 5..

perhaps a table problem again

[void frost]

I'd really just do it again, neatly writing this down

Then all of these table issues should be resolved

[topaz jasper]

4 is messed up?

[void frost]

yeah

[topaz jasper]

should i consider it to make another paper so it wouldn't messed up?

wait is 5 also wrong?

[void frost]

Yeah, it's just one page of calculation anyways and you have the time, so why not

[topaz jasper]

its midnight xd

[void frost]

Oh, lol

Calculator says 9 for 5th decile

You say 9.5

Probably a table issue

[topaz jasper]

may i ask what calculator are you using?

[void frost]

https://www.calculatorsoup.com/calculators/statistics/quartile-calculator.php, https://www.calculatorsoup.com/calculators/statistics/percentile-calculator.php

Using 1, 2, 3, 3, 3, 4, 4, 5, 5, 6, 6, 6, 7, 7, 8, 8, 8, 9, 9, 9, 9, 9, 10, 10, 10, 11, 11, 12, 12, 12, 12, 13, 13, 13, 14, 14, 14, 15, 15, 15 as data

[topaz jasper]

does it solve all math problem no?

[void frost]

How would anything be able to do that

Of course not, lol

I just found these online right now, I've never used these before, just to check your answers

[topaz jasper]

fr?

dang I've searched around and haven't find like that thing

but anyways i thank you alot i finally solved this misery

thanks

[void frost]

np

[topaz jasper]

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[void frost]

Well you shouldn't rely on any calculator anways, in exams, you won't have it

Though not sure how it is in statistics exams, I'm a math major

[topaz jasper]

i dont actually use calculator alot

[topaz jasper]

they alow us to use it so why not

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[lethal dawn]

Heyo

[lethal dawn]

Here i saw that ecb=dec

And Found DC as 8

Dont know how to continue though

Nvm

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[barren urchin]

please help me with part c, in the textbook it says 2^n is not a sol but my answer says it is. what did i do wrong?

[barren urchin]

^ my sol

[potent igloo]

What is 4-4?

[barren urchin]

oh fk

i dont know why i factored out 2^n

mb

thank you @potent igloo

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[brazen agate]

can some one help with taking a common factor for this equation to make it (1 + tan^(2) x) ( 1+ tan^(2) y)

[teal drift]

You already have it in orange

So:
[1+t²(x)] + t²(y)•[1 + t²(x)]

[brazen agate]

my book saidto take (1 + tan ^(2) x) idk its maybe a mistake i have bee ntrying to figger out for like 40m thanks i understand it now 🙏🙏

[teal drift]

Which is the same as:
[1+t²(x)]•[1 + t²(y)]

[teal drift]

It's equivalent, you can do it both ways

You still get the same factorisation

[brazen agate]

i am trting to solve this dy/dx = (1 + t^2(x) + t^2(y) + t^2(x) t^2(y)

[wraith robin]

Factorise and separate the variables and then integrate

[brazen agate]

ok thank you

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@brazen agate Has your question been resolved?

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[timber bison]

answer after rounding: 2.42 x 10^4

problem: HOW?!?! I'm mega lost on what to do

[untold compass]

the three triangles with the dotted lines are all right triangles with 74.3 as the diagonal, so you can divide by sqrt(2) for the shorter side

you can either use the formula 1/3*base*height (volume for any 3d shape between a base and a point) or 1/6*cube volume

[timber bison]

can you show me how I would plug in for the first formula?

I understood the divide by sqrt(2) part

[untold compass]

so 74.3/sqrt(2) is the side length, I'll just call it x

base is 1/2*x^2 since it's a triangle, and height is x

so 1/3*1/2*x^2*x

which gives that 2.42x10^4

[timber bison]

ohh the area of the base

I get it now

I'll watch some videos on that formula, ty for the help

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[candid glade]

using reimann sums?

[candid glade]

How to do it pls

[grand glacier]

Have you seen the taylor series for ln(1+x)?

[slender tide]

Help

[candid glade]

no

[candid glade]

im stuck here

[calm coral]

@candid glade Has your question been resolved?

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@candid glade Has your question been resolved?

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@candid glade Has your question been resolved?

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[versed pelican]

Can someone help me solve this

If 0 is positive

Would it be 60 degrees

because 2/6 left = 1/3

Or am i just completely wrong

It would go counter clock wise

yeah I have no idea

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@versed pelican Has your question been resolved?

[versed pelican]

Help please

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[upper pilot]

Hey