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[livid sage]

i.e. ab + bc + cd + ad - ad

[raw jetty]

er

oh right

(a+c)(b+d)-ad

[livid sage]

indeed

[raw jetty]

would it be possible to assume a+c=b+d=17?

[livid sage]

you have the simultaneous goals of maximizing (a + c)(b + d) while minimizing ad, yes?

[raw jetty]

yea

[livid sage]

this is an appropriate assumption then

[raw jetty]

ok

so 289-ad

cant ad be like 1 or smth where a=d=1

[livid sage]

that is indeed correct

[raw jetty]

so is the answer 288?

[livid sage]

i would say so yes

[raw jetty]

ok

ty

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[livid sage]

yw

[gritty bramble]

there was also another method

lagrange's multiplier i think

[livid sage]

is cumbersome and not very accessible at this level

not very applicable either because a, b, c, d are specifically natural numbers

[gritty bramble]

idk i remember doing it

and it worked

[livid sage]

if it worked then you made additional logical leaps

[gritty bramble]

ig

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[umbral tendon]

hiii! do i first do the exponents and then multiply 4a^2 into it or do i mulitply first into the parentheses and then convert the number to negative?

[prisma ravine]

id take the (4a)^-1 as 1/4a first

that makes it way easier

generally try to get everything that belongs on the numerator on the numerator, and vice versa

[alpine sable]

which standard math is this???????????

[prisma ravine]

8th or 9th

maybe 10th

its not that hard

[prisma ravine]

the (a^2)^-1 goes to the numerator asw

[umbral tendon]

so it doesnt matter what i do first?

[prisma ravine]

if you do it correctly, no

but do what you find easiest

i find it easiest to get everything where it belongs first

[umbral tendon]

okay thank you, i will try and see how it goes!

[prisma ravine]

no worries

all correct methods should get you to the right solution, its just that you dont wanna do more work than you need to

[umbral tendon]

thank u!

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[alpine sable]

exact grade?

8th 9th or 10th?

.

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[dull phoenix]

any idea on how to solve 15?

The standard equation is ax + by +cz = d right?

[small lance]

Can you take the question properly?

[dull phoenix]

a new picture?

[small lance]

Certainly

[gritty bramble]

$\text{find the general equation and a vector equation of the plane that passes through P(1,2,4), Q(1,-1,6) and R(1,4,8)}$

[ocean seal]

Astar777

[small lance]

Notice that with two points, you can determine a line

Do you agree?

[dull phoenix]

yes

I was going to set Xo to P

and then get V1 and V2

by doing Q-P

and R-P

[small lance]

What do they represent?

[dull phoenix]

Xo is the vector from origo to the point P

and V1 and V2 is the distance between P to Q, and P to R

[small lance]

what solution did you get after these move?

[dull phoenix]

(x1,x2,x3) = (1,2,4) + s(0,-3,2) + t(0,2,4)

s,t are part of the real numbers

this answer is wrong

according to the answer sheet

[small lance]

yk, you can find the exact normal vector to the plane

I'm referring to the direction

[dull phoenix]

how would you do that?

dont know how to do it with 3 points

ax + by + cz = d right?

[small lance]

do you know how to find the vector that is perpenculiar to two lines?

[dull phoenix]

two lines

no

[small lance]

how about two vectors?

[dull phoenix]

I dont think so

two vectors, do you refer to v1 and v2?

[small lance]

yes

[dull phoenix]

do you use one point to get the normal vector of one line

and then set the number you get on d

for the other line?

[small lance]

2

[dull phoenix]

no I dont know how to get the normal

[small lance]

have you learnt outer multiplication

[dull phoenix]

scalars?

(3,2,1) dot (x-3, y-4, z-2)

this?

[dull phoenix]

= 0

[small lance]

it's inner multiplication

I'm referring to the other one

[dull phoenix]

probably

but dont know how to use it

[small lance]

hm

If A and B are two vectors

one of thier outer multi would look like

AXB

[dull phoenix]

A = 1,2,3, and B = 1,2,3 , does AXB = 1x1 + 2x2 + 3x3

is this the thing?

[small lance]

no

that's inner multi

[dull phoenix]

okey but you sure you have to solve it with outer multiplication?

[small lance]

it appears that you haven't learnt outer multi

[small lance]

yeah

I don't know any other ways to solve it ngl

wait for the next helper

[dull phoenix]

well we had a lecture on matrix product

but thats a chapter after the exercise im looking to complete

[dim garnet]

do you mean cross product?

[small lance]

yes

[dull phoenix]

this chapter that exercise 15 comes from should be about planes and lines

[dim garnet]

outer product is sth different though

[small lance]

I don't know how it is called in English

I just translated it directly from my native tongue to eng

[small lance]

I see

mb

[dim garnet]

do you know the cross product?

[dull phoenix]

ive heard of it

but havent used it at all

maybe I should watch some videos about the topic

and return to the problem later

[dim garnet]

I mean did you do it in class

[dull phoenix]

a lecture

all I have is notes

is this the thing?

[dim garnet]

Yeah

[dull phoenix]

I see

alright I'll try to solve it

thanks for the help!

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[smoky crane]

Hello. I am confused by your message. Could you clarify please?

[lone heart]

@smoky crane Has your question been resolved?

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@smoky crane Has your question been resolved?

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[molten heart]

Hi

[molten heart]

What is the answer to a

I got 2 different answers on it when i got a bot to do it

[sour mica]

7 hours and 30 min is not 7.3 hours, it's 7.5 hours

[molten heart]

I fixed it before

That was my first answer

But I fixed it

See

Am I right

[sour mica]

The numbers you use are correct, so if you did the products and the sum correctly, the answers should be right

[molten heart]

Ok

Are these ones correct?

[sour mica]

The second one is wrong, you calculated the installments only for the interest. You also have to consider the principal

[molten heart]

Can u show calculations for b

[gritty bramble]

total amount to pay = 30,000 + 27,000 = 57,000
total months to pay = 12 x 5 = 60

monthly payment = 57,000/60 = 950

[molten heart]

Oh ok thank you

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[brittle void]

the figure i'm getting is a right semicircle radius 2 and a left semicircle radius 1, but the figure my friends are getting is a right ring of outer radius 2 and inner radius 1, i don't know what im doing wrong

[brittle void]

wait okay so there shouldn't be anything in the left side because it falls out of theta range so just the right semicircle?

I'm so confused

<@&286206848099549185>

[pseudo ice]

I'd agree with you, I think, e.g. r = -1 and theta = 0 is equivalent to r = 1 and theta = pi

[brittle void]

but what about the range of theta?that's still valid right

even with the whole reflection equivalence thingy

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@brittle void Has your question been resolved?

[proper ginkgo]

WTF

[brittle void]

what😭

thats not a very comforting response

<@&286206848099549185>

[undone glacier]

what your problem?

[brittle void]

.

[undone glacier]

I see

well it is def confusing that the radius can be negative

what does this mean in this convention?

[brittle void]

reflection across the pole, ideally

but that falls out of the given range for theta

[undone glacier]

well that would mean that the theta input would also be reflected on the pole, no?

I'd imagine you get a big semi circle on the right and a smaller one on the left?

[brittle void]

that's what I was initially getting

[brittle void]

but then this

[undone glacier]

ah ok i see

i got this

ignore the covariant stuff above

[brittle void]

already done, no idea what that is

[brittle void]

what about theta though

[undone glacier]

or just dont plot it in x,y but r, theta coords

[brittle void]

or does it not matter because its defined only for x and not when negative values are reflected

[undone glacier]

if a circle section with a negative radius means that you invert it on the pole, then that means you would also invert all input coords in the pole, hence also invert the theta area input on the pole

so then you get the left hand side

of the image

[brittle void]

okay okay i think i understood

thank you so so much

[undone glacier]

this is also a solution

this rectangle is a solution

because you plot it in r, theta coords

[undone glacier]

sure, my please

[brittle void]

right that makes sense

but im pretty sure we were meant to use the polar and not the cartesian system in this question

[undone glacier]

gotcha haha

Also I'd think for a sec about the other alternative

namely the possiblity where only the absolute value of the radius is preserved, but the theta range is still the same

well in that case your region would be smaller buldge, so a smaller semi circle with r=1 on the right

but this is already covered in the other approach

so either way I think you're safe

[brittle void]

wait why is it not r=2

[undone glacier]

But ask your professor / teacher about what negative radius means in your polar conventions

[undone glacier]

no we're subdividing the one problem into two smaller ones

regions add

we got r=2 covered already

now question is r=1

[brittle void]

oh right my bad

[brittle void]

we've got a quiz tomorrow on this, so maybe a while after that😭

[undone glacier]

Sry my cam didn't rotate

but in that case its already taking up the same area so its good

[undone glacier]

ah ok

well good luck, you got this

any other way I can help?

[brittle void]

but using your logic and convention, our initial answer would be more appropriate right

[undone glacier]

if you have any notes of the class or lecture, I recommend going through that

else, ask chatgpt 🤣

[undone glacier]

yay glad i could help

[brittle void]

doing that rn

[brittle void]

yesss the saviour

[brittle void]

tysm again

[brittle void]

no i think you've pretty much clarified my doubts

thank you!

[undone glacier]

my please

[brittle void]

.close

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[wind sapphire]

I have a small question about negation and propositional logic

[latent ore]

Yall

Do yall have sparx maths

[prime badge]

wrong channel, you were too late bad luck

[latent ore]

Huh

What do you mean

[wind sapphire]

alright sorry for the sloppy hand writing

and whatever formalities I may have missed I'm kind of a Noob.

but this summs up my question up decently I guess (or not)

I'm just wondering why in the correction of my text book, x strictly less than y goes unchanged when negated

[tiny sky]

there are like different rules to negating statements that you could memorise

but i think in general it's easier to work intuitively rather than blindly applying rules to random symbols

[fast lichen]

lets say P is being a cat and Q is being an animal. Then P implies Q. Lets consider a chair. A chair is both not a cat and not an animal, so not P and not Q. This does not say anything about our original claim that P implies Q. To actually show it is false we need to find some cat that is somehow not an animal, so P and not Q.

[tiny sky]

the first statement says that "f is a decreasing function"

so if f weren't a decreasing function

then there'll be some x < y where f(x) >= f(y)

hope that made sense

[tiny sky]

but ^ in general

[wind sapphire]

makes sense

although my brain is being abit fussy with it on a intuitive sense I'm sure it will come into its own

[wind sapphire]

yeah 👍

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[distant eagle]

Hello
Does anyone know about for a mathematichs engineering project?
i couldn t find any project
can anybody help me?
(for secondary school in TÜRKİYE)
<@&286206848099549185>

[lone heart]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

@distant eagle Has your question been resolved?

[distant eagle]

Hello
Does anyone know about for a mathematichs engineering project?
i couldn t find any project
can anybody help me?
(for secondary school in TÜRKİYE)
<@&286206848099549185>

[eternal root]

I do

For a mathematics engineering project aimed at secondary school students in Türkiye, there are a few ideas that can be both educational and fun, while incorporating elements of engineering and mathematics. Here are some project suggestions:

1. Bridge Building and Analysis:
   Concept: Build different types of bridges (e.g., truss, beam, arch) from materials like popsicle sticks or straws and analyze their strength using mathematical principles.
   Math Aspect: Students can learn about geometry, forces, and load distribution. They can calculate the angles, areas, and forces acting on the structure.
   Software: Use basic simulations, such as in GeoGebra or engineering software, to predict which structure will hold the most weight.
2. Catapult Project:
   Concept: Build a small-scale catapult and test its efficiency by launching objects.
   Math Aspect: This involves learning about projectile motion, angles, and distances. Students can calculate the optimal angles for launching, range, and velocity using trigonometry and basic physics formulas.
3. Math in Coding (Using Scratch or Python):
   Concept: Use simple coding environments like Scratch or Python to model mathematical problems.
   Math Aspect: Create simulations or games based on math problems like solving equations, geometrical transformations, or logic puzzles.
4. Solar Energy Project:
   Concept: Design a simple solar panel system that can power a small fan or light bulb.
   Math Aspect: Students can learn about energy efficiency, measure angles for optimal sunlight capture, and calculate the energy output of the solar panel.
   5

[distant eagle]

thank you but its need to be a like; more math little enginering

[lone heart]

@distant eagle Has your question been resolved?

[distant eagle]

<@&286206848099549185> please..... help me

[versed sluice]

I think you should try building basic calculators in Python

It wouls be impressive since its a high schoool project

I could a send a video link where i Learnt how to build basic calcualtor if u want

[distant eagle]

it can be nice

ty

(sorry my english is so bad)

[distant eagle]

.

[versed sluice]

https://youtu.be/rfscVS0vtbw?si=wmnSfA8TGEZT2ei_

Just search the chapters for the calculator building part

[distant eagle]

Thank you so much

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[desert narwhal]

where is my fault

[desert narwhal]

5/7 is right

next one... i think no

[tacit arch]

do you need to use this to solve for an N so that the inequality holds for all n > N?

[desert narwhal]

yes..?

[lone heart]

@desert narwhal Has your question been resolved?

[tacit arch]

yes indeed

[tacit arch]

first step could be to observe that for n >= 1, 49n^2 - 21 < 49n^2

then bound 22 / (49n^2 - 21) < 22/(49N^2) < eps. now just solve for N

[desert narwhal]

soo... my numbers is right? okk

[tacit arch]

yea i can't find any errors

[desert narwhal]

thanksss

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[verbal gyro]

hey! I have a small question, if X ~ Normal(0, 1), is aX ~ Normal (0, a)?

[naive valley]

yes, as long as a is positive

[verbal gyro]

thanks!!

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[tacit arch]

🤔

[verbal gyro]

wait

.reopen

[lone heart]

✅

[tacit arch]

Variance(kX) = k^2 Variance(X)

[naive valley]

oh yea, it depends on what your notation Normal (0, a) means

i assumed (mean, stdev)

if it's (mean, variance) then you need a square

[verbal gyro]

yes, mean stdev

hahahah

ok! thanks again ^^

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[formal wraith]

Hey there

[formal wraith]

oups

sorry mate

yeah

Thanks 🙂

Can someone give me a hint with this?

I already notice that there is a secondary triangle, and a rectangle, but I'm still struggling finding x

I show you my drawing:

[placid zinc]

In derivative notation, can you identify what the question is asking for?

[formal wraith]

I don't know what derivative notation is, but he is asking about the shadow of the men, which is represented as x

[placid zinc]

Oh lol mb I was thinking the question was different

[formal wraith]

This is what I got so far

[placid zinc]

So yeah, the hint is good. Similar triangles comes to win here

[formal wraith]

the rest I did is nosense

But I don't know how to relate the big triangle with the big one

I thinked about

If I get the full hypotenuse and the subtract by the hypotenuse of the big triangle, ill get the hypotenuse of the little triangle, and so I can apply pythagoras theorem and find the X

but when I did the calculus, it just give me random stuff

[formal wraith]

Is this thinking correct?

[placid zinc]

You should use similar triangles

Trying to find a source that shows it off for this kind of triangle

[formal wraith]

what

I mean, I can see that T is a similar triangle to t

right?

ouh

I see

Give me this hint: Do I have to cut the rectangle into triangles

?

[placid zinc]

No

There's two triangles you can make here, they are similar. You want to start by identifying their side lengths

[formal wraith]

2 triangles?

Im not seeing it

You mean the 2 triangles that are already drawed no?

[placid zinc]

The whole thing is a triangle

The small triangle is a triangle

[formal wraith]

Yeah

I got that

their side lengths is: 4 and 2

and their base is 10 and x

[placid zinc]

The bigger triangle has side length 6

[formal wraith]

isnt it 4?

oh

yeah

the whole thing

has a side length of 6

[placid zinc]

These are the two triangles

[formal wraith]

yeah

so the sides are 6 and 2

right?

[placid zinc]

Yeah exactly

The base of the red one is a little tougher

[formal wraith]

okay

and the base of the bigger one is 10+x

meanwhile the base of the little is x

[placid zinc]

Can you finish the thought from here?

[formal wraith]

mmm

give me 2 mins to think about this

xd

But I don't really know how to search for X

For applying pythagoras I need the hypotenuse

And I don't have any values like the Area for making a equation

[atomic rapids]

Those are the 2 triangles

[formal wraith]

yeah

So far I understand

But from here I don't know how to continue

[atomic rapids]

This is de key to solve it, pythagoras

[formal wraith]

but solving for H gives me; sqrt(2+x)

[atomic rapids]

Yea

Gime a second

[placid zinc]

You should use similar triangles. Do you remember the formula similar triangles gives you?

[formal wraith]

nope

well

If they are similar triangles

I can see that the big triangle is 3 times the small triangle

right?

[placid zinc]

Yes that's true!

But basically,
6/2 = (x + 10) / x

[formal wraith]

but how you know they are similar triangles?

x = 5

never heard of this similar triangles, Im new to math , im with the stewart precalc book, and they insert me this exercises without seeing nothing about triangles lol

[placid zinc]

If you are missing some of the words they are using, it can be a good idea to look up the word or check your book for it.

[formal wraith]

Okay

thank you both

btw

whats the meta role you have?

its something about a meta competition or something?

[placid zinc]

I can participate in talks about the server itself, like suggestions to improve it, for example.

... I don't participate, but I can!

[formal wraith]

fine!

thank you so much

to both of you

❤️

Have a nice day

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[tidal kiln]

Stuck on part a of this, I don’t know if this is right and if it is where to go after this to put it into a form I know the formulas for

[lone heart]

@tidal kiln Has your question been resolved?

[tidal kiln]

<@&286206848099549185>

[livid sage]

(4r^2 - 1) is not equal to (r - 1)(4r + 1)

[tidal kiln]

Oh yeah, it’s too late to be doing maths LMAO

Still not sure on where to go next

Nvm I got it, method of differences

.close

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[willow dagger]

$xy^{\prime} - y = x^ny^2$

[ocean seal]

syecko

[willow dagger]

n is a natural number

please ping if replying

.close

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[lament ginkgo]

why is the initial value incorrect and how do i find the domain

[ocean whale]

My guess is that the starting position, is for n = 0, so S(0) = 1

[lament ginkgo]

still incorrect

[ocean whale]

What did you put?

[lament ginkgo]

@ocean whale

[ocean whale]

Oh I didn't realize that was what you typed in. I thought it was asking what S(1) was and you answered 1

Could be the format, not sure

[lament ginkgo]

ill just ask my teacher

do you know how to find the domain

[ocean whale]

No sorry

[lament ginkgo]

oh

i would think its negative infinity to positive infinity

but idk how to put that in

nevermind i figured out all the parts

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[cold cedar]

i need some help pls

[stray socket]

so 14 painters make 35% thats the info you've been given

[cold cedar]

yes

[stray socket]

what are you trying to find in terms of percentage then

[cold cedar]

uh how maany painters there are in total?

[stray socket]

yep so 100%

[cold cedar]

OH

so would it be

100 - 35

65 %

?

[stray socket]

umm, not quite what the question is looking for

its looking for what number of painters represents 100%

[cold cedar]

how many painters in all worked on the job

oh

[stray socket]

you already know that 14 painters represent 35%

[cold cedar]

yeah

[stray socket]

so do you know how to get from 35% to 100% apart from adding or subtracting

[cold cedar]

no i don't

[stray socket]

ahh ok

one way you could do this question is using ratios i.e 14:35 then trying to get to x:100

have you done ratios yet?

[cold cedar]

yeah we did ratios in the last unit

we just started doing percent and calculating percents

[stray socket]

yep this questions links back to ratios i believe

[cold cedar]

im in gr 9

btw

[cold cedar]

Oh

[stray socket]

all good

are you allowed a calculator?

[cold cedar]

yes we are

wait let me get my calculator

one sec

alr im back

[stray socket]

you dont need it for this question but it can be useful when converting between equivalent ratios

[cold cedar]

my teacher said we can use it on the test

[cold cedar]

Oh i see

[stray socket]

do you know what equivalent ratios are?

[cold cedar]

I think i do

[stray socket]

could you give me some equivalent ratios?

just simple ones

[cold cedar]

would it be like

1:2 2:4

kinda like aa fraction

[stray socket]

yes

[cold cedar]

or

Yeah!

[stray socket]

you got it perfect here

so now imagine you have 14:35 and x:100

x being the number you are trying to find

[cold cedar]

would you put it as a fraction

eg

14
__
35

=

x
__
100

[stray socket]

yes exactly

now just find x

this can work for all percentage problems like the one above where you dont have any one value

[cold cedar]

how would i find x

do i divide?

[stray socket]

so you are trying to x on its own

you already established that 14/35 = x/100

how could you get smth like (a number) = x

[cold cedar]

would x be 40

[stray socket]

yes

[cold cedar]

because

14 divided by 35

has to be the same as X divided by 100

[stray socket]

yep

[cold cedar]

thats the way i did it or is there another way (meaning my method wont work for every question)

or is the method i did it the way ur supposed to do it?

[stray socket]

you could just do 14/35 times 100 to get 40

[cold cedar]

ohhhh

so if the question to be like

7/24

i would js do

7/24 times 100

?

[stray socket]

yep you could to find the total number

7 painters represent 24 %

how many total painters are there would be the question

so 7/24 = x/100

x = 29 (rounded to nearest whole number)

[cold cedar]

ohhh

tysm

i have another question if u dont mind helping me

would this just be

$12000

[stray socket]

yep

[stray socket]

i could have told you this method from the start but its way better if you understand where it comes from hope that helps

[cold cedar]

yeah it was very helpfull

[stray socket]

no problem

[cold cedar]

this question was easy cus there was no decimals but what if the money amount was 1377

how would i find the answer

[stray socket]

using your calculator and then rounding to the nearest whole number

[cold cedar]

oh yeah

thanks

[cold cedar]

how would i do this question?

[stray socket]

same way as we did the first question

[cold cedar]

ohhhh

how do we know when to use the method?

[stray socket]

you know three out of the four values

[cold cedar]

ohh

thanaksaa

thanks*

[stray socket]

total is 100

[lone heart]

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[coarse ermine]

can someone help me use integration by parts to prove the reduction formule

[coarse ermine]

never done a problem like this

.close

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[coarse ermine]

.reopen

[lone heart]

✅

[coarse ermine]

i lied i thought i had it

i did integration by parts

and i got

tan^n(x) * x - ∫xsec^2(x) dx

[muted frost]

Please help this poor soul <@&286206848099549185>

[tight pier]

Hint: [ \tan^nx = \tan^{n-2}x \cdot \tan^2x \textbf{ and } \tan^2x + 1 = \sec^2x ]

[ocean seal]

bacc

[coarse ermine]

hm

[coarse ermine]

so in this first tan^nx i wanna

make it equal to tann^-2x * tan^2x

[tight pier]

yes and it is

Then applying the identity with secans

[ocean seal]

bacc

[tight pier]

You can split this now into two integrals

[tight pier]

Then you got the first part

[ocean seal]

bacc

[coarse ermine]

this is going of fwhat i had with integration by parts right

[tight pier]

The problem is if you straight away use integration by parts

you have no chance

[coarse ermine]

ohh

so u have to

[tight pier]

you will be stuck with that x

[coarse ermine]

i see

[tight pier]

so we take advatange of the fact that tan²x + 1 = sec²x for a nice sub

and also we know we need I_n-2

[tight pier]

that's why we did this

[coarse ermine]

i see now

[tight pier]

tan^(n-2)xtan^2x

[coarse ermine]

mhm

how da heck do those equal each other

[ocean seal]

bacc

[tight pier]

Use the hint

[coarse ermine]

ok so i wanna get rid of that sec

alr

[tight pier]

I assume you know integration techniques

[coarse ermine]

only learnt about u sub and integration by parts

trig sub too

[tight pier]

u sub

what about that

seeing the hint

[ocean seal]

bacc

[tight pier]

forgot the integral sign

[coarse ermine]

ohh

i can make

sec2x du

[tight pier]

almost

du = sec²x dx

[ocean seal]

bacc

[coarse ermine]

yup

[tight pier]

So what happens

to the secans term

[coarse ermine]

goes away cuz u sub

[ocean seal]

bacc

[tight pier]

Now

[coarse ermine]

yup

[tight pier]

is it already clearing

[coarse ermine]

u get the anti derivative

yea

[tight pier]

why they are equal

[coarse ermine]

and

[tight pier]

q.e.d.

[coarse ermine]

its n-1

cuz

[tight pier]

yea

trivially

also you remind me of @silver urchin

[coarse ermine]

goat

oh

waht the jheck

same pfp

LOL

.close

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[austere charm]

Question f

Okay so first 4*64

256

So what multiplies to 256 but adds to -32

[elder forge]

-16 and -16?

[austere charm]

Yes

So

[elder forge]

Mhm

[austere charm]

Oh shit

We were suppose to take out 4 first

Since it's a common factor

[elder forge]

U can do that

Yes

[austere charm]

So we should have done

4(x^2-8x+16)

Then 1*16 is 16

What what multiplies to 16 but adds to 4

That would be -4 and -4

So 4(x-4)(x-4)

4(x-4)^2

Correct?

@elder forge

[elder forge]

Yes

Correct

[austere charm]

@elder forge let me try m now so first we take our 3, 3(x^3 - 5x^2 + 4x)

So now what multiplies to 4 but adds to 5

-4 and -1

[elder forge]

Mhm lemme look

[austere charm]

So 3(x-4)(x-1)

[elder forge]

3x*

Not 3*

u can just factor out 3x from whole thing

Rest is correct

[austere charm]

My teacher didn't factor out 3

Was she wrong

This is what my teacher did

[elder forge]

Nop

It completely depends on you

Whether u remove the 3 or not

[austere charm]

Okay

[elder forge]

Ur teacher is right

And u are also right

[austere charm]

Let's remove 3 and x

[elder forge]

Mhm

[austere charm]

3x(x^2-5x+4)

[elder forge]

X²*

Yep

[austere charm]

3x(x-1)(x-4)

Or no?

[elder forge]

Yes

[austere charm]

Okay final practice let's do k

So first 20*-18 is

-360

So multiples to that and adds to 9

[elder forge]

Mhm

[austere charm]

24*-15

(X+24)(x-15)

?

This is were I'm confused because the teacher got soemthing completely different

@elder forge ?

[elder forge]

Just a sec

[elder forge]

Yes this is wrong

[austere charm]

Oh I'm suppost to do the grouping tirnolms way

Idk how to spell but uejs

Anyways thanks for the help I think I understand good enough

[elder forge]

Anytime!

[austere charm]

.close

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[gritty bramble]

do you know what happens to the graph of a function when you multiply it by -1?

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[kindred obsidian]

i dont understand the

f'(x)g(x) = g'(x)f(x)

[swift musk]

the derivative of (\frac{f(x)}{g(x)}) is [\frac{f'(x) g(x) - g'(x) f(x)}{g(x)^2}]
so the derivative is zero whenever (f'(x) g(x) - g'(x) f(x) = 0) (and (g(x) \neq 0)), which is the same thing as [f'(x) g(x) = g'(x) f(x)]

[ocean seal]

Invariance

[kindred obsidian]

oh i see

thanknyou

.close

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[unborn burrow]

I think I may be understanding this wrong

[unborn burrow]

would the basis for the null space be [1 0]

or I mean

[2 , 0] or [1, 0]

one of the left two most columns

since the right most one would be the column space?

but by the rank nullity theorem the nullity is 2

which is the dismension of the null space

so the dimension of the null space is 2?

but that doesn’t make much sense to me

because [2, 0] and [1, 0] are linearly dependent

so only one would be in the basis of the null space, which would mean it would have a dimension of 1

where am I going wrong?

[mortal trellis]

the columns of the rref have nothing to do with spanning the nullspace

the nullspace is a subspace of R^3 so it has length 3 vectors as elements

for the nullspace you have to ask yourself "for which x,y,z is x+2y+z=0"

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@unborn burrow Has your question been resolved?

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[unborn burrow]

is part a asking whether B and C are in the row space of A or if they are in the row space of each other?

[hushed locust]

the most reasonable interpretation would be whether they're in the row space of A

[unborn burrow]

ok ill try solving for that

is there some simple way to do a?

besides setting it equal to A and then solving for both systems

[lone heart]

@unborn burrow Has your question been resolved?

[mortal magnet]

simplest way is to first do (b), which will show you that ||A's rowspace spans the entirety of R^4||
from there, (a) is ||yes for both of them since A spans the whole space||
and (c) is ||[1,0,0,0], [0,1,0,0], [0,0,1,0], [0,0,0,1]||

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[unborn burrow]

.reopen

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✅

[unborn burrow]

I got something different for c

Is reducing the original matrix and then using the pivot rows as the basis the wrong way

[mortal magnet]

to confirm, what did you get for (b)

[mortal magnet]

thats not correct

try redoing (b)

oh wait

Ive redone (b) and yea its correct, sorry

so is the (c)

[unborn burrow]

Oh ok

[mortal magnet]

from there, you can check (a) by seeing if theyre in the rowspace of R

in other words, see if you can represent [4, 1, 2, 5] as some linear combination of [1, 0, 0, 1], [0, 1, 0, 5], and [0, 0, 1, -2]

youll notice that since only one of the rows has a 1 for three of the columns, youre forced into a particular calculation

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[native oracle]

How do I model this?

[native oracle]

Like the diagram

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[strange hearth]

how do i find a hamilton cycle on this graph?

[strange hearth]

in and out didn't work

[lone heart]

@strange hearth Has your question been resolved?

[strange hearth]

<@&286206848099549185>

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@strange hearth Has your question been resolved?

[lost relic]

Could you elaborate in and out ?

@strange hearth

[strange hearth]

like 0, 9, 10, 1, 3, 12, 11, 2...

two steps in the inner ring

two steps in the outer ring

[strange hearth]

i found a hamilton cycle for the same graph with 14 nodes and it seems pretty random

[lost relic]

hmm

[lost relic]

decompose the graph into three different layers.

or three different sets to be precise.

[strange hearth]

and what should each set be?

also what do i do with the three layers to make a hamilton cycle?

[lost relic]

How do I download this image

lmfao

[strange hearth]

click, right click, and save?

idk lol

discord is hard 😔

[mortal magnet]

can you draw the hamilton cycle on this image?

[strange hearth]

ok one sec

[mortal magnet]

if you can, use the same idea to draw the hamilton cycle for this one

[strange hearth]

it looks like it gets split into 3 parts and that doesn't really work on the other one?

[mortal magnet]

not correct

look closer

do you see that 0 has symmetrical connections?

[strange hearth]

omg

tried symmetry but missed two 😔

[strange hearth]

the crossover has to be 17, 9, 10 right?

[mortal magnet]

redo the connections starting from 7

7, 16, etc

[strange hearth]

7 17 15 6 misses 5 and 14
7 16 15 14 5 doesn't connect to 6

??

[strange hearth]

pretty sure it needs to include this?

[mortal magnet]

yes you do

now draw 7 to 16

what connection comes next after 7 16?

[strange hearth]

16 15

[mortal magnet]

what about 2?

[strange hearth]

2 11

[mortal magnet]

then?

[strange hearth]

16 15

11 12

[mortal magnet]

what comes next?

[strange hearth]

15 14

12 13

then

14 5

13 4

i think?

[mortal magnet]

then?

[strange hearth]

omg

holy

shoot

im actually dumb

thank you so much

[mortal magnet]

np

[strange hearth]

4 6 and 3 5 cant believe i missed it

🙏

[mortal magnet]

lets go

[strange hearth]

thank you!

.close

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[inner pawn]

find tan²x+2tan²y if 2sinxsiny+3cosy+6cosxsiny=7
Chat pls help 😭

[lone heart]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

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[junior kraken]

How can I start to write a research paper?

[tight pier]

Maybe you could do a research about that

[junior kraken]

💀

I really need help 🥹

[tight pier]

I wish I knew 🥹

[junior kraken]

For undergraduate students*

[junior kraken]

Np 🫂

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[tight thunder]

This is illegal or?

wait wrong

I Thought

in denominator would remain ^3

and upper part would turn to simply 1, right?

[paper cipher]

hi

[tight thunder]

and then also cancel the other (x-1)^2 in numerator then in denominator remains x-1 simply

[paper cipher]

Take (x-1) common from numerator

then you can cut them

[tight thunder]

hmmmm

not very understand

first pic was wrong denoinator btw

(x-1)^6 is in denonminator

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[severe edge]

Without using the graph, solve the following quadratic inequalities by interval notation.
2. x ^ 2 - 6x + 8 >= 0

[gritty bramble]

factor it

[gritty bramble]

,rotate

[gritty bramble]

now put x <2, 2<x<4, x>4 in the equation

and see where its >= 0

[severe edge]

I don't get it

My math teacher randomly gave this to us for extra points

[gritty bramble]

like put x=1 (which is less 2, so x<2) and check if x^2-6x+8 is >= 0

if yes then for the interval (- infinity, 4], the inequality is true

[gritty bramble]

similarily check for the other 2

[severe edge]

What's wrong with this

[gritty bramble]

,rotate

[gritty bramble]

nothing wrong

[severe edge]

Then why does my classmate say it's wrong?

[gritty bramble]

the inequality sign is $\ge$ right

[ocean seal]

Astar777

[severe edge]

What that mean

[gritty bramble]

its x^2 -6x + 8 >= 0

and not x^2 -6x + 8 > 0

so 2 and 4 are counted too

because you get zero at x=2 and x=4

[severe edge]

They say the test point is wrong and my internal notation lacking

[gritty bramble]

oh

the question in that image is $x^2 -2x - 8 \ge 0$ while you are solving for $x^2 -6x + 8 \ge 0$

[ocean seal]

Astar777

[severe edge]

Yup that's an example

[gritty bramble]

yeah so ofc the solution wont be same

did they say your test points for x^2 - 6x + 8 >= is wrong?

[severe edge]

Yea idk anymore

[severe edge]

Yup and my internal notation is lacking

[gritty bramble]

test points arent wrong

as for interval

what did u write

[severe edge]

Idk none

[severe edge]

This is my answer

[gritty bramble]

u didnt write the final intervals for which the inequality holds

$(-\infty, 2] \cup [4, \infty)$

[ocean seal]

Astar777

[severe edge]

Ohhh

Can I make the test point different?

@gritty bramble I'm Abt to sleep can you answer?

[gritty bramble]

yes

but

like

for x<=2

you can take x anythiny from - infinity to 2

for 2<x<4, anything between 2 and 4

[severe edge]

I'ma js sleep ika just send this and say it's right

[gritty bramble]

aight

!done

[lone heart]

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[lone heart]

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

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

[edgy mantle]

hello

[edgy mantle]

can someone tell me what im doing wrong?

im not sure what the correct blanks would be

[ancient shuttle]

= instead of ||?

[edgy mantle]

i should change parralel to congruent?

this is the final check i did multiple wrong tries

@ancient shuttle

[ancient shuttle]

The length of arc AB = The length of arc CD?

[edgy mantle]

≅?

@ancient shuttle is that the only thing i should change

anyone?

hello

@robust light

is this correct now

<@&286206848099549185> kindly help

[finite linden]

makes sense

kinda

circles wasn’t my best

[edgy mantle]

wrong