help-0

[surreal meadow]

yeah but it’s an end point

[vestal talon]

then itd go under closed endpoints

[surreal meadow]

if we’re talking about open end points (1,2) is the only one

the second curve extends a little further left

[vestal talon]

wdym a little further left

[vestal talon]

why is (4,6) not open

[surreal meadow]

(4,6) is included in the graph, and is also not an endpoint

f(4)=6 which is part of the function

[vestal talon]

then that line has no endpoint?

just a closed one

[surreal meadow]

not on the right

but on the left it does

[vestal talon]

i guess i dont understand the difference why the left does and the right doesnt

is it because <1

is an endpoint

[surreal meadow]

the left line has 2 endpoints

[vestal talon]

but x>=1 isnt?

[surreal meadow]

because it has a start and end

the right line only has 1 end point

because it starts at x=1

but doesn’t end

it just keeps going forever

[vestal talon]

I think I understand

so then the left line shouldnt have an arrow continuing it?

i added one

[surreal meadow]

like look at the lines on the graph, and literally think about “end points”

[surreal meadow]

it should not

because we only defined it for -2<=x<1

[vestal talon]

so the answers are open (1,2) closed (-2,5),(2,4)

[surreal meadow]

no

(1,2) and (-2,4) are right

but your other closed end point is wrong

look again at the function definition

when does the right line start?

[vestal talon]

(2,4)?

[surreal meadow]

no

[vestal talon]

🤯

[surreal meadow]

the second line begins at x=1

look at your function definition

[vestal talon]

my right line doesnt include x=1

[surreal meadow]

yeah that’s what i was talking about here

the graph is missing that extra section

[surreal meadow]

looks good

[vestal talon]

so (-2,5) and (1,3)

[surreal meadow]

yes

[vestal talon]

and open is (1,2)

[surreal meadow]

yes

[vestal talon]

good lord okay i get it now

okay heres my last problem

,rotate

[vestal talon]

I dont think ive ever seen how to get the vertex before

so i need to do -b/2a to get the x of the vertex?

not sure that makes sense if i dont know a

cant be right

[surreal meadow]

that’s the x value for you vertex, yeah

[vestal talon]

then thatd be -.2

[surreal meadow]

you do know a

a =-1

just the coefficient in front of the quadratic term

[vestal talon]

so then its actually 2

[surreal meadow]

(-2)/(2*-1)

[vestal talon]

i hate math

okay so its just positive 1

oh i guess I kept squaring it instead of 2a

duh

that would make the vertex (1,7)

[surreal meadow]

yep

how about the intercepts

[vestal talon]

Well I would guess I need to do something with the vertex

[surreal meadow]

not necessarily

[vestal talon]

y-y1=mx-x1?

[surreal meadow]

we want the intercepts of the x and y axis

[vestal talon]

then make x 0

get y

then make y 0 get x

[surreal meadow]

yeah take your quadratic and make x 0

yep

but you don’t really need to use the vertex

you can just use the quadratic you’ve been given

[vestal talon]

2/1 and 6?

[surreal meadow]

hm

are those your zeroes?

[vestal talon]

well if you take the quadratic and make x zero its just f(x)=6

[surreal meadow]

mhm that’s your y intercept

(0,6)

what are your x intercepts

[vestal talon]

i dont go back and plug in 6 for x right?

[surreal meadow]

no

for the x intercepts solve 0 = -x^2 + 2x + 6

we want y = 0

[vestal talon]

uhmmm

I either did this wildly wrong or x = +-SQRT.3

[surreal meadow]

yeah i don’t think that’s right

how did you get that

[vestal talon]

subtracted 6 divided it by two divided it by negative and then square rooted

but that left an extra x aside from being wrong

,rotate

[surreal meadow]

yeah so that doesn’t work

one way to solve this is the quadratic formula

[vestal talon]

yep okay was about to suggest

i think thats -b+-SQRTb^2-4(a)(c)/2(a)

[surreal meadow]

with parentheses but yes

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

[ocean seal]

maximo

[vestal talon]

so do I make 6 my x

[surreal meadow]

no

you’re solving for x

we know a b and c

so just plug them in

[vestal talon]

a = -1 b = 2 c = 6

[surreal meadow]

yes

[vestal talon]

k

-2+-SQRT28/-2

[surreal meadow]

ye

[vestal talon]

so (-2+SQRT28/-2,0),(-2-SQRT28/-2,0)

are my x intercepts

[surreal meadow]

yes

[vestal talon]

Okay I think I have a better understanding of both of these problems hopefully after I do the review for the exam theyll be cemented in my head

thank you maximo

3

[surreal meadow]

np

[vestal talon]

.close

[lone heart]

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[alpine sable]

Hey

[lone heart]

Please don't occupy multiple help channels.

[alpine sable]

I need to find df/dx and df/dy

Partial differentiation

But the r is confusing me

<@&286206848099549185>

[keen plinth]

use the chain rule

[lone heart]

@alpine sable Has your question been resolved?

[alpine sable]

How

@keen plinth

I know how to use the chain rule but dk how to apply it to this instance

[keen plinth]

d(sinr/r)/dy = d(sinr/r)/dr dr/dy

[lone heart]

@alpine sable Has your question been resolved?

[lone heart]

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[raw void]

<@&286206848099549185>

[lone heart]

@raw void Has your question been resolved?

[lone heart]

@raw void Has your question been resolved?

[lone heart]

@raw void Has your question been resolved?

[lone heart]

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[tired drum]

How to do this?

It says r=1 just so u know

[mortal trellis]

induction maybe?

[tired drum]

Yeah i know

But i dont know how to by induction

[mortal trellis]

what have you tried

[red kindle]

induction might be the way

[tired drum]

Idk where to go from there

<@&286206848099549185>

[mortal trellis]

so the induction hypothesis says that for some $n$ you have $\sum_{r=1}^n r^2 2^{n-r} = 2^{n+3}-2^{n+1}-n^2-4n-6$

[ocean seal]

Denascite

[mortal trellis]

now you want to prove that $\sum_{r=1}^{n+1} r^2 2^{n+1-r} = 2^{n+1+3}-2^{n+1+1}-(n+1)^2-4(n+1)-6$

[ocean seal]

Denascite

[tired drum]

Yeah

[mortal trellis]

after that I don't really know what you are doing with the k's

[tired drum]

Don’t u have to assume n=k is true

Then prove n=k+1 is true for induction

[mortal trellis]

the sum on the LHS is $1^2 \cdot 2^{n+1-1} + 2^2\cdot 2^{n+1-2} + 3^2\cdot 2^{n+1-3} + 4^2\cdot 2^{n+1-4}+\ldots+n^2\cdot 2^{n+1-n} + (n+1)^2\cdot 2^{n+1-(n+1)}$

[ocean seal]

Denascite

[mortal trellis]

yes?

[tired drum]

Yeah

[mortal trellis]

ok. lets just consider the first n terms for the moment

$\sum_{r=1}^n r^2 2^{n+1-r} = \sum_{r=1}^n r^2 \cdot 2\cdot 2^{n-r}$

yes?

[ocean seal]

Denascite

[tired drum]

Yeah

[vital vapor]

Hello?

[mortal trellis]

good what can we do now

[mortal trellis]

hello? read #❓how-to-get-help if you need help with something

[vital vapor]

I actually do

[tired drum]

Erm not too sure

[vital vapor]

How to do Long Division?

I need to know just incase we need to learn that

[tired drum]

Go to a different channel

This one is already occupied

[vital vapor]

Which one?

[tired drum]

This channel

[vital vapor]

Ah alr

[tired drum]

@mortal trellis what do I do?

[mortal trellis]

first we pull the 2 out of the sum

then we have $2\cdot \sum_{r=1}^n r^2 2^{n-r}$

[ocean seal]

Denascite

[tired drum]

Right

[mortal trellis]

this sum now seems familiar. what can we do with it

[tired drum]

Replace it

Substitute that sum for the earlier value right

[mortal trellis]

yes

[tired drum]

What then?

[mortal trellis]

so now we have $\sum_{r=1}^{n+1} r^2 2^{n+1-r} = 2\cdot \sum_{r=1}^n r^2 2^{n-r} + (n+1)^2 = 2\cdot \left( 2^{n+3}-2^{n+1}-n^2-4n-6\right) + (n+1)^2$

[tired drum]

Right

[ocean seal]

Denascite

[mortal trellis]

and we want to show that this is equal to $2^{n+1+3}-2^{n+1+1}-(n+1)^2-4(n+1)-6$

[ocean seal]

Denascite

[mortal trellis]

and from here on its essentially just some manipulations I think

[tired drum]

Right

But like what are the manipulations

@mortal trellis

Sorry for ping

[mortal trellis]

well for starters multiply everything out

then combine the n^2 and n terms

and then you see that you have the same stuff on both sides

[tired drum]

By ,multiply everything out

Do u mean just expanding the brackets

@mortal trellis

[mortal trellis]

yes

[tired drum]

I expanded everything out

It doesnt equal the RHS

[mortal trellis]

what did you get

[tired drum]

Where’d i screw up here?

[mortal trellis]

ok and what did you get on the RHS?

[tired drum]

What u showed here

[mortal trellis]

well you are also supposed to multiply that out

[tired drum]

Ohh u multiply the RHS out aswell?

[mortal trellis]

well that's the easiest way to do this

multiplying stuff out is most of the time easier than trying to factor it

when you write up the stuff you can do it backwards if you want

but for finding the solution path multiplying the RHS out is easier

[lone heart]

@tired drum Has your question been resolved?

[lone heart]

@tired drum Has your question been resolved?

[lone heart]

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[hollow zephyr]

I don't get this graph

there are two t axes?

[ornate condor]

lol

the

number of t's

is insane

[hollow zephyr]

that is correct sadly

causing confusion

[alpine sable]

t is variable
T is a constant

[hollow zephyr]

how should the axes be labelled?

t and T?

[alpine sable]

t along x-axis and f(t) along y-axis

[hollow zephyr]

ahh

[ornate condor]

LOL

pain

[lone heart]

@hollow zephyr Has your question been resolved?

[hollow zephyr]

then what

😦

[lone heart]

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[blazing grail]

Can y pow 2 be simplified so pow is removed? Just sqrt the right side?

y^2 = x^3+0*x - 7

[ancient meteor]

Yes

[blazing grail]

.close

[lone heart]

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[barren shore]

i tried 1+a x 5+b = 28

but it doesnt rlly make sense

since b needs to = 11-a

[gray isle]

where's your equation coming from

[barren shore]

what you mean?

we have to prove that in the shed, b = 11-a

[gray isle]

what through process led to you writing

1+a x 5+b = 28

[barren shore]

yeahj i thought of that

oh,

i dont konw tbh

[barren shore]

i thoguht that 1+a * 5+b wo uld equal the amount of tiles

but then i thought more, and knew it was wrong

in that 1x5 area, 5 tiles will fit

[gray isle]

tiles are being used for the walls of the barn

the sum of those red parts will be 28m

[barren shore]

oh

so the perimeter in this case

so 1 + a + 1 + a +5 + b + 5 + b?

= 28

?

[gray isle]

no

[barren shore]

oh

what will it be then

[gray isle]

watch where i marked the read

you don't count the 1m and 5m that come from the house

[barren shore]

oh

OH

its because its with the house

right

thanks for the help

.close

[lone heart]

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[honest warren]

what is the radius of a cone with a diameter of 2 and a slant height of 6?? I can't figure it out

[vale wigeon]

diameter = 2
radius = ?

either you misstated the question or there is extraneous info

[honest warren]

I know that the radius is half of a diamter

Let me see

It's essentially a "how many oranges are in the truck" question

and im trying to find the radius of this cone

[golden canyon]

and the diameter is 2

which means...

[honest warren]

Yes I understand that, but its not correct

[golden canyon]

you want to find the height, radius was the first part of the problem

did you use the pythagorean theorem before?

[honest warren]

Yeah, but first I need to find the radius. Do I need to times the diameter by pie and then 1/2 first or do I just split it?

[honest warren]

yes but I didn't really care for it

<@&286206848099549185>

[lone heart]

@honest warren Has your question been resolved?

[lone heart]

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[native nebula]

help

[native nebula]

how to factorise this?

how this happened?

what's the procedure?

[gray isle]

if you don't know the factorisation identity, you could approach it using completing the square

$$r^4 + r^2 + 1 = \underbrace{r^4 + 2r^2 + 1}_{(r^2+1)^2} - r^2$$
followed by the factorisation of a difference of two squares

[ocean seal]

ℝamonov

[native nebula]

thanx

.close

[lone heart]

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[coarse socket]

here as we can see the squares on the left keep on going till infinite, and the radius of the semicircle is given as r=4, then what will be sum pf perimeters of all such squares

[coarse socket]

i’m guessing it’d be some sort of geometric progression

[trim wagon]

hi hi hi

uhh did you first found the side lengths of squares

use pythagorean theorem

like let length 1 be l_1

so $l_1 ^2 + \left(\frac{l_1}2\right)^2=4^2$

[ocean seal]

Deep.

[trim wagon]

that gives us

l_1=..?

[coarse socket]

$8/ \sqrt 5$

[trim wagon]

yes

now use that to find l_2

[coarse socket]

but how

[trim wagon]

similarly

pythag

red is l_2

blue is 4/sqrt(5)

green is 4

apply pythagorean

[coarse socket]

oo

[ancient meteor]

Ouch that hurts my eyes

[coarse socket]

is l2 = 4/ root(5)?

[trim wagon]

,w x^2 + (x+ 8/sqrt5)^2 =4^2

yes good job

now

similarly

find l_3

[coarse socket]

this might take some time

wait

[trim wagon]

sure

[coarse socket]

haha

[trim wagon]

hmm 🤔

i wonder if it is useful

[ocean seal]

Results provided by WolframAlpha

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

Requested by Deep.#2005

[trim wagon]

doesn’t look useful

[ancient meteor]

F

[coarse socket]

[trim wagon]

i can’t think of something useful

sowwy

[coarse socket]

o

its alright

[lone heart]

@coarse socket Has your question been resolved?

[coarse socket]

.close

[lone heart]

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[ruby gale]

Hey, I’ve got an exam in like two hours and I have couple questions i’m not getting. Can anyone help? It’s mainly the asymptotes that I don’t understand

[keen pasture]

What is the definition of cot(3x)?

[ruby gale]

1/tan3x ?

[keen pasture]

Can you plug in the tan 3x definition?

[ruby gale]

like Tan3x = (3tanx – Tan2x)/ (1- 3 Tan2x) ?

[lone heart]

@ruby gale Has your question been resolved?

[worldly igloo]

Another way to look at this is to just see what values of x give you a “1/0”

[lone heart]

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[fair dome]

Write the equation of the line which intercepts with the parable y = (1/4)x^2 - 1 in the points A and B, that have X coords respectively of 0 and 4. After that calculate the length of the segment AB and the Area of the triangle ABO, O being the origin of the axis.

I don't even know where to start so this is kind of a problem...

[fair dome]

<@&286206848099549185>

<@&286206848099549185>

This is absurd

.close

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[quiet quartz]

What is an equation for this sentence "The complement of an angle is 15 more than twice the angle. Find the angle and its complement."

[valid smelt]

well complement is just what you have to add to the angle to get to 90, so it's 90-x
90-x = 15+2x i think?

[quiet quartz]

yeah i was just checking

i got 2x+15 = 90-x

oh shoot

this one is more complicaated

The supplement of an angle is 20 more than twice its complement

[valid smelt]

just use the same logic
supplement = 180-x
complement = 90-x

180-x = 20 + 2(90-x)

[quiet quartz]

u meant 20 right

ok and just to make sure, five times the complement is 6 more than twice the supplement

that would be 5(90-x)=6 + 2(180-x)

@valid smelt one more question, would "An angle is one-third of its supplement" just be x = 1/3(180-x)

[valid smelt]

[quiet quartz]

ok

[valid smelt]

you got it

[quiet quartz]

.close

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[lone heart]

@eager fractal Has your question been resolved?

[lone heart]

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[chrome vortex]

hi 🙂

[chrome vortex]

is there a way we could model it for the otherside of the pulley

other side*

i did T-5g-10=3(a)

i think u can

i just did it and i was off by 0.2

which must be my inaccuracy of rounding

they just did it the other way because it looked easier ig

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[gloomy sigil]

can this be simplified anymore?

[vague coral]

no

[gloomy sigil]

would i only be able to simplify it if both the top and bottom were switched

[vague coral]

?

[gloomy sigil]

if i had

2/1+3j

then would I be able to simplify it?

[vague coral]

hell naw

[gloomy sigil]

i should have probably specified that

j = sqrt(-1)

[vague coral]

I know

still no

[gloomy sigil]

in my textbook they are working out 2/1+j3 fine tho

by taking the complex conjugate of the complex number and multiplying both sides

i was just curious if that only worked if there is a complex number in the denominator

[vague coral]

thats if you dont want to have j on the denominator that you take the conjugate and blabla

in general we don't work with irrational things on the denominator

[gloomy sigil]

this is valid still tho no?

[vague coral]

it is

[gloomy sigil]

so you can divide it if the complex number is in the denominator

I was just curious if you can divide if the complex number is in the numerator (top)

nvm

got it

thanks herels

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[valid smelt]

are the generating functions for (1+x)^n and (1-x)^n the same?

[tacit arch]

did you calculate them?

[valid smelt]

i'm not exactly sure how to go about calculating them kek

i know that (1+x)^n expands into nC0, nC1, nC2....
does (1-x)^n just go into (-1)^i nCi

[mortal trellis]

yes

[valid smelt]

alright, tyty

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[visual narwhal]

Trying to wrap my head around this concept: If 70% of 5 is 3.5, and 30% of 3 is 0.9, then why does the sum (3.5 + 0.9 = 4.4) represent the average of 5+3?

[visual narwhal]

any help in describing this concept is much appreciated

(5+3)/2 = 4

but if we're taking a larger percentage of 5 then, the average skews towards a higher number of 4.4

I don't know if I am thinking about this correctly

[valid smelt]

it's a weighted average

if you wanted a true average every single number would have to contribute an equal amount

so 5 would be 50% and 3 would also be 50%

2.5 + 1.5 = 4 which is the average of 5 and 3

i don't know how your school does it but many classes and schools have their grades based of a weighted average

[visual narwhal]

I see, so when we're taking an "average" or "mean", we're just taking an equal contribution of each value

[valid smelt]

yeah

[valid smelt]

for example, tests might be 60%, quizzes 20%, homework 15%, participation 5%

if you got a 100 on participation and a 0 on everything else, it wouldn't make sense for your class average to be 25%

[visual narwhal]

I see

thank you so much for teaching me something new today

much appreciated!

.close

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[full niche]

need help finding the length something will poke up when going up a ramp

[tacit arch]

do you have the full problem statement and your attempt?

[full niche]

yeah drawing the issue rn, i need it to help my dad and its hard to get all of the details

@tacit arch

[tacit arch]

idk how anyone can help you without details

[full niche]

@tacit arch Not what I meant, here:

Its a clearance problem

he's trying to figure out if the vehicle will fit in the "box"

sorry I'm not exactly an artist

ah wtv nvm

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[rotund copper]

Good day, I'm trying to practice solving fractions, Im confused on where to start on this problem

[silent parcel]

Multiply the 2y/6x by y/y, this is the same as multiplying by 1

so,
3/(3xy) + (2y/6x)(y/y)
= 3/(3xy) + (2y^2)/6xy

Divide (2y^2)/6xy part by (2/2) or multiply top and bottom by 1/2... (half the top and bottom...
...
= 3/(3xy) + (y^2)/3xy

You know have a common denominator and it is the least common denominator as it can't be factored further.

** = (3+ y^2)/3xy

[rotund copper]

much appreciated

[silent parcel]

np

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[zinc trench]

Assignment: 15 (Solve with quadratic function)

A zinc plate is 4 meters long and 20 cm wide. Both sides are bent to create a gutter with a rectangular cross-section. Investigate at which dimensions the volume of the gutter is maximum.

I really need help solving this quadratic function, ive been staring at it for literally 3 hours lmao

[jovial burrow]

I might be to

Well A.=0.8msquare

yea, IDk -what I'm doing

[zinc trench]

xdd No worries man

Ive been looking at the assignment for so long

I just cannot seem to think abstractly enough to grasp the assignment lmao

While in general quadratic functions are ez

Im starting to think its a trick question lol

[jovial burrow]

J have no the has quadratics are even involved

no clue#

splitv it into 6 difference areas of equal lengh so u get a cube

nd maybe tATS THE MAX

[gray isle]

have you tried drawing a diagram

[zinc trench]

Uhm

Not really, do you think that would be necessary?

[gray isle]

yes

[zinc trench]

dont think ive ever made a diagram

or maybe its translated differently in my language

[gray isle]

picture

sketch

visual representation of the problem

(to get a better idea of what's happening, and help show how you're defining your variable(s))

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[alpine sable]

||x + 4| - 7| < 5

[alpine sable]

i got -2 < x < 8

but it was wrong

<@&286206848099549185>

[summer trail]

You got part of the solution but there is another interval of solutions

Consider x=-8

[alpine sable]

ohhhhh

how do i find that other interval?

[summer trail]

Ummm im really bad at explaining absolute values lol

but it would be similar to how you solved for your interval that you found

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@alpine sable Has your question been resolved?

[alpine sable]

lol thats fine

[lone heart]

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[livid torrent]

Is this the correct way to write 1/∞ = 0? Started learning about limits recently. Thanks.

lim(1/x) x → ∞ = 0

[ornate condor]

well

its cool u r learningn limits not

so

its more like

lim x → ∞ (1/x) = 0

u write the val x is appraoching first

not after

tho normally u write it below..

idk its weird typing

[wary stream]

It's more like $\lim_{x \to \infty} \frac{1}{x} = 0$

[ocean seal]

dldh06

[ornate condor]

ye that

[livid torrent]

How would I properly write it without access to LaTex? lim(x → ∞) 1/x = 0?

I felt like lim(1/x) x → ∞ = 0 made more sense because we're saying "the limit of/on one over x as x approaches infinity is zero". Right?

I just want to be able to write it properly with just Unicode for my notes lol

[ornate condor]

well

u write what it goes to first

but alternatively

u can just use english

and say as x approaches inf

before or after i guess

[wary stream]

In words, it's the limit as x approaches infinity of the function 1/x equals 0

[ornate condor]

as x approaches inf, 1/x approaches 0

alternatively

1/x appraoches 0, as x approaches inf

[livid torrent]

lim x → ∞ (1/x) = 0 this mehbeh?

I wish my notepad was latex compatible but it isn't

unicode needs to step their game up imo

And are there any ways to use limits that don't involve infinity?

[lone heart]

@livid torrent Has your question been resolved?

[livid torrent]

If x is approaching infinity, then is x infinitely approaching infinity?

[ornate condor]

well

yes

u can have

as x approaches 0

x = 0

smt like that

[ornate condor]

well

if x is approaching inf

aka

x->

x-> inf

then its just approaching inf

thrs no such thing as

infinitely approaching inf

to my knowledge anw

[livid torrent]

huh. okay.

I mean if it is constantly approaching but never reaching then it must be infinitely approaching.

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@livid torrent Has your question been resolved?

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[lost coral]

whats question b asking for

[abstract fractal]

Question a is asking you to use as many decimal places as your calculator shows

Question b is sig fig shit

[lost coral]

significant figure shit?

[abstract fractal]

Yep

Unless your calculator only has 2 decimal places, your answer for question a is wrong btw

[lost coral]

what if the decimal is irrational

[lost coral]

why?

[abstract fractal]

Because the question asks for the full calculator answer

[lost coral]

so i write 3.732291521

[abstract fractal]

Yes

Then for question b, you round to the correct number of sig figs

[lost coral]

what is the correct number of sig figs?

[abstract fractal]

What do you think it should be?

[lost coral]

im guessing rounding to 2 significant figures?

so 3.7

[abstract fractal]

Yep

[lost coral]

oh so its 1 decimal place?

so the correct number of sig figs is 1 decimal place?

what if its 39.8325123

then it would be 40 if its rounded to 2 sig fig

[abstract fractal]

I'd put 40. just to emphasize that the 0 is significant

[lost coral]

ok

thanks

.close

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[vale wigeon]

you can approach sup(S) arbitrarily closely from within S

[ocean seal]

marc

marc

[vale wigeon]

then you can take the sequence (1, 1, 1, 1, ...)

prove that for every ε > 0 there exists a member of S that is greater than sup(S) - ε

[ocean seal]

marc

[vale wigeon]

then sup(S) - ε would be an upper bound for S.

not necessarily its supremum.

but yes, correct.

now try using this fact to show that sup(S) can be approached from within S by means of a well-chosen epsilon.

for every n there exists a member of S that is greater than sup(S) - 1/n

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[warped arrow]

hi

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✅

[vale wigeon]

no

well you're kind of overthinking it but

if it wasn't true that for every ε>0 there exists a member of S greater than sup(S)-ε,

then there would exist an ε>0 such that no member of S is greater than sup(S)-ε,

thereby establishing sup(S)-ε as an upper bound for S,

thereby violating the definition of sup(S) as the least upper bound of S.

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[vale wigeon]

you do not

and i did not suggest that you take that

for every n there exists a member of S that is greater than sup(S) - 1/n

yes

yes it does

$\sup(S) - \frac{1}{n} < s_n \leq \sup(S)$

[ocean seal]

Ann

[vale wigeon]

you can pick any other sequence you want instead of 1/n so long as it approaches zero and squeeze thm works

now you are definitely overthinking it

[ocean seal]

marc

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[timber burrow]

is PEMDAS still like the meta or is there a better way to do order of operations I swear I heard about something else or I'm just really dumb

[lone heart]

@timber burrow Has your question been resolved?

[surreal meadow]

it's the most common practice. if you see a vague arithmetic expression, PEMDAS should be assumed

[timber burrow]

ok ty friend

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[night ruin]

are the constant monomials 5 and 10 alike (do they have the same main part?)

[night ruin]

I believe they are alike because if we consider that all integers are monomial because of : 5x° = 5*1 = 5 so that makes 5 a monomial called steady which is 0 degree (all steady monomials are 0 degree) , but my question is: Are all steady monomials alike , I believe they are not because one’s main part could be y° and others would be x° so yeah

[keen plinth]

alike

[pallid scarab]

Even if 5 could stand for 5x⁰y⁰j⁰

It is not important which variables the monomial 5 stand for

It's always gonna be 5

If you look at another example of monomial, 3x²y⁴ and -4x²y⁴

I could say that 3x²y⁴ is actually a 3x²y⁴h⁰j⁰k⁰l⁰ in disguise

But the variables h,j,k,l don’t matter

[keen plinth]

if your question is just can you add them together

then you can

5 + 10 is just 15

[pallid scarab]

Nono

[night ruin]

I know but the question asks does 10 and 5 have the same main part( main part are all the variables) and I think that’s not true because as you said it can be anything as long as all the variables turn into a nice “1”

[pallid scarab]

Alike is they share the same variables at the same exponents

[keen plinth]

[pallid scarab]

It's a weird term ik

[keen plinth]

alike is definitely not terminology thats used everywhere

[pallid scarab]

Yep

[keen plinth]

i mean

its the same as

can you add them together

and you can

5 + 10 is always 15

5xy^2 + 10xy^2 is always 15xy^2

[pallid scarab]

It's a treacherous way of saying it, really you can add any polynomials between them

But that's the """idea"""

xd

[night ruin]

Are they alike or not ? 5 and 10 , could and couldn’t be if you think it from both sides , first of all I believe they couldn’t because 5 could have a million options of different variables and 10 different ones , second option is that they do because all off their variables will turn into a 1 after all so their main part is 1 and they are alike! Which of the two options is right?

[keen plinth]

i think you will need to provide a definition of what it means to be alike

[night ruin]

Or simply are they alike or not ? 😂

[pallid scarab]

The true thing is that a monomial doesnt truly depend on any variables

It's gonna stay constant anyway

[keen plinth]

depending on how you define alike, you might say different things

[night ruin]

I found it , it’s called literal part in English

Does 5 and 10 have the same literal part ?

[keen plinth]

no no like

these terms arent exactly

commonly used terms

they arent like "polynomial" or like "linear function" or whatever

so to answer the question of whether or not they have the same "literal part"

you will have to define what it means to be a "literal part"

[pallid scarab]

Yep

[night ruin]

It’s the

[keen plinth]

and then ofc you will have to define what it means to be "the same"

for all intents and purposes

10 and 5 can be added together to be 15

[night ruin]

To be alike they need to have equal literal part

[keen plinth]

so you would probably say that they are "alike"

i would think that any other classification would be completely unuseful

like if you had

[keen plinth]

would just simplify it down to

[keen plinth]

which is obviously -19

so for all intents and purposes they can be added together

[pallid scarab]

their literal part is gonna be equal to 1 no matter what variables they depend on

[night ruin]

So that means that their literal parts are alike?

Right

The same

[pallid scarab]

x⁰t⁰(johncena)⁰= 1

[night ruin]

I think I got my answer

Thank you guys for helping me

[pallid scarab]

No problem :)

[night ruin]

Enjoy your day

/close

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[keen plinth]

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[slender carbon]

Hi! How do I rearrange the equation below to make v the subject of the formula?

[slender carbon]

I should also note that v <= v_t

[surreal meadow]

take e^ both sides

then you should be able to multiply by (v+v_t), distribute, move the vs to the same side, factor, and isolate the v

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[shell ore]

hi, how do i solve for x in -ax^2 - bx = c?

[languid bolt]

move everything to one side and then use the quadratic formula, you can also try multiplying both sides by -1, do factoring if you have the values, etc.

[shell ore]

oo

[languid bolt]

-ax^2 - bx = c
-ax^2 - bx - c = 0
-1(ax^2 + bx + c) = 0
ax^2 + bx + c = 0

something like this

[shell ore]

does the quadratic formula work if its -ax^2 -bx -cx -d = 0?

[ornate condor]

yes

[shell ore]

oo

[ornate condor]

u can rewrite it as

a'x^2+b'x+c'=0

where a' = -a

b' = -b-c

c' = -d

[languid bolt]

b' should be - (b + c) no?

[ornate condor]

its

exactly the same

[languid bolt]

oh wait

yeah my bad

[shell ore]

okay thanks

[ornate condor]

yes

[shell ore]

.close

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[shell ore]

.reopen

[lone heart]

✅

[shell ore]

oh i have another question

if its

[ornate condor]

another one

[shell ore]

ax^2 + bx + cx + a + b = 0

u can still use quadratic right

[ornate condor]

yes

its

the same

[shell ore]

oo

so confusing

[ornate condor]

what is

[shell ore]

okay i try it out now

thanks

[ornate condor]

yes

np

[shell ore]

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[ornate condor]

new emote to me

nvr seen it

[languid bolt]

If the highest power is 2, its a quadratic

[ornate condor]

btw

y not name urself

Helpers

[languid bolt]

Cuz like

[ornate condor]

mhmmm

[languid bolt]

"I need help"

[ornate condor]

LOL

icic

[languid bolt]

can get some help

[ornate condor]

good one

[shell ore]

[languid bolt]

im not getting help

Stuff like that

[ornate condor]

there

[languid bolt]

And i lost my active role so i cant change it anymore

[ornate condor]

OH WOT

U CAN LOSE UR ACTIVE ROLE??

[languid bolt]

Yeah, ofc

[ornate condor]

wow

go spam

[languid bolt]

If ur inactive

[ornate condor]

:O

i thought

it was total msgs

or similar

[languid bolt]

t!rank @ornate condor

[quasi bramble]

Viewing rank card • [ MarveI#4096 ] • 2

[languid bolt]

Wew u have so much messages

[ornate condor]

hm

LOL

[gentle veldt]

ok so 24x2+25x−47=(−8x−3)(ax−2)−53 butt

thenn you multiply (−8x−3) and (ax−2) using FOIL.

and end up getting 24x2+25x−47=−8ax2−3ax+16x+6−53

soo

[ornate condor]

LOL

#❓how-to-get-help

go to a new channel

[gentle veldt]

no 1 will answer me...

[ornate condor]

well

this channel

shld close in like

a min or so

[gentle veldt]

omg ok

[lone heart]

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[wide rapids]

normal distribution btw
yo what should I do in this situation? is this like asking for the middle 95%/99%

[vale wigeon]

yes

you are tested on your knowledge of the 68-95-99.7 rule

[wide rapids]

wait wait so what kind of answer should I input?

[vale wigeon]

...not clear from the picture as given here

can you take a pic of the entire worksheet

[wide rapids]

heres a recreation of it (i dont have a full pic)
we were tasked to put our own data values so yeah theres literally quite nothing here

the data should be about temperatures though

[lone heart]

@wide rapids Has your question been resolved?

[wide rapids]

oh well guess I'll die

.close

[lone heart]

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[gilded vessel]

what have you tried

hmm

do you know how to calculate a ratio

do you know what a ratio is

what is a ratio

yeah sure

if you had the ratio 10:2

could you rewrite it

exactly

so you have the ratio of $1.5 \cdot 10^8: 4.5 \cdot 10^9$

[ocean seal]

dog.

[gilded vessel]

if it helps you can think of it as

$\frac{1.5 \cdot 10^8}{4.5\cdot 10^9}$

[ocean seal]

dog.

[gilded vessel]

then simplify ^^

you can break it into two fractions if it makes it easier

just worry about it at the end

nope

the ratio would be numerator to denominator

usually when they ask for it in the form 1:n the most simplified form should be like that

alright

good but dont cancel to -1

just leave a 10 at the bottom

then youll be left with

$\frac{1}{3}\cdot \frac{1}{10}$

[ocean seal]

dog.

[gilded vessel]

now combining back into one fraction..

can you see what n would be?

exactly

for the second question first start with the triangle area formula

heh

i dont choose whats on top

its a ratio

the thing on the left goes on top

if you swap them u need to remember to reswap

so its kind of complicating things

anyway for the second one

$A = \frac{1}{2}bh$

[ocean seal]

dog.

[gilded vessel]

right?

10^9 = 10^8 * 10

the two 10^8 cancel

you leave the ten on the bottom because generally both sides of the ratio should be whole numbers

if you make it 10^-1

then you get 0.1 on one side

0.1: 3

you would wanna rewrite that as 1:30 anyway

[gilded vessel]

oops sorr

@alpine sable

[gilded vessel]

first step rearrange this to find base length

uh

okay

do you want me to keep helping @alpine sable

thats the point of me helping you

incorrect

[gilded vessel]

this is the area formula

solve for b

you can start by getting rid of the fraction

$2A = bh$

[ocean seal]

dog.

[gilded vessel]

how could you get b on one side from here?

@alpine sable

very close

just need to keep the 2

[gilded vessel]

dividing h on both sides

$b = \frac{2A}{h}$

[ocean seal]

dog.

[gilded vessel]

okay for higher values of h will b, have smaller or larger values @alpine sable

note: we are dividing by h

correct

and the same way with smaller values of h, b will be higher

then if we want b to be big should we choose smaller or higher values for A @alpine sable

right

so in the question

we want b to be as big as possible cause we are looking for an upper bound

so plugging in the formula we should use the upper bound for A and the lower bound of h

so b is as big as possible

h = 3.2 cm
A = 5.2 cm^2

not give them anything

just find their bounds based on how they are rounded

and plug them in

so for h = 3.2 cm

its rounded to 1 decimal point

and we want it to be as low as possible

this is correct

not quite

remember we want A as big as possible

so the upper bound for A would be 5.25

exactly

then plugging those numbers into the formula for base height we can find its upper bound

do you have a way to check your answers for this btw?

alright

check if that gets you the right answer then

is it correct

lets go

no problem

[lone heart]

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[teal lagoon]

fx = x + 2

[teal lagoon]

gx = root((x+2)^2)

is fx = gx

[karmic rapids]

you mean f(x) and g(x)?

[teal lagoon]

ye

[karmic rapids]

no

[ancient meteor]

Different domains

[karmic rapids]

no

[teal lagoon]

thanks

[karmic rapids]

put in a big negative number

[teal lagoon]

.close

[lone heart]

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[ancient meteor]

Oh wait no sry

Im dumb

Mb

[teal lagoon]

.close

[tall topaz]

$g(x) = |x+2|$

[ocean seal]

Pure

[lone heart]

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[spring spruce]

Im not sure what identity to use for this. Or if there is some sort of weird substitution I should use.

[rigid smelt]

the substitution u=sqrt(1+cosx) works here, but if you dont see how that works, you can try the substitution u=cosx

the first substitution is faster but not as straight-foward

whereas the second one is not as fast but is straight-foward

[marsh rapids]

Turning the 1+cos x into 2cos^2(x/2), working with x/2 and then u-subbing sin(x/2) works as well I think