help-23

[obtuse jackal]

Don't forget to make it small enough

[zinc token]

|a-b| is rational right

[steep pewter]

Hmm so a and b are rational pi -3 plus a where a<b?

[obtuse jackal]

like (a+b)/2 + pi / 4 * (b-a)/2 should work

[zinc token]

you can kind of split the distance between a and b into an irrational chunk

[steep pewter]

But for that b-a has to be larger than pi-3

[zinc token]

something like this

[steep pewter]

Ok

U have a and b

U need to show a irrational number between a and b exists

b>a

b-a * 4/10

Gives u a irrational number

Thats pretty smaller than the gap between a and b

No where in particular

I juat dont want a+ the other thing to exceed b

[steep pewter]

*pi my bad

Anyways just do what i said bro

.

This is a gud enough answer

[zinc token]

Ok let me

[steep pewter]

For a<b

[zinc token]

suppose WLOG a<b

set d = b-a

now

let's "scale" an irrational number to d

so let's set r = sqrt(2)/2 *d

now

a < a+r < b

and thus you have found it

actually, it is

i guess that statement was nonsense but

but basically since we had a distance

which is rational

we'll just turn it irrational and split it up

so notice sqrt(2)/2 * d + (1-sqrt(2)/2) * d = d

it absolutely does since

I assumed nothing about a and b other than the fact that they are rational

notice we have not lost generality because

if b > a then just set a=b and b=a

[steep pewter]

Take the larger number as h and the smaller as p and use pythagorean theorem boom!

[zinc token]

this is a real analysis class right?

[steep pewter]

[zinc token]

is this your first proof-based course?

gotcha

wait what class is it

ohhhh

got it

so because a and b are arbitrary rational numbers

it holds true for all cases

the distance between a and b is rational

so I used that to construct an irrational number

and simply added it to a

no, r = sqrt(2)/2 * d

and the irrational number we found is a+r

im showing you that a < a+r < b

ok visually, think of d as a stick

i split the stick into a piece that is sqrt(2)/2 the length of the stick

sqrt(2)/2 < 1

I just didn't want to type it out

lmao

they're just useful variables

it helped us find r

which is the irrational number i added to a

we used d to ensure that this irrational number a + r doesn't become bigger than b

the rest are left as exercises

yup

you wanna work thru the proof yourself

as in

what I told you

write it down and try to understand the proof

[safe radish]

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[wet yoke]

i need help with taylor expansion of differencing to find truncation error

[wet yoke]

these are the expressions i need to expand

this is the full qs

[safe radish]

@wet yoke Has your question been resolved?

[wet yoke]

<@&286206848099549185>

[safe radish]

@wet yoke Has your question been resolved?

[safe radish]

@wet yoke Has your question been resolved?

[wet yoke]

.close

[safe radish]

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[finite crypt]

these are the points:

(-3,0),(-1,1),(1,2),(3,3)

answer: (0,0)

correct?

[lapis shadow]

yes

[finite crypt]

ty

.close

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[lean otter]

How do I solve Logx^3=-16, I want to understand how I get the x value, since my calculator only does base 10

[lean otter]

Now if im not mistaken, I boot off the -16 to the other side right

And 3 becomes the power of x

[pseudo scroll]

What's the original question

$\log_x (-16) = 3$

[flat frigate]

NEONPerseus

[pseudo scroll]

Is it this?

[lean otter]

Yes

[pseudo scroll]

It will become $x^3 = -16$

[flat frigate]

NEONPerseus

[pseudo scroll]

Although idk if you can have a negative number in the log at this point

[lean otter]

Yes but how do I find X? Or im being stupid, and not realizing thats the end point

[thin bridge]

a bit dodgy for there to be a negative in the argument

[pseudo scroll]

tbf the math works

[pseudo scroll]

just cube root both sides

[thin bridge]

was this really the very original question, or did you make adjustments

[lean otter]

THAT NEGATIVE, is like that little rock in your shoe

[pseudo scroll]

Maybe write it off as abuse of notation

[lean otter]

Original question

[lean otter]

In more simple way? This is my first year studying in english

Idk what u mean by cube

[pseudo scroll]

$\sqrt[3]{}$

[lean otter]

Or abuse of notation

[flat frigate]

NEONPerseus

[lean otter]

ah, but i would square root the 16 too right?

[pseudo scroll]

Yes, and it's a cube root not a square root

See the little three?

[lean otter]

Cube root!

yes! 2 = square, 3= cune

Cube*

It will be -2.519

That should be correct for x right?

Question the x is the base right @pseudo scroll

Or its base 10

[pseudo scroll]

You can have an x in the base no problem

But the - inside the log

[pseudo scroll]

Well thats the question right

[lean otter]

Yes

[lean otter]

What do you mean by that?

[pseudo scroll]

the negative inside the logarithm

[lean otter]

Well the end suppose to be
Logx^1=-2.519

[pseudo scroll]

negative numbers inside the log isn't well defined

[pseudo scroll]

what

$\log x = -2.519$

[flat frigate]

NEONPerseus

[pseudo scroll]

Is this the answer?

[lean otter]

If u cube root -16

Thats what u get

[pseudo scroll]

Yeah thats the value of x

not log x

[lean otter]

ah! So, if I, put that number in place of x

I should get -16 right?

YUP! But its positive

Oh wait nvm

I forgot to add - to the first calculation

Thanks brother, I appreciate it a lot

@pseudo scroll 🫶

[pseudo scroll]

np

[safe radish]

@lean otter Has your question been resolved?

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[lean otter]

HELP

[wet wigeon]

2nd equation is just like 1st equation where left hand side is multiplied by -2

in other words it can be written as , 4x-8y=-16 (1st equation)

our 2nd equation is 4x-8y=a

[lean otter]

hmm

[lean otter]

yes

[wet wigeon]

so if a=-16, its just the same line as previous right

that means every point of theirs intersect

since its the same line

if a=-16 it has infinitely many solutions

[lean otter]

wow

what about one solution then

[wet wigeon]

impossible to have one solution,

assume that a is not -16, then a is different then -16

it makes these two lines parallel because their slope is the same

parallel lines dont intersect, unless its the same line

same line case was where a was-16

[lean otter]

ohh

[wet wigeon]

slope of line is m=-a/b where equation of the line is ax+by+c=0

isolate the y, you would get y = (-c-ax)/b

[lean otter]

so whats for one solution

[wet wigeon]

well

[lean otter]

ohh

what about

no solution

[wet wigeon]

no solution is IR - {-16}, i showed above

[lean otter]

so

basically 16?

[wet wigeon]

no

[stray socket]

The set above means "any real number but -16"

[wet wigeon]

its subset of real numbers that doesnt include -16

[lean otter]

ty

[lean otter]

ty

[stray socket]

You've probably seen it as ${a \in \bR | a≠-16}$

[wet wigeon]

-16 but yeah same scenario

[stray socket]

Well it should be a, not x

And I forgot the negative

[flat frigate]

Umbraleviathan

[stray socket]

There

[lean otter]

ohh

tyty

wait

i have a new question

guysss

@wet wigeon @stray socket

hello?

[stray socket]

Vertical line test

It's a bit ambiguous there though

[wet wigeon]

you can draw a line that is perpendicular to x-axis to determine if its function or not, it should only cut 1 point for each x value

[lean otter]

so it is a function

wait no its not

cuz

two points hit one line

[stray socket]

I mean ngl this is one where it's kinda ambiguous

[wet wigeon]

yeah i cant tell if it intersects or not too

[stray socket]

It kinda looks like $y = \sqrt[3]{-(x+1)}$

[flat frigate]

Umbraleviathan

[lean otter]

prob

[stray socket]

,w graph (-x-1)^(1/3) from -4 to 4

Bruh

[flat frigate]

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[stray socket]

Its not showing

The full thing

The graph above is a function

Kinda looks like what you have

[lean otter]

oh

yea

it does

also

btw

how do i determine if a equation is linear or not @wet wigeon @stray socket

[stray socket]

Is it a line

[lean otter]

or if it follows y= mx+b?

[wet wigeon]

yeah

[stray socket]

Pretty much

[wet wigeon]

there are properties that linear functions satisfy too

[stray socket]

Constant slope

So if m = c

where c is a real number

[wet wigeon]

f(a+b) = f(a) + f(b)

[stray socket]

That too

[wet wigeon]

f(ca)=cf(a) (c is a real number)

you dont need this if you are high schooler most likely

[lean otter]

oohh alr

[wet wigeon]

didnt say it but, a and b were variables

[lean otter]

thanks

how about

this question

[wet wigeon]

do you know how to graph

[lean otter]

yea

but idk how to find range

[wet wigeon]

well if you know the domain

you would know the bounds of the equation

for 1-x, it says up there is that domain is (-inf,-2) and for 2x+3 it is [-2,inf)

plug the bounds domain into the function to find the range

domain's bounds of the func 1-x are -inf and -2 as you can see

if you plug -inf to 1-x, its just inf

plug in -2 to 1-x , you would get 3

so range is (3,inf) rigt

now find range of 2x+3 by pluggin its domain into the function

[lean otter]

but how is that done

[wet wigeon]

2x + 3 ---> 2.-2 + 3 = -1 ,

2x+3= 2inf+3 = inf

now take union of these ranges

range of 2x+3 where x>2 and x=2 is [-1,inf) from above (we include -1 because x can be 2 too)

union of these two ranges are (3,inf) U (-1,inf) which would give us (-1,inf)

[lean otter]

ohh

so the range is

(3inf) U (-1inf)

[wet wigeon]

you can see that domain is IR

[lean otter]

right

?

[wet wigeon]

in other words (-1,inf)

[lean otter]

okay

[wet wigeon]

inf is infinity btw if you didnt know

[lean otter]

tyty

one sec

[wet wigeon]

let me know if there is something that is not clear to understand

[lean otter]

nono

its very clear

[wet wigeon]

not the cleanest explanation

[lean otter]

but kinda understandable

another question tho

@wet wigeon

what is it asking me to do

[wet wigeon]

so it gave you to lines right

lines are typed in the following format

y=ax + b

[lean otter]

i dont understand what to solve

[wet wigeon]

first line is y=3x+3

and 2nd line is y=2x+1

[lean otter]

yea but

what am i solving

[wet wigeon]

to find where these 2 lines have the same y value

you just let y1=y2 (y1 is equation of first line where y2 is equatino fo 2nd)

you are finding the intersection point

[lean otter]

oohhh

alr

[wet wigeon]

as you can see from the graph they intersect at one point

[lean otter]

but how do i know

[wet wigeon]

well graph isnt the best for 1st one

but for 2nd question you can clearly tell where they intersect right

[lean otter]

yea

[wet wigeon]

just solve the equations

find the x value

plug it in any of the equations

then you will get the y value

since equation of line is y=ax+b here clearly

[lean otter]

alright

ty ty

[safe radish]

@lean otter Has your question been resolved?

[lean otter]

Ok

I have another question

@wet wigeon

this one is long

i know

[wet wigeon]

you can understand this better by drawing

[lean otter]

but draw what

graph?

[wet wigeon]

no

is it obvious that before the race starts you are also 40m away from the pond since t=0 during that

and you can see that your friend is 80m away

[lean otter]

ohh

so im faster

but

how do i solve the two

[wet wigeon]

no

your friend closes in

what i quite didnt understand is that

after reaching the pond, does your friend end at where you started , or where THEY started

regardless of which you can find how fast you finish

since you started 40m apart, and have to run back after reaching pond

just let your equation be =40

you will get the answers t=0 (the moment you start) and the moment where you end which is t=80/3

for your friend i will assume they will end it where they started too

and say that is t=160/3

so you finished 80/3 seconds earlier

if your friend will end it where you started (which is most likely the case here)

then lets find |-5t+80|=40

you get t=40/5 and t=120/5 , higher one is where they would finish

[lean otter]

so they finishes the race

but when do they catch up to me

[wet wigeon]

and we can see that in this case your friend finished earlier since 120/5 > 80/3 yeah

to find the time they reach you

you would have to find at which point both of you are equally far from the pond

so just let |-3t+40|= |-5t+ 80|

and solve

[lean otter]

do i solve for intersect or

[wet wigeon]

yeah

[lean otter]

t= 20, 15

is it right?

[wet wigeon]

yeah

[lean otter]

but i dont understand this

why is there two t answers

[wet wigeon]

only 1 is true but let me find out

you can actually plug in a value between 15 and 20

and see which one is further from the pond

[lean otter]

okay

[wet wigeon]

i believe its t=15 which is correct (turns out i was wrong)

because at point t=20 your friend is at the pond right

oh my bad your friend is at the pond at t=16

but you reach the pond at 40/13which is approx 13.33

oh nvm you reached the pond earlier

so its impossible for you and your friend to meet at t=15 because you are already returning and your friend is just barely about to make it to pond

they reach the pond at t=16

[lean otter]

so

thats when they catch up to me

[wet wigeon]

but at the time t=15 you both are equally far away from the pond

your friend catches up to you at t=20

[lean otter]

tyty

also this question

[young nexus]

s = number of small bags, l = number of large bags. -> s+l = 14 and 4s+8l = 80 -> solve.

[lean otter]

uhh

how

is it elimination

or

substitution?

[young nexus]

choose the method you like

[lean otter]

tyty

[lean otter]

@young nexus

what about this question

[young nexus]

are you dropping here your homework and we should solve it for you? if not, tell what you have done and what is your question?

[lean otter]

well i've stated the basic infos

$100 profit = x + 50profite y

300 = x + y

but idk

what to do next

this is why i kept asking cuz i do know the basic info but I just need that one push to get the answer

[young nexus]

how much profit do you make when you sell x digital cameras and y cell phones?

[lean otter]

well

u make 100 from x

50 from y

sooo

100x + 50y?

[young nexus]

exactly. the profit should be at least 300. write this as inequality

[lean otter]

wait wait

so

300 ≤ 100x + 50y?

[young nexus]

exactly. so you have written the inequality. next you should make a graph,

and then you should interpret two solutions, e.g. how many cameras you need to selll for 300 if you sell only cameras. then withncell phones

[lean otter]

3

[young nexus]

yes, and cell phones?

[lean otter]

hmm

6

[young nexus]

exactly. so you have everything done (if you drawed the graph)

[lean otter]

wait

what does the 6 and 3 do

where do i put it anyway

i know i have 300 ≤ 100x + 50y

[young nexus]

the last sentence was "identify and interpret two solutions of the inequality". ou have two solutions. Sell at least 3 cameras is enaough for 300 and sell at least 6 cell phones is enough for 300.

[lean otter]

ohhh

[lean otter]

but the actualy inequality is still this right

im supposed to graph this right?

[young nexus]

yes. take x and y-axis (you need only positiv) mark (3/0) (= 3 cameras, 0 phones) and (0/6) = 0 cameras and 6 phones and draw a line beetween.'this two points. upper/right you are above 300

[lean otter]

im a lil confused

[young nexus]

wait a moment

[lean otter]

Ohhh!

[young nexus]

x axis = number of cameras, y-axis number of cell phones

[lean otter]

so

which part is the shaded area

oh wait nvm

the above 300 right?

[young nexus]

what do you mean with "shaded area"?

[lean otter]

like

the inequality shades an area prob

like this

[young nexus]

above 300 means > 300, below 300 means < 300, and the line means = 300, so: yes

[lean otter]

okok

so the actual equation

[lean otter]

is it this?

like is the equation this and all done'

[young nexus]

yes, i would say so

[lean otter]

tytyty

but

i came across another question

this

[lean otter]

.closed

.close

[safe radish]

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[true estuary]

have i done part a correct?

[vast cloud]

your vector equation for part (a) is wrong

and cartesian

[true estuary]

how

[vast cloud]

youve built a line that passes thru (-1, 2) and (1, 3)

thats not the right line

[true estuary]

u ever had a moment

where u dont read the question properly

?

i just had that moment

[vast cloud]

happens

[true estuary]

charbtiz

t

is mehdi gone

lmao

havent seen him in time

[junior smelt]

I don't know, maybe so, maybe he might be busy or something

Hey snow 🙋‍♀️

[true estuary]

is my part b correct?

[junior smelt]

Looks like it to me

[true estuary]

ms did something diff

[junior smelt]

Equivalent, scaled direction vector

[true estuary]

kl

[true estuary]

need some help for c plz

i tried a different method but

its completely wrong

[junior smelt]

Solution is?

[true estuary]

why did my method not work

i basically did the parametric to cartesian way but backwards

[junior smelt]

Line equations aren't unique

Your direction vector is fine

[true estuary]

how do u it the old way

[junior smelt]

Remember you can take any point on the line and have that as your "constant"

[true estuary]

how do u do it this way?

[junior smelt]

I'd do it the way you did, but as before, line [parametric] equations aren't unique (in particular any non-zero scaling of the direction vector and any point of the line will give you something valid)

Just because it looks different doesn't mean you're wrong

They probably chose to find some point on the line that looks nicer

[true estuary]

wait so what did i do wrong

[junior smelt]

https://tenor.com/view/nothing-curtis-payne-house-of-payne-not-a-thing-none-gif-18823459

[junior smelt]

[true estuary]

wait

ios my answer

correct?

so this is correct?

dang

[junior smelt]

E.g. if you had the line y=x, and you gave me something like 5y=5x, then I wouldn't say you were wrong, right?

[true estuary]

depends

cuz if my uni is marking it

its wrong

[junior smelt]

Well your university is just weird and strange, whoever's responsible should go in the bin

All of them

Like unless they imply or say "give this in simplest form" or something

[true estuary]

i go to oxford btw

i told u before didnt i?

[junior smelt]

No I don't think so!

Still gonna say they're weird then

[true estuary]

im taking the piss

u really think i go oxford

lmao

🤣

[junior smelt]

You never know

If the guys there were that weird, I'd quit life that moment tbh

[true estuary]

my answer just looks

really diff to the mark scheme

getting doubts

[junior smelt]

Well yep it does, but remember I said that your direction vector and theirs are scaled

[true estuary]

so both my postion vector

and direction vector

is correct?

[junior smelt]

Your direction vector is fine, it just suffices to show that the you can get each position vector from the "opposite" line

,w solve {t - 2 = 0, t/3 + 1 = 5/3, t/2 - 2 = -1}

[flat frigate]

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[junior smelt]

,w solve {6t = -2, 2t + 5/3 = 1, 3t - 1 = -2}

[flat frigate]

Results provided by WolframAlpha

Click here to refine your query online
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Requested by chartbit#3171

[true estuary]

oh

so i am wrong?

shouldnt t be the same?

[junior smelt]

It doesn't matter, think of e.g. motorways

They may meet but their junction numbers aren't necessarily the same where they meet (nor their kilometerage)

[yes, GB motorways are measured in km, before someone @'s me]

[true estuary]

lmao

[junior smelt]

But otherwise you get the concept right?

Each line has an equivalent direction vector

[true estuary]

yh yh

main thing is that my answer is correct

which im happy with

[junior smelt]

And also that above shows that the two lines have two points in common

#euclidspostulates

[true estuary]

i goit one final question

for now

r my cartesian equations fine?

[junior smelt]

Look fine to me!

[true estuary]

got scalar product left to do

then im done with vectors

[junior smelt]

Scalar as in inner products?

Inner product spaces

[true estuary]

this kinda stuff

and angles

between lines

and whatnot

and orthogonal spaces

and whanot

[junior smelt]

Ah yeee fair fair

[true estuary]

is it easy?

i did most of this in further maths

if i rememeber

[junior smelt]

It's not too bad imo for the most part

[true estuary]

how does this mean modulus i = modulus j = 1?

[junior smelt]

[true estuary]

how does i dot j equal 0 then?

[junior smelt]

$\pmqty{1 \ 0} \cdot \pmqty{0 \ 1} = 1(0) + 0(1) = 0$

[flat frigate]

chartbit

[true estuary]

ok kl

makes sense

[safe radish]

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[halcyon basin]

I need to use Maclaurin Series to get a polynom of 14th degree, anybody has suggestion on how to do sin x^4 ?

[cosmic grove]

,w maclaurin series sin(x)

[flat frigate]

Results provided by WolframAlpha

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Requested by Ｈｅｒｅｌｓ#1420

[cosmic grove]

all the x becomes x^4

[halcyon basin]

Ooh, I see

Thanks

.close

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[pine fossil]

Correct?

[potent bay]

no

[pine fossil]

It can’t be that easy

I knew

[cosmic grove]

my life would have been so easy if it was true

[obtuse plover]

Common misconception

[pine fossil]

xd

im glas i asked

glad

any hints?

[potent bay]

on what

[pine fossil]

here

[potent bay]

what is the question

[pine fossil]

its just = ?

i think simplify

[potent bay]

no

it stays that

[pine fossil]

bruu so professor pranked me

thx

[potent bay]

do you know anything about a and b?

[pine fossil]

no

thats all

[potent bay]

generally u cant simplify this

[pine fossil]

only that the solution must be positive

ok thx i think its a prank because to this he has no solution

Correct?

and how did he got from root 18 to 3 root 2

[buoyant shadow]

root(ab) = root(a)root(b)
root(18) = root(2)root(9)

[pine fossil]

what

.close

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[lean otter]

Hello ? Please I need help with my maths assignment which is due on the 16th

[wooden garnet]

Ask the question

[potent bay]

sorry, we cant help with that, #rules

[lean otter]

A laboratory technician has been asked to make up 2000mls of a 60% concentration of a solution using 80% and 20% solutions. How much of the 80% and how much of the 20% solutions are required

I don’t need help on all the questions lol I’m not very confident with maths but I do need help

[plucky elk]

If you use x mls of the 80%, solution, then how many mls of the 20% solution are you using?

[nova creek]

Who says we can't help?

[potent bay]

u say its not academically dishonest if u help him solve problems in his assignment?

[nova creek]

We can't help on tests and exams. Homework and shit is fair game

[lean otter]

Yh I really just need help with the questions I’m not as good at maths and understanding questions.

[potent bay]

assignment=test with longer thinking time

but anyway

suppose u have half, and half what concentration do u get?

can you write an equation for that maybe?

[lean otter]

So far I got 2000 (0.6) = x (0.8) + y (0.2) already when I done it out

[potent bay]

yeah, and cuz u have 2 variables u need a second one too

[nova creek]

Wrong channel

[lean otter]

2000 (0.6) = x (0.8) + (2000 - x) (0.2)

That’s all I got then my brain completely went blank after

That’s how bad I am at maths

[potent bay]

yeah, but u need to write it down, that x+y=2000, because they are poured in a container together

[potent bay]

and this became a regular eqn, put all the x terms on one side and the others on the other side

[lean otter]

Can you explain more ? Sorry

So far I’ve gotten this

[potent bay]

do the multiplication

and move the numbers to the left side, and leave the x-terms on the right

(the note 🙂 )

[lean otter]

The first plate I got 1200

Part *

= 0.8

[potent bay]

1200=0.8x+400-0.2x

800=0.6x

800/0.6=x

[lean otter]

Ok I got that

But the last two equations?

[plucky elk]

Plug the x they gave you to solve for y

[lean otter]

So the 20% ?

[potent bay]

no, what is the value of x?

[lean otter]

800 ?

[potent bay]

no

[potent bay]

0.6 apple costs 800 gold, so how much is a full apple?

[safe radish]

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[amber bloom]

Hi if anyone can solve this with working steps i will send them 10 bucks on steam, Thank you so much

[hasty folio]

a. Differentiate your function and find when it equals 0
b. Solve what h is at this point.
c. Solve the original equation for when h = 0
d. Sketch your function

[safe radish]

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[gusty lark]

oddly a little confused, the line is throwing me off

[safe radish]

@gusty lark Has your question been resolved?

[hardy wedge]

well, it's not a line is it?

[safe radish]

@gusty lark Has your question been resolved?

[forest orchid]

Depends on what you're defining a line

It it linear? No

But you don't need if it's a line equation to answer the other part of the question

[safe radish]

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[ebon seal]

how can i make 36=x² into x²=36 without making the numbers negative

[ebon seal]

then making it the square root

because

u cant find the square root of a negative number

and i cant give a negative number as an answer

[peak estuary]

if a=b then b=a

[frank glen]

x^2 = 36 is the same as 36 = x^2

[junior smelt]

And are you saying that x can't be negative, or that x^2 can't be negative?

[frank glen]

Proof:
36 = x^2
36-x^2 =0
-x^2 = -36 | •-1
x^2 = 36

[ebon seal]

sorry

im dumb as hell

thanks

.close

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[jaunty flame]

how do i solve this shit

[thin bridge]

what have you tried so far?

[jaunty flame]

I tried logarithms

and exponentials

[thin bridge]

can you show your attempts

[jaunty flame]

how

im on pcn

pc not phone

[thin bridge]

taking a pic? or reproducing it in paint or sketch.io

[jaunty flame]

ok

[thin bridge]

or type it in latex if you're good enough

[sharp tiger]

u dont need logs

just need to arrange some exponents

write 12 as 3.4 and try to group up the 3

[jaunty flame]

where do i put the (a+b) power when i rearrange

[thin bridge]

apply the relevant power (of a product) law

[jaunty flame]

dont get it

[thin bridge]

you know that 12 = 3 * 4 right

[jaunty flame]

yeah

[thin bridge]

apply that you would have something in the form
$$(p\cdot q)^n$$
on the right side of the equation

[flat frigate]

ℝamonov

[thin bridge]

and then distribute that exponent

[jaunty flame]

so it becomes

$$p^n * q^n$$

[flat frigate]

Beepze

[thin bridge]

yes

[jaunty flame]

oh ok

ok i get

a = -b

and b = -a

[lean otter]

no no

if a = -b so 2^((a+b)/b) should be 1 because this is 2^((a - a)/b) = 2^(0/b) = 2^0 = 1

[jaunty flame]

the answer is 1/3

[lean otter]

yep

[lean otter]

u need to use this

[jaunty flame]

how

i already did that tho

[lean otter]

ok

what did u get?

in the factor form

,texsp $3^{\left(a-b\right)}=\left(3\cdot4\right)^{\left(a+b\right)}$

[flat frigate]

SamuDoki

[lean otter]

oh, u did wrong in the 3^(a-b)/3^(a+b)

because it will be 3^(a - b - (a + b))

[jaunty flame]

oh so its -2b?

[lean otter]

yep

u will have 3^(-2b) = 4^(a+b)

[jaunty flame]

ok and then what

now im stuck

[lean otter]

try to rewrite the "4"

in another form

[jaunty flame]

(3^0 * 4)^a+b

[lean otter]

no, no

for example

25 = 5^2

[jaunty flame]

oh its

2^2

[lean otter]

yep

so u will have 3^(-2b) = 2^(2(a+b))

[jaunty flame]

ok now its 3^-2b = 2^2a+2b

[lean otter]

remember the product law in power?

(a^(bc)) can be rewrite in another form

[jaunty flame]

a^b + a^c ?

[lean otter]

nop

[jaunty flame]

yeah i forgot

[lean otter]

(a^(bc)) = (a^b)^c

remember?

[jaunty flame]

yeah yeah

so now its 2^2^a+b

[lean otter]

$\left(3^{-b}\right)^{2}=\left(2^{\left(a+b\right)}\right)^{2}$

[flat frigate]

SamuDoki

[lean otter]

remember something?

about ||root||

[jaunty flame]

root removes square

[lean otter]

yep

try this

and after try to remove "-b" root

[jaunty flame]

• 1/-b

[lean otter]

so....

u are in the end

and its ^(1/-b) not *

[jaunty flame]

oh ok

so its 3 = 2^(-a/b -1)

2^((-a/b)-1)

[lean otter]

put the "-1" in the frac

3 = 2^(-(a+b)/b)

and elevate by -1 in both sides

[jaunty flame]

ok

so now what

im so bad at this

[lean otter]

what did u get?

[jaunty flame]

im sitll here

3 = 2^(-(a+b)/b)

[lean otter]

elevate -1 in both sides to take off this -(a+b)/b and have (a+b)/b

[jaunty flame]

so it becomes 1/3 = 2^((a+b)/b)

omg thanks

thank you for being patient with me bro

ur the best

[lean otter]

[jaunty flame]

.close

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[sand minnow]

.close

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[jaunty flame]

how do i find the range

[dusk vapor]

denominator =! 0

is that how u spell not equal to

[jaunty flame]

!=

like 1 is included in the set

bro why isnt this b

[wispy sentinel]

ln(3x-3) >= 0

3x-3 > 0

[woeful plume]

3x - 3 >= 1

[jaunty flame]

huh

its not 3x - 3

its 3x - 2

[hybrid wedge]

To find range you have to find the domain of inverse function

[safe radish]

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[hasty snow]

Anything wrong in my proof?

[safe radish]

@hasty snow Has your question been resolved?

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@hasty snow Has your question been resolved?

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@hasty snow Has your question been resolved?

[hasty snow]

<@&286206848099549185>

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@hasty snow Has your question been resolved?

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[finite sable]

can anyone teach me 6th grade lvl math?

[sharp fox]

Hello, I worked out these 3 questions, so I'm unsure if it is right or wrong. Please help out if I made a mistake.

[wheat cave]

we only provide help for specific problems, you are too vague

[finite sable]

oh

[misty coral]

Head over to khan academy or similar websites if u want broad education on specific areas

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[spare walrus]

,rotate

[spare walrus]

first thing you should do is plot your points on the graph

i don't see any that you did in the first place, so yes?

i have no idea, since they are erased and i cannot see them

@lean otter these points are correct if that's what you're asking

[safe radish]

@lean otter Has your question been resolved?

[spare walrus]

you only need a minimum of 2 points, but your teacher may require more. 3 is totally fine

[safe radish]

@lean otter Has your question been resolved?

[safe radish]

@lean otter Has your question been resolved?

[safe radish]

@lean otter Has your question been resolved?

[safe radish]

@lean otter Has your question been resolved?

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[split terrace]

is my mind fried or is this wrong? i keep getting it to 6/8

[lapis shadow]

$a^2-2b^2=\left(-1\right)^2-2\cdot\left(-\frac{1}{4}\right)^2$

[flat frigate]

SilverSoldier

[split terrace]

yes but isn't this 6/8? 1- 2/8

[lapis shadow]

how is it 2/8

[split terrace]

(-1/4)^2=1/8 and it's multiplied by two giving 2/8

wait

my mind is actually fried

4*4 aint 8 XD ohh god

sorry and thank you

.close

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[marsh cloud]

hi

[marsh cloud]

i need help

[dull sequoia]

[marsh cloud]

y +√y -20

how to sove these questions

solve

[quasi bison]

so far there is no question

only an expression, and the way it's written is ambiguous as well.

[sand robin]

y + square root of y -20 = 0

looks like a hidden quadratic to me

let y = x^2

work from there

[quasi bison]

@sand robin don't guess at what the problem says

that's not your job

it's the job of the person who wants help to provide the statement in full

[sand robin]

I'm not guessing, I am positive that's what he meant and am trying to help

If I'm wrong he can correct me

[safe radish]

@marsh cloud Has your question been resolved?

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[marsh cloud]

y +√y -20

[solemn orbit]

elaborate

[sly field]

what

is this

[marsh cloud]

expression

[potent shale]

nvm then

[solemn orbit]

can you talk in whole sentences and explain your problem

you were asked last time too

[thin bridge]

its just an expression
what exactly are you being asked to do with that

[solemn orbit]

$y+\sqrt{y}-20$?

[flat frigate]

Zabbx

[marsh cloud]

i need to find the solution of it wich is root y -4 and rooty -5

[marsh cloud]

yes

[quasi bison]

we aren't telepaths. "find the solution of it" is incredibly vague. nobody knows what you're talking about.

take a pic of the problem from your textbook

[sly field]

16

[solemn orbit]

so you want to solve $y+\sqrt{y}-20=0$?

[flat frigate]

Zabbx

[thin bridge]

were you being asked for the roots/zeroes
and/or are you oblivious to = 0

[marsh cloud]

yes

[sly field]

like solution means when it cut y axis?

[solemn orbit]

it does not

[solemn orbit]

they asked how to solve it, not the solution itself

[dull sequoia]

,w graph x + sqrt(x) - 20

[sly field]

it cut y axis on 16

[marsh cloud]

its not graph!

[flat frigate]

Results provided by WolframAlpha

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

Requested by Frosst#1111

[marsh cloud]

its not graph!

its factorize

[dull sequoia]

it does have a root

[solemn orbit]

you mean the x axis?

[thin bridge]

take a pic of the problem from your textbook

[quasi bison]

there is a lot of people here.

[thin bridge]

are you being asked for zeroes? roots? factorisation?
were you originally being presented with an equation?

can you post the original problem in its entirety with no modifications

[quasi bison]

i think the real problem is incommunicability.

[sly field]

nah

like if a equation is in terms of x then we find zero when it cut x axis

do we have to turn this y=f(x)

[sly field]

yeah

[marsh cloud]

?

i m not even able to understand

[dull sequoia]

take a picture of where you got the question from

[thin bridge]

can you post the original problem in its entirety with no modifications

[sly field]

like what you want to ask

[thin bridge]

because there are potentially multiply things going on here

[marsh cloud]

no textbook pic

[thin bridge]

no pic at all possible?

[marsh cloud]

ye

[thin bridge]

type out the problem in its entirety with no modifications

include all instructions

don't omit or add words

[solemn orbit]

im so sorry

https://tenor.com/view/why-isnt-it-possible-christian-bale-patrick-bateman-american-psycho-why-is-it-impossible-gif-18161290

[marsh cloud]

i pasted the question exact same factorize: y+ root y -20

[thin bridge]

you failed to mention factorise in the post,
you just typed out an expression with no instructions

[sand robin]

my brother in christ, I tried to save you last time

[thin bridge]

but you said yes when someone asked whether you wanted to solve an equation

so at that point we had no idea what you really wanted

[sand robin]

solve or factorise?

[thin bridge]

so just to be clear, the original question and what you want is \
factorise
$$y + \sqrt{y} - 20$$

[flat frigate]

ℝamonov

[thin bridge]

is that correct?

yes/no

[sand robin]

😂😂finished

[sly field]

@marsh cloud ???

[marsh cloud]

yea

[marsh cloud]

yes

@sand robin

you need to explain me this

[thin bridge]

note that y = sqrt(y) * sqrt(y)

[marsh cloud]

ok

[thin bridge]

this is a quadratic expression in sqrt(y)

you can do a substitution like
x = sqrt(y)
to get something more recognisable

doing that substitution would get you
$$x^2 + x - 20$$
which looks less intimidating to factorise

[flat frigate]

ℝamonov

[thin bridge]

and then you can sub your sqrt(y) back in after you're done

[marsh cloud]

is

anser root y-4 * root y+5 ?

answer*

[thin bridge]

um severe lack of parentheses there

what you've typed would be interpreted as
$$\sqrt{y} - 4\sqrt{y} + 5$$

[flat frigate]

ℝamonov

[marsh cloud]

yes

[thin bridge]

which isn't right

[marsh cloud]

k

[thin bridge]

and if you actually had that written on the page, it would be wrong

severe lack of parentheses there

[marsh cloud]

ok

[sand robin]

If you are still confused I could post the solutions
But I’m sure you can do it yourself

[thin bridge]

$(a+b)(c+d)$ a common factorised form is NOT the same as $a+bc+d$

[flat frigate]

ℝamonov

[marsh cloud]

@sand robin i am not able to understand quadratic equations because i dont have them in my syllabus and i dont understand how did y become x^2

[sand robin]

The method is called substitution

[thin bridge]

$x^2 + 3x + 2 \red\neq x + 1x + 2$

[flat frigate]

ℝamonov

[marsh cloud]

never heard of it before

[sand robin]

You don’t have quadratic equations in your syllabus?

[marsh cloud]

yes i dont have them