help-13

[elfin hemlock]

What is the definition of a function?

[silver fable]

You mean the domain?

[elfin hemlock]

No I mean the definition of a function

What does a function do?

[silver fable]

It assigns an output to all inputs

I guess..

[elfin hemlock]

Yes see the problem with this then?

[silver fable]

But u told me that i can put anything in the co domain

[elfin hemlock]

As long as it fits the definition of a function yes

[silver fable]

What would you put?

[elfin hemlock]

?

[elfin hemlock]

First of all answer this question

[silver fable]

No

[elfin hemlock]

What does f(-1) map to for example?

[silent pawn]

You can obviously define functions functions like f : R{0} - - - -> (1,3), but f(x)=1/x cannot have domains like this. If x=1/4, the output is not in the codomain (1,3) anymore.

[silver fable]

Yeah i got it

So what is the co domain then??

[silent pawn]

It can be anything depending on the domain.

Is your domain R{0} fixed?

[silver fable]

Yes

The function is 1/x

[elfin hemlock]

You already gave a codomain that works earlier

[silent pawn]

Well then R must be a subset of B.

[silver fable]

Oh i think i figured it out now

Thanks guys 🙏

[silent pawn]

But don't get range and codomain mixed up. The range is B=R\0.

[silver fable]

.close

[cedar kiln]

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[elfin hemlock]

No

[cedar kiln]

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[silver fable]

The range is the set of outputs

[silver fable]

Right?

[silent pawn]

The set of the outputs of all the inputs (which are elements of the domain).

[silver fable]

Yeah exactly

.close

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[cosmic temple]

i am asking again because the previous one was closed due to timeout

[cosmic temple]

pls help me with this question

[cedar kiln]

@cosmic temple Has your question been resolved?

[crimson sedge]

Are you asking to solve this non linear equation in x?

[cedar kiln]

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[cosmic temple]

.reopen

[cedar kiln]

✅

[cosmic temple]

yes

<@&286206848099549185>

[crimson sedge]

I can see a simple root, that is x = 0

I dont know about the complete resolution, I would solve it graphically or by Newton method

[cosmic temple]

i want to do it algebraically, because i can't make graphs in exams 🙂

[cedar kiln]

@cosmic temple Has your question been resolved?

[cosmic temple]

<@&286206848099549185>

[fickle trellis]

Hi

@cosmic temple

Is this an Inv. Trigo Mathematical problem

[cosmic temple]

yes

[fickle trellis]

@cosmic temple

Sorry took some time to type

I hope this helps

[cosmic temple]

oh thanks 🙂 , i didn't know this identity.

[cosmic temple]

yeah, its solved for me 🙂

[fickle trellis]

Oh, thats nice to hear... lemme have a look at the other problem you posted

[cosmic temple]

also , if you could help me with another question in #help-23 that would be great 😄

[fickle trellis]

Yeah

I was attempting it

[cosmic temple]

ohkie 🙂

.close

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[agile pumice]

Find the Length of AC. Then round two decimal places

[fickle trellis]

@agile pumice what is your level

[agile pumice]

my level?

[fickle trellis]

Yeah I mean your Maths level

[agile pumice]

or do u mean grade?

[fickle trellis]

Kind of

[agile pumice]

umm im not rlly good at math but i am in 10th grade

[fickle trellis]

Ohkay I got it

[agile pumice]

ya

[fickle trellis]

Its a Coordinate Geo. question

Isnt it

[agile pumice]

yes

[fickle trellis]

Hmm

Use distance formula

Lemme help you out

[agile pumice]

okay ty

(-4,2) (2,-2)

are the corrdinates i think

[fickle trellis]

This may help

@agile pumice

[agile pumice]

yes exactly that!

[fickle trellis]

Yeah! Have a great day!

I got a helper role, yeahhh!

[agile pumice]

yeahhh!! good jobb!!

.close

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[regal oak]

can anyone tell me infinite sum of a HP ?

[regal oak]

feel free to tag while replying

[cedar kiln]

@regal oak Has your question been resolved?

[regal oak]

<@&286206848099549185>

[dire nexus]

There is no formula for finding the sum of infinite terms of a harmonic progression. But if you want to find the sum then write the general term of the series and split the general term in such a way that if you put the term it will go on deleting automatically and the terms which get left will be your required sum.

.close

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[crimson sedge]

cant figure out how to do c

i thought maybe where dy/dx not equal to 0

for that i got where lambda is not equal to -tanx but i have no clue what that actually means lmao

oh wait does that mean -1 to 1

because range of tanhx is 1 to -1???

[cedar kiln]

@crimson sedge Has your question been resolved?

[cedar kiln]

@crimson sedge Has your question been resolved?

[crimson sedge]

<@&286206848099549185>

[astral crown]

I think it means that lambda is NOT -1 < lambda < 1

So you have no real solutions

[cedar kiln]

@crimson sedge Has your question been resolved?

[crimson sedge]

is not between -1 and 1? for no real?

i thought it means

if between -1 and 1, there is no real so answer to q is between -1 and 1

is that right or no?

[astral crown]

wait nvm you're right

1/lambda is not in -1 to 1

[cedar kiln]

@crimson sedge Has your question been resolved?

[tardy current]

turning point is just local max and min right

[cedar kiln]

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[sterile skiff]

would this be correct?

<@&286206848099549185>

<@&286206848099549185>

[fickle trellis]

Hi @sterile skiff

[cedar kiln]

@sterile skiff Has your question been resolved?

[cedar kiln]

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[lost kindle]

how do I go about proving that the difference of the absolute value of two numbers stays within the range of the two numbers? For example, the question I'm trying to figure out states: 1, 2, 3, 4, 5..., 100 are written on a chalkboard. We are to replace two of the numbers with the difference of the absolute value. It asked me to show that if we were to repeat the procedure until there is 1 number left, it will be between 1 and 99

[lost kindle]

It looks rather obvious, but how do I go about proving it?

[loud parcel]

1 - 100 <= x <= 100 - 1
99 - 100 <= x <= 100 - 99

This are the extrema

[lost kindle]

and also how do I generalize it for two numbers, x and y?

[lost kindle]

that's an acceptable enough proof? I do not need to show that repeating the procedure will give the same result?

[loud parcel]

When you will show it for [a, b] then you can try it with this:
(a - b) <= x <= (b - a)

((b - a) - b) <= x <= (b - (b - a))

• a <= x <= a

But I can't tell you if this is enough

[cedar kiln]

@lost kindle Has your question been resolved?

[lost kindle]

Wait, can you explain that a little bit more? I don't understand where you got the (b-a)-b

[loud parcel]

Let a = 1 and B = 100
((100 - 1) - 100) <= x <= (100 - (100 - 1))
(99 - 100) <= x <= (100 - 100 + 1)
-1 <= x <= 1

[lost kindle]

ah ok

that makes sense

[loud parcel]

and because you will use the abs, a must be the lower bound: |a| = |-a| and when a >= 0 then |a| = a

And now you can argue that, when a is the lower bound, b - a must be the upper bound so, the range for the difference must be in [a, b - a]

For all a >= 0 and b >= a

[lost kindle]

ok, how would you show that for repeated procedure, that the lower bound is always a and upper bound is b-a?

[loud parcel]

That should be enough because every difference must be in this bounds

[lost kindle]

ok so no proof is necessary?

[loud parcel]

You could assume that there is an d with |a - b| > |b - a|

[lost kindle]

ok

[lost kindle]

can you also guide me through this problem?

[obsidian coral]

Don't ask in an occupied channel
Read #❓how-to-get-help

There is literally a section that says "Math help (available)"

Use one of the channels under there

[loud parcel]

Draw a circle with r = 1
When ABP is 90°, then from point P must be a line of 45° into the circle

[lost kindle]

does the graph look like this?

[loud parcel]

when the angle is 45° we can form a square in the circle so we get PA' = PM + MA'
with pythagoras we get (PA')² = (PM)² + (MA')² <==> (PA')² = 1² + 1² <==> PA' = sqrt(1² + 1²)

So AB must be: (AB)² = (PA')² + (PB')²

[lost kindle]

sorry it took a bit for image to load

[loud parcel]

Yeah, this should be the thing

[lost kindle]

wdym by 45 degrees

[loud parcel]

because 90° / 2 = 45°

[lost kindle]

is that what you are saying?

[loud parcel]

I mean the angle between PMA

M is the middle of the circle

[lost kindle]

the center?

[loud parcel]

yes

[lost kindle]

Sorry I'm a bit confused, wouldn't that also be a 90 degree angle?

wait I think I get what you are tring to say

[loud parcel]

I mean the angle between A and M with P as point

Yeah sorry, my english is not so good

[lost kindle]

no it's all good

are you referring to angle map as 45

[loud parcel]

You see, AB must be 2r in this case

[lost kindle]

yes

wait

wouldn't that make ab just 2?

[loud parcel]

yep

[lost kindle]

the answer is 2 really?

[loud parcel]

Should be

[loud parcel]

https://www.geogebra.org/calculator/ambzb2hu

[prime garnet]

wouldnt it be the square root of 2?

assuming m to p is 1 and its from the center of the circle to the edge then m to a is 1 since its the radius

that would make p to a the hypotenuse

Pythagorean theorem a^2+b^2=c^2

so c = the square root of a^2 + b^2

a and b are both 1

so the square root of 2?

[loud parcel]

AB = M1P + M2P

[prime garnet]

sorry not and and b

oh shoot its for a to b not p to a or p to b

my bad

[loud parcel]

np, I'd first read it false also

[cedar kiln]

@lost kindle Has your question been resolved?

[cedar kiln]

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[sterile skiff]

hello

[sterile skiff]

could you check the problem i posted earlier?

[sterile skiff]

^

[fickle trellis]

This one?

[sterile skiff]

Yes that one

[fickle trellis]

Your level of Mathematics is?

[sterile skiff]

it is for discrete math

im a college freshman

these were my answers

[fickle trellis]

Oh, I'm a senior HS student, but hopefully I can solve it for you

We're finding relations

[sterile skiff]

Yes

[fickle trellis]

In the elements of cartesian product of Z

Integers, right/

?

[sterile skiff]

Yes

[fickle trellis]

Well

[wraith dagger]

Muhammad Hussaini
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[fickle trellis]

ZxZ

So well

See if its reflexive

Means, does it agree on every ordered pair (x,x)

[wraith dagger]

Muhammad Hussaini

[fickle trellis]

ZxZ

For symmetry

We know

|y|=|x|, so (y,x) would belong to Z x Z

And would belong to the relation R1

[sterile skiff]

i see

[fickle trellis]

For transitivity

You can check

[wraith dagger]

Muhammad Hussaini

[fickle trellis]

This shows an equivalence relation exists

[sterile skiff]

i see

[fickle trellis]

For R2, its defined on the set N,

You start N from 0 or 1

because some books start from 1, and some from 0

[sterile skiff]

so all equivalence relations are reflexive then

[fickle trellis]

See, the rule is:

If a relation is symmetric, reflexive and transitive, then it is an equivalence relation

So the converse is obviously true

[sterile skiff]

ah ok

[fickle trellis]

All three conditions must be met for a relation to be equivalence

Similarly, find for R2, R3

x has the same number of prime factors as x, so it is reflexive

If x has same number of prime factors as y, then y should have same number of prime factors as x, or not

Answer.

[sterile skiff]

but how would you tell that x = y for R2

[fickle trellis]

Well take two numbers

Who have the same number of prime factors

Give me such two numbers

[bold hinge]

3 and 5

[sterile skiff]

sure, 3 and 5

[fickle trellis]

Hmm, if (3,5) exists in the relation, such that 3 and 5 have same number of prime factors, i.e. one.
Then can (5,3) exist such that 5 and 3 have same number of prime factors
?

[sterile skiff]

yes

its just switched

[fickle trellis]

So, is it symmetric or not?

[sterile skiff]

yes

what would be a case where that wouldnt work?

like 2 and 5

[wraith dagger]

Muhammad Hussaini

[fickle trellis]

Well, 2 has how many prime factors?

[sterile skiff]

1

[fickle trellis]

How many prime factors does 5 have

[sterile skiff]

1

[fickle trellis]

then...?

[sterile skiff]

oh wait

ah i see

[fickle trellis]

Look, the thing is

The basic thing is

[sterile skiff]

so (x,y) all x prime factors have to be in y

and to be symmetric, it has to be vice versa with (y, x)?

[fickle trellis]

If (2,5) exists in the relation.... then (5,2) also exists... thats what is symmetric

YEAH

[sterile skiff]

okok

[fickle trellis]

I said, "if" the first one i.e (x,y) exists... and the second one (y,x) "also" exists

[sterile skiff]

ahh

[fickle trellis]

Then the relation is symmetric

If (x,y) doesnt exist in the first place, like take 4 and 18

4 has 1 prime factor

and 18 has 2 prime factors i.e (2,3)

So the ordered pair (4,18) does not exist for Relation 2

[sterile skiff]

ahh

so then it would be true for R2 symmetry

[fickle trellis]

Yeah

The transitive one is okay, if you want me to explain I can

[sterile skiff]

Could you please

also it would not be true for r3 Sym.

well it can be true

since theres a chance they could have the same number of prime factors

[fickle trellis]

Well, the R3 thing is a bit different.

See

I'll show you this way

It says, ordered pairs
$(x,y) $

such that no. of prime factors of x < no. of prime factors of y

Okay?

[sterile skiff]

well wouldnt it be =<

[fickle trellis]

Its actually
$n(x) <= n(y)$

Yeah

[wraith dagger]

Muhammad Hussaini

[sterile skiff]

got it

[fickle trellis]

Yeah

So

If I have to find symmetric

Then what should I say

[wraith dagger]

Muhammad Hussaini

Muhammad Hussaini

[fickle trellis]

So ordered pair (y,x) wont exist

And hence symmetry wont exist at all cases

[sterile skiff]

oh it has to be at every case

that makes sense

so only "=" works with symmetry

[fickle trellis]

Yeah

I said

[sterile skiff]

no less than, greater than, or equals to

got it

[fickle trellis]

The definition of symmetry was

[sterile skiff]

equivalent if switched

[fickle trellis]

**For all **(x,y) in R, there exists (y,x)

This is a symmetric relation

[sterile skiff]

ah right

[fickle trellis]

The relation is reflexive

[sterile skiff]

that makes sense

[fickle trellis]

Because

Every (x,x) exists

As n(x)=n(x)

and that satisfies the relation

[sterile skiff]

what case would that not work?

[fickle trellis]

Reflexive says

That every element is in relation to each other

For example

If I said

The number of prime factors of x is Less than the number of prime factors of y

Then I cant say equal

So I couldnt make a reflexive relation

[sterile skiff]

oh

so it is not for all

[fickle trellis]

I couldnt say that n(x) = n(x) so (x,x) is in R, or xRx exists

You have to see casewise

[sterile skiff]

ah i see

[fickle trellis]

Each relation is different

[sterile skiff]

understood

[fickle trellis]

Now transitive

[sterile skiff]

yes

[fickle trellis]

Lets say

In R3

[wraith dagger]

Muhammad Hussaini

Muhammad Hussaini

Muhammad Hussaini

[fickle trellis]

So

Can I say

n(x1) <= n(x3)

So can I also put (x1, x3) in the relation

[sterile skiff]

wait can you explain the "n"

[fickle trellis]

I defined n here as the number of prime factors

Just for ease

[sterile skiff]

ohh okay

[fickle trellis]

It isnt a definition

n of something usually means number of elements in a set

Now I just used it for no. of prime factors

To ease out the work

[sterile skiff]

ya makes sense

[sterile skiff]

yes

[fickle trellis]

Yeah

Thats what transitive is

[sterile skiff]

oh

[fickle trellis]

If (x1, x2) and (x2, x3) exist, then (x1, x3) also exist, then relation is a transitive relation

[sterile skiff]

couldnt u also do the same but replace less than or equal to with greater than or equal to

[fickle trellis]

The relation says the no of prime factors of x is not more than the prime factors of y

[sterile skiff]

ah okay

[fickle trellis]

If I use greater or equal to, then we cant put them in this relation

[sterile skiff]

oh right bc its specific to each relation

[fickle trellis]

We need to make a new relation

YEAH

its specific

You got it!

[sterile skiff]

yay

[fickle trellis]

Okay tell me now,

R3 is reflexive and transitive, but its not symmetric,
Is it equivalence?

[sterile skiff]

noo

has to be all 3

[fickle trellis]

Yeah

so its not equivalence

I could show u this stuff on a VC too

Well, finally we're done

[sterile skiff]

Yes so the correct version would be

WAIT

[fickle trellis]

I sent the question again so that i dont need to scroll my way up

Now wait, I need a little help from @obsidian coral for R1.

[sterile skiff]

okok

[fickle trellis]

Wait... R2... its equivalence na?

[obsidian coral]

I don't know this stuff

[sterile skiff]

but r2 wouldnt be transitive right?

[fickle trellis]

Why you wrote No

[sterile skiff]

accident

[obsidian coral]

So don't ping me

[bold hinge]

Are you sure that R3 is reflexive

[fickle trellis]

R2 is transitive...

[sterile skiff]

oh wait ur right\

i forgot the equal to

[fickle trellis]

Because it says not more.

So can be equal

Which satisfies reflexive nature

[bold hinge]

Can be but not necessarily

[sterile skiff]

that better

BRO

hold on

messed up again

[fickle trellis]

Wait a min

You messed up again

[sterile skiff]

wait

r2 is not symmetric i thought?

[fickle trellis]

R2 is reflexive, symmetric and transitive

Symmetric because its an equality there

[sterile skiff]

but it does now work for every case

[fickle trellis]

if n(x1)=n(x2)... that also means n(x2)=n(x1) so
(x2, x1) also exists

[fickle trellis]

The question doesnt say not equal to, it says not more than, which means can be equal or less than

[sterile skiff]

but can't there be a case where x has less prime factors than y

[fickle trellis]

Yeah

[sterile skiff]

im dumb

[fickle trellis]

Well

[sterile skiff]

for r2

[fickle trellis]

Imma show you in a Voice channel if you dont mind

[sterile skiff]

dont answer that i read it wrong haha

ya that's fine

[fickle trellis]

Come over to mathematics

[fickle trellis]

If and only if a=b or a=-b

|a| = |b|

[sterile skiff]

S : {(x, y) ∈ Z×Z ∣ |a| = |b|

[fickle trellis]

(a,b) ∈ Z x Z, implies that |a|=|b|,,

Also then |b|=|a|, so (b,a) ∈ S

So relation is symmetric

For all (a,a), we can say that |a| = |a|

So (a,a) ∈ S

And S is a reflexive relation

[sterile skiff]

For all a, b, c ∈ Z×Z

[fickle trellis]

(a,b) ∈ S

aSb

[sterile skiff]

(a,b) ∈ S

(b,c) ∈ S

(a,c) ∈ S

,rotate

[sterile skiff]

The inverse is: {(1, 1), (4, 2), (3, 3), (4, 3), (5, 4), (5, 5)}

[cedar kiln]

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[haughty python]

What is 69 + 69

[haughty python]

<@&286206848099549185>

[leaden otter]

69+69=69×1+69×1=69(1+1)=69×2

[livid hound]

2*70-2

[nova geyser]

@haughty python hi

,calc 69+69

[wraith dagger]

Result:

138

[nova geyser]

@haughty python done ?

[haughty python]

Yes

How do I close

[cedar kiln]

@haughty python Has your question been resolved?

[cedar kiln]

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[modest vapor]

I am currently stuck on the part that requires algebraic manipulatin

[wraith dagger]

Euclid31415

[modest vapor]

aye

that was proven earlier in my textbook

while i did see some similarities, i have no idea how to reconcile that sum into this problem

[wraith dagger]

Euclid31415

[modest vapor]

hmm

i can see some resemblance, but i am not sure exactly what is being connected

for example, why do they require that i subtract one sum from another?

[wraith dagger]

Euclid31415

[modest vapor]

ohhhh

and thats the sum shown in 1.10?

no wait

excuse me, its not

[leaden otter]

Yes the left side

[leaden otter]

Why?

[modest vapor]

well, my reasoning is that the sum u posted starts at n+1, and it increases by 1. so (2(n+1) - 1) + (2(n+2) -1) +....

[leaden otter]

Does it look the same as the sum they wrote?

[modest vapor]

its not the same for 1.10 bcs the series alternates by odd numbers. like (2n+3) + (2n +5)+ ...

thats what i think

[leaden otter]

(2(n+1)-1)+(2(n+2)-1)+...=(2n+1)+(2n+3)+...

[modest vapor]

omg

haha ur right

i just completely forgot abt DL lol

okay this makes sense

but i have no idea how to use this fact

the textbook generally showed that if a + b + c + ... + n = d(n), where a, b, c are arbitrary numbers in N, and d is some function of n,

then we can prove P(n+1) as a + b + c + ... + n + (n+1) = d(n+1)

subtracting n+1 from boths RHS and LHS should agree with our induction hypothesis

but i only know this trick and thats that

[wraith dagger]

Euclid31415

[modest vapor]

oooo

that's a very neat trick

its v analogous to addition of integrals with a common boundary

[wraith dagger]

Euclid31415

[modest vapor]

Ah, that is a beautiful result

so, i assume the question is asking me to use mathematical induction

[modest vapor]

nvm i think i got it

thanks @leaden otter

.close

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[glass crypt]

Can someone help me with this ?
Translation:
Let x be a positive real
Prove that

[fickle trellis]

Can you please translate your question...

[glass crypt]

i did

[fickle trellis]

Prove what

Is this your question:

[glass crypt]

this

1st line of my translation is the 1st sentence in the image, 2nd is 2nd

[fickle trellis]

What is the first equation then?

[glass crypt]

this

[fickle trellis]

Yeah what it is?

[glass crypt]

it's in the image i seent

[glass crypt]

wdym ?

[fickle trellis]

I mean what its for in the question?

[glass crypt]

we use the 1st equation to prove for the second one

[fickle trellis]

Oh I got it

[glass crypt]

i tried solving it by moving root(5) to the left and making dominator root(x) but and then remove it after having 0 on right, but that gives me 2 solutions

[fickle trellis]

Could you send your answer

[glass crypt]

okay hold on

[neat oak]

let sqrt(x)=y and solve for y in first equation

then sub into second one

[glass crypt]

ok i will try

but how do i solve it, we haven't studied the quadratic formula yet

this lesson is about operations and order in R btw

<@&286206848099549185> been >20mn

[fickle trellis]

Wait Im here now

Let me help

[glass crypt]

ok

[cedar kiln]

@glass crypt Has your question been resolved?

[fickle trellis]

@glass crypt

Can you come in a VC

[glass crypt]

yes ?

[fickle trellis]

Or should I solve it here

For you

[glass crypt]

yes but i won't be able to talk

[fickle trellis]

No prob

[glass crypt]

yes okay

[fickle trellis]

Just write down whatever you want here

Whilst I explain

[glass crypt]

ok tysm

[fickle trellis]

Join mathematics

[glass crypt]

which vc channel

hi

ok tysm

wait

yes i can

yes

-x or x

but it's x bcs x>0

ah yess

okay

yes i see

yes :D

y^2 + 2 + 1/y^2

=5

(a-b)^2

?

-4

on both sides

omg yess

i see now tysm

now replace y with root(x)

tysm <33333 for help i ve been stuck for 2 days lol

lol

yes <3

tysm again have a great day

.close

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[woeful marten]

I need to find x - y

[woeful marten]

Dont know where to start

[craggy coyote]

Extend line AC to the right and create a new right triangle ABD with a new point D

[dense wing]

D's already a point..

[craggy coyote]

E then

[woeful marten]

i dont rlly understand, can you please draw it out?

[dense wing]

well you can get 2 equations for x and y

[woeful marten]

whats the equation

[dense wing]

one is from the line segment AC, one is from the fact BD is common

[craggy coyote]

BD = h

h²+x²
h²+(10-x)²

[woeful marten]

10-x is y ?

[craggy coyote]

Yes

[woeful marten]

okay... then whats next

[craggy coyote]

Substract two equation from each other, so h² will disapeear and only x will remain with somenumber

[woeful marten]

but why subtract

[craggy coyote]

There are 2 unknowns, there we cant solve the equations

h²+x²=121
h²+100-20x+x²=81

h and x are both unknown

But if its only x, we can find the value x

[woeful marten]

oooh

[craggy coyote]

Or you can just say 121+100-20x = 81

[woeful marten]

so x = 7?

y = AC - x
y = 10 - 7
y = 3?

x - y
= 7 - 3
= 4

wah finally solved it

thx

.close

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[real bluff]

Hello

[real bluff]

<@&286206848099549185>

[cedar kiln]

@real bluff Has your question been resolved?

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[muted shell]

Hello

[muted shell]

I need help

I need to solve a 2x2 equation system using the replacement method

And i did it

But then i verified it and i dont understand what i did wrong

<@&286206848099549185>

[cedar kiln]

@muted shell Has your question been resolved?

[cedar kiln]

@muted shell Has your question been resolved?

[cedar kiln]

@muted shell Has your question been resolved?

[tardy current]

you need to isolate x

here you have 4x=10-3y

and when you substituted in for x in 7x-2y=30 as 7(10-3y)-2y=30

you really substituted in 4x

not x

[cedar kiln]

@muted shell Has your question been resolved?

[muted shell]

Ok, ill try doing it that way, thanks

[muted shell]

.close

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[sand ether]

For $u_n$, I got $u_n = 3^{n-1} + 2\sum_{r=1}^{n} 3^{r-1}(n-r+1)$

[wraith dagger]

azeem321

[sand ether]

I believe this expands out correctly, but it isn't correct according to the textbook...

Unless the summation can be written as a function of $n$ which I don't know how to

[wraith dagger]

azeem321

[sand ether]

.close

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[crimson sedge]

convert the polar equation to rectangular form and identify the graph r=-3 cos θ - 2 sin θ

[crimson sedge]

how to begin

[sand ether]

Rectangular form means in terms of $x,y$. $x^2+y^2 = r^2$ and $\sin(\theta) = \frac{y}{r}$ and $\cos(\theta) = \frac{x}{r}$

[crimson sedge]

what

[wraith dagger]

azeem321

[crimson sedge]

so

i rewrite it with x y and r

[sand ether]

only x,y

[crimson sedge]

heh

how

the thing u just showed me has rs

[sand ether]

multiply throughout by r

in your polar equation

[crimson sedge]

help

[sand ether]

$r^2 = -3r\cos(\theta) - 2r\sin(\theta)$.

[wraith dagger]

azeem321

[sand ether]

Do you see how now?

[crimson sedge]

ok but what does this do for me

[sand ether]

Try and see if you can rearrange the equations i gave you earlier to get rid of r and the trig functions

[crimson sedge]

r^2=-3x-2y?

@sand ether sorry for slowness

[sand ether]

Yeh, now just get rid of the r^2

[cedar kiln]

@crimson sedge Has your question been resolved?

[cedar kiln]

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[jade cargo]

Prove that for every variable x, A = x^4 - 8x^2 -x + sqrt(3x^2 - 6x + 9) + 15 is a non-negative number. Find x when A = 0 (algebra only)

[cedar kiln]

@jade cargo Has your question been resolved?

[jade cargo]

<@&286206848099549185>

[rough mango]

what problems have u encountered?

[jade cargo]

FInd the minimum value of A = x^4 - 8x^2 - x + sqrt(3x^2 - 6x + 9) + 2019

this was the original one

[rough mango]

U can just find x right? and prove that there are no negative numbers?

it's given when A=0

[jade cargo]

umm

and prove that it's non-negative

the A = 0 part i figured it out

[rough mango]

so what are the x's

[jade cargo]

x = 2 <=> A = 0

but i need the proof that A is non-negative for every x

[rough mango]

im lost, other than trial and error

[jade cargo]

oh

or simplifying it is also an option?

[rough mango]

yeah i guess you can simplify it

then trail and error

then you can x<A<x or smth

[jade cargo]

there is a fact that

i know the minimum value and the x

but i need proof

[rough mango]

I don't recall how to give proof lol, I thought it was just like given x's, therefore <x<

[jade cargo]

idk if this can help

but minA = 2004 <=> x = 2

[rough mango]

what does <=> mean? If?

[jade cargo]

if and only if

[rough mango]

oh ok,

is it minimum tho?

try x=1

[jade cargo]

umm

[rough mango]

wait what

doesnt make sense

when A=0, x=2

[jade cargo]

approximately 2013,44

[jade cargo]

this one

[rough mango]

ok

[jade cargo]

this happens

[rough mango]

wait what

when A=0 in that equation, the x's are 2013 and 44?

[jade cargo]

i mean

[jade cargo]

for this, minA = 2004 <=> A = 2

[jade cargo]

for this, A = 0<=> x = 2

these 2 are tecnically the same

i just subtracted 2004

[rough mango]

minA = 2004 if a=2?

[jade cargo]

x = 2 not a

[rough mango]

why are there 2 equations

that means a=0 what ever the x is the minimum so that's right then

[jade cargo]

Prove that for every variable x, A = x^4 - 8x^2 -x + sqrt(3x^2 - 6x + 9) + 15 is a non-negative number. Find x when A = 0 (algebra only)

ditch this

FInd the minimum value of A = x^4 - 8x^2 - x + sqrt(3x^2 - 6x + 9) + 2019
take this one

[rough mango]

yeah so this

then find if there are x's above a=0 that are negative, i doubt there is

[jade cargo]

x is allowed to be negative

any real one

[rough mango]

I meant

a that are negative

all x that are above the a=0,

i doubt there are real numbers greater than x <=> a = 0

[jade cargo]

i used a calculator

and it gave me that

[jade cargo]

this thing

um with x = 2 sorry

[rough mango]

what happens to 3?

x=3?

I mean did you check MinA=0?

cause the question was when A=0, what is x

[jade cargo]

let me explain

for the question that the A = 0 thing

[jade cargo]

it's the same thing as this one, but subtracted by 2004

that's why i get A = 0

[rough mango]

ye and when u plug in A = 0 you get x=2, in that ditched eqaution

so when u use A=0 in the "2019" equation,

what's the x?

[jade cargo]

it's also 2

[rough mango]

and isn't it the minA

[jade cargo]

it's all the same thing but subtracted by 2004

[rough mango]

so minA = 0 <=> x = 2?

[jade cargo]

yes

[rough mango]

so all real numbers above x = 2, is A positive?

unless there are negative answers, I doubt it

[jade cargo]

A is always positive and greater or equal to 2004

if x = 2 then A has the value of 2004

for every other x, A > 2004

[rough mango]

in any x's?

[jade cargo]

yes

[rough mango]

ok so that's the proof then

[rough mango]

.(?)

I think,

[jade cargo]

yes it's true

other than 2, every x leads to A > 2004

[rough mango]

that's proof then

[jade cargo]

i need the analysis of the thing

A = x^4 - 8x^2 - x + sqrt(3x^2 - 6x + 9) + 2019

this thing

simplify it so that we can prove that minA = 2004

u got the picture?

[rough mango]

I think so,

I think you can proive it by trial and error

i honestly don't know how to prove it other than trial and error which is vague

but wsince u said every other x is a>2004

then u did trial and error or smth

[jade cargo]

i tried simplifying it

i get this

sqrt(3(x-1)^2 + 6) + (x^2 - 4)^2 - (x-2003)

at this point i'm stuck

[rough mango]

how about this

[jade cargo]

why is it the 8th power? so complex :<

so what do we do next w/ that

[rough mango]

that's the simplified form

I think

OH WAIT

that's the "15" equation

you can simplify it by making A=0, then squaring everything except the 3x^2 equation

[jade cargo]

wdym by the"15" eq?

[rough mango]

there are 2 equations right?

the +15, and +2019

[jade cargo]

umm

didn't I tell u to ditch the 15 one?

[rough mango]

Yeah, I solved that before u told me

[jade cargo]

wait

my brain is loading

[rough mango]

forget about that pic

[jade cargo]

ok

can u simpilfy the 15 one

?

[rough mango]

this was the 15

[jade cargo]

i can take out 2004 from the A and from that simplify the 15 one

[rough mango]

yeah sure jsut add a 2004^2

[jade cargo]

i mean

u simplify the 15 eq

and prove that it's non-negative

from there my problem is finished

[rough mango]

yeah iguess, u can add the simplified equation then prove A>2004 with all real numbers of x

[jade cargo]

umm

can u help me simplify the 15 equation for me pls

[cedar kiln]

@jade cargo Has your question been resolved?

[cedar kiln]

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[golden atlas]

2x+3(4)=27
how do i do this using samdeb

[golden atlas]

2x+3-3(4)=27-3
2x4=24
2x4/4=24/4
2x=6
2x/2=6/2
x=3
is this correct

[glad kestrel]

no

[golden atlas]

its not

how do i do it

[glad kestrel]

i have no idea what samdeb is, but you just need to isolate x

[golden atlas]

its bedmas backwords

[nova geyser]

what its samdeb i dont get it

Oo

[golden atlas]

im in grade 8

[nova geyser]

,calc 3 * 4

[wraith dagger]

Result:

12

[glad kestrel]

still just need to isolate x

[nova geyser]

so $2x + 12 = 27$

[wraith dagger]

IVMC Gaming CH

[nova geyser]

find x

[golden atlas]

im lost

isebnt x 3

[obsidian coral]

It's $$2x + 3 \cdot 4 = 27$$

[wraith dagger]

dldh06

[obsidian coral]

You can simplify the 3(4) first

[glad kestrel]

huh

[golden atlas]

.close

[cedar kiln]

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[sand kayak]

why is this wrong

[sand kayak]

decide the equation for a tangent to the curve

i derived y

[short blade]

you need to find it at that point

so when x=2

[sand kayak]

its 0

[short blade]

actually i'm a little confused

what does the text say

[sand kayak]

it says what i said

[short blade]

i meant the question

[sand kayak]

i the point (2,4)

[short blade]

what does it say in english

[sand kayak]

i told you already

[short blade]

lright

then what you found is the slope of the tangent

now you need to use point slope form to find the equation

[sand kayak]

but y' is the equation of the tangent

[floral thistle]

no it’s not

[short blade]

y' is the slope of the tangent

[floral thistle]

it’s the slope of the tangent

[sand kayak]

so what do i do

[short blade]

.

take the point they gave you

and the slope when x = 2

and find an equation in the form y = f(x)

[sand kayak]

what

i dont get it

what am i suppoed to do

[floral thistle]

y = f’(a)(x-a)+f(a) is the equation of the tangent line to f(x) at (a,f(a))

[sand kayak]

what

[floral thistle]

what don’t you understand

[sand kayak]

like everything

[short blade]

you found y'(x)

[sand kayak]

i thought the derivative = tangent

[short blade]

which is the slope of the tangent line to y at some value of x

it is the slope of the tangent

if we plug in 2, you get that y'(2) = 0

and we're using x=2 since we're finding the equation of the tangent to the point (2, 4)

so now we can use point slope form to find the equation of the line:
y - 4 = 0 * (x - 2)
y - 4 = 0
y = 4

[sand kayak]

i dont get it

[short blade]

which part

[sand kayak]

every part

[short blade]

please be more specific

do you understand what y'(x) is?

[sand kayak]

all this time i have been taught by the book that tangent = derivative

we draw tangents to get derivatives ..

[short blade]

i can't speak as to what the book says

but what's important is that derivative = slope of tangent

[sand kayak]

i dont understand what to do with this method

i got y-4=0(x-2)

[short blade]

mhm

[sand kayak]

this isnt helpful at all

[short blade]

so what does y =?

[sand kayak]

i dont know

[short blade]

what's 0 * (x-2)?

[sand kayak]

0

[short blade]

then y - 4 = 0

so what is the value of y

[sand kayak]

4

[short blade]

so your line is y = 4

the equation for the tangent line for y = x + 4/x at (2,4) is y=4

[sand kayak]

i am lost again

[short blade]

which part

[short blade]

you calculated y'(1) incorrectly

[sand kayak]

How

[short blade]

what's 1^(-1/2)

[sand kayak]

1

[short blade]

and 1 * 1/2?

[sand kayak]

1/2

[short blade]

and 1 + 1/2?

[sand kayak]

1,5

[short blade]

then y'(1) = 1.5

[sand kayak]

I dont know how calculating y’ helps me get the equation

[short blade]

it's the slope of the line

you can now use 1.5 in point slope form:
y - y1 = m(x - x1)

so
y - 2 = 1.5(x - 1)

and solve for y

[short blade]

looks good

[cedar kiln]

@sand kayak Has your question been resolved?

Channel closed

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Use .reopen if this was a mistake.

[sand kayak]

how do i write 2^56 in basic power form

[cedar kiln]

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[sand kayak]

is e a constant?

[floral thistle]

yes

all numbers are constants

[sand kayak]

Is this correctly derived?

I need to calculated the speed at t=125 and i am getting the wrong answer

I get 0,03 or 0,16

[radiant topaz]

,rotate

[radiant topaz]

What's the speed tho

[cedar kiln]

@sand kayak Has your question been resolved?

[sand kayak]

0,031

In a nutrient-rich lake, aquatic plants can quickly cover the surface. One year, 250 m2 in a lake is covered by aquatic plants. 8 years later, 375 m2 is covered.

Assume that the increase is exponential and calculate

a) how many m2 are covered after 15 years.
Round to the nearest integer.

i got 1,05

[cedar kiln]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[sand kayak]

and its wrong

.reopen

[cedar kiln]

✅

[radiant topaz]

Show the question instead

[sand kayak]

i did

[radiant topaz]

For this

[sand kayak]

i solved it so there is no need

[radiant topaz]

Ok

[sand kayak]

why is 1,05 wrong

[radiant topaz]

Ok ill check the other

So we have $P(1)e^{7r} = P(8)$ So $e^{7r} = \frac{P(8)}{P(1)} = \frac{375}{250}$\

Now $P(15) = P(1)e^{14r} = P(1) \frac{375²}{250²} = \frac{375²}{250}$

[wraith dagger]

it's Sam

[sand kayak]

what is that

[radiant topaz]

That's how exponential increase work

[sand kayak]

why doesnt my solution work? because it should

i set up an exponential equation

[radiant topaz]

First of in your solution it's a⁷

And second you don't have to find a

[sand kayak]

its a^8

[radiant topaz]

You have to find 250 a¹⁴

[radiant topaz]

See in 1 year it is 250 and in 8 year it's 375
So in 7 years it increased by 250a⁷

[sand kayak]

i dont get it

[radiant topaz]

8 years - 1 year = 7 years

[sand kayak]

but a^1 wouldnt change anything

[radiant topaz]

Why not

a^1 is change in one year

If initial value is p then after one year it's pa

In two year it's pa²

And so on

Here at 1 year the amount is 250
And at 8 year it is 375
So it increased from 250 to 375 in 7 years

That's why 250 a⁷ = 375

[sand kayak]

okay

[radiant topaz]

And we want to find what happens after 15 years

So it will be equal to 250a¹⁴

And a⁷ = 375/250

[sand kayak]

i got a= 1,05

[radiant topaz]

No need to find a

[radiant topaz]

We get this right

And we want to find 250a¹⁴

So we get 250(375/250)²

Which is 375²/250

,w 375²/250

[wraith dagger]

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