help-0

[median oar]

Yeah you should know what 2ˣ looks like

Then just shift it down by 1

[alpine sable]

Yeah yeah i got it

I got confused while you kept saying it's not c

Ty for the help though

.close

[lone heart]

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[thorn echo]

Is absolute value a 1-1 inverse function?

[thorn echo]

@median oar sorry for the ping, ur the person that entertained my question connected to this earlier

[median oar]

No

[thorn echo]

I would really appreciate if u can help me again, that's if u know the answer

[thorn echo]

it's not?

[median oar]

One to one means if I tell you the output

You can tell me the unique input

If I tell you |x| = 3

You don’t know if x was +3 or -3

[thorn echo]

So if it the input is uncertain, then it's not 1-1?

[median oar]

Wait wdym 1-1 inverse function

[thorn echo]

Nvm the inverse part, i may have mixed it up

Mb

[median oar]

The the definition of a one to one function

A one to one function (injective) has the property

$(f(a) = f(b)) \implies a = b\quad \forall a,b \in \text{domain}(f)$

[ocean seal]

Frosst

[median oar]

All that is saying is

If I get 2 outputs and they are the same

Then the inputs must’ve been the same as well

For all possible inputs/outputs

So each output can be uniquely identified with an input

(If it exists)

[thorn echo]

Alright gets

So that means absolute value is not 1-1?

Like y=|x|

[lone heart]

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[pliant harness]

Hello everyone I hope you are well.

I am studying maths from a book and im quite confused.

Im stuck on question g.

The question is simplify the expression ab-ba.

Now I found all the otoher questions eg as all I did was collect the like terms.

However, I find ab - ba difficult because there are no like terms.

My answer to this question was its an unsolvable equation as there are no like terms.

It looks like a very simplle questioon but I havent studied maths for 4 years now so im a bit rusty

[echo socket]

a and b are numbers, right?

[pliant harness]

hi lonely bean

um its letter

wait actuaally

a and b are 1

i think

[echo socket]

Why would ab and ba be different then

[pliant harness]

because its not like terms i thought like terms have to be the same letter and same order.

ab and ba are switched around

im confused

[echo socket]

So you would regard $x \cdot x^2$ and $x^2 \cdot x$ as not similar terms even though they both are $x^3$?

[ocean seal]

A Lonely Bean

[pliant harness]

ah I see

they will both be x to the power of 3

[quasi scarab]

both have a in them, so you could do a(b-b) -> a*0 = 0 (or notice that ab=ba so they are the same)

[pliant harness]

thanks Spreeter.

so the answer will be 2a - 2b = 0

[echo socket]

Where's 2 coming from?

Weren't you asking aboyt ab - ba?

[pliant harness]

i thought I had to add up all the A and the B. So there are 2a and 2b in this expression ab - ba.

[quasi scarab]

thats wrong on many levels

since when can you do something like that?

[pliant harness]

Im not sure I havent done this maths in a long time so im very confused.

[quasi scarab]

2a-2b is only 0 if a=b , and that wasnt given anywhere

[pliant harness]

It seems like a very simple question but somehow I struggle

[quasi scarab]

i gave you the solution

[pliant harness]

both have a in them, so you could do a(b-b) -> a*0 = 0 (or notice that ab=ba so they are the same)

thank you for the solution Spreeter

[quasi scarab]

do you understand it?

[pliant harness]

to be honest no

so ab and ba are the samme

thats the answer

because they are the same letters just different order

[quasi scarab]

yeah

5 * 7 = 7 * 5 for example

and this goes with any numbers

its called commutative property

[pliant harness]

oh yay now I get it,

thank you Spreeter.

So if I were to right the answer.

All I would say is ab - ba = the same as it has the same letters just different order.

[quasi scarab]

ab-ba=ab-ab =0

[pliant harness]

Thanks Spreeter. Now I learned a new thing. That 0 means the same with expressions

I appreciate you being patient with me Spreeter

[pliant harness]

Thank you Lonely Bean as well for youre help

[quasi scarab]

0 means zero, as that expression will always give us 0 as its value

do you want me to check a-f too?

[pliant harness]

Oh yeah Just d please Spreeter.

I actually found d the second most difficult as it was the longest.

But I managed to do it

Was this correct Spreeter or did I miss a few things out?

I got 8r squared + s + rs - 3s squared

.

I didnt know how to write 8r squared on keyboard spreeter

[quasi scarab]

8r^2

[pliant harness]

Thank you Spreeter for teaching me that.

So my answer was 8r^2 + s + rs - 3s^2

yay.

youre the best Spreeter.

I was struggling most with g but you explained it in a way I understood at the end.

Thank you Spreeter

[quasi scarab]

you are welcome

[lone heart]

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[wise stirrup]

having trouble with relational rates outside of geometric shape examples

[lone heart]

@wise stirrup Has your question been resolved?

[wise stirrup]

I realized it’s a triangle but I don’t believe my set up to be correct because I still have dy/dt aaand dx/dt

"when dy/dt is zero"?

to solve for dx/dt?

[wise stirrup]

<@&286206848099549185>

[lone heart]

@wise stirrup Has your question been resolved?

[wise stirrup]

Where did I go wrong?

[lone heart]

@wise stirrup Has your question been resolved?

[wise stirrup]

.close

[lone heart]

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[alpine sable]

How can I show this using the derivation ?

[ivory pivot]

do derivatives and show that its equal to zero

then then function on LHS is constant

[alpine sable]

LHS?

[ivory pivot]

and compute is value for a particular x

left hand side

[alpine sable]

F’(x)=0

[ivory pivot]

yes

[alpine sable]

And I should find an x that f(x)= pi/2?

In R

[ivory pivot]

then lets try the value x=0 and find F(x)=pi/2

yes

try x=0

[alpine sable]

Yeah it works

[ivory pivot]

gg

[alpine sable]

But I have a question why if the function is constant it’s makes this true ? Like what is the relationship ?

[ivory pivot]

if you know that the function is constant you can exploit that by verifying the identity only in one point ...in general you have to verifying the identity for all x

because constant(ness) ensures you that the function has only one value

[alpine sable]

Oh I see now

Thank u so much

[ivory pivot]

np

[alpine sable]

.close

[lone heart]

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[sonic harbor]

Hello! 😄
Derivative approximation evaluation:
Why do I get 1.68 when using Richardson (Q0026) for the second function, cos(x^2-1) at x = 0?

[steady mantle]

cos(-1) != 0

[sonic harbor]

The assignment is to evaluate the derivative at x = 0

Q0026 is Richardson and Q0025 is numm_diff. Richardson being a method to approximate the derivative.

[ocean seal]

Result:

0.54030230586814

[sonic harbor]

-.-

[steady mantle]

oh wait nvm

!show

[lone heart]

Show your work, and if possible, explain where you are stuck.

[sonic harbor]

.close

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[ocean seal]

Nerdy_Coder

https://media.discordapp.net/attachments/359052581022203914/1165399064981016686/image0.jpg?ex=6546b575&is=65344075&hm=bc210c92783c6bd9500955f32043ce073a4fc7d3c028f06aeb92a05c01f041dd&=&width=800&height=958

How did they do this? $$ u\frac{\Delta v}{\Delta x} $$  $$ because Prove $$ \Delta a \cdot \Delta b = \Delta(ab)$$ $$ (a_2 - a_1) = a_2b_2 - a_1b_1 = a_2b_2 - a_2b_1 - a_1b_2 + a_1b_1$$
```Compilation error:```! Misplaced alignment tab character &.
l.57 ...165399064981016686/image0.jpg?ex=6546b575&
                                                  is=65344075&hm=bc210c92783...
I can't figure out why you would want to use a tab mark
here. If you just want an ampersand, the remedy is
simple: Just type `I\&' now. But if some right brace
up above has ended a previous alignment prematurely,
you're probably due for more error messages, and you
might try typing `S' now just to see what is salvageable.```

[lone heart]

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[native temple]

https://media.discordapp.net/attachments/359052581022203914/1165399064981016686/image0.jpg?ex=6546b575&is=65344075&hm=bc210c92783c6bd9500955f32043ce073a4fc7d3c028f06aeb92a05c01f041dd&=&width=800&height=958

How did they do this? $$ u\frac{\Delta v}{\Delta x} $$ because $$ \Delta a \cdot \Delta b = \Delta(ab) = $$ $$ ab_2 - ab_1 = a_2b_2 - a_1b_1 = a_2b_2 - a_2b_1 - a_1b_2 + a_1b_1$$

[ocean seal]

Nerdy_Coder

[pallid scarab]

$$" \Delta a \cdot \Delta b = \Delta(ab)"$$
This is where you go wrong

[hot bluff]

thats not how the delta works. you can't factor out a delta.

[ocean seal]

rafilou2003

[hot bluff]

delta is part of the notation of the variable.

[native temple]

Yea I know so how did my book do it?

You can see that when I evaulated Delta a times Delta b

It doesn't equal delta a times b

[pallid scarab]

$\Delta(ab) = \Delta a \cdot b + a \cdot \Delta b$

[ocean seal]

rafilou2003

[pallid scarab]

• some additional residue

[native temple]

So how can you factor the a

[native temple]

What?

That is imnportant

[pallid scarab]

and the additional residue is $\Delta a \Delta b$

[ocean seal]

rafilou2003

[native temple]

otherwise the chain rule is the residue

[pallid scarab]

Which goes to 0

[native temple]

Okay so can you convert that to

[ocean seal]

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

[pallid scarab]

?

[lone heart]

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[native temple]

...

what

[native temple]

/close.close

.close

[lone heart]

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[languid hinge]

Anyone can teach me math???????

[native temple]

.close

[lone heart]

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[cinder lion]

Can someone help me with percentage changes? I’m not sure how to answer this question

[cinder lion]

Any suggestions on how to do this would be appreciative. Is the question trying to achieve the original value?

[lone heart]

@cinder lion Has your question been resolved?

[cinder lion]

<@&286206848099549185>

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@cinder lion Has your question been resolved?

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[tulip oasis]

hi i am really having trouble with these odd power graphs
like x^5 and x^3
How does a horizontal shift to the left make it look like any of those options
ALso if anyone knows a video that can help explain this concept for these odd power functions plz
Thanks very much

[tulip oasis]

( i ask this earlier but went afk and lost the chat ) sry

LikI do realize it has a horizontal shift to the lift but these options are have such a dramatic shape change it is so odd to me

[lofty gorge]

part of the reason for the shape change is that the drawing itself is 'zoomed out' (look at the axes)

if you zoom in to where the function does the wiggle, does that help you see that the shape isn't changing that much?

[lone heart]

@tulip oasis Has your question been resolved?

[lone heart]

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[vapid steppe]

ignore the work

is the answer 98/25 ft/s?

.close

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[alpine sable]

This is the answer

I really don't understand how they got to that. x3 = 0 sure but they just came up with y3 = 0 from where?

This is theorem 1

how is their answer of "they have a coordinate of zero" sufficient to utilizing theorem 1?

[lofty gorge]

the labels x and y are arbitrary

[alpine sable]

i understand

but how did they check if its a nonempty subset?

[lofty gorge]

so if (y1, y2, y3) is in W, y3 = 0

they technically didn't but it's pretty obvious (0,0,0) is in W

[alpine sable]

afaik, (y1, y2, 0) being in W does not make it a nonempty subset

the 0 vector needs to be in W

[lofty gorge]

(really they should have at least mentioned it in an introductory linear algebra problem)

[alpine sable]

and you check that by plugging 0 into the equation

but i wasnt given an equation

[lofty gorge]

so the zero vector in R^3 is (0,0,0)

i.e. the vector (x1,x2,x3) where x1=0, x2=0, x3=0

[alpine sable]

how do you know the zero vector is in R^3 though

[vale crag]

x3 = 0, how is that not an equation ?

it depends on only one coordinate of the vector, sure

[alpine sable]

thats fine because (0) = 0 but

[alpine sable]

is this answer sufficient because it shows that their sum and scalar multiple has 0 in their vectors?

how

[vale crag]

yes

[alpine sable]

like how is that good enough though

[vale crag]

what do you mean by how

[alpine sable]

if you gave me that as a proof to prove u,v in W so u + v in W id be so confused

because it crosses the origin?

so it must be in there i guess?

[vale crag]

If take two vectors for which the third coord is 0, then the third coord of the sum is also 0

that's the translation of "u,v in W so u + v in W" in english

I'm not exactly sure what's confusing you

[alpine sable]

why does the third coord of the sum also being 0 prove its in W?

[vale crag]

cause that's what the criterion for being in W is

"W is the set of all vectors for which the third coordinate is 0"

[alpine sable]

thank you

i understand now

.close

[lone heart]

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[queen escarp]

yo

[queen escarp]

so i got a simple question

(2x+3)^2

that would equal 4a^2 +6x + 9 right

[white crater]

yes

well

[dusty radish]

No

[white crater]

wait

[queen escarp]

but when I expand it (2x+3)(2x+3)

[white crater]

no

[pallid scarab]

not true

[queen escarp]

you get 12x instead of 6

[white crater]

12x

yeah

[pallid scarab]

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

[white crater]

also not a btw

[proven leaf]

prolly just a typo tbh

[queen escarp]

but what about this

[white crater]

ye

just pointing it

[queen escarp]

yh mb=

[proven leaf]

that is generalized

[queen escarp]

so what doees that mean

i cant use it in this case?

[pallid scarab]

you can, you just used it incorrectly

[proven leaf]

no it means you can use it in more cases

[queen escarp]

how exactly

[proven leaf]

a=2x

[pallid scarab]

.

[proven leaf]

b=3

[queen escarp]

ohhh mb

i get it now

i forgot to multiply it by the 2

[proven leaf]

$(\textcolor{red}{2x}+\textcolor{blue}{3})^2=(\textcolor{red}{2x})^2+2(\textcolor{red}{2x})(\textcolor{blue}{3})+(\textcolor{blue}{3})^2$

[ocean seal]

PajamaMamaLlama

[queen escarp]

thanks guys

[proven leaf]

[pallid scarab]

no problem

[queen escarp]

[proven leaf]

[queen escarp]

exam tmmr so this helps tons

[proven leaf]

hope the colors did I've been using them a bit more recently and they seem to be helping more :)

[queen escarp]

thats insane

it takes a while doesnt it

good lad

[proven leaf]

well it's just copy and pasting but hope you have a good one chief, and good luck on ur test

[queen escarp]

thanks llama

one more thing,,,,

i sec

whats the diffference between these two

[pallid scarab]

well in the one on the left, you first do the difference, and then you square the whole thing.
In the one on the right, you square both terms and then compute their difference

[queen escarp]

well both start with a^2 - b^2

so why are they both different

[pallid scarab]

?

$(a-b)^2$ is not $a^2-b^2$

[queen escarp]

its not?

[ocean seal]

rafilou2003

[queen escarp]

so when theres no brackets u use right, but when there is you use left

[pallid scarab]

yes

[queen escarp]

okay thanks

[pallid scarab]

the one on the left is Difference, THEN square
the one on the right is square (both), THEN difference

[queen escarp]

ok

[lone heart]

@queen escarp Has your question been resolved?

[queen escarp]

probably

but i might have more

[lone heart]

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[sick quarry]

a convex hexagon ABCDEF with an interior point P is given. Assume that BCEF is a square and both ABP and PCD are right isoceles triangles with right angle at B and C, respectively. Lines AF and DE intersect at G. Prove that GP is perpendicular to BC

[sick quarry]

Can someone help me solve it

<@&286206848099549185>

plssss

[lone heart]

@sick quarry Has your question been resolved?

[sick quarry]

nah

[sick quarry]

<@&286206848099549185>

plssss

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[lone heart]

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[alpine sable]

how do you integrate e^(sinx)?

[lone heart]

@alpine sable Has your question been resolved?

[tacit arch]

,w int e^sin(x) dx

[ocean seal]

Results provided by WolframAlpha

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

Requested by rie.mann

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[charred echo]

how do i do questions from 9 - 10

[jagged cobalt]

what parts of them dont you understand

[charred echo]

sorry i just dont get any of it

<@&286206848099549185>

[lone heart]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[jagged cobalt]

were you able to do the prior questions

[charred echo]

my teacher told me to skip those ones

and only do 9-10

[jagged cobalt]

do you know what an x intercept it

[charred echo]

yea

[jagged cobalt]

how do you find it

[charred echo]

2x+10=0
2x=-10
x=-5

[jagged cobalt]

good, so what is the point E

[charred echo]

thats the part i dont know

[jagged cobalt]

youve done all the work for it, you just have to write the coordinate

you know x=-5, what does y=

[jagged cobalt]

remember its the x intercept, what did you set y= to when you did this

[charred echo]

y+10=0
y=-10

i did this

but the answer is meant to be 0

idk what to do

[jagged cobalt]

that would be the y intercept, youre overthinking it

when i asked you to find the x intercept, you set y=0 didnt you

[charred echo]

oh

[jagged cobalt]

so whats the point E

[charred echo]

wait i got it

so how do i do 10

[jagged cobalt]

bit more complicated, it may help to get both lines into the form y=mx+c

[charred echo]

yes

what do i do after

[jagged cobalt]

set them equal to each other

and solve for x

[charred echo]

okay

im just wondering and since 7a has a circle, how do you do that one

it looks more confusing

[jagged cobalt]

B C and D will all lie on the same line, C will be the midpoint of BD

that might help

[charred echo]

mhm

so how to work it out

[jagged cobalt]

if you have two points (a,b) and (c,d) then the midpoint is
$$\left(\frac{a+c}{2},\frac{b+d}{2}\right)$$

[ocean seal]

AℤØ

[charred echo]

what to do after

[jagged cobalt]

after what

[charred echo]

how to get the coordinates for d

[jagged cobalt]

this

C is the midpoint of B and D

[charred echo]

it doesnt give me the right answer

the answer is supposed to be D(9, 6) but how to get that

[jagged cobalt]

what did you do

[charred echo]

5+1=6/2=3
10+14=24/2=12
(3, 12)

[jagged cobalt]

C is the midpoint, not an endpoint

and you should have x's and y's

[charred echo]

so do u substitute

[jagged cobalt]

no?
the midpoint of (1,14) and (x,y) is (5,10)
this means (1+x)/2=5
and (14+y)/2=10

[jagged cobalt]

by use of this

[charred echo]

im still not getting the answer

[jagged cobalt]

how are you solving (1+x)/2=5

[charred echo]

idk how to solve that

[jagged cobalt]

$$\frac{1+x}{2}=5?$$

[ocean seal]

AℤØ

[charred echo]

so how to solve y

[jagged cobalt]

did you solve this

[charred echo]

yes

[jagged cobalt]

then solving y shouldnt be an issue

[charred echo]

ok

.close

[lone heart]

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[spiral elk]

4/x<x

Instructions are to resolve the nonlinear inequality

[lone heart]

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[subtle birch]

open a new channel

[spiral elk]

Okay

[subtle birch]

This one will close abdruptly

[spiral elk]

Thank you

.close

[subtle birch]

well, it is already closed and soon you won't be able to type

So no need to ".close"

[lone heart]

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[balmy patrol]

how do i do this ?

[vernal basin]

x+y = z+w
x+w = y+z

a bit of moving around with these equations leads to

y = wx+w
z = x

Meaning it is a 2D space, since of {x,y,z,w}, only x,w are linearly independent

[balmy patrol]

just give me minute

[vernal basin]

Alternative explanation: if x and w are already given, then y and z are also "already given" by the equations. Therefor you only really have 2 degrees of freedom

[balmy patrol]

yep, got it

[balmy patrol]

where can i find more questions like these ?

any idea ?

also this one ?

[balmy patrol]

isnt it 2 ? for this ?

no wait got it

.close

[lone heart]

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[alpine sable]

I got some math home work and I am struggling on this question in specific with factoring polynomials I think its because im struggling to get my multiplication tables down but I feel like I am not grasping this well enough

[alpine sable]

also this is on pearson which means I gotta put it in a specific way which is a pain in my butt sometimes

[vale wigeon]

they tell you exactly what to factor out

they want you to factor out $4a^{-17/4}$

[ocean seal]

AnnGhost

[vale wigeon]

$8a^{-17/4} = 4a^{-17/4} \cdot (???) \ -4a^{-5/4} = 4a^{-17/4} \cdot (???)$

[ocean seal]

AnnGhost

[alpine sable]

ohhhh

I got this tysm

.close

[lone heart]

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[golden ridge]

under which conditions do the two equations:
a1x1 + ... + anxn = b and a1'*x1 + ... + an'*xn = b'
describe parallel hyperplanes in R^n?

[mortal trellis]

think about their normal vectors

[lone heart]

@golden ridge Has your question been resolved?

[lone heart]

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[ionic sedge]

hi there i am having an issue with this exercise

[ionic sedge]

This is my solution

and this is the solution of the book

where did i go wrong???

i think its either with the cross product or with ab = b-a v ac=c-a

oh sorry, the pictures are rotated

Rotated pictures of my solution

[alpine sable]

To find the equation of a plane that passes through three non-collinear points, you can use the point-normal form of the equation for a plane. In this form, the equation of a plane is given by:

Ax + By + Cz = D

where (A, B, C) is a normal vector to the plane, and (x, y, z) are the coordinates of a point on the plane.

To find the normal vector, you can use the cross product of two vectors formed by the given points. Let's calculate it:

Point A(1, 1, -1)
Point B(2, 0, 2)
Point C(0, -2, 1)

Vector AB can be found as:

AB = B - A = (2 - 1, 0 - 1, 2 - (-1)) = (1, -1, 3)

Vector AC can be found as:

AC = C - A = (0 - 1, -2 - 1, 1 - (-1)) = (-1, -3, 2)

Now, to find the normal vector N to the plane, take the cross product of AB and AC:

N = AB x AC

N = (1, -1, 3) x (-1, -3, 2)

To calculate the cross product, use the determinant of the following matrix:

|i j k |
|1 -1 3 |
|-1 -3 2 |

N = (i(1 * 2 - (-1) * (-3)) - j(1 * 2 - 3 * (-1)) + k(1 * (-3) - (-1) * (-1)))
N = (i(2 + 3) - j(2 + 3) + k(-3 - 1))
N = (i(5) - j(5) + k(-4))
N = (5i - 5j - 4k)

Now you have the normal vector N, which is (5, -5, -4). You can use this normal vector and one of the given points (let's use point A) to form the equation of the plane:

5x - 5y - 4z = D

Now, substitute the coordinates of point A (1, 1, -1) into the equation:

5(1) - 5(1) - 4(-1) = D
5 - 5 + 4 = D
4 = D

So, the equation of the plane that passes through the points (1, 1, -1), (2, 0, 2), and (0, -2, 1) is:

5x - 5y - 4z = 4

[hushed locust]

is that ai generated? it's not correct

[alpine sable]

it is correct and yes it is ai genereted but i checked it

[pearl hamlet]

chatgpt cannot do math for the record

[ionic sedge]

yet a different awnser then my book :D, but my book could be wrong too

[alpine sable]

yah

[ionic sedge]

yeah it is ment for language things

but i think i did the cross product wrong

[hushed locust]

yes

[ionic sedge]

i see my error, i did -1*2 - -3x3 = -2-9 = -11

wait ill redo it 😄

[alpine sable]

oh

[ionic sedge]

wait i am confused now

[alpine sable]

hmmmmmm

[hushed locust]

(2, 0, 2), (0, -2, 1) both don't fall on that plane. did you only check one point?

[alpine sable]

hmmmm

lets see

[hushed locust]

the book's answer is correct. you should redo the cross product and go from there

[ionic sedge]

yeah i am a bit confused on what I am doing wrong

oh sorry rotated again

[hushed locust]

should be (-1*2) - (-3*3), not +

[ionic sedge]

ohhh

okay that makes more sense

ty

,close

[hushed locust]

use .close

[ionic sedge]

.close

[lone heart]

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[lone heart]

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[vague turret]

hello

[vague turret]

i need to know how

became

Its about ''Find the LCD''

[swift grail]

multiply the denominators together

then change the top accordingly

[vague turret]

3x x x?

and then 3 x 5

[swift grail]

the lowest common denominator is (3x+3)(x+5)

[vague turret]

what do i get it

where*

[swift grail]

wdym where

[swift grail]

from the denominators here

[vague turret]

how do i knoe the least common denominator

[swift grail]

you just need the denominator to be the same really

just that in this case, to get the same denominator you multiply both the denominators together

[vague turret]

who do i multply by who

[swift grail]

for instance, 3/5 + 6/11, how would u solve this?

[vague turret]

15 + 66 ?

oh wait

[swift grail]

no

not how it works

your denominator must be the same right?

[vague turret]

18/55 ???

[vague turret]

ye

[swift grail]

ok

so how did u get 55

[vague turret]

i multiplied 5 and 11

[swift grail]

exactly

[swift grail]

.

its the same thing

[vague turret]

ok wait so i multiply 3x and x and the other 3 times 5

like that?

[swift grail]

you multiply (3x+3) and (x+5) together

what do u get from this ^

[vague turret]

oh wait

3x and 3 is multipled and x and 5 is multiplied?

so 9x and 5x?

[swift grail]

no what

do u know how to multiply (3x+3)(x+5)

[vague turret]

i dont think so

if i do

3x(2) then 15

wait

yeah

[swift grail]

u gets?

[vague turret]

wait a min

is that the foil method?

[swift grail]

i dont know what foil is but

uh sure

i think so

[vague turret]

damn i am so confused

the lines are weird

[swift grail]

do u happen to use your calculator to do algebra

[vague turret]

is the x times 5?

[vague turret]

no

[swift grail]

ok well do u know how foil works?

[vague turret]

yeah F= First
O=outer
I=inner
L=Last

[swift grail]

yeah idt its a good way to learn this ngl

[swift grail]

so what this is, is that u take 3x and multiply by (x+5), then take 3 and multiply by (x+5)

do it seperately

so 3x(x) + 3x(5) + 3(x) + 3(5)

whats what the lines are

no help remembering an acronym if you dont know what it means though

[vague turret]

ok i get it nw

how bout the lsat one?

[swift grail]

wdym the last one

[swift grail]

??

just add it together

15 + 3 = 18

[vague turret]

oh so only add the middle one?

[swift grail]

yeah you cant add x^2 to x

nor can you add 18x to 15

i suggest you work on your algebraic fundementals before working on questions like this

its important you master working with the basics first

[vague turret]

what are the basics?

[swift grail]

addition

multiplication

subtraction

division

[vague turret]

i know those

[deep mauve]

@swift grail is he trying to solve product of binomial?

[swift grail]

no hes just asking how to make the denominator the same

but hes not sure how to multiply

(3x+3)(x+5) < this

[vague turret]

ok how bout the upper part

[vague turret]

ik this now

[swift grail]

well what did you do when u got this?

[vague turret]

multiply them

[swift grail]

ok

[vague turret]

wont i get 4x?

[deep mauve]

What's this?? Rational algebraic expression?

[vague turret]

find the lcd

[vague turret]

yeah idk

[swift grail]

its just addition of fractions

you find a common denominator

then add it together

so if you multiplied your denominator by (x+5) you need to multiply the numerator by (x+5) as well

right @vague turret

[vague turret]

yeah

[swift grail]

so what is 2x(x+5) - 2(3x+3)

[vague turret]

2x(2) 10x then

[swift grail]

?

[vague turret]

what

[swift grail]

what is your answer to this

send in your full answer and not your partial working

[vague turret]

2x(2) and 10x for the other one with the negative -6x and -6

[swift grail]

so how would u write that

on a piece of paper

[vague turret]

2x^(2)+10x-6x-6

[swift grail]

good

and if you simplify that what would u get

[vague turret]

do i add the middle one?

[swift grail]

yes

do you know why?

[vague turret]

why?

[swift grail]

because they both have x

do u know what is 3x+2x?

[vague turret]

5x

[swift grail]

uh huh

yeagh

[swift grail]

same goes for this

[vague turret]

so 2x(2)-16x-6 ???

[swift grail]

where is your x

noo

why is it 16

+10x-6x

[deep mauve]

@vague turret as long as they have the same variable, just remember that

[swift grail]

^ this

[vague turret]

16x(2) ?

[swift grail]

no

ur getting colder

try again

[vague turret]

i have to add the middle one or subtract?

[swift grail]

.

positive 10x minus 6x

what is that

[vague turret]

4x

[swift grail]

ok

so what is your answer to the question

[vague turret]

2x^(2)+4x-6

[swift grail]

and is that the same?

[vague turret]

yeah

oh ok i get it now

[vague turret]

wait is this

find the LCD

oh wait its not

[vague turret]

is this the same to this

[swift grail]

yes

so what are your steps

[vague turret]

oh ok

the demaninator is

x times x and x times 1

since there is x number 1 wont show up

[swift grail]

which is?

[vague turret]

x(2) + x

[swift grail]

ok

well the correct way to write it is x^2

x^2 = x squared

[vague turret]

idk where to get that pointy thing

[swift grail]

x(2) is 2x

its on the 6 key

shift 6

[vague turret]

oh so thats what t is

kinda big

6^

okay

what bout

[swift grail]

ok whats your numerator

[vague turret]

upper one

[swift grail]

i explained earlier

[vague turret]

lemme read back

[swift grail]

if u multiply the denominator you need to do the same for numerator

[vague turret]

ok so 5(x+1) and x(x)

[swift grail]

ok

so adding those together

what do u get

[vague turret]

5x+5+x^2

is that it?

[swift grail]

look

[vague turret]

if this is correct 5x+5+x^2+X^2

[swift grail]

does it match your answer?

the answer given?

[vague turret]

no

[swift grail]

means its not correct then

[swift grail]

does this match?

[vague turret]

no

wait

[swift grail]

why not?

[vague turret]

does not feel like it

[vague turret]

because 5x is first and on the answer 2^2 is first

oh i think i have to reverse them wait

x^2+5x+5

is that it?

oh wait it is

did i do it right?

[swift grail]

yes

[vague turret]

oh ok

feels cool

thanks

btw

is this the same equation

or problem

welp that goes it ig

.close

[lone heart]

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[median oar]

but multiplication is not distributive over multiplication

we have that (a + b) * c = a * c + a * b

we don't have (a * b) * c = a * c * b * c

[alpine sable]

instead of just x * x + 3 * -4

[median oar]

but we have addition/subtraction inside the brackets

here we have multiplication inside the brackets

[median oar]

this property here is distributivity

this doesn't apply to multiplication with multiplication

[alpine sable]

alright so I should just remember when it's adding inside of a bracket you multiply the like-terms of both brackets and when its multiplication you times everything inside the first bracket by the second?

thank you very much for your help i didnt realise it differed cuz the * and +

m.close

.close

[lone heart]

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[atomic creek]

How many three-digit numbers become bigger when their digits are reversed?

[lone heart]

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

[atomic creek]

1

[exotic canopy]

let $\overline{abc}$ be your number where $a, b, c$ are your digits. when is $\overline{abc} > \overline{cba}$?

[ocean seal]

artemetra

[exotic canopy]

try expanding the numbers in terms of hundreds, tens and units

[prime badge]

around half of them

i find it really hard to explain

[atomic creek]

how do I know how many are there

[pallid scarab]

First, how do a and c compare, given the digits of hundreds ?

[atomic creek]

is there any simple solution ?

[prime badge]

yes

i'm trying to explain it, but no luck

[pallid scarab]

Look at the hundreds digits in "abc > cba", what do we know for sure about a and c?

[prime badge]

basically there's three types of numbers, they either become smaller, or bigger or neither

and it's impossible that there's more numbers of the second type than the first

[exotic canopy]

you will get that $100a + c > 100c + a$, then you simplify this and test all pairs of what digits $a$ and $c$ can be, since $a, c \in {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}$

[exotic canopy]

also a good approach

[ocean seal]

artemetra

[prime badge]

because all numbers from the first group become numbers from the second group when reversed, and 2 different numbers don;t become the same number

there, it's incomprehensible

[exotic canopy]

the set of all triple digit numbers is partitioned between those 3 groups

let me attempt to explain too lol

either they become bigger when reversed, smaller when reversed, or don't change

so for the first group, the action of reversal maps a number to a bigger one, and since the original number is smaller than the new one, it means that reversing can only happen from the set of smaller numbers to bigger numbers (and vice versa)

thus, the amount of all 3-digit numbers that become bigger when reversed is (#{all triple digit numbers}-#{all numbers that don't change})/2

$\frac{#{\text{all triple digit numbers}}-#{\text{all numbers that don't change}}}{2}$

[ocean seal]

artemetra

[exotic canopy]

where #{...} means 'number of'

@prime badge @atomic creek here you go, hopefully this is comprehensible

[atomic creek]

I just needed to know the answer 😭

[exotic canopy]

we don't give out answers here

[atomic creek]

oh alr

[exotic canopy]

i technically did most of the reasoning work for you, here's a formula

[atomic creek]

ok ok

[lone heart]

@atomic creek Has your question been resolved?

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[alpine sable]

does this simplify to 2(x-11)
or do you divide it by two and it becoems 2x-11

[drifting beacon]

Notice that 4x=2(2x) and 22=2(11)

Combine the like terms

[swift saffron]

2(2x-11)

[alpine sable]

wait wouldnt it be 2(2x-11)

fully simplified

or is 4x-11 already simplified

[swift saffron]

Yea

I don’t know why I forgot that 2

So 2(2x-11)

[alpine sable]

but that can be expanded

so wouldnt 4x-22

already be

the simplest form

[swift saffron]

I think you will want you answer factored out

[lone heart]

@alpine sable Has your question been resolved?

[lone heart]

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[plucky canyon]

<@&286206848099549185> In a word jumble, there are 8 consonants and 5 vowels given. Find out in how many ways can we form a 5-letter word having three consonants and 2 vowels?

[idle crane]

this is

hello

@plucky canyon

so to think about it

[plucky canyon]

Yes

[idle crane]

6720

[plucky canyon]

8C3 x 5C3 right

[idle crane]

is the answer

[plucky canyon]

8P3 x 5P3

[idle crane]

so by using a formula

[plucky canyon]

5P2

Bro pls don't leave

[lone heart]

@plucky canyon Has your question been resolved?

[digital basalt]

probability formula

[lone heart]

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[marsh rapids]

according to the phrasing of the question, you should definitely not assume that f(2) = -2, and look for the tangent equation and solve for the right starting point

but given the possibility of poor phrasing, and the time it takes to check that

just check anyways

yes, the general method always yields the right answer

it's just that if you can actually easily find a, it's faster to skip the general step

the phrasing is pretty clear to me, as otherwise it would be the tangent at point P

[lone heart]

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[fervent ferry]

i don't understand why K and B are splitted. based on what?
why B is amplitude of input signal and not kB ? k seems to be part of the gain.
are k and B constant or functions or it doesnt matter?

[lone heart]

@fervent ferry Has your question been resolved?

[lone heart]

@fervent ferry Has your question been resolved?

[glad socket]

it seems to make the solution nicer

so you can define g=k/sqrt(k^2+w^2) and have it gBcos..

instead of B/sqrt(k^2+w^2)

you could probably also do wB and then g = w/sqrt(k^2+w^2) instead of wB/sqrt(k^2+w^2)

more likely its to do differentiate between the gain of the system (k) and the amplitude of the input signal @fervent ferry

[fervent ferry]

sorry where the w comes from here?

the say the gain is the whole k/(sqrt)
i don't get whats the diff between k and B

[glad socket]

$\omega B = B' $

[glad socket]

just instead of writing B'cos(..) you do wB(..)

like youre saying to do with k

[glad socket]

itll be helpful if you gave the context to the question

[fervent ferry]

you lost me here

[glad socket]

i assume its about springs?

[glad socket]

ignore it honestly its not really a good explanation

is it a physics problem or math problem lol

[fervent ferry]

no, it's just about explaining the Amplitude, Phase, Gain and Bode Plots https://ocw.mit.edu/courses/18-03sc-differential-equations-fall-2011/resources/mit18_03scf11_s9_3text/

[glad socket]

do you have any knowledge about circuits/physical systems/practical uses?

[fervent ferry]

some but not enough. the mentioned it briefly, and explained why the diff equations have the input , the system, response..

[glad socket]

basically theyre factoring out k since k is related to the system itself

as you can see in the LHS

x'+kx = (something)

[fervent ferry]

ok

[glad socket]

so its more natural to think of the k as the gain of the system

[fervent ferry]

okok

this makse sense

ok

[glad socket]

and b would be the gain of the input

you are correct though, its possible to just write $x'+kx = Acos(\omega t - \phi)$

[fervent ferry]

ok question here. This is in case the input is sinusoidal? or it's just some scalar in front of the input function?

[ocean seal]

nadav

[glad socket]

the input is probably the cos(..)

although since its an ODE given to you from nowhere there isnt really an input

if you really want to think of an input id say its the whole kBcos(..) in this case

[glad socket]

the solution would be $\frac{A}{/\sqrt{k^2+\omega^2}} \cos(\omega t - \phi)$ which is just a less neat version of what they wrote

[fervent ferry]

are you trying to answer this or are you expanding on previous?

not sure where the A comes from now but ok.

[ocean seal]

nadav

[glad socket]

sorry for the confusion. ill answer that:
in the case of an abstract ODE like this there isnt really an input. usually though an ODE like this will come from a physical system where it makes sense to think of somehting as an input. in that case its more plausible that only the Bcos(wt-phi) part is the input
if you need to think of something as an input in the abstract ODE, probably the whole RHS including the k

the rest was expanding on why they took out the k

[glad socket]

instead of writing ...=kBcos(wt) you define A=kB then get Acos(wt) as per your original question

sorry for getting mixed up lol

[fervent ferry]

ok thanks for clarifying everything up for me

really appreciate it

[glad socket]

np haha

lmk if you need anything else

[fervent ferry]

.close

[lone heart]

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[remote agate]

why can't the square root of 4 be equal to -2?

[ivory pivot]

by definition

square root its always >=0 in reals

[remote agate]

is there a reason?

[molten relic]

The square root of 4 is defined as the non-negative value that, when multiplied by itself, equals 4, which is 2.

[ivory pivot]

is the more natural choice to do i think ...

[median oar]

we just chose it like that

you can use it however you want you just need to define it

[remote agate]

thanks everyone

[remote agate]

what do you mean by that

[stark path]

Both 2 and -2 can be the answer of √4

[ivory pivot]

you can't use a function tha you didn't define ...so that is the definition of sqrt and now you can use it

[median oar]

$\sqrt{\cdot}:\mathbb{R}^{0+}\to \mathbb{R}^{0+}$

[ocean seal]

Frosst

[median oar]

depends on what √ means to you

[stark path]

Yea obviously

Like while doin mensuration u arent advised to get minus answer

[median oar]

by convention we use it to mean that √(x) = the positive real number whose square is x

[remote agate]

but generally the answer of a square root is positive

ok thanks

[median oar]

if you want a different answer you can define √ differently

you could go √(x) = the negative real number whose square is x

then √(x) will always return a non-positive real number

but then it's just up to definition at that point

[remote agate]

Ok now I understand, thanks to everyone who answered :)

.close

[lone heart]

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[strong pivot]

for question 6

[strong pivot]

is that the only way to prove it
because i disproved it by saying LHL=5 and RHL=-1/3
therefore LHL != RHL and the limit does not exist

[lone heart]

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[hollow oriole]

Hi guys! I'm new to the community, I'd like to know where is the correct channel where I can get some help in partial differential equations?

[ivory pivot]

here!

[echo socket]

Or #odes-and-pdes and #advanced-pdes

[ivory pivot]

there are a lot of experts in this discord

[gray isle]

make sure to also read #rules #❓how-to-get-help for guidelines on how to behave / what the expect

[hollow oriole]

First place, sorry about the confuse text book. My doubt is: why does the integral of the ~g function ( here named as h(x) ) is not -4x - x^2/20?

[last part of the image]

Context: d'Alambert solution for wave equation, where there is no initial displacement

This is the problem. I'm stuck since yesterday in only this integration, I would aprecciate a lot if you could give any help

[hollow oriole]

I understand why there is (40 + x) instead of x in the negative part of the domain, and (40 - x) in the positive part. Bu where does the 40 comes from in both cases?

[lone heart]

@hollow oriole Has your question been resolved?

[lone heart]

@hollow oriole Has your question been resolved?

[lone heart]

@hollow oriole Has your question been resolved?

[lone heart]

@hollow oriole Has your question been resolved?

[hollow oriole]

<@&286206848099549185>

[harsh swallow]

have you integrated g tilde?

from 0 to x?

when x > 20

just on your own?

[lone heart]

@hollow oriole Has your question been resolved?

[lone heart]

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[weary abyss]

need help with this

[weary abyss]

i know what the principle is but I need help going over this step by step

[vale wigeon]

@weary abyss first find the following two quantities:

• the count of integers from 1 to 2386 divisible by 3
• the count of integers from 1 to 2386 divisible by 13
  then also do part ii first, recognizing that a number is divisible by 3 and by 13 at once if and only if it's divisible by ___

it might help to make a Venn diagram for this.

[weary abyss]

are the discs 3 and 13

2 discs right

[vale wigeon]

the disks will represent the sets of numbers divisible by 3 and 13 respectively.

you can call them $A$ and $B$, but for more descriptive names, you could do $D_3$ and $D_{13}$ perhaps.

[ocean seal]

AnnGhost

[vale wigeon]

and yes there are two of them

[weary abyss]

should I write 2386 at the top

[vale wigeon]

enclose them in a rectangle

wait actually no

you are not interested in what's outside both of them anyway

find |D_3| now

[weary abyss]

how

my professor had a video on this but didnt actually do any real examples

[vale wigeon]

ok let's try a simpler example

how many numbers from 1 to 30 are divisible by 5

if you aren't sure how to count, list them all out

[weary abyss]

5, 10, 15, 20, 25, 30

[vale wigeon]

and how many is this?

[weary abyss]

6

[vale wigeon]

right

so you see how they go from 1*5 up to 6*5 right

[weary abyss]

yes

[vale wigeon]

ok

so you might think that the count of numbers from 1 to n that are divisible by d is n/d

[weary abyss]

yes

[vale wigeon]

except there's a caveat: you need to round that division down to the nearest integer

do you see why

[weary abyss]

so if n/d = a decimal, just do n * only the whole number?

[vale wigeon]

no that is not what i said also bad anyway

the count of numbers from 1 to n divisible by d equals floor(n/d).

[weary abyss]

what is floor

[vale wigeon]

the floor function

the function that rounds its input (a real number) down to the nearest integer

[weary abyss]

yeah

[vale wigeon]

do you have anything else you want to ask me at this point

[weary abyss]

no im just following along for now

[vale wigeon]

ok then find how many numbers from 1 to 2386 are divisible by 3

[weary abyss]

i did 2386/3

got a non integer

rounded it to an integer

[vale wigeon]

did you round it down?

[weary abyss]

yes

[vale wigeon]

ok

how much did you get

[weary abyss]

i then multiplied it by 3

got 2385

[vale wigeon]

[weary abyss]

and divided by 3

and got 795

[vale wigeon]

what was the point

[weary abyss]

lmao isee

i could have just skipped multiplying

[vale wigeon]

of multiplying it by 3 then dividing by 3 again

do you maybe wanna do that seventeen more times

795 yes

that's |D_3|

now find |D_13|

[weary abyss]

2386/13

183/13

is that wrong

[vale wigeon]

wrong notation

why are you dividing 183 by 13 again

[weary abyss]

is 183 the answer

[vale wigeon]

i'd ideally like you to not rely on me so much to tell you what is and isn't the answer.

i gave you a formula, but you are for some reason not able to apply it properly. you make these missteps both times.

[weary abyss]

n/d

my n = 2386 and d = 13

[vale wigeon]

like yeah 183 is |D_13| yes, but you could have and should have known this without having to ask me

having divided 2386 by 13 and rounded down

and not hallucinated some other step i didn't tell you to do

[weary abyss]

ok great

[vale wigeon]

ok

now i know this may appear hard to you

but we're gonna need to do some thinking

and i'm gonna send you a sentence

and i will want you to fill in the blanks

and send me back the same sentence with the blanks filled