help-28

[west mango]

It's a preuniversity problem for entering a national uni in my country

[cyan kelp]

[west mango]

<@&286206848099549185>

[west mango]

<@&286206848099549185>

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@west mango Has your question been resolved?

[west mango]

4x, -3, -3

Those are the factors that can be used

[sick prawn]

I gave this a go but I couldn’t really make it work tbh

[cyan kelp]

i mean the bad but still possible way is (ax + by + c)(dx + ey + f) = adx^2 + (ae + bd)xy + (af + cd)x + (bf + ce)y + bey^2 + cf

now equate coefficients

and enjoy the beautiful system of 6 equations

you know that a,b,c,d,e,f are integers at least

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@west mango Has your question been resolved?

[west mango]

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@west mango Has your question been resolved?

[gentle totem]

8x²+4xy+18x-12y²-33y-18
8x²+(12xy-8xy) + (24x-6x) - 12y² + (-24y-9y) - 18

=> (8x²+12xy) + (-8xy-12y²) + (-6x-9y) +24x-24y-18

=> 4x(2x+3y) -4y(2x+3y)-3(2x+3y) + 6(4x-4y-3)

=> (2x+3y)(4x-4y-3) + 6(4x-4y-3)
=> (4x-4y-3)(2x+3y+6)

[west mango]

Ok

Now what

[ancient sandal]

damn renato lol.

the myth, the legend.

[short bridge]

Now, you got the factorisation that you need. What else?

[short bridge]

How do you even begin to manipulate there? Basically, did it just come to you intuitively or did you notice some pattern?
It's amazing that you found it. Unless, you reverse engineered it. Lol

[west mango]

Yes

[west mango]

How we proceed after this, we have to get the prime factors

[short bridge]

These are the prime factors. You just have to give the sum of constant terms.

[west mango]

how so? may you elaborate further¡

[west mango]

What is the answer 3? -3+6=3

[short bridge]

Umm... You had a polynomial in two variables - it is now factorised. I don't know what exactly prime means.

[short bridge]

Yes

[west mango]

We need to find the sum of the constant terms of the prime factors

[short bridge]

Yes. These are the factors. 🤦‍♂️

You can't get actual numbers. Can you?

They can't even be factorised further since
GCD(4x-4y-3, 2x+3y+6) = 1

[west mango]

I'm trying to understand this guy's factorization method

Wtf

[west mango]

<@&286206848099549185>

8x²+4xy+18x-12y²-33y-18
8x²+(12xy-8xy) + (24x-6x) - 12y² + (-24y-9y) - 18

=> (8x²+12xy) + (-8xy-12y²) + (-6x-9y) +24x-24y-18

=> 4x(2x+3y) -4y(2x+3y)-3(2x+3y) + 6(4x-4y-3)

=> (2x+3y)(4x-4y-3) + 6(4x-4y-3)
=> (4x-4y-3)(2x+3y+6)

how is this happening?

can someone link me a website or something?

[tall finch]

So you don't understand all of this?

[west mango]

what am saying is that I understand grouping the polynomial for enabling factoring, but how did he came up with this?

[tall finch]

Well to be honest I have no clue how he did the first line, but I kind of see a pattern. The rest of it makes sense to me.

[tall finch]

The first two lines, shows him splitting up the 4xy, 18x, and -33y

He splits the 4xy into 12xy -8xy
Splits 18x into 24x-6x
And -33y into (-24y-9y)

[west mango]

<@&286206848099549185>

[torn jolt]

wsg

[west mango]

<@&286206848099549185>

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[high aurora]

is this correct?

[high aurora]

.close

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[versed granite]

hi guys

[versed granite]

what

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@versed granite Has your question been resolved?

[gritty rose]

Don't open a help channel without a question

.close

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[gritty rose]

!show

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@wind rose Has your question been resolved?

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[violet minnow]

I wrote this correctly right? for some reason i put it as = the next fraction in my notes but i know both are supposed to be the answer

[violet minnow]

also this

[neon basin]

Can you show your entire work?

[neon basin]

1/a^-n is 1/(1/a^n) or just a^n, in this case i^27, which isn't i, but is ...

[violet minnow]

-i

[neon basin]

Yes.

[violet minnow]

and this would be 1

[violet minnow]

alright

[neon basin]

No, -1 is correct

[violet minnow]

oh ok

so why is -27 -i

[neon basin]

It's equal to 1/i^26. i^26 = -1, so 1/-1 = -1

[violet minnow]

ok

[neon basin]

You didn't have i^-27

You had i^27

Or 1/i^-27

Same thing

[violet minnow]

ohhhhhh

I see

[neon basin]

I don't understand what you are trying to do here

[neon basin]

Do you know how to divide complex numbers?

[violet minnow]

idk

i did a similar problem this way tho

was 12+3i/12-3i and i got 15/17=8/17i

[neon basin]

You multiply the numerator and denominator by the denominator's conjugate

So you get (a+bi)*(a-bi) which equals a^2 + b^2

[violet minnow]

right

[neon basin]

So what do you multiply with here

[violet minnow]

hol on ima redo it

[neon basin]

If you want to, you can factor out 2 first as well to make it slightly easier

[violet minnow]

oops

17/19

[neon basin]

I still have no idea what you are doing

[violet minnow]

what's your way of doing it

[neon basin]

Okay. Let's factor out two

What do we get?

[violet minnow]

1

[neon basin]

What

[violet minnow]

6

[neon basin]

Factor out 2 in the numerator and the denominator

[violet minnow]

12i/12

[neon basin]

Okay, let's take this: 3x + 21x

What would you factor out here

[violet minnow]

7

i mean 3

[neon basin]

What about x?

[violet minnow]

x

[neon basin]

It's present in both terms

So you get what

[violet minnow]

3x

[neon basin]

Times ...

[violet minnow]

3x(3x)

9x^2

[neon basin]

Huh

[violet minnow]

🐶 💔

[neon basin]

Are you trolling

[violet minnow]

I'm so serious

[neon basin]

I don't think you are

[violet minnow]

guess i'm not

[spare ermine]

This makes sense

[neon basin]

😂

[violet minnow]

so this is wrong

[neon basin]

That's what you get for trying to help people

[violet minnow]

[neon basin]

Well, enjoy

Knowing 17/19 = 6/19 * i

😂

[violet minnow]

thas what i asked in the first place if it was supposed to be a plus or not

.close

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@covert light Has your question been resolved?

[covert light]

<@&286206848099549185>

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@covert light Has your question been resolved?

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@covert light Has your question been resolved?

[thorn ridge]

Honestly, don't know much about accounting. But, I think you've gotta take into account the fact that it's at year end, and so your wages payable for Jan 04 should only be the stuff that was during December, not the stuff during January. So ... it should be $240 I think.

Then the Wages Expenses will be $720 because .... I don't know.

I'm just working backwards from this post I found on brainly.

[covert light]

okay thanks for the help

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[covert wing]

...

What

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[wide sundial]

what have you tried

[haughty fulcrum]

DNE and inf

[ornate bobcat]

First we want to go to
π/3 from left which means we want to try values values less than it and go to it

$$\cot(3x) = \frac{\cos(3x)}{\sin(3x)}$$

[glossy valve]

Sherif Player

[ornate bobcat]

If you tried and substitute
π/3
It will give you

$$-\frac10$$

[glossy valve]

Sherif Player

[ornate bobcat]

Which states for - infinity

@haughty fulcrum

[haughty fulcrum]

oh thank you

[ornate bobcat]

Do you understand it ?

[haughty fulcrum]

yea it makes sense but -1/0 doesnt not equal -infinity

[rain sundial]

It's not exactly-1/0, it is -1/0.00000000.....1 which approaches-inf

[ornate bobcat]

It is not equal to - infinity
But mean as the value of x approaches to π/3
Cos(3x) approaches more to -1
And the sin(3x) approaches to 0
Which means that the number is getting bigger and bigger in the negative direction
Which gets to -infinity

[haughty fulcrum]

oh so it tends to -infinity

[ornate bobcat]

Yeah

[haughty fulcrum]

thank you

[ornate bobcat]

You're welcome

[haughty fulcrum]

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[true ingot]

i need help

[onyx glen]

to know where you went wrong you need to show your work

[torn jolt]

how did you turn [
\s{3 + \s 5} \tss{to} \s 3 + \s 5
]

[torn jolt]

and the same for the second term

[onyx glen]

^

$\sqrt{3 + \sqrt{5}} \neq \sqrt{3} + \sqrt{5}$

[glossy valve]

Ann

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[coarse sinew]

I have a question

[coarse sinew]

<@&286206848099549185>

[hot herald]

!15m

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[hot herald]

also be more specific about waht you need help with

including which quesiton you're on and what you've tried

[coarse sinew]

Well let's start with rh first question

I have no idea what to start with

Cuz there are many options and area of triangle is 12base x h

1/2

@hot herald

[hot herald]

the idea behind these questions is to apply the formula for area in two different ways

[coarse sinew]

But there is only one area formula for triangle

[hot herald]

note that in the area formula,
bh represent
b: a base
h: its respective/corresponding altitude

[coarse sinew]

Iknow

[hot herald]

each b has its own associated height/altitude

[coarse sinew]

So for the first one for example no h is given

[hot herald]

look for the perpendiculare segments

pick one of the labelled sides

[coarse sinew]

1/2 8 x 10

40

[hot herald]

no

[coarse sinew]

1/2 8 x 6

[hot herald]

that's better

[coarse sinew]

24

So h is also 6 ?

[hot herald]

no

[coarse sinew]

Why

[hot herald]

using 8cm for the Base, its respective Height is 6cm (this isn't the h you want to find)

and using the area formula in this way, gives you a numerical expression for the area, in this case 24(cm^2)

the area formula can be applied using another side as the base,

here the other available base is 10cm, and its respective height is that unknown h

(as those are perpendicular)

[coarse sinew]

So 2.4

[hot herald]

what's 2.4 supposed to be

and where is it coming form

[coarse sinew]

H

From other available

Base

[hot herald]

how are you getting that value

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[solid kelp]

i need help i dont know what to do with this

[tacit granite]

i guess question is wrong

cause no limits, or delta x given

[tacit ferry]

limits usually supposed to delta x ---> 0

usually

[tacit granite]

that's what I'm saying

[solid kelp]

Uhm

[tacit ferry]

yea i got that

[solid kelp]

delta is used as a variable

in this

apparently

if that helps

idk what delta even is

[tacit granite]

delta = change

if the question was supposed to ask the limits

[solid kelp]

were not learning limits yet

[tacit granite]

then substitute delta x

like f(x + delta x) = 2(x+ delta x) + 3

[tacit ferry]

yea that

was bouta say

[solid kelp]

wait can u explai why

i dont get it

[tacit ferry]

f(x) is a function f in terms of x

[solid kelp]

yes

[tacit ferry]

so f(delta x +x) is a function f in terms of delta x+x

so put delta x+x in place of x

[solid kelp]

ohh x + deltax is the x

[tacit ferry]

yes

[solid kelp]

okay

then multiply 2

[tacit ferry]

i got the answer

its very simple

[solid kelp]

uhm

im at 2x+2deltax+3

=f(x+deltax)

[tacit ferry]

ye thats correct

now put that in the original question

[solid kelp]

like

2x+2deltax+3 - f(x)

/ deltax

I GOT IT NOW NVM

TY

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[ocean sand]

cos(-beta) is

[tacit granite]

cos(beta)

even function

[ocean sand]

| cos(a) -sin(a) 0 |
| sin(a) cos(a) 0 |
| 0 0 1 |

if here if we replace a with -b then

according to ques we have to prove that f(a) + f(-b) = f(a-b)

but if I take cos(-b) as cos(b) then I am getting f(a+b) not f(a-b)

@tacit granite

[ocean sand]

it is a matrix question

[tacit granite]

how f(a)+f(-b)=f(a-b)?

[ocean sand]

of sorry

f(a) . f(-b) = f(a-b)

now check

[ocean sand]

..

what bout -sina when replacing a with -b

will it be sin b?

[tacit granite]

-sin(b)

[ocean sand]

oh

[tacit granite]

sorry, gtg

[ocean sand]

wait 1 min

just 1

[ocean sand]

-sina then -sin(-b)

wont it be sinb

@tacit granite

just tell this one

tomorrow is my maths exam

<@&286206848099549185>

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[ocean sand]

.close

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[harsh elk]

what is the sequence for 0, 5, 15, 30, 50?

[chrome dock]

i dont really know where to start w thi

[harsh elk]

.close

[chrome dock]

the answer sheet says its E)

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[chrome dock]

.reopen

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[past cloak]

In 2022, a company launched the production of batteries for electric vehicles. That year, 60,000 units were manufactured.
The director has set a goal of increasing production by 4% each year. Assuming that this rate is maintained, determine in what year the millionth battery will leave the production line.

[past cloak]

im just stuck, idk what i should be doing

but im sure that i cant just add number of batteries made year by year until i get to 1000000

[rugged harbor]

You could! But there are nicer and fancier ways
Try to write the quantities of battery produced per year (without simplifying it, don't find the actual numerical value), and then try to find a more general function

Try to see if you notice any patterns

[past cloak]

well yea

i started by doing that and couldnt find anything

then thought gp would work but it doesnt

[rugged harbor]

Can you show your work? Geometric progressions could work here

[past cloak]

topic is digital sequences btw

[rugged harbor]

???

[past cloak]

dunno if it makes sense

[rugged harbor]

How did 1 - 1.04 (denominator) become 1/4

[cloud moon]

you need parentheses in there

o i missed those nvm

@past cloak explain what you wrote and why step by step

[rugged harbor]

(you set it up right, I think you got the denominator wrong here)

[past cloak]

good question

[past cloak]

ohh i wanted to multiply (1-1,04^n) by 1/(1-1,04)

[past cloak]

well basically i was doing gp but instead of it being a just a random sum its 1000000

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[rugged harbor]

So did you solve it?

[past cloak]

well not really im stuck on the same step 🥹

[rugged harbor]

so did you fix the denominator?

[visual sun]

can you isolate to get 1.04^n on one side and use log

[past cloak]

dont think i can isolate in this case

[visual sun]

can you not?

[past cloak]

maybe if i just do 1000000/60000=1,04^n

[visual sun]

no that's not the equation tho

also you can isolate it i think

[past cloak]

Ohh

i can

do i just log everything?

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[lofty heath]

what is this question even asking?

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@lofty heath Has your question been resolved?

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@lofty heath Has your question been resolved?

[glacial pasture]

the acceleration has 2 components now

and you need the resultant

[lofty heath]

how so? are they using the same term 'speed' to refer to different things?

[glacial pasture]

in (a) its assuming there was no acceleration in the direction of v (perp to the centripetal force)
in (b) its saying that now there was always an acceleration of 1.15 in the direction of v

so youre finding the resultant at that exact same time frame, but now assuming that 1.15 was always there

so your answer to a is still in use, but you need to combine it with the 1.15 by sqrt(a^2+b^2) where a and b are the respective accelerations

[lofty heath]

so basically this question is atrociously worded? it saying constant speed at the top of both part a and b seems to imply it applies to both parts, then it goes on to contradict that in part b saying the speed is now changing, but then asks for when the speed is the same as the supposed constant speed?

[glacial pasture]

its saying in b that its speed now increases at some rate
the centripetal force had no bearing on the speed of the bird

therefore theres an acceleration in the direction of v

where there wasnt before

[lofty heath]

surely that means its using the term speed to mean 2 different things for part a and b

[glacial pasture]

its not

its just 2 different situations

in a speed is constant

in b its increasing

[lofty heath]

i'm still completely lost

let me try to re-word the question and you can let me know if you think its asking the same thing

A hawk flies in a horizontal arc of radius of 12.5 m.

a: Find the centripetal acceleration of the bird when it travels at a constant speed of 3.65 m/s
b: Starting from rest, the bird starts to increase its speed at a rate of 1.15 m/s. Find the acceleration at the moment the hawk's speed is 3.65 m/s.

[glacial pasture]

it doesnt have to be starting from rest, but yes its the same

[lofty heath]

surely it makes no sense for it to already be flying at 3.65 m/s and have some tangential acceleration causing it to go 3.65 m/s

[glacial pasture]

it doesnt say that

[lofty heath]

how so

[glacial pasture]

it is just at the time its at that speed,
the bird could have been flying at 1m/s for an hour before this event happens it doesnt matter
youre just supposing its now flying on the arc path with this tangential velocity and that its accelerating still at this time

the conditions before this moment dont really matter at all

it could have just reached an acceleration of 1.15 in the same situation as a

[lofty heath]

the question states in global scope that the bird is moving at 3.65 m/s, can we agree on that? i don't know about you but any words prefacing a multi-part question apply globally to all parts as far as im concerned

[glacial pasture]

but it hasnt had time to change speed at all yet

[glacial pasture]

sure

[lofty heath]

then in part b it asks for when the bird is moving 3.65 m/s.. but it already was? thats where i'm coming from

[glacial pasture]

it could suddenly have a tangential force acting on it that gives it this acceleration, but if we suppose this happens at the exact same instant of time as in (a) (though we dont need to) then it has this acceleration, but its speed has not changed yet

[lofty heath]

so basically.. this is a illogical hypothetical question asking a question if a sudden force appears out of nowhere within the span of 0 seconds?

i think i understand what it actually is wanting me to do now but i still completely disagree that this is a well worded question

[glacial pasture]

sure, but thats being pedantic

i get your issue, but i dont think it was that difficult to get the gist of

[lofty heath]

i'm of the opinion that if they're trying to teach me physics in my first ever physics course they should be more careful about wording stuff like this so it makes logical sense

[glacial pasture]

its just a snapshot in time where we are saying, okay now suppose that this force/acceleration is acting

[lofty heath]

before we end this debate, can we at least agree that the following is an improved wording of the same question:

A hawk flies in a horizontal arc of radius of 12.5 m.

a: Find the centripetal acceleration of the bird when it travels at a constant speed of 3.65 m/s.
b: The bird increase its speed at a rate of 1.15 m/s. Find the acceleration at the moment the hawk's speed is 3.65 m/s.

[glacial pasture]

sure

[lofty heath]

ok, i'd say that ends on a good note. thank you for helping me understand this question

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[torn jolt]

How should I go about solving this problem?

[severe linden]

I would attempt binomial formula and expressing the term using sin and cos.

[full forum]

@torn jolt Has your question been resolved?

[torn jolt]

okay ty

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[light iris]

I’m feel like there’s a simpler way, but I’m forgetting something.

[keen shoal]

were you trying to use the limit def of the derivative?

[light iris]

Yes

[keen shoal]

are you familiar with the rules of differentiation?

[light iris]

At this point I’m not sure.

[keen shoal]

the power rule, quotient rule, etc.

[light iris]

Ah yes

[keen shoal]

ok, so why dont you try differentiating $x^{- \frac12}$ instead of $\frac{1}{\sqrt{x}}$

[light iris]

You see I would

[glossy valve]

saad

[keen shoal]

using the power rule

[light iris]

But my teacher said there are going to be problems where he wants to see the intermediate steps.

[keen shoal]

if he didn't specifically tell you to use the limit def then you dont need to, and itll be $\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$

[glossy valve]

saad

[keen shoal]

solve that and then just input 1/4

[light iris]

Ok

Am I going in the right direction or going way off course?

The function is f(x) = 1/sqrt(x)

I’m sorry I’m just not so sure about this problem. I’ll drop it for rn. Of course using the quotient rule would get this over with. Thank you for your time.

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[torn jolt]

What am i doing wrong!!!

[full forum]

@torn jolt Has your question been resolved?

[torn jolt]

<@&286206848099549185>

[night mural]

Hint: (n/n+1)^n can also be written as (1/(n+1/n))^n

[torn jolt]

I can't even read taht

huh

wait

$$(\frac{1}{\frac{n+1}{n}})^n$$

[glossy valve]

What should I do

[full forum]

@torn jolt Has your question been resolved?

[full forum]

@torn jolt Has your question been resolved?

[lean sphinx]

Can you think of any way to simplify (n+1)/n further

[torn jolt]

nope

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[night echo]

I dont know what i did worng here

[night echo]

There is no x=1 as the answer

Its only x=e

[next sail]

again

the problem is when you go from the second line to the third

[night echo]

Isnt the log e to e equal to 1

[next sail]

this is wrong

you have c = a * b

then you take the log of both sides

so that should become ln(c) = ln(a) + ln(b)

it should be log_e(e) + log_e(x ^ (1 + log_e(x)))

on the RHS

so it should be like 1 + log_e(x ^ (1 + log_e(x)))

[night echo]

Like this?

[gleaming panther]

No ln e is 1 not log e

[next sail]

ln(e) is the same as log_e of e

[gleaming panther]

Yeah I didn't see that to e

Mb

[night echo]

So log_e of e is 1 then??

[next sail]

one sec

i'll do the fancy stuff

[night echo]

Np

Thx

[next sail]

$x^3 = e x^{1 + \text{ln}(x)}$\
$3\text{ln}(x) = \text{ln}(e) + (1 + \text{ln}(x))\text{ln}(x)$

[glossy valve]

Kaisheng21

[next sail]

it should be a plus

[night echo]

What did u do

U multiplied everything my ln?

[next sail]

ok fine i'll do it simpler

i'll do it with log_e like you're doing instead of ln

$x^3 = e x^{1 + \text{log}_e(x)}$\
$3\text{log}_e(x) = \text{log}_e(e) + \text{log}_e(x^{1 + \text{log}_e(x)})$

[glossy valve]

Kaisheng21

[next sail]

this is how it works

[night echo]

So what did u do?

[next sail]

i did what you did

i took log_e of both sides

you did this

[night echo]

Oh its a plus

[next sail]

yeah

if you have c = ab

then log(a) = log(a) + log(b)

[night echo]

Im so tired and its quite hard to think

Thx

[next sail]

np

log rules

classic, classic

[night echo]

❤️

No homo

😀

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[night jungle]

Each image belongs to a set A or B , if neither it belongs to C , its abstract reasoning and I have no clue for this one

[night jungle]

Not math but for logical people

[next sail]

is this a test

[night jungle]

Practice test

For interviews

[blissful nova]

looks like A is ordered and B is more chaotic

from left to right the first 3 are pretty clear, the last two are weirder

left to right: probably A, i would say A but maybe C, B, C?, id say A

[next sail]

hah

[night jungle]

I was thinking the same thing , I’ve never had a question with 2 C’s before

Probably what threw me off

Anyway thanks

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[normal thicket]

Sum of series

[normal thicket]

First one correct?

[willow sedge]

yes

,w lim x tends to 1 1/(x^2+x-2)-x/(x^3-1)

[glossy valve]

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[willow sedge]

@normal thicket

[normal thicket]

Yoooo beard bhai

[willow sedge]

rom rom bhai saryane

question (b) ke liye dekho dono polynomials factorize ho jyenge

[normal thicket]

Kon factor karega ab

Seeda hi lhopital karenge

2/9

[willow sedge]

hmm wo b kr skte

[normal thicket]

Tum batao fact koroge toh time na lagega?

Hindustan m time na h

[willow sedge]

Mujhko to jyada nhi lgta

[normal thicket]

Acha karke batana

Kaise karega factor random value put karega?

[willow sedge]

ruk

nhi.

Aise questione me saaf dikhta 3 factor hoga

taki limit exist kre

agar 3 dono ka factor nhi hoga to limit exist nhi krega

Aur hmko factorize krne ka itna aadat ho chuka ki long division ka jroorat nhi pdta

(x^3-7x^2+15x-9)=(x-3)(x^2-4x+3)

[gleaming panther]

phir next Wale ko middle term krdo

[willow sedge]

usko krne ka kaam nhi

bas (x-3) ko eliminate krna h

baaki me value put krdo

[gleaming panther]

Ha woh bhi sahi hai

[willow sedge]

(x^4-5x^3+27x-27)=(x-3)(x^3-2x^2-6x+9)

[normal thicket]

9-12+3/27-18-18+9
0/0 form hua h abhi bhi@willow sedge

X-3 se kam na chalega

[willow sedge]

to ek baar aur factor krdo

l'hopital better hai tb i guess

[normal thicket]

Bat ghum k vhi aata h yar

[willow sedge]

is case me

[normal thicket]

Ha

@gleaming panther

U say

[gleaming panther]

Ruk vai krne de

Ha hogya

[gleaming panther]

Is Wale ko middle term kro (x-3)² (x-1) ayega

[willow sedge]

axha.
(x^3-7x^2+15x-9)=(x-3)^2(x-1)
(x^4-5x^3+27x+27)=(x-3)^2(x^2+x-3)

[gleaming panther]

Ha

Ab value put krdo

l'hopital better bhai factorise krne mei time lagega bohot

[worn matrix]

l'hopital

?

oh okay

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[normal thicket]

Hint

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[serene sage]

I tried substituting values but that isn't working

[glacial pasture]

what makes you think it wont work

youve only tried -1

[normal thicket]

B

[serene sage]

Im meant to try each and every value?

Is there a more efficient way?

[normal thicket]

Yes

It will not take long

[serene sage]

What?

[short bridge]

Are you familiar with terminology "roots, polynomials, or zeros"?

[serene sage]

Yep

[short bridge]

Good. So, what's root of a polynomial/function?

[serene sage]

the x intercept

when y is 0

[short bridge]

Ah. Kinda. Well, basically all the values of x for which value of function is zero.

So, what are the roots of your function according to given table?

[serene sage]

-1 & 1

[short bridge]

Yes.

That means that your function g(x) should be zero for both x=1 and -1.

You can check which one is that manually.

[serene sage]

do i set both the equations to 0?

like x^2 + 1 = 0
(2x+2) = 0

[still oak]

x^2 + 1 has no solution

[serene sage]

yah so thats eliminated

[still oak]

sub x value into the following functions

until it satisfies all the values

and to me right now that function is B

[serene sage]

so i sub the roots? or do I set the functions to 0?

[still oak]

g(x) is basically your polynomial expression

so if g(x) = 3, (x+1)(x-3) = 3 for example

g(x) is a polynomial

[serene sage]

i dont get it :(

[still oak]

ok lets take 2x + 1

[serene sage]

How can b be correct

[still oak]

what

ok here look at it this way

(x-1)(x-3) and i say x = 1

what is the resultant value

[serene sage]

0?

[still oak]

yeah

so why would you say its not B?

ok take a step back

how did you get 0?

for the equation just mentioned

[serene sage]

Substitute x is 1

[still oak]

and..?

[serene sage]

And then solve the equation..?

[still oak]

so in the picture above, (0) x (-5) is still 0, no?

it satisfies g(x)

you forgot about the zero product

[serene sage]

Wdym?

[still oak]

you saw above you tested both roots

one was 0

the other was -5

but g(x) is a product of your values

0 and -5 make 0

[serene sage]

Aren't the roots -1 & 1? 😭

[still oak]

for?

[serene sage]

According to the table of values

Cuz g(x) is 0 at those points

[still oak]

yes it is

give me a set of values we are testing

x =
g(x) =

[serene sage]

X=1
G(x) = 0

[still oak]

ok

in (x^2 - 1)(2x + 3)

put 1 for x

(0)(5) = 0

[serene sage]

Yep

OHHHHH

I think I get what I'm doing wrong

I was treating both of them as diff equations

It makes sense now why u were asking me to multiple -5 with 0

Thank uu

[still oak]

mhm

no problem

[serene sage]

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[onyx glen]

$\sum_{n=2}^{\infty} \frac{\log^2(n)}{n^2}$

[glossy valve]

Ann

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@normal thicket Has your question been resolved?

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[next basalt]

In a workplace is a pool of 8 printers. One of the printers stops working. In how many possible ways can 19 print outs in the que be divided between the 7 functional printers if:

a) at least 1 printjob goes to each printer
b) at most 4 prints go to every printer

[next basalt]

Which area in math is this question related? Is it the dove principle?

[tacit granite]

statistics, probablity

[next basalt]

I think i solved a but i need some confirmation and help with b)

7 printers = n
19 print outs = k

((n+k-1) / k)

[tacit granite]

so, a and b are independent?

[next basalt]

Yes

[elfin stream]

not probability i think

combinatrics to be specific imo

[tacit granite]

right

[next basalt]

My final answer on a is 177100 different ways

[tacit granite]

umm

[next basalt]

however i'm not sure if that is correct

i used the dove principle on a

[tacit granite]

ok, so

[elfin stream]

since atleast 1 printjob goes to every printer (assuming we need permuations), ur answer should 15P7

[next basalt]

15P7

is that the same as

(15 | 7) 15 over 7

[tacit granite]

a and b are independent

[elfin stream]

i know

[next basalt]

and why P? Isn't it C?

[tacit granite]

doesn't that mean, that some printers may be empty?

[next basalt]

Since we dont care about the order?

[elfin stream]

no its atleast 1

wait not 15P7 btw

i meant 12P7

[elfin stream]

am confused on that

idk if its 12C7 or 12P7

[next basalt]

C

[tacit granite]

i still don't get it

[elfin stream]

yeh then $\frac{12!}{5! \cdot 7!} = \frac{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8}{120} = 11 \cdot 9 \cdot 8 = 792$

[glossy valve]

ItzKraken

[next basalt]

How did you come up with that?

[elfin stream]

uhm n choose k = $\frac{n!}{k! \cdot (n-k)!}$

[glossy valve]

ItzKraken

[elfin stream]

and n choose k is equivalent to nCk

[next basalt]

Does that take into account that each printer gets at least 1 job each?

Which iirc uses the dove principle

[elfin stream]

idk what i used, but by logic, if all of them get atleast 1, then there are only 12 jobs left to distribute between 7 printers

i am doing a btw

so 12C7

[tacit granite]

how atleast 1 is printed in each printer?

a is for atleast 1 printout

b is for at most 4

[elfin stream]

yes so am doing a

[tacit granite]

they are independent

[elfin stream]

Ik

[tacit granite]

wasn't he asking for b?

[elfin stream]

he asked which area

[tacit granite]

this

[elfin stream]

his answer seems wrong to me

show work

[next basalt]

hmm i think your answer is more probable to be right

But isn't a) supposed to use the dove principle?

[elfin stream]

what is dove principle??

wait is it pigeonhole principle?

[next basalt]

Sorry in eng it's called

pigeonhole principle

[elfin stream]

ah that thing.

sorry but i havent studied it yet

[tacit granite]

oh, it's not

[next basalt]

Yeah alright

But 12C7

seems very plausible

[next basalt]

isn't this wrong?

12C7 is

(12! / (7! (12-7)!)

[elfin stream]

12-7 is?

and also

$12! = 12 \cdot 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7!$

[glossy valve]

ItzKraken

[elfin stream]

and $5! = 120$

[glossy valve]

ItzKraken

[next basalt]

Yeah i get 792 also

[elfin stream]

and if its not C and its P

then

,w 12!/5!

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[next basalt]

What about b)

[elfin stream]

ah thats a little tough

[next basalt]

isn't that

19C4 + 19C3

[elfin stream]

well i do casework-thingy (idk what its called again)

consider the first functional printer

it will have either 0,1,2,3,or4 prints

if it has 0

then we would have 19C6 combinations

if it has 1

then we would have 18C6 combi. if we continue like this

so this becomes 19C6 + 18C6 + 17C6 + 16C6 + 15C6

,w 19C6 + 18C6 + 17C6 + 16C6 + 15C6

[glossy valve]

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[elfin stream]

BRUH

,w 19!/(6! * 13!)

[glossy valve]

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[elfin stream]

,w 18!/(12! * 6!)

[glossy valve]

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[gentle totem]

writing 'choose' instead of C should work

[elfin stream]

,w 17!/(11! * 6!)

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[elfin stream]

ah k

,w 19choose6 + 18choose6 + 17choose6 + 16choose6 + 15choose6

[glossy valve]

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[elfin stream]

..

brackets ig?

[gentle totem]

that's crazy i think you should've put some parenthesis lol

[elfin stream]

,w (19choose6) + (18choose6) + (17choose6) + (16choose6) + (15choose6)

[glossy valve]

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[elfin stream]

@next basalt this is what am getting for b

tho this makes no sense..!

a cant be < b

[next basalt]

I have no idea

If they can max have 4

shouldn't it be 19C4

[elfin stream]

wait what i did doesnt put that restriction on the rest

[elfin stream]

let me think...no i dont think so

do u have an answer key btw @next basalt

[next basalt]

No

[elfin stream]

😭

how do we check if the answer is correct 😐

[gentle totem]

im getting 74613 for some reason

do u know what the answer is

[elfin stream]

thats more than mine..

[gentle totem]

wait no actually 5005

[elfin stream]

aight quora to the win

[gentle totem]

i assumed empty printers in the first one

[next basalt]

https://www.quora.com/In-a-large-workplace-there-is-a-pool-of-8-printers-Suddenly-one-of-the-printers-fails-On-how-many-19-prints-that-were-on-its-printer-are-distributed-between-the-other-7-working-printers-if-a-maximum-of-4-print-jobs

[elfin stream]

exactly what i was reading

[next basalt]

I found the exact question online

[gentle totem]

if its atleast one in each then 5005

[next basalt]

I'm completely lost

i have none to ask and none knows what to do

this is so fucked

[tacit granite]

ok

are you doing a or b rn?

[next basalt]

b

still not sure if a is right but i assume it is

[gentle totem]

for b, is it atleast 1 for each printer?

[tacit granite]

so, given is that sum of 7 is 19

[tacit granite]

no

[gentle totem]

yeah you can do number of solutions for that i think

a+b+c+d+e+f+g=19

now there was this formula for number of non negative solutions

yeah no its definitely wrong for b if we go by this

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[normal thicket]

How to tackle log here?

.close

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[desert musk]

y*tan(y) = sqrt(-y^2 + lambda)

[desert musk]

what is a good way that there is a solution in 0 < y < √λ

[desert musk]

<@&286206848099549185>

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@desert musk Has your question been resolved?

[full forum]

@desert musk Has your question been resolved?

[desert musk]

<@&286206848099549185>

is anyone here?

2 hours is fair waiting time i feel

[small barn]

What we need to find?

[plush egret]

looks like what the intersection of a quarter circle and the curve xtanx?

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@desert musk Has your question been resolved?

[desert musk]

no

we need to see if the they intersect

the two functions

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[digital fractal]

Hello, I am working on this problem, where I must find the slope of the graph.

I solved a by finding f(1) = 2 and f(1+h) = 2 + 4h + 2h^2

[digital fractal]

Plugging everything into the formula, I get 2 + 4h + 2h^2 - 2 = 4h + 2h^2/ h where h =/= 0

factor out the numerator for 2h ( h+2) / h and we get 2(h+2) = 4 + 2h which is the slope of the secant line

I am a bit confused at what B is asking for.

Do I use the slope from A and plug in 1, f(1)) to solve?

and if so, what would the equation look like?

y = 4 + 2h + b, y = 4 + 2h x + b?

if 4 + 2h = slope, then it is m or a (whichever you use), my formula should be y = (4 + 2h) x + b, but that can't be solved

[torn jolt]

yes it's good

[digital fractal]

but I can't solve the slope equation

since I have 2 variables in one and 3 in the other

I can replace x and y, but not h and b

[wheat jewel]

If u know derivative of a function u can use that

[torn jolt]

you should to use the quotient rule

[digital fractal]

derivative of 4 + 2h?

it would just be 2

[torn jolt]

yes it's awesome

[digital fractal]

b = 0?

but it's asking for the slope at 1, f(1)

I'm confused how to find it if I already have the slope

[torn jolt]

Well, you should study the theory then...

[digital fractal]

I'm not sure how I'm supposed to proceed. You tell me to find the derivative. I did that

I'm just not sure what the question is asking for, because I already have the slope

I tried entering 2, which I obtained by deriving 4 + 2h

[torn jolt]

you have to apply the simson formule

it's fine right ?

[digital fractal]

that's an integral, I am not at that part yet

I am only doing derivatives and limits

<@&286206848099549185> Can someone explain what I am not seeing please? I already found the slope and now it is asking for the slope at a given point

[torn jolt]

what do you search

?

[digital fractal]

I found the slope = 4 + 2h, but now it is asking for the slope at (1,f(1))

you're telling me "simpson rule", which is an integral and I have not learned it yet

[digital fractal]

I can speak french too if that helps

[torn jolt]

tu parles français ?

[digital fractal]

Oui je parle français

Mais pas aussi bien que mon anglais

[torn jolt]

oh bah c'est plus simple comme ça

[digital fractal]

Donc question B ca demande la pente au points (1,f(1))

F(1) = 2

Donc (1,2) pour x and y

[torn jolt]

tu dois t'abstiner à executer la règle de la pente de la tengente

[digital fractal]

J’ai appris tout les maths que je connais en anglais, je ne comprend pas que ve dire “executer la regle de la pente de la tangente”

[torn jolt]

Alors il faut effectuer la limite, lorsque x tend vers l'infini pour trouver la concavité de la fonction, et trouver alors la pente de la tangente.

n'oubliez pas de regarder en + l'infini et en - l'infini pour la pente

tu as enfin compris ? c'était pas si dûr hein

@digital fractal

[digital fractal]

Juste un moment pardon

[torn jolt]

dis moi quand tu auras terminé, j'espère que je t'ai aidé

si tu as d'autres questions n'hésite pas à me les poser

[digital fractal]

ok je suis de retour

donc + infini de la droite and - infini de la gauche

[torn jolt]

Oui exactement, tu es sur la bonne voie

de la réussite

[digital fractal]

mais la formule que je dois utiliser est f(a + h) - f(a) / h

et la limite de h qui sapproche a 0

et la question c'est la pente de la graphe au points (1, f(1)) ou (1, 2)

.close

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[torn jolt]

no

i will help you

je vais t'aider

mais vraiment tu devrais essayer la formule de simpson

c'est la plus adaptée

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[swift fulcrum]

Hello

[swift fulcrum]

I have a math related coding question

How can I create a function that returns a whole number?

For instance, if we input 3.455, the return value will be 3

3.893 -> 3
-2.424 -> -3

I am limited to using castings and math libraries

[wicked aurora]

this is the floor function

in python: math.floor(x) if I am not mistaken

But one warning: Some libraries use a weird variant of the floor function that outputs wrong numbers for negatives.

Check documentation

[swift fulcrum]

I am using c++

[wicked aurora]

ok, seems like std has floor

oh no

it is in cmath

like, you could make your own function, but it would be much slower.

[swift fulcrum]

Am aware

How convenient would it be if I could use int() or round()

Apparently I can not use them

[wicked aurora]

Why these limitations

[swift fulcrum]

Never mind

.close

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[silk nexus]

I am trying to do partial integration

[silk nexus]

Question 19

But I can’t seem to find the values of A,B,C,D when doing its decomposition

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[formal scarab]

^

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[regal forge]

Can i get some help with this?

[full forum]

@regal forge Has your question been resolved?

[regal forge]

<@&286206848099549185>

[wary walrus]

what's the issue

[regal forge]

I'm confused about (c)

[wary walrus]

regarding which property ?

or just the set itself ?

[regal forge]

how can we tell which property it is

yeah

the set itself

[wary walrus]

hmm

xy = 1 means y = 1/x

think of C as R^2

that should help

[regal forge]

Could you please walk me through the steps?

[wary walrus]

well

do you visualize the graph of f(x) = 1/x

[regal forge]

yes

[wary walrus]

formally it's the set of points (x, y) such that y = 1/x

[regal forge]

yes

[wary walrus]

and that's basically S3

provided (x, y) is identified to x + iy

[regal forge]

isn't this in 3d?

how can you know it is open?

[wary walrus]

wdym

[regal forge]

like z, y,x.

[wary walrus]

no it's in the complex plane

so it's 2 dimensional

[regal forge]

wait nevermind, you jiust visualise the function xy = 1

[wary walrus]

yep

[wary walrus]

exactly

[sharp dawn]

Gio d'où t'a appris the terminology? T'es doué mdr?!

[wary walrus]

l'internet regorge de ressources

[regal forge]

So its closed and a region right?

[wary walrus]

yep

[regal forge]

is it a region?

[sharp dawn]

I'm too lazy to searche

[regal forge]

Isn't it a line?

[wary walrus]

wdym by region

[sharp dawn]

Ensemble ?