help-28

[supple jay]

which means it will then

[potent vine]

I don't follow

[supple jay]

be - - = +

so its actually +1 on exponent

of p^1/3

[quaint prawn]

This is multiplying by p

[supple jay]

hmmmm

i would say its multiplying by 1/p

[potent vine]

from this you get $p^{4/3} = 9 \cdot 4^{1/3}$ now find p

[quaint prawn]

It’s not

[glossy valve]

exophades

[supple jay]

oh

i just did that yes

wait

woah

huge enlightenment

so if divide by something that is negative exponent

its actualy multiply

cos its like 1/something

so divide reciprocates it

[potent vine]

slow down haha

[quaint prawn]

Yeah, you are dividing by a fraction. It just flips and becomes multiplication

[supple jay]

https://tenor.com/view/israel-jerusalem-state-of-israel-flag-of-israel-independence-day-in-israel-gif-14951064

israel coffe for you people

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[proud marten]

this all ik

not sure if helps

[bright heron]

thank you, but no. This is stuff from algebra 2

[proud marten]

oh okay

yeah I just alg 3

[simple ridge]

the name it is under doesnt rlly matter what matters is that your answer can be found using what AblueA sent

[bright heron]

oh, ok

[simple ridge]

isnt there something like 180 deg rotation about y=-x

[proud marten]

idk

[simple ridge]

oh ok

i gtg now cya all

have a nice day/night everyone

[exotic otter]

u too

[bright heron]

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[lean coral]

@torn jolt , the one in the middle need to find the missing side

[exotic otter]

I got this one

.status

.status

The bot down or smth?

[lean coral]

🤷‍♀️

[exotic otter]

Ok anyways

[jagged saddle]

that means parallel

[exotic otter]

That means parallel

[lean coral]

😭no way I’m so dumb sorry thanks

I wasn’t sure because it originated as 2 separate triangles forces

[exotic otter]

Bro just answered his question by asking his question

[exotic otter]

Did u get it?

Can u do it now

[lean coral]

Wait, I’m not sure it is parallel, but I’m going to hand it in as it is, becuase those double arrows are on the one next to it too, but thanks

[exotic otter]

Aight

[lean coral]

It originates from the 2 triangles on the left

[jagged saddle]

i brute forced it with law of sines

[exotic otter]

Why are u making it so complicated

Anyways he got it

[lean coral]

I’m going to go along with it being parralell, it’s just a daily question we’ll go through in lesson

[exotic otter]

Just remember that symbol means parallel

[lean coral]

However thanks. But it’s annoying, that angle at the bottom really throw me off.

[lean coral]

Yeah I know, normally does but I couldn’t get my head around why it would be . Cheers tho

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[cosmic canopy]

evaluated this and got the wrong answer don't know why

oh wait i evaluated it wrong hold on

i got $\frac{61}{12}$ could someone check that

[glossy valve]

ed

[cosmic canopy]

nvm

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[turbid thunder]

could someone help me understand why the derivative of arccot is not the same as the reciprocal of the derivative of arctan?

[stable plover]

is the derivative of 1/x the same as the reciprocal of the derivative of x

also arccot and arctan are not reciprocals

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[digital fractal]

For a) if I plug in -2 in x, I get 0/0, which is IF.

so I factor the denominator giving me (x+2) (x-2), which I can cancel one set of (x+2) from the numerator. I now get (x+2) / (x-2)

plugging in -2 for x, I get 0/-4, which is 0.

Would that mean that the limit does not exist or is it 0?

[mighty ridge]

if the limit is 0/-4, it's equal to 0 yeah

[digital fractal]

ok so I didn't do anything wrong

I must have misheard my professor, because I thought is the limit = 0, means there is no limit

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[digital fractal]

.reopen

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✅

[digital fractal]

sorry you were typing

[mighty ridge]

you definitely misheard yeah, might want to clarify it with your prof

[digital fractal]

so is it possible that I heard for example

16/0 = DNE that the limit does not exist then?

[mighty ridge]

yes

[digital fractal]

ok maybe that's what I got confused by

thanks

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[stiff swan]

when someone asks me to add this, does it mean they expect me to transpose either A or B, or the answer is just DNE?

[left bone]

DNE

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[vernal ocean]

whats does the letter "d" before the sigma motion does (to clarify I want to know in general what a letter before a sigma motion does)

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[tulip citrus]

someone please help

i almost gotta go and this is due tommorow

[light sonnet]

What exactly are you stuck on? Those are just like all the other equation problems you've done before

[tulip citrus]

well

could we see since i have a hard time remembering

also i think this is different from the equation im used to

[light sonnet]

Then look back at your message history to see what was done before

[tulip citrus]

isnt that gonna take ages?

also i dont know which help it was on

[light sonnet]

The message history takes you back to that message

[tulip citrus]

oh wait i thought you meant scroll to it

[light sonnet]

Like this problem you did before #help-20 message

[tulip citrus]

could we review atleast 4 tho

number 4

[light sonnet]

You want to solve for w, what's the first step you should do?

[tulip citrus]

+4 - 4

cancels out

and 52 + 4

56

but what do i do with the fraction

thats where i get stuck

[light sonnet]

What's the equation you have now?

[tulip citrus]

56 = 8/5W

[light sonnet]

$56 = \frac{8}{5}w$

[glossy valve]

dldh06

[light sonnet]

That's the same as $56 = \frac{8w}{5}$

[glossy valve]

dldh06

[light sonnet]

Do you see what you can do now?

[tulip citrus]

uh

8/5W * 5?

[light sonnet]

Yes

Now you have?

[tulip citrus]

W is left

56 * 5

280

W = 280

so then i gotta check my answer

[light sonnet]

No

You weren't done

[light sonnet]

No

[tulip citrus]

wait what

[light sonnet]

You have $56 = \frac{8w}{5}$, if you multiply by 5, yes that cancels out with that 5 in the fraction but what about the 8?

[glossy valve]

dldh06

[tulip citrus]

now that

thats where i get very confuse

confused*

[light sonnet]

Well if you multiplied by 5, the 5's cancel like is said, so you end up with 56 * 5 = 8w, right?

[tulip citrus]

ah

so then 56 * 5 = 280

280 = 8w?

[light sonnet]

Yes

[tulip citrus]

alright then

ill try doing these now

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[twilit pebble]

i am having trouble finding the exponents for the numerator and the denominator

[full forum]

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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[twilit pebble]

2

[narrow helm]

What's causing you trouble?

[twilit pebble]

tbh idk how to explain it

i cancelled out a

i found the exponent for b

but for problem the answer is a fraction with b in the numerator and the denominator and it confuses me

sry for bad explanation

[narrow helm]

what did you get?

[twilit pebble]

1/b^16

[narrow helm]

you know that b^(0) = 1?

[twilit pebble]

ohHhhHHh

i forgor

my bad

ty

[narrow helm]

do you need help with anything else?

@twilit pebble

[twilit pebble]

no im good

thx

[narrow helm]

type .close to close the channel.

[twilit pebble]

ok

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[full forum]

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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[sharp nebula]

y' = 3a x^2 + 2bx + c

How do I go about this?

[torn jolt]

what does a local maximum entail

[sharp nebula]

When y'=0
?

[torn jolt]

yes

[tacit granite]

you answered the question

[torn jolt]

you know y' = 0 for x = 2

[sharp nebula]

Yes

[torn jolt]

and y = 4 for x = 2 as well

[sharp nebula]

Uhhh

y= 4 when subbed into the original

[tacit granite]

so 3ax^2+2bx + c = 2a*4+2b*2+c = 0

[sharp nebula]

Y is it 2bx4

[narrow helm]

first of all, the inflection point lies at the origin

so that means the curve passes through the origin (0,0)

so you can get d from this

[tacit granite]

x = *

** = italic

[sharp nebula]

Yep

[sharp nebula]

Why 4

X=2

[tacit granite]

cause x^2

oh

[narrow helm]

the maximum point is (2,4)

[sharp nebula]

Ye

[narrow helm]

so the curve passes through (2,4)

so x=2 should yield y = 4

you'll get a relation in a,b and c from there

then f'(2) = 0

you'll get another relation

then what about point of inflection?

[sharp nebula]

Is this one of it?

[sharp nebula]

y ' '

[narrow helm]

y'' what?

=?

[sharp nebula]

6ax + 2b

[narrow helm]

i mean when does POI occur?

[sharp nebula]

(0,0)

[narrow helm]

ahh

the condition

[sharp nebula]

Eh?

[narrow helm]

like maximum occurs when f'(x) = 0

so when does POI occur?

[sharp nebula]

y" = 0

[narrow helm]

yes

exactly

so do it, and that should give you the value of b

[torn jolt]

its not just that

[narrow helm]

then solve those equations you got earlier

[sharp nebula]

B is 0

Now I only hv
8a + 4b + 2c = 4
B=0

[narrow helm]

you have another

f'(2) = 0

[sharp nebula]

Ahhh

Ok I see it now

Thanks so much y'all

:))

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[charred geode]

Guys what are the number of ways i can arrange these letters :
A m b a t a k u m

[onyx glen]

are the A and a different or the same

[charred geode]

Different

[onyx glen]

ok so then you have 2 letters repeated twice each (a and m) and 5 unique letters (Abtku)

[charred geode]

Yes sir

[onyx glen]

don't call me "sir" please.

[charred geode]

Sry didn't know

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[dim wing]

i need help about radicals and surds

[dim wing]

more specifically this :

<@&286206848099549185>

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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.

[dim wing]

ok

[swift fulcrum]

!status

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1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[swift fulcrum]

!show

[full forum]

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

[dim wing]

i am stuck in trying to simplify them

i dont where to begin

basically

?

@dim wing

[torn jolt]

whyd u ping yourself

[torn jolt]

just multiply by the conjugate to get rid of sqrts

[dim wing]

what?

[dim wing]

?

[dim wing]

idk

[torn jolt]

(a+b) x (a-b) = a^2 - b^2

[dim wing]

ah ok

[dim wing]

the rest?

[torn jolt]

its all the same?

[dim wing]

how?

[torn jolt]

have you solved any 1 of them yet?

[dim wing]

i tried

i dont understand it at all

[torn jolt]

OK SO

see this as an example

theres a square term in the denom, and we dont like and thats what the question is asking us to do too, remove it, to remove a sqrt we need to square the square root (3 - sqrt 2) (3 + sqrt 2) = (a -b ) (a+b)

[torn jolt]

apply this

[dim wing]

ah ok

[torn jolt]

and you dont have the sqrt anymore 🙂

[dim wing]

ok thx

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[chilly mason]

hey guys can u help me with these problems thanks

[wide wind]

Are you familiar with sin, cos and tan?

[tacit siren]

Didn't u already put in help forum

[narrow helm]

this is not the only problem

there's more

so I redirected them here

[chilly mason]

help me these one too

[chilly mason]

yes cus u told me to be fast, so i just post one first

[chilly mason]

yea but we r not allowed to use calculator in these problems

is there any ways we do these problems without calculator?

[wide wind]

Nothing's coming up in my mind.

[chilly mason]

what about the other problems?

[steep flame]

are you 100% sure about that first problem? It says round to the nearest degree and I'm goign to assume they didn't make you memorize a whole trig table

[wide wind]

Hmm, I'm not familiar enough with trigonometry. But the second one, Billy's jogging distance, what would you do if those terms were singular? Like x and y.

[narrow helm]

1st one cant be done without a calculator, unless you know those values too

[chilly mason]

wait i been thinking these long, i think we r allowed to use calculator. Cuz it seems impossible without calculator

[chilly mason]

yes i m confused about the sentence itself, it said jogging and the end its say howlong bill cycled? is there any misundertanding

[wide wind]

Probably, but it seems to be solving x.

[chilly mason]

do u know the solution?

just share it, i have the answers

[wide wind]

x=7/2

Does that seem right?

[full forum]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[narrow helm]

for the first one, do you know the trigonometric ratios?

@chilly mason

[chilly mason]

Yes i do, i know how to do it, i thought we r not allow to ise calculator so i post it but after thinking so long i think its impossible to not use calculator

[narrow helm]

so you can do the first one?

[chilly mason]

Wait, i gotta sleep, i will continune tmrw, remind me tho, its late in my country here, i m supposed to sleep like 2-3 hours ago

[chilly mason]

Yep

[narrow helm]

ok, cool

then which ones are left now?

[chilly mason]

The under one

[chilly mason]

.

[narrow helm]

jogging?

[chilly mason]

Yea i dont even understand the sentence itself how come it become cycled after he say he jogs

[narrow helm]

yeah, question seems incomplete.

or just 0? because he just jogged?

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[latent aspen]

How do I change bounds from polar to cartesian

After drawing the graph, the region of integration is between circle of radius 2 and radius of 4 in the 2nd quadrant.

Im p sure the x bounds should be from [-4,0] but I am conflicted on how to write the y bounds.

I have tried splitting it into 2 parts: but im not sure this is correct

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@latent aspen Has your question been resolved?

[latent aspen]

<@&286206848099549185>

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@latent aspen Has your question been resolved?

[severe linden]

looks ok

[latent aspen]

oh really?

[severe linden]

yes?

why the doubts

[latent aspen]

🤷‍♂️

[severe linden]

[latent aspen]

haha thanks for the confirmation

[severe linden]

you are welcome

[latent aspen]

how do i close?

[severe linden]

.close

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[tough scaffold]

Is this method 100% legit?

[copper cape]

yes

[quaint prawn]

100% legit

[copper cape]

yes

[quaint prawn]

@supple jay can you verify?

[tough scaffold]

@supple jay can you verify?

[noble pivot]

@supple jay can you verify?

[torn jolt]

Legit

[noble pivot]

how is it legit?? the answer is 3/10

[torn jolt]

Non por favor, the answer is 0.3

[noble pivot]

seniõr after further calculations I arrived at 30%

[torn jolt]

After further information from the FBI i have reached the conclusion that the answer is
The division of the smaller root by the larger root, in the polynomial
x² - 13x + 30

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[torn jolt]

I need help with 2D motion

[woven knoll]

Send problem and what you exactly need help with

[torn jolt]

I believe the answer to this problem is c for $(a)$ and $ - \pi$ for $(b)$

wait

[glossy valve]

Replaced by new brandon H

[torn jolt]

am i correct?

I don't really know what they mean by

airplane's displacement during this period

I think it's just the change in position

This seems to easy, that's why i'm confused

[woven knoll]

This is Displacement

[short bridge]

Direction won't just be an angle. You need to find the unit vector across the displacement vector.

[torn jolt]

huh

so

the direction is a vector

so

just

$$\vec{d} = -c \hat{j}$$

[glossy valve]

Replaced by new brandon H

[torn jolt]

oh

it's

$-\hat{j}$

[glossy valve]

Replaced by new brandon H

[torn jolt]

wait

i

j is up

Ok

TY everyone

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[torn jolt]

If i wanted to solve this problem would I just visually integrate then divide by time?

[torn jolt]

but its a vector

😭 too many steps

[spiral vigil]

take them one at a time

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.close

ty

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[left fossil]

could anyone give their suggestion as to what topic about functions I can use and study for this

[silver thorn]

you gotta give more than that? i dont understand whats going on

what are you doing exactly

[exotic otter]

Can u send the whole question, I cant really make out what ur asking

[full forum]

@left fossil Has your question been resolved?

[left fossil]

My Teacher is basically asking us to come up with any topic be it a real life situation or just anything at all as long as it relates to functions

[left fossil]

this picture is really all he gave us

[silver thorn]

ok but whats the task?

are you writing an essay?

doing a powerpoint?

writing a textbook?

assuming its for a indivudal research topic, It kinda of depends on your level of math knowledge

if this si a grade 12 precalc course, you could try to learn about integration or derivatives

if its algebra you could talk about the trig functions or logarithms

i mean you could even talk about computer programs and how you write very abstract function all the time when coding

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[tawny birch]

https://gyazo.com/952b32faa4ae9f4dd9ec1ef4f51c5990

[tawny birch]

Hey II was thinking the answer was c because

using m=e and b=0

when they are set equal to each other they are the same

and the dervatives are the same on both sides...

am i thinking correctly?

I just wanted to make sure its correct and im not going down the wrong path LOL

[atomic blade]

Yup you're fine

[tawny birch]

appreciate it!

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[torn jolt]

hi

.reopen

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[normal thicket]

Can I write it like matrix?

[spiral vigil]

sure? matrix times ⟨a,b,c⟩ column vector

[quaint prawn]

@normal thicket ⚔️

[normal thicket]

a1,0,0
b1+c1, b2+c2, b3+c3
a1+b1+c1, a2+b2+c2 a3+b3+c3

Two rows are the same so many of them will be 0

Am I on the right path?

, rotate

[normal thicket]

@spiral vigil

[spiral vigil]

wait

i don't understand what's happening at all here

is there more to this question?

[normal thicket]

Nothing

More

Maybe I wrote wrongly in matrix

[spiral vigil]

trying to figure out what $[\vec a.\vec b.\vec c]$ even means

[glossy valve]

hayley!

[normal thicket]

Ohh wait

I got this

2[abc]

Sorry for a messy work

Mo it's 0

@spiral vigil

Any short method?

[spiral vigil]

you might be able to rearrange it so stuff cancels

i don't really know this operator

[normal thicket]

What??

[normal thicket]

Any short except this one?

@spiral vigil

<@&286206848099549185>

[quaint prawn]

🫳 🛡️

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[pseudo cape]

Does anyone have a {\em hint} on proving Menelaus' theorem? \ It states that in the geometric construction below, we have that [\overline{AZ} \cdot \overline{BX} \cdot \overline{CY} = \overline{AY} \cdot \overline{BZ} \cdot \overline{CX}.]

[pseudo cape]

@polar valve suggested to draw perpendiculars from A, B and C and then use the intercept theorem

But wouldn't it basically look like this?

We can't really use the intercept theorem (or similar triangles) on that, can we?

[tacit granite]

use similarity

[pseudo cape]

On what?

[tacit granite]

on triangles

[pseudo cape]

Of course, but which two triangles do you mean that should be similar?

[tacit granite]

I don't really remember

Because I'm not in geometry

but Helpers should know i guess

[pseudo cape]

<@&286206848099549185>

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[rotund birch]

So far, i showed that this is true when $n=2$, since $2 \in P$ and cant be expressed as $ 2 = ab$ for $a,b \in S$
\
\
Then i assumed that for an arbitrary integer $k \leq 2$ that i is either in $P$ or can be expressed as $i=xy$ for every integer i with $2 \leq i \leq k$
\
\
Then i attempted to show that either $k+1 \in P$ OR $k+1=x_1y_1$ for some $x_1,y_1 \in S$
\
\
I begun with showing that when $k=2$, this holds true because $k+1 =3 \in P$ (and it cant be expressed as $k+1= x_2y_2$ for some
$x_2,y_2 \in S$
\
Similiarly, i showed that this is true for k=3
\
\
So i assumed that $k \leq 4$
Therefore $ 2 \leq k \leq k$
\
\
And because of this, either $k \in P$ or $k = sr$ for some $s,r \in S$
\
\
From here i considered two cases:

Case 1: Let $ k \in P$ where $4 \leq k$
\
Since 2 is the only even prime and $k \leq 4$, k must be odd.
\
Since k is odd, k+1 is even and can be expressed as $k+1= a_1b_1$ for some $ a_1b_1 \in S$
\
\
Case 2: Let $ k=sr$ for some $s,r \in S $
\
Well since $s\leq 2$ and $r \leq 2$, it follows $k \leq 4$ without a need for our assumption.
If either s or r is odd, k+1 will be even, which can be expressed as multiplication of two elements from S
\
How can i show k+1 is true for the cases where k+1 is odd?

[glossy valve]

Cyrenux

LaTeX source sent via direct message.
```Compilation error:```! Missing number, treated as zero.
<to be read again> 
                   !
l.63 ...1y_1$ for some $x_1,y_1 \in S$   \hspace (
                                                  1)
A number should have been here; I inserted `0'.
(If you can't figure out why I needed to see a number,
look up `weird error' in the index to The TeXbook.)```

[proper hawk]

umm

Since k is odd, k+1 is even and can be expressed as $k+1= a_1b_1$ for some $ a_1b_1 \in S$

[glossy valve]

WhereWolf(ping if needed)

[proper hawk]

you need to show a1,b1 are in P, not S

[rotund birch]

Not both a1 and b1 are in P if k+1=8=a1.b1 though

Also bruh this is so painful to edit in mobile

With all the low fps stuff

Resending my work if needed to be read again:

[proper hawk]

oh I meant your end goal is to prove they can be factored in primes

showing they are in S is not enough

I worded that poorly sry

[rotund birch]

So far, i showed that this is true when $n=2$, since $2 \in P$ and cant be expressed as$ 2 = ab$ for $a,b \in S$
\
\
Then i assumed that for an arbitrary integer $k \geq 2$ that i is either in $P$ or can be expressed as $i=xy$for every integer i with $2 \geq i \geq k$
\
\
Then i attempted to show that either $k+1 \in P$ OR $k+1=x_1y_1$ for some $x_1,y_1 \in S$
\
\
I begun with showing that when $k=2$, this holds true because $k+1 =3 \in P$ (and it cant be expressed as $k+1= x_2y_2$ for some
$x_2,y_2 \in S$
\
Similiarly, i showed that this is true for k=3
\
\
So i assumed that $k \geq 4$
Therefore $ 2 \geq (k) \ geq k$
\
\
And because of this, either $k \in P$ or $k = sr$ for some $s,r \in S$
\
\
From here i considered two cases:

Case 1: Let $ k \in P$ where $4 \geq k$
\
Since 2 is the only even prime and $k \geq 4$, k must be odd.
\
Since k is odd, k+1 is even and can be expressed as $k+1= a_1b_1$ for some $ a_1b_1 \in S$
\
\
Case 2: Let $ k=sr$ for some $s,r \in S $
\
Well since $s\leq 2$ and $r \leq 2$, it follows $k \leq 4$ without a need for our assumption.
If either s or r is odd, k+1 will be even, which can be expressed as multiplication of two elements from S
\
How can i show k+1 is true for the cases where k+1 is odd?

[glossy valve]

Cyrenux

[rotund birch]

That was way harder than it should have been

Brb switching to pc

[rotund birch]

so to show that k+1 can be factored in primes, i need to assume that k (or some lower value) is factored in primes?

also bruh i just realized this is the first work i sent, where i confused geq with leq , let me fix that again

So far, i showed that this is true when $n=2$, since $2 \in P$ and cant be expressed as $ 2 = ab$ for $a,b \in S$
\
\
Then i assumed that for an arbitrary integer $k \geq 2$ that i is either in $P$ or can be expressed as $i=xy$for every integer i with $2 \leq i \leq k$
\
\
Then i attempted to show that either $k+1 \in P$ OR $k+1=x_1y_1$ for some $x_1,y_1 \in S$
\
\
I begun with showing that when $k=2$, this holds true because $k+1 =3 \in P$ (and it cant be expressed as $k+1= x_2y_2$ for some
$x_2,y_2 \in S$
\
Similiarly, i showed that this is true for k=3
\
\
So i assumed that $k \leq 4$
Therefore $ 2 \leq (k) \leq k$
\
\
And because of this, either $k \in P$ or $k = sr$ for some $s,r \in S$
\
\
From here i considered two cases:

Case 1: Let $ k \in P$ where $4 \leq k$
\
Since 2 is the only even prime and $k \leq 4$, k must be odd.
\
Since k is odd, k+1 is even and can be expressed as $k+1= a_1b_1$ for some $ a_1b_1 \in S$
\
\
Case 2: Let $ k=sr$ for some $s,r \in S $
\
Well since $s\leq 2$ and $r \leq 2$, it follows $k \leq 4$ without a need for our assumption.
If either s or r is odd, k+1 will be even, which can be expressed as multiplication of two elements from S
\
How can i show k+1 is true for the cases where k+1 is odd?

[glossy valve]

Cyrenux

[rotund birch]

(repost)

[proper hawk]

actually you don't have to do two cases

k+1 is either prime or can be factored into smaller numbers

and those number are either prime or can be factored into even smaller numbers

they keep decreasing, but you eventually hit 2 and 3

thus a prime factorization is always possible

try to write that rigororously in set language

[rotund birch]

(btw im not afk, im thinking as i havent expressed prime numbers in set language before)

and well i wont always hit 2 and 3 , since nor a nor b can be 1 on definition of set S, 5 is element of P as well

since 5 cannot be expressed as 5=ab for some a,b in S

but i get your point

[proper hawk]

I mean the numbers can't be composite forever, as it's getting smaller every time we factor it

[rotund birch]

so i should have assumed that either k is a prime number OR k is a composite number which can be expressed as multiplication of prime numbers p1,p2,p3...pn (for n>1)

instead of just assuming n=ab for some a,b in S

right

[proper hawk]

umm

k+1

since you assumed k

[rotund birch]

oops yeah k=ab for some a,b in S

[rotund birch]

wait what do yo mean k+1

im talking about my assumptions currently, not what im showing

i think my assumption is wrong/not enough

[proper hawk]

h

oh

yeah that's the assumption

[proper hawk]

this looks good

[rotund birch]

splitting this into 2 cases is so free though lol

since i picked $k \geq 4$ , if k is prime then it must be odd, thus k+1 is even

[glossy valve]

Cyrenux

[proper hawk]

umm

[rotund birch]

but what do i do now

[proper hawk]

it really doesn't matter k+1 is even or odd

[rotund birch]

oh

so i should abort this thinking style

[proper hawk]

yeah

[rotund birch]

but then i have no idea how to show that k+1 is composite

[proper hawk]

you don't

k+1 is prime or composite

either

prime or composite

[rotund birch]

oh i meant, i dont know how to show that it can be expressed as multiplication of set P's elements

[proper hawk]

.

didn't I just told you

k+1 is either prime or composite

[rotund birch]

thats not what im looking for though

[proper hawk]

if it's prime then we're done

if k+1 is composite, then it's product of two smaller numbers a and b

[rotund birch]

i need to show that k+1 is either element of a prime or can be expressed as multiplication of P's elements

[proper hawk]

since a and b is smaller then k, a and b can be written as product of primes.

QED

[rotund birch]

ohh

i thought you need to show more symbolically

🤦

but yeah if i knew i couldnt show more i would have tried to type something like this, thanks

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[rotund birch]

(i havent done many proofs involing primes so i wasnt sure if that would have been enough)

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[outer tree]

Hello, can someone help me with this question please?

[outer tree]

When I try to do it I’m not getting the same values, this is my work

On the back of the textbook , the table for the interval and avg velocity is

[unreal fiber]

Are you using radians?

Yep seems like youre using degrees instead of radians

[outer tree]

ohhhh, that's why the steps weren't working. Thank you!

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[nimble crane]

!status

[full forum]

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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[nimble crane]

1

I'm having a really hard time trying to prove that 1+1 = c

or that 1+c = 0

I can prove that c*c = 1 if I can prove either of these facts

[ornate bobcat]

Ok so it is like you have a 3 base number system

0,1,c

In fields like this the first element usually is the place holder which is zero in this case

Meaning saying
0 + 0 will be 0

The next number which is 1 is the first number that we can use operations with

Saying
0 + 1 = 1

Now imagine if we are on the binary field

If we said 1+1

What would be the result

It will be 10

Because there's no other element in the binary system except
(0,1)

But if there is another element in a field
Like that one
{0,1,c}
We can say
1+1 = c
In that field

[nimble crane]

I'm not exactly seeing how that helps me prove it though.

[ornate bobcat]

Hmm I don't know also how can I prove it using math language

I thought that this the way the we should prove it

1+1 = C
Because adding one to a term in any field takes us to the next term if it is not the last term

As the field also has a finite number of elements
Adding 1 to the final term in it will take you all the way to the first element meaning
1+c = 0

[nimble crane]

I don't believe this is rigorous enough?

<@&286206848099549185>

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@nimble crane Has your question been resolved?

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@nimble crane Has your question been resolved?

[nimble crane]

<@&286206848099549185>

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@nimble crane Has your question been resolved?

[nimble crane]

.close

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[frank grove]

hi

[frank grove]

I would want some help

so

heres an equation :

log3(27) + log4(64)

So I can convert this to

3 + 3 right?

and then so

log3(27) + log4(64) = 6

am i correct?

[hot herald]

not an equation but
yes

[frank grove]

k

one more

question

so

a+b = 2

a+c = 3

b+c = 4

so whats a+b+c = ?

so is this right?

so first i do

a+b = b+c

and then I subsitude

[hot herald]

is already wrong

[frank grove]

how do I do it?

wait

let me try sometihng

a+b = 2 right?

k

so

2 + c = ?

im confused

[hollow sable]

When you have cyclic sums or products you should always ||try to add or multiply everything up||

[frank grove]

so like what?

[hollow sable]

Try it

[frank grove]

2+3+4?

im not getting it

[hollow sable]

Ok what about the variables

[frank grove]

a+b+b+c+a+c?

[hollow sable]

Simplify

[frank grove]

a+a+b+b+c+c

2a+2b+2c

[hollow sable]

Keep going

Yep

[frank grove]

then?

2a+2b+2c = 9?

[hollow sable]

Yes

Now what do you do

[frank grove]

then divide all of them?

[hollow sable]

Yes

[frank grove]

so divide it by 8?

[hollow sable]

What

[frank grove]

like

[hollow sable]

Why 8???

[frank grove]

I summed the 2's up

2 x 2 x 2 = 8

so i dont need to divide 2 three times

so 9 divided by 8 is 1.125

so that means a+b+c = 1.125

[hollow sable]

No

[frank grove]

?

[hollow sable]

If 2a=1 what is a

[frank grove]

1/2

a = 1/2

[hollow sable]

If 2a+2b=2 what is a+b

[frank grove]

2/4

wait no

[hollow sable]

I think you are confused

[frank grove]

Wait

Im dumb

2/2

1/2

k

1/2

[hollow sable]

No

[frank grove]

??

don't you subsitude

[hollow sable]

That’s only when you multiply

[hot herald]

2/2? 1/2? which one?

[frank grove]

2a/2 = a

though

so then you do it for the other side

[hollow sable]

So if I had 2a * 2b=4, I would divide by 2 * 2

[frank grove]

2/2

[hot herald]

you should divide based on factors, not how many times a number happens to appear on one side

[frank grove]

$ 3x + 4 = 28 $

[hollow sable]

But if I had 2a+2b=4 I would divide by 2 since 2a+2b=2(a+b)

[frank grove]

$3x+4 = 28$

[glossy valve]

Agent78

[frank grove]

bro

k

[frank grove]

heres what i mean

$3x = 24$

[glossy valve]

Agent78

[frank grove]

dont you divide

both sides?

[hot herald]

you divide by what's being multiplied to what you're trying to isolate

which is 3 in this case

[frank grove]

but why can't you divide for 2a+2b+2c = 9?

were figuring out whats a+b+c

[hot herald]

$\boxed{\text{what}?} \times (a+b+c) = 2a + 2b + 2c$

[glossy valve]

ℝam()n()v

[frank grove]

2?

[hot herald]

yes

[frank grove]

$\boxed{\text{what}?} \times (a+b+c) = 2a + 2b + 2c$

[glossy valve]

Agent78

[frank grove]

hm

so your saying

9 = 2(a+b+c)

[hot herald]

$2a + 2b + 2c = 9 \
2\underbrace{(a+b+c)}_{x} = 9$

[glossy valve]

ℝam()n()v

[frank grove]

so

9 = 2(a+b+c)

[hot herald]

you divide by what's being multiplied to what you're trying to isolate
which is 2 here

[frank grove]

so then a+b+c = 4.5*

[hot herald]

no

[frank grove]

i meant 4.5

[hot herald]

yes

[frank grove]

hm

so this is practice

k

$3a + 7b + 9c = 25
2\underbrace{(a+b+c)}_{x} = 25$

$3a + 7b + 9c = 25 \

2\underbrace{(a+b+c)}_{x} = 25$

[glossy valve]

Agent78
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[frank grove]

k thanks

.close

[full forum]

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[deep hull]

the goal in this question is to find d using only 5 angle theorems

[deep hull]

im struggling to even know where to start

ive spent a lot of time looking at it

the quadrilateral that contains 58 degrees and d looks like a parallelogram but none of the sides are marked as parallel or congruent

theres a lot of other questions like this one that ive already done

im just struggling with this one

[tacit siren]

The dashed lines are marking congruent sides I think

[deep hull]

im aware yeah

but it doesnt help me at all

as far as i can tell

[tacit siren]

uhh are you allowed to use the fact that isosceles triangles have equal base angles??

[deep hull]

i can mark the angles as congruent but it doesnt acomplish anything

yeah thats ITT

[tacit siren]

yeh it soes

[deep hull]

what does it do

[tacit siren]

OK so

Label the angles of the isosceles triangles as theta1 and theta2 first or w/e you wanna call them ur choicw

[deep hull]

uh huh

[tacit siren]

What is theta1+d+theta2

Label them in ur diagram as such if it helps

Oops

I forgot to say "base" anglea

We're just calling the base angles of one triangle theta1 and of the other theta2

[deep hull]

like this?

[tacit siren]

yeee

x and y are good choices

[deep hull]

yeah x + d + y = 180

actually wait does it

[tacit siren]

Ye

[deep hull]

i mean it looks like it but what if the diagram isnt to scale

theres no markings for it being 180

[tacit siren]

They sum up to a line's angle tho

well hmm ic now

[deep hull]

what if this is more accurate

[tacit siren]

Yeh guess we can't assume it but we don't need it

[deep hull]

and the diagram just isnt accurate

[tacit siren]

Actually nvm can't think of a way of doing this without assuming it's a line

[deep hull]

well you can only use angle theorems anyway

[tacit siren]

which ones are they exactly?

[deep hull]

sorry for phone shadow

those are all the ones they provide

[tacit siren]

Oh we'd be using supplementary angles if we used the line thing

[deep hull]

supplementary is just 2 tho, right?

[tacit siren]

Description says "2 or more"

[deep hull]

ah i see

[tacit siren]

Anyways if we can assume SA then you can use SATT to get the third angles of the isosceles triangles in terms of x and y

And those angles all belong to a bigger triangle that has the 58 degree angle

So u can solve for (x + y) in terms of a nunber using SATT again

And from x + y +d = 180 u get d

[deep hull]

wdym by "in terms of x and y"

what would be the equation for the angles in that case

[tacit siren]

Like for instance saying <ABC = x + 90 (ABC are just random letters of a triangle)

And for the other angle saying it equals y + 90

Those aren't the right equations btw

Just examples

[deep hull]

uhh ok

[deep hull]

but we cant assume sa, right?

[tacit siren]

I think we have to assume they drew an unambiguous line 🙃

[deep hull]

thats fair i guess

[tacit siren]

I mean in the illustrations they drew for the theorems they don't always tell you that they're lines either

[deep hull]

ok so im still a little confused on the equations

x + y + d = 180

but how do we turn that into angles in terms of x and y

[tacit siren]

Well the idea is you can solve for (x +y) in some other way

[deep hull]

how would that be

[tacit siren]

And to start that off you calculate what the third angles are on the isosceles angles as an expression of x for one of them, and as an expression of y for another

Cus eventually

We want to use the labeled angle

[deep hull]

so its like, 180 - 2x?

[tacit siren]

And those angles belong to a larger triangle containing the 58

Yeeee

[deep hull]

ok so (180 - 2y) + (180 - 2x) + 58 = 180

[tacit siren]

Yup

[deep hull]

i just have a feeling this may be the wrong route

[tacit siren]

whys that

[deep hull]

other questions right before dont require any algebra or labeling or anything like that

you start from an angle and work your way around

the whole point is that you're just using angle theorems and nothing else

[tacit siren]

hmm

I mean we are using just the theorems

I guess here first we do isosceles twice

Then SATT three times and then SA one time so that might be too much unless they're not counting by number of timea

[deep hull]

ur meant to use 5

not 5 different

just 5 total

[tacit siren]

rip

[deep hull]

yeah i think you need a different approach

[white flicker]

You got this solved?

[deep hull]

no not yet

[white flicker]

Ok

Let me help you

[deep hull]

alright sure

[white flicker]

Where is the question

[deep hull]

in pins

[deep hull]

these are the list of angle theorems you're supposed to use

[white flicker]

This is very ez though

[deep hull]

well uhh

idk what to say to that

its hard for me

ive spent a lot of time trying to figure it out

[white flicker]

But i see you just figured it out

[tacit siren]

we need to use 5 theorems and repeats count

[deep hull]

in what way

[tacit siren]

But we used 6

[white flicker]

wait

you filled out the blank

which it means your issue is solved

[deep hull]

what blank

[white flicker]

Right?

[tacit siren]

Those are different problema

[tacit siren]

They were example problems

[white flicker]

Hmmm

[deep hull]

yeah those are different problems

[tacit siren]

Scroll all the way up

[white flicker]

Oh

I didn't see that

[deep hull]

i was trying to show that the methodology didnt seem right

cuz other questions used a different method

[tacit siren]

Problem is figuring it out in 5 or less theorems rn

My method used 6

[white flicker]

This one?

[deep hull]

yes

[tacit siren]

And again repeats count

[white flicker]

Lol

[white flicker]

This is very easy brother

I mean bro

[deep hull]

what do i say to that

[white flicker]

Let me fill those answer for you

[deep hull]

i find it hard

[tacit siren]

No the answer isn't the problem

[white flicker]

This is like middle school geometry

[tacit siren]

See that paper of theorems?

We need to use it 5 timea

[white flicker]

Ye

[deep hull]

specifically its grade 10 curriculum

[white flicker]

I don't need it though

[tacit siren]

No like

[deep hull]

you do tho

[white flicker]

I still remember all the theory of triangles

[deep hull]

its required

oh i see

[tacit siren]

No omg just go away

[white flicker]

I did questions probably hundreads of times like this

Lol

[deep hull]

ok we get it how do you do it

[tacit siren]

The problem is that he needs to use it 5 times for his homework and he's restricted to those

Not that we dont know them

[deep hull]

yeah

[tacit siren]

Hell we already figured out the solution too but it took 6 stepa

That's the only problem we need a solution of 5 or less steps :p

[white flicker]

ok

i get it

you will only and have to do 5 steps to get the angle of D

[deep hull]

yes

[white flicker]

Lemme do it

Hold my beer

[deep hull]

uh ok sure

[white flicker]

wait

where is the term though

[deep hull]

wdym

[white flicker]

is that all you got?

[deep hull]

yes

[white flicker]

bruh

[deep hull]

you arent given anything more

[white flicker]

😰

[tacit siren]

no parallel lines labeled but maybe we can prove them

[white flicker]

ez

hold my beer

[deep hull]

but how do we prove parallel lines using only angle theorems

[tacit siren]

Actually nvm not sure if we're allowed to assume the converse of those angle theorems for the parallel lines

[deep hull]

yeah

[tacit siren]

yeah the theorems only say "if parallel then (thing" not if thing then parallel

[deep hull]

almost all of the solutions so far have involved drawing/continuing an extra line then using that to solve

so thats why i drew a bunch of lines

they havent helped tho lol

[white flicker]

Tbh

[deep hull]

yeah

[white flicker]

You sure there is no any term for this question?

No way you can do this quesiton without any other angles

[tacit siren]

U can but in 6 steps :p

[white flicker]

yea

i mean by that

[deep hull]

i mean this is part of a 41 question package

so its not a big deal if i dont do it

[tacit siren]

when the side boss tougher than the final boss

[deep hull]

but its one of the last questions i need to do

[deep hull]

i mean it is literally the very last question in the package

so it probably is the final boss

[tacit siren]

Oh kek

[white flicker]

can you show me how do you get the 6 steps

[tacit siren]

First use ITT twice

[white flicker]

I'm wondering no angles no parallel

[tacit siren]

Then on the isosceles triangles use SATT twice

Then use SATT à third time on the bigger triangle containing the 58 degree angle and the third angle of the 2 other triangles

[deep hull]

oh new person typing

this could be big yall

[tacit siren]

And finally supplementary angles and solve for d

So like you basically make two equations and you can consider x+y as a single variable