help-23

[warped roost]

so

to show the full derivation

[little glacier]

thats a word i like

[warped roost]

the amount of profit you make with a single instance of machine C is given by [
(0) + (0 + 1) + (0 + 1 + 2) + (0 + 1 + 2 + 3) + \dots = \sum_{m=1}^{n} (0 + 1 + \dots + m-1) = \sum_{m=1}^{n} \sum_{k=0}^{m-1} k] which is

[= \sum_{m=1}^{n} \frac{(m-1)m}{2}]

[little glacier]

im going to save this image

what is n, and m

and k

[warped roost]

m and k are just indices of summation

n is the total number of time steps that have elapsed

[flat frigate]

Namington

[warped roost]

now youre trying to solve [\sum_{m=1}^{n} \frac{(m-1)m}{2} \geq 100n] since machine C' produces 100 dollars every second, times $n$ seconds

[flat frigate]

Namington

[warped roost]

i would strongly recommend just doing this with a calculator

but if you wanted to do it by hand

you get [\sum_{m=1}^{n} (m-1)m \geq 200 n \implies \sum_{m=1}^{n} m^2 - m \geq 200 n] and we can break up the sums to [\sum_{m=1}^{n} m^2 - \sum_{m=1}^{n} m \geq 200 n]

[flat frigate]

Namington

[warped roost]

now $\sum_{m=1}^{n} m^2 = \frac{n(n+1)(2n+1)}{6}$ and $\sum_{m=1}^{n} m = \frac{n(n+1)}{2}$ are both well-known formulae

[flat frigate]

Namington

[warped roost]

substituting that in gives [\frac{n(n+1)(2n+1)}{6} - \frac{n(n+1)}{2} \geq 200n]

[flat frigate]

Namington

[warped roost]

from here its just algebra

[little glacier]

((x(x+1)(2x+1))/6)-((x(x+1))/2)

sorry judt needed something to put in

[warped roost]

[n(n+1)(2n+1) - 3n(n+1) \geq 1200n] and thus expanding and simplifying [2n^3 - 2n \geq 1200n] which is [2n^3 - 1202n \geq 0]

[flat frigate]

Namington

[warped roost]

from here we can assume $n \neq 0$ and divide out by $n$ to get $2n^2 - 1202 \geq 0$, i.e. [n^2 \geq 601] which gives $n \geq \sqrt{601}$

[flat frigate]

Namington

[warped roost]

now $24^2 < 601$ while $25^2 > 601$

[flat frigate]

Namington

[warped roost]

so 25 is the minimum number (amount of time) such that buying machine C is better than buying C'

[little glacier]

damn

DAMN

youre a beast

[warped roost]

for machine B the math is similar but much easier (fewer steps, less indices to track)

youd get 20 at the end

[little glacier]

i will mark this as unoccupied?

eh before that ill just leave with a

more realistic problem that would be in a roblox game

[warped roost]

oh whoops youd get 21

slight indexing error on my part

just cause i was printing t-1 instead of t

close enough though

its literally just solving (n-1)n/2 >= 10n

which gives n² - n ≥ 20n, i.e. n ≥ 21

anyway i would not solve these algebraically

i'd write a simulation

much nicer

[warped roost]

or a spreadsheet if you prefer those

advantage of spreadsheets is that theyre easier to check for errors (like the indexing errors i made)

since they display info more visually

[little glacier]

Machine A makes 1$/t
Machine B makes 1A/t
Machine C makes 1B/t and decreased the tickspeed by 2 ms
Machine D makes 1C/t and multiplies the money Machine A makes by 1.1x
Machine E makes 1D/t

a tick is a one second. it can only go down to 500ms before it stops.

Machine E¹ makes 100000/s

When does the first set of machines, starting with Machine E have a more total money than Machine E¹?

problem 2(definitely a problem for a computer to brute force)
Machine A makes 1/s, and costs 10$, scaling to be about 1.1x more each time.
Machine B makes 1A/s, costs 75$, and scales 1.15x each time.
Machine C makes 1B/s, costs 500$, and scales 1.2x each time.
Machine D makes 1C/s costs 5000$, and scales 1.25x each time.
Machine E makes 1D/s costs 75000$, and scales 1.3x each time.

Machine E¹ makes 1 Million per second.

How fast can you have more money than Machine E¹?

closing ticket

[warped roost]

ok yeah this is the kind of problem i'd 100% just write a program for

i can do that quick

"more money than E¹" meaning like

[little glacier]

yes like total money

[warped roost]

how fast does the amount of money generated by purchasing a single E exceed the amount generated by purchasing a single E¹?

or are you re-investing your money

[little glacier]

oh oops

for problem 2

you start with 0$ and have a single Machine A

and Machine E¹ is always just there

.close

[safe radish]

Channel closed

Closed by @little glacier

Use .reopen if this was a mistake.

[safe radish]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[dull sequoia]

does the counting process count as a sequence? it would be $N:\mathbb R^+ \to \mathbb N_0$ which looks like it has a nicely well ordered domain

[flat frigate]

Frosst

[obtuse jackal]

A sequence is a function whose domain is N (except possibly a few first values)
Potentially a countable domain
But not R

[obtuse jackal]

Also, what's the counting process ?
Cause that sounds ungoogleable to me

[dull sequoia]

https://ocw.mit.edu/courses/6-262-discrete-stochastic-processes-spring-2011/3a19ce0e02d0008877351bfa24f3716a_MIT6_262S11_chap02.pdf

im on page 70

[obtuse jackal]

Scrolls down to page 35, then to page 2

[dull sequoia]

perhaps they define it more in another chapter but i dont see it in this chapter

haha

i found this on another page

[obtuse jackal]

You could consider the sequence tn of the arrivals
Assuming they're infinite (or accept finite sequences)

[dull sequoia]

well that's the point

it's saying there are 3 ways to look at this arrival process

1 is to look at the time of each arrival in a sequence

1 is to look at the interarrival time between each occurence in a sequence

[toxic stratus]

a counting process is defined on $\Omega \by [0, \infty)$

[dull sequoia]

and the last is to consider the counting process

i suppose here we are just considering Ω to have only 2 events

[toxic stratus]

what

no

Ω is usually very rich

[obtuse jackal]

It's the set of all possible sequences of arrivals

[toxic stratus]

i mean

it doesn't have to be

[dull sequoia]

$N:\Omega\times [0,\infty) \to \mathbb N_0$ where $(A,t) \mapsto N(A,t)$? where A is how much you count the even that occured?

[flat frigate]

Frosst

[dull sequoia]

from 0 to time t?

[toxic stratus]

why do you choose A

[dull sequoia]

well where's this thing from Ω

[obtuse jackal]

A is the reality/distribution you consider

[dull sequoia]

idk just some symbol

[toxic stratus]

,, (\omega, t) \mapsto N(\omega)(t)

[dull sequoia]

like what do you do with the input from Ω

[dull sequoia]

i can count different events happening?

[toxic stratus]

say you take Ω to be the canonical space

,, \Omega = \set {, \omega : [0, \infty) \to \Z_{\ge0} \where \text{$\omega$ is non-decreasing},}

[dull sequoia]

oh i see, this w depends on what you're actually counting

[toxic stratus]

then you can just take the canonical process [ N : \Omega \to \Omega, \quad N(\omega) = \omega ]

[toxic stratus]

ω is then one particular path taken

[dull sequoia]

right so if say im standing by the road

one ω could be how many cars pass me

[toxic stratus]

N(ω)(t) = ω(t) would be what we typically write as N_t

[dull sequoia]

another could be how many times the light turns green

and it would just depend on what im counting to determine which ω im looking at

[toxic stratus]

i mean each ω is just abstractly some possible count of something over time

[dull sequoia]

right

[toxic stratus]

and the probability measure you put on Ω will decide what the distribution of the canonical counting process is

[dull sequoia]

that sounds like it makes sense

[toxic stratus]

[dull sequoia]

hold up, why does the counting process have a distribution

[toxic stratus]

i mean it must right

[dull sequoia]

im told processes are not RVs and dont have a distribution

[toxic stratus]

well

[dull sequoia]

im told processes are just a sequence of RVs

[toxic stratus]

depends on how you interpret it

[toxic stratus]

this N is a measurable map (Ω, F) -> (Ω, F), for some sigma algebra F on Ω of your choosing

you should assign probabilities to this measurable space to know how N behaves

each outcome is a path itself

[dull sequoia]

right, depending on how i choose the measure, the measure of N would change

[toxic stratus]

no not the measure of N

the "distribution" whatever that means

[dull sequoia]

and that's the part where we go P(N(t) = n) which is asking what's the measure of the preimage of n under N at some t

[toxic stratus]

you can probably relate this to finite dimensional distributions or something

[dull sequoia]

oh it wouldn't be the measure of N itself but rather the preimage of the outcome

oh and then f_N(n) = P(N(t) = n) would make that a distribution and most definitely depend on P?

[toxic stratus]

you can also view N as a map [ (\Omega \by [0, \infty), \FF \by \BB([0, \infty)) \to (\Z_{\ge0}, \PP(\Z_{\ge0})) ]

[toxic stratus]

no like

[toxic stratus]

in this view

each N(ω) is an entire path

so you'd be talking about [ P(N \in A) ] where $A$ is a collection of paths

[toxic stratus]

like for example

[ P(N(t) = n)] has [ A = \set {, \omega : [0, \infty) \to \Z_{\ge0} \where \omega(t) = n,} ]

[dull sequoia]

right it's all the different paths that can also result to having n counts at time t

[toxic stratus]

if you take this view, you're thinking of each outcome of the sample space given you a single value of N at time t

[dull sequoia]

is the F x B part the events and some time interval

[toxic stratus]

like that's just some suitable sigma algebra on the sample space

the x isn't actually cartesian product

it's some weird sigma algebra product thing

[dull sequoia]

oh right

kinda like the product topology?

[toxic stratus]

yeah

[dull sequoia]

so just all the events and time interval we can measure

[toxic stratus]

pretty much

[dull sequoia]

so as you said just some suitable sigma algebra right

[toxic stratus]

well it depends on what you want to measure i guess

there's probably a very canonical one for the Ω i defined

which will just be the borel sigma algebra

since Ω is actually a metric space

[dull sequoia]

how do you know it has a metric

[toxic stratus]

uh

you can put one on it

can't remember what it's called

some uniform metric

[dull sequoia]

so are counting processes RVs then, they go from a sample space to a measurable space N_0

[dull sequoia]

this looks mighty like a measurable map that might as well be an RV

[safe radish]

@dull sequoia Has your question been resolved?

[dull sequoia]

im also kinda confused what this set notation is

{N(t) ≥ n} is a set, it contains numbers bigger than or equal to n, so it could look like {n, n+2, n+3, n+5}?

or is this more like ${N(t) \geq n} = \text{preim}_{N}([n, \infty))$

[flat frigate]

Frosst

[dull sequoia]

but where did the t go, i can't put N(t) at the bottom or that would become the value in the codomain and not a map

[dull sequoia]

ok i've resolved this problem

thanks!!

.close

[safe radish]

Channel closed

Closed by @dull sequoia

Use .reopen if this was a mistake.

[safe radish]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[drifting skiff]

why does the x just randomly get there?

3 --> 3x

[peak estuary]

well thats how you integrate constants

[drifting skiff]

oh wait lmao

youre right XD

the heck

.close

[safe radish]

Channel closed

Closed by @drifting skiff

Use .reopen if this was a mistake.

[safe radish]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[brazen parrot]

helo

[brazen parrot]

how i do dis?

helo?

<@&286206848099549185>

[safe radish]

@brazen parrot Has your question been resolved?

[brazen parrot]

is this right

[safe radish]

@brazen parrot Has your question been resolved?

[safe radish]

@brazen parrot Has your question been resolved?

[safe radish]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[dull sequoia]

so im looking at poisson processes, and im up to the proof of this theorem

[dull sequoia]

https://ocw.mit.edu/courses/6-262-discrete-stochastic-processes-spring-2011/3a19ce0e02d0008877351bfa24f3716a_MIT6_262S11_chap02.pdf

page 73

i've got this as the given definition of memoryless

[flat frigate]

Frosst

Frosst

[dull sequoia]

but how do i do that

[flat frigate]

Frosst

[dull sequoia]

or at least how can i tell from just the information that X is a memoryless RV

[safe radish]

@dull sequoia Has your question been resolved?

[safe radish]

@dull sequoia Has your question been resolved?

[covert yoke]

Why X_1 vs X?

Where is X_1 coming from exactly @dull sequoia

[covert yoke]

These are obviously not independent. Because if we give concrete values, like, for instance z = 3 and t = 2, we have if X_1 > z then it's always greater than t.

[safe radish]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[craggy island]

ive applied the sin 2x identity and the cos 2x identity

[craggy island]

i got 2sinxcosx-1+2sinx^2=1+sinx-cosx

whats the next step?

[safe radish]

@craggy island Has your question been resolved?

[safe radish]

@craggy island Has your question been resolved?

[safe radish]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[safe radish]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[dense wadi]

i dont get it whats wrong iwth my asnwer

first one is good but second is wrong

[thin bridge]

how are you getting thje 120/126

[dense wadi]

its 100 mb

.close

[safe radish]

Channel closed

Closed by @dense wadi

Use .reopen if this was a mistake.

[safe radish]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[next flower]

Help

[next flower]

How do I find X, Y, and Z

The two shapes are supposed to be similar

ABCDE~KLMNP

[shadow sparrow]

do you know what is the scaling factor?

[next flower]

no

I think I figured it out tho

[lean otter]

it is 2:3 i think

[shadow sparrow]

good job

[next flower]

Is x

2/3

[lean otter]

no

[next flower]

I mean 3/2

[lean otter]

yes

[next flower]

Can I find y by doing 2/3 = 3 square root 2/y

[lean otter]

yes that looks right

[next flower]

But how do I multiply

3 and 3 2

3 square root 2

[lean otter]

so 2/3 = 3*sqrt(2)/y

so multiply both sides by 3 to get 2 = 9*sqrt(3)/y

[next flower]

.done

.close

[safe radish]

Channel closed

Closed by @next flower

Use .reopen if this was a mistake.

[safe radish]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[viscid spire]

nvm

it shouldve been 2in

not 2i

[safe radish]

Channel closed

Channel closed due to the original message being deleted.
If you did not intend to do this, please open a new help channel,
as this action is irreversible, and this channel may abruptly lock.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[granite cobalt]

5c

can someone explain how to do part c

[safe radish]

@granite cobalt Has your question been resolved?

[safe radish]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[granite cobalt]

.reopen

[safe radish]

✅

[twilit spindle]

Are you familiar with the Cauchy Inequality?

[granite cobalt]

no

[twilit spindle]

$\left(\sum_{i}a_ib_i\right)^2\leq\left(\sum_{i}a_i^2\right)\left(\sum_{i}b_i^2\right)$

[granite cobalt]

how do i use that

is there a video

[twilit spindle]

sqrt(x,y,z)^2=x,y,z

[granite cobalt]

i dont really get what you mean

[twilit spindle]

oh wait I messed up the inequality

[flat frigate]

PajamaMamaLlama

[twilit spindle]

This is the corrected one

[twilit spindle]

In this case, $(\sqrt{x}^2+\sqrt{y}^2+\sqrt{z}^2)\left(\left(\frac{1}{\sqrt{x}}\right)^2+\left(\frac{1}{\sqrt{y}}\right)^2+\left(\frac{1}{\sqrt{z}}\right)^2\right)\geq\left(\sqrt{x}\cdot\frac{1}{\sqrt{x}}+\sqrt{y}\cdot\frac{1}{\sqrt{y}}+\sqrt{z}\cdot\frac{1}{\sqrt{z}}\right)^2$

[flat frigate]

PajamaMamaLlama

[safe radish]

@granite cobalt Has your question been resolved?

[drowsy karma]

hint for you:

[safe radish]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[frigid coyote]

do not include similarity

congruency stuff rela ted

[lean otter]

oh this i responded to this in an older channel

so basically

MNS ~ RTS right?

since it’s perpendicular bisector TN = NS

so then since TS = (TN + NS) = 2NS the scale factor is 2

then reverse this argument to prove that RM = MS therefore M is the midpoint of RS so it is a median

oh wait do you know what similarity is?

oh lol i didn’t see that

ok

so

you have to prove that RM = MS

if your geometry textbook has some kind of theorem for proving this with right angles then use that

[frigid coyote]

well uhh

how do we know if

RS = 2MS

...

we are using congruency

so either add angles or add segments

[lean otter]

yeah

[frigid coyote]

try to make somehow angles 45 degrees

[lean otter]

sorry i just learned similarity lol

[frigid coyote]

sorry about that, but we are not using similarity

[lean otter]

yeah

[frigid coyote]

congruency over

proving a right triangle is like

if two angles are supplementary and congruent

both angles are right

@lean otter

yout here

[lean otter]

yeah

is the pythagorean theorem allowed?

nvm

is there trig?

so to prove that <RMT ~= <SMT will prove it

[frigid coyote]

no

no trig or pythagorean

theorem

[lean otter]

yeah i thought so

ooh <TRM and <NMR must be supplementary

so

ooh i think i’ve got it

and <NMR and <NMS must be supplementary too

so therefore <TRM and <NMS are congruent

so the angles of the triangles are 45 degrees

[frigid coyote]

okay I get it but

how do ou get that

these angles are supplementary

[lean otter]

oh wait

[frigid coyote]

and NMR is congruent to itself

[lean otter]

yeah nvm i made a mistake srry

[frigid coyote]

oops

basically

if MN is perpendicular bisector of TS

TN = NS and MN is perpendicular to TS making

<TNM and <SNM

[lean otter]

both congruent and right

[frigid coyote]

yes

[lean otter]

if i remember correctly 2 angles supplementary to the same angle are congruent?

[frigid coyote]

the supplement theorem

yes

[lean otter]

ok

yeah so <TRM is 45 degrees

[frigid coyote]

wait wait

we got angle TNM and SNM

only

[lean otter]

yeah but here’s why

[frigid coyote]

they have to forma linear pair

to be supplementary

[lean otter]

so the angles of a quadrilateral must add up to 360 degrees have you learned that? if not i can fix it to not use that

[frigid coyote]

nope

[lean otter]

ok

[frigid coyote]

so

what now

[lean otter]

oh ok i got an idea

so <NMS ~= <TRS right?

[frigid coyote]

idk

how do you prove it

[lean otter]

they are supplementary to the same angle, <NSM

[frigid coyote]

wait wait wait

Can we start from the beginning

[lean otter]

ok

[lean otter]

wait can i finish it first

[frigid coyote]

sure

then we do it again

[lean otter]

yeah

wait what if <X is complementary to <Y and supplementary to <Z does this mean something?

[frigid coyote]

so what is X, Y and Z

also it doesnt

not in the lesson

[lean otter]

ok

[frigid coyote]

@lean otter

help

[lean otter]

i’m working on it

[frigid coyote]

okay

that is due tomorrow

[lean otter]

i’ll try what time

[frigid coyote]

basically tomorrow for me

its like

25 hours

left

but actually its 12

[lean otter]

i’m in eastern us time so i can help u tmr (for me)

[frigid coyote]

but tmr

is my school

[lean otter]

ye

[frigid coyote]

i have to pass it

[lean otter]

i’ve proved that <TRM sup. < NMR

[frigid coyote]

how?

[lean otter]

so basically <TRM and <S are sup. and <NMS and <S are sup so <TRM ~= <NMS then <NMS and <NMR are sup. so <TRM sup. <NMR

[frigid coyote]

hang on, isnt <NMS and <S complementary

it doesnt look supplementary

[lean otter]

wait a second TM = MS

they are because of a higher level theorem

[frigid coyote]

TM = MS now I am listening

probably my future lesson

[lean otter]

that i have converted into simpler form

anyway

[frigid coyote]

but I need to prove RM = mS

[lean otter]

yeah

so now we have to prove that RM = TM or <MRT ~= <MTR

isoceles

YESS I HAVE IT

ok so <TRM + <RMT + <TMR = 180

[frigid coyote]

wait wai twait

okay

[lean otter]

repost the problem pls

[lean otter]

ty

then <TMN + <MTN = 90

and <TRM + <RMT + <TMR = 180

[frigid coyote]

wait

how did you get TMN and MTN = 90

[lean otter]

but finally <MTN + <MTR = 90 substituting should prove that <TRM ~= <MTR i think

[lean otter]

its a right triangle

the angles of a triangle sum to 180 degrees therefore the other 2 angles of a right triangle sum to 90

[frigid coyote]

so you can prove that

they are 45 degrees:/

[lean otter]

no not that

[frigid coyote]

well they can be congruent

[lean otter]

ok i’m gonna try to construct a proof outline now

<R + <S = 180 given
<NMS + <S = 180 given
<R + <NMS = 180 substitution
<TMN + <MTN = 90 right triangle other angles are sup.
<MTN + <MTR = 90 given
<TMN ~= <MTR substitution
RM = TM isoceles
TM = MS “given”
RM = MS

rough draft

[frigid coyote]

ok

i will just try

[lean otter]

ask me if you have any questions

?

wdym

[safe radish]

@frigid coyote Has your question been resolved?

[frigid coyote]

@lean otter

where did you get angle R and angle S

and angle NMs nad angle S

[safe radish]

@frigid coyote Has your question been resolved?

[safe radish]

Channel closed

Closed by @frigid coyote

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[primal condor]

how do i calculate the gradient and hessian?

[drowsy karma]

the second partial derivatives, needed for the Hessian, can be calculated analogously

[safe radish]

@primal condor Has your question been resolved?

Channel closed

Closed by @primal condor

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[wet gust]

can i ask physics 1 type questions here

[wet gust]

in an action-­‐adventure film, the hero is supposed to throw a grenade
from his car, which is going 90 km/h, to his enemy’s car, which is going 110 km/h.
The enemy’s car is 30 m in front of the hero’s when he lets go of the grenade. If the
hero throws the grenade so its initial velocity relative to him is at an angle of 45o
above the horizontal, what should the magnitude of the initial velocity be relative to
the hero?

so i did the r = vo^2sin2theta

and solved for Vo but not sure how to get the magnitude of Vo

or is magnitude just Vo ? so confused

[safe radish]

@wet gust Has your question been resolved?

[safe radish]

@wet gust Has your question been resolved?

[safe radish]

@wet gust Has your question been resolved?

[safe radish]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[safe radish]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[white mirage]

just wanted to check if I have done this correctly as the textbook doesnt have answers for graphing questions

[exotic sky]

its right

[white mirage]

tyty

.close

[safe radish]

Channel closed

Closed by @white mirage

Use .reopen if this was a mistake.

[safe radish]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[white mirage]

how would one go about doing this question?

[versed wave]

the only way for a number to reduce in term of digits count is 10, 100, 1000..., and its prime so we can ignore -1

the question is asking how many digits in 2^74207281

when we say y = log10(x), what does x equal?

(for sake of convenience ill call log10 as just log)

[safe radish]

@white mirage Has your question been resolved?

[white mirage]

why can we ignore the -1 because its prime?

[versed wave]

read again

[white mirage]

oh

it wont change the number of digits

[white mirage]

x= 10^y?

[versed wave]

correct

so when y = 1, x = 10

y = 2, x = 100

y =3, x = 1000

conversely, x = 999, y = 2.999

from this we can infer the pattern that floor(y) + 1 = number of digits of x

are you following?

[white mirage]

ok

yes

[white mirage]

ohhhhhhhhh

but we have base 2

[versed wave]

now lets consider not x=10^y, but rather x=b^n

[white mirage]

ok

[safe radish]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[white mirage]

.reopen

[safe radish]

✅

[versed wave]

that is, b and n are any number

so we got y = log(b^n)

and what does that equal?

@white mirage

[white mirage]

as in like y=nlogb?

[versed wave]

correct

putting this all together, the number of digits of b^n is 1+floor(nlogb)

now just plug in all the numbers and there you have it

[white mirage]

ah i see

[versed wave]

note, this only works if b and n are positive integers

otherwise its a different story, but thats far from what the original question is asking here

[white mirage]

okok ty

I don't seem to be getting the correct answer

its close

because i typed it in wrong lol

[white mirage]

and im assuming because of this we always round down

[safe radish]

@white mirage Has your question been resolved?

Channel closed

Closed by @white mirage

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[lean otter]

I am studying Pythagoras theorem and came across this problem,

[west haven]

,rotate

which one

[lean otter]

6

The answer is 466.18

But I don’t know how to solve

[west haven]

what did you try

[lean otter]

Idk really

I tried finding the distance from the 80m cable and the base

Like the distance between the base of the radio tower and the end of the 80m cable

[west haven]

and what did you get?

[lean otter]

60m is the base distance

I think

Oh wait

I think it’s 78

M

<@&286206848099549185>

[safe radish]

@lean otter Has your question been resolved?

[lean otter]

<@&286206848099549185>

<@&286206848099549185>

[polar jay]

man, that question is confusing me... i cant get a straight image of what that question is tryna convey...

[lean otter]

I can try draw if u want

One minute

[polar jay]

ight

[lean otter]

The diagram on the bottom

The X and a are unknown

[polar jay]

ah ok

is a the entire green line?

[lean otter]

Yes

But I’m pretty sure the part of the left is 30m

Because the cable is 60m

And it connect the the tower 30m high

[polar jay]

is that 80 the length of red or the height of it?

[lean otter]

Height

[polar jay]

ah ok

[lean otter]

Trying to figure out length of X

If it helps I know the answer

It’s in the back of the book

Just not the solution

[polar jay]

is the 60m the length of the green line too?...

[lean otter]

No 60m is the length of the cables on the bottom

[polar jay]

oh

[lean otter]

The one circled is 60m

[polar jay]

ight sorry man, i couldnt solve it. good night

[lean otter]

Good night

<@&286206848099549185>

[shadow aurora]

what

[polar jay]

i think x is 160m

because the smaller triangle and the bigger triangles are similar

[shadow aurora]

i don’t understand the hand writing

[lean otter]

I can do better handwriting

[polar jay]

the ratio between the small triangle and the big one is 1:2.666, since 80/30 gives us 2.666. so you just have to multiply the length 60 by 2.666 and it gives us the length of the bigger triangles hypotinuse.

[lean otter]

That’s what I tried before

[polar jay]

you were correct then...

or were you?

[lean otter]

But it says the answer is 466.18

In the back of the book

[polar jay]

huh...

[lean otter]

If I times 160 by 3

And then an extra 180 for the bottom cables

It’s 660

[polar jay]

well im out, cya man

[lean otter]

Bye

,rotate

[lean otter]

@shadow aurora

[safe radish]

@lean otter Has your question been resolved?

[safe radish]

@lean otter Has your question been resolved?

[safe radish]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[misty ember]

"Consider the region bounded by the curves y= x/(x^2 + 3/4) and y = x

a. These two curves intersect three times; find the three points of intersection.
b. Find the area of the region bounded by the two curves in the first quadrant."

Posting this again because in the last channel I didn't get any answer for an hour, hopefully that's okay. I'll send the work I've done in a second.

[stoic dune]

,w graph x/(xx + 3/4) and x

[flat frigate]

Results provided by WolframAlpha

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

Requested by kaynex

[stoic dune]

Yeah your work looks good so far

[misty ember]

Thank you!

.close

[safe radish]

Channel closed

Closed by @misty ember

Use .reopen if this was a mistake.

[safe radish]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[shadow grove]

How do I do this question?

[tame raft]

if it were an equation could you solve it

[teal skiff]

factorise it, draw simple graph, solve from there

[shadow grove]

Is it a b^2-4ac equation?

[hidden mist]

yeh

[shadow grove]

Ohh

[hidden mist]

wait tho

[shadow grove]

Wait

[hidden mist]

why do u need b^2 - 4ac

u can factorize it in the easier way

but either is fine'

[shadow grove]

Yeah sorry that equation is for something else I think

So I factorise it

And then and draw a graph?

[hidden mist]

why do u need a graph

u gotta find x right

[shadow grove]

X?

[hidden mist]

u jsut take the intersection of the values of x

the two values

[shadow grove]

I don’t think you are supposed to solve for x

[teal skiff]

drawing graph here would be really helpful to solve the problem

[hidden mist]

tho u can jsut look at the roots and take intersection

[shadow grove]

If I can remember you do the b^2-4ac equation

[hidden mist]

its the same as factorization

[teal skiff]

the discriminant will only tell you how many roots there are

or like

[hidden mist]

the formula is

[teal skiff]

how many real roots

[shadow grove]

Is that one way to do it?

[teal skiff]

not for this case no

[shadow grove]

Right

[teal skiff]

if you're talking about the entire quadratic equation

[hidden mist]

-b + or - rootB^2-4ac / 2a

u can use that

[teal skiff]

yeah that could be used to find the 2 roots

to graph

but factorising it is quicker if you're familiar with how to factorise a non-monic polynomial

and you'll see the equation is just

(2x-5)(x+1)

and then graph that

and just see for what x values is the graph above the x axis aka (> 0)

and solve the inequality that way

once you plop it on, you can clearly see it holds true for x < -1 and x > 2.5

[hidden mist]

how did u get that

[teal skiff]

i do criss cross method

[hidden mist]

Its wrong tho

its (x-3)(2x+1)

[teal skiff]

yeah i wrote down 3 on paper but i read it as a 5 lol

mb lol

wait hold up

yeah shit mb lol-

[hidden mist]

yeahh

[teal skiff]

no i completely fucked it up

[hidden mist]

oof-

[teal skiff]

yea sorry

(2x+1)(x-3)

but after that same shit remains

[shadow grove]

Still sorta confused

[hidden mist]

u see its a quadratic equation right

[teal skiff]

ah snowflake can explain im going bed

[hidden mist]

degree 2

so theres basically gonna be 2 values of c

x

and the question is with the sign less than 0

so u basically find the common thing in the 2 values of x

[vapid cypress]

this works ig but wavy-curve method is easier

[hidden mist]

Im not sure if he knows it

thats the easier one ik

u just put the plus minus stuff and get the negative sign here cuz les than

[vapid cypress]

yea

[shadow grove]

What am I supposed to do?

[hidden mist]

but they way hes reacting on quadratic I thought of going with normal method

[hidden mist]

lemme ask

do u know wavy curve method?

[shadow grove]

No clue

[hidden mist]

something used to solve inequalities

wait lemme show u how is it done

gimme 2 mins

@shadow grove here

[shadow grove]

Ah yeah I see it now

[hidden mist]

theres a whole lot of concept here

u gotta learn it as well

u want some slight idea about it?

[shadow grove]

That’s what u thought you do

[hidden mist]

u dont use that formula usually in inequalities to find roots

[shadow grove]

Ah right

[hidden mist]

thats for like finding the location of graph

do u know what this thing means

[shadow grove]

Dunno

[hidden mist]

its

belongs to

x belongs to -1/2 till 3

the brackets are open

[shadow grove]

The graph is definitely a sad face yeah?

[hidden mist]

well for this one

what was the equation

[shadow grove]

Alright cool

[hidden mist]

its a smily I think

[shadow grove]

Ah is it?

Okay

[hidden mist]

@shadow grove

check it out the guy showed it

[shadow grove]

Got it

How can you tell when it is a smiley one or sad one?

[hidden mist]

if the leading coefficient

that is the coefficient of x^2 is positive its a smily

if it is negative its a sad

[shadow grove]

Right okay thanks

I think I got it solved now

[hidden mist]

Good

[shadow grove]

Thanks

[hidden mist]

Np

[shadow grove]

Ohhhh

I just realised

Was getting confused with using b^2-4ac with this question

[hidden mist]

?

whats the confusion

do u know the conditions for delta

aka discriminant

I cant understand that-

[shadow grove]

Do you know what a b and c are for this question?

[hidden mist]

Yes its clear

the coefficients

[shadow grove]

Yes

What are they?

[hidden mist]

a = 4 b = 8-4k and c = 4-k

[shadow grove]

Thanks

[hidden mist]

np

[safe radish]

@shadow grove Has your question been resolved?

[safe radish]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[finite iris]

why is this integral equal to 0

[finite iris]

surely computing this becomes: [x] upperbound = 4, lowerbound = - infinity

so you get 4 - (-infinity) which goes to infinity no?

[stoic dune]

You can't plug infinity into an equation

[tame raft]

integral 0 is just 0

it's an exception

[stoic dune]

You can take a limit to infinity though!

[tame raft]

like, definite integral 0 is just 0

well ok

i shouldn't say it's an exception really but yeah

[finite iris]

i see, thanks

[finite iris]

ikik, i was just being lazy haha

[tame raft]

like it's not integral 1

integral 1 would be x

but integral 0 is just 0

[finite iris]

I've been taught to substitute R and say R tends to infinty/negative infinity

OH wait

yeah i had a brain fart moment lmao

.close

[safe radish]

Channel closed

Closed by @finite iris

Use .reopen if this was a mistake.

[safe radish]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[kind beacon]

can someone tell me what I did wrong here

[plucky elk]

Very first equation

,tex .wrong root

[flat frigate]

riemann

[kind beacon]

oh it dosent?

lol omg

[plucky elk]

Try x=4 and y=9

[kind beacon]

oh yea

Dang

Originally I was thinking of doing trig sub, but I wanted to see where simplifying takes me but guess it didn’t work out

ty

.close

[safe radish]

Channel closed

Closed by @kind beacon

Use .reopen if this was a mistake.

[safe radish]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[lament echo]

can someone help me understand this

[lament echo]

so i have the 3 areas

im just trying to understand how dividing by L then differentiating finds the optimum value

[safe radish]

@lament echo Has your question been resolved?

[lament echo]

.close

[safe radish]

Channel closed

Closed by @lament echo

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[glossy cypress]

I noticed a strange pattern when calculating various functions in the form of (x+1)^n. each one is has the derivative of the term with the greatest value added to it, along with a strange sequence. is there a reason that (x+1)^n has the second term of the polynomial as the derivative of x^n?

[tame raft]

do you know the binomial expansion

[glossy cypress]

yes

[tame raft]

well it's just the second coefficient

[glossy cypress]

but why is the derivative there

[tame raft]

so n choose 1

so n

so it's just nx^(n-1)

it just happens to be there

i mean ok you could consider (x+1)^n - x^n as sorta a really crappy derivative

((x+1)^n - x^n)/1

that should be approximately nx^(n-1)

so it makes sense

[glossy cypress]

wait did i just accidentally make a shitty derivative

lmaoo

the weird pattern i don't get though

depending on the number term it is, there's a relation to it's derivative

if you add the nth term's derivative, and divide it by n, you get the following term

[tame raft]

i think it's just a coincidence

[glossy cypress]

for example: x^{5}+5x^{4}+10x^{3}+10x^{2}+5x+1
d/dx (x^5) = 5x^4.
(d/dx(5x^4))/2= 10x^3

[tame raft]

no

ok wait yeah

sure

but that's just because of the binomial thing again

i mean

i guess this is the statement that (x+1)^n is equal to its taylor series

or

x^n is equal to its taylor series expanded around x = -1, rather

or something like that

[glossy cypress]

that's probably where i needa look

i don't really have a good understanding of taylor series

so

that's what im going to do

ty

.close

[safe radish]

Channel closed

Closed by @glossy cypress

Use .reopen if this was a mistake.

[safe radish]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[tall bough]

The exponents have a common base

[eternal carbon]

one way you could continue is by noting that…

3^n = 3^(n - 1 + 1) = 3^(n - 1) * 3^(1) = 3 * 3^(n - 1)

[safe radish]

Channel closed

Closed by @river gate

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[plain spear]

what did i do wrong here: The question is 4 3/7 - 3 2/3. I first converted it to improper fractions which is 31/7 and 11/7, and then I minused it which is 20/7. which equals to 2 6/7. Now what

[final halo]

how did you get 11/7?

[plain spear]

cause when u convert 3 2/3 to improper you get 11/7

[obsidian oracle]

uh no

[plain spear]

oh fuck I mean

[obsidian oracle]

why would the denominator change?

[plain spear]

/3

[plain spear]

yeah just realized my dumbass

thanks bruh

idk how i missed that

thanks g

.close

[safe radish]

Channel closed

Closed by @plain spear

Use .reopen if this was a mistake.

[plain spear]

.reopen

@obsidian oracle

[safe radish]

✅

[plain spear]

wait I have another quewtion

question

would the answer

be 16/21

?

[obsidian oracle]

Yes I got that

[safe radish]

@plain spear Has your question been resolved?

[safe radish]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[safe radish]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[wooden oyster]

how do i prove that the sqrt(x) -> 0 when x -> 0

[wooden oyster]

with the episolon delta proof

[tranquil tendon]

technically the (two-sided) limit of sqrt(x) as x approaches zero is undefined

this is because sqrt(x) is not defined for any negative x and so there are no open intervals containing 0 on which sqrt(x) is defined everywhere except possibly at 0

[wooden oyster]

i meant for x^+

[tranquil tendon]

ok, so it's a limit from above?

[wooden oyster]

or like

0_+

the limit where it can work

where it gets to 0.00000001 and not -0.00000000001

[tranquil tendon]

so the difference in a onesided limit is that instead of using $0 < |x-a| < \delta$ you use $a < x < a+\delta$. otherwise the approach is exactly the same as a two-sided delta-epsilon proof

[flat frigate]

rome of oxtrot

[wooden oyster]

mhh ok and then what

this one isnt the same as normal

[tranquil tendon]

how so?

[wooden oyster]

it wirtte ndifferently?

[tranquil tendon]

only in that one step, otherwise it follows the same pattern

[wooden oyster]

a is the point we are going to and delta is positively small

[tranquil tendon]

intuitivitely, this limit is clearly 0

so what you need to do is show that for any $\epsilon > 0$, there exists a $\delta$ such that if $0 < x < \delta$ then $|\sqrt x| < \epsilon$

[flat frigate]

rome of oxtrot

[wooden oyster]

is the play here to say delta = epislon^2

[tranquil tendon]

seems to me that that would work

[safe radish]

@wooden oyster Has your question been resolved?

[safe radish]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[lilac drift]

How do i solve this ?

[rocky kraken]

if $$ RV = SV $$ then $$ 2x+8 = 6x - 4 $$ therefore $$ 6x - 2x = 8 + 4 $$ therefore $$ 4x = 12 = > x = 3 $$

[flat frigate]

The Mad Pirate

[lilac drift]

Ok but what about for angle RVU ?

[rocky kraken]

$$ \tan(m/2) = RV/V = \frac{2*3+8}{3} = \frac{14}{3}$$

[flat frigate]

The Mad Pirate

[rocky kraken]

now

$$ m = 2*\atan(\frac{14}{3})$$

[flat frigate]

The Mad Pirate

[rocky kraken]

the result I get for m is 155,81048584597579749513266277056

[lilac drift]

What about this one ?

[rocky kraken]

@lilac drift first you need to find x, which will give you the oppsite lenght on the trianfle BEC, therefore cos(BEC/2) = BE/x , you can find BE as x*cos(BEC/2)

finding follows the same procedure than before, equate both linear equations and solve for x

you can check the result with the ,w solve command

[lilac drift]

!w solve

w solve

[rocky kraken]

,w solve 3x+6 = 5x-2

[flat frigate]

Results provided by WolframAlpha

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

Requested by themadpirate

[rocky kraken]

,w (3*4+6)*cos(54/2)

,w (34+6)cos(54/2*pi/180)

,w (34+6)cos(54/2pi/180)

[flat frigate]

Results provided by WolframAlpha

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

Requested by themadpirate

[rocky kraken]

to get m is exactly the same way than the previous excercise

$$ m = 2*\atan(\frac{18}{4})$$

[flat frigate]

The Mad Pirate

[rocky kraken]

.w 2*atan(18/4)

,w 2*atan(18/4)*180/pi

[flat frigate]

Results provided by WolframAlpha

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

Requested by themadpirate

[rocky kraken]

@lilac drift are you getting the idea ?

[lilac drift]

.close

[safe radish]

Channel closed

Closed by @lilac drift

Use .reopen if this was a mistake.

[rocky kraken]

Good

[safe radish]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[fiery mural]

We learned a way to do this one but I forgot how to do it

[lean otter]

ok is it ok if you solve it in a different way? bc idk which way you’re supposed to solve it my way involves a system of linear equations i think

if not wait for someone with better knowledge

[fiery mural]

I mean go ahead

[lean otter]

ok

so you are looking for a_1 and d

such that

(a_1 + (4 - 1)d) + (a_1 + (5 - 1)d) = 75

and (a_1 + (5 - 1)d) + (a_1 + (6 - 1)d) = 93

[fiery mural]

ohhhhh yeah, it was this

[lean otter]

solve this system note that there is a duplicated term which probably simplifies it i think

[fiery mural]

so for the 11th term, i got 52.125

[lean otter]

what about a_1 and d?

[fiery mural]

ummm, a_1=29.625, d=2.25

[lean otter]

it does not appear to be correct

[fiery mural]

ok a_1=6 and d=9

.close

[safe radish]

Channel closed

Closed by @fiery mural

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[safe radish]

Channel closed

Channel closed due to the original message being deleted.
If you did not intend to do this, please open a new help channel,
as this action is irreversible, and this channel may abruptly lock.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[pearl vale]

ive figured out the first part how do I do the second part

[pearl vale]

showing theres no minimum with the constraint

[safe radish]

@pearl vale Has your question been resolved?

Channel closed

Closed by @pearl vale

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[rare dragon]

How would i know to use the discriminant

helpers

<@&286206848099549185>

bruh

<@&286206848099549185>

i know the method

of applying b2-4ac

however

how do i know by just the fact that x is real i am able to use the discriminant and prove it

because i swear u can have a real number and not have it cross the x axis at all

<@&286206848099549185>

[slow dagger]

Hm

[rare dragon]

yo do u have answer?

oh i think i get it

because x cant be real if it doesnt touch the x axis

LETS GO

IT CLICKED

ALHAMDULLIAH

GOD GIVE ME EVERYTHING

GOLDEN AGE OF ISLAM

.CLOSE

.close

[safe radish]

Channel closed

Closed by @rare dragon

Use .reopen if this was a mistake.

[safe radish]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[swift vale]

quick question is log (xy) = log x + log y

[timid bolt]

ye

[swift vale]

thanks

.close

[safe radish]

Channel closed

Closed by @swift vale

Use .reopen if this was a mistake.

[safe radish]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[fading iris]

How would I solve this?

[round rover]

what did you try

[fading iris]

It’s transformations

But I don’t know this, especially when it’s crossing the y axis

[round rover]

there's nothing special about the y axis in a sense

[fading iris]

Idk

[round rover]

in this case, i recommend looking at where each point goes

G goes to G'